geometry
geometric transformation
homothety
power of a point
radical axis
geometry proposed

Points $ A$ and $ B$ lie on the circle with center $ O.$ Let point $ C$ lies outside the circle; let $ CS$ and $ CT$ be tangents to the circle. $ M$ be the midpoint of minor arc $ AB$ of $ (O).$ $ MS,\,MT$ intersect $ AB$ at points $ E,\,F$ respectively. The lines passing through $ E,\,F$ perpendicular to $ AB$ cut $ OS,\,OT$ at $ X$ and $ Y$ respectively.
A line passed through $ C$ intersect the circle $ (O)$ at $ P,\,Q$ ($ P$ lies on segment $ CQ$). Let $ R$ be the intersection of $ MP$ and $ AB,$ and let $ Z$ be the circumcentre of triangle $ PQR.$
 Prove that: $ X,\,Y,\,Z$ are collinear.

This is a nice problem, compliments to your friend!

First, we prove that the circumferences with centres X,Y and radii XS and YT touch the line AB and the main circumference.
This picture explains why, whatch it.
[img]8326[/img]

Now, here is a simplified picture to solve the problem:
[img]8327[/img]

First, we prove that the quadrilaterals $TPRF,SPRE$ are cyclic. To prove this, consider the homotety with centre T bringing F to M, it brings AB to the tangent t at M, and using directed angles it immidiately follows that $\angle RFT = \angle (AB,MT) = \angle (t,MT) = \angle MPT = \angle RPT$, the same for the other circles.

Now, let's observe that $CT^{2}= CP \cdot CQ = CS^{2}$ because of power of points, and also $MF \cdot MT = MR \cdot MP = ME \cdot MS$ because of power of points and the ciclicy proven above.

But then both C,M lie on the radical axis of all the three red circles taken in pairs, and then the three circles are coaxial, and their centres (X,Y,Z) are coollinear!  :)

It is problem 1 of China TST 2007.
It is proposed by XiongBin.

[quote="edriv"]This is a nice problem, compliments to your friend!

First, we prove that the circumferences with centres X,Y and radii XS and YT touch the line AB and the main circumference.
This picture explains why, whatch it.
[/quote]

Don't get it... how can the picture explain?

Ok, I'll explain why, then. I forgot to name the points: let M be the midpoint of the arc determined by r, $C = MP \cap r$, D the intersection of OP with the perpendicular to r through C.

Note that OM is perpendicular to r, and also DC is. Therefore, OM and DC are parallel. Then, triangles $PDC$ and $POM$ are similar. But $POM$ is isosceles! Then $PDC$ is isosceles, too: $PD = DC$.
Therefore there is a circle centered at D that passes through P and C. But since P,D,O are on a line, this circle is tangent to the big circle. And DC is perpendicular to r: our circle touches also r.

Is it clear?

[quote="April"][color=darkred]Points $A$ and $B$ lie on the circle with center $O.$ Let point $C$ lies outside the circle; let $CS$ and $CT$ be tangents to the circle. $M$ be the midpoint of minor arc $AB$ of $(O).$ $MS,\,MT$ intersect $AB$ at points $E,\,F$ respectively. The lines passing through $E,\,F$ perpendicular to $AB$ cut $OS,\,OT$ at $X$ and $Y$ respectively.
A line passed through $C$ intersect the circle $(O)$ at $P,\,Q$ ($P$ lies on segment $CQ$). Let $R$ be the intersection of $MP$ and $AB,$ and let $Z$ be the circumcentre of triangle $PQR.$
 Prove that: $X,\,Y,\,Z$ are collinear. [/color][/quote]
[color=red]I don't know Chinese but I think the solution's to seem like:[/color]

[color=indigo][b]Proof.[/b] We have:
$\blacktriangleright\quad \begin{cases}XE\parallel OM\\ \triangle OMS\;\text{is isosceles}\\ \overline{OXS}\perp CS\end{cases}\Longrightarrow\begin{cases}E,S\subset (X,\,XS) \\ (X)\;\text{is tangent to}\; (O)\end{cases}$

$\blacktriangleright \quad \angle MAB=\angle MPA=\angle MSA\Longrightarrow\begin{cases}\triangle MAR\sim\triangle MPA\\ \triangle MAE\sim \triangle MSA\end{cases}\Longrightarrow \begin{cases}MA^{2}=MP\cdot MR\\ MA^{2}=MS\cdot ME\end{cases}$ $\Longrightarrow MP\cdot MR=MS\cdot ME\Longrightarrow M\;\text{lies on the radical axis of two circles}\; (PQR)\;\text{and}\; (X)\quad (1)$

$\blacktriangleright\quad CP\cdot CQ=CS^{2}\Longrightarrow C\;\text{lies on the radical axis of two circles}\; (PQR)\;\text{and}\; (X)\quad (2)$

From $(1)$ and $(2)$ we have: $MC\perp XZ$
Similar $MC\perp YZ$

And hence $X,\,Y,\,Z$ are collinear[/color]

Let $\mathcal{P}(X, \omega)$ denote the power of a point $X$ w.r.t. a circle $\omega.$ Animate the point $P$ on $(O)$ and let $\Gamma\left(O_1\right)$ be the circumcircle of $\triangle PQR.$ We claim that as $P$ varies, $O_1$ varies along a fixed line.

To see this, note that the inversion with center $M$ and radius $MA = MB$ swaps $AB$ and $(O).$ Thus $P, R$ are swapped under this inversion, implying that $\mathcal{P}(M, \Gamma) = MP \cdot MR = MA^2$ is fixed. Meanwhile, note that $\mathcal{P}(C, \Gamma) = CP \cdot CQ = \mathcal{P}(C, (O))$ is fixed as well. Hence, it follows from the Power of a Point Formula that $\mathcal{P}(M, \Gamma) - \mathcal{P}(C, \Gamma) = O_1M^2 - O_1C^2$ is fixed. Since $M$ and $C$ are fixed, it is then well-known the locus of $O_1$ is a line. To determine this line, we will consider two special cases.

First, note that $OM \perp AB$ because $M$ is the midpoint of arc $\widehat{AB}.$ Therefore, $OM \parallel XE$, implying that $\triangle SOM \sim \triangle SXE \implies XS = XE.$ Then when $P \equiv S$, we have $Q \equiv S$ as well, implying that $\Gamma$ and $(O)$ are tangent at $S.$ Hence $O_1 \in OS$, but we also have $O_1S = O_1E$, whence we deduce that $O_1 \equiv X.$ Similarly, when $P \equiv T$ we find that $O_1 \equiv Y.$ Therefore, $XY$ is the fixed line upon which $O_1$ varies. $\square$

We can consider that $M,A,B,S$ and the circle are fixed and $P$ varies along the circle. It's clearly shown that $X=Z$ and $Y=Z$ when $P=S$ and $P=T$ respectively.
It suffice to show that locus of $Z$ is a line as $P$ varies along the circle. Let $N=MQ \cap AB$. $MQ \cdot  MN=MP \cdot MR$. Hence, $P,Q,R,N$ are concyclic.
Let $P_1,P_2$ are points lie on the circle. We define $Q_1,Q_2,R_1,R_2,N_1,N_2,Z_1,Z_2$ similarly.
Since, $CQ_1 \cdot CP_1=CQ_2 \cdot CP_2$ and $MQ_1 \cdot MN_1=MQ_2 \cdot MN_2$. $MC$ is radical axis of $\odot P_1Q_1R_1$ and  $\odot P_2Q_2R_2$.
Hence, $Z_1Z_2\bot MC$. Locus of $Z$ is a line perpendicular to $MC$.

Solved with [b]VJ27[/b] \\
Let $\omega$ be the circle centered at $X$ with radius $XE$ ; $\gamma$ be the circle centered at $Y$ with radius $YF$ and $MQ$ intersect $AB$ at $D$. $S$ and $T$ are the touchpoints of $(ABPQ)$ & $\omega$ ; $(ABPQ)$ & $\gamma$ respectively coz of homothety. $DRPQ$ is cyclic coz $MD*MQ=MB^2=MR*MP$. $M$ lies has equal power w.r.t. $\omega$, $DRPQ$ & $\gamma$ since $MT*MF$=$MB^2$=$ME*MS$=$MD*MQ$ .Also $C$ has equal power with respect to the 3 circles coz of radax theorem on $ABPQ$ , $DRPQ$ & $\omega$ and $CS=CT$. So the three circles are coaxial having radax $CM$.Hence $X$ , $Y$ and $Z$ are collinear.

Nice problem!

Let $\overline{AB}$ and $\overline{MQ}$ intersect at $R'$. By the shooting lemma, $PQR'R$ is cyclic. Since $\triangle SOM \sim \triangle SXE$ and $\triangle TOM \sim \triangle TYR$, we have $XE=XS$ and $YR=YT$. Let $\omega$ be the circumcircle of $PQR'R$, let $\omega_1$ be the circle centered at $X$ passing through $E$, and let $\omega_2$ be the circle centered at $Y$ passing through $F$. Notice that $CS^2=CT^2=CP \cdot CQ$, $C$ has equal power with respect to $\omega$, $\omega_1$, and $\omega_2$. By the shooting lemma, $M$ has equal power with respect to $\omega$, $\omega_1$, and $\omega_2$. Thus, $\omega$, $\omega_1$, and $\omega_2$ are coaxial, so $X$, $Y$, and $Z$ are collinear.


inequalities
algebra
polynomial
inequalities solved

Well, this is indeed strange, since it is obvious that 1-b>0, 1-bc>0 and so on, so everything is clear. Now I check if it is true for any a,b,c>0

You are right. Sorry for my mistake. The inequality is obvious for 0<a,b,c<1  but it is false for a,b,c real positive numbers so it would be very interesting to find the better value of x ( 0<a,b,c<x)  for which it is true.

i'm not even sure that we can find such a value ...

What do you use to find those constants :-/ ?


limit
linear algebra
matrix
vector
algebra proposed
algebra

Sequences $(x_n)_{n\ge1}$, $(y_n)_{n\ge1}$ satisfy the relations $x_n=4x_{n-1}+3y_{n-1}$ and $y_n=2x_{n-1}+3y_{n-1}$ for $n\ge1$. If $x_1=y_1=5$ find $x_n$ and $y_n$.

Calculate $\lim_{n\rightarrow\infty}\frac{x_n}{y_n}$.

$3y_n=2x_n+5\Longrightarrow \frac 32=\frac{x_n}{y_n}+\frac{5}{2y_n},\ y_n\rightarrow\infty\Longrightarrow \frac{x_n}{y_n}\rightarrow \frac 32\ .$

Could You Rust or anyone write something about finding the general method for finding  $ (x_n) (y_n)$

Because You wrote about characteristic equation. Could You explain more clearly what are the properties and other interesting concludes if we will compute that?

This is just linear algebra; what we have is

$ \left[ \begin{array}{c} x_n \\
y_n \end{array} \right] \equal{} \left[ \begin{array}{cc} 4 & 3 \\
2 & 3 \end{array} \right] \left[ \begin{array}{c} x_{n \minus{} 1} \\
y_{n \minus{} 1} \end{array} \right]$

hence, letting the above matrix be $ \mathbf{M}$ and $ \mathbf{x}_n$ denote the column vector whose entries are $ x_n, y_n$, we have

$ \mathbf{x}_n \equal{} \mathbf{M}^n \mathbf{x}_0$

which we can compute by [url=http://en.wikipedia.org/wiki/Diagonalizable_matrix]diagonalizing[/url] $ \mathbf{M}$.


Prove that for all $a,b,c,d$ be positive real then
                  $a^4b+b^4c+c^4d+d^4a \geq abcd(a+b+c+d)$

Since the RHS is symmetric, we can assume that $a\leq b\leq c\leq d$. It follows that $a^2\leq b^2\leq c^2\leq d^2$ and $bcd\geq acd\geq abd\geq abc$. Hence by rearrangement ineq. we have that

\[
\begin{eqarray*}
abcd(a+b+c+d) \& \& = a^2\cdot bcd+b^2\cdot acd+c^2\cdot abd+d^2\cdot abc\\
                   \& \& { }\leq a^2\cdot abc+b^2\cdot bcd+c^2\cdot acd+d^2\cdot abd\\
                       = a^3bc+b^3cd+c^3ad+d^3ab
\end{array*}
\]
Since $b\leq d$ and $c^3a\leq d^3a$ then
\[
c^3a\cdot d+d^3a\cdot b\leq c^3ab+d^4a
\]
Hence
\[
a^3bc+b^3cd+c^3ad+d^3ab\leq a^3bc+b^3cd+c^3ab+d^4a
\]
Now since $a^3\leq c^3$ and $ab\leq bc$ then
\begin{eqnarray*}
a^3bc+c^3ab &=& a^3\cdot bc+c^3\cdot ab\\
                     &  & \leq  a^3\cdot ab+c^3\cdot bc\\
                    &=& a^4b+c^4b
\end{eqnarray*}
So
\[
a^3bc+b^3cd+c^3ab+d^4a\leq a^4b+c^4b+b^3cd+d^4a
\]
Now since $b^3\leq c^3$ and $bc\leq cd$ then

\[
c^3\cdot bc+b^3\cdot cd \leq c^3\cdot cd+b^3\cdot bc
= c^4d+b^4c
\]

Combine all the above, we get
\[
abcd(a+b+c+d)\leq a^4b+b^4c+c^4d+d^4a
\]

Any body have an idea why the eqnarray doesn't work perfectly????

It's wrong, the LHS is not symmetric, you cannot assume that $a\leq b\leq c\leq d$ or $a\geq b\geq c\geq d$.

[quote="harazi"]Strange, this problem appears in a romanian contest some 25 years ago, proposed by a well-known (to the romanian public!) inequalities creator, Mihai Onucu Drambe. The only solution I know uses AM-GM and it's really not interesting. I hope someone will find something smarter.[/quote]

If my calculations are not wrong then by AM_GM $\sum(23a^4b+7b^4c+11c^4d+10d^4a)\ge 51\sum a^2bcd$

Hmm i thought this one was very easy if one knows holder's inequality

divide both sides of the inequality by $abcd$ then we get

$\frac{a^3}{cd}+\frac{b^3}{da}+\frac{c^3}{ab}+\frac{d^3}{bc}$ $\geq$ $a+b+c+d$

we will prove the inequality above.

By Holder's inequality we have

$(\frac{a^3}{cd}+\frac{b^3}{da}+\frac{c^3}{ab}+\frac{d^3}{bc})(c+d+a+b)(d+a+b+c)\geq(a+b+c+d)^3$

dividing both sides of the inequality above by $(a+b+c+d)^2$ and we get the desired result.

Am i wrong?

Or it's very easy?

Muirhead doesn't work because the LHS isn't symmetric.

AFAIK Murihead is only for symmetric inequalities.
For example, take this one:
$a^3b + b^3c + c^3a \geq a^2b^2 + b^2c^2 + c^2a^2$
The LHS is cyclic (isn't it?) and $(3,1) \succ (2,2)$ but the inequality isn't true for all positive reals.

[quote="Detective"]$(\frac{a^3}{cd}+\frac{b^3}{da}+\frac{c^3}{ab}+\frac{d^3}{bc})(c+d+a+b)(d+a+b+c)\geq(a+b+c+d)^3$[/quote]

Is this true  :huh:

[quote="yassin88"][quote="shyong"]yassin88 : It is holder inequality so should be true  :)[/quote]

I do not know how he applies Holder inequality.   :([/quote]

$(a^3/cd+b^3/ad+c^3/ab+d^3/bc)(c+d+a+b)(d+a+b+c)\geq \left(\sqrt[3]{a^3/cd\cdot c\cdot d}+\sqrt[3]{b^3/ad\cdot a\cdot d}+\sqrt[3]{c^3/ab\cdot a\cdot b}+\sqrt[3]{d^3/bc\cdot b\cdot c}\right)^3$  ;)

Yeah, I know that; what I do not know is where he gets the right side 3 exponent from.

The power is $\alpha=\beta=\gamma=\frac{1}{3}$ :)

[quote="Xamog"]It could be interesting to show how to construct an example where TomciO's inequality is false.[/quote]

Take $c \rightarrow 0$ and $b>a$. But IIRC the inequality is true for $a, b, c$ being sides of a triangle.

$a^4b + b^4c + c^4d + d^4a \geq abcd(a+b+c+d)$ ; $\frac{a^3}{bc} + \frac{b^3}{cd} + \frac{c^3}{da} + \frac{d^3}{ab} \geq a+b+c+d$.
We can order $a,b,c,d$,and we have $\frac{a^3}{bc} + \frac{b^3}{cd} + \frac{c^3}{da} + \frac{d^3}{ab} \geq$ $\frac{a^3}{ac} + \frac{b^3}{bd} + \frac{c^3}{ca} + \frac{d^3}{db} \geq$ ${\frac{a^3}{a^2} + \frac{b^3}{b^2} + \frac{c^3}{c^2} + \frac{d^3}{d^2} = a + b + c + d}$,for (rearrangment?I can't remember inequality's name... :blush:  :blush:  :blush: ).


I don't think you're supposed to post alcumus problems into the forum.

D.

There are $ 16!$ ways to put on the socks and shoes with no restrictions.

The chance of the sock being put on before the shoe for any leg is $ \frac{1}{2}$ and this event is independent of what order the socks and shoes are put on on the other legs. So multiply by $ \frac{1}{2}$ for each leg.

If you are trying to save karma, please don't.


Here's a good problem, but I'm not sure if it's on a beginner level.

Find number so that when the leftmost digit is moved to the rightmost position, the resulting number is three times the original. as well as me, I know joml88 and gauss202 can both help you with this problem.

[hide]0[/hide] is my first guess, though I'm not sure that is what you are looking for.

no, that's not it--sorry, i forgot to say natural number.

[hide]0 isn't what he is looking for. [/hide]

[hide]0 is not the answer[/hide]

You remember this one from the ARML trip, don't you, joml88?

Yes I remember.  I posted right when you were posting so we both responded to nr 's statement.

This is a nice problem, however it belong more appropriately in the intermediate forum. I posted it there a while ago in this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12201

Sorry--I didn't know that it should be on intermediate or that it was already posted there.

That's okay. The thread had been dead for a while. You should try the extensions to the problem that I mentioned there.


algebra function domain limit symmetry calculus derivative

Find the minimum value of $ f(x)\equal{}x^{\frac{3}{2}}\plus{}x^{\minus{}\frac{3}{2}}\minus{}4(x\plus{}x^{\minus{}1})$.

With the square roots, in there, we'll assume that the domain of the function is $ x\in(0,\infty).$

$ x^{\frac32}$ and $ x^{\minus{}\frac32}$ are the dominant terms at the ends of the intervals, we have $ \lim_{x\to0^\plus{}}f(x)\equal{}\lim_{x\to\infty}f(x)\equal{}\infty.$ Hence there must be a global minimum, since $ f$ is continuous.

Since $ f(x)\equal{}f\left(\frac1x\right),$ symmetry shows that $ f$ has a critical point at $ 1.$ We have $ f(1)\equal{}\minus{}6.$ Are there any other critical points? Clearly yes, since the function is concave downward at $ x\equal{}1.$

OK, I've found the answer. The minimum value is $ \minus{}10,$ which occurs when $ x\equal{}\frac{7\pm3\sqrt{5}}2.$

Some of the details: First, let $ y\equal{}\sqrt{x}$ and write the thing to be minimized as $ y^3\minus{}4y^2\minus{}4y^{\minus{}2}\plus{}y^{\minus{}3}.$ The derivative of that is $ 3y^2\minus{}8^y\plus{}8y^{\minus{}3}\minus{}3y^{\minus{}4}$ and this derivative can be factored as

$ y^{\minus{}1}(y\minus{}y^{\minus{}1})(3(y\plus{}y^{\minus{}1})\minus{}1)(y\plus{}y^{\minus{}1}\minus{}3).$

The critical points of interest happen when $ y\plus{}y^{\minus{}1}\equal{}3,$ which we solve to get $ y\equal{}\frac{3\pm\sqrt{5}}2.$ The rest follows from chasing these points.

You can change $ f(x)$ into a cubic function by setting something.

In particular, if $ z\equal{}x^{1/2}\plus{}x^{\minus{}1/2},$ then $ f(x)$ becomes $ z^3\minus{}4z^2\minus{}3z\plus{}8,$ with $ z$ restricted to $ [2,\infty).$

That's right. I was planned to post the problem in Intermidate forum for calculus beginner, but by rule I have posted the problem here.


Solve for x:
8sinxcosx - (sqrt6)sinx - (sqrt6)cosx + 1 = 0

[hide="hint"]
$\sin^2{x}+\cos^2{x}=1$
$(\sin{x}+\cos{x})^2=1+2\sin{x}\cos{x}=1+\sin{2x}$
Then just solve $\sin2x$[/hide]

[hide="solution"]$8 \sin x cos x - \sqrt {6} \left (\sin x + \cos x \right) + 1 = 0$

so then bring all the terms except +1 to the right side, we have

$\sqrt{6}\left (\sin x + \cos x \right) - 8 \sin x cos x = 1$
$\sqrt{6}\left (\sin x + \cos x \right) - 4\left (2 \sin x cos x \right) = 1$
$\sqrt{6}\left (\sin x + \cos x \right) - 4\left (\sin 2x \right) = 1$
$\sqrt{6}\left (\sin x + \cos x \right) = 1 + 4\left (\sin 2x \right)$

square both sides:

$6\left (\sin x + \cos x\right)^2 = \left (1+4 \sin 2x \right)^2$
$6\left (1 + \sin 2x \right) = \left (1+4 \sin 2x \right)^2$

let's say $\sin 2x = A$. THEN:

$6 + 6A = \left (4A + 1 \right)^2$
$16A^2 + 2A - 5 = 0$
$\left (8A + 5 \right) \left (2A - 1 \right) = 0$
$A = \sin 2x = \frac {1}{2} or \frac {-5}{8}$.

i would throw out the latter solution because 1) if it was AIME, i would not have time or the calculator to find the arcsin of $\frac {-5}{8}$ 2) the first one gives easy answers (.....).

SO, $2x =  \frac {\pi}{6} or \frac {5 \pi}{6}$

yielding $x = \frac {\pi}{12} or \frac {5 \pi}{12}$[/hide]

Sorry with all the LaTexing late at night... actually early 2 in the morning. I have no idea why i am still up using computer.

i forgot to mention, in my solution, i actually ommitted the part where i was supposed to show the equation that shobber was talking about. sorry about that.


(a+b+c)^2 < 3a^2 +3b^2+ 3c^2

This inequality appears in the proof of every book that prove the existence result concerning SDEs.
Examples of usage of this are in Oksendal, Lecture notes of Evans and Elliott & Kopp
I could prove it with 2 instead of 3 in the coefficients, so the inequality should be true but I don't see a use for the 3.
Is it trivial and I missing something?

Thanks.

$ 3(a^2 \plus{} b^2 \plus{} c^2) \minus{} (a \plus{} b \plus{} c)^2 \equal{} (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \geq 0$.

The inequality is obviously no longer true if you replace the coefficient $ 3$ with the coefficient $ 2$; take $ a\equal{}b\equal{}c$ to see this.

Indeed you're right.
Tomorrow I will check my calculations to see what I've done wrong.

Thank you.


limit
trigonometry
integration
function
real analysis
calculus
calculus computations

prob1:
compare 2  expression :
 $ A \equal{} 1 \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {36}}$
 and$ B \equal{} 1 \plus{} \frac {1}{\sqrt [3]{2}} \plus{} \frac {1}{\sqrt [3]{3}} \plus{} ... \plus{} \frac {1}{\sqrt [3]{27}}$
prob2 :  find limit
$ \lim_{n \to \infty} sin \pi  \sqrt {(n^2 \plus{} n)}$

1. $ B>A$
2. limit don't exist events members $ \to 1$, odd's $ \to \minus{}1$.

$ \sqrt{n^2\plus{}n}\equal{}n\left(1\plus{}\frac1n\right)^{\frac12}\equal{}n\plus{}\frac12\plus{}O(n^{\minus{}1})$

Hence, $ \sin\left(\pi\sqrt{n^2\plus{}n}\right)\equal{}\sin\left(\frac{(2n\plus{}1)\pi}2\plus{}O(n^{\minus{}1})\right).$

But $ \sin\left(\frac{(2n\plus{}1)\pi}2\right)\equal{}(\minus{}1)^n.$

Rust is correct and pankajsinha is not correct. The limit does not exist.

[quote="hsiljak"]pankajsinha probably thought about this (well-known) limit:

$ \lim_{n \to \infty} \sin \pi \sqrt {(n^2 \plus{} 1)}$

:D[/quote]

that doesn't converge either though
unless you restrict n to integers or something

If you restrict $ n$ to integers (which was being tacitly understood in all of the other posts in this topic), then it does tend to zero.

[quote="Hong Quy"]why A<B
explain for me[/quote]

Quite simply, the second expression has lower values in the denominator, thus 1/that value becomes higher.

ex/ 0.0005 is very small but 1/0.0005 is humongous.

[quote="Kent Merryfield"]$ \sqrt {n^2 \plus{} n} \equal{} n\left(1 \plus{} \frac1n\right)^{\frac12} \equal{} n \plus{} \frac12 \plus{} O(n^{ \minus{} 1})$

Hence, $ \sin\left(\pi\sqrt {n^2 \plus{} n}\right) \equal{} \sin\left(\frac {(2n \plus{} 1)\pi}2 \plus{} O(n^{ \minus{} 1})\right).$

But $ \sin\left(\frac {(2n \plus{} 1)\pi}2\right) \equal{} ( \minus{} 1)^n.$
[/quote]

i have a doubt regarding this sum (the second one)
unless i'm mistaken, $ O(n) \equal{} \frac {a_1}{n^2} \plus{} \frac{a_2}{n^3} \plus{} \frac{a_3}{n^4}.... (\infty$ terms)
where $ a_i$ are all some constant numerals.

so wen $ n\to \infty$ , $ O(n)$ will be of the form $ 0 * \infty$
and hence,
$ \lim_{n \to \infty}$ is [b]indeterminate[/b]

please clarify :)

I have no idea what you're trying to say. Your whole post makes no sense to me.

To say that $ a_n\equal{}O(n^{\minus{}1})$ is to say that there is some constant $ C$ such that $ |a_n|\le\frac{C}{n}$ for sufficiently large $ n.$ That's all it means - nothing more. And of course, anything that is $ O(n^{\minus{}1})$ tends to zero.

i'm sorry if i wasnt clear. :maybe:
but let me get this cleared first:
$ O(n^{\minus{}1})$ has infinite terms right?

so when n tends to infinity, each term of $ O(n^{\minus{}1})$ tends to 0.

[quote]$ \sin\left(\pi\sqrt {n^2 \plus{} n}\right) \equal{} \sin\left(\frac {(2n \plus{} 1)\pi}2 \plus{} O(n^{ \minus{} 1})\right)$.[/quote]
according to what you have posted in your solution to that sum, you neglect $ O(n^{\minus{}1})$ in the next step as it takes the value 0.

but my argument lies in that step.
In $ O(n^{\minus{}1})$, aren't you adding 0 + 0 + 0... infinite times?
so it will be of the form $ 0* \infty$ right?

[quote="hash_include"]$ O(n^{ \minus{} 1})$ has infinite terms right?[/quote]
No, it does not mean that. Read what I wrote, and don't invent meaning that isn't there. You are obsessing about a nonexistent issue.

I realize that there's a subtle point to my argument that I didn't address, namely that I was relying on the function $ \sin x$ to be not merely continuous but [i]uniformly[/i] continuous.

Theorem:

Suppose $ f(x)$ is uniformly continuous on $ \mathbb{R}$ and $ a_n$ and $ b_n$ are sequences such that $ b_n$ tends to zero. Then the sequence $ f(a_n\plus{}b_n)$ has the same behavior as the sequence $ f(a_n).$ That is, if $ f(a_n)\to L$ for some limit $ L,$ the $ f(a_n\plus{}b_n)\to L,$ and if $ \lim f(a_n)$ does not exist, then $ \lim f(a_n\plus{}b_n)$ does not exist.

This theorem would be false without the word "uniformly" in the hypothesis.

sir, my argument is not about the theorem.

i'm sorry if i'm a bit annoying here, but could you pls expand
$ O(n^{\minus{}1})$ ?

i think my confusion lies there because all this while i was under the impression that $ O(n^{\minus{}1})$ has $ \infty$ terms  :maybe:

I already explained this to you. The notation is simple, and you are trying to invent a complex meaning that it does not have.

To say that $ f(x)\equal{}O(g(x))$ is to say there is some constant $ C$ such that $ |f(x)|\le C|g(x)|.$ (Or at least, that is "eventually" true as you approach some limit.)

That's all it means. Do not try to read anything else into it.

I think hash_include is under impression that $ \left(1\plus{}\frac 1n\right)^{1/2}$ is being decomposed into its Taylor series here. This is not what's going on. Kent just uses a finite Taylor approximation with the remainder in the Peano form.

[quote="fedja"]I think hash_include is under impression that $ \left(1 \plus{} \frac 1n\right)^{1/2}$ is being decomposed into its Taylor series here. This is not what's going on. Kent just uses a finite Taylor approximation with the remainder in the Peano form.[/quote]

aah that was what i was confused about!
thanks fedja!  :)

[quote="JBL"]Obviously the preceding post is irrelevant to the problem.

$ \sum_{i \equal{} 1}^{36} \frac {1}{\sqrt {i}} < \int_{0}^{36} \frac {dx}{\sqrt {x}} \equal{} 12 \equal{} \int_{1}^{27}\frac {dx}{\sqrt [3]{x}} < \sum_{i \equal{} 1}^{27} \frac {1}{\sqrt [3]{i}}$.  (This is not a great estimate, but it's the one the problem was clearly designed for.  The LHS is a lot closer to $ 1 \plus{} \int_{1}^{36} \frac {dx}{\sqrt {x}} \equal{} 11$, so the difference between the two sums is larger than 1.)[/quote]
$ \int_{1}^{27}\frac {dx}{\sqrt [3]{x}}  < \sum_{i \equal{} 1}^{27} \frac {1}{\sqrt [3]{i}}$
 i think  it's wrong

You're entitled to think whatever you want, but in the future it would be nice if you actually provided a reason for your thoughts -- that way, I can explain to you why you're wrong instead of just telling you that you obviously have made a mistake and should go check your work.


calculus
integration
function
logarithms
calculus computations

How do you correctly evaluate the following improper integral:

integral (from 0 to oo) of ( ((e^(-3x)) - (e^(-7x))) / x ) dx

[quote="Cezanne123"]$\int_{0}^{\infty}\frac{e^{-3x}-e^{-7x}}{x}\,dx$
[/quote]
Quote to see the code.

We have $\frac{e^{-3x}-e^{-7x}}{x}=\frac1x\int_{3x}^{7x}e^{-t}\,dt=\int_{3}^{7}e^{-sx}\,ds$ by the substitution $t=sx$.
Therefore, $I=\int_{0}^{\infty}\frac{e^{-3x}-e^{-7x}}{x}\,dx=\int_{0}^\infty \int_{3}^{7}e^{-sx}\,ds\,dx$
The function is positive, so we can reverse the order of integration (Tonelli's theorem):
$I=\int_{3}^{7}\int_{0}^\infty e^{-sx}\,dx\,ds=\int_{3}^{7}\frac1s\,ds=\ln 7-\ln 3$.


An 59-kg woman stands in an elevator. What force does she exert on the floor of the elevator....

a) The elevator rises with a constant velocity of 2.0m/s

I knew the answer to be mg which is 578.2
is this correct????? acceleration is constant so it should be 1
and the force would be stationary here...right?

b) The elevator accelerates upward at 1.8 m/s^2

ok this would be different from the previous problem. F=ma
so i thought it would be first 59x1.8=106 and the add mg  to it which is got above which will give me 684.2 N correct??

c) The elevator goes down with a constant velocity of 4.0m/s

here the velocity is negative, although acceleration is constant, so it wouldnt be different from #1.same as a)right?

d) The elevator descends with a downward acceleration 2.8 m/s^2

This problem shows a= -2.8. Since N=ma+w    ma=-162.4
i did that  +  58g   which gave me 406. right?

i know this sounds weird but can you check my work?

I didn't check the numbers, but I agree your approach is sound and won't give way under you (like an elevator decending).  (He .. he .. he.)


inequalities
trigonometry
inequalities proposed

Let $ B$ and $ C$ numbers such that $ B \plus{} C\le{\pi}$ (less strict) ($ B$ and $ C$ two angles in a triangle)
and $ B\geq{C}$
Prove:
$ \frac {\sin{(B \minus{} C)}\sin{(B \plus{} C)} \plus{} \sin{B}\cos^2{C} \minus{} \cos^2{B}\sin{C}}{(1 \plus{} \sin{C})(1 \plus{} \sin{B})}\leq{\sin{(B \minus{} C)}}$

Please if possible give a trigonometric proof

i assume that $ B,C>0$ since you say $ B,C$ are angles of a triangle... note that

$ \sin B\cos^2C\minus{}\sin C\cos^2B\equal{}(\sin B\minus{}\sin C)(1\plus{}\sin B\sin C)$

it's easy to prove that $ \sin B\minus{}\sin C\leq \sin (B\minus{}C)$, so, it's enough to prove that

$ \dfrac{\sin (B\plus{}C)\plus{}(1\plus{}\sin B\sin C)}{(1\plus{}\sin B)(1\plus{}\sin C)}\leq 1$

which is equivalent to $ \sin (B\plus{}C)\leq \sin B\plus{}\sin C$,  that is clearly true, and we're done.  :D

$ \sin B\minus{}\sin C\leq\sin (B \minus{} C) \Leftrightarrow 2\sin\left(\dfrac{B \minus{} C}{2}\right)\cos\left(\dfrac{B \plus{} C}{2}\right)\leq 2\sin\left(\dfrac{B \minus{} C}{2}\right)\cos\left(\dfrac{B \minus{} C}{2}\right)$

and the last one is equivalent to $ 4\sin\dfrac{B}{2}\sin\dfrac{C}{2}\sin\left(\dfrac{B \minus{} C}{2}\right)\geq 0$ which is clearly true...


function
algebra
polynomial
limit
calculus
derivative
abstract algebra

could anyone help me with this equation? i've tried it for 2 days and i haven't been able to finish it.
let $f: \mathbb{R}\to\mathbb{R}$ be a differentiable funcion such that $f(x)=f(\frac{x}{2})+\frac{x}{2}f'(x)$. show that f is a polynomial funcion with degree smaller than 2.

Take $x_{0}\in\mathbb{R}$. By using Lagrange's Theorem  on $\left[\frac{x_{0}}{2},x_{0}\right]$ we get $\frac{x_{0}}{2}<x_{1}<x_{0}$ such that $f'(x_{1})=\frac{f(x_{0})-f(\frac{x_{0}}{2})}{\frac{x_{0}}{2}}=f'(x_{0})$.

By repeating this step we obtain a sequence $x_{0}>x_{1}>x_{2}>\dots >x_{n}>\dots$ such that $f'(x_{0})=f'(x_{i})$ for every $i$. The sequence is decreasing so it has a limit $l=\lim_{n\to\infty}x_{n}$. From the hypothesis we know that $f'$ is continuous on $\mathbb{R}\setminus\{0\}$ so $f'(x_{0})=f'(l)$. If $l\neq 0$ we get another number $l'<l$ such that $f'(l')=f'(x_{0})$. So the sequence can only be bounded by $0$, therefore $f'(x_{0})=f'(0)$.

But the only functions with constant derivative are linear. This can be seen by considering the function $g(x)=f(x)-f'(0)x$.

that's the exact approach i had, but i think it is wrong to conclude that $f'(l)=f'(x_{0})$ if $l=0$ because we don't know wether $f'$ is continuous at $0$ or not. if you are sure it is ok, please tell me. thanx

Actually $f'$ is continuous at $0$ as well because $\lim_{x\to 0}f'(x)=\lim_{x\to 0}\frac{f(x)-f(x/2)}{x/2}=f'(0)$ (the first equality is due to the functional equation and the second is true for any function $f$ differentiable at $0$).

[quote="Thor"] If $l\neq 0$ we get another number $l'<l$ such that $f'(l')=f'(x_{0})$. So the sequence can only be bounded by $0$, therefore $f'(x_{0})=f'(0)$. [/quote]
I do not understand what you mean by this. Please explain! I think it is an interesting problem.

I think that we are to apply the same procedure to $\ell$ (the one Thor applied to $x_{0}$).

I still don't see why the sequence of $l,l',l'' ...$ must converge to $0$. I think that is what Thor means.

For any $x_{0}>0$ consider a set $E=\{x\in[0,x_{0}]\colon f'(x)=f'(x_{0})\}$. Let $l=\inf E$ then $l=0$. If we had $l>0$ then we would be able to construct $l'\in E$ such that $l'<l$ what contradicts the definition of $l$. The set $E$ is closed for $f'$ is continuous. So $0=\inf E\in E$ i.e. $f'(x_{0})=f'(0)$.

[quote]I still don't see why the sequence of $l, l', l'', \dots$ must converge to $0$. I think that is what Thor means.[/quote]

lofar made it more rigorous, but it's the same idea: $l, l', l'', \dots$ is convergent, but only $0$ can bound any such sequence.


[quote="blahblahblah"][quote="Jimmy"]--- Thanks.[/quote]

Yes, Jimmy, that proof is fine.  I just don't like using calculus where I don't have to, just like I don't like using proof by contradiction where I can come up with a constructive proof instead.  It's just a personal thing - the other way involves limsups and liminfs and other tools of analysis which are a bit more complex.[/quote]

I know, I took your advice and started reading [u]Priciples of Mathematical Analysis[/u], and that's where I learned the "formal" proof.

Good!  How are you liking Rudin?  Though you know this, you should be trying to do all the exercises you can.  Rudin's exercises are quite difficult as far as textbooks go, and if you can solve the last few questions in every chapter, you're in great shape.

It's great, a little hard at first but not impossible.  And yes, I agree his exercises are much harder than my usual homework assignments, but I can get most of them.  Thanks for the recomendation. :D

By the way, since you need someone who knows the solutions to get the most out of any text (but especially an analysis text) I don't mind looking over some of your solutions if you ever have problems with them or would just like to see another approach or whatever.

Just don't ask me to write up 3.24 for you - that has to be the single most annoying problem of all time.

[quote="blahblahblah"]
Just don't ask me to write up 3.24 for you - that has to be the single most annoying problem of all time.[/quote]

But also one of the most important problems! :lol:

Completions are very, very important, especially once you begin with Lebesgue measure.  That is why you should leave it for 4 chapters and do 7.24 instead - it makes way more sense and achieves the same thing (that is, you find an isometry from $(X,d)$ which may or may not be complete, to $(\mathcal{C} (X),||\cdot||_{\infty})$ which you have already proven to be complete).

Both problems are valuable. They present two different approaches to the problem - one more intuitive and one more advanced. I think the first problem is good for developing an intuition for what completion really means in a metric space. The completion of metric spaces is also used in number theory, for the study of things like the p-adic numbers.

Btw, sorry to everyone for having hijacked the thread with this discussion. Perhaps a moderator can move these last few posts somewhere else.

I'm not saying that both problems aren't valuable.  It's just that 3.24 is significantly more difficult, especially given where it appears in the text.  Kolmogorov and Fomin devote 3.5 pages to that exact problem, and if anything, it's a more advanced text than Rudin is!

Edit: Sorry, we'll take it to PM in the future.

Ok, I split and moved it so now you are free to talk as much as you like about it  :gathering:


induction
inequalities
irrational number
number theory proposed
number theory

Can it be existed postive integers $ n$ such that there are integers $ b$ and non zero integers $ a_i\ (i\equal{}1,\ 2,\ \cdots n)$ for rational numbers $ r$ which satisfies $ r\equal{}b\plus{}\sum_{i\equal{}1}^n \frac{1}{a_i}?$

If we can choose $ n$ dependent on $ r$, it's trivial:
If $ r \equal{} \frac cd$, take $ b \equal{} 0, n \equal{} c$ and all $ a_i \equal{} d$.

If not, it's not true.

So how is this problem meant to be¿

I think that these problem is:
Exist (or not) natural n, suth that $ \forall r\in Q$ exist $ \in Z,$ exist $ _1,a_2,...a_n\in N$, suth that $ r\equal{}b\plus{}\sum_{i\equal{}1}^n \frac{1}{a_i}$.
It is not hard constuct cotrexample $ \forall n\in N$ exist $ r(n)\in Q$, suth that exist not $ b,a_i$.

[color=red]Last edited[/color]

Do there exist positive integers $ n$ satifying the following condition?

Condition: There exist integers $ b$ and non zero integers $ a_1,\ \cdots a_n$ for arbiterary rational numbers $ r$ such that

$ r\equal{}b\plus{}\frac{1}{a_1}\plus{}\cdots \plus{}\frac{1}{a_n}$.

[quote="indybar"]So how to do it?

And where is problem 2 of this competition?  :)  :D[/quote]

Here is.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1136355#1136355[/url]

take r = (k! - 1) / k!
where k = very large value depending on n ( say 2 ^ 2 ^ 2 .. n times)

No!

More general result:

Let a1,a2,..., ak,.... --> 0 as k tends to infinity. For any number n, the set  of sums of n elements from this sequence is nowhere dense

Let $ S_n = \{r\in\mathbb{Q}: \exists b\in\mathbb{Z} \textrm{ and non - zero integers }$ $ a_1,a_2,\dots,a_n \textrm{ such that } r = b + \sum_{i = 1}^n\frac {1}{a_i}\}$

I think we can prove that $ \mathbb{Q}\backslash S_n\not = \emptyset$ for every $ n$ by induction, hence the answer is $ \textrm{NO}$. :D

[u]Proof:[/u]
$ n = 1$ $ \Rightarrow \left(0;\frac {1}{2}\right)\subseteq \mathbb{Q}\backslash S_1$.
Let for $ n = k$ there exist $ x$ and $ y$ such that $ (x;y)\subseteq \mathbb{Q}\backslash S_k$.

Let us we look $ n = k + 1$.

Let us suppose every number in $ (x,y)$ can be presented as element of $ S_{k + 1}$.
We can take one of these numbers:
$ x < b + \sum_{i = 1}^{k + 1}\frac {1}{a_i} < y$.

If we fix one index $ t\in\{1,2,\dots,k + 1\}$ then there occurs one of the following inequalities:
$ - b + \sum_{i = 1,i\not = t}^{k + 1}\frac {1}{a_i} < x$
$ - b + \sum_{i = 1,i\not = t}^{k + 1}\frac {1}{a_i} > y$.

Hence either $ \frac {1}{a_t} > \frac {y - x}{3}$ or $ \frac {1}{a_t} < \frac {x - y}{3}$. $ \Rightarrow \frac {3}{x - y} < a_t < \frac {3}{y - x}$. Therefore, there are a finite count of possible values for $ a_t$. From these reasonings, we conclude there aren't infinitely many rational numbers in $ (x,y)$. And, of course, from this follows that there exists a non-empty set $ (x_1,y_1)\subseteq \mathbb{Q}\backslash S_{k + 1}$. :)

(methode analytique)

[suppose that $ ( \exists n )$ such that for all $ r \in Q$ $ \exists b,|a_1| \le...\le |a_n|$

$ r = b + \frac {1}{a_1}....\frac {1}{a_n}$

you can took a rational sequence (r_n) such that $ lim (r_n) = e$ and e is irrational number

so $ ( \exists m )$  for all $ n\in N$ $ \exists b,|a_{1,n}|\le,...\le |a_{m,n}|$ such that

$ r_n = b + \frac {1}{a_{1,n}}....\frac {1}{a _ {m,n}}$

 know i will prove that $ lim |a_{n,m}| = + \infty$

suppose that there $ \exists M > 0$ such that for all $ n$ ($ \exists b_n > n$)

$ | a_{m,b_n}| < M$ so for all  $ 0 \le i \le n | a_{i,b_n}| \le M$

so the set of $ (r_{b_n})_n$   is fini what is imposible because $ lim (r_{b_n} )= e \in R - Q$

so $ lim  a_{n,m}$=+$ \infty$ so $ lim \frac {1}{a_{n,m}}$=0 so

$ lim(b + \frac {1}{a_{1,n}}....\frac {1}{a_{m - 1,n}}$=$ e$

by same methode prove $ lim | a_{n,m - 1}| = + \infty$ ...$ lim | a_{n,1}| = + \infty$ so $ lim (b) = b = e$ (absurde)

Thank you very much for your nice proof,mohamed-01-01 :)

And I will rewrite a solution based on what you thought to make it clear. :)
[b]Lemma[/b] for a given positive integer$ n$ and $ \all x,x$is irrational then $ \exists c,c > 0$ so that for every positive

integers $ b,a_1,...,a_n$ then $ |x - b - \frac {1}{a_1} + ... + \frac {1}{a_n}| > c$

Proof of the lemma:

Use induction on $ n$, the foundation is trivial,Let us assume it is true for all integers $ < n$.

Now turn to the case $ n$

Denote $ min\{\{x\},\{ - x\}\} = x'$  Then  pick a positive real $ M$ such that $ M = \frac {2n}{x'}$(of course $ x'$ doesn't

equal to $ o$)

Now we divide it into two cases

Case 1:If $ |a_1|,...|a_n|$ are all bigger that $ M$

Then $ |x - b - \frac {1}{a_1} + ... + \frac {1}{a_n}| \ge |x - b| - |\frac {1}{a_1} + ... + \frac {1}{a_n}| > x' - \frac {x'}
{2n}*n = \frac {x'}{2}$ as desired

Case 2:$ \exists a_i$ such that $ |a_i| \le M$

WLOG we assume $ i = n$

Then We only have a finite choose for$ a_n$

For a certain $ a_n = b_i,i = 1,2,...,s$Denote $ x_i = x - \frac {1}{a_n}$ of course $ x_i$ is irrational

Then use induction htpotheis $ \exists c_i > 0$ such that for every choose of $ b,a_1,...a_{n - 1}$

$ |x_i - b - \frac {1}{a_1} + ... + \frac {1}{a_{n - 1}}| > c_i$holds

This indicate  $ |x - b - \frac {1}{a_1} + ... + \frac {1}{a_n}| > c_i$

Denote $ c = min\{c_1,...c_k, \frac {x'}{2} \}$

$ c$ is a perfect choose for our lemma

So lemma is proved.

Return to original,assume exist a $ n$ so that for every $ r\in Q$ $ \exists b,a_{1},a_{2},...,a_{n}\in Z$ and $ r = b + \frac
{1}{a_{1}} + ... + \frac {1}{a_{n}}$

Choose $ e$ as our irrational number and for a certain $ n$ $ \exists c > 0$ such that  $ |e - b - \frac {1}{a_1} + ... + \frac
{1}{a_n}| > c$

Then pick a positive integer $ m$ so that $ \frac {1}{10^m} < c$

Choose a rational number $ r_m$ such that $ |e - r_m| < \frac {1}{10^m}$

So $ |r_m - b - \frac {1}{a_1} + ... + \frac {1}{a_n}| =|e-b-\frac{1}{a_1}+...+ \frac{1}{a_n}+r_m-e|> |e - b - \frac
{1}{a_1} + ... + \frac {1}{a_n}| - |r_m - e| > c - \frac {1}{10^m} > o$ a contradiction!

So there exists no $ n$ satisfy the condition.

QED

We'll prove that for a fixed $n$ and a constant $c$, the possible values of $f(a_1,a_2,\dots,a_n)=c+\sum_{1\leq i\leq n}\frac{1}{a_i}\in(0,1)$ have a maximum, which will conclude the problem since $\mathbb{Q}$ is dense.

We proceed by induction. If $n=1$ the result is trivial. Now, suppose that there is an infinite sequence of sequences $(A_k)_{k\geq1}=(a_1^{[k]},a_2^{[k]},\dots,a_n^{[k]})_{k\geq1}$ with $a_1^{[k]}\leq a_2^{[k]}\leq\dots\leq a_n^{[k]}$ that satisfies $f(A_k)>f(A_{k-1})$ and
$$\lim_{k\to\infty} f(A_k)=1.$$
Since $f(A_k)\leq c+\frac{n}{a_1^{[k]}}$, since this must approach $1$ (or even get bigger than $1$), there must be some constant $c_1$ such that $a_1^{[k]}\leq c_1$ for all large $k$.
Therefore, there is some value $b_1$ for which $a_1^{[k]}=b_1$ infinitely many times. Now, we only look at this subsequence of $(A_k)$ that satisfies $a_1^{[k]}=b_1$. Now taking $c\to c+\frac{1}{b_1}$ we get $n\to n-1$ and the problem is solved. $\blacksquare$

I think that, this question is not properly framed.

My framing: Can there exist countably many positive integer $n$ such that every rational number $r$ can be expressed as a sum of an integer with sequence of numbers of the form $(\frac{1}{a_i})_{n} $, where each $a_i's$ are integers(obviously non-zero) .

Answer to my framing to this question is affarmative that is yes such configuration exist for any rational,because if some $n$ works then $n+1$ also works for each rational.
Key idea to this framming is simple,
$\frac{1}{a+1}+\frac{1}{a^2+a}=\frac{1}{a}$ where we can take $a$ integer.


In a certain class, 3 out of 24 students are in student organizations. What is the ratio of students in student organizations to students not in student organizations?

[hide]First of all, I know that 3/24 is equivalent to 1/8. In the form with colons, the ratio would be 1:7 (right?), because there'd be 1 part for every student in an organization and 7 for every student not in the organization for a total of 8 (or rather any positive multiple of 8). But what I never really learned was how exactly I'm supposed to write the ratio if I put it in fraction form (and fractional ratios were the only choices). I thought that in fraction form, the ratio would be 1/8. But I was told that even in fraction form, the answer was 1/7. Why is this?[/hide]

Ratio form and fraction form are basically the same thing when dealing with ratios. So, for example, 4:5 is the same as 4/5

Ratios can be expressed in two ways:

colon form, such as $ 1: 7$

and fraction form, such as $ \frac{1}{7}$

Why is this? Ratios and fractions are both ways of comparing things. If you were finding probability, then it would be $ \frac{1}{8}$. But we're comparing things, so our answer is $ 1: 7$ or $ \frac{1}{7}$.

Ratios mean:
1 out of 7
Fractions mean:
1 part out of 7
These are basically the same thing.  :)

Actually, ratios mean 1 [b]to[/b] 7. Fractions are just another way of expressing them.

[color=red][size=200]read![/size][/color]
1:7 mean 1 in a student organization for every 7 [b]which are not in it[/b]
1/7 is different it means 1 in a student organization for every 7[b] total[/b] students

if there were 56 kids and we used the first way then there would be 7 in it and 49 not so for each person in it there are 7 who arn't in it

using the second way there would be 8 people in it (1 in it for every seven total kids)

Actually, I believe that's inaccurate.

Fractions can have different meanings, depending on the context.

If it's ratios, then a:b is equivalent to a/b. (Meaning there are "a" parts of one type for "b" parts of another.)

In probability, though, a/b means a out of b parts.


Hey guys, can someone explain to me the differences between the IMC test for undergraduates and the Putnam exam? Which one is more difficult, what are the prizes and which one is administered where? Thanks

I can't speak too much about the IMC: what I gleaned from their web site is that it's held in the summertime in Europe, and you have to actually go to the host site to take it (so if you're in North America, you have to go to Europe in the summer to take it, which is not a bad proposition in and of itself!)

I know a lot about the Putnam.

- You must be enrolled in a college or university in the US or Canada.  You can only take it 4 times.

- You take the exam at your home university.  It's the first Saturday in December, 3 hours in the morning and 3 hours in the afternoon, with 6 problems during each 3-hour session.

- There are both individual and team cash prizes (each university has a 3-person team).

More details at [url]http://math.scu.edu/putnam/[/url]

AoPS is hosting a Math Jam for the Putnam on September 15; see the Math Jam page for details.


[quote="happyface10"]Will someone still be able to be accepted into a top 10 university if they completed undergrad in a school ranking in the thirties or less?[/quote]

Of course, you can but there are probably some other factors whether you will. But I guess you need to offer some more credentials to judge whether you will. Do you want to do math? The for instance I assume Yale is much easier to get into than say Harvard or MIT...


Sorry if this is an easy question, but can you make a cone from any sector of a circle?

Never mind. Sorry. I just thought that they assumed the sector was 270 degrees in this AIME Solution:

http://www.artofproblemsolving.com/Wiki/index.php/2004_AIME_II_Problems/Problem_11


geometry
geometric transformation
reflection
3D geometry
prism
geometry theorems

Let $ ABC$ be a triangle, and let $ X$, $ Y$, $ Z$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Denote by $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ the reflections of these points $ X$, $ Y$, $ Z$ in the midpoints of the segments $ BC$, $ CA$, $ AB$, respectively. Prove that $ \left\vert XYZ\right\vert \equal{}\left\vert X^{\prime}Y^{\prime}Z^{\prime}\right\vert$.

I remember a paper by Mario Dalcin, [i]Isotomic inscribed triangles and their residuals[/i], Forum Geometricorum, 2003. See [url]http://forumgeom.fau.edu/FG2003volume3/FG200314index.html[/url]. However, no synthetic solution there.

The points $ X, Y, Z, X', Y', Z'$ lie on the same conic.

Let me present the simplest solution I have in mind.

$ \bullet$ We have $ |CX'Y'| \equal{} |AX'Y| \equal{} |AX'C| \minus{} |CX'Y|$. Similarly, $ |AY'Z'| \equal{} |BY'A| \minus{} |AY'Z|$ and $ |BZ'X'| \equal{} |CZ'B| \minus{} |BZ'X|$. Hence, we conclude that
$ |CX'Y'| \plus{} |AY'Z'| \plus{} |BZ'X'|$ $ \equal{} |AX'C| \plus{} |BY'A| \plus{} |CZ'B| \minus{} |CX'Y| \minus{} |AY'Z| \minus{} |BZ'X|$ $ (1)$

On the other hand, $ |CXY| \equal{} |BX'Y| \equal{} |BYC| \minus{} |CX'Y|$. Similarly, $ |AYZ| \equal{} |CZA| \minus{} |AY'Z|$ and $ |BZX| \equal{} |AXB| \minus{} |BZ'X|$. It implies
$ |CXY| \plus{} |AYZ| \plus{} |BZX|$ $ \equal{} |BYC| \plus{} |CZA| \plus{} |AXB| \minus{} |CX'Y| \minus{} |AY'Z| \minus{} |BZ'X|$ $ (2)$

Otherwise, $ X'C \equal{} XB$, $ Y'A \equal{} YC$, $ Z'B \equal{} ZA$, so $ |AX'C| \equal{} |AXB|$, $ |BY'A| \equal{} |BYC|$, $ |CZ'B| \equal{} |CZA|$ $ (3)$

From $ (1)$, $ (2)$ and $ (3)$, we get $ |CX'Y'| \plus{} |AY'Z'| \plus{} |BZ'X'| \equal{} |CXY| \plus{} |AYZ| \plus{} |BZX|$, it imlies $ |ABC| \minus{} |X'Y'Z'| \equal{} |ABC| \minus{} |XYZ|$, and it follows $ |XYZ| \equal{} |X'Y'Z'|$, as desired.

Although the above solution by dear [b]Son[/b] is very beautiful, however it is useful to all of us in the future, to know that the proposed problem is a porism of the below nice formula presented lately [url=http://www.mathematica.gr/forum/viewtopic.php?f=22&t=16521][color=red]Here[/color][/url], by a friend of mine [b]Stathis Koutras[/b].

[b][color=DarkBlue]THEOREM. A triangle $\triangle ABC$ is given and let $D,\ E,\ Z$ be, three arbitrary points on its side-segments $BC,\ AC,\ AB,$ respectively.
Prove that $(DEZ) = \frac{(DB)(EC)(ZA) + (DC)(EA)(ZB)}{(BC)(AC)(AB)}\cdot (ABC)$[/color][/b]

Kostas Vittas.


function
probability
algebra unsolved
algebra

I have this formula...

r = z(1,2(y/g))

and given the values of z, y and g.

The z, y and g are no problem but what do I do with the 1,2 to get an answer?

I have tried searching but I don't actually know what the term for 1,2 is in this case.

Help would be appreciated.

What is the context?
It looks to me like one of a few things.
1. Are you sure z is an actual number, and not a function with two arguments?
2. (1, 2(y/g)) could be a pair of co-ordinates or some other two-dimensional object. If that's so, you multiply both by z.
3. Sometimes (a,b) without a function indicated is used to denote gcd(a,b), but that seems highly unlikely if one of the variables is 1.

Thanks for coming back, Saul.
I thought I was simplifying the equation to ask the question.

It is a preamble leading to a Bayesian probability analysis I'm struggling with. I wish to end up with a 'longhand' explanation that my 60+ year non-math brain can understand and code into the program.  Maybe I'm in the wrong forum so please be gentle with me!

The formula  ..

rg = min(1,2(good(w)/G))   and   rb = min(1,bad(w)/B)

,     Pr|w = max(0.01,min(0.99,rb/(rg + rb)))

where Pr is the probability
           w is the token whose probability is being calculated
            good and bad are hash tables of data
            G is the number of good input data
            B is the number of bad input data
            ( rg and rb would be subscript g and b)

Now the author of the formulae was pretty dismissive when I asked for clarification but I'm willing to learn if only I can be pointed in the right direction.

So what does the 1,2 mean and while I'm asking, please clarify Pr|w if you would be so kind.

Ah. min and max aren't numbers - they're functions. Min means you choose the smallest number in the bracket. For example, min(7, 5) = min(5, 7) = 5.

So you have
rg = 1 [if 2(good(w)/G)) is more than 1]
rg = 2(good(w)/G)) [if it is less]
Of course, if neither one is less then they're the same and it doesn't matter which one you choose.

Max is the same, but the other way round. So max(7, 5) = max(5, 7) = 7.

As for Pr|w, I'm not sure - maybe probability of w?


modular arithmetic
induction
Divisibility Theory

Let $a, b$, and  $c$ be integers such that $a+b+c$ divides $a^2 +b^2 +c^2$. Prove that there are infinitely many positive integers $n$ such that $a+b+c$ divides $a^n +b^n +c^n$.

Just use the identity $ 2((x \plus{} y)^4 \plus{} x^4 \plus{} y^4) \equal{} ((x \plus{} y)^2 \plus{} x^2 \plus{} y^2)^2$.

So if $ x \plus{} y \plus{} z \equiv 0 \equiv x^2 \plus{} y^2 \plus{} z^2 \pmod{a \plus{} b \plus{} c}$ it follows from the identity that:

$ x^4 \plus{} y^4 \plus{} z^4 \equiv (y \plus{} z)^4 \plus{} y^4 \plus{} z^4 \equiv (y \plus{} z)^2 \plus{} y^2 \plus{} z^2 \equiv \\ x^2 \plus{} y^2 \plus{} z^2 \equiv 0 \pmod{a\plus{}b\plus{}c}$

Continuing with induction, if $ a^{2^k} \plus{} b^{2^k} \plus{} c^{2^k} \equiv 0 \pmod{a \plus{} b \plus{} c}$ for all $ k \leq n$ then:

$ a^{2^{n\plus{}1}} \plus{} b^{2^{n\plus{}1}} \plus{} c^{2^{n \plus{} 1}} \equiv (b^{2^{n \minus{} 1}} \plus{} c^{2^{n\minus{}1}})^4 \plus{} (b^{2^{n \minus{} 1}})^4 \plus{} (c^{2^{n\minus{}1}})^4 \equiv \\ (b^{2^{n \minus{} 1}} \plus{} c^{2^{n \minus{} 1}})^2 \plus{} (b^{2^{n \minus{} 1}})^2 \plus{} (c^{2^{n \minus{} 1}})^2 \equiv a^{2^n} \plus{} b^{2^n} \plus{} c^{2^n} \equiv 0 \pmod{a\plus{}b\plus{}c}$

WLOG we can assume that $(a,b,c)=1$ then if $(a,a+b+c)=(b,a+b+c)=(c,a+b+c)=1$

we can easily see that for $\forall t\in N: n=t\times \varphi(a+b+c)+2$ has the property because

 $a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2} \equiv a^2+b^2+c^2 \equiv 0 \mod (a+b+c)$

now we prove that the condition above always happens. let $a+b+c=\Pi p_i^{\alpha_i}$ if $p_i$ is a prime factor of one of $a,b,c$, for

example $a$ then because $(a,b,c)=1$ we get $p_i \nmid b,c$ and $p_i \mid b+c, p_i\mid b^2+c^2 \implies p_i \mid 2bc \implies p_i=2$

now because $a+b+c$ is even it means one of them is even and two others must be odd

note that three of them cannot be even because $(a,b,c)=1$ $\implies a^2+b^2+c^2 \equiv 6$ or $2 \mod 8$

which implies $2 \mid a+b+c, 4\nmid a+b+c$ and then because $2 \mid a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2}$

and $a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2} \equiv a^2+b^2+c^2 \equiv 0 \mod p_i^{\alpha_i}$

implies that $a+b+c \mid a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2}$ and we're done!


function
calculus
derivative
real analysis
real analysis unsolved

Let $f: \mathbb R^{2}\to \mathbb R$ be differentiable. For each $x\in \mathbb R$ define $g_{x}(y): \mathbb R\to\mathbb R$ by $g_{x}(y)=f(x,y)$. Suppose that for each $x$ there is a unique $y$ so that $g_{x}'(y)=0$; let $c(x)$ be this $y$.
a) if $D_{2,2}f(x,y)\neq 0$ for all $(x,y)$ show that $c$ is differentiable and \[c'(x)=-\frac{D_{2,1}f(x,c(x))}{D_{2,2}f(x,c(x))}\] b) Show that if $c'(x)=0$ then for some $y$ we have \[D_{2,1}f(x,y)=0\]  \[D_{2}f(x,y)=0\]

I think $f$ should be differentiable more than once. The problem is really about its partial derivative in $y$. Call it $g$. The solution of the equation $g(x,c(x))=0$ is differentiable provided that $g$ is and $D_{2}g\ne 0$ -- this is just the Implicit Function theorem. Part (b) seems to follow from (a) and the definition of $c(x)$.


abstract algebra
algebra
polynomial
advanced fields
advanced fields unsolved

Hi all,

I'm trying to show that if $ I$ is the ideal in $ k[x_1, \ldots, x_n]$ generated by a set $ T$, then $ Z(I)\equal{}Z(T)$ (where $ Z(T)\equal{}\{P \in \mathbb{A}^n |f(P)\equal{}0 \mbox{ for all } f \in T\}$, with $ \mathbb{A}^n$ being affine $ n$-space over $ k$).

I think this is probably a mere exercise in basic algebra, but I have so far been unable to prove one inclusion.  Showing $ Z(I) \subset Z(T)$ seems easy enough.  To show the reverse inclusion, I tried to say that since $ k[x_1, \ldots, x_n]$ is noetherian, $ I$ is finitely generated, say $ I\equal{}(f_1, \ldots, f_r)$, and so since it's the smallest ideal containing $ T$, we must have that $ (f_1, \ldots, \hat{f}_i, \ldots, f_r)$ does not contain $ T$.  To me this seems to imply that $ Z(T)$ is 'sandwiched', so to speak, between $ Z(f_1, \ldots, \hat{f}_i, \ldots, f_r)$ and $ Z(I)$ forcing it to actually equal $ Z(I)$.  Maybe I'm not explaining well what's going on in my head.  Any direction at all would be very much appreciated.

Thanks in advance.

If $ f(P) \equal{} 0$ and $ g(P) \equal{} 0$, then $ (f\plus{}g)(P) \equal{} 0$. If $ f(P) \equal{} 0$, and $ g$ is any polynomial, then $ (gf)(P) \equal{} 0$. Therefore $ Z(T) \subset Z(I)$, since $ I$ consists of finite sums $ g_1f_1 \plus{} \dots \plus{} g_nf_n$, where $ f_i \in T$ and $ g_i$ are any polynomial.

That's what I understood the definition of an ideal generated by a set to be, but revisiting Dummit and Foote, they defined the ideal $ (T)$ generated by the set $ T$ to be the smallest ideal containing $ T$.  Therefore, $ (T)\equal{}\cap_{T \subset J} J$ over all such ideals $ J$.  I'm having trouble seeing that these two definitions are equivalent.  But thank you so much for your answer to my question.

These are the same, and it's not too hard to see that: if $ g,h \in I$, for any ideal $ I$, then so is $ ag \plus{} bh$ (where $ a,b$ are in your ring). So any ideal containing $ T$ must contain these formal sums. Furthermore, the set of these formal sums itself forms an ideal.


trigonometry
geometry
angle bisector
trig identities
Law of Sines
geometry unsolved

Let $ ABC$ be a triangle, and $ P$ a point on the internal bisector of angle $ \angle BAC$. It is known that $ \angle ABP\equal{}30^{\circ}$ and $ CP\equal{}BC$. Find the measure of the angle $ \angle APC$.

Let
$ X\in AB, PX\perp AB$
$ Z\in PB, PZ \equal{} PX$
$ Y\in AC, PY\perp AC$

$ \angle XPZ \equal{} 90^0 \minus{} \angle XBP \equal{} 90^0 \minus{} 30^0 \equal{} 60^0$
$ PX \equal{} PZ\implies \Delta PXZ$ is equilateral.
$ \angle BXZ \equal{} 90^0 \minus{} \angle PXZ \equal{} 90^0 \minus{} 60^0 \equal{} 30^0 \equal{} \angle XBZ$
$ \implies ZX \equal{} ZB$
By $ AAS\implies \Delta AXP\equiv \Delta AYP$
$ PX \equal{} PY$
$ PY \equal{} PX \equal{} PZ \equal{} ZX \equal{} BZ$
$ CZ\perp PB$
By $ RHS\implies \Delta CPZ\equiv \Delta CPY$
$ \angle APC \equal{} \angle APY \plus{} \angle YPC \equal{} \frac {1}{2}\angle XPY \plus{} \frac {1}{2}\angle YPZ \equal{} \frac {1}{2}(\angle XPY \plus{} \angle YPZ) \equal{}$
$ \equal{} \frac {1}{2}(360^0 \minus{} \angle XPZ) \equal{} \frac {1}{2}(360^0 \minus{} 60^0) \equal{} \boxed{150^0}$

Let the intersection of angle bisector of angle A and BC be K. Extend BP to meet AC at M. Extend CP to meet AB at L. Let angle BAK = angle CAK = $ x$. Since triangle BPC is isosceles let angle BPC = angle PBC = $ y$.
Simple angle chasing yields angle ACL = $ y\minus{}30\minus{}2x$. Also angle BCP = $ 180\minus{} 2y$ and Angle APC is equal to $ 210 \plus{} x \minus{} y$. Now we applying Trignometric form of Ceva's theorem to cevians AK, CL and BM we get
$ (sinx/sinx)(sin30/siny)(sin(180\minus{} 2y)/sin(y \minus{} 30 \minus{}2x)) \equal{} 1$
Simplifying we get $ sin( y\minus{} 30 \minus{} 2x) \equal{} cosy (i)$. From triangle BPC we know both $ y$ and $ y \minus{} 30\minus{} 2x$ are acute. From $ (i)$ we now get $ y\minus{} 30 \minus{} 2x \equal{} 90 \minus{} y$.
From this we get $ y \minus{} x \equal{} 60$. But angle APC = $ 210 \plus{} x \minus{} y \equal{} 210 \minus{}( y\minus{} x)$.
But $ y \minus{} x \equal{} 60$. This implies angle APC = $ 210 \minus{} 60 \equal{} 150$
$ QED$

I will use Akashnil picture:) $ PX \equal{} PY$ so $ BP \equal{} 2PC*sin\angle PCY$  let $ \angle PBC \equal{} x$ so from triangle BPC by law of sines($ CB \equal{} CP$) we get $ BP \equal{} 2cosx*PC$ or $ cosx \equal{} sin\angle PCY$ and $ \angle XBP < \frac {\pi}{2}$ so $ \angle PCY \equal{} \frac {\pi}{2} \minus{} x$
$ \angle ABC \plus{} \angle BAC \plus{} \angle BCA \equal{} \pi$ or $ \angle BAC \equal{} 2x \minus{} \frac {2\pi}{3}$ so $ \angle PAC \equal{} x \minus{} \frac {\pi}{3}$ and from triangle APC $ \angle APC \equal{} \frac {5\pi}{6}$


Two sequences [tex]x_n[/tex] and [tex]y_n[/tex] are defined recursively by

[tex]\displaystyle{x_1=y_1=\frac{1}{5}}[/tex]

[tex]\displaystyle{x_n= \frac{x_{n-1}+5y_{n-1}}{2} \mbox{ for }x>1}[/tex]

[tex]\displaystyle{y_n= \frac{x_{n-1}+y_{n-1}}{2} \mbox{ for }x>1}[/tex]

Show that [tex]x_n[/tex] never takes integer values whereas [tex]y_n[/tex] takes integer values infinitely often (i.e. there are an infinite number of values of n for which [tex]y_n[/tex] is an integer).

[hide]By induction we have :

. a(n) = p(n)/5 with p(n) integer

. b(n) = q(n)/5 with q(n) integer

. p(n+2) = p(n+1)+p(n), p(1) = 1, p(2) = 3

. q(n+2) = q(n+1)+q(n), q(1) = 1, q(2) = 1

So :

. q is in fact the famous Fibonacci sequence which have many many many properties including 5 | Fib(5*n) for all n. b(5n) is an integer for all n.

. (p(n) mod 5) has a very short period,  it is 1,3,4,2,1,3,4,2,1,.....

Hence p(n) is never divisible by 5 so a(n) is never an integer.

[/hide]

Belenos

Yes, well done - you spotted the [hide]Fibonacci [/hide]number link straight away.

But can you explain your very first induction step in more detail ?

If x(n)=p(n)/5 and y(n)=q(n)/5 with p(n) and q(n) integers, can we definitely conclude that x(n+1)=p(n+1)/5 and y(n+1)=q(n+1)/5 with p(n+1) and q(n+1) also integers ?

What if p(n) is even and q(n) is odd (or vice versa) ?

Although the conclusion is correct - p(n) and q(n) are indeed integers for all n - I couldn't see how to prove this by induction.

Yes you're right, there is something wrong in what I wrote.

It should be :

. a(n) = p(n)/5
. b(n) = q(n)/5
. p(n+2) = p(n+1)+p(n), p(1) = 1, p(2) = 3
. q(n+2) = q(n+1)+q(n), q(1) = 1, q(2) = 1

So the p(n), q(n) are integer !

Okay, so once you spot the re-arranged recursion formulae then you are home and dry.

But how do you get to the re-arranged recursion formulae ?

For example, what if we had

p(1)=1; q(1)=3

p(n+1) = (3p(n) + q(n))/2

q(n+1) = (13p(n) + 3q(n))/2

How do you get from here to second-order recursion formulae that show that p(n) and q(n) are always integers for all n ?

Well I haven't actually worked out the details in this case, but the standard way of doing that would be [hide] use the first formula to work out q(n) in terms of p(n) and p(n+1), and then substitute into the second formula everywhere a q is. [/hide]


Find five integers, the sum of the squares of every four of which is a square.

Solution of the question is only possible when they are all equal to zero or one of them is zero two will be 1,2 and other will be there negative.
Solution-(-2,-1,0,1,2) or ( 0,0,0,0,0)


Let $ a_1,a_2,\ldots ,a_n$ be $ n$ complex numbers. Define $ \xi \equal{} \displaystyle \max \left\{\frac {|a_k|}{k} , 1\leq k \leq n \right \}$. Prove that all the roots of $ p(x) \equal{} \xi x^n \plus{} a_1x^{n \minus{} 1} \plus{} \ldots \plus{} a_n$ satisfy the relation $ |z| < \frac {3 \plus{} \sqrt {5}}{2}$

It's a straightforward application of Rouche's theorem. If $ |z|\equal{}r$, $ \left|a_kz^{n\minus{}k}\right|\le k\xi r^{n\minus{}k}$, and $ \sum_{k\equal{}1}^n \left|a_kz^{n\minus{}k}\right|\le \xi r^n\sum_{k\equal{}1}^n kr^{\minus{}k}< \xi r^n\sum_{k\equal{}1}^{\infty}kr^{\minus{}k}\equal{}\xi r^n\frac {\frac1r}{(1\minus{}\frac1r)^2}$

We want $ \frac{\frac1r}{(1\minus{}\frac1r)^2}\le 1$, or equivalently $ \frac{r}{(1\minus{}r)^2}\le 1$ and $ 1\minus{}3r\plus{}r^2\ge0$. This occurs when $ r\ge \frac{3\plus{}\sqrt{5}}{2}$.

On any circle with radius $ \ge \frac{3\plus{}\sqrt{5}}{2}$, the first term strictly dominates all other terms, and there is no zero. All zeros must therefore be strictly inside that disk.

Is rouche's theorem a standard Olympiad technique? Or are you simply using it from complex analysis?


Two cyclists, $ k$ miles apart, and starting at the same time, would be together in $ r$ hours if they traveled in the same direction, but would pass each other in $ t$ hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:

$ \textbf{(A)}\ \frac {r \plus{} t}{r \minus{} t} \qquad\textbf{(B)}\ \frac {r}{r \minus{} t} \qquad\textbf{(C)}\ \frac {r \plus{} t}{r} \qquad\textbf{(D)}\ \frac {r}{t} \qquad\textbf{(E)}\ \frac {r \plus{} k}{t \minus{} k}$

[hide="Click for solution"]
Let $ f$ and $ s$ denote the speeds of the faster and slower cyclists, respectively.  The distances (measured from the starting point of the faster cyclist) at which they meet are:

$ fr\equal{}k\plus{}sr$ when they travel in the same direction, and

$ ft\equal{}k\minus{}st$ when the slower one comes to meet the other.

From the second equation, $ f\plus{}s\equal{}\frac{k}{t}$, from the first we have $ f\minus{}s \equal{} \frac{k}{r}$.  Thus, $ 2f\equal{}\frac{k}{r}\plus{}\frac{k}{t}\equal{}\frac{k(r\plus{}t)}{rt}$ and $ 2s\equal{}\frac{k}{t}\minus{}\frac{k}{r}\equal{}\frac{k(r\minus{}t)}{rt}$.  Thus, $ \frac{f}{s}\equal{}\frac{r\plus{}t}{r\minus{}t}$, or $ \boxed{\textbf{(A)}}$.
[/hide]

Let the faster cyclist have speed $ f$ and the slower have speed $ s$.  Every single hour, the faster cyclist is getting $ f \minus{} s$ closer to the slower cyclist.  They will meet, therefore, after $ \frac {k}{f \minus{} s}$ hours.  If they're going in the same direction, they will meet after $ \frac {k}{f \plus{} s}$ hours.   Setting our times to their respective variables, we get that:
$ (1) \frac {k}{f \minus{} s} \equal{} r$ and
$ (2) \frac {k}{f \plus{} s} \equal{} t$.
Solving these equations, we find that $ \frac {f}{s} \equal{} \boxed{\frac {r \plus{} t}{r \minus{} t}}$.


vector
superior algebra
superior algebra unsolved

Let $ R$ be a left primitive ring and V be a faithful simple R-module. Let K be the division ring $ End (_{R}V)$. Then if R is a left artinian, then $ n: \equal{}dim (_{K}V)$ is finite, and R is isomorphic $ M_{n}(K)$.
Proof
Assume $ dim (_{K}V)\equal{}n$, we know that R is isomorphic $ M_{n}(K)$. Why R is left artinian???

[quote="wiosna"]Let $ R$ be a left primitive ring and V be a faithful simple R-module. Let K be the division ring $ End (_{R}V)$. Then if R is a left artinian, then $ n: \equal{} dim (_{K}V)$ is finite, and R is isomorphic $ M_{n}(K)$.
Proof
Assume $ dim (_{K}V) \equal{} n$, we know that R is isomorphic $ M_{n}(K)$. Why R is left artinian???[/quote]

A left ideal $ I$ of $ M_n(K)$ is also a left $ K$ vector subspace of $ M_n(K)$ and $ \dim_K I < \infty$ because $ \dim_K M_n(K) < \infty.$ So every descending (or ascending) chain of left ideals

eventually stabilizes.


inequalities
Cauchy Inequality
inequalities unsolved

$ a,b,c$positive numbers
Show that
$ \frac {3 \plus{} \sqrt {3}}{9} \geq \frac {abc(a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2)}}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}$

We have
$ a\plus{}b\plus{}c \le \sqrt{3(a^2\plus{}b^2\plus{}c^2)}$
It suffices to prove that
$ \frac{\sqrt{3}}{9}(ab\plus{}bc\plus{}ca)\sqrt{a^2\plus{}b^2\plus{}c^2} \ge abc$
By applying Cauchy inequality for 3 positive real numbers, it leads to
$ ab\plus{}bc\plus{}ca \ge 3\sqrt[3]{a^2b^2c^2}$
and $ a^2\plus{}b^2\plus{}c^2 \ge 3\sqrt[3]{a^2b^2c^2}$

We are done! :)

$ \frac {abc(a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2)}}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}$
$ \equal{} \frac{3}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} } \frac {a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}}{3(a^2 \plus{} b^2 \plus{} c^2)}$
$ \leq\frac{a\plus{}b\plus{}c}{3} \frac {(a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2)}}{3(a^2 \plus{} b^2 \plus{} c^2)}$
$ \equal{}\frac{1}{9}(\frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{a\plus{}b\plus{}c}{\sqrt{a^2\plus{}b^2\plus{}c^2}})$
$ \leq\frac{1}{9}(3\plus{}\sqrt{3})\equal{}\frac {3 \plus{} \sqrt {3}}{9}$


abstract algebra
geometry
circumcircle
trigonometry
3D geometry
calculus
derivative

Omdat ik weet dat het voor veel mensen hier een moeilijk begin is op MathLinks om direct in het Engels te beginnen problemen oplossen, dacht ik dat dit woordenboek misschien wel een handig hulpje zou kunnen zijn. Ik zal al beginnen met wat algemene termen, [b]wie wil meehelpen met een bepaalde tak uit te spitten: meer dan welkom![/b] :)
[i](of als ik per ongeluk 'Vlaamse' woorden gebruik waar er in Nederland andere zijn, vertel ook gerust)[/i]

[hide="Engels - Nederlands"]
[size=167][color=darkblue]A[/color][/size]
abelian = abels
absolute value = absolute waarde
acute angle = scherpe hoek
acute-angled triangle = scherphoekige driehoek
altitude = hoogte
angle = hoek
approximation = benadering
arc = cirkelboog
area = oppervlakte
arithmetic mean (AM) = rekenkundig gemiddelde
arithmetic sequence = rekenkundige rij
average = gemiddeld, ongeveer
axis = as
[size=167][color=darkblue]B[/color][/size]
base = basis, grondgetal
bisector = bissectrice
bounded = begrensd
[size=167][color=darkblue]C[/color][/size]
cancel = schrappen van beide zijden
centroid = zwaartepunt (v.e. driehoek)
circumcenter = middelpunt van de omgeschreven cirkel
circumcircle = omgeschreven cirkel
circumference = omtrek
circumscribed = omgeschreven aan
closure = sluiting
common denominator = gemeenschappelijke noemer
common multiple = gemeenschappelijk veelvoud
concurrence = het snijden (van rechten/krommen)
concurrent = snijdend (van rechten/krommen)
conjugate = toegevoegd
connected = samenhangend
convex polygon = veelhoek waarbij ieder lijnstuk tussen 2 punten binnen de veelhoek ligt
coprime = onderling ondeelbaar, relatief priem
coset = nevenklasse
cosine = cosinus
crap = afval, van lage kwaliteit
cube = kubus / derde macht
cube root = derdemachtswortel
curl = rotor
curvature = kromming
curve = kromme
cyclic = cyclisch
cyclic quadrilateral = koordenvierhoek
[size=167][color=darkblue]D[/color][/size]
decagon = tienhoek
decompose = ontbinden in factoren
degree = graad
denominator = noemer
derivative = afgeleide
difference = verschil
digit = cijfer
directed = gericht
divisible = deelbaar
divisor = deler
dodecagon = twaalfhoek
[size=167][color=darkblue]E[/color][/size]
edge (combinatorics) = tak (van een graaf)
edge (geometry) = zijde
equal = gelijk
equation = vergelijking
equidistant = op gelijke afstand
equilateral = gelijkzijdig
even = even
expand = uitwerken (ook: ontwikkelen van bv. een reeks)
extension = uitbreiding
exterior angle = buitenhoek
[size=167][color=darkblue]F[/color][/size]
fool's proof = nepbewijs
face = zijvlak
factorial = faculteit
FLT = kleine stelling van Fermat
fraction = breuk
[size=167][color=darkblue]G[/color][/size]
geometry = meetkunde
geometric mean (GM) = meetkundig gemiddelde
geometric sequence = meetkundige rij
geometric series = meetkundige reeks
graph (analysis) = grafiek
graph (combinatorics) = graaf (grafentheorie)
greatest common divisor (GCD) = grootste gemene deler
[size=167][color=darkblue]H[/color][/size]
harmonic mean (HM) = harmonisch gemiddelde
heigth = hoogte
heptagon = zevenhoek
hexagon = zeshoek
hypotenuse = hypotenusa (schuine zijde van rechthoekige driehoek)
[size=167][color=darkblue]I[/color][/size]
incenter = middelpunt van de ingeschreven cirkel
index = volgnummer
inequality = ongelijkheid
infinity = oneindig(heid)
inscribed = ingeschreven
integer = geheel getal
intersecting lines = snijdende rechten
inverse = omgekeerde
irregular = onregelmatig
isosceles = gelijkbenig
iteration = herhaling
[size=167][color=darkblue]J[/color][/size]
(nog geen woorden)
[size=167][color=darkblue]K[/color][/size]
(nog geen woorden)
[size=167][color=darkblue]L[/color][/size]
least common multiple (LCM) = kleinste gemene veelvoud
leg = been, rechthoekszijde
lemma = lemma (hulpstelling)
line = rechte
locus = meetkundige plaats
[size=167][color=darkblue]M[/color][/size]
map = afbeelding
mean = gemiddelde
median (in geometry) = zwaartelijn
median (in statistics) = mediaan
midpoint = midden
multiple = veelvoud
multiplication = vermenigvuldiging
[size=167][color=darkblue]N[/color][/size]
number base = basis van het getalstelsel
numerator = teller
number = getal
[size=167][color=darkblue]O[/color][/size]
obtuse angle = stompe hoek
obtuse-angled triangle = stomphoekige driehoek
octagon = achthoek
odd = oneven
opposite angles = overstaande hoeken
orbit = baan (oa in groepentheorie)
origin = oorspong (van assenstelsel)
orthocenter = snijpunt van de hoogtelijnen
[size=167][color=darkblue]P[/color][/size]
pair = paar, tweetal
pentagon = vijfhoek
perfect square = volmaakt kwadraat
perpendicular = loodrecht
perpendicular bisector = middelloodlijn
plane = vlak
polygon = veelhoek
polynomial = veelterm
power = macht
probability = waarschijnlijkheid / kans
proof = bewijs
Pythagorean triple = pythagorees drietal
[size=167][color=darkblue]Q[/color][/size]
quadratic = kwadratische vorm
quadratic mean (QM) = kwadratisch gemiddelde
quadrilateral = vierhoek
Q.e.D (quod erat demonstrandum) = wat te bewijzen was
[size=167][color=darkblue]R[/color][/size]
radical = symbool voor worteltrekking
radical axis = machtslijn
radius = straal
random = willekeurig, toevallig
range = bereik, waardenbereik
ratio = verhouding
reciprocal = omgekeerde breuk
reflection = spiegeling
regular = regelmatig
relatively prime = relatief priem, onderling ondeelbaar
remainder = rest (bij deling)
representation = voorstelling
right angle = rechte hoek
right-angled triangle
rigorous = wiskundig correct
root = wortel (van een veelterm)
row = platte rij in een rechthoek of matrix
[size=167][color=darkblue]S[/color][/size]
segment = lijnstuk
sequence = rij
series = reeks
set = verzameling
similar = gelijkvormig
sine = sinus
solution = oplossing
solvable = oplosbaar
space = ruimte
sphere = bol
square = vierkant / kwadraat
square root = vierkantswortel
squarefree = kwadraatvrij, niet deelbaar door een kwadraat
subtraction = aftrekking
surface = oppervlak
statement = verwoording
[size=167][color=darkblue]T[/color][/size]
tangent = raaklijn; rakend; tangens
theorem = stelling
trace = spoor
trapezoid = trapezium
triangle = driehoek
trigonometry = goniometrie
[size=167][color=darkblue]U[/color][/size]
unique = uniek, enig
unit = eenheid
[size=167][color=darkblue]V[/color][/size]
vertex (combinatorics) = knoop (van een graaf)
vertex (geometry) = hoekpunt
vertices = meervoud van vertex
[size=167][color=darkblue]W[/color][/size]
WLOG (without loss of generality) = zonder verlies van algemeenheid
[size=167][color=darkblue]X[/color][/size]
(nog geen woorden)
[size=167][color=darkblue]Y[/color][/size]
(nog geen woorden)
[size=167][color=darkblue]Z[/color][/size]
(nog geen woorden)[/hide]
Nederlands - Engels zal wel komen eens dit wat gegroeid is :)

[hide="Vlamingen Opgepast!"]Wij gebruiken (zoals vaker) de franse conventies terwijl hier de amerikaanse gebruikt worden. Dus:
negative = strikt negatief
positive = strikt positief
increasing = strikt stijgend
decreasing = strikt dalend
non-increasing = gewoon dalend
non-decreasing = gewoon stijgend
nonnegative = gewoon positief
nonpositive = gewoon negatief, maar wordt eigenlijk niet gezegd[/hide]

Mooie werkje!

Ik zei [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=293702#p293702]gisteren[/url] nog:
[quote]...there is very little material in my language, Dutch. The fact that you have only training materials in English, discourages people to train themselves.[/quote]
Dit kan zeker de drempel voor Nederlandstalige mensen om het Mathlinks-forum te verkken verlagen  :) (y)

[quote="Stijn"]Mooie werkje![/quote] Het is wel lang nog niet af. En eigenlijk de bedoeling dat jullie een beetje helpen aanvullen, ik kan niet aan alle woorden denken... :oops:

Er bestaat een woordenboek Nederlands-Engels-Frans-Duits voor wiskundige termen, dat mijn moeder heeft. Ik weet nu helaas niet precies meer hoe het heet, dus ik kan geen goede verwijzing geven, maar ik heb het wel een keer in de boekwinkel zien staan voor ca. 5 euro, dus iedereen die geinteresseerd is zou daarnaar op zoek kunnen gaan.
  Ik zoek nog wel de precieze verwijzing een keer op.

[quote="Arne"]Hm, ik zie dat er trouwens ook nog veel meetkundige termen ontbreken :D Zal daar werk van maken wanneer ik daar de tijd voor vind.[/quote]
That's what I was pushing for with the [b]bold[/b] phrase ;) :P

[quote="Peter VDD"]That's what I was pushing for with the [b]bold[/b] phrase ;) :P[/quote]

Euhm... we zitten hier nog altijd op het Nederlandstalige forum Peter? :D

Een beetje talenflexibiliteit kan nooit kwaad. :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=84952[/url]
Op mijn verzoek hebben de mensen van het Franstalige topic een woordenlijstje Engels-Frans gemaakt.

Het is gebaseerd op deze lijst, men heeft gewoon wat hier in het Engels stond vertaald naar het Frans.
Kan altijd handig zijn!
Toch fijn van hen he?

[quote="Arne"]Hm, ik zie dat er trouwens ook nog veel meetkundige termen ontbreken :D Zal daar werk van maken wanneer ik daar de tijd voor vind.[/quote] Denk je hier nog eens aan, Arne?

Bij deze zitten we aan 174 begrippen.

Dat moet je me nu eens écht niet in de examenperiode vragen hé Peter...

Ik zeg ook niet dat het NU moet he. :P

Maar als je wat meetkundedocumenten schrijft deze vakantie, mss die woorden effenaan toevoegen ofzo?

Een poging tot meer structuur : het is een beetje absurd om met een lijst Frans naar Nederlands, Nederlands naar Frans, Frans naar Engels enz... bezig te zitten.

Ik zou willen alles in excel zitten. Als ik het goed begrijp kan je in een excel file sorteren op eender wat.  Ik wil wel de geslachten niet verwaarlozen, en als ik zo le/la/une enz ervoor zet loopt dat mis.

Ik ga het toch proberen.  Maar alle repliek is welkom.

Okay laat me es weten wat jullie hiervan denken.
Schuchtere eerste poging!!

Hmm... om een of andere reden heb ik deze post nooit gezien.

Anyway, ja, andere talen kun je in een excel-file doen, goed idee.

Engels-Nederlands en Nederlands-Engels zou ik hier wel direct zo houden, da's toch het meest gebruikte.


Sue has 122 crates of the same size. Each crate contains at least 90 apples and at most 100 apples. Crates containing the same number of apples are stacked one on top of another in their own stack. Not all of the stacks have the same number of crates. What is the least possible number of crates in the tallest stack of crates?

How do you solve this prob?

There are 11 possibilities for the number of apples in a crate. For there to be the lowest number of crates in the highest stack, we use the pigeonhole principle. The minmum is 12 in which case there are 10 stacks of 11 and 1 stack of 12.

If you want the least number of crates in the tallest stack, you're going to ideally want all of the stacks to have the same number of crates, except for the tallest, which will have one more.  That is, suppose there are $ p$ stacks.  You want $ p\minus{}1$ of the stacks to have $ n$ crates, and the other to have $ n\plus{}1$ crates in the ideal scenario.

This is pretty much the same thing as the above post.


limit
integration
logarithms
trigonometry
calculus
calculus computations

[b]Ascertain :[/b]

$1\blacktriangleright\lim_{n\to\infty}I_{n}$ , where $I_{n}=\frac{1}{n}\cdot\int_{0}^{n}\ln (n+[x])\ dx-\ln n\ .$

$2\blacktriangleright\lim_{n\to\infty}\frac{a_{n}}{b_{n}}$ , where $a_{n}=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\arctan nx\ dx$ and $b_{n}=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\arcsin nx\ dx\ .$

$3\blacktriangleright I=\int_{0}^{2n\pi}\min \{sin x\ ,\ \arcsin (\sin x)\}\ dx\ .$[hide="Answer :"]$n\cdot\left(\frac{\pi^{2}}{4}-2\right)\ .$[/hide]

$4\blacktriangleright J=\int_{0}^{\pi}x[\sin^{2}(\sin x)+\cos^{2}(\cos x)]\ dx\ .$[hide="Answer :"]$\frac{\pi^{2}}{2}\ .$[/hide]

$5\blacktriangleright K=\int_{0}^{\pi}x\sin(\cos^{2}x)\cos(\sin^{2}x)\ dx\ .$[hide="Answer :"]$\frac{\pi^{2}}{4}\cdot\sin 1\ .$[/hide]

Hi,
Here is my approach:
1) $I_{n}=\frac{\ln n+\ln (n+1)+\cdots+\ln (2n-1)-n\ln(n)}{n}$, which converges to $2\ln(2)$ via Cesaro Stolz lemma.

2) $a_{n}=(\frac{1}{n}-\frac{1}{n+1})\arctan{n\theta_{1}}$ for some $\theta_{1}\in (\frac{1}{n+1},\frac{1}{n})$
$b_{n}=(\frac{1}{n}-\frac{1}{n+1})\arcsin{n\theta_{2}}$ for some $\theta_{2}\in(\frac{1}{n+1},\frac{1}{n})$.
Thus we get that
${\frac{a_{n}}{b_{n}}=\frac{\arctan{n\theta_{1}}}{\arcsin{n\theta_{2}}}\rightarrow \frac{\frac{\pi}{4}}{\frac{\pi}{2}}}=2$.
Didilica

$1\blacktriangleright I_{n}=\frac{1}{n}\cdot\sum_{k=1}^{n-1}\ln (n+k)-\ln n=$ $\frac{1}{n}\cdot \left(\sum_{k=1}^{n-1}\ln (n+k)-\sum_{k=1}^{n-1}\ln n-\ln n\right)=$ $\frac{1}{n}\cdot\sum_{k=1}^{n}\ln \left(1+\frac{k}{n}\right)-\frac{\ln n}{n}-\frac{\ln 2}{n}\rightarrow \int_{0}^{1}\ln (x+1)\ dx=\boxed{\ 2\ln 2-1\ }\ .$

$2\blacktriangleright$ Prove easily that $a_{n}\rightarrow 0$ , $b_{n}\rightarrow 0\ .$ From Cauchy's theorem obtain that for any $n\in N^{*}$ exists $c_{n}\in \left(\ \frac{1}{n+1}\ ,\ \frac{1}{n}\ \right)$

so that $\frac{a_{n}}{b_{n}}=\frac{\arctan nc_{n}}{\arcsin nc_{n}}\ .$ But $\frac{n}{n+1}<nc_{n}<1\ ,\ n\in N^{*}\ .$ Therefore, $nc_{n}\rightarrow 1$ and $\frac{a_{n}}{b_{n}}\rightarrow \frac{\frac{\pi}{4}}{\frac{\pi}{2}}=\boxed{\ \frac{1}{2}\ }\ .$

4) Use the substitution $\pi-x=y$ ang get that
$J=\frac{\pi}{2}\int_{0}^{\pi}[\sin^{2}(\sin(x))+\cos^{2}(\cos(x))]dx$.
$\int_{0}^{\pi}[\sin^{2}(\sin(x))+\cos^{2}(\cos(x))]dx=\int_{0}^{\frac{\pi}{2}}[\sin^{2}(\sin(x))+\cos^{2}(\cos(x))]dx+\int_{\frac{\pi}{2}}^{\pi}[\sin^{2}(\sin(x))+\cos^{2}(\cos(x))]dx$.
Use the substitution $\pi-x=y$ in the second integral , add the two integrals and we get that $J=\frac{\pi^{2}}{2}$.
I hope that there are no computational mistakes( this is the idea).
5) Using the same substitution as above we get that
$K=\frac{\pi}{2}\int_{0}^{\pi}\sin(\cos^{2}(x))\cos(\sin^{2}(x))dx$.
Now $\sin(1)=\sin(\sin^{2}(x)+\cos^{2}(x))=\sin(\sin^{2}(x))\cos(\cos^{2}(x))+\cos(\sin^{2}(x))\sin^{2}(\cos^{2}(x))$
Thus,
$K=\frac{\pi}{2}[\int_{0}^{\frac{\pi}{2}}\sin(\cos^{2}(x))\cos(\sin^{2}(x))dx+\int_{\frac{\pi}{2}}^{\pi}\sin(\cos^{2}(x))\cos(\sin^{2}(x))dx]$.
In the second integral make the substitution $x-\frac{\pi}{2}=y$ and we get that:
$K=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\sin(1)dx=\sin(1)\frac{\pi^{2}}{4}$.
I hope that there are no mistakes.
Nice integrals. Who proposed them?

I have a solution to the 3rd one.
the given integral canbe written as,
${I=n\int_{0}^{2\pi}\min\{sinx,arcsin(sinx)}$$dx$
or $I=n[\int_{0}^{\pi}sinxdx+\int_{\pi}^{2\pi}arcsin(sinx)dx]$
(from graph of both)
or $I=n$ $[$ $\frac{\pi^{2}}{4}-2$ $]$ (sorry I intented here yo print the final answer but not getting it rightly)


quadratics
algebra
polynomial
quadratic formula
special factorizations

One of my homework assignments is an extended response. I have to derive the [color=blue]quadratic formula[/color] from the [color=red]standard general form[/color] explaining how and why i took each step.

[color=red]Standard form:[/color]ax^2+bx+c=0

[color=blue]Quadratic Formula:   [img]http://trevorpythag.files.wordpress.com/2008/02/qdrtc2.gif[/img]
[/color]
Please help me!! :huh:

Go to the AoPSWiki. It's one of the buttons at the left, and it contains links to learn about LaTeX. (Please learn to use it! I beg of you!) :)

You are given the standard form of a quadratic, which is $ ax^2 \plus{} bx \plus{} c \equal{} 0$. What you want to do is something called completing the square, which gives you a perfect square polynomial (you'll understand in one of the steps below).

First, we assume that the quadratic has $ a\neq0$, by its definition. Dividing both sides by $ a$, we get $ x^2 \plus{} \dfrac{bx}{a} \plus{} \dfrac{c}{a} \equal{} 0$.

Subtract $ \dfrac{c}{a}$ from both sides to get $ x^2 \plus{} \dfrac{bx}{a} \equal{} \minus{}\dfrac{c}{a}$.

Add $ \dfrac{b^2}{4a^2}$ to both sides to get $ x^2 \plus{} \dfrac{bx}{a} \plus{} \dfrac{b^2}{4a^2} \equal{} \dfrac{b^2 \minus{} 4ac}{4a^2}$. This step is called completing the square, as you can see in the next step.

Now we see that the left side factors as $ \left(x \plus{} \dfrac{b}{2a}\right)^2$.

Taking the square root of both sides yields $ x \plus{} \dfrac{b}{2a} \equal{} \dfrac{\pm\sqrt{b^2 \minus{} 4ac}}{2a}$.

Subtract $ \dfrac{b}{2a}$ from both sides to get $ \boxed{x \equal{} \dfrac{\minus{}b\pm\sqrt{b^2 \minus{} 4ac}}{2a}}$. This is the standard-form quadratic formula.

my parents are always pushing me to find different proofs for things. i found that one, but im wondering, are there any more?

If you're just looking to prove the quadratic formula works without deriving it, you can always plug it back in. :wink:

see this link for Alternative Derivation of the Quadratic Formula:
[url]http://www.karlscalculus.org/quadrtic.html[/url]

and [url]http://en.wikipedia.org/wiki/Quadratic_equation#By_Lagrange_resolvents[/url] using Lagrange resolvents...

Actually, if you know how to complete the square, then it shouldn't be that hard. Just use standard completeing the square.

i agree with grinchlygreen...a simple solution to a simple problem is always the best solution...

besides that method, I think that there were at least 5 others, because last year, my math teacher made us solve a lot of quadratics and gave us a packet on solving them. completing the square, i think she said, was the one that was always useable?

What about this method?
[hide]$ ax^2 \plus{} bx \plus{}c \equal{} 0$
$ ax^2 \plus{} bx \equal{} \minus{}c$
Multiplying by 4a throughout, we get
$ 4{a}^2{x}^2 \plus{} 4abx \equal{} \minus{}4ac$
$ {(2ax)}^2 \plus{} 4abx \plus{} b^2 \equal{} b^2 \minus{} 4ac$
$ {(2ax\plus{}b)}^2 \equal{} b^2 \minus{} 4ac$
$ 2ax \plus{} b \equal{} \pm \sqrt{b^2 \minus{} 4ac}$
$ 2ax \equal{} \minus{}b \pm \sqrt{b^2 \minus{} 4ac}$
$ {x} \equal{} \frac{\minus{}b \pm \sqrt{b^2 \minus{} 4ac}}{2a}$ which is the quadratic formula[/hide]

If you have Intro to Algebra, then the proof of the Quadratic Formula is in Chapter 13 along with full detail of completing the square.


Let a,b,c are real numbers. Prove that
$ a^4\plus{}b^4\plus{}c^4\plus{}ab^3\plus{}bc^3\plus{}ca^3\geq 2(a^3b\plus{}b^3c\plus{}c^3a)$
 :D
I hope a nice solution!

[quote="thegod277"]Let a,b,c are real numbers. Prove that
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} ab^3 \plus{} bc^3 \plus{} ca^3\geq 2(a^3b \plus{} b^3c \plus{} c^3a)$
 :D
I hope a nice solution![/quote]
$ \sum_{cyc}(a^4\plus{}a^3c\minus{}2a^3b)\equal{}\frac{1}{2}\sum_{cyc}(a^2\minus{}b^2\minus{}ab\plus{}bc)^2.$
I don't found the link.

$ a^4\plus{}b^4\plus{}c^4\plus{}ab^3\plus{}bc^3\plus{}ca^3\minus{}2(a^3b\plus{}b^3c\plus{}c^3a)\geq 0$
 $ \sum (a^3(a\plus{}c\minus{}2b))\geq 0$
 $ \sum (a^3(a\minus{}b)\plus{}a^3(c\minus{}b))\geq 0$
 $ \sum (a\minus{}b)(a^3\minus{}c^3)\geq 0$
 $ \sum(a^2\plus{}ac\plus{}c^2)(a\minus{}b)(a\minus{}c)\geq 0$
 It is easy to prove $ \sqrt{a^2\plus{}ac\plus{}c^2}\plus{}\sqrt{b^2\plus{}ba\plus{}a^2} \geq \sqrt{c^2\plus{}cb\plus{}b^2}$
 thus, apply The Cuon Schur the inequality is proved. :D

This is The Cuon Schur inequality:
Let a,b,c,x,y,z are real numbers. If $ |x| \plus{} |z|\geq |y|$ and $ a\geq b\geq c$ or $ a\leq b\leq c$, prove the inequality
$ x^2(a \minus{} b)(a \minus{} c) \plus{} y^2(b \minus{} c)(b \minus{} a) \plus{} z^2(c \minus{} a)(c \minus{} b)\geq 0$ :D
It is easy...? :)


[quote="moldovan"]If $ a,b,c$ are positive real numbers, prove the inequality:

$ \frac {a^2}{b} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^4}{a^3} \ge \minus{} a \plus{} 2b \plus{} 3c.$[/quote]
Oh, if $ a\equal{}b\equal{}c\equal{}1$, this inequality is wrong ?

[quote="moldovan"]If $ a,b,c$ are positive real numbers, prove the inequality:

$ \frac {a^2}{b} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^4}{a^3} \ge \minus{} a \plus{} 2b \plus{} 2c.$

P.S. Thank you for the observation. I've already corrected it.[/quote]

i think we can add 3 inequalities :

$ \frac {a^2}{b} \plus{} b \ge 2a$

$ \frac {b^3}{c^2} \plus{} c \plus{} c \ge 3b$

$ \frac {c^4}{a^3} \plus{} a \plus{} a \plus{} a \ge 4c$

 :maybe:


modular arithmetic
number theory unsolved
number theory

I don't know if this problem has been posted before, I couldn't find it.

given integers $ a,n$, show that there are infinitely many primes $ p$ for which $ a$ is a perfect $ n$-th power $ \pmod{p}$

I have a solution for the $ n$ odd case using Dirichlet's theorem, but don't know how to complete the even case...

as we know the equation $ a \equiv x^n \pmod{p}$ has an answer if and only if $ a^{\frac{p\minus{}1}{d}} \equiv 1 \pmod {p}$.
where $ d\equal{} \gcd(n,p\minus{}1)$.
as u mentioned the case in which $ n$ is odd is easily solved since the primes $ p$ with $ p \equiv \minus{}1 \pmod {n}$ are infinite.i think that there exists a more elementary proof for this property and we can avoid using Dirichlet's theorem.
anyway,for these primes we get $ d\equal{}1$ and the problem will be solved.
now consider the case in which $ n$ is even.
note that if $ p \equiv \minus{}1 \pmod {n}$ then $ d\equal{}2$ so
it is sufficient to find infinitely many primes for which $ p \equiv \minus{}1 \pmod {n}$ and $ \left(\frac{a}{p} \right)\equal{}1$.
if $ a\equal{}q_1^{a_1}...q_k^{a_k}$ then $ \left(\frac{a}{p} \right)\equal{}\prod_{i\equal{}1}^{k} \left(\frac{q_i}{p} \right)^{a_i}$.
now we choose the prime $ p$ such that $ p \equiv \minus{}1 \pmod {8. lcm(a,n)}$
and obviously there are infinitely many such primes.
now we can prove that
$ \left(\frac{a}{p} \right)\equal{}1$
just note that for any $ q_i$ we have $ p \equiv \minus{}1 \pmod{q_i}$ and since $ p \equiv \minus{}1 \pmod {8}$ then $ \left(\frac{2}{p} \right)\equal{}1$ and also $ 2 \nmid \frac{p\minus{}1}{2}$.
let's suppose that $ 2 |a$ then if $ q_1\equal{}2$

$ \left(\frac{q_1}{p} \right)^{a_1}\equal{}1$

and if $ q_i$ is odd

$ \left(\frac{q_i}{p} \right)\equal{} (\minus{}1)^{\frac{q_i\minus{}1}{2}} \times \left(\frac{p}{q_i} \right) \equal{} (\minus{}1)^{\frac{q_i\minus{}1}{2}} \times \left(\frac{\minus{}1}{q_i} \right) \equal{} (\minus{}1)^{\frac{q_i\minus{}1}{2}}  \times (\minus{}1)^{\frac{q_i\minus{}1}{2}}\equal{}1$
which implies
$ \left(\frac{q_i}{p} \right)^{a_i}\equal{}1$
so by multiplying these equalities we get
$ \left(\frac{a}{p} \right)\equal{}\prod_{i\equal{}1}^{k} \left(\frac{q_i}{p} \right)^{a_i}\equal{}1$
so we are done(i hope)!
 :)

here's an elementary solution without QR that i just came up with...

Assume that there are only finitely many primes that satisfy the condition: $ p_1,...,p_k$. Let $ x_1^n \equiv a \pmod{p_1}$, ..., $ x_k^n \equiv a \pmod{p_k}$. choose $ r_1, ...,r_k$ such that $ r_i^n \not\equiv x_i^n \pmod{p_i}$ for every $ i$. now by the Chinese Remainder Theorem, there exists an integer $ x$ such that, for every $ i$, $ x\equiv r_i \pmod{p_i}$. then for all $ i$, $ x^n \equiv r_i^n \not\equiv x_i^n \equiv a \pmod{p_i}$. thus, $ x^n\minus{}a$ is not divisible by any $ p_i$, which means that there is a prime $ p$ different from all the $ p_i$ that divides it. thus we have found a new prime that also satisfies the condition -- a contradiction.


3. Let a, b, c be positive real numbers. Prove that (a/b+b/c+c/a)^2 >=(a+b+c)(1/a+1/b+1/c) [url]http://www.bmoc.maths.org/home/bmo2-2005.pdf[/url] if you're interested in the paper.

Can you prove this by using Chebyshev's sum inequality? Im sure you can prove this with Cauchy-Schwartz but im more intrigued by a possible proof involving Chebyshev's sum inequality.  Is order a factor that may prevent such a proof?

In LATEX form:

Prove that $ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ where $ a$, $ b$, $ c$ $ \in R^{+}$

a solution usng am/gm is quite simple.

\[ RHS=(a+b+c)(\frac{bc+ac+ab}{abc})\]
\[ LHS=\frac{(a^{2}c+b^{2}a+c^{2}b)}{abc}^{2}=\frac{(a^{4}c^{2}+b^{4}a^{2}+c^{4}b^{2}+a^{3}b^{2}c....)}{abc}^{2}\]

now subtracting the terms n the LHS which equal RHS leaves a^4c^2+b^4a^2+c^4b^2>=3(abc)^2

which is trivial by am/gm

i am however now very tired and may have made errors though i belive this souloution to be correct. i assume that it is obvious what the dots are. appologies for my lack of neatness. if you dont follow please post back and i shalll repost. sorry i cant do the method you suggest

Nice. Though like you said, you've made a couple of errors in your steps, which didn't seem to have stop you from achieving your answer. :)

Maybe it will be a little bit more useful if i move this to the algebra>>inequality section... :|


In order to draw a graph of $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, a table of values was constructed. These values of the function for a set of equally spaced increasing values of $ x$ were $ 3844$, $ 3969$, $ 4096$, $ 4227$, $ 4356$, $ 4489$, $ 4624$, and $ 4761$. The one which is incorrect is:

$ \textbf{(A)}\ 4096 \qquad\textbf{(B)}\ 4356 \qquad\textbf{(C)}\ 4489 \qquad\textbf{(D)}\ 4761 \qquad\textbf{(E)}\ \text{none of these}$

[quote=hunter117][quote=AIME15][hide="Click for solution"]
We are told that the values of $ f(x)$ listed correspond to $ f(x),f(x\plus{}h),f(x\plus{}2h),\ldots,f(x\plus{}7h)$.  Observe that the difference between successive values is given by $ f(x\plus{}h)\minus{}f(x)\equal{}a(x\plus{}h)^2\plus{}b(x\plus{}h)\plus{}c\minus{}(a^2\plus{}bc\plus{}c)\equal{}2ahx\plus{}ah^2\plus{}bh$.  Since this difference is a linear function of $ x$, it must change by the same amount whenever $ x$ is increased by $ h$.  But the successive differences of the listed values are $ 125,127,131,129,131,135,137$ so that, if only one value is incorrect, $ 4227$ is that value.  This corresponds to answer choice $ \boxed{\textbf{(E)}}$.
[/hide][/quote]

I'm really confused. How are the differences linear function?[/quote]
$\Delta f(x)=f(x+h)-f(x)=(2ah)x+(ah^2+bh)=Mx+B$ where $M=2ah$ and $B=ah^2+bh$.

@AIME15: Also the successive differences should be $125, 127, 131, 129, 133, 135, 137$, otherwise there would be more than one incorrect value.

[quote=hunter117][quote=AIME15][hide="Click for solution"]
We are told that the values of $ f(x)$ listed correspond to $ f(x),f(x\plus{}h),f(x\plus{}2h),\ldots,f(x\plus{}7h)$.  Observe that the difference between successive values is given by $ f(x\plus{}h)\minus{}f(x)\equal{}a(x\plus{}h)^2\plus{}b(x\plus{}h)\plus{}c\minus{}(a^2\plus{}bc\plus{}c)\equal{}2ahx\plus{}ah^2\plus{}bh$.  Since this difference is a linear function of $ x$, it must change by the same amount whenever $ x$ is increased by $ h$.  But the successive differences of the listed values are $ 125,127,131,129,131,135,137$ so that, if only one value is incorrect, $ 4227$ is that value.  This corresponds to answer choice $ \boxed{\textbf{(E)}}$.
[/hide][/quote]

I'm really confused. How are the differences linear function?[/quote]

Search up Finite differences


For what positive integer $ n$ is $ n^2 \equal{} 2^6$?

There are 2 solutions for this.
First, you can solve 2^6, which equals 64 and then take the square root, so the answer is 8.

Also, using exponent rules, (2^3)^2=2^(3*2)=2^6.
Thus, 2^3=8.

Or you could find the square root of both sides, leaving $ n \equal{} 2^3$.

simply find two ^ 6 is 64 and then find the square root of 64 which is 8

we know that in order for the two to be equal the mantissa(i'm not sure if this is right) must be larger. Since n^2 has 2^6 times less that means n must be 2^3 times more so n is 8.


hey,
could anyone by any chance explain the game of Schnapps of me?
its the one where you click a sequence of clicks, say semi-random things, and convey to someone else that knows how it works the name of a person/thing.
where i live you're meant to work it out yourself so im pretty confused ..
- thanks, epitomy01.

Funny think that you brought this up because i just learned about it. Heres how it goes: you play with 3 people A, B, C, let you be A. You start off with B telling you the name of something (team, city etc.) let us use for example Paris.
C hasent a clue what the city is and A has to convey to him the city. First A says a sentence with something to the effect of "Lets go visit a [u][b]P[/b][/u]yramid" because p is the first letter in Paris. Then after that you would snap once because the next letter is a vowl and its an a. ( using a e i o u you snap once for a, twice for e, etc.). the you make up another sentence and snap if you need to. You keep doing this until you are done with the word and C should be able to figure it out.

Hope i was clear enough.

thanks for that. it seems to make sense enough, but i still dont understand how the following instance of 'schnapps' works.

Schnapps is the name of the game, * * *, OK?
Be careful with the clues, *, but don't rush yet, * * * * * * *, alright?
* * * *, now can you guess what it is, * *, OK?
Maybe this will help, * * * *, recognise it yet? * *
This should help, * * *, Maybe not, * *.
[ Done. and a * denotes a click. that was the conveyal of 'britney spears']

Schnapps is the name of the game, ##, OK?
Can you guess what it is
Let's see if this helps, ####, can you guess now, know what it is yet?
See if this helps. Alright
[ Done. and a # denotes a click. that was a conveyal of 'coldplay' the musical band]

Schnapps is the name of the game, ####, OK?
Don't guess yet, ###, #######, be more patient, #, alright?
[ Done. and a # denotes a click. that was bush george junior]
 so yeah, explanations for those would b every helpful.

I don't exactly know though if you do or have found out, please email me at astantine@gmail.com or reply to this post. I am really interested...thanks


Two numbers have a ratio of 4:3. The greater number is 60 more than 8 times their difference. What could the two numbers be?

that's what I got, but it depends if the problem means positive difference...


Let $H$ be the orthocenter of acute triangle $\triangle{ABC}$. The tangents from
$A$ to the circle with diameter $BC$ touch the circle at $P$ and $Q$. Prove
that $P$, $Q$ and $H$ are collinear.

you posted this b4 in the proposed problems section  :?
http://www.mathlinks.ro/viewtopic.php?t=5850
you just use the fact that the intersection of the diagonals of the cyclic quad lies on the polar of the other intersection of side.


Could anyone give me a hint on the following AOPS Volume 2 Problem:

The magnitudes of the sides of triangle ABC are a = BC, b = AC, and C = AB with c less than or equal to b and b less than or equal to a. Through interior point P and the vertices A,B,C, lines are drawn meeting the opposite sides at X, Y, and Z respectively. Prove that AX + BY + CZ < 2a + b.

[hide="Hint"]Prove $ CZ<a$, $ BY<a$, and $ AX<b$. [/hide]


geometry
circumcircle
trigonometry
incenter
trig identities
Law of Sines
angle bisector

Let $A', B', C'$ be the mid-points of arcs $BC,CA,AB$ of the circumcircle of triangle $ABC$ not containing points $A,B,C$. Sides $BC,CA,AB$ intersect the pairs of segments $(C'A', A'B'), (A'B', B'C')$ and $(B'C', C'A')$ at the pairs of points $(M,N), (P,Q)$ and $(R,S)$, respectively.

Prove that $MN=PQ=RS$ if and only if $\triangle ABC$ is equilateral.

I found (not-very nice) trigonometric solution - I don't want to post many computations so I just point out main things in solution.

It is sufficient to prove that $BM+CN=CP+AQ$. Do some angle-chasing and it is easy to see that $CN=CP$. Now make use of law of sines in triangles $(A'BM)$ and $(A'CN)$ to express $BM$ as product of sines and $CN$; analogically with $AQ$. Now what you are left with is to prove some easy trigonometric equation.

Hope it was at least a bit helpful  :)

Try to prove, for example, that $PS\|BC$ and $RQ$ is perpendicular to the angle bisector of $\angle A$.It will follow that $PQ=RS\iff AB=AC$.

Since it is so easy, let me post a complete solution (you know, I am lazy) :) :

First we remember some well-known properties of the points A', B', C': Mainly, if I is the incenter of triangle ABC, then the points A', B', C' lie on the lines AI, BI, CI and are the circumcenters of triangles BIC, CIA, AIB; the lines B'C', C'A', A'B' are the perpendicular bisectors of the segments AI, BI, CI. [This was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=6095 post #2, where the points A', B', C' were named D, E, F.]

Since the point R lies on the line B'C', which is the perpendicular bisector of the segment AI, we have AR = IR; thus, the triangle ARI is isosceles, so that < RAI = < RIA. On the other hand, since I is the incenter of triangle ABC, the line AI is the angle bisector of the angle CAB, so that < RAI = < QAI. Hence, < RIA = < QAI. Thus, IR || CA. Similarly, IN || CA. Consequently, the points R and N both lie on the parallel to the line CA through the point I; thus, RN || CA.

We know that the line C'A' is the perpendicular bisector of the segment BI; thus, $BI\perp C^{\prime}A^{\prime}$, or, equivalently, $BI\perp SM$. Thus, in the triangle SBM, the line BI is the altitude from the vertex B. But on the other hand, since I is the incenter of triangle ABC, the line BI is the angle bisector of the angle ABC, i. e. the angle bisector of the angle SBM. Hence, in the triangle SBM, the angle bisector of the angle SBM is the altitude from the vertex B. This yields that the triangle SBM is isosceles with SB = BM. Hence, we get the equivalence $RS=MN\Longleftrightarrow\frac{RS}{SB}=\frac{MN}{BM}$. But by Thales, we have $\frac{RS}{SB}=\frac{MN}{BM}\Longleftrightarrow SM\parallel RN$, and since RN || CA, we have $SM\parallel RN\Longleftrightarrow SM\parallel CA$. Again, by Thales, $SM\parallel CA\Longleftrightarrow\frac{AB}{SB}=\frac{BC}{BM}$. But since SB = BM, we have $\frac{AB}{SB}=\frac{BC}{BM}\Longleftrightarrow AB=BC$.

Combining this all, we see that $RS=MN\Longleftrightarrow AB=BC$. Similarly, $MN=PQ\Longleftrightarrow BC=CA$ and $PQ=RS\Longleftrightarrow CA=AB$. Hence, $MN=PQ=RS\Longleftrightarrow BC=CA=AB$, so the problem is solved.

Quite similarly you can show that $QR=SM=NP\Longleftrightarrow BC=CA=AB$, by the way.

  Darij