Consider a deck of 12 cards which includes the king, queen, and jack from each suit.
A hand of 3 cards is randomly selected.
Find the prabability that at least [b]2[/b] cards are of the same rank.
Solution 1:
1 - P(all different ranks)= 1- 64/220 = 156/220
Solution 2:
P(choose 2 or 3 same ranks)= [3*C(4,2)*C(10,1)]/C(12,3)=9/11
What is wrong with the solutions?

[quote="linj"]Consider a deck of 12 cards which includes the king, queen, and jack from each suit.
A hand of 3 cards is randomly selected.
Find the prabability [sic] that at least 3 cards are of the same rank.[/quote]
[hide="Think algorithmically."]
I can choose the rank in $ \binom{3}{1}$ ways. I can choose the 3 cards of the chosen rank in $ \binom{4}{3}$ ways. I can choose 3 arbitrary cards in $ \binom{12}{3}$ ways. Therefore the probability is:

$ \frac {\binom{3}{1}\binom{4}{3}}{\binom{12}{3}} \equal{} \boxed{\frac {3}{55}}$
[/hide]

[quote="linj"]
Solution 1:
1 - P(all different ranks)= 1- 64/220 = 156/220[/quote]
You are including those hands that have two kings and a jack, etc..

[quote="linj"]
Solution 2:
P(choose 2 or 3 same ranks)= [3*C(4,2)*C(10,1)]/C(12,3)=9/11
[/quote]
I don't understand why you are interested in finding two cards of the same rank. The problem asks for the probability of all three cards having the same rank.

=====

After reading your solutions, you may have intended that only two cards of the chosen three need be of the same rank. In this case, your first solution is correct. Your second overcounts the solutions in which all three cards are the same rank; you would need to divide the counts into either two of the same rank or three of the same rank:

[hide="Solution."]
[b]Case 1:[/b] Two cards of same rank.

$ \binom{3}{1}\binom{4}{2}\binom{8}{1} \equal{} 144$

[b]Case 2:[/b] Three cards of same rank.

$ \binom{3}{1}\binom{4}{3}\equal{}12$

Thus there are $ 144\plus{}12\equal{}156$ possible hands, and the probability becomes:

$ \frac{156}{\binom{12}{3}}\equal{}\frac{156}{220}\equal{}\boxed{\frac{39}{55}}$
[/hide]

[quote="jvenezuela716"][quote="linj"]Consider a deck of 12 cards which includes the king, queen, and jack from each suit.
A hand of 3 cards is randomly selected.
Find the prabability [sic] that at least 3 cards are of the same rank.[/quote]
[hide="Think algorithmically."]
I can choose the rank in $ \binom{3}{1}$ ways. I can choose the 3 cards of the chosen rank in $ \binom{4}{3}$ ways. I can choose 3 arbitrary cards in $ \binom{12}{3}$ ways. Therefore the probability is:

$ \frac {\binom{3}{1}\binom{4}{3}}{\binom{12}{3}} \equal{} \boxed{\frac {3}{55}}$
[/hide]

[quote="linj"]
Solution 1:
1 - P(all different ranks)= 1- 64/220 = 156/220[/quote]
You are including those hands that have two kings and a jack, etc..

[quote="linj"]
Solution 2:
P(choose 2 or 3 same ranks)= [3*C(4,2)*C(10,1)]/C(12,3)=9/11
[/quote]
I don't understand why you are interested in finding two cards of the same rank. The problem asks for the probability of all three cards having the same rank.

=====

After reading your solutions, you may have intended that only two cards of the chosen three need be of the same rank. In this case, your first solution is correct. Your second overcounts the solutions in which all three cards are the same rank; you would need to divide the counts into either two of the same rank or three of the same rank:

[hide="Solution."]
[b]Case 1:[/b] Two cards of same rank.

$ \binom{3}{1}\binom{4}{2}\binom{8}{1} \equal{} 144$

[b]Case 2:[/b] Three cards of same rank.

$ \binom{3}{1}\binom{4}{3} \equal{} 12$

Thus there are $ 144 \plus{} 12 \equal{} 156$ possible hands, and the probability becomes:

$ \frac {156}{\binom{12}{3}} \equal{} \frac {156}{220} \equal{} \boxed{\frac {39}{55}}$
[/hide][/quote]
I am so sorry!
I type the question wrong.
It should be 2 not 3.

[quote="linj"]
I am so sorry!
I type the question wrong.
It should be 2 not 3.[/quote]
[quote="jvenezuela716"]
After reading your solutions, you may have intended that only two cards of the chosen three need be of the same rank. In this case, your first solution is correct. Your second overcounts the solutions in which all three cards are the same rank; you would need to divide the counts into either two of the same rank or three of the same rank:

[hide="Solution."]
[b]Case 1:[/b] Two cards of same rank.

$ \binom{3}{1}\binom{4}{2}\binom{8}{1} \equal{} 144$

[b]Case 2:[/b] Three cards of same rank.

$ \binom{3}{1}\binom{4}{3} \equal{} 12$

Thus there are $ 144 \plus{} 12 \equal{} 156$ possible hands, and the probability becomes:

$ \frac {156}{\binom{12}{3}} \equal{} \frac {156}{220} \equal{} \boxed{\frac {39}{55}}$
[/hide][/quote]


function
real analysis
real analysis unsolved

Let f:(0, \infty ) such that f(x)>0 for any x. Suppose f is bijective , increasing and derivable. Also, suppose that f'(x)/x^2 (f' is the derivative)tends to infinity when x tends to infinity. Prove that any positive number is a limit point of the sequence x_n=f^-1(n)*{f^(-1)(n)}, where f^(-1) is the inverse of f and {} is the fractional part.

[quote="harazi"]Let f:(0, \infty ) ... .[/quote]
Hehe sadly to say this function is quite sad for it is weirdly defined with a sad face. :( and : ( Hehe good one harazi : D oops  :D


The fourth number in an arithmetic sequence is 7 and the eleventh term is 49.

(a) What is the common difference in the sequence?
(b) What is the first term in the sequence?
(c) What is the formula for the nth term in the sequence?
(d) What is the sum of the first 100 terms in the sequence?

Answer as many as you are able.

[quote]The fourth number in an arithmetic sequence is 7 and the eleventh term is 49.

(a) What is the common difference in the sequence?
(b) What is the first term in the sequence?
(c) What is the formula for the nth term in the sequence?
(d) What is the sum of the first 100 terms in the sequence?
[/quote]

a) 6
b) -11
c) 6n-17
d) 583

now on to my solutions (im soooo tired ackkk).

a) 7 numbers from 4-11, the difference is 42, so 42/7=6. which is the difference d
b)basically subtract 18 from 7 cause there are 3 numbers before the 4th term. so 7-3x6=-11
c)common difference is 6, so using the first term. 6(1)-y=-11
solving the equation y=17.
so to find the nth term its 6n-17
d)*yawn* so tired. oh well, on i go. the sum of an arithmetic sequence is the first term [n1+(k-1)d]. where k is the number of terms and d is the common difference.
so [-11+(100-1)6]=[-11+99*6]=-11+594=583

hope i helped
hope i was right
any questions about it, just ask

Sincerely,
High Inquisitor Syntax Error


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14278[/img]

 :(

in these kinds of problems,one of the  ideas is to suppose the problem as solved,and according to this,we would try to find out some simple relations beween angles,ok let's do that:
if $ A'B'C'D'E'$ is cyclic,then $ \angle A'B'D'\equal{} 180^{\circ} \minus{} \angle E'\equal{}$constant
and $ \angle B'A'D'\equal{} 180^{\circ} \minus{} \angle C'\equal{}$constant
it means that points $ B,D$ should lie on the given circle$ (O)$ with  property that $ \triangle ABD$ is simiar to a given $ \triangle
A'B'D'$.to find out where should we put the points,we use a simple homotechy,
draw a $ \triangle A''B''D''$,which is similar to $ \triangle A'B'D'$,then draw its circum circle$ (O'')$,points $ C'',E''$ will lie on $ (O'')$ in which $ \angle A''E''D''\equal{}\angle A'E'D'$ and $ \angle B''C''D''\equal{}\angle B'C'D'$,then we could draw a pentagon which is cyclic ,and the circumscribed circle is $ (O'')$,with radius $ r$.
but the problem asked us to draw the pentagon in a given circle$ (O)$,with radius for example $ R$,
the thing which we do is, to find the length according to the radius of the homotechy $ \frac{R}{r}$,and every thing is ok :wink:


trigonometry
function
algebra
domain
conics
parabola
absolute value

:rotfl:
Ok, the problem is at the point when you get $\sqrt{3x+1}=x-3$ and you square this relation, you actually can obtain a solution more, because $1-3=-2$ which is negative, but in absolute value is same as $\sqrt{3\cdot 1 + 1}$.
Anyway, for college playground and Open Questions, this is  :spam:
But since you are new on this forum,  :welcome:
I just like using emoticons  :yup:

can you pls tell me why square root cannot be negative

you can have -8 * -8 and you get 64 ... so why can't it be negative

[quote]can you pls tell me why square root cannot be negative [/quote]
Convention. The symbol $\sqrt{4},$ by agreed-upon convention, means $2$ and does not mean $-2.$ The symbol $\sin^{-1}\left(\frac12\right)$ means $\frac{\pi}6$ and does not mean $\frac{5\pi}6.$ It's nothing deeper than that - taking a function that isn't one-to-one, cutting down its domain to a portion on which it is one-to-one, and defining an agreed-upon inverse on that set.

And as moderator, I'll find somewhere else to move this topic.

[quote="unhappy"]i have this equation:

3 + SQR(3x+1) = x

i solved it like this:

SQR(3x+1) = x-3 /^2

3x+1 = x^2 - 6x +9

x^2-9x+8 =0

Discriminant = 7

2 solutions {8, 1}

My question is. Why is 1 not a valid solution. I tested it and I know it doesn't work but why. Even my book says that 8 is theonly solution. Can anyone explain to me what is the problem.[/quote]

it is not a part of solution because u can check by plugging in 1 to the original equation, that $3+\sqrt{4} =5  \not = 1$

[quote="xxreddevilzxx"]it is not a part of solution because u can check by plugging in 1 to the original equation, that $3+\sqrt{4} =5  \not = 1$[/quote]
He was asking why it doesn't doesn't work when it is plugged in.  ;)

is there a way to solve radical equation by which we won't get the extra solutions ... or do you have to simply check

*massages temple*
*takes a deep breath*

Okay.

When you square both sides of an equation, funny things happen. It's like magic! You get a solution that doesn't work. It's because the square root of a negative expression turns positive.

There is no algebraic way to find the solutions without finding the false solutions, but checking is a very easy process.  If you graph the respective functions, I think it helps give a clearer idea where the false solution comes from: graph the parabola $y^2 = 3x + 1$ and the line $y = x-3$ and you will see two intersections, one on the half of the parabola given by $y = \sqrt{3x + 1}$ and one on the half given by $y = -\sqrt{3x + 1}$.  Such a problem may have two solutions, one solution (in three different ways, where the difference is somewhat clear depending on when the "extra" solution disappears, but can be made even more clear on a graph) or zero solutions (in a number of different ways).


1. Consider the following AC circuit.

(a) Find the total impedance of this circuit.

(b) Compute the power dissipated on the circuit.

I am stuck up here. Please check whether this is correct.
[hide]Let $ i_{L}$ be current through $ L_{2}$. Then $ i_{L}$ satisfies:
$ (L_{eq}\plus{}L_{2})i_{L}\plus{}\frac{ L_{eq}L_{2}}{R}\frac{di_{L}}{dt}\equal{}\frac{V_{0}\cos\omega t}{\omega}$; $ L_{eq}\equal{}\frac{L_{1}L_{3}}{L_{1}\plus{}L_{3}}$[/hide]

Explain how you arrive there. Anyway, there is no need for a differential equation.

First, I replace $ L_{1}$ and $ L_{3}$ by $ L_{eq}$.
As $ R$ and $ L_{2}$ are in parellel, we have that $ i_{R}R\equal{}L_{2}\frac{di_{L}}{dt}... (1)$
Next, by loop law, we have $ i_{R}R\plus{}L_{1}\frac{di}{dt}\equal{}V_{0}\sin\omega t$, where $ i$ is the current in the circuit.
Integrating and putting initial condition, $ L_{1}i\plus{}R\int_{0}^{t}i_{R}dt\equal{}\frac{V_{0}\cos\omega t}{\omega}$
Using $ (1)$ and initial condition, we get my result.

That seems ok. So, what are your answers?

2. Consider the following AC circuit.

(a) Find the total impedance of the circuit.

(b) What is the phase difference between the source voltage and the circuit current?

3. The following RLC circuit is composed by two resistors ($ R_{1}\equal{} R_{2}\equal{} 100\Omega$), two capacitors ($ C_{1}\equal{} C_{2}\equal{} 10\mu F$), two inductors ($ L_{1}\equal{} L_{2}\equal{} 1H$), and one AC source with frequency 50 Hz and unknown effective voltage.

(a) Find the total impedance of the circuit.

(b) Knowing that the effective current through the circuit is 0.2 A, find the effective voltage of the source and its phase shift relative to the current.

4. Consider the following AC circuit.

(a) Find the total impedance of the circuit.

(b) Calculate the average power supplied by the source to the circuit.

5. An AC circuit is composed by a resistor, an inductor, a capacitor, and a source with angular frequency 2000 rad/s and amplitude 24 V.

(a) Find the total impedance of the circuit.

(b) What is the value of the effective current through the circuit?

[hide="Answer 4"]
(a) $ Z\equal{}(\frac{1}{R}\plus{}\frac{1}{|(\omega_{d}L\minus{}\frac{1}{\omega_{d}C})|})^{\minus{}1}\equal{}1k\Omega$
(b) $ P_{avg}\equal{}\frac{\frac{1}{2}V_{0}^{2}\cos\phi}{Z}$[/hide]

6. In the AC circuit showed bellow the source have frequency $ f \equal{}\frac{2000}{\pi}\, Hz$ and effective voltage 0.2 V. Find the effective current through capacitor $ C_{1}$. What is the effective value of the voltage on its terminals?

7. Consider the following AC circuit.

(a) Find the total impedance of the circuit.

(b) Compute the power dissipated in each element of the circuit.

Shreyas_patankar, your answer to 4.(a) don't seem to be right.

I took $ f$ to be $ \omega$ :oops:
$ Z\equal{}400\Omega$
I hope only the numerical answer is wrong.

That still seems to be wrong. Why don't you give the complex impedance to see where is (your or my) error?

Let $ I$ be current in the circuit, $ I_{R}$ be current through resistor, $ I_{S}$ through L&C and $ i^{2}\equal{}\minus{}1$
$ V\equal{}I_{R}R\equal{}i\omega LI_{S}\minus{}\frac{iI_{S}}{\omega C}$
$ I\equal{}I_{R}\plus{}I_{S}$ and $ Z\equal{}\frac{V}{I}$
Hence, $ \frac{V}{Z}\equal{}\frac{V}{R}\plus{}\frac{V}{i\omega L\minus{}\frac{i}{\omega C}}$
Or, $ Z\equal{}(\frac{1}{R}\minus{}\frac{i}{\omega L\minus{}\frac{1}{\omega C}})^{\minus{}1}$

There is no need to complicate things. In circuit analysis it is common practice to use $ j \equal{}\sqrt{\minus{}1}$, because the $ i$ letter is used for the current intensity. The inductor and the capacitor are in series, and so its equivalent impedance is

$ Z_{1}\equal{} j\omega L\minus{}\frac{j}{\omega C}\equal{} j\left(\omega L\minus{}\frac{1}{\omega C}\right)$.

Substituting by the given values ($ \omega \equal{} 2000\pi$ rad/s) we get $ Z_{1}\equal{} 398j\,\,\Omega$.

Now this impedance is in parallel with the resistor, and so the total impedance $ Z_{eq}$ will be

$ Z_{eq}\equal{}\frac{Z_{1}\cdot Z_{R}}{Z_{1}\plus{}Z_{R}}\equal{}\frac{400\cdot 398j}{400\plus{}398j}\equal{} (199\plus{}200j)\,\,\Omega$.

How about part b)?

We know that $ \left\langle P\right\rangle \equal{} V_{rms}I_{rms}\cos\phi$, where phase $ \phi\equal{}\tan^{\minus{}1}\frac{X_{L}\minus{}X_{C}}{R}$
Hence, $ \left\langle P\right\rangle \equal{}\frac{V_{0}^{2}\cos\phi}{2Z}\equal{}1.1\times10^{\minus{}1}W$

No, that's not the correct value. I don't understand your phase formula.

In the mean time, here's a nicer problem.

8. Find current I in the following circuit:

$ X_{L}$,$ X_{C}$ are inductive and capacitative reactances respectively.
$ X_{L}\equal{}\omega L$ and $ X_{C}\equal{}\frac{1}{\omega C}$

Ok, then correctly $ \phi \equal{} 45^{o}$. Your power formula is correct but your numerical answer isn't.

Could you post solution for (1) ?
This is quite an extensive (indeed, almost exhaustive) collection for AC circuits.
Could you describe the source?

[hide="Solution for 1"](a) Inductors L1 and L3 are in parallel, and because they have the same inductance then (also $ \omega \equal{} 1000\,\,rad/s$)

$ Z_{13}\equal{}\frac{Z_{1}}{2}\equal{}\frac{j\cdot 1000\cdot 0.6}{2}\equal{} (300j)\,\,\Omega$.

Inductor L2 and the resistor are also in parallel, and so

$ Z_{2R}\equal{}\frac{Z_{2}Z_{R}}{Z_{2}\plus{}Z_{R}}\equal{}\frac{j\cdot 1000\cdot 0.5\times 1500}{1500\plus{}500j}\equal{} (150\plus{}450j)\,\,\Omega$.

Hence, $ Z_{eq}\equal{} 300j\plus{}(150\plus{}450j) \equal{} (150\plus{}750j)\,\,\Omega$.

(b) The only circuit element that dissipates energy is the resistor, for which $ P_{diss}\equal{}\frac{1}{2}RI_{m}^{2}$, where $ I_{m}$ is the current amplitude through the resistor. Let's start by finding the total current I through the circuit:

$ I \equal{}\frac{V}{Z_{eq}}\equal{}\frac{10cis(0^{o})}{765cis(79^{o})}\equal{} 0.013cis(\minus{}79^{o})$.

Now let's apply the current division principle to R (and L2):

$ I_{R}\equal{}\frac{Z_{L_{2}}}{Z_{R}\plus{}Z_{L_{2}}}I \equal{}\frac{500j}{1500\plus{}500j}\times 0.013cis(\minus{}79^{o}) \equal{} 0.32cis(72^{o})\times 0.013cis(\minus{}79^{o}) \equal{} 4.2\times 10^{\minus{}3}cis(\minus{}7^{o})$.

So $ I_{m}\equal{} 4.2\times 10^{\minus{}3}$ and $ P_{diss}\equal{}\frac{1}{2}\cdot 1500\cdot (4.2\times 10^{\minus{}3})^{2}\equal{} 0.013\,\,W$.
[/hide]

[quote="Shreyas_patankar"]Could you describe the source?[/quote]

The source is indicated on top of each post.


trigonometry
geometry
area of a triangle
Heron\u0027s formula

Given a quadrilateral ABCD with sides AB = 4 cm, BC = 8 cm, CD = 6 cm , DA = 7 cm and BD = 10 cm.
Find the length of AC. ( Answer should be expressed in surd form.)

Why don't you try it yourself?
shobber gave you a hint in pre-olympiad forum already.

Please don't post the same question in more than one forum.  Thanks.

[quote="pkothari13"]try ptolemy's theorem[/quote]
The quadrilateral is not cyclic.

indeed, my hint was nearly useless... :P
but at least, we can do some trig calculations. The law of Cosine and Sum formulas can solve this problem.

$\angle BCD=\frac{\pi}{2}$,remarking $|ABC|+|ACD|=|ABD|+|BCD|$.Using Heron's formula......

[quote="luimichael"]I know the solution of this question because I set this question by myself.
If someone has answered this question already, I won't post it twice.
The hint given in pre-olympic is not crucial to the solution of this question.
Indeed, even there isn't a right angle in the quadrilateral, we can still get the solution in surd form.
I just want to see if someone can answer this elementary question with simple tools.[/quote]

Applying Heron's formula for several times and solve a monstrous equation, then u finish it.  :P

Let $x$ the length of side AC, $\angle BCD=\frac{\pi}{2}$,remarking $|ABC|+|ACD|=|ABD|+|BCD|$,using Heron's formula, we have

$|ABC|+|ACD|=|ABD|+|BCD|$

$\Longleftrightarrow \frac{\sqrt{(144-x^2)(x^2-16)}}{4}+\frac{\sqrt{(169-x^2)(x^2-1)}}{4}=30+\frac{7\sqrt{39}}{4}$

I think the area of the whole quadrilateral should be $24+\frac 74 \sqrt{39}$.
Using Heron's formula to attack the problem is a natural trial but the giant equation is really difficult to solve.
Here is the length of AC ( in cm) :
[hide]$\frac25\sqrt{21\sqrt{39}+232}$[/hide]
In fact, there is an elegant formula to express the area of a quadrilateral in terms of the four sides and the two diagonals:
$Area=\frac14\sqrt{4p^2q^2-(a^2-b^2+c^2-d^2)^2}$
Here p and q are the lengths of the diagonals and a, b, c , d are the lengths of the sides placing around the quadrilateral in either clockwise sense or anti-clockwise sense.
Note that this formula can be regarded as a generalization of the Heron's formula.


function
topology
limit
Functional Analysis
real analysis
real analysis unsolved

1 .$ 1 < p < \infty$, and $ f_i$ is a bounded sequence in  $ L^p$. It is well known that there is a sub-sequence that converge weakly to $ f_0$ in $ L^p$.

My question is: Is $ \|f_0\|_p$ is bounded too?

2. If $ f_i$ is a bounded sequence in  $ L^{\infty}(R)$.
Prove: there is a function  $ f_0 \in L^{\infty}(R)$ and there is a subsequence $ f_k$ such that $ f_k \minus{}> f_0$ weakly in $ L_{loc}^2 (R)$.

1) What does it mean for a number to be bounded? The norm $ \|f_0\|_p$ is finite - that's a part of the statement you quoted. More precisely, $ \|f_0\|_p\le \liminf_{i\to\infty}\|f_i\|_p$. Indeed, bounded linear functionals are continuous in the weak topology, which means $ \varphi(f_0)\equal{}\lim_{i\to\infty}\varphi(f_i)$ for all $ \varphi\in (L^p)^*$. Take the supremum over all $ \varphi$ with unit norm. (This proof works in any Banach space.)

2) "converges weakly in $ L_{\textrm{loc}}^2(\mathbb R)$" means that the sequence converges weakly in $ L^2[\minus{}n,n]$ for any $ n$. Since $ f_i$ is a bounded sequence in $ L^2[\minus{}n,n]$, it has a convergent subsequence. Use diagonal selection to choose one subsequence that works for all $ n$.

my first question should be:
1 .$ 1 < p < \infty$, and if $ \sup_i  \|f_i\|_p<\equal{} M$. It is well known that there is a sub-sequence that converge weakly to $ f_0$ in $ L^p$.

My question: is it true $ \|f_0\|_p<\equal{}M$?

I checked banach space book, there is a thm:

A banach space is relexive iff the unit ball is weakly compact.

So the answer is yes.


inequalities
calculus
integration
function
calculus computations

[quote="aktyw19"]Let $ f(x) \equal{} Ax^3 \plus{} Bx^2 \plus{} Cx \plus{} D$  and  $ f(x) \ge 0\; ,\; \forall x\in [0,1]$ ,Prove
$ \int\limits_{0}^{1} xf(x)\; dx \le K\int\limits_{0}^{1}f(x)\; dx \;\;\; \; \mbox{where} \; \; \; \; K \equal{} \frac {6 \plus{} \sqrt {6}}{10}$[/quote]
[hide="Hint"]
Supposed $ f$ is any continuous function that satisfies $ f \geq 0$ in $ [0,1]$ we will prove that :
$ \int\limits_{0}^{1} (x\minus{}K)f(x)\; dx \leq 0$ where $ K\equal{}\frac {6\plus{}\sqrt {6}}{10}$
Proof :  By mean value theorem for integral, because $ \int\limits_{0}^{1} (x\minus{}K)\; dx\equal{}\minus{}\frac {\minus{}1\plus{}\sqrt {6}}{10}$, that is not equal to $ 0$ we know that there exist $ a$ such that $ \int\limits_{0}^{1} (x\minus{}K)f(x)\; dx \equal{}f(a)\int\limits_{0}^{1} (x\minus{}K)\; dx\equal{}\minus{}f(a)\frac {\minus{}1\plus{}\sqrt {6}}{10}$. Because $ f(a) \geq 0$ for all $ a \in [0,1]$, then the last expression is smaller than $ 0$. I am sorry if I am wrong.
[/hide]

[quote="Qruni"]
Proof :  By mean value theorem for integral, because $ \int\limits_{0}^{1} (x \minus{} K)\; dx \equal{} \minus{} \frac { \minus{} 1 \plus{} \sqrt {6}}{10}$,
[/quote]

to use mean value theorem we need that $ (x \minus{} K)$ don't chaning it sing on $ [0,1]$.  :(

[quote="aktyw19"]Let $ f(x)\equal{}Ax^3\plus{}Bx^2\plus{}Cx\plus{}D$  and  $ f(x) \ge 0\; ,\; \forall x\in [0,1]$ ,Prove
$ \int\limits_{0}^{1} xf(x)\; dx \le K\int\limits_{0}^{1}f(x)\; dx \;\;\; \; \mbox{where} \; \; \; \; K\equal{}\frac{6\plus{}\sqrt{6}}{10}$[/quote]

i think, it is direct


Functional Analysis
real analysis
real analysis unsolved

Let $ V$ be a real Banach space.
Can a bilinear form on $ V$ be continuous (as map from $ V\times V \to \mathbb{R}$) but not bounded (i.e. $ B(u,v)\leq M\|u\|\|v\|$ for some $ M\in\mathbb{R}$)?

I know that for Hilbert space continuity <=> boundedness because every continuous bilinear form is of the form $ \langle Au, v\rangle$ for some bounded operator $ A$.

For any fixed $ u$ we have a continuous linear functional $ v\mapsto B(u,v)$; call it $ T_u$. Fixing $ v$ instead, we find that $ \sup_{\|u\|\le 1}|T_u(v)| < \infty$ for any $ v$. The [url=http://en.wikipedia.org/wiki/Uniform_boundedness_principle]uniform boundedness principle[/url] tells us that $ \sup_{\|u\|\le 1}\|T_u\| < \infty$. Thus the form is bounded.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be any group and let $ g \in G$ have order $ mn$ where $ (m, n) \equal{} 1$. Prove that there is some power $ g^k$ so
that the order of $ g^k$ is $ m$ and the order of $ g^{k\minus{}1}$ is $ n$

There exist integers $ a, b$ such that $ na \minus{} mb \equal{} 1$.  Then let $ k \equal{} na$.

But doesn't order imply that it has to be the least power such that $ g^x \equal{} e$?
How do you prove that?

Let's $ H \equal{} <g>$ the subgroup generated by $ g$, $ |H| \equal{} mn$ and $ \forall t \in \mathbb{Z}, g^t \in H$. So there is a finite number of powers.

P.S: [b]Bézout's identity:[/b]
$ \forall a,b,n \in \mathbb{Z},\ \exists s, t \in  \mathbb{Z}$ such that $ sa \plus{} tb \equal{} n(a, b)$
http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity

$ na \minus{} mb \equal{} 1$ implies that $ \gcd(a, m) \equal{} \gcd(n, b) \equal{} 1$.  Then the order of $ g^{na}$ is clearly $ m$ and the order of $ g^{mb}$ is clearly $ n$.  (Note that here we are considering $ g$ as a generator of a subgroup as vict85 pointed out.)


Kristin has 5 different pairs of earrings. What is the probability
that two randomly-selected earrings will be a matching pair?
Express your answer as a common fraction.

After picking one earring, one out of the other earrings will match it. $ \boxed{\frac19}$


trigonometry
geometry
parallelogram
trig identities
Law of Sines
geometry proposed

Let $ABCD$ be parallelogram (with not equal sides) and $\angle BAC< 90^{\circ}$. Let $E$ and $F$ be feet of perpendiculars from $A$ to $CB$ and $CD$ respectively. If perpendicular from $A$ to $AC$ meet $BD$ at $P$ prove that $P$, $E$, and $F$ are collinear.

[Moderator edit: This topic contained 2 more problems, but now they are both moved, together with solutions, to different topics where they already have been discussed. - Darij]

Let's make a little change of notation: interchange $E$ and $F$ (I have already solved this problem some time ago, and I'm lazy to change my previous notation :D).

Let $\angle BCA=\theta$, $\angle BAC=\epsilon$. We have $\angle CAD=\theta$, $\angle ACD = \epsilon$, $\angle ADE= \angle ABF$. Now observe that
\[BF=AF \cot (\theta+\epsilon), FC=AF\cot \theta \Rightarrow \frac{BF}{FC}=\frac{\cot(\theta+\epsilon)}{\cot \theta}\]

\[CE=AE\cot \epsilon,\ ED=AE\cot (\theta+\epsilon) \Rightarrow \frac{CE}{ED}=\frac{\cot \epsilon}{\cot (\theta+\epsilon)}\]
Now comparing the areas of triangles $ADP$ and $ABP$ we get
\[\frac{(ADP)}{(ABP)}=\frac{DP}{PB}=\frac{\sin(\pi/2-\theta)}{\sin (\pi/2+\epsilon)}\cdot \frac{AD}{AB}=\frac{\cos \theta}{\cos \epsilon}\cdot \frac{\sin \epsilon}{\sin \theta}\]
The last equality follows from the law of sines in triangle $ABD$: $\frac{AD}{AB}=\frac{\sin \epsilon}{\sin \theta}$. Hence in triangle $BCD$ we have
\[\frac{BF}{FC}\cdot \frac{CE}{ED}\cdot \frac{DP}{PB}=\frac{\cot(\theta+\epsilon)}{\cot \theta}\cdot \frac{\cot \epsilon}{\cot (\theta+\epsilon)}\cdot \frac{\cos \theta}{\cos \epsilon}\cdot \frac{\sin \epsilon}{\sin \theta}=\frac{\cot \theta \cot \epsilon \tan \epsilon}{\cot \theta}=1 \]
And by Menelaus' theorem we conclude that $P$, $E$ and $F$ are collinear, as desired ;)


For some reason, I can't rate posts when using Safari. I can using Firefox and IE, just not Safari.


Diophantine equation
number theory solved
number theory

here's the problem
Find the smallest positive integer n such that 19n + 1 and 95n + 1 are both integer squares.
this cute problem was proposed by titu andreescu.
best regards

oh i thought that the romanians (or the other geniuses on this group,for that matter )would devour this problem in no time.it took me 3 days :D .neways,i used pells equation to do this one

there are a lot of problems like these ones in The American Mathematical Monthly:

1) Find out all natural numbers n so that 2n+1 and 3n+1 are simultaniously perfect squares.

2) Find out all natural numbers n so that n+1 and 3n+1 are simultaniously perfect squares.

all done with pell!

cheers!

2n+1=x 2
3n+1=y 2
=>3x 2 -2y 2 =1
put x=u+2v and y=u+3v to reduce to u 2 -6v 2 =1
which has infinitely many solutions in positive integers
QED
cheers
similar method for the 2nd i guess


Hello,

Let the function $ f$ defined by : $ f(x)\equal{} \arctan \left(\sqrt{\frac{1\minus{}\cos(x)}{1\plus{}\cos(x)}}\right)$ and $ D_f$ is domain of $ f$.

Prove that : \[ \forall x\in D_f , f(x) \equal{} \frac{\left|x\minus{}2\pi \lfloor \frac{x}{2\pi}\plus{}\frac{1}{2}\rfloor\right|}{2}\]

[quote="linda2005"]Hello,

Let the function $ f$ defined by : $ f(x) \equal{} \arctan \left(\sqrt {\frac {1 \minus{} \cos(x)}{1 \plus{} \cos(x)}}\right)$ and $ D_f$ is domain of $ f$.

Prove that :
\[ \forall x\in D_f , f(x) \equal{} \frac {\left|x \minus{} 2\pi \lfloor \frac {x}{2\pi} \plus{} \frac {1}{2}\rfloor\right|}{2}
\]
[/quote]

$ \frac{1\minus{}cos x}{1\plus{}cos x}\equal{}\frac{1\minus{}(1\minus{}2sin^2(\frac{x}{2}))}{1\plus{}(2cos^2(\frac{x}{2})\minus{}1)}\equal{}\frac{sin^2 (\frac{x}{2})}{cos^2 (\frac{x}{2})}\equal{}tan^2 (\frac{x}{2})$

This problem isn't all that clear. What if $ x\equal{}2\pi \plus{}1$? Then the given functions are not equal.

hi,
we have
$ \forall x \in ]\minus{}\pi,\pi[ , f(x)\equal{}\frac{|x|}{2}$
and $ f$ is $ 2\pi$-periodic, so $ \forall x \in D_f,  f(x)\equal{}f(x\minus{}2\pi n)\equal{}\frac{|x\minus{}2\pi n|}{2}$ , $ \exists! n \in\mathbb{Z}$ and $ x\minus{}2\pi n \in ]\minus{}\pi,\pi[$
this is because $ \frac{x}{2\pi}\minus{}\frac{1}{2}<n<\frac{x}{2\pi}\plus{}\frac{1}{2}$


Could someone prove to me that a linear function is bijective iff it has an inverse?

For any function, bijectiveness and inversitivity are equivalent (forgive my coining words).

Yes I know, it was presented in class today as a theorem without a proof, and I would like to see a formal proof if possible.

Let $ f : A \to B$.

1. Assume $ f$ bijective. Define $ g : B \to A$ thusly: for $ b \in B$ there exists $ a \in A$ such that $ f(a)\equal{}b$, since $ f$ is surjective; moreover $ a$ is unique with this property, since if we also had $ f(a')\equal{}b$, then $ f(a') \equal{} f(a)$, whence $ a'\equal{}a$ from $ f$ being injective. Therefore we can define $ g(b)\equal{}a$. It is easy to check that $ g \circ f \equal{} \textrm{id}_A$ and  $ f \circ g \equal{} \textrm{id}_B$, i.e. $ g \equal{} f^{\minus{}1}$.

2. Assume $ f$ has an inverse $ g : B \to A$. Then from $ f(a') \equal{} f(a)$ follows $ a' \equal{} g \circ f(a') \equal{} g \circ f(a) \equal{} a$, so $ f$ is injective; while for any $ b \in B$ we have $ f(g(b)) \equal{} f \circ g(b) \equal{} b$, so $ f$ is surjective.


inequalities
function
vector
algebra unsolved
algebra

Let $ a_1$, $ a_2$, ..., $ a_n$ and $ b_1$, $ b_2$, ..., $ b_n$ be $ 2 \cdot n$ real numbers. Prove that the following two statements are equivalent:

[b]i)[/b] For any $ n$ real numbers $ x_1$, $ x_2$, ..., $ x_n$ satisfying $ x_1 \leq x_2 \leq \ldots \leq x_ n$, we have $ \sum^{n}_{k \equal{} 1} a_k \cdot x_k \leq \sum^{n}_{k \equal{} 1} b_k \cdot x_k,$

[b]ii)[/b] We have $ \sum^{s}_{k \equal{} 1} a_k \leq \sum^{s}_{k \equal{} 1} b_k$ for every $ s\in\left\{1,2,...,n\minus{}1\right\}$ and $ \sum^{n}_{k \equal{} 1} a_k \equal{} \sum^{n}_{k \equal{} 1} b_k$.

The intention must be that (i) is true for all possible increasing n-tuples $ x$, otherwise the problem is false.  Also, the inequality in (ii) must go the opposite way, as follows from the proof below.  Here is the corrected version:

[b][Moderator edit: I have corrected Orl's original post above according to this. $ \minus{}$ Darij][/b]

This problem is very interesting, because (ii) is the condition for $ b$ to majorize $ a$, but [i]without[/i] the usual condition that $ a$, $ b$ are ordered.   So it looks like a majorization condition for the class of increasing functions in place of convex ones.  Therefore, there should be several more equivalent conditions such as

(iii) Sequence $ b$ can be obtained from $ a$ by a series of transformations of the following type: given indices $ i < j$ and a positive real number $ p \geq 0$, change $ a_i \to a_i \minus{} p$ and $ a_j \to a_j \plus{} p$.

[b]Question [/b]:  what is the analogue of the convex majorization criterion using double stochastic matrices (row and column sums = 1, Birkhoff polytope)?

Anyway, the proof for the problem:

Inequality (i) holds for all n-tuples $ x$ if and only if it holds for constant n-tuples $ (c,c,c,...,c)$ and all n-tuples of the form $ (0,0,...,1,1,1,...)$. This is because any increasing sequence $ x$ can be written as a sum (with non-negative coefficients) of such vectors.
(i) for constant vectors is the condition $ \Sigma a_i \equal{} \Sigma b_i$.
(i) for the vector with $ x_i \equal{} 0$ for $ i < k$ and $ x_i \equal{} 1$ for $ i \geq k$  is the condition $ \sum^{n}_{i \equal{} k} a_i \leq \sum^{n}_{i \equal{} k} b_i$.
That's all.

Here is an easy way to help solve inequaility (ii).
If we let S(k)= (a1+a2+^^^+ak)-(b1+b2+^^^bk).
We will get a(k+1)-b(k+1)=S(k+1)-S(k)
Noticing that S(m) is greater than or equal to 0. for every positive integer m such that 1≤m≤n-1, and S(n)=0
we can get the proof easily by computing the difference between the two sides of the inequality (ii).


Find all integer solutions of the equation
$x^6 + x^3 + 1 = y^4$

This is equation for beginner section, isn't it?

Well, not beginner, but it probably should be in a lower level forum - I originally had them there but moved it up since there were some posts about national-olympiad-level problems being in another forum.. didn't want to risk arguments.


email
combinatorics unsolved
combinatorics

We have a integer sequence: $ a_{1}$, $ a_{2}$, $ a_{3}$, .... and we have $ a_{1}$=1, $ a_{2}$=2.   $ a_{n}$=$ 3a_{n\minus{}1}$+$ 5a_{n\minus{}2}$         n=3,4,5....    . Does exist integer number $ x\geqslant 2$ such that $ a_{x}$ is a divisor of $ a_{x\plus{}1}$$ a_{x\plus{}2}$  ??

Sorry for my english

This is problem 3 from 1st series of 1st round of 59th Polish Mathematical Olympiad. The 1st series of the competition ends at 8.10.2007. Moderators, please lock the topic.

[quote="Rand0m"]Look. I don't know that this question is placed in polish mathematical olympiad. I am from sweden and I am studying in Englang - i am not in high school, and I an not from Poland. I like such a questions, so don't close the topic ! If it is really from POL math olympiad please don't place any answer here, just tell me  on private message. If somebody think that he know the answer just make a conversation with me by private message.[/quote]
I am sorry, but what you said is doubtful. Your e-mail is [...] , so it is registered at a Polish server. Moreover, the problems from the Polish MO appeared in the Internet today and today you posted six of them. How would you explain this?

Man, what do you want from me? My nationality is polish, I lived in sweden now I live in england. If you are irritated it is your problem kid.
PEOPLE; DO NOT PLACE ANSWERS HERE, MAKE A CONVERSATION WITH ME BY PRIVATE MESSAGING.

I'm just curious... Is that Internet competition? If no, I don't see the reason to lock the topic.

Nope, this is not an internet competition.... There are always little kid with stupid arguments... Why, and for what this he wants topic to be closed?? I don't get it. Questions are open, people, try to make this question..- If somebody thinks that answering here is a cheat (LOL) write to me private message. Thanks

[quote="Rand0m"]If somebody thinks that answering here is a cheat (LOL) write to me private message. Thanks[/quote]
Obviously, answering here is not cheating, but it can help cheaters: anyone that takes part in the competition can read the solution which can help him to qualify to the second round. I can not prove that you are going to cheat in such a way, but you seem to do so.
[quote="Rand0m"]There are always little kid with stupid arguments... [/quote]
I am not a little kid, but even if I were, it would not matter.  As far as my arguments are concerned, they are not stupid. All what you write is really weird, especially when you apologize for your English and later you state that you study in England.

[quote="djuro"]I don't understand something: is that competition HELD or is NOT HELD YET?[/quote]
It is held.
[quote="Rand0m"]Rzeszut, can we stop arguing, ok?[/quote]
All that should be said is already said, so we can stop arguing.

[quote="djuro"]If it's already held, how can he cheat?[/quote]
I thought that [i]a competition is held[/i] means that it is currently running.

[quote="Rand0m"]Yes it is running, but the rules of 1 stage ALLOW to help in thease OPEN questions.[/quote]
The rules allow students to ask questions in order to clear doubts (they are [url=http://www.om.edu.pl/regulamin.html]here[/url], unfortunately only in Polish).

What the hell? This guy is posting all problems from the Polish national olympiad, which is cheating, obviously. To moderator: please close all his topics and ban him asap, since he's posted all problems from the Polish MO and he argues that it's all right.

I have edited the email address so that spammers won't get it (we don't want to help cheaters, but we shouldn't help spammers either).

@Rand0m: The person showing "little kid" behaviour most perfectly in this topic is you...

  darij


On the January results page, it says 840 points were needed to be promoted to Silver. Does that mean that anyone who scored 840 or higher on that single contest was promoted? Or is it an average of 840 over all the contests so far? Or cumulative 840 from all the contests??


Will there be a fall session for intro to counting and probability?

We currently do not expect to offer Intro C & P until spring, 2009 (probably starting in February).

What level of probability knowledge do you think would take this class?
And I am also thinking about intermediate c & p so could you post the level for both?

The class pages in the Online Classes portion of the web page have diagnostic tests for both classes -- these will give you the best indication of whether or not the class is right for you.  We assume no prior knowledge of probability for the Intro class, and we expect you to know the Intro material for the Intermediate class.


search
number theory unsolved
number theory

Let $ X_{n}$ be the sum of first $ n$ prime numbers. (ex $ X_{4} \equal{} 2\plus{}3\plus{}5\plus{}7$)
Prove that there exists a square between $ X_{n}$ and $ X_{n\plus{}1}$, for all $ n \equal{} 1,2,\cdots$.

[quote="shoki"]search before u post
http://www.mathlinks.ro/viewtopic.php?t=150552[/quote]

Er...Pen... I missed it. Sorry. :(


I can't access the list of blogs, either by typing in the address in the address bar or clicking AoPS blogs. Is it just my browser? I'm using Seamonkey 1.0.

We are having some problems with the server right now ...


dorood
biaid to in marathone az har mabhasi khastim soal bedim faghat ghanoonesh in bashe ke hagh nadarim vaghti soale hal nashodeii hasti saol post konim  .

Inam soale avval :
$ a,b,c > 0$
sabet konid
$ (a^5 \minus{} a^2 \plus{} 3)(b^5 \minus{} b^2 \plus{} 3)(c^5 \minus{} c^2 \plus{} 3) \ge (a\plus{}b\plus{}c)^3$

salam . manam moafegham .
soale namosaviam injoori hal mishe :
avval tavvajoh konin ke baraye har $ a > 0$ darim , $ (a \minus{} 1)^{2}(a \plus{} 1)(a^{2} \plus{} a \plus{} 1)\geq 0$ ke natije mide :
$ a^{5} \minus{} a^{2} \plus{} 3\geq a^{3} \plus{} 2$
pas kafie sabet konim $ (a^{3} \plus{} 2)(b^{3} \plus{} 2)(c^{3} \plus{} 2)\geq (a \plus{} b \plus{} c)^{3}$ ke inam mishe injoori nevesht :
$ (a^{3} \plus{} 1 \plus{} 1)(1 \plus{} b^{3} \plus{} 1)(1 \plus{} 1 \plus{} c^{3})\geq (a \plus{} b \plus{} c)^3$ ke inam hamoon namosavie holdere . pas masale halle .

manam ye soale hendese mizaram :
$ H$ va $ G$ be tartib markaze ertefayi va markaze seghle mosalas $ ABC$ hastand . az $ A$ , $ B$ va $ C$ be tartib bar $ AG$ , $ BG$ va $ CG$ se amood kharej mikonim . ta in se khat mossalase $ A'B'C'$ ra tashkil dahand . neshan dahid markaze seghle mosalase $ A'B'C'$ rooye khate $ HG$ gharar darad .

vali avval soalaye ghabli ro hal konim baaaaaaaaaad :(

hendese he ham rahate vali hal nadaram benevisam :!:  :idea:

besyar harekate jalebi bood  :idea:
modat ha bood ke in anjoman taghriban faliatesho az dast dade bood

khob in soale hendeseye ghashangie vali berahati ba tahlili ya mokhtalet hal mishe ke yekam mohasebatesh ziade ba in hal man daram donbale rahe hendesish migardam va fek nemikonam kheili sakht bashe :wink:

[quote="Toxic_Frog"]
$ a,b,c > 0$
sabet konid
$ (a^5 \minus{} a^2 \plus{} 3)(b^5 \minus{} b^2 \plus{} 3)(c^5 \minus{} c^2 \plus{} 3) \ge (a \plus{} b \plus{} c)^3$[/quote]

fek konam male amrica 2004 e in.

javabe soale khashi :

avsate azlaee $ AB$,$ AC$ va $ BC$ ro be tartib $ D$ , $ E$ va $ F$ benamid .

markaz seghle $ A'B'C'$ ro $ G'$ benamid .
ebtedaan sabet mikonim amoodhayi ke az $ A'$ , $ B'$ , $ C'$ be tartib bar $ BC$  , $ AC$ va $ AB$ vared mishavand mianehaye $ A'B'C'$ hastand pas mahalle barkhordeshan markaze seghle $ A'B'C'$ ast .

 paye amoode vared az $ A'$ bar $ BC$ ro $ F'$ minamim darim :

$ {{\sin \angle F'A'B'} \over {\sin \angle F'A'C'}} \times {{A'B'} \over {A'C'}} \equal{} {{\sin \angle GCB} \over {\sin \angle GBC}} \times {{A'B'} \over {A'C'}} \equal{} {{\sin \angle GCB} \over {\sin \angle GBC}} \times {{\sin \angle BGA} \over {\sin \angle AGC}} \equal{}$

$ {{\sin \angle GCB} \over {\sin \angle GBC}} \times {{\sin \angle BGF} \over {\sin \angle CGF}}$

aknoon tavajjoh konid ke dar mosallase $ GBC$ darim :
$ {{\sin \angle GCB} \over {\sin \angle GBC}} \times {{\sin \angle BGF} \over {\sin \angle CGF}} \equal{} 1$

pas natije mishe ke $ {{\sin \angle F'A'B'} \over {\sin \angle F'A'C'}} \times {{A'B'} \over {A'C'}} \equal{} 1$ yani $ A'F'$ miyaneye nazire zele $ B'C'$ ast .

be hamin tartib sabet mishe ke $ B'G' \bot AC$ va $ C'G' \bot AB$

*****

hala be rahati mishe did ke gharineye $ G$ nesbat be $ F$ markaz ertefayii $ A'BC$ ast(yani gharineye $ G$ nesbat be $ F$ rooye $ A'G'$ gharar dare) pas vazeh ast ke amoodmonaseffe $ BC$ az vasate $ GG'$ migzare be hamin tartib ham sabet mishe amoodmonaseffe $ AB$ az vasate $ GG'$ migzare banabarin yani mahalle taghatoe amoodmonassef haye mosallase $ ABC$ dar vasate $ GG'$ gharar dare yani $ GG'$ az $ O$ migzare ya be ebarate digar $ OG$ az $ G'$ migzarad .

-------------------------
inam ye soale dige ( nobati ham ke bashe nobate nazaryias ) :

farz konid $ a_1,a_2,...,a_n$ adadi sahih hastand va bozorgtarin maghsoom alayhe moshtarake anha barabare $ 1$ ast .farz konid $ S$ majmooei az adad sahih ba in 3 vijegi bashad :
1-baraye har $ i\equal{} 1 ,2, ... , n$ darim : $ a_i \in S$
2-baraye har $ i,j \equal{} 1,2,... , n$ ( na lozooman motemayez) darim : $ a_i \minus{} a_j \in S$
3-baraye har $ x,y \in S$ darim : $ x\plus{}y \in S  \Rightarrow  x\minus{}y \in S$

sabet konid $ S$ majmooeye adade sahih ast .

ina soalaye maroofian hame,javabe ino pain minivisam vase inke marathon kharab nashe,age ghaboole begid ta man soale badio benevisam.

[hide]

[url]http://www.mathlinks.ro/viewtopic.php?p=17440#17440[/url]

[/hide]

fek nakonam in soal ham maroof bashe :
$ X$ yek majmooeast . $ P$ majmooeii shamel tedadi az zir majmooehaye $ 2k$ ozvi $ X$ ast . har zir majmooe az $ X$ ba hadde axar $ (k\plus{}1)^2$ ozv ya hich zir majmooeiii dar $ P$ nadarad ya inke hameye zir majmooehayii az in majmooe ke dar $ P$ gharar darand ozv moshtaraki darand . sabet konid hameye azaye $ P$ ba ham eshterak darand ( ozvi vojod darad ke dar hameye zir majmooeha amade ast ) .


One 110kg linesman is running to the right at a velocity of 2.75m/s while another 125kg lineman is running directly toward him at 2.60m/s.

a)what are the magnitue and direction of the net momentum of these to athletes

this is what i done

$ m_aV_{1a}+m_bV_1b=P$
P=22kgn/s to left

$ m_aV_{1a}+m_bV_1b=V_2(m_a+m_b)$

making V_2 the subject and subst. the values

b)How do u do this?

$ m_aV_{1a}+m_bV_1b=V_2(m_a+m_b)$

making $ V_2$ the subject and subst. the values

KE=$ \frac{(m_a+m_b){V_2}^2/2}$

??
is this procedure correct?

[quote="bos1234"]One 110kg linesman is running to the right at a velocity of 2.75m/s while another 125kg lineman is running directly toward him at 2.60m/s.

a)what are the magnitue and direction of the net momentum of these to athletes

this is what i done

$ m_aV_{1a} + m_bV_1b = P$
P=22kgn/s to left

$ m_aV_{1a} + m_bV_1b = V_2(m_a + m_b)$

making V_2 the subject and subst. the values

b)How do u do this?

$ m_aV_{1a} + m_bV_1b = V_2(m_a + m_b)$

making $ V_2$ the subject and subst. the values

KE=$ \frac {(m_a + m_b){V_2}^2/2}$

??
is this procedure correct?[/quote]

I believe the a) part should be: $ 22.5\quad \text{kg}\cdot\text{m/s  to the left}$.

As for the b) part, I'm not sure there is any point to this but you can do it by figuring,
\[ KE =\frac{P^2}{2(m_a + m_b)} =\frac{(506.25)}{2\cdot(235)} \approx 1.077\quad \text{J}\]


Let be given two matrices A, B from M_n(R) such that  A^2001 = 0 and AB = A+B. Show that det(B) = 0.

<i>Source VUMC 2001</i>

Namdung

I like undergraduate problems

AB = A+B => (A-I)(B-I)=I => (B-I)(A-I)=I =>BA=AB
A^2001=0 => A is nipoltent detA=0
detB=det(A+B)=det(AB)=detA.detB=0

It was talked before
lemma: If X,Y commute, Y nilpotent
then det(X+Y)=detX
Valentin gave a proof

Am I wrong or is it trivial?

You have det(A) = 0 since A is nilpotente.
In another  hand, you have B = A(B-I), then det(B) = det(A)det(A-I) = 0.

Pierre.

Aaargh...you are too fast Moubi!

Pierre.

AB = A+B => (A-I_n)B=A => det(A-I_n)det(B)=det(A)=0. We need det(A-I_n) != 0 but that is easy since A^(2001)-I_n = - I_n = (A-I_n)(...)


inequalities
geometry
incenter
circumcircle
inequalities unsolved

[quote="quykhtn-qa1"]let G,I,O are,in order,the centroid ,incenter and circumcenter of a triangle ABC. put BC=a,CA=b,AB=c.prove that:
2OI^2>=2OG^2+IG^2+((a-b)^2+(b-c)^2+(c-a)^2)/4. :lol:[/quote]
Must be : $ 2OI^2 \le 2OG^2\plus{}IG^2\plus{}((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)/4$

$ \ \left\|\begin{array}{ccc} OI^2\equal{} R^2\minus{}2Rr \\
\ OG^2\equal{} R^2\minus{}\frac{2}{9}(s^2\minus{}r^2\minus{}4Rr)\\
\ IG^2\equal{}\frac{1}{9}(s^2\plus{}5r^2\minus{}16Rr) \\
\ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \equal{} 2(s^2\minus{}3r^2\minus{}12Rr)\end{array}\right\|$

So we must prove : $ 12Rr\plus{}3r^2 \le s^2$ which is obviously true ( $ 16Rr\minus{}5r^2 \le s^2 , 2r \le R$ ) .


Let $0<f(1)<f(2)<f(3)...$an increasing sequence of natural numbers. the $n$-th number which doesn`t figure in the sequence is $f(f(n))+1$. Find $f(n)$.

Hi Rust,
you are right, I am wrong.
My formula for f is false (34/21~1.619, a~1.618,......)

you can see here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=61020[/url] the full proof  :oops:

Hi Pohoatza,   splendid proof.
   A remark: the formula for f(x) is special if x is in the Fibonacci sequence: f(F(2*k))=F(2*k+1) and f(F(2*k-1))=F(2*k)-1.
   A question: have you a proof of f(n)=[a*n] ???


Amsmath makes it easy - make sure you have \usepackage{amsmath} in your document. Then
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
gives $\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$
You can also use bmatrix, Bmatrix, vmatrix and Vmatrix to get
$\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ $\begin{Bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{Bmatrix}$ $\begin{vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ $\begin{Vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{Vmatrix}$

The order of matrices he (stevem) displayed is bmatrix, Bmatrix, vmatrix, Vmatrix. His order in the text was slightly off, so I wanted to make sure you weren't confused.

Thanks for spotting that and I will change it. What's worse is that I copied it from another post which I will have to find and change :(


induction
combinatorics proposed
combinatorics

In a house There are some rooms separated with some walls.On every wall

 between rooms there are some doors >=0 .A room is labled as $S$ and

another as $T$ so that $S$ has only one door.
For every sequence ${x_n}$ of letters $R,L$, we define a path in the

house this way:we start from $S$. In $i$th step if $x_i=L$ we go from

the left door of the room (with respect to our entrance) and if $x_i=R$

we go from the right one.
Obviousely if a room has just one door, then $R$ isn't different from

$L$.And the first letter is not important.
Prove that there exist a sequence ${x_n}$ such that for any house with

any $S,T$ if we start from $S$, then after finite steps we arrive at

$T$.
Example of a path:
	RLRLRLLRRLLRRL

i remember that i state this problem and i was persuaded to send the solution in this section :D

i find it
http://www.mathlinks.ro/Forum/viewtopic.php?t=5197

Let me get this straight: what are the rooms? Can they be considered polygons with a door on each side? It's not very clear :?

Which stage/year did this problem take place ?

I wonder if this solution is without mistakes:
Consider a graph with rooms as vertices and  doors as edges. We prove that if the graph is connected, then from every room  we can reach every door. The proof is by induction of $n$ the number of rooms. We can consider a room suppose the room $r_n$ s.t. if we remove it the graph stays connected. We replace a door $d$ through $r_n$ by the pair of doors $d,d'$ where $d'$ is the door next to $d$ clockwise in $r_n$. We can apply now induction hypothesis and easily perform induction step (am I right?).

Now we can prove that for a fixed house, we can get from every room $i$ to every room $j$.
Now for a fixed house and a fixed starting room $1$ we can go from $1$ to $2$,$2$ to $3$ etc.  then backwards,
at the end   producing a sequence of steps s.t. somewhere during the sequence we go from room $i$ to room $j$ for any $i,j$.
Now we prove we can find for a fixed house $H$a sequence of steps s.t. no matter what room we start from, during the sequence we somewhere go from $i$ to $j$ for any $i$,$j$.(*)
Indeed, if the room is $1$, we've already constructed it before(*): sequence $s_1$. Else, if the start is $2$ we see where we went following the previous sequence of steps, suppose we reached room $k$, for $k$ we perform (*) to produce a sequence $s_2$ and so on. The sequence $S_H=s_1,s_2,\cdots$ is our desired sequence (**).
Now we easily prove for any $u=(n,k)$ there are finitely many non-isomorphic connected houses with $n$ rooms and $k$ doors. Let $H_1, \cdots H_m$ be all possible houses. The the sequence $T_u=S_{H_1},S_{H_2},\cdots $ satisfies for any connected house  with $n$ rooms and $k$ doors, no matter where we start from somewhere we get from room $i$ to room $j$ for any $i,j\leq n$.
Now since $N\cdot N$ is countable, we enumerate the pairs $u_1,\cdots u_n,\cdots $.
Then the sequence $T=T_{u_1},T_{u_2},\cdots $ is our desired sequence.

QED if I am not mistaken again.

Your solution is almost right.But for proving that in a connected house we can go from any room to any another,there is a simple way.Consider $G$ to be the planar graph with rooms as vertices and doors as edges.We assume that $G$ is connected.Evidentally if we go through an edge of a face, we can go around it so we can entre any face adjacent to it.Now consider the dual graph of $G$ as $G'$.$G'$ is connected so we can reach any vertice by a sequence of adjacent faces.So the assertion is proved.

It's similiar to problem of Russia 1998 the God and Evil problem

And also to Sam it's Iran TST 1996 0r 1997


algebra polynomial function calculus derivative trigonometry vector

Depends what type of equation you are talking about. Polynomial equation? If so, how many variables? Differential equation? Linear recurrence?

And this is the definitely the wrong place to post this question.

err im sry but i donno where to post ...

For differential equations, can someone plz explain what is homogeneous equation?

Hello. In the context of differential equations, the term "homogeneous" has two meanings. A first order differential equation, $\frac{dy}{dx} = f(x, y)$, is homogenous if $f(x, y)$ can be written as a function of y/x or x/y. Linear differential equations are also labeled as homogeneous if the term that is not multiplied by y or its derivatives equals zero.

These two entirely different - indeed, incompatable - uses of the word "homogeneous" plague every first differential equations textbook or first course using that textbook. My reaction is usually to greatly downplay (maybe even ignore) the first meaning in mathisfun1's post, and make a very big deal out of the second meaning. Getting at that second meaning:

A differential operator $T$ is a mapping which takes as input a function, produces as output a function, and involves one or more derivatives of that functiion.

$W(u)=c^2\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial t^2}$ is a differential operator. Since it is a partial differential operator, it's kind won't be seen much in an course on ODE's. The rest of my examples will be ordinary differential operators.

$T(y)=(y')^2+2xy$ is a nonlinear differential operator.

$L(y)=x^2y''+2(\sin x)y'-\frac yx$ is a linear differential operator. So also is

$L(y)=3y'''-2y'+4y$

The definition of "linear" is exactly the same as the definition of linear operator or linear transformation in a linear algebra course: $L$ is linear iff $L(c_1y_1+c_2y_2)=c_1L(y_1)+c_2L(y_2)$ for all (nice enough) functions $y_1$ and $y_2$ and all constants $c_1$ and $c_2.$

Now suppose $L$ is a linear differential operator. Then $L(y)=0$ is a [b]homogeneous[/b] linear differential equation. Its solution space is a vector subspace. This includes the fact that the zero function is a solution. A practical method of solving such an equation amounts to a procedure for finding a basis for that vector subspace.

But if $g(x)$ is not the zero function, $L(y)=g$ is a [b]nonhomgeneous[/b] linear differential equation. Here, we need to find just one "particular" solution $y_p$ to $L(y_p)=g$ and then the general solution is of the form $y=y_p+y_c$ where $y_c$ is any solution of the corresponding homogeneous equation $L(y_c)=0.$

This last use of the word "homogeneous" is exactly the same as its use in a linear algebra course. The first use - the procedures appropriate to an equation like $y'=\frac{y^2-x^2}{y^2+x^2},$ for instance - is useful, but the use of the same word is unfortunately confusing.


[quote="ckck"][quote="Quevvy"]What is the length of a space diagonal of a rectangular prism with edge lengths $5\sqrt{3},5\sqrt{11},5\sqrt{11}$?[/quote]
[hide]The formula is $\sqrt{x^{2}+y^{2}+z^{2}}$. Solving for this the space diagonal is 25.[/hide][/quote]

For anyone who is interested in why this works...

the diagnol on the face of any of the face with side lengths a and b, is $\sqrt{a^{2}+b^{2}}$ by phythag.

so to get the space diagnol, we connect this length to side c, to create a right triangle where the hypotenuse is the space diagnol.  so $\sqrt{\sqrt{a^{2}+b^{2}}^{2}+c^{2}}=\sqrt{a^{2}+b^{2}+c^{2}}$

[quote="Quevvy"]What is the length of a space diagonal of a rectangular prism with edge lengths $5\sqrt{3},5\sqrt{11},5\sqrt{11}$?[/quote]

[hide="Complete Solution"]Let the sides be $a$, $b$, and $c$.

We are interested in $\sqrt{a^{2}+b^{2}+c^{2}}$.

But we are given:

$a^{2}={(5\sqrt{11})}^{2}= 275$

$b^{2}={(5\sqrt{11})}^{2}= 275$

$c^{2}={(5\sqrt{3})}^{2}= 75$

So, adding up all three equations, $a^{2}+b^{2}+c^{2}= 625$.

So the answer is $\sqrt{a^{2}+b^{2}+c^{2}}= \sqrt{625}= 25$.

The answer is $25$.[/hide]

[quote="vishalarul"][quote="Quevvy"]What is the length of a space diagonal of a rectangular prism with edge lengths $5\sqrt{3},5\sqrt{11},5\sqrt{11}$?[/quote]

[hide="Complete Solution"]Let the sides be $a$, $b$, and $c$.

We are interested in $\sqrt{a^{2}+b^{2}+c^{2}}$.

But we are given:

$a^{2}+b^{2}={(5\sqrt{11})}^{2}= 275$

$b^{2}+c^{2}={(5\sqrt{11})}^{2}= 275$

$b^{2}+c^{2}={(5\sqrt{3})}^{2}= 75$

So, adding up all three equations, $a^{2}+b^{2}+c^{2}= 625$.

So the answer is $\sqrt{a^{2}+b^{2}+c^{2}}= \sqrt{625}= 25$.

The answer is $25$.[/hide][/quote]
What?

This is confusing......

[quote="vishalarul"][quote="Quevvy"]What is the length of a space diagonal of a rectangular prism with edge lengths $5\sqrt{3},5\sqrt{11},5\sqrt{11}$?[/quote]

[hide="Complete Solution"]Let the sides be $a$, $b$, and $c$.

We are interested in $\sqrt{a^{2}+b^{2}+c^{2}}$.

But we are given:

$a^{2}+b^{2}={(5\sqrt{11})}^{2}= 275$

$b^{2}+c^{2}={(5\sqrt{11})}^{2}= 275$

$b^{2}+c^{2}={(5\sqrt{3})}^{2}= 75$

So, adding up all three equations, $a^{2}+b^{2}+c^{2}= 625$.

So the answer is $\sqrt{a^{2}+b^{2}+c^{2}}= \sqrt{625}= 25$.

The answer is $25$.[/hide][/quote]
Four things:

1. You misunderstood the problem. We are given the side lengths, not the lengths of the diagonals.

2. You have $b^{2}+c^{2}$ twice.

3. When you added the three equations, you were supposed to have $2$ as a coefficient for each variable in the LHS.

4. Somehow, your answer is correct...

[hide]$\sqrt{\left(5\sqrt{3}\right)^{2}+\left(5\sqrt{11}\right)^{2}+\left(5\sqrt{11}\right)^{2}}=\sqrt{625}=\boxed{25}$[/hide]

[hide="suggestion for less computation"]
Factoring out the five, you can shrink this down to a fifth of the dimensions then multiply that answer by 5[/hide]


geometry
geometric transformation
reflection
function
LaTeX
analytic geometry

[hide="tedious solution"]
(the more you think about the graph, you will get a general idea of how it looks, but for me it was so hard to prove it...) :(

Since $0 \le \Big|\big||x|-2\big|-1\Big| \le 1$ and $0 \le \Big|\big||y|-2\big|-1\Big| \le 1$
$-1 \le \big||x|-2\big|-1 \le 1$
$0 \le \big||x|-2\big| \le 2$
$-2 \le |x|-2 \le 2$
$-4 \le x \le 4$
Also $-4 \le y \le 4$.

Define $f(a)=\Big|\big||a|-2\big|-1\Big|$.
If $0 \le a \le 1$:
$f(3+a)=\Big|\big||3+a|-2\big|-1\Big|=a$
$f(3-a)=\Big|\big||3-a|-2\big|-1\Big|=a$
$f(3+a)=f(3-a)$

If $0 \le a \le 2$:
$f(2+a)=\Big|\big||2+a|-2\big|-1\Big|=a-1$
$f(2-a)=\Big|\big||2-a|-2\big|-1\Big|=a-1$
$f(2+a)=f(2-a)$

If $0 \le a \le 4$:
$f(a)=\Big|\big||a|-2\big|-1\Big|=a-3$
$f(-a)=\Big|\big||-a|-2\big|-1\Big|=a-3$
$f(a)=f(-a)$

So the graph of $y(x)$ at $3 \le x \le 4$ is symmetric to $y(x)$ at $2 \le x \le 3$ (reflected over the line x=3)
And the graph of $y(x)$ at $2 \le x \le 4$ is symmetric to $y(x)$ at $0 \le x \le 2$ (reflected over the line x=2)
And the graph of $y(x)$ at $0 \le x \le 4$ is symmetric to $y(x)$ at $-4 \le x \le 0$ (reflected over the line x=0)
[this is also true for horizontal reflection, with $3 \le y \le 4$, etc]

So it is only necessary to find the length of the function at $3 \le x \le 4$ and $3 \le y \le 4$:
$\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1$
$x-3+y-3=1$
$y=-x+7$
(Length = $\sqrt{2}$)

This graph is reflected over the line y=3, the quantity of which is reflected over y=2,
the quantity of which is reflected over y=0,
the quantity of which is reflected over x=3,
the quantity of which is reflected over x=2,
the quantity of which is reflected over x=0..

So a total of $6$ doublings = $2^6$ = $64$
Total length = $64 \cdot \sqrt{2} = a\sqrt{b}$
$a+b=64+2=\boxed{66}$

[/hide]

[quote="minsoens"][hide="tedious solution"]
(the more you think about the graph, you will get a general idea of how it looks, but for me it was so hard to prove it...) :(

Since $ 0 \le \Big|\big\parallel{}x| \minus{} 2\big| \minus{} 1\Big| \le 1$ and $ 0 \le \Big|\big\parallel{}y| \minus{} 2\big| \minus{} 1\Big| \le 1$
$ \minus{} 1 \le \big\parallel{}x| \minus{} 2\big| \minus{} 1 \le 1$
$ 0 \le \big\parallel{}x| \minus{} 2\big| \le 2$
$ \minus{} 2 \le |x| \minus{} 2 \le 2$
$ \minus{} 4 \le x \le 4$
Also $ \minus{} 4 \le y \le 4$.

Define $ f(a) \equal{} \Big|\big\parallel{}a| \minus{} 2\big| \minus{} 1\Big|$.
If $ 0 \le a \le 1$:
$ f(3 \plus{} a) \equal{} \Big|\big\parallel{}3 \plus{} a| \minus{} 2\big| \minus{} 1\Big| \equal{} a$
$ f(3 \minus{} a) \equal{} \Big|\big\parallel{}3 \minus{} a| \minus{} 2\big| \minus{} 1\Big| \equal{} a$
$ f(3 \plus{} a) \equal{} f(3 \minus{} a)$

If $ 0 \le a \le 2$:
$ f(2 \plus{} a) \equal{} \Big|\big\parallel{}2 \plus{} a| \minus{} 2\big| \minus{} 1\Big| \equal{} a \minus{} 1$
$ f(2 \minus{} a) \equal{} \Big|\big\parallel{}2 \minus{} a| \minus{} 2\big| \minus{} 1\Big| \equal{} a \minus{} 1$
$ f(2 \plus{} a) \equal{} f(2 \minus{} a)$

If $ 0 \le a \le 4$:
$ f(a) \equal{} \Big|\big\parallel{}a| \minus{} 2\big| \minus{} 1\Big| \equal{} a \minus{} 3$
$ f( \minus{} a) \equal{} \Big|\big\parallel{} \minus{} a| \minus{} 2\big| \minus{} 1\Big| \equal{} a \minus{} 3$
$ f(a) \equal{} f( \minus{} a)$

So the graph of $ y(x)$ at $ 3 \le x \le 4$ is symmetric to $ y(x)$ at $ 2 \le x \le 3$ (reflected over the line x=3)
And the graph of $ y(x)$ at $ 2 \le x \le 4$ is symmetric to $ y(x)$ at $ 0 \le x \le 2$ (reflected over the line x=2)
And the graph of $ y(x)$ at $ 0 \le x \le 4$ is symmetric to $ y(x)$ at $ \minus{} 4 \le x \le 0$ (reflected over the line x=0)
[this is also true for horizontal reflection, with $ 3 \le y \le 4$, etc]

So it is only necessary to find the length of the function at $ 3 \le x \le 4$ and $ 3 \le y \le 4$:
$ \Big|\big\parallel{}x| \minus{} 2\big| \minus{} 1\Big| \plus{} \Big|\big\parallel{}y| \minus{} 2\big| \minus{} 1\Big| \equal{} 1$
$ x \minus{} 3 \plus{} y \minus{} 3 \equal{} 1$
$ y \equal{} \minus{} x \plus{} 7$
(Length = $ \sqrt {2}$)

This graph is reflected over the line y=3, the quantity of which is reflected over y=2,
the quantity of which is reflected over y=0,
the quantity of which is reflected over x=3,
the quantity of which is reflected over x=2,
the quantity of which is reflected over x=0..

So a total of $ 6$ doublings = $ 2^6$ = $ 64$
Total length = $ 64 \cdot \sqrt {2} \equal{} a\sqrt {b}$
$ a \plus{} b \equal{} 64 \plus{} 2 \equal{} \boxed{66}$

[/hide][/quote]

Is there any better solution for this problem?

[quote="joml88"]Let $ S$ be the set of points in the Cartesian plane that satisfy
\[ \Big|\big|{|x| \minus{} 2}\big| \minus{} 1\Big| \plus{} \Big|\big|{|y| \minus{} 2}\big| \minus{} 1\Big| \equal{} 1.
\]
If a model of $ S$ were built from wire of negligible thickness, then the total length of wire required would be $ a\sqrt {b},$ where $ a$ and $ b$ are positive integers and $ b$ is not divisible by the square of any prime number.  Find $ a \plus{} b.$[/quote]

LaTeX fixed. Remember that || is interpreted as "parallel to" and is its own symbol.

easy method,

you can easily realize 64root2


Hi, i need some urgent help in finding the determinant of a matrix. I'm okay with finding it for  2 x 2 and 3 x 3 matrices, but anything more, and it's confusing me. I've got the matrix here as follows:

5 6 3 1
5 7 9 3
3 5 8 2
3 4 6 7

Can someone help me in solving this step by step? i've been trying for hours and hours, and i'm getting no where. I've used an online calculator which shows me step by step, which i partly understand, and partly dont. The calculator i've been using is http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi .

Please help :(

the way to proceed for a 3x3 matrix is equal to the one for all nxn matrix. So, to calculate the determinant of your matrix you take one row or one coloumn, call it for example $r_i$ and you sum the product of all elements of $r_i$ for them algebraic complemets, [b]with their sign[/b]. So, if the element is in a place $\{i,j\}$, its sign is $(-)^{i+j}$.

what the calculator says to you, is that you can use 2 important properties of matrices:

1) if you multiplicate one row or one coloumn for $k\neq 0$, the determinant is multiplicated by $k$.
2) if you substitute to one row (coloumn) a linear combination of that row (coloumn) and another, the determinant doesn't change.

this properties can help you to "create" some zeros in the matrix without changing the value of the determinant, so your calculus will be easier. For example, if you substitute to $r_1$ the row $r_2-r_1$, you will have a 0 instead of 5 in place $\{1,1\}$ of the matrix. Going on to use those properties, you can arrive to have a lot of zeros in your matrix without change determinant's value


algebra
polynomial
superior algebra
superior algebra unsolved

Easy and interesting

Find conditions on $k\in Z$ such that there exist $Q\in Z[X]$ with the conditions
$Q(10)=400$
$Q(14)=440$
$Q(18)=k$

I'm not very sure... but I think it's enough $32|k$.


calculus
integration
function
geometry
3D geometry
sphere
trigonometry

Let $d(x,y,z)$ be the shortest distance from $(x,y,z)$ to the qube
\[\mathcal{Q}= \{ (x,y,z)\,;\, |x|\leq 1,\, |y|\leq 1,\, |z|\leq 1 \}\]
Determine
\[\iiint_{\mathbb{R}^{3}}e^{-d(x,y,z)}dxdydz \]

[hide="answer"]
$32+20\pi$
[/hide]

Since we are integrating in $\mathbb{R}^{3}$ we can think of this as blowing up a sphere: $0\leq \phi \leq \pi \, , 0\leq \theta \leq 2\pi \, , \rho = N$ and making $N\to+\infty$.

So using the substitution $x = \rho \cos \theta \sin \phi , y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$ we have,
$\iiint_{\mathbb{R}^{3}}\exp (-d(x,y,z)) dV = \\ \int_{0}^{\rho}\int_{0}^{\phi}\int_{0}^{\theta}\exp(-d(\rho \cos \theta \sin \phi, \rho \sin \theta \sin \phi,\rho \cos \phi)) \cdot \rho^{2}\sin \phi d\theta \, d\phi \, d\rho$
And then take the limit as $\rho \to \infty$.

The difficulty I think is determining what that function $d(x,y,z)$ is. I am thinking about it, and will update my post if I find it. I just wanted to be the first to respond even though I have an incomplete solution.

[b]sylow_theory[/b]: It is easier to divide the region of integration, since en different regions, the function $d$ is easily determined. For example it is 0 inside the qube. Now a point far away has the distance equal to the distans to the nearest corner, and so on.

[b]sos440[/b]: Thanks for writing so much  :) . I dont have time to check it now ,perhaps you missed some region. I'll look at it later.

Did you skip the integral inside the cube?

[hide="A solution"]

If $(x,y,z)\in D_{1}$ where $D_{1}=\{(x,y,z); \, y>1 \, x>1 \, z>1 \}$, then $(1,1,1)$ in the corner of the qube $\mathcal{Q}$ the nearest point, and therefore $d(x,y,z)=\sqrt{(x-1)^{2}+(y-1)^{2}+(z-1)^{2}}$. By symmetry all other [b]8[/b] regions
$\{(x,y,z); \, \pm y>1 \, \pm x>1 \, \pm z>1 \}$ give the same contribution to the integral as $D_{1}$.

If $(x,y,z)\in D_{2}$ where $D_{2}=\{(x,y,z); \, y>1 \, x>1 \,-1<z<1 \}$, then $(1,1,z)$ on the
side of the qube is the nearest point in $\mathcal{Q}$ and our dunction here is thus
$d(x,y,z)=\sqrt{(1-x)^{2}+(1-y)^{2}}$. Again by symmetry the [b]12[/b] regions
$\{(x,y,z); \, \pm y>1 \, \pm x>1 \,-1<z<1 \}$
$\{(x,y,z); \, \pm x>1 \, \pm z>1 \,-1<y<1 \}$
$\{(x,y,z); \, \pm y>1 \,\pm z>1 \,-1<x<1 \}$
 gives the same contribution as $D_{2}$

If $(x,y,z)\in D_{3}$ where $D_{3}=\{(x,y,z); \, x>1 \,-1<y<1 \,-1<z<1 \}$, then $(1,y,z)$ on the cube surface is the nearest point, and we end up this time with the function
$d(x,y,z)=x-1$. By symmetry all [b]6[/b] regions
$\{(x,y,z); \, \pm x>1 \,-1<y<1 \,-1<z<1 \}$
$\{(x,y,z); \, \pm y>1 \,-1<x<1 \,-1<z<1 \}$
$\{(x,y,z); \, \pm z>1 \,-1<y<1 \,-1<x<1 \}$
gives the same contribution to the integral as $D_{3}$

Finally as mentioned above by [b]sos440[/b] if $(x,y,z)\in \mathcal{Q}$ then $d(x,y,z)\equiv 0$

Now the integral becomes
\[\iiint_{\mathbb{R}^{3}}e^{-d(x,y,z)}dxdydz= 8\iint_{D_{1}}e^{-\sqrt{(x-1)^{2}+(y-1)^{2}+(z-1)^{2}}}dxdydz+\]

\[+12\iiint_{D_{2}}e^{-\sqrt{(1-x)^{2}+(1-y)^{2}}}dxdydz+6\iiint_{D_{3}}e^{-(x-1)}dxdydz+\iiint_{\mathcal{Q}}e^{0}dxdydz= \]

\[=\ldots=32+20\pi \]
[/hide]


Is it possible to find $100$ positive integers not exceeding $25000$ such that all pairwise sums of them are different?

[b]Lemma 1[/b] There exist $ p\minus{}1$ positive integers not exceeding $ 2p^{2}$ whose pairwise sums are different.
[b]Proof[/b] We will take the integers $ a_{1},a_{2},...,a_{p\minus{}1}$ where for $ i \equal{} 1,2,3,...,p\minus{}1$, $ a_{i}\equal{} 2pi\plus{}\{i^{2}\}$ where by $ \{k\}$ we denote the residue of $ k$ $ \pmod p$.
Assume we can find two distinct pairs of integers with equal pairwise sums, ie $ a_{x}\plus{}a_{y}\equal{} a_{z}\plus{}a_{t}$. Then:
$ 2px\plus{}\{x^{2}\}\plus{}2py\plus{}\{y^{2}\}\equal{} 2pz\plus{}\{z^{2}\}\plus{}2pt\plus{}\{t^{2}\}$    (*)
Because $ 2px\plus{}2py\leq a_{x}\plus{}a_{y}< 2p(x\plus{}y\plus{}1)$ as $ 0\leq\{x^{2}\}\plus{}\{y^{2}\}< 2p$, then $ a_{x}\plus{}a_{y}\equal{} a_{z}\plus{}a_{t}\Rightarrow x\plus{}y \equal{} z\plus{}t$, so $ x\plus{}y\equiv z\plus{}t\pmod p$
Now, substituting $ x\plus{}y\equal{}z\plus{}t$ into (*) we get $ \{x^{2}\}\plus{}\{y^{2}\}\equal{}\{z^{2}\}\plus{}\{t^{2}\}$.
But then, $ x^{2}\plus{}y^{2}\equiv z^{2}\plus{}t^{2}\pmod p$, and so:
$ 2(x^{2}\plus{}y^{2})\minus{}(x\plus{}y)^{2}\equiv 2(z^{2}\plus{}t^{2})\minus{}(z\plus{}t)^{2}\pmod p$ so:
$ (x\minus{}y)^{2}\equiv (z\minus{}t)^{2}\pmod p$.
Then, $ x\minus{}y\equiv z\minus{}t\pmod p$ or $ x\minus{}y \equal{} t\minus{}z\pmod p$, and if we use this and $ x\plus{}y\equiv z\plus{}t\pmod p$ we get  $ (x,y) \equal{} (z,t)$ or $ (x,y) \equal{} (t,z)$ because $ x,y,z,t\in\{1,2,...,p\minus{}1\}$. But this contradicts the assumption that the two pairs of integers are distinct, hence there do not exist two distinct pairs with equal pairwise sums, and lemma 1 follows.

Applying lemma 1 for $ p \equal{} 101$ and because $ 2*(101)^{2}\equal{} 20402 < 25000$ we get the result.


for the sake of completeness of the resources box at jbmo (the only missing is 2006) i will add here the links so that Valentin could put them there:

- problem 1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=99348[/url]

- problem 2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=99349[/url]

- problem 3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=99350[/url]

- problem 4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=99351[/url]

PS: the BMO box is incomplete too, tell me if you want to put the problem/links of the missing bmo, and i will conform :)


[quote="arqady"][quote="2424"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. prove that
 $ 1. \frac {1}{4 \minus{} ab} \plus{} \frac {1}{4 \minus{} bc} \plus{} \frac {1}{4 \minus{} ca} \leq \frac {1}{abc}$
 $ 2. \frac {1}{1 \plus{} 2ab} \plus{} \frac {1}{1 \plus{} 2bc} \plus{} \frac {1}{1 \plus{} 2ca} \geq 1$[/quote]
After substitution $ a \equal{} \frac {2x}{y \plus{} z},$ $ b \equal{} \frac {2y}{x \plus{} z}$ and $ c \equal{} \frac {2z}{x \plus{} y},$
where $ x,$ $ y$ and $ z$ are positive numbers, we obtain:
$ (1)\Leftrightarrow\frac {(x \plus{} y)(x \plus{} z)(y \plus{} z)}{2xyz}\geq\sum_{cyc}\frac {(x \plus{} y)(x \plus{} z)}{x(x \plus{} y \plus{} z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3y \plus{} x^3z \minus{} 2x^2yz)\geq0,$ which is obvious.
$ (2)\Leftrightarrow\sum_{cyc}\frac {x^2 \plus{} xy \plus{} xz \plus{} yz}{x^2 \plus{} xy \plus{} xz \plus{} 9yz}\geq1,$ which is true because
$ \sum_{cyc}\frac {x^2 \plus{} xy \plus{} xz \plus{} yz}{x^2 \plus{} xy \plus{} xz \plus{} 9yz}\geq\frac {\left(\sum(x^2 \plus{} 3xy)\right)^2}{\sum(x^2 \plus{} xy \plus{} xz \plus{} yz)(x^2 \plus{} xy \plus{} xz \plus{} 9yz)}\geq1,$
where the last inequality is equivalent to $ \sum_{cyc}(x^3y \plus{} x^3z \minus{} 2x^2yz)\geq0.$ Done![/quote]
Nice solution,arquady! :P
So,you can't prove this problem:http://www.mathlinks.ro/viewtopic.php?t=284038? :P


prove that the value of
\[ \prod_{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor}\left(3+2\cos\frac{2k\pi}{n}\right)\]
is n-th fibonacci number

here you have some nice links

http://en.wikipedia.org/wiki/Root_of_unity

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

http://planetmath.org/encyclopedia/RootOfUnity.html

[quote="Anaxerzia"][hide="Idea?"]
Try roots of unity.
[/hide][/quote]

I don't see it............could you write a sketch of solution?

[hide="Hint 1"] $ x^n \minus{} 1 \equal{} \prod_{k \equal{} 0}^{n \minus{} 1} (x \minus{} e^{i \frac {2 \pi k}{n}} )$ [/hide]
[hide="Hint 2"] $ (x \minus{} e^{i \frac {2 \pi k}{n} })(x \minus{} e^{ i \frac {2 \pi (n\minus{}k)}{n} }) \equal{} x^2 \minus{} 2 \cos \frac {2 \pi k}{n} \plus{} 1$ [/hide]
[hide="Hint 3"] Let $ \phi, \varphi$ be the two roots of $ x^2 \equal{} x \plus{} 1$.  Then

$ F_n \equal{} \frac {\phi^n \minus{} \varphi^n}{\phi \minus{} \varphi}$ [/hide]

[quote="Leonard"]prove that the value of
\[ \prod_{k \equal{} 0}^{\left\lfloor \frac {n \minus{} 1}{2}\right\rfloor}\left(3 \plus{} 2\cos\frac {2k\pi}{n}\right)
\]
is n-th fibonacci number[/quote]

This isn't true??

or

math has me crazy??

 :|  :maybe:

Indeed, it is false: for $ n \geq 1$, the product is actually $ 5f_n$ (with $ f_1 \equal{} f_2 \equal{} 1$).

yes, my mistake... this product should start from 1, not 0
sorry guys :oops: and thanks for hints :)


algebra
polynomial
quadratics
number theory unsolved
number theory

Show that

n^4+4

is composite for all positive integers n greater than 1

If n=5

n^4+4=629

and 5 doesn't divide 629.

I don't know how show that is a composite in general...

The polynomial $n^{4}+4$ factors into two quadratic polynomials.

@intejita: the identity of Sophie Germain states that
$a^{4}+4b^{4}=(a^{2}+2ab+2b^{2})(a^{2}-2ab+2b^{2})$ $\forall a,b$.


how do u cross the entrance to the fortress of the palace in the island which u get into first after straying on to the island...

uuhh... I'm not stranded on an island, unless the Americas make kup an island...

[quote="1=2"]I am. I don't know what game you are talking aobut, but please tell the NAME of the game at least....[/quote]
sigh..
the NAME of the game is the NAME of the thread...

thanks the name of the game is prince of persia warrior within . tried googling no  use


algebra
polynomial
linear algebra
linear algebra unsolved

I did not know wether to post this problem in Combinatorics section or here. So I posted here.
Prove that there are finitely many coonected graphs $G$ that eigenvalues of $G$ are in set $\{\lambda_{1},\lambda_{2},\dots,\lambda_{k}\}$.

This isn't exactly true (just consider the graphs with only isolated vertices; they all have all eigenvalues equal to $0$). Maybe the graphs we're interested in are supposed to be connected?

The graph should be connected.
I edited the problem.

what does it mean to say " eigenvalues of a graph G"? sorry i havent heard of it before, but i'm interested in knowing it. can someone explain please?

Eigenvalues of a graph are eigenvalues of its adjacency matrix.

thanks for the link. i'll try the problem

One may use the characteristic polynomial $\chi (G,\lambda)=\lambda^{n}+c_{2}\lambda^{n-2}+\dots+c_{n}$ to prove that $-c_{2}$ is equal to the number of the edges in a graph. Here since the graph is connected we're done.

[quote="Omid Hatami"]That's not right. Of course $-c_{2}=\mbox{tr}(A(G))=0$[/quote]$-c_{1}=\mbox{tr}(A(G))=0$ and I've taken it into account and not even mentioned it in the sum. My statement is based on the minors of the adjacency matrix.
(PS I've seen this problem before. Give me some time and I hope I'll modify my post later.)


First note that
$ \frac 1 9 \equal{} 0.111111111.....$ recurring decimal.
And also note that 0.999999999......( again recurring decimal ) is in fact equal to 1.
Then we may do a division by 9 as follows:
Suppose we have to calculate 87654 / 9:
Write
8888.8........
+777.7......
+  66.6......
+    5.5......
+    0.4......
Then add up to 9,739.333333333......( recurring decimal)

[b]Remark[/b]
First of all:
Remember to write tidily and align the corresponding digits and the decimal points correctly!
Secondly:
 The recurring part is obatained in this way:
8 + 7 + 6 + 5 + 4 ---> 30 = 9 + 9 + 9 + 3 ----> Carry 3 to the adjacent digit, that is the digit on left of the decimal point, and the recurring part should be 0.333333333........

uh...
what's this for??
any problems based on this??

and how is 0.9999..... equal to 1?
rounded up?

$ x\equal{}0.\bar{9}$.
$ 10x\equal{}9.\bar{9}$
Subtracting, $ 9x\equal{}9$, or $ x\equal{}1$.

hm... i might get it

x = 1
and,x = 0.99...
but using transitive property...
1 = 0.999....

that does not work

It's weird :wink:
A useful pattern for Mathcounts is that if the digits after the decimal start repeating after the kth digit, the fraction is the block of digits between  the decimal point and the kth digit, inclusive, and the denominator is k 9s.

So, if k was 5, and we have $ 0.\overline{12345}$, then it can be written as $ \frac {12345}{99999}$.

Explanation:

\begin{tabular}{crcr}&100000x&=&12345.\overline{12345}\\
-&1x&=&0.\overline{12345}\\\hline
&99999x&=&12345.00000
\end{tabular}

The same works for all other recurring decimals.

the denominator of a repeating decimal is,
x 9's and y 0's where x is the number of repeting decimals and y is the number of nonrepeating decimals. For example the denominator of 0.1(64repeating) is 990.

you know what
instead of thinking too hard about it
i'll just memorize it
put it in my brain......


[quote="heartwork"]Can you find a rational number $q$ such that:
$\sqrt{\frac{1+(1-4q^{2})\sqrt{1-q^{2}}}{2}}$
is also rational.
There is an infinte such set of rationals?[/quote]
No,there isn't any.
$\sqrt{1-q^{2}}$is rational $\rightarrow q=\frac{a^{2}-b^{2}}{a^{2}+b^{2}},or,q=\frac{2ab}{a^{2}+b^{2}}$where $(a,b)=1,a+b\equiv1(\mod2)$
case1$q=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
then $\sqrt{\frac{1+(1-4q^{2})\sqrt{1-q^{2}}}{2}}\rightarrow%Error. "display" is a bad command.
ab(5b^{2}-3a^{2})$is a prerfect square.$\rightarrow%Error. "display" is a bad command.
5a^{2}-3a^{2}$is a perfect square.
if $3\mid a$then $3\mid b$ a contradution.
otherewise$5a^{2}-3b^{2}\equiv 2 (\mod 3)$ which implies $5a^{2}-3b^{2}$cannot be a perfect square.
case 2
$q=\frac{2ab}{a^{2}+b^{2}}$
then $\sqrt{\frac{1+(1-4q^{2})\sqrt{1-q^{2}}}{2}}\leftarrow%Error. "display" is a bad command.
\frac12(a^{2}-b^{2})(a^{2}+b^{2}-8ab)$is a pefect square.but$\frac12(a^{2}-b^{2})(a^{2}+b^{2}-8ab)$ is not even a integer.

[quote="Hawk Tiger"]No,there isn't any.
case 1
$q=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
then $\sqrt{\frac{1+(1-4q^{2})\sqrt{1-q^{2}}}{2}}\rightarrow%Error. "display" is a bad command.
ab(5b^{2}-3a^{2})$is a prerfect square.
case 2
$q=\frac{2ab}{a^{2}+b^{2}}$
then $\sqrt{\frac{1+(1-4q^{2})\sqrt{1-q^{2}}}{2}}\rightarrow%Error. "display" is a bad command.
\frac12(a^{2}-b^{2})(a^{2}+b^{2}-8ab)$is a pefect square.but$\frac12(a^{2}-b^{2})(a^{2}+b^{2}-8ab)$ is not even a integer.[/quote]

I think you've made some misscomputations at least in the case 1, can you check again...

Yes ,I am so careless that I made a stupid mistake.


conics
hyperbola
Asymptote
analytic geometry
graphing lines
slope

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is:

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 2\text{ or }3 \qquad\textbf{(C)}\ 2\text{ or }4 \qquad\textbf{(D)}\ 3\text{ or }4 \qquad\textbf{(E)}\ 2,3,\text{ or }4$

I think it is E because if the lines are the asymptotes of the hyperbola it will be 2 intersections, if 1 is the asymptote it will be 3 intersections, and if both then 4.

@maththemonster: How can the lines intersect the hyperbola if they are the asymptotes of it? Is the definition of an asymptote different for a hyperbola? I thought the asymptote was a bound which is never reached.

[hide="My Solution"]
If the lines intersect inside the hyperbola, the total number of intersections will be $ 4$. If both lines are the same line, then there will be $ 2$ intersections. So the answer is $ \boxed{(D)}$.
[/hide]

Looking at this problem again, I see that I am wrong.

[hide="Corrected Solution"]
You can have two intersection points when one line is the transverse axis of the hyperbola and the other is the conjugate axis. You can have three interesection points when 1 line interesects at two points and another line intersects at one point (this occurs when the line has a slope very close to $ 0$). Finally, you can have four intersection points when the two lines intersect the hyperbola at two points each. Therefore, we have $ \boxed{(E)}$ as the answer.
[/hide]


parameterization
inequalities
inequalities proposed

x^6-6x^5+12x^4+ax^3-6x+1>=0 such that this equality, find the a parametr?

[quote="chinzoo"] $ x^6\minus{}6x^5\plus{}12x^4\plus{}ax^3\minus{}6x\plus{}1\geq0$ such that this equality, find the a parametr?[/quote]
 $ x\in R$?

I think that you forgot $ \plus{} 12x^2$!

This problem(with $ \plus{} 12x^2$) is from Bulgarin Spring Math Contest  1996.

Is the inequality holds for any real numbers $ x$, or for some $ x$?


Do AP Courses cover the same materials that are tested in SAT II? I am debating whether to take a SAT II test next January so there will one less thing to worry about when the AP Exams come along. If AP courses pretty much covers everything in SAT II, I may as well take them together. I'm specificaly talking about US History and Physics/Chemistry. Thanks for any suggestion.

i'm pretty sure that AP courses cover above and beyond what SAT II subject test cover. i haven't taken a real SAT II test but i've seen the material on the SAT II Bio test and it seems almost identical to the concepts covered in Hon. Bio. (I think...)

You probably want to wait and take them close to the same time if you plan to do the APs anyway. SAT II covers the subjects at a high school level, and AP covers at an introductory college level. So your AP preparation will help you on the SAT II. For some subjects, there are more topics on the AP than on the corresponding SAT II. For Physics, the SAT II and AP Physics B exam will cover more or less the same topics. But the AP Physics C exams cover fewer topics than the SAT II (but in much greater depth).

AP Chem just built on H. Chem, so it was only natural that it prepared me well for the Chem SAT II.  If you understand all the concepts after an H. Chem course and you do not plan on taking AP Chem, it would be reasonable to take the SAT II immediately following H. Chem.  Yeah, AP Chem does help with the SAT II, but I also second texas137's advice:

[quote="texas137"]You probably want to wait and take them close to the same time if you plan to do the APs[/quote]

[quote="anhnguyen"]I am debating whether to take a SAT II test next January so there will one less thing to worry about when the AP Exams come along.[/quote]

I'd recommend this (I did it, and it worked).
Preparing for the January SAT II's helped gloss over most of the material for the AP (at least superficially) so that there wasn't anything totally unfamiliar that I had to learn in April. The breadth of the material, I think, isn't much different in SAT II's and AP's--for example, the PR SAT II Physics book is completely identical to the MC portion of the PR AP Physics book.

And especially if you're taking many AP's, I'd recommend that you don't take the SAT's in May. Multiple standardized tests in a row can be very very tedious....

[quote="sky9073"]And especially if you're taking many AP's, I'd recommend that you don't take the SAT's in May. Multiple standardized tests in a row can be very very tedious....[/quote]

This year, I took an SAT II (USH) in May (I only had one left), right after the AP (everything was still fresh in my mind), and I took the SAT I in June.


Does anyone have the answer key for 1999 chapter competition?

I don't think anyone has the test itself, let alone the answer key.

I do...
Not necessarily the answer key file, just the answers is okay

Where do you get the 1999 Competitions? I could only find competitions going back to 2000.


I'm a mathematics student and past IMO/IOI medalist in Italy looking at pursuing a quantitative finance career.
Based on various sources, it seems to me that the financial firms that would value mathematical and computational skills the most (and thus presumably allow a mathematically gifted person to be most successful and have the best job satisfaction and compensation) would be hedge funds using trading strategies mainly based on mathematical modelling (like D.E. Shaw or Jane Street Capital, which are often mentioned here), as opposed to investment banks or less quantitatively based hedge funds.

Unlike other companies recruiting there in the financial sector (such as big investment banks), the firms I mentioned and other top places tend to be based only in the United States and seem to mostly look at US candidates, and I cannot find on their websites any information about opportunities for non-US students.

So I'd really like to know whether pursuing a career in a top quantitatively-oriented hedge fund would be realistic in the future for a foreign undergraduate student and how policies concerning them differ between internship candidates, candidates holding only non-doctoral degrees and candidates holding PhDs in recruiting decisions and immigration arrangements.
Is there anyone who is working or worked there who is knowledgeable about this?

Thank you very much in advance,
Luca

Why in the name of god do you want to go for a US job ?
Aside from having to deal with the Creationist controlled US immigration system, you have the legal right to move to the centre of world finance, which your nation founded some 2000 years ago. Later enough Italians financiers fled to London that the symbol for the British pound is actually Italian, and that chunks of the financial districts are named after places like Lombardy.

You have the legal right to work in London which has Europe's top 3 financial centres, (City, Canary Wharf and Mayfair), and is a train ride away from #4, Edinburgh.
If you're that good, there are plenty of top tier firms in London. If NY still appeals to you, then after a year, the banks will pay the appropriate fees to explain to the Creationists that an Italian Catholic is unlikely to be a member of AlQaeda using short words.

I'm a headhunter, so I have a professional interest in this.


Suppose that positive numbers x,y,z and xyz=1,we can get
∑1/√1+a≤3/√2
But how to prove the following with the same condition:

So sorry,I posted a wrong problem as I'm thinking about other problems...I have corrected now.

We need the condition $ \lambda \le \frac {5}{4}$ ;) Choose $ x \equal{} y \equal{} t,z \equal{} \frac {1}{t^2},$ then let $ t\rightarrow 0.$ ;)

P/s: To prove this inequality in this condition, you can try with Cauchy Schwarz. ;) :)

Could someone help me to solve this problem (the first picture)?


probability
number theory
prime factorization

[b]Positive integers from 1 to 30 inclusive are selected without replacement.  What is the least number of integers that must be selected to guarantee that the product of at least one pair is a perfect square?

 :huh: [/b]

[hide="all I have so far"]
the answer is greater than 10 because what happens if you get 2,3,5,7,11,13,17,19,23, and 29?
[/hide]

[hide]
Let's find the worst-case scenario.
Obviously, we can choose all the primes, leaving us with
$\{1,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30\}$to choose from.

Now let's look at the perfect squares in our list.  We can't choose more than one square, otherwise, if we multiply them we'll get a square, which is bad because we're trying to get the most possible numbers for ou worst cae scenario.  Let's just choose $1$, because it doesn't change much.

Now we have all the noncomposite numbers in our chosen set, while the possible remaining numbers are
$\{6,8,10,12,14,15,18,20,21,22,24,26,27,28,30\}$.
Look at $6$.  If we take $6$ into our set, we won't be able to take any other numbers with an odd number of $2$'s and an odd number of $3$'s and no other primes in its prime factorization, because when multiplied by $6$ it will give a perfect square.  So, we take $6$ and must eliminate $24$ from our list of possible takers. (we could take $24$ and leave $6$, instead.  no matter.)
Next, we can't take $8$ because of $2$.
$10$ is okay to take, there are no other numbers with an odd number of $2$'s and $5$'s.
$12$? Nope, we already have have $3$ in our set.
$14$ is good to take. $15$ too.
$18$ is a no-no because of $2$.
$20$ is bad because of $5$.
$21$ :) , $22$  :) , $26$  :) , $27$  :( , $28$  :) $30$  :)

So our worst case scenario set is
$\{1,2,3,5,6,7,10,11,13,14,15,17,19,21,23,22,26,28,29,30\}$
and if we add one more then guaranteed we get a square with some pair.  So our answer is
$\boxed{21}$.[/hide]

[quote="archimedes"]So our worst case scenario set is
$\{1,2,3,5,6,7,10,14,15,21,22,26,28,30\}$
and if we add one more then guaranteed we get a square with some pair.  So our answer is
$\boxed{15}$.[/quote]

Are you sure thats right?  What if we add 11..I dont see how that can affect it..or even 29 or soemthing like that.

Oh wow... heh...I forgot to put a bunch of primes in... :blush:


Prove that there is no natural number n, 99999<n<1000000 such that [n + (the number created by reversing the digits of n)] is prime.

[b]Lemma:[/b]
[i]For odd [b]k[/b],[/i]\[ 11|10^k\plus{}1\]
[b]Proof:[/b]
Working mod 11, $ 10^k\equiv(\minus{}1)^k\equiv\minus{}1$. Thus $ 11|10^k\plus{}1$.

Okay: back to the problem. The range you describe consists of all six digit numbers (if I read correctly that is). So:
\[ \ \  \ \text{abcdef}\\\plus{}\underline{\text{fedcba}}\]\[ N\]
where $ N$ is the number we are trying to prove something about.
In fact, \[ N \equal{} (a\plus{}f)(1\plus{}10^5) \plus{} (b\plus{}e)(10\plus{}10^4) \plus{} (c\plus{}d)(10^2\plus{}10^3)\]
Each number of the form \[ 10^a\plus{}10^{5\minus{}a}\equal{}10^a(1\plus{}10^{5\minus{}2a})\equiv 0\mod{11}\]
using the lemma. Thus, $ 11$ must divide $ N$. As $ N>>11$, $ N$ is not prime.
(Clearly this generalizes to any range with an even number of digits.)


Hi there,

I'm trying to get a fairly complex (and fairly large) tabular to fit on a single page without latex thinking that it's too big for the page. I've tried the following:

\setlength{\paperheight}{20in}
\setlength{\topmargin}{-1.25in}
\setlength{\textheight}{20in}
\textheight = 20in

I've also tried working with \table, \tabular, \tabular*, \longtable, and \minipage.

In every case, however, latex seems to think the table is too big for the page, and proceeds to push it back a page, thus creating a blank page in front of it, which is quite annoying. If I cut stuff off from the tabular to make it short enough there's no problem. Does anyone have any experience with this?

I've been working with this for ages, and would really, REALLY appreciate any help I can get.

Thanks!

Jordan

Try \small or \scriptsize or \tiny or one of those. Always works for me. And my tables tend to be quite long.
SSN.

Thanks SSN,

Making the text smaller does allow me to fit the tabular on the first page, and this may have to be the solution, but if possible I'd really like to keep the text size normal. What I want to do basically is extend the top and bottom margins for this one page (have maybe .2 inch margins, instead of the normal 1 inch). I need LaTex to recognize that the table actually DOES fit on the page that it's on.

Any other suggestions?

Jordan


[quote="knexpert"]I submitted my solutions on 10/7/06 for the first round. How long approximately will it take to log my solutions in your database?[/quote]
It can take as long as a week (though it usually doesn't) - don't worry about it until then.

We are in the process of sorting and logging them.  Our fax machine alone has gone through several reams of paper, not to mention all the emailed and mailed solutions.  Hopefully they will all be logged this week.

It has been almost a week and my solutions have still not been logged. Should I be worried?

Nope, don't worry about it.  Mine still hasn't been logged either. It takes a while to handle the approx 500 entries for the first round.  Give it at least another week before you start worrying.

Yeah... I forgot to put the right format in the subject line of the e-mail. I included most of the necessary info, but I forgot to use my ID number.

I hope that doesn't affect it in any way.


Is there a way to find out if the entry form you submitted was received? Because I think that mine might not have been since I faxed it in a couple days before the 2nd round deadline, and emailed the solutions then, but haven't gotten it graded, or the green check mark  :blush:

An admin can probably help you with that problem.  Just fyi though, you should always check for the checkmark or confirmation e-mail, because it's easier to fix the problem when it's caught early than late.

Thanks for your help. Yeah, I should have checked earlier but I sort of forgot about the contest, so I haven't even looked at Round 3 yet  :oops:


Prove neatly

 ((ab+bc+ca)/3)^(1/2) \geq (abc)^(1/3)

for a,b,c \geq 0

Come on Manlio, this is really too easy. One time AMGM and you are done.

Isn't this a well-known inequality already


Yes, that holds.

Let $ f(x)\equal{}x^2\minus{}x\minus{}\ln x\ (x>0),$ we have $ f'(x)\equal{}\frac{2x\plus{}1}{x}(x\minus{}1)$, yielding $ f(x)\geq f(1)\equal{}0.$
Thus $ f(x)\plus{}f(y)\plus{}f(z)\geq 0.$

$ \frac {2}{3}x^2 \plus{} \frac {1}{6}y^2 \plus{} \frac {1}{6}z^2\ge x^\frac {4}{3}y^\frac {1}{3}z^\frac {1}{3} \equal{} x^2$ by weighted AM-GM. Add up cyclically and get your inequality. =)

[quote="giannis"]can anyone tell me if holds $ x^2 \plus{} y^2 \plus{} z^2 > \equal{} x \plus{} y \plus{} z$ for $ xyz \equal{} 1$ ?

Thank you![/quote]

$ \frac {x^2 \plus{} y^2 \plus{} z^2}{x \plus{} y \plus{} z}\ge \frac {x \plus{} y \plus{} z}{3}\ge \sqrt[3]{xyz} \equal{} 1$

Why does everyone think so hard these days?! :roll:

But x,y,z are real numbers , why use AG-GM .

Well, exactly zero or two of the terms can be negative, and so we can replace them by their negative counterparts which makes the LHS stay the same and increases the RHS so the ineq. still holds. :)


function
integration
real analysis
real analysis unsolved

Given $ a < b$, show that there is an $ \psi\in C^{\infty}(\mathbb{R})$ such that $ \psi(x) \equal{} 0$ for $ x\leq a$, $ 0 < \psi(x) < 1$ for $ a < x < b$, and $ \psi(x) \equal{} 1$ for $ x\geq b$.

There is a hint, which says to consider $ \psi(x) \equal{} c\int_{ \minus{} \infty}^{x}\varphi$, where $ \varphi$ is a function such that $ \varphi(x) \equal{} 0$ for $ x\notin (a,b)$ and $ \varphi(x) > 0$ for $ x\in (a,b)$.

$ e^{\frac {1}{x^2 \minus{} 1}}$ for $ |x| \leq 1$, 0 outside. It is not hard to show that it is $ C^\infty$ and does what you want. It's a "bump" function. (To think of this particular example one usually first would think of the example $ e^{\frac { \minus{} 1}{t^2}}$ for $ t > 0$ and 0 otherwise. And by multiplying two such you get what's above.)

There is something similar in the idea of partitions of unity, as Kalle knows. That example is not exactly what you are looking for but it can be modified into a piecewise function satisfying what you seek.

At Kalle, are you choosing that "bump function" as the $ \varphi$ or $ \psi$?  Also note that my interval is not $ [\minus{}1,1]$, but $ (a,b)$.  How would I account for that?

Here is another approach. First we take a piecewise linear function which is 0 if $ x\le a\plus{}\epsilon$, 1 if $ x \ge b\minus{}\epsilon$. Then we can take the convolution to make that function smooth as desired.

@MegaRockman: convolution of your ( piecewise linear ) function with what ? :maybe:  And also notice that JRav what's it to be $ C^{\infty}$, I don't see how can you achieve that by convolution. I would be grateful if you could clarify.

Let $ \eta$ be a test function. $ \eta(x)\equal{}c.e^{\frac{1}{x^2\minus{}1}}$ if $ |x|<1$ and equals to 0 otherwise. The constant $ c$ can be chosen such that $ \int\eta(x)\,dx\equal{}1$. Such function is smooth.
For each $ \epsilon>0$, set $ \eta_\epsilon(x)\equal{}\epsilon^{\minus{}1}\eta(\frac{x}{\epsilon})$. We will take the convolution with this function.
First, choose $ f$ such that $ f$ is piecewise linear. So that, $ f$ is $ 0$ if $ x\le a\plus{}\delta$, $ \frac{1}{b\minus{}a\minus{}2\delta}x\minus{}\frac{a\plus{}\delta}{b\minus{}a\minus{}2\delta}$ if $ a\plus{}\delta\le x\le b\minus{}\delta$, and $ 1$ if $ x\ge b\minus{}\delta$
Then, we take the convolution of $ f$ with $ \eta_\epsilon$. Choose $ \epsilon$ small enough to ensure that the function after convolution satisfies the required properties (maybe $ \epsilon<\delta$. )


Find the remainder when this number is reduced modulo 2004.
   $1!.5 + 2!.11 +...+k!(k^2 + 3k + 1) + 200!.40601$
    Astyan

maybe this can help ?

[hide]
$k!(k^2+3k+1)=(k+2)!-k!$
[/hide]


let a,b,c>0 such that ab+bc+ca=1  prove that
$ \sum\frac{1\plus{}b^2c^2}{(b\plus{}c)^2} \geq \frac{5}{2}$

[quote="tranquoc"]let a,b,c>0 such that ab+bc+ca=1  prove that
$ \sum\frac {1 \plus{} b^2c^2}{(b \plus{} c)^2} \geq \frac {5}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57683