function
inequalities
inequalities proposed

[quote="peine"]what about $ k$ is it a natural number?[/quote]
Yeah, forgot to add: $ k\in\mathbb{N}$. :oops:

[quote="Potla"]Let $ x_i > 0$ be positive reals $ (i \equal{} 1,2,\cdots n)$ such that $ \sum_{i \equal{} 1}^{n} a_i \equal{} n$. Find $ P_{min}$ where:
$ P \equal{} \frac {x_1^k}{x_n \plus{} x_1} \plus{} \frac {x_2^k}{x_1 \plus{} x_2} \plus{} \cdots \plus{} \frac {x_n^k}{x_1 \plus{} x_{n \minus{} 1}}$.
 :blush:
Hope you will enjoy it! :)
Happy Bengali New year to all mathlinkers!(It might be a bit late) :lol:
[hide="P/s"] I have to be offline for 7 days or so from today as my internet session is to expire today midnight[/hide][/quote]
I think $ \sum_{i\equal{}1}^{n}x_i\equal{}n$.If so,use AM-GM or Holder   :wink:  :)

$ P \equal{} \frac {x_1^k}{x_n \plus{} x_1} \plus{} \frac {x_2^k}{x_1 \plus{} x_2} \plus{} \cdots \plus{} \frac {x_n^k}{x_1 \plus{} x_{n \minus{} 1}}$

Maybe, we could use the convex function   $ f(x)\equal{}\frac{1}{x}$  and Jensens ineq.

$ P\geq (\sum {{x}_{i}}^{k})(\frac{1}{\frac{\sum 2{x}_{i}}{n}})\equal{}\frac{ (\sum {{x}_{i}}^{k})}{2}$

If k=1 this results in the minimum n/2. Assume that k>1 Now, using jensen inequality on the function  $ f(x)\equal{}{x}^{k}$

   $ \frac{ (\sum {{x}_{i}}^{k})}{2}\geq \frac{n}{2}$

    Which gives the minimum if all the variables are equal.

[quote="fredrik1232009"]$ P \equal{} \frac {x_1^k}{x_n \plus{} x_1} \plus{} \frac {x_2^k}{x_1 \plus{} x_2} \plus{} \cdots \plus{} \frac {x_n^k}{x_1 \plus{} x_{n \minus{} 1}}$

Maybe, we could use the convex function   $ f(x) \equal{} \frac {1}{x}$  and Jensens ineq.

$ P\geq (\sum {{x}_{i}}^{k})(\frac {1}{\frac {\sum 2{x}_{i}}{n}}) \equal{} \frac { (\sum {{x}_{i}}^{k})}{2}$

If k=1 this results in the minimum n/2. Assume that k>1 Now, using jensen inequality on the function  $ f(x) \equal{} {x}^{k}$

   $ \frac { (\sum {{x}_{i}}^{k})}{2}\geq \frac {n}{2}$

    Which gives the minimum if all the variables are equal.[/quote]
Thanks for your nice solution, but I think my solution is easier than this..... :oops:
I do hope that Kunny will give us a solution....... :blush:

I think that the right enunciation of your problem is the following:

Let $ x_i > 0$ be positive real numbers $ (i \equal{} 1,2,...,n),$ where $ n\in N, n\geq 1$, such that  $ \sum_{i \equal{} 1}^{n}x_{i} \equal{} n$.
For $ k > 0$, find $ P_{min}$, where:
$ P \equal{} \frac {x_{1}^{k}}{x_{n} \plus{} x_{1}} \plus{} \frac {x_{2}^{k}}{x_{1} \plus{} x_{2}} \plus{} \cdots \plus{} \frac {x_{n}^{k}}{x_{n \minus{} 1} \plus{} x_{n}}.$

[u]My solution[/u]:

$ P \equal{} \frac {x_{1}^{k}}{x_{n} \plus{} x_{1}} \plus{} \frac {x_{2}^{k}}{x_{1} \plus{} x_{2}} \plus{} \cdots \plus{} \frac {x_{n}^{k}}{x_{n \minus{} 1} \plus{} x_{n}}\geq$
$ \geq \frac {(x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n)^k}{n^{k \minus{} 2}\cdot\left[(x_n \plus{} x_1) \plus{} (x_1 \plus{} x_2) \plus{} \cdots \plus{} (x_{n \minus{} 1} \plus{} x_n)\right]} \equal{} \frac {n^k}{n^{k \minus{} 2}\cdot2\cdot n} \equal{} \frac {n}{2}$
Thus: $ P_{min} \equal{} \frac {n}{2}$

I'm sure you know that:

If: $ a_i\geq0,b_i > 0,i \equal{} 1,2,...,n,$ and $ n\in N,n\geq1$, then, for $ k > 0,$ we have:
$ \frac {{a_1}^k}{b_1} \plus{} \frac {{a_2}^k}{b_2} \plus{} \cdots \plus{} \frac {{a_n}^k}{b_n}\geq\frac {(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^k}{n^{k \minus{} 2}\cdot(b_1 \plus{} b_2 \plus{} \cdots \plus{} b_n)}$

[quote="marin.bancos"]
I'm sure you know that:

If: $ a_i\geq0,b_i > 0,i \equal{} 1,2,...,n,$ and $ n\in N,n\geq1$, then, for $ k > 0,$ we have:
$ \frac {{a_1}^k}{b_1} \plus{} \frac {{a_2}^k}{b_2} \plus{} \cdots \plus{} \frac {{a_n}^k}{b_n}\geq\frac {(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^k}{n^{k \minus{} 2}\cdot(b_1 \plus{} b_2 \plus{} \cdots \plus{} b_n)}$[/quote]
Yes :rotfl: that is Hölder:

$ \left(\frac {{a_1}^k}{b_1} \plus{} \frac {{a_2}^k}{b_2} \plus{} \cdots \plus{} \frac {{a_n}^k}{b_n}\right)(b_1 \plus{} b_2 \plus{} \cdots \plus{} b_n)(1\plus{}\cdots\plus{}1)^{n\minus{}2}$ $ \ge {(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^k}$


calculus
integration
inequalities
function
logarithms
calculus computations

[quote="Gibbenergy"]From my Math Club:

f(x) is a continuous positive function on [0..1] Prove that
$\int_{0}^{1}f(x) dx \geq e^{\int_{0}^{1}ln(f(x)) dx}$
Thanks[/quote]

I think it follows from $\ln{x}$ being concave and the continuous version of Jensen's Inequality (I'm not sure of the details of this, though).

[quote="Gibbenergy"]From my Math Club:

f(x) is a continuous positive function on [0..1] Prove that
$\int_{0}^{1}f(x) dx \geq e^{\int_{0}^{1}ln(f(x)) dx}$
Thanks[/quote]

By Jensen inequality
$\frac{1}{n}{\sum_{k=1}^{n}ln(f(\frac{i}{n})) \leq ln[ \frac{1}{n}{\sum_{k=1}^{n}f(\frac{i}{n})]}}$

Left side approaches to:
$\int_{0}^{1}lnf(x) dx$
and right side approaches to:
$ln \int_{0}^{1}f(x) dx$

Here's one way to phrase Jensen's inequality: Suppose that $\varphi$ is a convex function on a certain interval, and suppose that $X$ is a random variable taking values in that interval. Suppose also that all of the expectations in this theorem are absolutely convergent. Then

$E(\varphi(X))\ge\varphi(E(X)).$

(For a concave function, reverse the order of the inequality.)

I like this form, as it tends to unify the various other ways of writing Jensen. The proof is essentially the same as for all of the others, just a little more streamlined. Here's the proof:

Let $x_{0}=E(X).$ There is a "supporting line," a function $L(x)=ax+b$ such that $L(x)\le\varphi(x)$ and $L(x_{0})=\varphi(x_{0}).$

Then $E(\varphi(X))\ge E(L(X))=L(E(X))=L(x_{0})=\varphi(x_{0}) =\varphi(E(X)).$

The key step, and the place where we need that this is a random variable (that is, that we are integrating over a set of total mass 1) is that $E(L(X))=L(E(X).$ In more detail, $E(L(X))=E(aX+b)=aE(X)+E(b)=aE(X)+b=L(E(X)).$

For the case of this problem, note that if $X$ is the random variable that is uniformly distributed on $[0,1]$ and $Y=f(X),$ then we are asked to show that $E(Y)\ge e^{E(\ln(Y)),}$ which is the same as $\ln(E(Y))\ge E(\ln(Y))$ and is true by Jensen (with the sign reversed because $\ln$ is concave.)


If i hadn't made any mistakes, sum (x_i)^3-sum (x_i)^4=sum (1<=i<j<=n) x_i*x_j*((x_i)^2+(x_j)^2), which is the IMO 1999 problem.


quadratics
number theory
number theory unsolved

pro) Prove that for all primes $p$ of the form $4k-1$, we have $\left(\frac{p-1}{2}\right)!\equiv \left(-1\right)^m \mod p$, where $m$ is the number of quadratic residues modulo $p$ between $\frac{p}{2}$ and $p$.

$ \frac{p\minus{}1}{2} \equal{} 2k\minus{}1$

$ p\minus{}i \equiv \minus{}i (\mod p)$

Multiplying the equation for $ i\equal{}1,2,..,2k\minus{}1$, and multiplying (2k-1)! to both sides, we arrive at
$ \minus{}(4k\minus{}2)! \equiv ((2k\minus{}1)!)^2 (\mod p)$

Wilson's Theorem yields $ (4k\minus{}2)! \equiv \minus{}1 (\mod p)$

Therefore, the above equation is equivalent to $ 1 \equiv ((2k\minus{}1)!)^2 (\mod p)$.

This implies $ (2k\minus{}1)! \equiv 1$ or $ \minus{}1 (\mod p)$

Using the legendre symbol, we easily arrive at the conclusion;

(i) $ (2k\minus{}1)! \equiv 1 (\mod p)$ <=> $ ( \frac{(2k\minus{}1)!}{p} ) \equal{} 1$ and $ (2k\minus{}1)! \equiv \minus{}1 (\mod p)$ <=> $ ( \frac{(2k\minus{}1)!}{p} ) \equal{} \minus{}1$

(ii) $ (\frac{(2k\minus{}1)!}{p}) \equal{} (\minus{}1)^a$, where a is the number of quadratic nonresidues of p; (this is trivial)

(iii) Observe that $ (\frac{\minus{}1}{p}) \equal{} \minus{}1$, and therefore $ (\frac{\minus{}k}{p}) \equal{} \minus{}(\frac{k}{p})$

This implies that a number i=1,2,..,2k-1 is a quadratic nonresidue iff the number p-i is a quadratic residue.

This yields a=m, and this finishes the proof.
(The number of quadratic nonresidues between 1 and $ \frac{p}{2}$ is equal to the number of quadratic residues between $ \frac{p}{2}$ and p.)


Find the exact value of $ \csc{20^{\circ}}\minus{}\cot{40^{\circ}}$ without a calculator

[quote="tinytim"]Find the exact value of $ \csc{20^{\circ}} \minus{} \cot{40^{\circ}}$ without a calculator[/quote]

I don't understand: $ \boxed{\csc{20^{\circ}}}$

csc=reciprocal of sin

sec= reciprocal of cos

cot=reciprocal of tan

Hint for my problem:

$ \csc{20^{\circ}}\minus{}\cot{40^{\circ}}\equal{}\frac{1}{\sin{20^{\circ}}}\minus{}\frac{\cos{40^{\circ}}}{\sin{40^{\circ}}}$


calculus
induction
algebra unsolved
algebra

[quote="mcdat"]Let $ (U_n)$ be a sequence given recursively as follows:

$ \left \{\begin{array}{l}U_0 \equal{} 2 \ ; \ U_1 \equal{} 10 \\
U_{n \plus{} 2} \equal{} 8U_{n \plus{} 1} \minus{} 2U_n \end{array} \right.$

Determine the maximum possible value of n such that $ \ U_n \leq 2009^2$[/quote]

According to me, the simpliest methedod is rude calculus (though not very clever)

It's immediate to show with induction that $ u_n$ is increasing
Then :
$ u_2\equal{}76< 4036081\equal{}2009^2$
$ u_3\equal{}588< 4036081\equal{}2009^2$
$ u_4\equal{}4552< 4036081\equal{}2009^2$
$ u_5\equal{}35240< 4036081\equal{}2009^2$
$ u_6\equal{}272816< 4036081\equal{}2009^2$
$ u_7\equal{}2112048< 4036081\equal{}2009^2$
$ u_8\equal{}16350752> 4036081\equal{}2009^2$

So the required value is $ n\equal{}7$

[quote="pco"][quote="mcdat"]Let $ (U_n)$ be a sequence given recursively as follows:

$ \left \{\begin{array}{l}U_0 \equal{} 2 \ ; \ U_1 \equal{} 10 \\
U_{n \plus{} 2} \equal{} 8U_{n \plus{} 1} \minus{} 2U_n \end{array} \right.$

Determine the maximum possible value of n such that $ \ U_n \leq 2009^2$[/quote]

According to me, the simpliest methedod is rude calculus (though not very clever)

It's immediate to show with induction that $ u_n$ is increasing
Then :
$ u_2 \equal{} 76 < 4036081 \equal{} 2009^2$
$ u_3 \equal{} 588 < 4036081 \equal{} 2009^2$
$ u_4 \equal{} 4552 < 4036081 \equal{} 2009^2$
$ u_5 \equal{} 35240 < 4036081 \equal{} 2009^2$
$ u_6 \equal{} 272816 < 4036081 \equal{} 2009^2$
$ u_7 \equal{} 2112048 < 4036081 \equal{} 2009^2$
$ u_8 \equal{} 16350752 > 4036081 \equal{} 2009^2$

So the required value is $ n \equal{} 7$[/quote]

Thank  :)  :)


trigonometry
function
geometry
trig identities
Law of Cosines
area of a triangle
Heron\u0027s formula

a triangle with sides of a,b,c in that order is a=75

b=53
c=47

angle y is between the 75 and the 53
 while angle x is between 53 and 47

Show your work and the formula u used

find y and x

[quote="CheeseIsGood"]What are you supposed to do?  :huh:[/quote]

I think you just find the missing angles, since they are deemed variables.

Law of Cosines...?

[hide]The Law of Cosines is sqrt(a^2 + b^2 - 2abcosc), where a and b are the sides adjacent to an angle c in a triangle. It finds the side opposite from the angle.
Using the law, you can get this equation:
75 = sqrt(47^2 + 53^2 - 2(47)(53)cosx)
By squaring, subtracting, and dividing, you can isolate it to cosine of X, which you can undo by using the cosecant function or cos^-1.
Use the same method for Y.
Hope it works....[/hide]

you get rated..... im so stupid.....i didnt thought about that

but tell me the answer for x and y so i can check

Alternate way of getting the answer which may or may not be quicker...

Use Heron's Formula to find the area, then use the formula $ \frac{ab\sin \theta}{2}$, and equate it to what you got for Heron's and simply solve for $ \theta$

[quote="ZzZzZzZzZzZz"]Why can't you calculate it yourself?[/quote]

i said for checking....i got the answer already :football:

Hmm. Your new name shall be Sharkman.

[/spam]

AoPS is not a place where you check your work. Really.


$ a, b, c$ real positive numbers, $ abc \equal{} 1$. Prove that $ \frac {a \plus{} b}{2(a^7 \plus{} b^7 \plus{} c)}$ +$ \frac {b \plus{} c}{2(b^7 \plus{} c^7 \plus{} a)}$+ $ \frac {c \plus{} a}{2(c^7 \plus{} a^7 \plus{} b)}$<=1


Find the function f = f (x), defined and continuous on R+ = {x | 0 ≤ x < ∞}, that satisfies f (x+1) = f (x)+x on R+ and f (1) = 0.

I think you can define an arbitrary continuous function $ g : [0,1] \rightarrow [0,\infty)$ with $ g(0) \equal{} g(1) \equal{} 0$, and extend it to an $ f$ that works. It's easy to show $ f$ is continuous inductively with the definition $ \lim_{\epsilon \rightarrow 0} f(x\plus{}\epsilon) \minus{} f(x) \equal{} 0$.

Alternately, let $ h(x) \equal{} f(x) \minus{} x(x\minus{}1)/2$, and the given equation reduces to $ h$ having period $ 1$ (and some range oddities from $ f \geq 0$), with no other information.


trigonometry
geometry
geometric transformation
reflection
circumcircle
power of a point
radical axis

In a square $ABCD$, let $P$ and $Q$ be points on the sides $BC$ and $CD$ respectively, different from its endpoints, such that $BP=CQ$. Consider points $X$ and $Y$ such that $X\neq Y$, in the segments $AP$ and $AQ$ respectively. Show that, for every $X$ and $Y$ chosen, there exists a triangle whose sides have lengths $BX$, $XY$ and $DY$.

Since both AP, AQ are different from the bisector AC of the right angle $\angle DAB$, we have

$BX < DX \le DY + XY,\ \ DY < BY \le BX + XY$

It remains to show that XY < BX + DY. Denote a = AB = BC = CD = DA the square side, 0 < p = BP = CQ < a, and $\beta = \angle PAB < 45^\circ,\ \delta = \angle QAD < 45^\circ$.

$\tan \beta = \frac{BP}{AB} = \frac p a,\ \ \tan \delta = \frac{DQ}{DA} = \frac{a - p}{a}$

$\tan(\beta + \delta) = \frac{\tan \beta + \tan \delta }{1 - \tan \beta \tan \delta} = \frac{\frac p a + \frac{a - p}{a}}{1 - \frac{p(a - p)}{a^2}} = \frac{1}{1 - \frac{p(a - p)}{a^2}} > 1$

Consequently, $\beta + \delta > 45^\circ,\ 2\beta + 2\delta > 90^\circ.$ There are other ways to show this, for example, see the solution of problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81900]Square[/url], but it is not worth the effort. Let B', D' be reflections of the square vertices B, D in the lines AP, AQ, respectively. The quadrilateral BD'B'D is cyclic with the circumcenter A, because the perpendicular bisectors AP, AQ, AC of BB', DD', BD meet at the vertex A. Since $\angle BAB' = 2 \beta,\ \angle DAD' = 2\delta,$ we have $\angle BAB' + \angle DAD' > 90^\circ$ and consequently, the points B, D', B', D follow on the minor arc BD of the circumcircle (A) in this order. As a result, any circle (X) centered on the segment (AP) and with the radius BX = B'X contains the point D' in its interior and any circle (Y) centered on the segment (AQ) and with radius DY = DY' contains the point B' in its interior. Thus any 2 such circles (X), (Y) intersect, which means that

$XY < B'X + D'Y = BX + DY$

Let $ \omega_1$ and $ \omega_2$ be the circumferences centered at $ X,Y$ with  radii $ XB$ and $ YD,$ respectively. It suffices to prove that $ \omega_1$ and $ \omega_2$ are secant. Let $ M \equiv BQ \cap AP$ and $ N \equiv DP \cap AQ.$ From $ \triangle ABP \cong \triangle BCQ$ and $ \triangle ADQ \cong \triangle DCP,$ it follows that  $ AM \perp BQ$ and $ AN \perp DP$ $ \Longrightarrow$ $ T \equiv BQ \cap DP$ is the orthocenter of $ \triangle APQ.$

$ TP^2 \minus{} TQ^2 \equal{} AP^2 \minus{} AQ^2 \equal{} BP^2 \minus{} DQ^2 \Longrightarrow TX^2 \minus{} XB^2 \equal{} TY^2 \minus{} YD^2$

$ \Longrightarrow p(T,\omega_1) \equal{} p(T,\omega_2),$

which means that radical axis $ \zeta$ of $ \omega_1$ and $ \omega_2$ passes through $ T.$ Since $ BQ > BC \equal{} BA,$ then we have $ \angle BAQ > \angle AQB$ $ \Longrightarrow$ $ \angle DTQ > \angle TDQ$ $ \Longrightarrow$ $ QD > QT,$ $ YD > YT.$ Hence, $ T$ is always inside $ \omega_2$ $\Longrightarrow$ $ \zeta$ cuts  $ \omega_2,$ say at $ U,V$ $ \Longrightarrow$ $ \omega_1$ and $ \omega_2$ meet at $ U,V$ and proof is completed.

Here I have a super beautiful solution by the argentinian Carlos di Fiore. Enjoy:

Let $ PB=QC=a $ and $ BA=CP=b $ (obviously $ BA=AD=a+b $).
Now consider in $ \mathbb{R}^3 $ the points $ A'=(a+b, 0, 0) $, $ P'=(0, a, 0) $ and $ Q'=(0, 0, b) $ (and $ O=(0,0,0) $).
Since $ \angle PBA = 90^\circ =  \angle P'OA' $, $ BA=a+b=OA' $ and $ PB=a=P'O $ we have that $ \triangle PBA \equiv \triangle P'OA' $. In the same way  $ \triangle QDA \equiv \triangle Q'OA' $ and $ \triangle PCQ \equiv Q'OP' $. From that we have $ PA=P'A' $, $ QA=Q'A' $ and $ PQ=P'Q' $ so $ \triangle AQP \equiv \triangle A'Q'P' $.
Now we consider $ X' $ and $ Y' $ in $ A'P' $ and $ A'Q' $ respectively such that $ AX=A'X' $ and $ AY=A'Y' $. Obviously $ \triangle AXB \equiv \triangle A'X'O' $, $ \triangle AYD \equiv \triangle A'Y'O $ and $ \triangle AXY \equiv \triangle A'X'Y' $ so $ BX=OX' $, $ DY=OY' $ and $ XY=X'Y' $. But it's obvious that $ \triangle X'OY' $ is a triangle, and has sides $ BX $, $ DY $ and $ XY $, hence the solution is complete $ \Box $.


[b][size=150][color=red]PA[/color][color=white]RA[/color][color=blue]GUAY[/size][/b][/color]

[b][i][color=red]Tit[/color][color=white]ula[/color][color=blue]res[/color][/i][/b]
1. Iván Hazevich
2. Jesús Torres
3. Rodrigo Benitez
4. Mauricio Maluff

[b][i][color=red]Sup[/color][color=white]len[/color][color=blue]tes[/color][/i][/b]
1. Osmar Quiñónez
2. Nair Aguilera

[b][size=150][color=red]PE[/color][color=white]RU[/color][color=red] 2007[/color][/size][/b]

[b]Titulares:[/b]

- Angles Larico, Tomás Miguel
- Cuenca Lucero, Cesar Augusto
- Vélez Lee, Luis Enrique
- Rey Rodriguez, Diego Eduardo

[b]Suplentes:[/b]

(Primer suplente) Guerra Rios, Percy Augusto
(Segundo suplente) Leiva Torres, Juan Luis

Saludos,

[color=blue]Me extraña que no se presenten los seleccionados de los demás países…[/color]

[color=blue]Quizá por vez única, una olimpiada del Cono Sur tiene sitio Web propio (blog, en este caso). La dirección es [url]http://www.conosur2007.org/[/url].

Lastimosamente, no aparecen ni los resultados ni los enunciados. Si alguien tiene alguno de ellos… ya sabe qué hacer (preferiría recibir los enunciados).

Saludos a todos.[/color]

Los seleccionados chilenos son:

    * Jaime Soza Parra
    * Francisco Orlando Garrido Valenzuela
    * Sebastian Zuñiga Alterman
    * Cristobal Carvajal Hausdorf

Suerte a ellos.
Dato obtenido de http://www.fmat.cl

[b][size=150][color=red]PE[/color][color=white]RU[/color][color=red] 2007[/color][/size][/b]

[b]Titulares:[/b]

- Angles Larico, Tomás Miguel -> ORO 51 PUNTOS.
- Cuenca Lucero, Cesar Augusto -> ORO 53 PUNTOS.
- Vélez Lee, Luis Enrique -> ORO 50 PUNTOS.
- Rey Rodriguez, Diego Eduardo -> ORO 50 PUNTOS.

Saludos.

Ya pueden ver en mi blog
[url]http://selectivos-peru.blogspot.com[/url]
las dos pruebas tomadas, y próximamente con soluciones  :D

$Tipe$

se encuentran en algun lado los resultados?
supe que la delegacion Chilena ha salido quinta, pero no veo los resultados en ninguna parte.

[quote="tipe"][quote="tipe"]Ya pueden ver en mi blog
[url]http://selectivos-peru.blogspot.com[/url]
[/quote]

Ahí también puse los resultados por países.

$Tipe$[/quote]

Sí, gracias, de ahí obtuve la información, soloq ue me gustaria saber a nivel individual.

Saludos


Give an explicit bijection between $ N$, the set of natural numbers, and $ N^{2}$.

does the word explicit have any significance?

I guess it means give your function explicitly in mathematical notation.

It means that, given any integer $ n$, I should be able to compute from your answer the pair $ (p, q)$ that it corresponds to, and vice-versa.

Here's an example of a proof that those two are of the same cardinality without an explicit bijection:

Given an integer $ n \in \bf N$, assign it the ordered pair $ (n, 0)$.  This is an injection from $ \bf N$ to $ {\bf N} \times {\bf N}$.
Given an ordered pair $ (p, q)$, assign it the integer $ 2^p 3^q$.  This is an injection from $ {\bf N} \times {\bf N}$ to $ \bf N$.
Whenever we have an injection from $ A$ to $ B$ and an injection from $ B$ to $ A$, there exists a bijection between $ A$ and $ B$.  (I can't remember the name of this theorem.)  Thus $ \bf N$ and $ {\bf N} \times {\bf N}$ are in bijection, as desired.

I'm actually having a hard time coming up with a non-constructive proof.  Does anyone have a proof that doesn't actually hand you a bijection?

Heh, I was adding one in while you posted.  What's the name of that theorem?

But the only proof I know of CBS is constructive - given two injections, it explicitly constructs the corresponding bijection.  Is the standard proof non-constructive?

Well, okay, I would say that invoking CBS gives a non-constructive proof but that any such proof can be "constructivised" by replacing CBS with the proof of CBS.  This is obviously just a matter of personal taste, though.  Does the proof of CBS actually work "in practice," i.e., can you write down the bijection that results if you start with the two injections that I listed?


algebra
polynomial
induction
complex numbers
algebra solved

Let $\, P(z) \,$ be a polynomial with complex coefficients which is of degree $\, 1992 \,$ and has distinct zeros. Prove that there exist complex numbers $\, a_1, a_2, \ldots, a_{1992} \,$ such that $\, P(z) \,$ divides the polynomial \[ \left( \cdots \left( (z-a_1)^2 - a_2 \right)^2 \cdots - a_{1991} \right)^2 - a_{1992}.  \]

I was asked for a solution of this, so let me post one.

Since the zeros $z_{1}$, $z_{2}$, ..., $z_{1992}$ of the polynomial P(z) are distinct, the polynomial P(z) divides the polynomial $Q\left(z\right) = \left( ... \left( \left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{1991}\right)^{2}-a_{1992}$ if and only if $Q\left(z_{i}\right)=0$ for every $i\in\left\{1;\;2;\;...;\;1992\right\}$. So it is enough to show that for any 1992 complex numbers $z_{1}$, $z_{2}$, ..., $z_{1992}$, there exist 1992 complex numbers $a_{1}$, $a_{2}$, ..., $a_{1992}$, such that the polynomial $Q\left(z\right) = \left( ... \left( \left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{1991}\right)^{2}-a_{1992}$ satisfies $Q\left(z_{i}\right)=0$ for every $i\in\left\{1;\;2;\;...;\;1992\right\}$. This can be generalized:

[color=blue][b]Theorem.[/b] Let n be a positive integer, and $z_{1}$, $z_{2}$, ..., $z_{n}$ be n complex numbers. Then, there exist n complex numbers $a_{1}$, $a_{2}$, ..., $a_{n}$ such that the polynomial $Q\left(z\right) = \left( ... \left( \left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{n-1}\right)^{2}-a_{n}$ satisfies $Q\left(z_{i}\right)=0$ for every $i\in\left\{1;\;2;\;...;\;n\right\}$.[/color]

[i]Proof.[/i] We use induction over n.

For n = 1, the theorem is trivial, since $Q\left(z\right)=z-a_{1}$, so we can take $a_{1}=z_{1}$, and then the polynomial $Q\left(z\right)=z-z_{1}$ clearly satisfies $Q\left(z_{1}\right)=0$.

The interesting part is the induction step: Let k be a positive integer. Assume that for n = k, the theorem is proven, i. e. for any k complex numbers $z_{1}$, $z_{2}$, ..., $z_{k}$, there exist k complex numbers $a_{1}$, $a_{2}$, ..., $a_{k}$ such that the polynomial $Q\left(z\right)=\left(...\left( \left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{k-1}\right)^{2}-a_{k}$ satisfies $Q\left(z_{i}\right)=0$ for every $i\in\left\{1;\;2;\;...;\;k\right\}$. We have to prove the theorem for n = k + 1, so we have to prove that for any k + 1 complex numbers $z_{1}$, $z_{2}$, ..., $z_{k+1}$, there exist k + 1 complex numbers $b_{1}$, $b_{2}$, ..., $b_{k+1}$ such that the polynomial $R\left(z\right)=\left(...\left( \left(z-b_{1}\right)^{2}-b_{2}\right)^{2}...-b_{k}\right)^{2}-b_{k+1}$ satisfies $R\left(z_{i}\right)=0$ for every $i\in\left\{1;\;2;\;...;\;k+1\right\}$.

In fact, after our assumption, there exist k complex numbers $a_{1}$, $a_{2}$, ..., $a_{k}$ such that the polynomial $Q\left(z\right)=\left(...\left( \left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{k-1}\right)^{2}-a_{k}$ satisfies $Q\left(z_{i}\right)=0$ for every $i\in\left\{1;\;2;\;...;\;k\right\}$. We want to construct our polynomial R from this polynomial Q.

In fact, let $b_{i}=a_{i}$ for every $i\leq k-1$, then let $b_{k}=a_{k}+\frac12Q\left(z_{k+1}\right)$ and $b_{k+1}=\frac14\left(Q\left(z_{k+1}\right)\right)^{2}$. Then,

$R\left(z\right)=\left(...\left( \left(\left(z-b_{1}\right)^{2}-b_{2}\right)^{2}...-b_{k-1}\right)^{2}-b_{k}\right)^{2}-b_{k+1}$
$=\left(...\left( \left(\left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{k-1}\right)^{2}-\left(a_{k}+\frac12Q\left(z_{k+1}\right)\right) \right)^{2}-\frac14\left(Q\left(z_{k+1}\right)\right)^{2}$
$=\left(\left(...\left( \left(\left(z-a_{1}\right)^{2}-a_{2}\right)^{2}...-a_{k-1}\right)^{2}-a_{k}\right)-\frac12Q\left(z_{k+1}\right)\right)^{2}-\frac14\left(Q\left(z_{k+1}\right)\right)^{2}$
$=\left(Q\left(z\right)-\frac12Q\left(z_{k+1}\right)\right)^{2}-\frac14\left(Q\left(z_{k+1}\right)\right)^{2}=Q\left(z\right)\left(Q\left(z\right)-Q\left(z_{k+1}\right)\right)$.

Hence, for $i\in\left\{1;\;2;\;...;\;k\right\}$, we have $R\left(z_{i}\right)=Q\left(z_{i}\right)\left(Q\left(z_{i}\right)-Q\left(z_{k+1}\right)\right)=0$ because $Q\left(z_{i}\right)=0$, and $R\left(z_{k+1}\right)=Q\left(z_{k+1}\right)\left(Q\left(z_{k+1}\right)-Q\left(z_{k+1}\right)\right)=0$. Thus, $R\left(z_{i}\right)=0$ holds for every $i\in\left\{1;\;2;\;...;\;k+1\right\}$, and this proves the theorem for n = k + 1. Hence, the induction step is complete, and the Theorem is proven.

  Darij

Here's a different approach:

Lemma: Consider the equation $ p((f\minus{}a)^{2}) \equal{} 0$, where $ p$ is a polynomial, $ a$ is a constant, and $ f$ is a variable. Let's say we can choose the polynomial $ p$ to have any set of $ k\minus{}1$ distinct roots for an integer $ k\ge 2$ and that we can choose $ a$ to be any complex number. Then for any set of $ k$ distinct complex numbers $ f_{1}, ..., f_{k}$, we can choose a value of $ a$ and a set of $ k\minus{}1$ distinct roots for $ p$ such that $ p((f\minus{}a)^{2}) \equal{} 0$, when viewed as a polynomial in $ f$, has $ f_{1},...,f_{k}$ as roots.

Choose $ a \equal{}\frac{f_{1}\plus{}f_{k}}{2}$ and let $ p$ have roots $ (f_{1}\minus{}a)^{2}, ... , (f_{k\minus{}1}\minus{}a)^{2}$. Then $ p((f\minus{}a)^{2})$ is clearly $ 0$ for $ f \equal{} f_{1},...,f_{k\minus{}1}$. For $ f_{k}$, we have $ p((f_{k}\minus{}a)^{2}) \equal{} p((\minus{}(f_{1}\minus{}a))^{2}) \equal{} p((f_{1}\minus{}a)^{2}) \equal{} 0$, so $ f$ has all of the possible values $ f_{1},...,f_{k}$.
[hide="Remark"]
Geometric intuition is key here. The idea is that if we can choose any $ k\minus{}1$ possible values for $ (f\minus{}a)^{2}$, that's the same as choosing $ 2k\minus{}2$ complex numbers for $ (f\minus{}a)$ that come in pairs such that the midpoint of each pair is the origin. Based on this, we can get any set of $ k\minus{}1$ possible values of $ f$ by choosing each of the $ k\minus{}1$ pairs to have a desired root. This, however, completely determines the other number in each pair. We can then choose $ a$ to move the midpoint of each pair from $ 0$ to $ a$. If we choose $ a$ to be at the midpoint of two of the desired possible values of $ f$, one of the $ k\minus{}1$ pairs can produce two possible values of $ f$, leaving the other $ k\minus{}2$ pairs to produce the other desired $ k\minus{}2$ values for $ f$.
[/hide]

We now use induction to show that $ (...((z\minus{}a_{1})^{2}\minus{}a_{2})^{2}...\minus{}a_{n\minus{}1})^{2}\minus{}a_{n}\equal{} 0$ can have any given $ n$ distinct roots for $ z$ if we appropriately choose $ a_{1},...,a_{n}$.

For $ n \equal{} 1$ we choose $ a_{1}$ to be the one distinct root since our equation is $ z\minus{}a_{1}\equal{} 0$.

Then let's assume it's true for $ n\minus{}1$. Consider $ (...((z\minus{}a_{1})^{2}\minus{}a_{2})^{2}...\minus{}a_{n\minus{}1})^{2}\minus{}a_{n}\equal{} 0$. By the induction hypothesis we can get any $ n\minus{}1$ distinct possible values for $ (z\minus{}a_{1})^{2}$ with an appropriate choice of $ a_{2},...,a_{n}$. By the lemma, we can thus choose $ a_{1}$ so that we have the desired $ n$ possible values of $ z$, and the induction is complete.

If we replace $ n$ with $ 1992$, we see that we can achieve any desired set of $ 1992$ roots in $ z$, and if we choose those to be the roots of $ P(z)$, it follows that $ P(z)$ must divide this polynomial.

QED

What was Darij's motivation behind letting:
$b_k=a_k+1/2Q(a_k+1)$

Let $z_1, z_2, \cdots$ be the roots of $P$. We prove the case for general $n$ via induction. The base case is trivial; take $a_1 = z_1$.

Now suppose that there exist constants $b_1, b_2, \cdots, b_n$ such that setting $a_i = b_i$ for each $i$ creates a valid polynomial. Then, this means that$$(\dotsb ((z_i - b_1)^2 - b_2)^2 \dots - b_{n-1})^2 = k$$for some constant $k$ for each $1 \leq i \leq n$. Furthermore, let the value of this expression for $z_{n+1}$ be $r$. Then, we need to pick $b_n, b_{n+1}$ such that$$(k-b_n)^2 = (r - b_n)^2 = b_{n+1}.$$As the roots of the quadratic$$(x-a)^2 = b$$can be any pair of complex numbers by choosing suitable $a$ and $b$, such $b_n$ and $b_{n+1}$ clearly exist, as required.


function
trigonometry
geometry
geometric transformation
reflection
analytic geometry
graphing lines

Hello everyone,

I can express $ \tan (90 \minus{} x)$ as $ \frac {1}{\tan x}$ using algebra, but how would I figure out $ \tan (90 \minus{} x)$ out using the unit circle and comparing the two points to the tangent line?

Thank you. My diagram is below (Grey lines mark the x and y axis):

---

[img]http://img178.imageshack.us/img178/3271/tan90xfy2.jpg[/img]

The line with angle $ 90^{\circ} \minus{} x$ is the reflection of the line with angle $ x$ across $ y \equal{} x$ (the line with angle $ 45^{\circ}$).  Another way to phrase this reflection is the transformation $ (x, y) \mapsto (y, x)$.  What are the slopes of these lines?

Alternatively, you can use the properties of triangles.  You know that the two right triangles (formed by connecting each of the points on your diagram to the origin and dropping a perpendicular to the x-axis) are congruent because they have the same hypotenuse (the radius of the circle) and angles (they each have angles of measure $ 90$, $ x$, and $ 90 \minus{} x$).  From there, you can compute the tangent of each in terms of length and height.

Thank you very much for your replies. I understand the concept now.


there was a mistake here and I have fixed it, try this ineq now! :D

cheers! :D :D :cool:

just a small latex tip: you need not use $ x_k * y_k $ now, just use $x_ky_k$  or $x_k \cdot y_k$ the central dot's command is \cdot.


[quote="mdk"]Suppose a musical group has 11 weeks to prepare for opening night, and they intend to have at least one rehearsal each day. However, they decide not to schedule more than 12 rehearsals in any 7-day period, to keep from getting burned out. Prove that there exists a sequence of successive days during which the band has exactly 21 rehearsals.[/quote]

[hide="Solution"]
Let $ a_{i}$ mean the number of total rehersals upto the $ i$-th day.

Consider the numbers,
$ a_{1},a_{2},...,a_{77},a_{1}+21,a_{2}+21,...,a_{77}+21$ ... $ (*)$

Now the most number of rehershals possible is $ 11\cdot 12 = 132$.
So, the numbers $ a_{i}$ and $ a_{i}+21$ are among $ \{1,2,...,143\}$.

By pigeonhole (*) has at least two elements being equal. Now it cant be $ a_{i}=a_{j}$ for $ i\not = j$ because they place [b]at least one[/b] rehershal each day. And likewise it cant be $ a_{i}+21 = a_{j}+21$ for $ i\not =j$. So it must mean that $ a_{i}= a_{j}+21$. And so $ a_{i}-a_{j}= 21$. Which means from $ j+1,..,i$ the band played exactly $ 21$ games.
[/hide]


Un número $p$ se dice perfecto si la suma de sus divisores, exceptuando al propio $p$, da como resultado $p$. Sea $f$ una función tal que:

$f(n) = 0$ si $n$ es perfecto
$f(n) = 0$ si la cifra de las unidades de $n$ es $4$
$f(a.b) = f(a) + f(b)$

Calcular $f(1988)$

Lo he resuelto pero no me sirve el primer dato, lo cual me hace pensar q me he equivocado :blush:

[quote="mathmanman"]$f: \mathbb{N} \to \mathbb{N}$ ? :)[/quote]

si

Ok, que piensas de eso? :
$f(1988)+f(8) = f(15904) \Longrightarrow f(1988)=0$

O, por ejemplo :
$f(2006)+f(4) = f(8024) \Longrightarrow f(2006)=0$

:)

Pienso que podriamos modificar la tercera condicion :
$f(a)+f(b)=f(a \cdot b)$. $\forall a$, podemos hallar $b$ tal que el cifro de las unidades de $a \cdot b$ sea 4. Entonces $f(a)=0\ \ \forall a$.

correccion:
[quote="mathmanman"]
... podemos hallar $b$ tal que el cifro de las unidades de $a \cdot b$ sea 4... [/quote]
[quote="yo"]
... podemos hallar $b$ tal que [b]la cifra[/b] de las unidades de $a \cdot b$ sea 4... [/quote]
----------
respecto a la idea.........
esto ya esta dada en la condicion...

Carlos Bravo  ;)
Lima - PERU

Ok, gracias [b]carlosbr[/b] por la correccion!  :D

It's cool that you correct me so I can improve my poor spanish. ;)

Sí, el resultado es por supuesto 0, pero es un poco raro que se haga fácilmente el problema sin hacer referencia a números perfectos.

[quote="mathmanman"]Ok, que piensas de eso? :
$ f(1988) \plus{} f(8) \equal{} f(15904) \Longrightarrow f(1988) \equal{} 0$

O, por ejemplo :
$ f(2006) \plus{} f(4) \equal{} f(8024) \Longrightarrow f(2006) \equal{} 0$

:)

Pienso que podriamos modificar la tercera condicion :
$ f(a) \plus{} f(b) \equal{} f(a \cdot b)$. $ \forall a$, podemos hallar $ b$ tal que el cifro de las unidades de $ a \cdot b$ sea 4. Entonces $ f(a) \equal{} 0\ \ \forall a$.[/quote]
hay un pequeño error... toca calcular f(8) para poder hallar f(1988) de esa forma.
Sin embargo esto es muy facil:
f(8)=f(4)+f(2)
4 termina en 4, 2 es perfecto, entonces f(8)=0, entonces f(1988)=0
ahí se usa la primera condición :)

En cuanto a el problema con la condición modificada:
f(k)=f(k)+f(2)=f(2k)
f(2k)=f(2k)+f(2)=f(4k)
.
.
.
$ f(k) \equal{} f(2^n k)$
eventualmente, para algun n, $ f(2^n k)$ va a terminar en 4, a menos que 5 divida a n; esto es fácil de comprobar por tanteo.

$ 2$ es perfecto?
Pero $ \sigma(2)\equal{}1\plus{}2\equal{}3$.


If $ a,b,c$ are positive numbers, then

$ \frac a{\sqrt{2a+b}}+\frac b{\sqrt{2b+c}}+\frac c{\sqrt{2c+a}}\le \sqrt{a+b+c}\le \frac a{\sqrt{a+2b}}+\frac b{\sqrt{b+2c}}+\frac c{\sqrt{c+2a}}$.

Dear VASC! The right part is an exercises of secrets in Inequality. We use Holder to prove it. The left part follows Cauchy and the famous one
\[ \sum \frac{a}{2a+b}\le 1. \]

[quote="hungkhtn"]The right part is an exercises of secrets in Inequality. We use Holder to prove it. The left part follows by Cauchy and the famous one
\[ \sum \frac{a}{2a+b}\le 1. \]
[/quote]
Indeed, these are the solutions. The left inequality can be generalized as follows:

If $ a_{1},a_{2},...,a_{n}$ are positive numbers, then

\[ \frac{a_{1}}{\sqrt{(n-1)a_{1}+a_{2}}}+\frac{a_{2}}{\sqrt{(n-1)a_{2}+a_{3}}}+\frac{a_{n}}{\sqrt{(n-1)a_{n}+a_{1}}}\le \sqrt{a_{1}+a_{2}+...+a_{n}}. \]

[quote="Vasc"]If $ a,b,c$ are positive numbers, then

$ \frac a{\sqrt{2a+b}}+\frac b{\sqrt{2b+c}}+\frac c{\sqrt{2c+a}}\le \sqrt{a+b+c}$.[/quote]
This one for $ a,b,c>0$ and $ k\ge 2$ holds too:
\[ \frac a{\sqrt{ka+b}}+\frac b{\sqrt{kb+c}}+\frac c{\sqrt{kc+a}}\le \sqrt{\frac{3(a+b+c)}{k+1}}\]


function
logarithms
calculus
derivative
inequalities
calculus computations

Is (1+x/n)^n an increasing function of n for x real and n>0. If not, what if we restrict n>1.

Take logarithm and differentiate it wrt n. The derivative is $ \log(1\plus{}x/n)\minus{}\frac{x/n}{1\plus{}x/n}$. So we need the inequality $ \log(1\plus{}t)\ge \frac{t}{1\plus{}t}$, $ t\ge 0$. This runs in the opposite direction to the familiar $ \log(1\plus{}t)\le t$, but it's easy to turn a logarithm around: $ \log(1\plus{}t)\equal{}\minus{}\log \frac{1}{1\plus{}t}\equal{}\minus{}\log\left(1\minus{}\frac{t}{1\plus{}t}\right)\ge \frac{t}{1\plus{}t}$.


Condition: $x\ge-1$
Put $t=x\sqrt{x+1}$ then the inequation becomes:
$t^{2}+3t+2>0\Leftrightarrow t<-2$ or $t>-1$
From that we can easy to continue with variable $x$


[b]1.[/b]  An equilateral regions of side length $ 2cm$ is cut from a corner of an equilateral triangular region with side length $ 6cm$. What is the number of centimeters in the perimeter of the resulting solid?

[b]2.[/b] In the figure below, the smaller circle has the radius of two feet and the larger circle has the radius of four feet. What is the total area of the four shaded regions? Express your answer in terms of[i] pi[/i].

[b]3. [/b]Joe purchases a mirror that fits exactly in the frame. The outer perimeter of the frame measures $ 60cm$ by $ 80cm$. The width of the frame measures 10cm. What is the area of the mirror?

[b]4.[/b] In square $ ABCD$, point $ M$ is the midpoint of the side $ AB$ and point $ N$ is the midpoint of side $ BC$. What is the ratio of the area of the triangle $ AMN$ to the are of the square $ ABCD$? Express your answer as a common fraction.

[quote="garfielddisco123"][hide="Question 1"]The resulting polygon is a regular hexagon with sides two, because each side is subtracted 4. Perimeter is, thus, [b]12[/b].[/hide]

Where is the picture for 2?

[hide="Question 3"]If the frame is 10 cm thick, then 20 is subtracted from the length and height which would be 40cmx60cm mirror. The are of the mirror is [b]2400cm^2[/b].[/hide]

[hide="Question 4"]I do not how to explain this, and I wish I have LaTeX, but AMN/ABCD = [b]1/8[/b]. I kind of need a diagram to explain it, so sorry.[/hide][/quote]

Oh, my bad. Here, omit #3, will you?

Sorry!  :blush:

Well I can't take it out because I can't edit my post anymore so... yeah.


[b]Suppose that $n$ is a given integer ( $n\ge 2$ ). When the number $a^{n}+b^{n}$ may be written as a product of two distinct integers $\ge 2$ ?[/b]
  [color=green][b]It's possible to find all such pairs $(a,b), a\ne b ,$ with $a^{n}+b^{n}=cd$ where $c,d$ , $c\ne d$ , are in $\{2,3,...\}$   ?[/b][/color]
For instance, in case $n=2$ ; \[\begin{array}{lcl}2^{2}+5^{2}&=& 29 \\ 3^{2}+5^{2}&=& 2\times 17 \\ 4^{2}+5^{2}&=& 41\\ \vdots &\vdots & \vdots \\ \end{array}\]

If $n$ has an odd prime divisor we can use $a^{qm}+b^{qm}=(a^{m}+b^{m})(a^{(m-1)q}-a^{(m-2)q}b+-...+b^{(m-1)q})$ for odd $q$.

For the case of $n=2^{k}$, we can reduce it to $k=1$.
In this case, exactly the primes $\equiv 1 \mod 4$ are possible to be written as sum of two squares, so this are all cases.


Each successive term of the sequence 1, 3, 7, 15, 31, ... is obtained
by adding one to twice the previous term. The 10th term is 1023.
What is the value of the 9th term?

If the $ \ 10th$ term is $ \ 1023$, then the $ \ 9th$ term must be $ \ 1023 \minus{} 1 \equal{} 2x$
$ \ x \equal{} \boxed{511}$

Note that the $ n^\text{th}$ term is $ 2^n\minus{}1$, so the $ 9^\text{th}$ term is $ 2^9\minus{}1\equal{}\boxed{511}$.


let $ a$ , $ b$ , $ c$ , $ d$ $ \in$ $ {\mathbb R}^{\plus{}}$ , prove that :

$ (a\plus{}b\plus{}c\plus{}d)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d})$  $ \geq$ $ \frac{a\plus{}b}{c\plus{}d}\plus{}\frac{b\plus{}c}{d\plus{}a}\plus{}\frac{c\plus{}d}{a\plus{}b}\plus{}\frac{d\plus{}a}{b\plus{}c}$

hi , if we expand the first member i think it becomes easy to see :
First member = 4 + a/b + a/c + a/d + b/a + b/c + b/d + c/a + c/b + c/d + d/a + d/b + d/c
but for any x,y,z and t > 0 :
x/z + y/t >= x+y / z+t  (difference positive)
hence :
a/c + b/d >= a+b / c+d
b/a + c/d >= b+c / a+d
c/a + d/b >= c+d / a+b
and
d/c + a/b >= d+a / b+c
the result follows if no mistake.

[quote="y-a-s-s-i-n-e"]thank's , but i think that you prof is wrong[/quote]
i dont think so yassin!; why you think it  wrong?

When are LHS and RHS equal?
I don't know.

This is a trivial  and not a good inequality.
You have in $ RHS$ $ 16$ terms of degree $ 0$ and in $ LHS$ only $ 4$ terms of degree $ 0$.As Huynsoo Park saw, we have no equality case.If we equal terms on both sides, the ineq. should be:
$ (a\plus{}b\plus{}c\plus{}d)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d})\geq4(\frac{a\plus{}b}{c\plus{}d}\plus{}\frac{b\plus{}c}{d\plus{}a}\plus{}\frac{c\plus{}d}{a\plus{}b}\plus{}\frac{d\plus{}a}{b\plus{}c}).$
But this is not true.Try $ a\equal{}b\equal{}c\equal{}1, d\equal{}5$.
So,y-a-s-s-i-n-e, sorry,but you have to improve your inequality. :(

why it isn't true with  $ a \equal{} b \equal{} c \equal{} 1$ and $ d \equal{} 5$ you find after the calculation $ 19.2 \geq 6.666666667$

try to start from $ \frac {a \plus{} b}{c \plus{} d} \plus{} \frac {b \plus{} c}{d \plus{} a} \plus{} \frac {c \plus{} d}{a \plus{} b} \plus{} \frac {d \plus{} a}{b \plus{} c}$

hint
prove that
$ \frac {a \plus{} b \plus{} c \plus{} d}{a} \geq \frac {a \plus{} b}{c \plus{} d}$

[quote="y-a-s-s-i-n-e"]
hint
prove that
$ \frac {a \plus{} b \plus{} c \plus{} d}{a} \geq \frac {a \plus{} b}{c \plus{} d}$[/quote]
I think, your hint is confusing.
Try $ c \equal{} d\rightarrow0^ \plus{} .$ :wink:
By the way, the kprepaf's proof is true and nice.

[quote="y-a-s-s-i-n-e"]i didn't understand any thing in kperpaf's proof can you explaine please ?

is my inequality true ?[/quote]
Your inequality is true and nice.
Maybe kprepaf meant the following reasoning:
$ \frac{x}{z} \plus{}\frac{ y}{t}\geq\frac{ x\plus{}y}{ z\plus{}t}$ is true for positive $ x,$ $ y,$ $ z$ and $ t$ since,
$ \frac{x}{z} \plus{}\frac{ y}{t}\equal{}\frac{x^2}{xz}\plus{}\frac{y^2}{yt}\geq\frac{(x\plus{}y)^2}{xz\plus{}yt}\geq\frac{ x\plus{}y}{ z\plus{}t}.$ :wink:

thank's arqady
 sorry but if we assume we have :

$ \frac{2a}{c}\plus{}\frac{2c}{a}\plus{}\frac{2b}{d}\plus{}\frac{2d}{b}$ $ \geq$  $ \frac {a \plus{} b}{c \plus{} d} \plus{} \frac {b \plus{} c}{d \plus{} a} \plus{} \frac {c \plus{} d}{a \plus{} b} \plus{} \frac {d \plus{} a}{b \plus{} c}$

please  explaine :?:

[quote="y-a-s-s-i-n-e"]thank's arqady
 sorry but if we assume we have :

$ \frac {2a}{c} \plus{} \frac {2c}{a} \plus{} \frac {2b}{d} \plus{} \frac {2d}{b}$ $ \geq$  $ \frac {a \plus{} b}{c \plus{} d} \plus{} \frac {b \plus{} c}{d \plus{} a} \plus{} \frac {c \plus{} d}{a \plus{} b} \plus{} \frac {d \plus{} a}{b \plus{} c}$

please  explaine :?:[/quote]
But kprepaf explained it already.

yes

i did a little mistake when i assume

sorry kprepaf  :ush:  thank's a lot you proof is very nice  :D


A bag have 9 chips, 3 tan, 2 pink and 4 violet, if drawn randomly with out replacement, what is the probability in such way that 3 tan chips are consecutively, 2 pink consecutively and 4 violet consecutively, but not necessary in the tan-pink-violet order?

Find the normal probability without replacement then multiply by 3! for the 6 different order of the colors.

3!(3/9)(2/8)(1/7)(2/6)(1/5)(4/4)(3/3)(2/2)(1/1)

You can work it out to get the final answer.

[quote="Big M"]Why some time ask 8 red marble, 5 blue marble, the probability of getting 3 red marble should like this:

[(8*7*6)/3!]/[13*12*11/3!)

Can someone tell me why this needs to divide 3!, but the problem do not need to divide byanything?

Thank you![/quote]

Actually, you are multiplying by $3!$:

\[\frac{\frac{8*7*6}{3!}}{\frac{13*12*11}{3!}}\]
\[=\frac{8*7*6}{3!}\cdot\frac{3!}{13*12*11}\]


I also find this a little difficult
* means exponent
-2x*3 + 3x*2 + 5x - 6 > 70  ;)

this means exponent   ^
for example 2^3
is $2^3$

Factor the expression and those are your answers


Putnam
calculus
derivative
conics
hyperbola
inequalities
college contests

Find the minimum calue of
$(u-v)^{2}+\left(\sqrt{2-v^{2}}-\frac{9}{v}\right)^{2}$

for $0<u<\sqrt{2}$  and   $v>0$

well obviously taking the derivative isnt going to help too much.  :P
so i am think to try and relate it to points on the plane but that isnt going to well.
so far i have that $0<v\leq\sqrt{2}$ also i am certain that $u=v$ for some reason.  And i let $d(L)$ be the distance between the 2 points.
so the expression given is $d(L)^{2}$ and the points are
$(u,\sqrt{2-v^{2}})$ and $\left(v,\frac{9}{v}\right)$

but from here i am stuck, any ideas on what i should do next?

You are to find the mimimum value of the distance of two points on the circle with radius $\sqrt{2}$ and the hyperbola $y=\frac{9}{x}.$

ok that is what i thoug...so i guess then that $v$ and $u$ do not have to be equal, darn. :blush:  :mad:

http://www.mathlinks.ro/Forum/viewtopic.php?t=99494

And you said taking derivatives won't help much, but that is how the solution at Kalva starts out I believe.

i went to the link but was not able to understand the solution given there...sry

If you think of it as the distance between a circle and hyperbola, it's obvious: since points on the circle are always the same distance from the origin, the best you can do is when the point on the hyperbola is the closest to the origin and the point on the circle is on the same line through that point on the hyperbola and the origin (draw a pic to see what I mean).

The points are (1, 1) and (3, 3). That makes the answer 8. I can't follow langtupanda's solution either, but obviously 8 can be obtained, so there must be a mistake. (hopefully it is not mine :P)

[hide="remark"]
Actually, I think we were all assuming it was a typo in your problem statement, but as you wrote it, there's only a $u$ in the first part, so we just need to minimize the second part in $v$
$(\sqrt{2-v^{2}}-\frac{9}{v})^{2}$
which is surely best when $v=\sqrt{2}$ (we can't make $v$ bigger) and then we pick $u=v$ to obtain the answer $\frac{81}{2}$.
[/hide]

[hide="another remark"]
Most of the time if you have an inequality that's minimized or maximized by all of the variables being equal, it'll be symmetric in those variables. I'm sure it can happen at other times, though, too. [/hide]


http://www.kingdomofloathing.com

Fun game, it's a web RPG thingy with so many pop culture references and double entrendes that it isn't even funny (well, it is).

My username is Archbishop.

[quote="MysticTerminator"]secondbest?[/quote]

world of warcraft

[quote="Ignite168"]Versions of IdleRPG on most large IRC networks is so much better.[/quote]

i'm gonna smite you with my Dalnet sword and my Usenet hammer


Choose a 4-digit number $abcd$ called N.(at least two digits different)
Arrange the numbers in decreasing order to get $N_{d}$.
Arrange the numbers in increasing order to get $N_{i}$.
Then subtract: $N_{d}-N_{i}=N'$
Continue the above described procedure on $N'$
You will eventuelly end up with $6174$ if you did not choose a very inappropriate
$N$. [color=red]WHY?[/color]

Maybe because $6174$ is a fix point for your operation.

6174 is the only fixed point of this operation, and starting from a certain 4-digit number not divisible by 1111, after a finite number of steps you get 6174.
It was discovered by Kaprekar, so the number 6174 is called his name.

There are many similar properties,
the following one was discovered by Polish mathematician A. Porges in 1945:
Let's take a certain natural number, and count the sum of squares of its digits.
Do the same with new number etc.
After a finite number of steps you get : either 1, or 4 what leads to infinite loop 4->16->37->58->89->145->42->20->4.


A pilot flew from the equator always on a bearing North East. How far had he traveled when he reached the north pole? Take the radius of the earth to 4000 miles

[hide]
northeast = north + east.
north=4000 mi.
therefore east =4000 mi.
therefore traveled = 4000 \sqrt{2} mi.

(right?)
[/hide]

radius = 4000 miles
circumference = 8000 pi miles

If the plot flew directly North, then he would have to travel 1/4 of the circumference, or 2000 pi miles, to reach the pole.

Flying North - East means for every mile he flies north, he also flies exactly one mile east. Note that flying east does not get him any closer to the pole. So in the end he still has to fly 2000 pi miles to the North, in addition to 2000 pi miles to the East.

Use Pythagorean theorem to find the total distance:
sqrt((2000 pi miles to North)^2 + (2000 pi miles to East)^2) = 2000 sqrt(2) pi miles


This is my thought, but have no idea whether it is true :(
Consider any real sequence that is convergent, is any rearrangement of this sequence convergent too?

I think it is correct and it has the same limit as the original sequence. I mean I see that no matter how we arrange some infinite numbers of the rearrangement will lie around the limit. But I think it is a delicate topic. Some body might create a monster sequence that does not agree to me

Any ideas appreciated,
Thanks :lol:

It is true, just apply the definitions (although you did not provide your rearrangement definition), e.g., let $ x$ be the limit of the sequence $ (x_n)_{n\equal{}1,2,\ldots}$ and $ (y_n)_{n\equal{}1,2,\ldots}$ be a [i]rearrangement[/i], then we have the following:

$ \forall \epsilon >0: \exists N_{\epsilon}$ such that $ \forall n>N_{\epsilon}: |x\minus{}x_n|<\epsilon$ holds. Now for a fixed arbitrary $ \epsilon >0$ determine $ y_{k_1},\ldots,y_{k_{N_{\epsilon}}}$ in the rearrangement, which correspond to $ x_1,\ldots,x_{N_{\epsilon}}$. Let $ M_{\epsilon}\equal{}\max\{k_1,\ldots,k_{N_{\epsilon}}\}$. Then we have $ \forall n>M_{\epsilon}: |x\minus{}y_n|<\epsilon$ and since $ \epsilon$ was arbitrary $ y_n\rightarrow x$

I had the same idea, but i thought I might be missing something very minute in the argument. Since in series, a conditionally convergent series behaves strangely, I thought a sequence might suffer correspondingly. But my argument showed otherwise, so wanted to clarify  :P
Thanks  :lol:


We all know our instructors are never wrong, so these might as well be the official answers from U of N in Lincoln.

This is for those who missed the Math Jam

[hide](I'm not too great at taking notes so you might want to check the Math Jam transcripts anyways.)[/hide]

1. 592

2. 441

3. 384

4. 927

5. 766

6. 408

7. 293

8. 054

9. 973

10. 913

11. 625

12. 134

13. 484

14. 108

15. 593

isn't R on the AMC Advisory Panel?? Doesn't he have hte answers????

I dropped off all test-related duties when I started this site, so I don't have access to the tests anymore until they're administered.


let a_n be a sequence of real number and we define the sequence u_n by:
u_(n+1)=u_n+sqrt(u_n^2 + a_n)
Prove that the sequence u_n converges iff the series sum(a_n,n) converges.

Hmmm... Am I missing something? What if $a_n=0$ and $u_0=1$?. $u_n$ clearly doesn't converge... Or does it? Another thing that should be cleared up: is it $u_{n^2}$ or $(u_n)^2$?


function
linear algebra
matrix
calculus
integration
real analysis
real analysis unsolved

Firstly, it's not a problem, just some dicussions.
Really this is very known but I' fond at its ideals. All we know that an ODE like as: $\frac{dy}{dt}=A(t)y$, $y(0)=y_{0}$ where $A$ is function on an interval $I$ with $n\times n$ matrix-values. If we could find a bound $M$ of $A$, namely $\sup|A(t)|<+\infty$ (sup takes on $t\in I$) then the local solution can be extended to all $I$. Now let me assume that $\sup|A(t)|=+\infty$,
a) Is there such $A$ so that all local solution can be extend to all $I$?
b) Is there such $A$ so that all local solution can not be extend to all $I$?

It could go either way. At least for scalars, there's really very little difference between this and the equation $y'=A(t)$. If the integral of $A$ converges, solutions can be extended. If it doesn't, solutions can't be extended.[/i]

All right, here's a problem:

Construct an explicit matrix-valued function $A(t)$ on an interval $I$ such that
(1) The (improper) integral $\int_{I}A(t)\,dt$ converges.
(2) Some solutions to the differential equation $y'(t)=A(t)y(t)$ on subintervals of $I$ cannot be extended to all of $I$.

[hide="Hints"]If $\int_{I}\|A(t)\|\,dt$ converges, this is impossible. We need conditional convergence to make this work.[/hide]


$ ABCD$ is convex quadrilateral with $ [AB]$ is not parallel to $ [CD]$.
$ E$ and $ F$ be the midpoints of $ [AD]$ and $ [BC]$ respectively. $ |CD|\equal{}12$ , $ |AB|\equal{}22$ and $ |EF|\equal{}x$.
What are the possible integer values of $ x$ ?

We can use coordinates.
We have A(a,b) B(c,d) C(12,0) D(0,0)
Let $ x \equal{} d \minus{} b$, $ y \equal{} c \minus{} a$
From the given, we have
$ x^2 \plus{} y^2 \equal{} 484$
We have to find the possible values of the square root of
$ \frac {x^2 \plus{} (y \plus{} 24)^2}{4} \equal{} 265 \plus{} 12y$
But y can range from 0 to 22, so the possibilities are 265-520, the integer square roots ranging from 17-22.


topology
function
vector
advanced fields
advanced fields unsolved

What are the extreme points of the unit ball of $ C(X)$, where $ X$ is a compact Hausdorff space ?
What about $ l^p$.

The extreme points are the functions such that the $ |f(x)|\equal{}1$ (identically). In the real (connected) case, this is not very interesting. In the complex case however, it turns out that these functions are dense in the unit ball.

For $ l^p$, the extremal boundary of the unit ball is the topological boundary: every unit vector is an extreme point.

[quote="phils"]these functions are dense in the unit ball.[/quote]
Do you mean: the convex combinations of these functions are dense?

[quote="phils"]For $ l^p$, the extremal boundary of the unit ball is the topological boundary: every unit vector is an extreme point.[/quote]
Unless $ p\equal{}1$ or $ p\equal{}\infty$.

Thanks for the corrections, mlok. I'm embarrassed by my semi-literate guesses - it's been a while since I've last seen that stuff.


Samantha deposited $ \$100$ in a bank account that has earned $ 5\%$ interest during this year. Audrey deposited $ \$50$ more than Samantha deposited but earned only $ 4\%$ interest during this year. Now, Audrey has how much more money than Samantha?

Samantha gets 5 dollars of interest, while Audrey gets 6 dollars of interest. So Audrey has 51 more dollars than Samantha.


Let $ABC$ be a triangle with $BC>CA>AB$ and let $G$ be the centroid of the triangle. Prove that  \[ \angle GCA+\angle GBC<\angle BAC<\angle GAC+\angle GBA . \]
[i]Dinu Serbanescu[/i]

[quote="indybar"]Let $ABC$ be a triangle with $BC>CA>AB$ and let $G$ be the centroid of the triangle. Prove that
\[\angle GCA+\angle GBC<\angle BAC<\angle GAC+\angle GBA . \]
[i]Dinu Serbanescu[/i][/quote]

it is false?? :?:


inequalities
function
real analysis
real analysis unsolved

Let $H(x)$ be a differentiable function on $R$  , $H(x_0) = 0$ and

        $H'(x) > 2cxH(x)$, $c >0$ for all $x > x_0$.

Prove that $H(y)$ is not equal to 0 for all $y>x_0$


linear algebra
matrix
algebra
binomial theorem
superior algebra
superior algebra unsolved

$ \mathcal R$ is a commutative ring with $ 1$ in which $ 2$ has inverse. Prove that for each $ n\geq 2$, and $ A\in\mathcal M_n(\mathcal R)$, there are 3 matrices $ A_1,A_2,A_3$ such that
\[ A \equal{} A_1^2 \plus{} A_2^2 \plus{} A_3^2
\]

I have the solution (found by my father, actually), but right now I can only provide a

[hide="rough sketch."]
First, for the $ n\equal{}2$ case, use

\[ \begin{pmatrix} 1\plus{}a & x \\ 0 & 1\minus{}a \end{pmatrix}^2 \plus{}
 \begin{pmatrix} 1\minus{}b & 0 \\ y & 1\plus{}b \end{pmatrix}^2 \plus{}
 \begin{pmatrix} 0 & \minus{}(1\minus{}a)^2\minus{}(1\minus{}b)^2 \\ 1 & 0 \end{pmatrix}^2 \equal{}
 \begin{pmatrix} 4a & 2x \\ 2y & 4b \end{pmatrix}
\]

and observe that since 2 is invertible, the result can be made anything we want.

For $ n \geq 3$, first consider matrices with 0 on the main diagonal, 1's below the diagonal, and variables above the diagonal. Square, and prove that you can make the main diagonal of the result anything you want (disregarding the results outside of the main diagonal).

Next, consider an upper triangular matrix with $ 1/2$ on the main diagonal, 0 below the diagonal, and variables $ x_{ij}$ above the diagonal. Square, and you'll notice that each entry in the result looks like $ x_{ij} \plus{}$ stuff involving variables on lower diagonals. Conclude that the square of such a matrix can be fully controlled above the main diagonal, while the main diagonal is of course forced to be $ 1/4$ and anything below it remains 0.

Now do the same for lower triangular.

The last step is easy: start with any matrix, use the first step to transform the main diagonal into all $ 1/2$, and then use the second step to find suitable upper and lower triangular matrices that can fix the result off the main diagonal.
[/hide]

Wow, that's an impressive idea, though I don't understand what you are doing for $ n \equal{} 3$.
Here is a complete proof based on your idea.

First I'll need the two formulas
$ \begin{pmatrix}{x \minus{} y \plus{} 1\over 2} & 2y \minus{} {(x \minus{} y \minus{} 1)^2\over 2} \\
{1\over2} & {x \minus{} y \minus{} 1\over 2}\end{pmatrix}^2 \equal{} \begin{pmatrix}x & * \\
* & y\end{pmatrix}$
$ \begin{pmatrix}0 & {x \plus{} y \minus{} z\over 2} & {x \minus{} y \plus{} z\over 2} \\
1 & 0 & { \minus{} x \plus{} y \plus{} z\over 2} \\
1 & 1 & 0\end{pmatrix}^2 \equal{} \begin{pmatrix}x & * & * \\
* & y & * \\
* & * & z\end{pmatrix}$
for any $ x,y,z\in\mathcal{R}$.

Then take any $ n\times n$-matrix $ A$ over $ \mathcal{R}$ with $ n\ge 2$.
Using the above formulas we can find some $ n\times n$-matrix $ A_1$, s.t. $ A_1^2$ and $ A \minus{} 2\cdot I_n$ have the same diagonal,
by putting some $ 2\times 2$ and $ 3\times 3$-matrices as blocks on the diagonal of $ A_1$ and otherwise $ 0$.

This gives $ A \minus{} 2\cdot I_n \minus{} A_1^2 \equal{} N_1 \plus{} N_2$ with some nilpotent triangular matrices $ N_1,N_2$ (upper and lower),
which leaves to show, that $ I_n \plus{} N$ is a square for any nilpotent $ n\times n$-matrix $ N$ over $ \mathcal{R}$.

But using [url=http://en.wikipedia.org/wiki/Binomial_theorem]Newton's generalized binomial theorem[/url] we get
$ M^2 \equal{} I_n \plus{} N$ with $ M: \equal{} \sum_{k\ge 0}\binom{1/2}{k}N^k$, where $ N^k \equal{} 0_n$ for large $ k$
and the binomial coefficient $ \binom{1/2}{k}: \equal{} {{1\over 2}({1\over 2} \minus{} 1)\cdots ({1\over 2} \minus{} k \plus{} 1)\over k!}$ is well-defined in $ \mathcal{R}$ for all $ k\ge 0$
using $ \binom{1/2}{0} \equal{} 1,\binom{1/2}{1} \equal{} {1\over 2}$ and the recursive formula $ \binom{1/2}{k} \equal{} \minus{} {1\over 2}\sum_{0 < l < k}\binom{1/2}{l}\binom{1/2}{k \minus{} l}$ for all $ k\ge 2$,
which comes from comparing coefficients of $ 1 \plus{} X \equal{} (\sum_{k\ge 0}\binom{1/2}{k}X^k)^2$ in $ \mathbb{Z} [ [X] ]$.

PS this shows, that $ 4^kA$ is a sum of three squares over any ring with $ 1$ (not just commutative) and $ k\in\mathbb{N}$ large enough.
PPS explicitly we get $ 4^{4n \minus{} 5}A$ a sum of three squares over any ring with $ 1$ and any $ n\times n$-matrix $ A$ with $ n\ge 2$.

[quote="olorin"]$ 1 \plus{} X \equal{} (\sum_{k\ge 0}\binom{1/2}{k}X^k)^2$ in $ \mathbb{Z} [ [X] ]$[/quote]
Sorry, I meant in $ \mathbb{Q} [ [X] ]$ the ring of [url=http://en.wikipedia.org/wiki/Power_series_ring#The_ring_of_formal_power_series]formal power series[/url] over $ \mathbb{Q}$.


calculus
integration
trigonometry
logarithms
function
calculus computations

Evaluate $ \int_{\frac{\pi}{2}}^{\pi} x^{\sin x}(1\plus{}x\cos x\cdot\ln x \plus{}\sin x)\ dx$.

The first guess worked:
$ (x^{\sin x\plus{}1})'\equal{}(e^{(\sin x\plus{}1)\ln x)})'\equal{}((\sin x\plus{}1)\ln x)' e^{(\sin x\plus{}1)\ln x}\equal{} (\cos x \ln x \plus{} \frac{\sin x \plus{}1}{x}) x^{\sin x\plus{}1}\equal{}x^{\sin x}(1\plus{}x\cos x\cdot\ln x \plus{}\sin x)$
Hence, the integral equals to $ \pi ^{\sin \pi \plus{}1} \minus{} (\frac{\pi}{2})^{\sin\frac{\pi}{2}\plus{}1}\equal{}\frac{4\pi \minus{}\pi ^{2}}{4}$

[hide]$ I \equal{} \int_{\pi/2}^{\pi}x^{\sin x}(1 \plus{} x \cos x \ln x \plus{} \sin x)\,dx \equal{} \int_{\pi/2}^{\pi}\left(x^{\sin x \plus{} 1}\right)'\,dx \equal{} \left[x^{\sin x \plus{} 1}\right]_{\pi/2}^{\pi}$

$ \equal{} \pi \minus{} \left(\frac{\pi}{2}\right)^2$.[/hide]

Carcul, your solution is same as bambaman's. :wink:

[quote="bambaman"]The first guess worked:
$ (x^{\sin x \plus{} 1})'$[/quote]
What made you think about it? Just a trial and error way of resolution or there were some "clues" in the integral? Your solution starts with a function that seems completely pointless in order to calculate the integral and then gets a very simple way to solve it.

Goodbye
GM


For any straight line, let $m$ and $b$ represent its slope and y-intercept, respectively. Consider all lines having the property that $2m+b = 2000$. These lines all have the specific point $(x_1,y_1)$ in common. Find this common point $(x_1,y_1)$.

Solve this problem without use of calculator.

[hide]
Isolating, we get:

$m=\frac{2000-b}{2}$.  In solpe intercept point:

$y=\frac{2000-b}{2} x+b$.  The only value (x,y) which is constant regaudrless of b is $(2,2000)$[/hide]

I hid your answer.

Guys, please hide your answer. ;)


calculus
integration
function
algebra
domain
LaTeX
complex analysis

Hi all,

This is a basic theoretical question I am not quite sure about.

Suppose you have an analytic function F defined on some domain contained in the open upper half plane. You then decide to define say F* on the "conjugate" domain D* in the lower half plane, as F*(x) = conjugate(F(conjugate(x))

Now I want to integrate F* around some oriented curved contained in D*, can I say that the integral is equal to the conjugate of the integral of F over the reversely-oriented conjugate curve? This seems obvious... yet I am a bit concerned regarding the umm symantics of proving this because when I want to make the substitution u = conjugate(z), but then I am concerned that the dz becomes d(conjugate(u)), which I don't know how to deal with, and I want to integrate wrt to u not its conjugate...

Anyways hope that makes sense, sorry I don't know latex to make that clearer, but if you can drop some knowledge regarding this I would greatly appreciate it, I am sure this is pretty basic, just not sure how to get around the issue of integrating wrt. the conjugate. Thanks.

Conjugation is not a conformal map, so $ f(z) \equal{} \bar{z}$ is not complex-differentiable.


[quote="Teki-Teki"]My advice: Brute-force it.  But first, check your posting.  Should it be (a^4 + b^4 + c^4) / 4, NOT divided by 2?  I'm just observing the pattern.[/quote]

Did you try to brute-force it? :-)

This question is not very difficult to deal with by brute force.
If you don't like brute force, here is an alternative.

Denote $A = ab + bc + ca, B = abc$.
Then $a, b, c$ are the roots of $x^3+Ax-B =0$.

Define $S_n=a^n+b^n+c^n$ for $n = 0, 1, 2, 3,......$
Then $S_0=3$, $S_1=0$,$S_2=(a+b+c)^2-2(ab+bc+ca)=-2A$.

Note that $S_{n+3}=-AS_{n+1}+BS_n$ for $n = 0, 1, 2, .....$
$S_3=-AS_1+BS_0=3B$,
$S_4=-AS_2+BS_1=2A^2$,
$S_5=-AS_3+BS_2=-5AB$, and
$S_7=-AS_5+BS_4=7A^2B$.

Direct verification shows that $\frac{S_3} 3 \cdot \frac{S_4} 2 = \frac{S_7} 7 = \frac{S_2} 2 \cdot \frac{S_5} 5$, which is equivalent to what we need to show.


$x+2y=-17$
$2x+y=-16$

the first equation can be written as $2x+4y=-34$
subtracting the second equation from this gives
$3y=-18$, so $y=-6$.  we can plug this into any of the equations to find $x=-5$.

$x+2y =-17$
$2x+1y =-16$

Eliminate either y or x from either equation. Let's choses to eliminate $x$ from the second equation. This can be done by adding to the second equation -2 time th e first equation to get a new second equation.

      $2x+1y =-16$
(+) $-2x-4y = 34$
----------------------------
             $-3y = 18$
                $y =-6$

Try finding $x$ now =P

[quote="Imperial Effect"]
             $-4y = 18$
[/quote]
check your subtraction again :wink:
also, $-2\cdot4\ne18$

Daermon,

I just pm'd you?  Is that alright?

a general method when given a system of equations
$a_{1}x+b_{1}y=w\\ a_{2}x+b_{2}y=z$ is to multiply the first equation by $a_{2}$ and the second by $-a_{2}$, then adding the equations together.  The coefficient of $x$ in the new equation will be $a_{1}a_{2}-a_{1}a_{2}=0$, so we will only have one variable to deal with: $y$ :D

Another (relatively) painless method with linear systems is [url=http://en.wikipedia.org/wiki/Cramer%27s_rule]Cramer's Rule[/url], which involves determinants of matrices (which aren't that hard to do by hard). This method is also a bit easier to do than the substitution method for a larger system of equations.

[quote="jenny221"]Does anyone know this?

Solve using elimination method x + 2y = -17, and 2x + 1y = -16[/quote]
[hide]
elimination is getting rid of a variable. An easy variable to get rid of is x (though y is just as easy).
2x+4y=-34
2x+y=-16
subtract them
3y=-18
y=-6
x=-5[/hide]

[hide="formula"]

we have

$ax+by=c$
$dx+ey=f$

$aex+bey=ce$
$dbx+eby=fb$
____________
$(ae-db)x=(ce-fb)$

$x=\frac{ce-fb}{ae-db}$

$ax+by=c$
$dx+ey=f$

$adx+bdy=cd$
$adx+aey=af$
_____________
$(bd-ae)y=(cd-af)$

$y=\frac{cd-af}{bd-ae}$

$(x,y)=\left(\frac{ce-fb}{ae-db},\frac{cd-af}{bd-ae}\right)$[/hide]

I recommend to NOT memorize the formula, but to see why it works.

[quote="Trisha"]I need help with the following problem.

2/3x-3/5y=-17/15
8/5x-7/6y=-3/10

Please solve by elimination and give the steps.  Thanks![/list][/quote]
Please do not revive a 2 month old topic.


Give a,b,c are real numbers.
Prove that : a(a+b)^3+b(b+c)^3+c(c+a)^3 >= 0

This problem was created by vasile Cartoaje long time ago and it eas already posted on this forum.

Move it from Algebra to Inequalities. Maybe some one will give us a proof :)

A stronger inequality was posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113218[/url].


integration
logarithms
calculus
calculus computations

Help me to check the solution, thanks.

$ \text{*  }I\equal{}\int\limits_{1}^{e}{\frac{\log \left( 1\plus{}\log x \right).dx}{x}}$

$ \text{Since for }x\in \left[ 1,e \right]\text{, we have }0\le \log x\le 1\Rightarrow$

$ I\equal{}\int\limits_{1}^{e}{\frac{dx}{x}.\sum\limits_{n\equal{}1}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{n\plus{}1}}.{{\left( \log x \right)}^{n}}}{n}}}\equal{}\sum\limits_{n\equal{}1}^{\infty }{\frac{{{\left( \minus{}1 \right)}^{n\plus{}1}}}{n}}\int\limits_{1}^{e}{{{\left( \log x \right)}^{n\minus{}1}}.d\left( \log x \right)}\text{, }$

$ \int\limits_{1}^{e}{{{\left( \log x \right)}^{n\minus{}1}}.d\left( \log x \right)}\equal{}\left[ \frac{{{\left( \log x \right)}^{n}}}{n} \right]_{1}^{e}\equal{}\frac{1}{n}\Rightarrow I\equal{}\sum\limits_{n\equal{}1}^{\infty }{\frac{{{\left( \minus{}1 \right)}^{n\plus{}1}}}{{{n}^{2}}}}\equal{}\frac{1}{2}\zeta \left( 2 \right)\equal{}\frac{{{\pi }^{2}}}{12}$
and

$ \text{Put }y\equal{}1\plus{}\log x\Rightarrow dy\equal{}\frac{dx}{x}\Rightarrow I\equal{}\int\limits_{1}^{2}{\log y.dy}\equal{}\left[ y\left( \log y\minus{}1 \right) \right]_{1}^{2}$

$ \equal{}2.\left( \log 2\minus{}1 \right)\plus{}1\equal{}2.\log 2\minus{}1$

what a pity!
The first answer is wrong,and the second is right.
you wrong at second "="
it shoud be (ln x)^n,not (ln x)^(n-1).


Is the proper divisor including or not including 1?
Also, I can't exactly understand what you're asking about 1/4 of the proper divisors.
Does this mean that we're looking for numbers that have 4 proper divisors?

[quote="pianoforte"]An random integer is choosen from the first 500 natural numbers. What is the probability that the integer picked is such that there is a 1/4 chance of picking any one of its positive, proper divisors?[/quote]

Hm. Interesting problem.  :P
[hide]I think the answer is $\frac{179}{500}$. I found that only numbers that can be expressed in the form $x\cdot y$ where $x$ and $y$ are prime numbers have a $1/4$ chance of picking any one of its positive, proper divisors. If $x=2$, there are 53 primes that can replace y and still be under 500. I did that until 500/x<x. So then there are 53+38+25+19+14+11+10+9 $x\cdot y$ combinations out of 500, or $\frac{179}{500}$?[/hide]

It would be the second case: "the probability of picking a number such that the probability of randomly selecting a specific proper divisor is 1/4"

I included negative divisors as divisors...because that's what I thought the problem was inferring...

[quote="samath"]Is the problem asking for the probability of getting a number such that the probability of then picking a number from 1-500 and getting it's divisors is $\frac{1}{4}$, or the probability of picking a number such that the probability of randomly selecting a specific proper divisor is $\frac{1}{4}$?  In the first case, it's obviously 0-no number under 500 has 125 factors.  For the second one, I believe that a "proper divisor" excludes one and itself, so we would be looking for numbers with 6 factors.  There are 2 numbers of the form $x^5$ and 57 numbers of the form $xy^2$ because if y=:
2: x= Any prime from 3-125: 29 (check that)
3: x=Any prime from 2-55 (other than 3):15
5: x=Any prime from 2-20 (other than 5): 7
7: x= Any prime from 2-10 (other than 7): 3
11: x=Any prime from 2-3: 2
13: x=2: 1

Then I get a probability of $\frac{59}{500}$[/quote]

I believe that a proper divisor just excludes itself, cause I remember that doing a problem saying to find the sum of the proper divisors of 256, and the answer was 255, so that didn't include itself.


Is there any way I can size the font and size within the texts in the table with LaTeX?

Also, is there anyway I can change the color within each cell of the text? I could always use paint, but that takes a while. Thanks!

You can use the usual size commands for text: \tiny, \scriptsize, \footnotesize, \small, \normalsize, \large \Large, \LARGE, \huge, \Huge. Similarly for font colour use \color $ \text{\Large \color{green} Large green text}$.

But if you mean that you want to change the background colour of a cell then the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=colortbl]colortbl package[/url] will do that for you.

$ ``I\ am\ just\ testing," said\ the\ man\ .$


Hello, I hope I've started this thread in the appropriate category. :)

I'm trying to find a solution to this particular difference equation:

$ a_r^2 \minus{} a_{r\minus{}1}^2 \equal{} 1\\a_0\equal{}2$

I haven't found anything useful in the Internet, no matter how have I searched. So any help would be grateful. :)
Thanks in advance. :)

[quote="modulo"]$ a_r^2 \minus{} a_{r \minus{} 1}^2 \equal{} 1 \\
a_0 \equal{} 2$[/quote]
Isn't it too obvious :?:  The square of each term is one more than the square of preceding term. Thus the sequence which has terms equal to the squares of the terms of the original sequence is a sequence of consecutive numbers. So $ a_n\equal{}\sqrt{4\plus{}n}$ :?:
I might be seriously wrong. I apologize for that in advance.

Here's one way to put it (I couldn't find a way using standard recurrence solving methods- z-transforms make a mess, solving corresponding differential equation does work (by separating variables), but it may be too complex. That's why I'll present an elementary proof:

$ a(n)^2\equal{}1\plus{}a(n\minus{}1)^2\equal{}1\plus{}1\plus{}a(n\minus{}2)^2\equal{}...\equal{}n\plus{}a(0)^2$. Hence, $ a(n)\equal{}\pm\sqrt{n\plus{}4}$

(I know, Akashnil already posted the same thing, while I was typing ;) )

Uh... yes. It should be $ \pm \sqrt{4\plus{}n}$
[quote="hsiljak"]...standard recurrence solving methods...[/quote]
I always longed for a tutorial on this... Do you know any link or something on the internet?

hello, i think, it is $ a_n \equal{} \pm\sqrt {n \plus{} 4}$.
Sonnhard.

Oh, indeed:

$ \left(\pm \sqrt {4 \plus{} n}\right)^2\minus{}\left(\pm \sqrt {4 \plus{} (n\minus{}1)}\right)^2\equal{}(4 \plus{} n)\minus{}(3\plus{}n)\equal{}1$

Can't believe it was THAT easy... how can I mind get stuck sometimes. Thank you everyone for your precious help. :)  :)  :)  :)


modular arithmetic
induction
number theory unsolved
number theory

a) Prove  every number can be expressed as alternating sum of powers of two. (for example: 2=-2+4,3=1-2+4,4=-4+8,5=1-4+8, etc.)

b)for a given natural number $n$, how many diffrent ways can we express $n$ as alternating powers of 2? (for example: 5=1-4+8, 5=-1+2-4+8

let's try with a)
[hide]

Let's start with pairs number, we know that if a number is pair but has not a power of 4, we have: $2n\equiv 2 \pmod 4$ so it's easy to show that taking the last power of 2 $2^n < 2n$ we got $2n -2 ^n = 2n'$, know we can do the same thing as we did first we see if it's a power of two if not we take the last power $2^m < 2n'$ and we $2n'- 2^m=2n''$ until we got 2.In this way we obtain our number with powers of two.
Now we will consider the odds number, if a number is odd it gives upon division by 4 a rest of 1 or 3, since $1= 2^0$ and $3= 2^0 - 2^1 + 2^2$ we could write, following the same method used for the pair numbers, every odd number. Q.E.D.
  [/hide]

Hope not to have missed anything


if a and b are two positive integers,
What is the smallest possible value of a+b given that $6a^3 = 5b^5$ ?
Corny Title, I know

Quote:If a and b are two positive integers,
What is the smallest possible value of a+b given that  ?[hide]

Let a = 2p*3q*5r, and b = 2s*3t*5u.

Then we have: (2)(3)(2p*3q*5r)3 = 5(2s*3t*5u)5

And: 23p+1*33q+1*53r = 25s*35t*55u+1

Hence: 3p + 1 = 5s, 3q + 1 = 5t, 3r = 5u+1

For the smallest values: p = 3, s = 2, q = 3, t = 2, r = 2, u = 1

Therefore: a = 23*33*52 = 5400, b = 22*32*51 = 180 . . . and a + b = 5580[/hide]


These are the questions of Iranian pre-TST first round
Question 1- http://www.mathlinks.ro/Forum/viewtopic.php?p=185850#p185850
Question 2- http://www.mathlinks.ro/Forum/viewtopic.php?t=30002
Question 3- http://www.mathlinks.ro/Forum/viewtopic.php?p=185853#185853
And the PDF version of questions:

Thank you - can you tell us when the next rounds will take place?  :D And how many rounds are there ?

Well there have been 3 exams week ago, 2 weeks ago, 3 weeks ago. I was in Tehran for the IMO Preparation program. Now because of Iraninan new year that is 4 days later we have vacation. I'm home in Isfahan and I have time for posting the questions. I'll post the second pre-TST soon

Questions of second Round:
1- http://www.mathlinks.ro/Forum/viewtopic.php?p=187969#187969
2- http://www.mathlinks.ro/Forum/viewtopic.php?p=187974#187974
3- http://www.mathlinks.ro/Forum/viewtopic.php?p=187975#187975
And PDF version:

Omid Hatami: When you will post the third test?  :)

Well for some reasons I couldn't send the problems of 3rd exam, But these are the problems of 4th exam that are from Silk Road Math Olympiad 2005
Problem 1- http://www.mathlinks.ro/Forum/viewtopic.php?p=205676#205676
Problem 2- http://www.mathlinks.ro/Forum/viewtopic.php?p=205678#205678
Problem 3- http://www.mathlinks.ro/Forum/viewtopic.php?p=205679#205679
Problem 4- http://www.mathlinks.ro/Forum/viewtopic.php?t=33037


I propose a contest where each person makes a picture. The winner will be the one who has the most creative drawing. Although it is not a very exciting prize the winner will be awarded the first consideration for any other contest I run. Now here are some restrictions:

There are up to 31 players(I will join in too).
There must be at least 8 players
If any new player wants to join the game he/she will post on the forum and be considered.
At the beginnig of the game 5 judges will be elected to work out the winner(after the game) .
The picture can be made however you like.
It is not permitted to copy someone else. If this happens then the person who posted the copy will be thrown out.
there will be three rounds.

Let the tournament begin!!!
 :starwars:


1. What is the average of all [i]4[/i] digit numbers with distinct digits fomred by choosing out of the digits 1,3,4,6.

2. What is the average of all [i]3[/i] digit numbers with distinct digits fomred by choosing out of the digits 1,3,4,6.

3. What is the average of all 3 digit [b]even[/b] numbers with distinct digits formed  by choosing out of the digits 1,3,4,6.

i dont know of any easy way to do these type of problems so i would love to learn one cause i see them all of the time on mathcounts and such.

#1
[hide]Ok so there are 24 numbers, and every number is in every position 6 times. so we have

   1(6) 1(6) 1(6) 1(6)
   3(6) 3(6) 3(6) 3(6)
   4(6) 4(6) 4(6) 4(6)
 +6(6) 6(6) 6(6) 6(6)
____________________
 9   3    3     2     4

When this is divided by 24 to get the averege we have $ \boxed{3888.5}$[/hide]


symmetry
trigonometry
real analysis
real analysis unsolved

One day a hunter lost in the forest.He doesn't know where the road is,but he knows that the road is somewhere on a distance equal to $ 1$,from the place where he initially stands.Find the trajectory in which he must go to find the road,so that the length of this trajectory would be minimized.Also find the length of this trajectory.

Do we know anything about the road?  For example, if the road is known to be perfectly straight, then a path of length $ 2 \plus{} \sqrt {2} \plus{} \pi < 1 \plus{} 2\pi$ is possible (maybe even shorter).  If at least the road is not 'too wavy' in some sense (e.g. the curve is smooth and its curvature is sufficiently small), it may be possible to get a bound below $ 1 \plus{} 2\pi$.  But if we know nothing about the road then it's effectively like searching for a point, so kenn4000's observation would seem to be correct.

[quote="kenn4000"]making a lot of assumptions here..  most importantly

1) said hunter cant "see" the road from any distance except when he stands on it?

then any path he takes out to the circle of radius 1 will be of length 1.  then he would walk around in either direction the circumfrence of 2 pi. (at most)

the symmetry of the unknowns seems to make this the only choice right?[/quote]
Nope,there exist a shorter way.
Answering to JoeBlow's question : Yes,we know that the road is a straight line.
If he could see the road from some distance then I would definitely mention it in the statement,so we may assume that our hunter is blind. :)
What do you mean by "symmetry of the unknowns"?
It seems to me that $ \sqrt {2} \plus{} 2 \plus{} \pi$ is correct answer,but I am not sure,maybe you would be so kind to post the image or anything else,for us to check it out.(at least a description)

Red is the set of points where the road may be, green is the hunter's path.
[asy]draw(Circle((0,0),10),red+linewidth(1));
draw(Arc((0,0),10,0,180),green+linewidth(1));
draw((-10,0)--(-10,-10),green+linewidth(1));
draw((10,0)--(10,-10),green+linewidth(1));
draw((0,0)--(-10,-10),green+linewidth(1));
label("$\sqrt{2}$",(-5,-5),SE);
label("$1$",(-10,-5),W);
label("$1$",(10,-5),E);
label("$\pi$",(0,10),N);[/asy]
I feel fairly confident this is the shortest path, but not sure how one might go about proving that.

Your example is almost correct,but not completely...

Now it seems to be correct.
I have no idea how it can be proved,but I'm pretty sure that this is the only right example.
I doubt if there exist any easy solution,I think its proof kinda messy and uses a lot of calculus.


function
combinatorics unsolved
combinatorics

A class of 25 students and 2 teachers collect money for charity. Each of the teachers donates 0, 3 or 6 coin, and each student 0 or 1. Let $a_{n}$ be the number of ways in which n coin can be collected

a) Find generating function $G_{\bar{a}}(x)$ for the sequence $\bar{a}$.

b)Determine $a_{20}$.

About the part b) it could be 25,584,855.

The teachers can collect 0, 3 or 6 coins, so there are 9 different ways:

0, 0
0, 3
0, 6
3, 0
3, 3
3, 6
6, 0
6, 3
6, 6

And the students have to collect the other coins:

20 coins
17
14
17
14
11
14
11
8

So the different ways to collect 20 coins are
$(25,20)+2*(25,17)+3*(25,14)+2*(25,11)+(25,8)$


probability
analytic geometry
graphing lines
slope

As the diagram shown , the gird-table is formed by 25 small squares.

When we get any $ 2$ points from the grids beside point $ A$ , what is the

probabilty of forming a straight line linked with $ A$?

 :D

Just a junior problem-solver. I used organized brute force (though you were probably looking for something more creative) [hide]sum of all possible cases where three points are collinear/total no. of possible choices=8/119, or maybe I screwed up, I am so tired![/hide] If anyone finds a way other than organized brute force and casework, please tell me.

[hide="Solution"]On the first diagonal, $ \binom{5}{2}$

On the other, $ \binom{4}{2}$.

On the two vertical lines, $ \binom{5}{2}$

Summing, $ 36$ ways.

Answer: $ \frac{36}{\binom{24}{2}}\equal{}\boxed{\frac{3}{23}}$[/hide]

I can't think of a better way to do it than those cases.

[quote="grn_trtle"][hide="Solution"]On the first diagonal, $ \binom{5}{2}$

On the other, $ \binom{4}{2}$.

On the two vertical lines, $ \binom{5}{2}$

Summing, $ 36$ ways.

Answer: $ \frac {36}{\binom{24}{2}} \equal{} \boxed{\frac {3}{23}}$[/hide]

I can't think of a better way to do it than those cases.[/quote]

there are 4 more lines, forming the slope of 2,-2,1/2,-1/2 around point A
:)
and the total # of ways of picking 2 points should be 35*34/2! i think

Wait, the method you guys are using doesn't take into account the [u]order[/u] of the different ways through which two points may be collinear with $ A$. For example, if $ X$ and $ Y$ are collinear to A, there are actually $ 2$ ways to pick points $ X$ and $ Y$ (if order counts), not 1 way as $ 2!/2!0!\equal{}1$ shows. Or perhaps I'm wrong and order isn't taken into account at all...

It doesn't matter so long as you use the same method when counting the number of "good" ways and the number of total ways.  (Switching between the two methods is the same as multiplying by $ \frac{2!}{2!}$.)

I don't see the other 4 lines... I looked for other lines, but...
A line with slope of 2 can't fit three collinear points in that square, can it?

:huh:  Let B be the point two up and one right from A and let C be the point two down and one left from A.  Then CAB is a straight line.

We're not choosing squares, we're choosing vertices of squares.


Given an ordered sequence of N elements, what is the minimum number of random relations of the form "A is before B" that will completely fix the order of the sequence, no matter what that order is?

The relations are required to be distinct, i.e. no two relations can deal with the same two elements of the sequence.

I don't have a solution either, so I'm opening it to general discussion.

[hide="I think this works"]
In order to completely fix the order of the set, I believe the minimum number would have to concern every element's relation to two other elements unlessthe element lies at the beginning or end, in which case only one relation is noted, so, if given the correct relations, the order of the set can be defined by $n-1$ relations. However, the problem says "random relations" so it may mean that for any two randomly selected elements, the relation is given. I believe the maximum to define the order is $n$ choose $2$, but the problem is somewhat unclear to me...[/hide]

If we know the relation of every number to each other except the biggest two: that gives us $1$ less than the maximum and you still can't determine the order, so you need all of them to fix it.

By the way, here's a clearer way of putting the problem:

Given an ordered sequence with N elements, whose order you know nothing about. How many different questions of the form "Does element A come before element B?" would you need to ask before you're [i]guaranteed[/i] to know the order (no matter what that order is)?

[quote="Laplace's_Demon"]Actually, you don't.

Say the order is A, B, C, D, E,...

Then the subset "A is before B," "B is before C," "C is before D," etc. of the set of all (n choose 2) possible relations is obviously sufficient to determine the order. There are only n - 1 relations in this set, and the rest are all unnecessary.

The problem is, of course, that if you're randomly given n - 1 different relations, they're likely not to be exactly those, as sapphyre pointed out...[/quote]

I thought the problem was talking about the worst-case scenario (i.e. the maximum number of of relations such that the order is not fixed). So you need all of them to guarantee you can order the sequence.

Sorry if my problem statement remains somewhat unclear.

It asks for the [i]minimum[/i] number of relations (given at random) that you would need before you could determine the order of any N-element sequence with certainty. So the "worst-case scenario" is where you have the largest possible set of [i]logically independent[/i] relations that still doesn't fix the order. If you're given (say) "A is before B" and "B is before C," then "A is before C" would [i]not[/i] be in this set because that can be inferred from the given relations.

As you see, for an ordered sequence A, B, C, D,... with N elements, you would have to be given at least N-1 relations (these being specifically "A is before B," "B is before C," etc.) before you can determine the order. But we assume you're given these relations randomly, so the "worst-case" set of relations--the one we're interested in--would have quite a bit more than N-1 relations. But it is not the set of all (N choose 2) relations, because those include precisely the N-1 we considered earlier that fix the order of the sequence on their own.

Therefore the worst-case set of relations, which

1) are logically independent of each other (none of them can be inferred from the others),
2) do not fix the order of the sequence, and
3) form the largest set that satisfies (1) and (2),

has strictly between N - 1 and (N choose 2) elements. If we add any relation to this set that cannot be inferred from those already in the set, then the order of the sequence must be fixed. The number of relations in this last set, in terms of N, is the answer to the question.


Consider two distinct positive integers $a$ and $b$ having integer arithmetic, geometric and harmonic means. Find the minimal value of $|a-b|$.

[i]Mircea Fianu[/i]

Clearly $a$ and $b$ must have the same parity so let $a = k+m$ and $b = k - m$ for integers $k > m > 0$.  We have $|a-b| = 2m$.  Note that $ab = k^2 - m^2 = n^2$ is a perfect square.  Thus, $(n,m,k)$ is a Pythagorean triple.

Now, $\left( \frac{1}{2a} + \frac{1}{2b} \right)^{-1} = \left( \frac{k}{k^2-m^2} \right)^{-1} = \frac{k^2-m^2}{k}$.  Thus, $k | m^2$ and $k | n^2$.  Starting with a primitive Pythagorean triple $(r,s,t)$ we want $t | r^2$ and $t | s^2$ so we must scale everything by $t$ since $(r,s,t) = 1$.  From this idea, we use $(20, 15, 25) = (n,m,k)$.

Finally, $a = 40, b = 10$ and $|a-b| = 30$.  We can check $\frac{a+b}{2} = 25$, $\sqrt{ab} = 20$, and $\frac{2ab}{a+b} = 16$!

Edit:  ThAzN1, your solution agrees with mine but you seem to have a slightly wrong formula for harmonic mean.  :)

$A = \frac{a+b}{2} \in \mathbb{N} \implies a \equiv b \bmod 2$

$G = \sqrt{ab} \in \mathbb{N} \implies ab = G^2$.  Suppose WLOG that $a > b$.  Then we can write $a = kG, b = \frac{G}{k}$ for some integer $k | G$, and hence $a = bk^2$.  Furthermore, $A = \frac{b(k^2+1)}{2}$ and $G = bk$.

$H = \frac{2ab}{a+b} \in \mathbb{N} \implies Hb(k^2 + 1) = 2b^2 k^2 \implies H = \frac{2bk^2}{k^2 + 1}$.  Since $k^2, k^2 + 1$ share no common divisors, it follows that $(k^2 + 1) | 2b$.

We know that $k \neq 1$.  Now, substitute $c = k^2 - 1$.  We are trying to minimize the product $|bc| = |a-b|$, and we can do this by rewriting in terms of $b$ and $c$:

$A = \frac{b(c+2)}{2}$
$G = b \sqrt{c+1}$
$H = \frac{2b(c+1)}{c+2}$

We divide into two cases.

Case 1:  $c$ is even.  Write $c = 2d$.  We have the condition $2d+2 | 2b \implies b = e(d+1)$.  Now, $c = 3, 8, 15, 24, ...$, so $d = 4, 12, ...$ will guarantee that $G \in \mathbb{N}$.  Because $c+2 = 2(d+1)$ is even, we have guaranteed that $A \in \mathbb{N}$, and because $c+2 | 2b$ we have guaranteed that $H \in \mathbb{N}$.  The minimum value here is therefore attained when $b = e(d+1) = 1(4+1) = 5$, and $bc = \boxed{40}$.

Case 2:  $c$ is odd.  We require that $c+2 | 2b$ - hence, $b = f(c+2)$.  In order that $A \in \mathbb{N}$ we require $f = 2g$, so $b = 2g(c+2)$ guarantees $A, G, H \in \mathbb{N}$.  The minimum value here is therefore attained when $b = 2g(c+2) = 2(3+2) = 10$ and $bc = \boxed{30}$.

OH oops yeah I'll edit it.. good thing that factor of 2 doesn't change anything  :)

A dumb solution: Let $ab=k^2$ and $a+b=2\ell$. Then $(a-b)^2 = (a+b)^2 - 4ab=4 \left( \ell^2 - k^2 \right)$, so we have to minimize $\left| \ell^2 - k^2 \right|$. Now, the dirty work: we check some cases, which will lead to the correct answer, I think.

[quote="t0rajir0u"]$G = \sqrt{ab} \in \mathbb{N} \implies ab = G^2$.  Suppose WLOG that $a > b$.  Then we can write $a = kG, b = \frac{G}{k}$ for some integer $k | G$,[/quote]

Are you sure ? Counter example : $225 = 9 \times 25$. But, $\frac{15}{9}$ is not integer.

Now, that this thread has been revived ... true! a rare occasion t0rajir0u has erred. Notice also that post #5 leads verbatim to the solution of post #2; the best shortcut, in my view. From $ab=k^2$ and $a+b=2\ell$, we need minimize $|a-b|$, or equivalently, minimize $(a-b)^2 = (a+b)^2 - 4ab = 4(\ell^2 - k^2)$. But $a,b$ are the roots of $\lambda^2 - 2\ell\lambda + k^2$, so $a,b = \ell \pm \sqrt{\ell^2 - k^2}$, whence we need $\ell^2 = k^2 + m^2$, a Pythagorean triple. This yields all instances where $\frac {a+b} {2}$ and $\sqrt{ab}$ are integer, with the minimum for $|a-b|$ equal to $6$, given for $5^2 = 4^2 + 3^2$, when $\{a,b\} = \{2,8\}$. The [b]third condition [/b]on $\frac {2ab} {a+b} = \frac {k^2} {\ell}$ to be integer now forces $\ell \mid k^2$ and $\ell \mid m^2$, thus the minimum for $|a-b|$ equal to $30$, given for $(5\cdot 5)^2 = (5\cdot 4)^2 + (5\cdot 3)^2$, when $\{a,b\} = \{10,40\}$.

Set $a=da_1$ and $b=db_1$ with $(a_1,b_1)=1$. Since $\sqrt{ab}\in\mathbb{N}$, we obtain $a_1,b_1$ are perfect squares, set $a_1=u^2$ and $b_1=v^2$. With this, $(a+b)/2\in\mathbb{N}$ implies $2\mid d(u^2+v^2)$. Furthermore, the harmonic mean of $a$ and $b$ arranges to
\[
\frac{2}{\frac{1}{du^2}+\frac{1}{dv^2}} = \frac{2du^2v^2}{u^2+v^2}.
\]
As $(u,v)=1$, we have $(u^2v^2,u^2+v^2)=1$, yielding $u^2+v^2\mid 2d$. Now if $u^2+v^2$ is odd, then $d=2s(u^2+v^2)$ for some arbitrary $s\in\mathbb{N}$, yielding $|a-b| = 2s|u^4-v^4|$, whose smallest value (under $u,v$ being odd) is clearly obtained for $\{u,v\}=\{1,2\}$ and $s=1$, that is $d=10$ and $(a,b)=(10,40)$. The second case, namely $u^2+v^2$ being even, is handled analogously; hence the smallest value is $|a-b|=|10-40|=30$.


I posted this once but forgot and can't find the old topic.

How do I do floor and ceiling function in $ \text{\LaTeX}$?

The topic is right [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=278238405&t=298226]here[/url]


In TeX writeup:

[tex]\large \frac{\sqrt[k]{a_{1}a_{2}...a_{k}}+\sqrt[k]{a_{n-k+1}a_{n-k+2}...a_{n}}}{\binom{n}{k}}\leq \sqrt[k]{\frac{a_{1}a_{2}...a_{k}+a_{n-k+1}a_{n-k+2}...a_{n}}{\binom{n}{k}}}[/tex].

In order to have the inequality valid you need the extra condition 0 < k < n.

Then, the inequality is trivial: Call $\sqrt[k]{a_{1}a_{2}...a_{k}}=a$ and $\sqrt[k]{a_{n-k+1}a_{n-k+2}...a_{n}}=b$; then we have to prove

[tex]\large \frac{a+b}{\binom{n}{k}}\leq \sqrt[k]{\frac{a^{k}+b^{k}}{\binom{n}{k}}}[/tex].

But since 0 < k < n, we have $\binom{n}{k}\leq 2$, so that

[tex]\large \frac{2}{\binom{n}{k}}\leq 1[/tex]

and

[tex]\large \frac{2}{\binom{n}{k}}\leq \sqrt[k]{\frac{2}{\binom{n}{k}}}[/tex],

and finally

[tex]\large \frac{a+b}{\binom{n}{k}}=\frac{a+b}{2}\cdot \frac{2}{\binom{n}{k}}\leq \frac{a+b}{2}\cdot \sqrt[k]{\frac{2}{\binom{n}{k}}}\leq \sqrt[k]{\frac{a^{k}+b^{k}}{2}}\cdot \sqrt[k]{\frac{2}{\binom{n}{k}}}=\sqrt[k]{\frac{a^{k}+b^{k}}{\binom{n}{k}}}[/tex],

where we have used the Power Mean inequality. Proof complete.

But I guess you meant a different inequality:

[tex]\large \frac{\sqrt[k]{a_{1}a_{2}...a_{k}}+...+\sqrt[k]{a_{n-k+1}a_{n-k+2}...a_{n}}}{\binom{n}{k}}\leq \sqrt[k]{\frac{a_{1}a_{2}...a_{k}+...+a_{n-k+1}a_{n-k+2}...a_{n}}{\binom{n}{k}}}[/tex]

(summation over all possible products of k of the n numbers). This one is, however, even more trivial using power means :-)

  Darij


This is another one of those problems that is easy, but I forgot how to do it...

When Kelly is not on the moving sidewalk, she can walk the length of the sidewalk in 3 minutes. If she stands on the sidewalk as it moves, she can travel the length in 2 minutes. If Kelly walks on the sidewalk as it moves, how many minutes will it take her to travel the same distance? Assume she always walks at the same speed.

 :oops:

Kelly's rate $ \equal{} \ r_1$ and sidewalk rate $ \equal{} \ r_2$
Distance of sidewalk $ \equal{} \ d$. Recall $ dr \equal{} t$ where $ d$ is distance, $ r$ is rate, and $ t$ is time.
You're given $ 3 r_1 \equal{} d$ and $ 2r_2 \equal{} d$. You want to find some number $ t$ that makes $ t(r_1 \plus{} r_2) \equal{} d$ true.

Multiply the first equation by $ 2$ and the second by $ 3$ to get: $ 6r_1 \equal{} 2d$ and $ 6r_2 \equal{} 3d$ and then add them to get
$ 6r_1 \plus{} 6r_2 \equal{} 5d \Rightarrow 6(r_1 \plus{} r_2) \equal{} 5d \Rightarrow \frac {6}{5}(r_1 \plus{} r_2) \equal{} d \Rightarrow \boxed{1 \ minute \ 12 \ seconds}$

Thanks... But I solved a bit differently...

 :thumbup:  :blush:  :play_ball:

this problem is from mathcounts, and should not be posted in the high school basics forum. in the future, post mathcounts problems in the [b]mathcounts forum[/b].


Derek just went out and bought a 4 GB (gigabyte) iPod Nano for 250 dollars while Eustace bought a 512 MB (half of a gigabyte) iPod Shuffle for a certain amount of money. Given that Eustace's 512 MB iPod Shuffle cost 2.24 times as much per gigabyte than Derek's iPod Nano, how much did Eustace's iPod Shuffle cost?

[hide="Answers"] [b]70[/b] dollars. By the way, these prices are the exact same current retail prices, so if you've been to your local electronics store and seen the "Retail Price: 69 dollars" sign on a 512 GB iPod shuffle, this problem should be pretty easy :p .
[/hide]

On a side note, Iron Fist has a 4 GB iPod Mini. It's just in the middle - too small to have a lot of memory and too big to fit in an Orbit gum pack. >_>

[hide]$70$[/hide]
Is there anyone else here besides me who is music deprived?
I have no MP3 Player and crappy internet and only like 3 albums on an itunes that freezes up my computer when played too long.

I have a nano. But i just got it and there isn't much on it. I'm bringing it to nats anyway.

[quote="ehehheehee"]I have nothing  :P[/quote]

Me too.
Then again, I have an alarmclock/radio.

[quote="mbabbitt2003"]I have TWO ipods.[/quote]

That's great, if you give me half of an iPod we'll both have the same amount.

If YOU give me a half of an ipod, I will have 5 times as many as you.

Haha.

If mbabbit gives treething 1/2 of a jPod, they will have the same amount.
If treething gives mbabbit 1/2 of a jPod, mbabbit will have 5 times as many jPods as treething.
How many jPods does...[hide]Klebian[/hide] have?

[hide]Klebian[/hide] has 0 [b]j[/b]pods.

woah, you're good.
and yes, i did jpod on purpose, in case you were "subtly" implying something.

No, I was just pointing out the j to people who didn't notice.

um i got $1400.00 anyone have any ideas as to how that happened because im severly confused
 :huh:

[quote="MonkeyBoy"]um i got $\$1400.00$ anyone have any ideas as to how that happened because im severly confused
 :huh:[/quote]
First, I thought you were talking about real life, as in "I just got a whole bunch of money and I don't know how I got it." :rotfl: And sorry, I don't know how you got 1400. It should be $\frac{.5}{4}*2.24*250=\boxed{70}$.


[b]Let[/b] $ A,B,C \in S_{n}(R)$ [b]such that [/b]: $ AB\equal{}BA,BC\equal{}CB,CA\equal{}AC$
[b]Slove the system of matrix equations :[/b]
           $ \begin{cases} e^{A}\minus{}e^{A\minus{}B}\equal{}B \\ e^{B}\minus{}e^{B\minus{}C}\equal{}C \\ e^{C}\minus{}e^{C\minus{}A}\equal{}A \end{cases}$
[b]Here[/b] $ S_{n}(R)$ [b]are set real symmetric nxn matrices .[/b]
------------------------------------------------
DVG : OOO  :wink:

One more time: Note that the matrices involved are simultaneously diagonalizable.

ps: Is that homework of a linear algebra class? I must admit that it smells like ...  :P


If $p$ and $q$ are the roots of the equation $(x-a)(x-b)-2x+1=0$, then what are the roots of the equation $(x-p)(x-q)+2x-1=0$?

(a)$a$, $b$
(b)$-a$, $-b$
(c)$\displaystyle\frac{1}{a}$, $\frac{1}{b}$
(d)$-\frac{1}{a}$, $-\frac{1}{b}$
(e) None of the above

$p+q=b+a+2$

and for the second equation, the sum of the roots is  $q+p-2=b+a$, so the roots are $a,b$

That seems reasonable.  Could you please explain to me how you got $p+q=b+a+2$ though?


Let $ a, b, c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove o disprove that

$ \sum\frac{4ab \plus{} 3}{c \plus{} 1}\le 8$ .

__________
 :roll:

[quote="Ligouras"]Let $ a, b, c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove o disprove that

$ \sum\frac {4ab \plus{} 3}{c \plus{} 1}\le 8$ .

__________
 :roll:[/quote]

My solution. :)

ineq $ \Leftrightarrow\sum(4ab\plus{}3)(b\plus{}1)(a\plus{}1)\leq 8\prod(a\plus{}1)\Leftrightarrow 4\sum a^2b^2\plus{}3\sum ab\leq 20abc\plus{}1\Leftrightarrow 4q^2\plus{}3q\leq 28r\plus{}1$
Case $ 0\leq q\leq\frac{1}{4}\Rightarrow LHS\leq 1;RHS\geq 1$
Case $ \frac{1}{4}\leq q\leq\frac{1}{3}:$
$ 4q^2\plus{}3q\minus{}28r\minus{}1\leq 4q^2\minus{}\frac{85}{9}q\plus{}\frac{19}{9}\equal{}(q\minus{}\frac{1}{4})(4q\minus{}\frac{76}{9})\leq 0$
We are done :)

After expand, it's becomes:
$ 4\sum\ a^2b^2 \plus{} 4\sum\ ab(a\plus{}b) \leq\ (a\plus{}b\plus{}c)^3\plus{}4\sum\ ab\plus{}8abc$
Obviously because:
$ (a\plus{}b\plus{}c)^3\plus{}8abc \geq\ 4\sum\ ab(a\plus{}b)$   (Schur)
And $ 4 \sum\ a^2b^2 \leq\ \sum\ ab(a\plus{}b)^2 \leq\ \sum\ ab(a\plus{}b\plus{}c)^2 \equal{} ab\plus{}bc\plus{}ca$
We have done.  :)


Prove that in any triangle $ABC$ the following inequality holds:

                $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq\frac{3}{2}\cdot\frac{s}{r(4R+r)}$

                                                                                standard notations.

We have : $ ab+ab+bc = s^2+4Rr+r^2 ; abc =4sRr $
 Change the problem into the expression of $ s,R,r $ and use the famous ineq : $ s^2 \geq 16Rr -5r^2 $  ;)


[quote="KostasGT"]Γειά σας και από μένα. Συγχαρητήρια σε όλα τα παιδιά που ασχολούνται παθιασμένα με τα μαθηματικά, πιστεύω ότι πιθανότατα κάποιοι από εσάς στο μέλλον θα είστε ερευνητές πρώτης γραμμής. Θα προσπαθώ όποτε βρίσκω το χρόνο (ελέω φροντιστηρίων) να ποστάρω κάποιες ασκήσεις. Ξεκινώ με μια ανισότητα και ελπίζω να μην την σνομπάρουν οι έμπειροι λύτες ανισοτήτων μιας και είναι εφαρμογή γνωστού λήμματος.

Αν a, b, c πραγματικοί και ισχύει abc=1 τότε να αποδείξετε ότι:

 $ \frac {1}{a^{6}} \plus{} \frac {1}{b^{6}} \plus{} \frac {1}{c^{6}} \geq a^{3} \plus{} b^{3} \plus{} c^{3}$

Επίσης, αν a, b, c είναι και θετικοί τότε να βρείτε την μικρότερη τιμή του πρώτου μέλους.[/quote]

Kanoume omonuma ta klasmata kai pollaplasiazoume me 2 kai pairnoume oti $ \sum_{sym}a^{6}b^{6}\geq \sum_{sym}a^{3}$. Pollaplasizaoume kathe oro tou deuterou melous me $ a^{3}b^{3}c^{3}$ kai paratiroume oti isxuei apo Muirhead me triades tis $ (6,6,0)\succ (6,3,3)$

you killed it! ego tha ekana to eksis $ LHS\equal{}\sum{\frac{b^3c^3}{a^3}}$ kai meta $ {\frac{b^3c^3}{a^3}}\plus{}{\frac{a^3c^3}{b^3}}\geq 2c^3$ prostheoume kai telos. to extremum pou zitas einai apli efarmogi am-gm.

Μια άλλη λύση: Από αναδιάταξη, $ \frac{1}{a^6} \plus{}  \frac{1}{b^6} \plus{} \frac{1}{c^6}  \geqslant \frac{1}{a^3b^3} \plus{} \frac{1}{b^3c^3} \plus{} \frac{1}{c^3a^3} \equal{} c^3 \plus{} a^3 \plus{} b^3$

Πολες λυσεις, οριστε αλλη μια:

$ LHS \equal{} \sum{a^6b^6} \equal{} \sum{(b^3a^3)^2} \geq \sum{a^3b^3b^3c^3} \equal{} a^3b^3c^3\sum{a^3} \equal{} \sum{a^3} \equal{} RHS$

[quote="Γιώργος"][quote="Nick Rapanos"]you killed it! ego tha ekana to eksis $ LHS \equal{} \sum{\frac {b^3c^3}{a^3}}$ kai meta $ {\frac {b^3c^3}{a^3}} \plus{} {\frac {a^3c^3}{b^3}}\geq 2c^3$ prostheoume kai telos. to extremum pou zitas einai apli efarmogi am-gm.[/quote]

telika i lusi einai sosti i oxi? mepiase anxos tora :P[/quote]

xaxax i didnt go through the calculations but i trust you! :P

Θαυμάσιες σκέψεις όλες. Η άσκηση κατασκευάστηκε με το σκεπτικό του NickNafplio, δηλαδή βασίζεται στο:

$ a^{2} \plus{} b^{2} \plus{} c^{2} \geq ab \plus{} bc \plus{} ca$

Καλό βράδυ σε όλους.