Wow, how crazy :). And where exactly is this vertical wall? Or do you think it doesn't matter?

The problem doesn't mention where the vertical wall is, but if you want, you can consider some special positions.

The new height of $C.G.$ is

\[ \dfrac{ \varrho_{w} 2a^2 (a-x)^2 + \varrho_{Hg} 2a^2 x^2 } { \varrho_{w} 2a^2 (a-x) + \varrho_{Hg} 2a^2 x } \] where $x$ is the horizontal distance between one edge and vertical wall on $Hg$ side . The initial $C.G.$ is at a height = $a$

[quote="perfect_radio"]We have a parallelipiped (on Earth) of dimensions $a \times a \times 2a$ ($\textrm{length} \times \textrm{width} \times \textrm{height}$). It has $2$ compartiments of equal volumes, separated by a vertical wall. In one of them there is $H_2 O$ and in the other $Hg$. What happens with the system after we remove the separating wall? What happens with the center of gravity of the $2$ liquids? (If it modifies, calculate the distance between the old and the new one -- on vertical that is)

Furthermore, what happens if we move the system (with the separating wall) in space? How does it look? And if we remove the wall, what then?[/quote]Given that the water and the mercury each occupy equal volumes there is only one position for the vertical wall. Further with such a wall in place, given that the water and the mercury are each homogeneous, the center-of-mass is at the vertical midpoint of the parallelipiped, a distance $a$ up from the bottom.

Removing the wall allows the water and mercury to mix. In steady state the water will float on top of the mercury. The water's COM will be at $\frac32a$, and the mercury's COM will be at $a$. The COM of the mixture, $y_{cm} =\dfrac{m_{H_2O}\left(\frac32a\right) + m_{Hg}\left(a\right)}{m_{H_2O}+m_{Hg}}$

Since they occupy equal volumes $\dfrac{m_{Hg}}{m_{H_2O}}=\dfrac{\rho_{Hg}}{\rho_{H_2O}} = 13.5: 1$.

$\therefore y_{cm} =\dfrac{1.5m_{H_2O} + 13.5m_{H_2O}}{m_{H_2O}+13.5m_{H_2O}}a =\dfrac{30}{29}a$.

[b]Edit:[/b] It occurs to me that I did not solve the 2nd part of the problem. In outer space, without gravity, no longer subject to Archimedes' Principle, both fluids are free to mix randomly. In doing so they shall come into an overall uniform distribution. Orignally the fluids were partitioned with each neatly occupying one half the total volume. If we stipulate then that $\rho_{H_2O} = 1$, the total mass, $M_{total} = 14.5a^3$. The total volume, $V_{total} = 2a^3$ and therefore, $\rho_{mix} = \dfrac{M_{total}}{V_{total}} = \dfrac{14.5a^3}{2a^3} = 7.25$.

Then $y_{cm_{new}} =\dfrac{1}{M_{total}}\int_{0}^{\nu} \rho_{mix}y\; \mathrm{d}V = \dfrac{a^2}{M_{total}}\int_{0}^{2a} 7.25 y\; \mathrm{d}y =\dfrac{7.25a^2}{14.5a^3}\frac12y^2\vert_{0}^{y = 2a}$

$\Rightarrow y_{cm_{new}}= \dfrac{14.5a^4}{14.5a^3} = a$.

[quote="perfect_radio"][quote="Dr. No"]Removing the wall allows the water and mercury to mix. In steady state the water will float on top of the mercury. The water's COM will be at $\frac32a$, and the mercury's COM will be at $a$.[/quote]

I think that the mercury's COM will be at $\frac{a}2$. Am I wrong?[/quote]
No, I am wrong. Originally I had integrated over the whole volume, not just the half that the mercury occupied at first... shame on me.  :oops:

It's not such a big mistake.
I just wanted to be sure I understood your solution. (It has happened many times to not understand a solution and then, after some time -in this case almost a year-, to be able to follow it easily.)

I remember that my teacher said something about a sphere and minimal potential energy. But I'm not really sure.

[quote="Vader"]Hi, No, i disagree with the solution in space.

Dr. No, You did everything humanly possible to solve that problem, but
remember: the intermolecular forces between the mercury and the water is not of comparable magnetude.

Thus, the mercury will form a ball somewhere in the container, and the water will form the rest.  This position is however relatively random,

Proof for this :  Look at the effect of spilt goods (Such as beverages) in spacecrafts, this doesn't mix with the controled enviroment but rather make a small ellipsoid like figure.[/quote]
Lord Vader,

I concur. I had simply not thought to account for surface tension (Van der Waals forces) on the liquids in question.


solve the following equation for positive integers x and y.

$x^{3}-y^{3}=xy+79$

[quote="DiscreetFourierTransform"][hide="Answer?"](sorry for no solution, I just graphed and guessed (not a rigorous solution whatsoever)) $x = 14$, $y = 14$?[/hide][/quote]

Ummm...I don't think that is the right answer...

Put $x = y+a$.  Expand.  It becomes $(3a-1)(y)(y+a) = 79-a^{3}$.  Then the LS is positive, so $a < 5$.  Also $a > 0$.  Checking a = 1,2,3,4; there is no solution.

[hide="Solution"]Since RHS is positive, the LHS must be too, therefore $x>y\implies x-y\geqslant 1$.

$xy+79=(x-y)(x^{2}+xy+y^{2})=(x-y)[(x-y)^{2}+3xy]$

Substituting $a=x-y,b=xy$ we get

\begin{eqnarray*}a(a^{2}+3b)=b+79\iff a^{3}+3ab=b+79\iff b(3a-1)=79-a^{3}\iff b={79-a^{3}\over 3a-1}\end{eqnarray*}

Since the denominator is positive, the numerator must be too, hence $a\leqslant 4$. Plugging $a=1,2,3,4$ we get an integer $b$ only for $a=1$, giving $b=39$, but the system $x-y=1, xy=39$ has no solution in integers.

Therefore, the initial equation has no solution.[/hide]


This appears to be equal to the multinomial coefficient

$ \binom{2^{2n}\plus{}2^{m\plus{}n}\plus{}2^{2m}}{ 2^n, 2^n, \ldots 2^n, 2^m, 2^m, \ldots 2^m }$

where on the bottom, the $ 2^n$ appears $ 2^n \plus{} 2^{m\minus{}1}$ times and the $ 2^m$ appears $ 2^m \plus{} 2^{n\minus{}1}$ times. Since $ 2^n(2^n \plus{} 2^{m\minus{}1}) \plus{} 2^m(2^m \plus{} 2^{n\minus{}1}) \equal{} 2^{2n} \plus{} 2^{m\plus{}n} \plus{} 2^{2m}$, everything adds up.

And multinomial coefficients are always integers so we're done.


search
number theory proposed
number theory

Find $(a,b,c)$ positive integers such that $a^2, b^2$ and $c^2$ are in arimetic progression in that order.

[quote="t0rajir0u"]Posted before, search for it (no solutions, right?).[/quote]
$a^2,b^2,c^2,d^2$ in nontrivial progression has no solutions.


inequalities
function
inequalities unsolved

[i]Find the greatest value of the expression[/i]
[i] T[/i] = $ \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {b^2 \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {c^2 \plus{} 1}{a^2 \plus{} 1}$
   [i]where a,b,c are nonnegative real numbers such that[/i] $ a\plus{}b\plus{}c\equal{}1$

  [i]       Love An Forever[/i]

[quote="nkht-tk14"][i]Find the greatest value of the expression[/i]
[i] T[/i] = $ \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {b^2 \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {c^2 \plus{} 1}{a^2 \plus{} 1}$

  [i]       Love An Forever[/i][/quote]
Try $ a\rightarrow\plus{}\infty$ and $ b\equal{}c\equal{}0.$ :wink:

Well, I think that with $ a\rightarrow \infty$ there would be a problem with $ a\plus{}b\plus{}c\equal{}1$ dear [b]arqady[/b] :wink:

[quote="polskimisiek"]Well, I think that with $ a\rightarrow \infty$ there would be a problem with $ a \plus{} b \plus{} c \equal{} 1$ dear [b]arqady[/b] :wink:[/quote]
Dear polskimisiek, nkht-tk14 has fixed it.

Aaah, yeah, sorry [b]arqady[/b]. Returning to the original problem, it seems harder to find the maximum, than to find minimum (which is clearly 3). Any ideas?

Let $ a\leq b\leq c$.
Alsow, let $ x \equal{} a^2 \plus{} 1$, $ y \equal{} b^2 \plus{} 1$ and $ z \equal{} c^2 \plus{} 1$.

[b](*)[/b] So $ 1\leq x\leq y\leq z\leq 2$

$ T(x,y,z) \equal{} \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}$

We have:
$ \frac {x}{y} \plus{} \frac {z}{x}\leq \frac {1}{y} \plus{} z$. This is true if $ 1\leq x\leq yz$. This is obviosly true because of [b](*)[/b]

$ T(x,y,z)\leq \frac {1}{y} \plus{} \frac {y}{z} \plus{} z$

Second, easy to prove that:
$ \frac {1}{y} \plus{} \frac {y}{z}\leq 1 \plus{} \frac {1}{z}$ This inequality is true because $ 1\leq y\leq z$.

$ T(x,y,z)\leq \frac {1}{y} \plus{} \frac {y}{z} \plus{} z\leq 1 \plus{} \frac {1}{z} \plus{} z$

Now finally we have: $ \frac {1}{z} \plus{} z\leq\frac {5}{2}$. This is true, because $ z\leq2$.

$ T(x,y,z)\leq \frac {1}{y} \plus{} \frac {y}{z} \plus{} z\leq 1 \plus{} \frac {1}{z} \plus{} z\leq 1 \plus{} \frac {5}{2} \equal{} \frac {7}{2}$

Equality holds if $ \{x,y,z\} \equal{} (1,1,2)\rightarrow \{a,b,c\} \equal{} (0,0,1)$.

Answer:
\[ \boxed{\frac {a^2 \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {b^2 \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {c^2 \plus{} 1}{a^2 \plus{} 1}\leq \frac {7}{2}}
\]

[i]You can not assume $ a \leq b \leq c$
  The ineq is not symetric[/i] :maybe:

I don`t understand why we can`t write $ a\leq b\leq c$  :oops:

[color=green]@Yuriy Solovyov : If $ T\equal{}f(a,b,c)$ then $ f(a,b,c)\not\equal{} f(a,c,b)$ for all $ b,c$ so we can assume $ a \leq b \leq c$.[/color]

I see, thanks. If for example $ T(a,b,c) \equal{} ab \plus{} c$ and $ a \plus{} b \plus{} c \equal{} 1$. We can`t say that $ a  \leq b\leq c$. :)

But you need only minor modifications to save your proof:

The above function $ T(a,b,c) , a,b,c \in [0,1]$ attains its maximum when $ (a,b,c)$ is
a permutation of $ (0,0,1)$.

Proof: $ T$ is not symmetric, but cyclic, so w.l.o.g. we may assume $ a \equal{} \min(a,b,c)$.
Then it is easy to see that $ T(a,b,c) \le T(0,b,c)$.

If $ b \le c$, then it is easy to see that $ T(0,b,c) \le T(0,0,c) \le T(0,0,1) \equal{} 7/2$.
Similarly, if $ c \le b$, then it is easy to see that $ T(0,b,c) \le T(0,b,0) \le T(0,1,0) \equal{} 7/2$.

Nice problem!

Yes, we just must to se two cases for $ a \equal{} \min\{a,b,c\}$: $ b\leq c$ and $ c\leq b$. :-)


number theory
least common multiple
parameterization
relatively prime
number theory unsolved

there are infinitey many $ x,y,z$ such that
\[ x^2\plus{}y^2\equal{}z^2\]

there are no $ x,y,z$ such that
\[ x^n\plus{}y^n\equal{}z^n\]
where $ n\ge 3$

natural questions: what about equations $ x^2\plus{}y^2\equal{}z^n$ or $ x^n\plus{}y^n\equal{}z^2$ (of course $ n\ge 3$) ?

for which $ n$ there are infinitely many solutions,
for which $ n$ there are finitely many solutions but at least one,
for which $ n$ there are no solutions.

is it known?

Two relevant conjectures are the [url=http://mathworld.wolfram.com/Fermat-CatalanConjecture.html]Fermat-Catalan conjecture[/url] and [url=http://en.wikipedia.org/wiki/Beal%27s_conjecture]Beal's conjecture[/url].

For most equations of this type it is straightforward to find many solutions.  For example, it's straightforward to characterize all numbers of the form $ x^2 \plus{} y^2$, hence the first question is trivial (there are infinitely many pairwise relatively prime solutions for all $ n$).  A large class of answers to a generalization of the second is the following

[b]Proposition:[/b]  Suppose $ a, b, c$ are positive integers such that $ c$ is relatively prime to $ \text{lcm}(a, b)$.  Then there are infinitely many "nontrivial" solutions to $ x^a \plus{} y^b \equal{} z^c$.

[i]Proof.[/i]  By Bezout's theorem there exist integers $ p, q$ such that $ p \cdot \text{lcm}(a, b) \plus{} 1 \equal{} qc$.  Now for any positive integers $ u, v$ set

$ x \equal{} u(u^a \plus{} v^b)^{\frac {p \cdot \text{lcm}(a, b)}{a} }$
$ y \equal{} v(u^a \plus{} v^b)^{\frac {p \cdot \text{lcm}(a, b)}{b} }$
$ z \equal{} (u^a \plus{} v^b)^{q}$.

A similar construction is possible if $ b$ is relatively prime to $ \text{lcm}(a, c)$.  The case $ \gcd(a, b, c) \ge 3$ is handled by Fermat's last theorem, but I don't know what happens if $ \gcd(a, b, c) \equal{} 2$ or $ \gcd(a, b, c) \equal{} 1$ but $ \gcd(a, b), \gcd(b, c), \gcd(a, c) > 1$.

[b]Edit:[/b]  Note that it's impossible for there to be at least one but finitely many solutions, since if $ (x_0, y_0, z_0)$ is a solution then so is $ (t^{bc} x_0, t^{ac} y_0, t^{ab} z_0)$ for any integer $ t$.  You can either require relative primality (which is an extremely strong condition; I don't think much is known about this case) or the weaker condition that the above simplification isn't possible.

Thanks for interesting answer. Yes, I know which numbers are sums of two squares so the first one is completely solved.
For the second one, if we know some (here even infinitely many) solutions, there is a natural question about all solutions - is every nontrivial solution of the form you wrote (for some $ u,v$) ?

The solutions listed in the article on the Fermat-Catalan conjecture are relatively prime, and so aren't of this form.

so is it an open problem ?
 (I mean all nontrivial solutions to $ x^n \plus{} y^n \equal{} z^2$, the special case of http://mathworld.wolfram.com/Fermat-CatalanConjecture.html for $ p \equal{} q>2$ and $ z \equal{} 2$)

Ah.  Well, one can perturb the argument I implicitly used to arrive at the parameterization I gave to give a slightly more comprehensive one.  Suppose $ n$ is odd and $ x, y$ have a common factor $ f$, so write $ x \equal{} fx', y \equal{} fy'$ and

$ x'^n \plus{} y'^n \equal{} \frac {z^2}{f^n}$.

Suppose further that $ z \equal{} f^{\frac {n \plus{} 1}{2} } z'$ (this doesn't cover all cases if $ f$ is composite, but remember that we're not trying to get all solutions here); then

$ x'^n \plus{} y'^n \equal{} f z'^2$.

The original parameterization I gave comes from the case $ z' \equal{} 1$.  For more general solutions, let $ x', y'$ be arbitrary and coprime such that $ x'^n \plus{} y'^n$ isn't squarefree and pick $ z'$ to be maximal.  This gives infinitely many solutions not of the form I gave if and only if there exist infinitely many numbers of the form $ x'^n \plus{} y'^n$ for coprime $ x', y'$ which aren't squarefree.  Heuristically this should be the case.

For example, $ 2^5 \plus{} 3^5 \equal{} 275 \equal{} 5^2 \cdot 11$, so setting $ z' \equal{} 5, f \equal{} 11$ we obtain the solution $ (x, y, z, n) \equal{} (22, 33, 5 \cdot 11^3, 5)$.

Of course any solution of this form relies crucially on the fact that $ f > 1$, which is why the relatively prime case is so hard.  (It's also why Fermat's Last Theorem is so hard.)


search
special factorizations
number theory unsolved
number theory

Intergers $ x, y, z$ satisfies the condition $ x^3\plus{}y^3\equal{}2z^3$.

Does $ x\equal{}y\equal{}z$ is always true? Why or why not?

No, not always true.

Take $ (x,y,z) \equal{} (1,\minus{}1,0)$.

Hi;

Looks like there are an infinite number of solutions of the form:

{x = n ,y =  - n , z = 0 }

Take the affine slice $ x \plus{} y \equal{} 1$; this gives the curve $ 1 \minus{} 3x(1 \minus{} x) \equal{} 2z^3$.  Completing the square, this gives $ (6x \plus{} 3)^2 \plus{} 3 \equal{} 24z^3$, and letting $ u \equal{} 24^2 (6x \plus{} 3), v \equal{} 24z$ this gives the [url=http://en.wikipedia.org/wiki/Mordell_curve]Mordell curve[/url]

$ u^2 \equal{} v^3 \minus{} 3 \cdot 24^3$.

Rational points on this curve tell us all the integer points on the original curve, and to determine the rational points on this curve you should find a computer algebra system that will tell you what the rank of this elliptic curve is.  If it's zero, all of the rational points are torsion, and then you can apply [url=http://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem]Nagell-Lutz[/url].  If the curve has positive rank then things are more difficult.

Oh, I am sorry. $ x, y, z$ are positive intergers. :blush:

WLOG suppose (x,y,z)=1,
substitute x=a+b , y=a-b (where a>b)

it is easy to show 0<a-b=y<z<x=a+b<2a----(*)

The equation becomes
a^3+3ab^2=z^3
a(a^2+3b^2)=z^3
=> a divides z

By (*) we can deduce z=a, b=0
we are done.

[quote="KWTLEO"]...The equation becomes
a^3+3ab^2=z^3
a(a^2+3b^2)=z^3
=> a divides z
....[/quote]

No, you can only conclude $ a|z^3$ and not $ a|z$. For example $ 9|15^3$ does not imply $ 9|15$

Well, is it too hard? Can somebody help me?

According to the search of internet, I learned that the only solution is $ x\equal{}y\equal{}z$.

Please help me...

I am thinking about the equation $ x^3 \plus{} y^3 \equal{} 2$, where $ x, y$ are rational numbers.

If $ (a, b)$ is a solution of this equation that satisfies $ a\ne b$, I just found out that $ (\frac {a(a^3 \plus{} 2b^3)}{a^3 \minus{} b^3}, \minus{} \frac {b(2a^3 \plus{} b^3)}{a^3 \minus{} b^3})$ is a solution too.

Hmmm, I'm not sure if you've seen it but this site [url]http://www.multimagie.com/English/SquaresOfCubes.htm[/url] says there's no solutions to $ x^3 \plus{} y^3 \equal{} kz^3$ for k=1,2,3,4,5 and I think it might be pretty hard since it was first shown by Euler.

I have been searching for a solution of this problem for a long time. Also, recently a post was made asking this question but it wasn't answered. I couldn't find the thread but instead found this one. So here goes the solution. Peace.
Faustus
 :yinyang:


What a block shot!!!!!

http://www.youtube.com/watch?v=N0NvKyWYAA0

that's pretty SICK...haha I remember some of those like dwyane wade's block on amare two years ago.

The video was awesome... Just awesome...

A note: The song does use some language, so you have been warned.

Yeah this is a great video.
To those who do not want to listen to bad language, just turn off ur speakers and enjoy the video. :wink:

Woah! I LOVED the song! And the action made me want to play water polo-too bad I have to wait...till tonight...

i remember watchign that dwayne wade one last year. ownage ;O


Determine the vector $ \vec {v}$, orthogonal to the vector $ \vec {u} \equal{} (2, \minus{} 3, \minus{} 12)$, and  collinear  to the vector $ \vec {w} \equal{} ( \minus{} 6,4, \minus{} 2)$

[quote="brunA"]Determine the vector $ \vec {v}$, orthogonal to the vector $ \vec {u} \equal{} (2, \minus{} 3, \minus{} 12)$, and hills to the vector $ \vec {w} \equal{} ( \minus{} 6,4, \minus{} 2)$[/quote]
What does "hills" mean?

I was thinking it meant parallel, but then like...any vector of the form $ (\minus{}6k,4k,\minus{}2k)$ would work, so I think it means something else.


calculus
derivative
vector
function
Functional Analysis
advanced fields
advanced fields unsolved

Do you know a book or other source which deals with optimization problems in a general (finite dimensional) Hilbert space? I'm interested in problems like the one I put below.
A, by the way, what books which cover the Frechet differentiation in a Hilbert space do you know?

Let $ H$ be a Hilbert space and $ a_{1},...a_{n}$ some unit vectors from $ H$.
a) Prove that the minimum value of the function $ f: H\rightarrow R$, $ f(x)\equal{}\left\Vert x\minus{}a_{1}\right\Vert ^{2}\plus{}...\plus{}\left\Vert x\minus{}a_{n}\right\Vert
^{2}$ is attained for $ x\equal{}(a_{1}\plus{}...\plus{}a_{n})/n.$
b) What about the function $ f: H\rightarrow R$, $ f(x)\equal{}\left\Vert
x\minus{}a_{1}\right\Vert \plus{}...\plus{}\left\Vert x\minus{}a_{n}\right\Vert ?$
c) Find the minimum of the function $ f$ from point a) under the constraint
that $ \left\Vert x\right\Vert \equal{}1.$

Please, if anyone knows a book/course notes on the web. I'm quite in a hurry.


Consider a collection $X$ of $k$ integer arithmetic progressions $\{a_i  +nd_i \mid n \in \mathbb{Z}\}$. It is obvious that if $M$ is the least commong multiple $[d_1,\ldots,d_k]$ of the differences of these progressions, and if $X$ covers some $M$ [i]consecutive[/i] integers (i.e. there is an $x$ such that each of the numbers $x+1,\ldots,x+M$ belongs to some progression in $X$), then $X$ is a covering system, i.e. it covers [i]all[/i] integers.

Now the remarkable problem: [i]prove that if $X$ covers some $2^k$ consecutive integers, then it covers all integers.[/i]

This is remarkable because the bound $2^k$ depends only on $k$ and [b]not[/b] on the $d_i$!

Enjoy!  :P

--Vesselin

[i]Hint: [/i] You may find it helpful to look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19262]here.[/url]

Indeed, this is truly amazing, but what is even nicer is that using the hint you gave the problem has a few-lines solution!!! If the following is true, than really I should ask: how on earth could Erdos miss that? Suppose that we have $k$ progressions $a_{i}Z+d_{i}$ and for some integer $x$ all $x+t$ for $0\leq t\leq 2^{k}-1$ are covered. Then we know that for all such $t$ we have
\[\prod_{1\leq j\leq k}{(1-e^{\frac{2i\pi}{d_{j}}(x+t-a_{j})})}=0.\] Develop this product and observe that you can write the condition in the form $\sum_{I}{\alpha_{I}\cdot e^{(x+t)\beta_{I}}}=0$, where the sumation is made after all subsets of ${1,2,...,k}$ (including the void set, in case of which the term is $1$), $\alpha_{I}$ and $\beta_{I}$ depend only on the cover itself. We need to prove that the same relation holds for any integer $y$ instead of $x$. Let $z_{I}=e^{\beta_{I}}$. Then we know that $\sum_{I}{\alpha_{I}z_{I}^{t+x}}=0$ for all $0\leq t\leq 2^{k}-1$. Let $u_{n}$ be $\sum_{I}{\alpha_{I}z_{I}^{n}}$. Then $u_{n}$ satisfies a linearly recurent relation of order $2^{k}$, the coefficient of $u_{n}$ being nonzero (that is, $u_{n+2^{k}}+A_{2^{k}-1}u_{n+2^{k}-1}+...+A_{1}u_{n+1}+A_{0}u_{n}=0$ for some nonzero $A_{0}$, which equals up to a sign the product of the $z_{I}$) and we know that $2^{k}$ consecutive terms of this sequence vanish. Thus all of them vanish and this finishes the proof. Where is the mistake???


geometry
trigonometry
Pythagorean Theorem
projective geometry

[quote="srulikbd"]thanks a lot!  :)
0 I didn't understand
[hide="1"]pythagorean theorem in four triangles.[/hide]
[hide="2"]the same as well as the previous one, or you want us to prove it for non convex quadrilateral?[/hide]
'waht does a.s.o stand for?[/quote]

Problem 0 (It's really strange to say problem 0):
Let $M$ be a constant such that $LA^{2}-LB^{2}=M^{2}$. Then let N be a point on AB such that $AN^{2}-BN^{2}=M^{2}$, then by Pythagorean theorem, the locus of L is just the line through N and perpendicular to AB.

So we have to solve the above problems in this thread?

srulikbd wrote pythagorean theorem in four triangles.
This is only to show that if $AC \perp BD \Rightarrow AB^{2}+CD^{2}=AD^{2}+BC^{2}$
Let diagonals $AC$ and $BD$ intersect in $O$ and $\angle AOB=\delta$ then $AB^{2}= AO^{2}+BO^{2}-2AO*BO \cos \delta$
$DC^{2}=DO^{2}+OC^{2}-2DO*OC \cos \delta$ so $AB^{2}+CD^{2}=AO^{2}+BO^{2}+DO^{2}+OC^{2}-2(AO*BO+DO*OC)\cos \delta$.
Simillary $AD^{2}+BC^{2}=AO^{2}+DO^{2}+BO^{2}+OC^{2}-2(AO*DO+BO*OC)\cos (180-\delta$). And as $AB^{2}+CD^{2}=AD^{2}+BC^{2}$ this means $-2(AO*BO+DO*OC)\cos \delta = 2(AO*DO+BO*OC)\cos \delta$ which is possible only when $\delta = 90^\circ$.

Actually, the locus in problem 0 implies the $\Longleftarrow$ part of problem 1.

In problem 0, are points  $A,B,L$ in the same plane? or is this in 3D?

EDIT: here's a proof of number 2:
Let $AB$ intersect $CD$ at $O$. $AB\perp CD$, so $\angle{AOC}= \angle{COB}=\angle{DOB}=\angle{AOD}=90^{o}$.Using Pythagorean theorem, obtain:
$CA^{2}-AD^{2}=CO^{2}-OD^{2}=CB^{2}-BD^{2}$, qed.
NOw to prove the $\Leftarrow$ part, we assume $AB$ intersects $CD$ at $O$ and then use Cosine Law in triangle $AOC,BOC,BOD,DOA$ and then we will get:
$(AC^{2}-AD^{2})-(CB^{2}-BD^{2})=0$; simplyfying, obtain:
$2AO(OC)cos(COA)+2BO(OD)cos(COA)+2BO(OC)cos(COA)+2AO(OD)cos(COA)=0$; this is true iff $cos(COA)=0$, ie $\angle{COA}=90^{o}$, so $AB\perp CD$. QED

[quote="rem"]In problem 0, are points  $A,B,L$ in the same plane? or is this in 3D?[/quote]
I think that when we do not say space geometry in the problem statement, we are often talking about plane geometry.

[color=darkblue]Sorry, [b]Rem, you are right ![/b] I didn't mentionated that here all problems belong to the plane geometry. If you wish, you can solve in space. For example, in the 0-problem the locus is a plane which is perpendicularly on the line $AB$.[/color]

i apologize for being off topic, but does harmonic division have anything to do with harmonic means or sequences? :blush:

EDIT: nevermind, i found it has nothing to do with either of those. For anyone as confused as i was:
http://en.wikipedia.org/wiki/Harmonic_division
Sorry for the interuption, please continue with the marathon. :)

Proof of 6:
We will use $T.2.$
$MA^{2}+NB^{2}=MN^{2}\Longleftrightarrow MN^{2}-MA^{2}=BN^{2}$;
$BN^{2}=IN^{2}-BI^{2}=IN^{2}-IA^{2}$
So $MA^{2}+NB^{2}=MN^{2}\Longleftrightarrow MA^{2}+NB^{2}=IN^{2}-IA^{2}$
From $T.2.$ obtain $MA^{2}+NB^{2}=IN^{2}-IA^{2}\Longleftrightarrow IM\perp AN$, so $MA^{2}+NB^{2}=MN^{2}\Longleftrightarrow IM\perp AN$, qed(first part)
Also, $MA^{2}+NB^{2}=MN^{2}\Longleftrightarrow MN^{2}-BN^{2}=MA^{2}$
Have $MA^{2}=IM^{2}-IA^{2}=MI^{2}-BI^{2}$
Therefore $MA^{2}+NB^{2}=MN^{2}\Longleftrightarrow MN^{2}-BN^{2}=MI^{2}-BI^{2}$
From $T.2.$ obtain $MN^{2}-BN^{2}=MI^{2}-BI^{2}\Longleftrightarrow IN\perp BM$ , so $MA^{2}+NB^{2}=MN^{2}\Longleftrightarrow IN\perp BM$, qed(second part).
EDIT: This is the proof for number 6 btw

[color=darkred]I am waiting the proofs of the all problems. Success !.
[/color]

This should be made into a sticky so doesn't keep getting moved down. Also, its a marathon! Post people! :D
I will follow my advice soon, i hope.

What is wrong? The posted problems and solutions disappeared somewhere... :maybe:

yeah the second page. People just aren't up to the challenge. :o

[quote="M4RI0"]when did the solutions dissapear :?:[/quote]
I don't know...Virgil's problems disappeared too. Edited 59 times... :maybe:

I don't know exactly. I remember that Virgil Nicula posted the solutions last week or the week before that, then I never visited this thread. And today when I decided to return to it, all the solutions Virgil Nicula posted and the problems disappeared...

i think virgil can explain, this is wierd.

[quote="Inspired By Nature"] this is wierd.[/quote]

Weird your "wierd"! (waiting M.Nicula...) :maybe:

[quote="Inspired By Nature"][quote="mstoenescu"][quote="Inspired By Nature"] this is wierd.[/quote]
Weird your "wierd"![/quote]
 :?: What do you mean? I was refering to the missing text.[/quote]

He was referring, I believe, to your misspelling.

[quote="Inspired By Nature"]Virgil, you should post some news so everyone knows.[/quote]
I don't think he really visits tis topic since he has not replied. I suggest just pming him. :wink:

I did, a long time ago.
[quote="Virgil Nicula"]Yes, I know and I am very aggrieved I worked many years for this topic. I don't know how it disappeared. I am sorry ! I communicated to Arne about this and not even he doesn't know. [/quote]
Well, there you have it. I hope VV can help when he comes back. :(

yes, I see. This is just some bug that appears sometimes. I once saw in another topic that one post was edited something like 5000 times, which is definitely a bug...
Hopefully, VV will be able to restore the disappeares posts. :roll:


Let be $ x,y,z,t\in \mathbb{Z}_ \plus{}$ such that $ t \equal{} \frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy}$ . Prove that $ t^2\minus{}4t\plus{}3\equal{}0$

If $ t^2\minus{}4t\plus{}3\equal{}0$, then $ t\equal{}3$ or $ t\equal{}1$. Put x=y=z=2, and then $ t\equal{}\frac32$.  :wink:

$ t\in \mathbb{Z}_\plus{}$ . You must take only the cases when $ t \in \mathbb{Z}_\plus{}$


quadratics
trigonometry
algebra
polynomial
algebra proposed

Let $ u_0 \equal{} 1, u_1 \equal{} \minus{} 1$ and $ u_{n \plus{} 1} \equal{} ku_n \minus{} u_{n \minus{} 1}$ for every $ n \geq 1$. Find every rational number $ k$ such that $ \{u_n\}$ is periodic.

[hide]
My approach to this problem:

Let $ a,b$ be the two roots of the quadratic equation
\[ x^2 \minus{} kx \plus{} 1 \equal{} 0\]
where $ a,b$ can be complex number. We notice that
\[ u_{n\plus{}1} \minus{} au_{n} \equal{} b(u_{n} \minus{} au_{n\minus{}1})\]
and since $ \{u_n\}$ is periodical, so is $ \{u_{n\plus{}1} \minus{} au_{n}\}$ and thus $ b^n$ is periodcial, which means $ |b| \equal{} 1$.
The same argument yields $ |a| \equal{} 1$. Now the left part is computing.
[/hide]

[quote="Mashimaru"]Let $ u_0 \equal{} 1, u_1 \equal{} \minus{} 1$ and $ u_{n \plus{} 1} \equal{} ku_n \minus{} u_{n \minus{} 1}$ for every $ n \geq 1$. Find every rational number $ k$ such that $ \{u_n\}$ is periodic.

[hide]
My approach to this problem:

Let $ a,b$ be the two roots of the quadratic equation
\[ x^2 \minus{} kx \plus{} 1 \equal{} 0\]
where $ a,b$ can be complex number. We notice that
\[ u_{n \plus{} 1} \minus{} au_{n} \equal{} b(u_{n} \minus{} au_{n \minus{} 1})\]
and since $ \{u_n\}$ is periodical, so is $ \{u_{n \plus{} 1} \minus{} au_{n}\}$ and thus $ b^n$ is periodcial, which means $ |b| \equal{} 1$.
The same argument yields $ |a| \equal{} 1$. Now the left part is computing.
[/hide][/quote]

1) $ k\equal{}\minus{}2$ and so $ u_n\equal{}(\minus{}1)^n$ is periodic

2) $ k\equal{}2$ and so $ u_n\equal{}1\minus{}2n$ is not periodic

3) $ k\notin\{\minus{}2,2\}$ and so the equation $ x^2\minus{}kx\plus{}1$ has two distinct roots (maybe complex) $ z_1$ and $ z_2$

Then $ u_n\equal{}\frac{(z_2\plus{}1)z_1^n\minus{}(z_1\plus{}1)z_2^n}{z_2\minus{}z_1}$

Periodicity would then mean $ u_T\equal{}1$ and $ u_{T\plus{}1}\equal{}\minus{}1$ for some $ T>0$

So $ u_T\plus{}u_{T\plus{}1}\equal{}0$ and so $ (z_1\plus{}1)(z_2\plus{}1)(z_1^T\minus{}z_2^T)\equal{}0$

3.1) $ z_1\equal{}\minus{}1$ and so $ z_2\equal{}\minus{}1$ (since $ z_1z_2\equal{}1$) and so $ k\equal{}\minus{}2$ and we got the case 1)

3.2) $ z_2\equal{}\minus{}1$ and so $ z_1\equal{}\minus{}1$ (since $ z_1z_2\equal{}1$) and so $ k\equal{}\minus{}2$ and we got the case 1)

3.3) $ z_1^T\equal{}z_2^T$ and so, since $ z_1\ne z_2$ in this paragraph 3 :

$ z_1$ and $ z_2$ can not be real numbers, else we would have $ z_1\equal{}\minus{}z_2$ and $ z_1z_2\equal{}1$, impossible

So $ z_1\equal{}e^{it}$ and $ z_2\equal{}e^{\minus{}it}$ and $ e^{2itT}\equal{}1$ and so $ t\equal{}\frac{n\pi}T$

And it is easy to check back that $ z_1\equal{}e^{r\pi}$ and $ z_2\equal{}e^{\minus{}r\pi}$ indeed are solutions for any rational $ r\notin\mathbb Z$ (else $ z_1\equal{}z_2$, forbidden in this paragraph 3)

And so $ k\equal{}2\cos(r\pi)$ for any $ r\in\mathbb Q\backslash\mathbb Z$

It remains to consider the constraint $ k\in\mathbb Q$ and it is a well known result that the only rational $ \cos(r\pi)$ are $ 0,\pm 1,\pm\frac 12$

Hence the result : $ \boxed{k\in\{\minus{}2,\minus{}1,0,1\}}$

[quote="Dumel"][quote="pco"]it is a well known result that the only rational $ \cos(r\pi)$ are $ 0,\pm 1,\pm\frac 12$[/quote]I can't find the proof of this fact. Coud you help me?
thx[/quote]

Here is a method to show this :

Problem : show that $ \forall n\in\mathbb N$ and $ \forall k\in\{0,1,...,n\}$ : $ \cos(\frac{k\pi}n)\in\mathbb Q$ $ \implies$ $ \cos(\frac{k\pi}n)\in\{\minus{}1,\minus{}\frac 12,0,\frac 12, 1\}$

[u][b]Lemma [/b][/u]: Let $ P(x)$ a monic polynomial $ \in\mathbb Z[X]$ whose degree is $ n\ge 1$. Then, if $ P(x)\equal{}0$ for some $ x\in\mathbb Q$, then $ x\in\mathbb Z$
Consider $ x\equal{}\frac pq$, with $ \gcd(p,q)\equal{}1$ and multiply the equation $ P(x)\equal{}0$ by $ q^n$. You get $ q|p^n$ and so $ q\equal{}\pm 1$
Q.E.D.

Let then the sequence of polynomials $ P_n(x)$ defined as :
$ P_0(x)\equal{}2$
$ P_1(x)\equal{}x$
$ P_{n\plus{}2}(x)\equal{}xP_{n\plus{}1}(x)\minus{}P_n(x)$

This is a sequence of monic polynomials $ \in\mathbb Z[X]$ and it is easy to show that $ P_n(2cos(\theta))\equal{}2\cos(n\theta)$

So, if $ \cos(\frac{k\pi}n)\in\mathbb Q$ $ P_n(u)\equal{}\pm 2$ with $ u\equal{}2\cos(\frac{k\pi}n)\in\mathbb Q$

And so, using lemma above, $ 2\cos(\frac{k\pi}n)\in\mathbb Z$. but $ 2\cos(\frac{k\pi}n)\in[\minus{}2,2]$

Hence the result $ \cos(\frac{k\pi}n)\in\{\minus{}1,\minus{}\frac 12,0,\frac 12, 1\}$


Let $n \ge 2$ be an integer. Prove that if $k^2 + k + n$ is prime for all integers $k$ such that $0 \leq k \leq \sqrt{\frac{n}{3}}$, then $k^2 +k + n$ is prime for all integers $k$ such that $0 \leq k \leq n - 2$.

[quote="Peter"]Let $a\ge0$ be the smallest integer such that $a^{2}+a+n$ is composite, having smallest prime divisor $b$. Assume for the sake of contradiction that $a\le n-2$.

If $b\ge 2a+1$, we have $a^{2}+a+n>a^{2}>b^{2}\be(2a+1)^{2}$, so $n>3a^{2}$ so $a<\sqrt{n/3}$, contradiction. So $b<2a+1$, then since $(a^{2}+a+n)-(c^{2}+c+n)=(a-c)(a+c+1)$, there is a positive integer $c<a$ for which one factor equals $b$. Now $b|(a^{2}+a+n)-(c^{2}+c+n)$ and $b|a^{2}+a+n$ so $b|c^{2}+c+n$, since $c<a$ this means $c^{2}+c+n=b$. Now an easy estimation gives $a-c<a+c+1<(n-2)+c+1<c^{2}+c+n=b$, thus $b$ is prime but $b$ can equal none of these factors, contradiction.[/quote]

Just a side note:
Using heavier artillery, we can show that the requirements are fulfilled for $ n\equal{}2, 3, 5, 11, 17, 41$ only.


can anyone please explain me about the functions[/b][/i]

Well, functions are kinda hard to cover in one post. Essentially, a funcion is a "machine" with some variables. If you put some numbers in it, the other end will result with something different.

That is kinda in simple terms, and you should do more research on your own time.

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200]Introduction to algebra, Art of problem solving series[/url]
[url=http://en.wikipedia.org/wiki/Function_(mathematics)]Wiki page[/url]
[url=http://library.thinkquest.org/2647/algebra/funcbasc.htm]A page I found on google[/url]


modular arithmetic
function
calculus
number theory proposed
number theory

You just won't believe how short and beautiful the original solution is to the following exceptional problem; I encourage you to ponder it independently.

Problem. (M. Zhang, 1989) A (finite) system $a_i \pmod{b_i}, 1 \leq i \leq k$ of residues and moduli is called "covering" if every integer belongs to at least one of the residue classes (i.e. satisfies at least one of the congruences $x \equiv a_i \pmod{b_i}$). For instance, it is easy to see that
\[
0 \pmod{2}, 2 \pmod{3}, 1 \pmod{4}, 1 \pmod{6}, 3 \pmod{12}
\]
is a covering system. Show that, for any such system, there exists a non-empty subset $I \subseteq \{1,\ldots,k\}$ such that
\[
\sum_{i \in I} \frac{1}{b_i} \in \mathbb{N}
\]
is an integer!!! (however strange it might seem.)

For example, for the system above, we have $1/2 + 1/3 + 1/6 = 1 \in \mathbb{N}$.

Enjoy!

--Vesselin

Ok, I agree that it is very difficult to find a solution for this problem  ;) , so I'll post it myself. This problem (especially its incredible solution) is an absolute classic.

Start by observing that a system $a_i \pmod{b_i}, 1 \leq i \leq k$ is covering if and only if
\[
\prod_{1 \leq j \leq k} \big( 1 - e^{2\pi i(n+a_j)/b_j} \big) = 0
\]
for $n=1,2,3,\ldots$. [This follows from the fact that $e^{2\pi i c} = 1$ iff $c \in \mathbb{Z}$.] For any fixed $n$, expand the product to obtain that
(*)
\[
1 + \sum_{I} d_I e^{2\pi i n c_I} = 0, \quad n=1,2,3,\ldots
\]
for some non-zero constants $d_I$, where the sum is over all non-empty subsets $I \subseteq \{1,\ldots,k\}$, and
\[
c_I := \sum_{i \in I} 1/b_s.
\]

The following lemma, which is proven via the well-known properties of the Riemann zeta function, is the crucial trick in Zhang's solution. I'll state it without proof, not because it is trivial nor because it is very difficult; it is a standard analytical exercise, and I leave the calculus-crunching forum members the pleasure of proving it  :P .

Lemma. For a real number $c$, the series
\[
\sum_{n=1}^{+ \infty} \frac{e^{2\pi i cn}}{n}
\]
diverges if and only if $c$ is an integer.

Once you prove this lemma, it's easy  :) ! Just multiply, for each $n$, the equation (*) by $1/n$, and sum over all $n$. Since the harmonic series diverges, it cannot be that all series
\[
\sum_{n \geq 1} \frac{e^{2\pi i c_I n}}{n}
\]
converge. That is, by the lemma, at least one of the constants $c_I$ is an integer, which is what we were asked to prove!!!

 :D  :D  :D  :D  :D  :D :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

So simple yet so insightful, huh?

--Vesselin

And here is another striking result which is easily proven with the same idea:

Problem. (Zhi-Wei Sun, 1992) A system $a_i \pmod{b_i}$ of residues and moduli is called an "exact $m$-cover" if every integer belongs to *exactly* $m$ of the residue classes. [Note that, obviously, $\sum_j 1/b_j = m$ for any such system.] Show that, for every such system and for every integer $n=1,2,\ldots,m$, there are at least $\binom{m}{n}$ subsets $I \subseteq \{1,\ldots,k\}$ such that $\sum_{i \in I} 1/b_i = n$.

Hint. Use the identity
\[
\prod_{1 \leq j \leq k} \big( 1 - x^{1/b_j} e^{2\pi i a_j/b_j} \big) = (1-x)^m,
\]
which is valid for every $x \geq 0$.

--Vesselin

It was really a magic proof.

[quote="vess"]The following lemma, which is proven via the well-known properties of the Riemann zeta function, is the crucial trick in Zhang's solution. I'll state it without proof, not because it is trivial nor because it is very difficult; it is a standard analytical exercise, and I leave the calculus-crunching forum members the pleasure of proving it  :P .

Lemma. For a real number $c$, the series
\[
\sum_{n=1}^{+ \infty} \frac{e^{2\pi i cn}}{n}
\]
diverges if and only if $c$ is an integer.[/quote]
It is a direct application of Abel-Dirichet's theorem, isn't it?


If 3 to the Xth power=7 find the value of 3 to the power of (2X+1)

I need solutions without logs, thx in advance  :lol:

You don't need to solve for $ x$ to find the answer.  :)  Just use [hide="Exponent rules"] $ 3^{2x \plus{} 1} \equal{} 3 \cdot 3^{2x} \equal{} 3 \cdot \left( 3^x \right)^2$ [/hide]

[quote="Ihatepie"][hide]

so would it be 147?

[/hide][/quote]Yes. It's a matter of substitution after applying the exponent rules suggested by t0rajir0u


Solve:

Two fine cardboard pieces of the same size with form of regular hexagon are had. To divide each hexagon in three pieces by means of two straight cuts so that with the 6 pieces it is possible to be armed, without hollows nor superpositions, an equilateral triangle. To explain porqu when reacomodar properly the 6 pieces obtains indeed an equilateral triangle.

it is translation of:

Se tienen dos piezas de cartulina del mismo tamao con forma de hexgono regular. Dividir cada hexgono en tres pedazos mediante dos cortes rectos de modo que con los 6 pedazos se pueda armar, sin huecos ni superposiciones, un tringulo equiltero.

Explicar porqu al reacomodar convenientemente los 6 pedazos se obtiene efectivamente un tringulo equiltero

Also discussed here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16302[/url]


geometry incenter circumcircle inradius ratio

A circle of center $ O$ of radius 48 is inscribed in an equilateral triangle $ ABC$. Another smaller circle of center $ P$ is tangent to the circle of circle $ O$and to $ AB$ and $ BC$. There are three such circles. Let us suppos that $ P$ is closer to $ B$ than $ A$ or $ C$. Find the radius of the center of circle $ P$.

[hide="Spoilers Contained =D"]Circle O is obviously an incircle of the equilateral triangle.  Since it has a radius of 48, the side of one triangle must be $ 96\sqrt{3}$.  The length from the incenter to a vertex is the radius of the circumscribed circle, which is 96.  If we draw in the radii that are tangent to the sides of the triangle of both Circle O and Circle P, then we see that there are similar triangles.

But first: We need to find the distance from the center of Circle P to B.  This is simply the circumradius minus the inradius minus the radius of the small circle, namely $ 48\minus{}r$.  However, this must also be equal to $ 2r$ because of the 30-60-90 right triangle.

Thus, $ r\equal{}16$.
Or, you could then set up the ratio: $ \frac{r}{48}\equal{}\frac{48\minus{}r}{96}$[/hide]


linear algebra
matrix
combinatorics unsolved
combinatorics

Call a binary string of length n 'good' if it has exactly k ones for a given integer k. m good binary strings are arranged one below one so as to form a matrix of numbers. Find the least value of m such that however the m strings are chosen, we can always find j columns in the matrix with all entries equal to one (j<k)

I would love it if anyone could generalize or direct me to similar problems
Thankyou

Dear lajanugen,
This is a very nice problem.
_______________________

Consider there exists a 4-tuple $ (n,k,m,j)$ so that the system described in the problem exists. Lets consider a n-tuple with exactly k ones. Consider another (lets call it the second) n-tuple with exactly k ones. When these two tuples are placed as a rows one below the other(the first tuple is placed above the second tuple), we obtain $ 2 \times n$ matrix. There will be at least $ k\minus{}(n\minus{}k)$ columns in this matrix completely filled with ones. By constructing the $ m\times n$ matrix in the similar procedure, we conclude that there shall be at least $ k\minus{}(m\minus{}1)(n\minus{}k)$ columns that contain only ones. Our system exists iff

\[ k\minus{}(m\minus{}1)(n\minus{}k)\equal{}j\]

the minimum value of m can be obtained from this.


How many grams of fluorine are there in a 3.5grams sample of phosphorus pentafluroide gas?

This is so hard for me.....

Step 1. Write the molecular formula of your compound.

Step 2. Using the Periodic Table, find its molar mass M.

Step 3. In M grams of your compound, what's the mass of fluorine?

Step 4. I think you can finish from here.


calculus
integration
trigonometry
function
calculus computations

Hi I have another integral question that I need help with.  Thanks.

Question: Evaluate $ \displaystyle\int_0^{\infty}\frac{(x\cos x\minus{}\sin x)^2}{x^6}dx$

Simply by repeated integration by parts,

$ \int_{0}^{\infty }\frac{\left( x\cos x\minus{}\sin x\right) ^{2}}{x^{6}}dx$

$ \equal{}\allowbreak \frac{2}{5}\int_{0}^{\infty }\frac{\sin ^{2}x\minus{}x\cos x\sin x}{x^{4}}dx$

$ \equal{}\allowbreak \frac{2}{15}\int_{0}^{\infty }\left( \frac{\frac{1}{2}\sin 2x\minus{}x\cos 2x}{x^{3}}\right) \,dx$

$ \equal{}\allowbreak \frac{2}{15}\int_{0}^{\infty }\frac{\sin 2x}{x}\,dx$

$ \equal{}\allowbreak \frac{\pi }{15}$

Also, to answer your original question in this post, is as follows:-

Starting off with integration by parts,

$ \int_{0}^{\infty }\frac{x^{2}\plus{}1}{x^{6}}\left( x\cos x\minus{}\sin x\right) ^{2}dx$

$ \equal{}\allowbreak 2\int_{0}^{\infty }\left( \sin ^{2}x\minus{}x\sin x\cos x\right) \left( \frac{1}{3x^{2}}\plus{}\frac{1}{5x^{4}}\right) \,dx\allowbreak$

$ \equal{}\frac{2}{3}\int_{0}^{\infty }\frac{\sin ^{2}x}{x^{2}}dx\minus{}\frac{1}{3}\int_{0}^{\infty }\frac{\sin x}{x}dx\plus{}\frac{2}{5}\int_{0}^{\infty }\frac{x\sin ^{2}x\minus{}\sin x\cos x}{x^{4}}dx$

It is well known that $ \int_{0}^{\infty }\frac{\sin ^{2}x}{x^{2}}dx\equal{}\int_{0}^{\infty }\frac{\sin x}{x}dx\equal{}\allowbreak \frac{\pi }{2}$

Again, via repeated integration by parts

$ \int_{0}^{\infty }\frac{\sin ^{2}x\minus{}x\sin x\cos x}{x^{4}}dx$

$ \equal{}\allowbreak \frac{1}{3}\int_{0}^{\infty }\left( \frac{\frac{1}{2}\sin 2x\minus{}x\cos 2x}{x^{3}}\right) \,dx$

$ \equal{}\allowbreak \frac{1}{3}\int_{0}^{\infty }\frac{\sin 2x}{x}\,dx$

$ \equal{}\frac{\pi }{6}$

Hence $ \int_{0}^{\infty }\frac{x^{2}\plus{}1}{x^{6}}\left( x\cos x\minus{}\sin x\right) ^{2}dx\equal{}\frac{2}{3}\left( \frac{\pi }{2}\right) \minus{}\frac{1}{3}\left( \frac{\pi }{2}\right) \plus{}\frac{2}{5}\left( \frac{\pi }{6}\right) \equal{}\allowbreak \frac{7}{30}\pi$

Thanks for your solution IndoChina.

Another method:  If we consider the function $ f(t)\equal{}1\minus{}t^2$ for $ |t|<1$ and $ f(t)\equal{}0$ elsewhere,
then I believe the Fourier transform of $ f(t)$ is $ \displaystyle F(x)\equal{}\minus{}4\left(\frac{x\cos x\minus{}\sin x}{x^3}\right)$.
Since $ \displaystyle F(x)\equal{}2F_C(x)$, the Fourier cosine transform, then $ \displaystyle F_C(x)\equal{}\minus{}2\left(\frac{x\cos x\minus{}\sin x}{x^3}\right)$.
According to Parseval's identity, $ \displaystyle\int_0^{\infty}[f(t)]^2dt\equal{}\frac{2}{\pi}\int_0^{\infty}[F_C(x)]^2dx$
Hence, $ \displaystyle\int_0^1(1\minus{}t^2)^2dt\equal{}\frac{2}{\pi}\int_0^{\infty}4\frac{(x\cos x \minus{} \sin x)^2}{x^6}dx$
Therefore, $ \displaystyle \frac{8}{15}\equal{}\frac{8}{\pi}I\Rightarrow I\equal{}\frac{\pi}{15}$.

See MAZ technic

[url= https://artofproblemsolving.com/community/c7h607212_d_integral_2] see here [/url]
https://artofproblemsolving.com/community/c7h607212_d_integral_2


Let $p(n)$ be the number of partitions of $n$, and let $p(n, m)$ be the number of partitions of $n$ containing $m$ terms. Show that
\[p(n) = p(2n, n).\]

[hide]
Let $n = a_{1}+a_{2}+\cdots+a_{k}$ be any partition of $n$ (by the definition of a partition, we have $a_{i}\ge 1$). Rewrite this as

$n = a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}+\cdots+a_{n}$

where $a_{i}= 0$ for $i > k$. Then

$2n = (a_{1}+1)+(a_{2}+1)+\cdots+(a_{k}+1)+(a_{k+1}+1)+\cdots+(a_{n}+1)$

is a valid partition for $2n$ with exactly $n$ terms. Clearly, the reverse works as well, giving us a one-to-one correspondence.[/hide]


I'm not sure if this is in the right forum...  :blush:

$ 3i\plus{}4>2i\plus{}2$ but how do we compare $ 3i\plus{}4 \text{  and  } 2i\plus{}5$? can we compare them with ><=?

Thank you...

we can compare only their moduls, the comparison of complex numbers as $ a\plus{}ib>c\plus{}id$ is not defined for complex numbers

does this mean that it is implied that $ x \in \mathbb{R^\plus{}}$, whenever we look for $ x$ in inequalities like $ \sqrt{x}>2x$?

http://www.cut-the-knot.org/do_you_know/complex_compare.shtml
here you've got a short article about it but generally you simply don't compare them


Find all $ n$ such as $ \{ 1,2,\ldots,n \}$ can be partition into
subsets of size $ d$
with equal sum
for all $ d$ with $ d|n$ and $ d>1$.

It is not difficult to prove that $ n\equal{}p$ with $ p$ prime works and
$ n\equal{}2007$ works too from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=324147[/url]

Somewhat similar question was asked before: http://www.mathlinks.ro/viewtopic.php?t=239933


How can I make a kind of box like this one

[img]http://img72.imageshack.us/img72/9850/boxwi9.png[/img]

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancybox]fancybox package[/url] with the \shadowbox command.

A related question:

I've seen we can also put color into the box, how do you do that?

\usepackage{color}. The documentation is [url=http://tug.ctan.org/tex-archive/macros/latex/required/graphics/grfguide.pdf]here[/url], section 3.6
\colorbox{red}{text} gives [img]http://sixthform.info/gblog/pictures/cc5cc522e0cf4c0f53d1b26396690da2.gif[/img]
\fcolorbox{red}{yellow}{text} gives [img]http://sixthform.info/gblog/pictures/b248ed5cdf5b3157edf26c143b0f7a6e.gif[/img]

Thanks a lot Steve.

One more question:

How can I put color when using \shadowbox?

Put one inside the other. Something like
\setlength{\fboxsep}{0.5pt}
\shadowbox{\colorbox{red}{some text}}
though you may need to experiment.


I need to differentiate the following equation:

x(t)=0.5e^(0.1t)sin(t)

I got the wrong answer using the following method:

let u=0.5e^(0.1t)           let v=sin(t)
    u'=0.05e^(0.1t)             v'=cos(t)

Use product rule: x'(t)=(v*u')+(v'*u)
                                =0.05e^(0.1t)sin(t)+0.5e^(0.1t)sin(t)

This answer is incorrect. I used graphmatica to solve the problem and it gave me:

y=0.5*e^0.1*x*cos x + 0.5*(sin x)*e^0.1

How did graphmatica arrive at this answer?

Thanks in advance for any help.

This is correct...

[quote="kieran_17"]

let $u=0.5e^{0.1t}$ let $v=\sin(t)$
$u'=0.05e^{0.1t} , v'=\cos(t)$

Use product rule:
$x'(t)=(vu')+(v'u)$
[/quote]

but $v'= \cos(t)$ therefore the answer is

$x'(t)=0.05e^{0.1t}\sin(t)+0.5e^{0.1t}\cos(t)$

 :)


Let a sequence $x_{n}$  with  $x_{1}=2$  and
$x_{n+1}=\frac{2+x_{n}}{1-2x_{n}}$  , n=1,2,....

Prove that
(a) $x_{n}\neq 0$

(b)$x_{n}$ is not periodic

$x_{n}=A^{n-1}{x_{1}\choose 1}$, were A is matrice 2*2  $A={1 \ \ 2\choose-2 \ 1}$ and ${a \ b \choose c \ d }{x\choose y}=\frac{ax+by}{cx+dy}$.

[quote="Megus"]There's an easier way to express $x_{n}$:

$x_{n}= \tan n \alpha$ where $\alpha = \arctan 2$  :)[/quote]
Yes your method is better to write, when $A={1 \ \ x\choose-x \ 1},x_{n}=\tan{n\arctan x }$. But it is not better for calculate.

Hi! I had some trouble with Latex. And that´s because i didn´t post the fully solution.  :lol:


If $ f(x) \equal{} 5x^2 \minus{} 2x \minus{} 1$, then $ f(x \plus{} h) \minus{} f(x)$ equals:

$ \textbf{(A)}\ 5h^2 \minus{} 2h \qquad \textbf{(B)}\ 10xh \minus{} 4x \plus{} 2 \qquad \textbf{(C)}\ 10xh \minus{} 2x \minus{} 2 \\
\textbf{(D)}\ h(10x \plus{} 5h \minus{} 2) \qquad \textbf{(E)}\ 3h$

wheeee plug and chug

$ f(x\plus{}h)\minus{}f(x)\equal{}5(x\plus{}h)^2\minus{}2(x\plus{}h)\minus{}1\minus{}(5x^2\minus{}2x\minus{}1)$
$ \equal{}5(x^2\plus{}2xh\plus{}h^2)\minus{}2x\minus{}2h\minus{}2\minus{}5x^2\plus{}2x\plus{}1$
$ \equal{}10xh\plus{}5h^2\minus{}2h$
$ \equal{}h(10x\plus{}5h\minus{}2)$
$ \equal{}(D)$

hello, we have by inserting and simplifying $ 5(x\plus{}h)^2\minus{}2(x\plus{}h)\minus{}1\minus{}(5x^2\minus{}2x\minus{}1)\equal{}5h^2\plus{}10hx\minus{}2h\equal{}h(5h\plus{}10x\minus{}2)$
So the answer is $ \text{D}$.
Sonnhard.


geometry
incenter
geometric transformation
reflection
parallelogram
angle bisector
geometry unsolved

Let $ ABC$ be a triangle where $ AB<AC$. In the opposite ray of $ BA$ and $ CA$, respectively pick E and F such that $ BE\equal{}CF\equal{}BC$. Let $ AD$ be angle bisector of $ \large \widehat{BAC}$, $ O$ is the intersection of $ CE$ and $ BF$ and $ (d)$ be the line which passes through $ O$ and paralel to $ AD$. $ (d)$ intersects $ AC$ at M. Prove that $ CM\equal{}AB$.

[img]http://i227.photobucket.com/albums/dd72/TPFC/3.png[/img]
Let $ P,I$ be midpoint of $ BC$ and incenter of triangle. Let $ OM\cap BC \equal{} Q$. We have:
Reflect over $ P$: $ D_P: B\rightarrow C,BI\rightarrow CO,CI\rightarrow BO$ (Because $ BI$ parallel with $ CO$,$ CI$ parallel with $ BO$ )
Then $ D_P: I\rightarrow O,AD\rightarrow OM$(Because $ AD$ parallel with $ OM$)
Now $ D_P: D\rightarrow Q$.
Have: $ \frac {CQ}{CM} \equal{} \frac {CD}{AC} \equal{} \frac {BD}{AB}$
But $ CQ \equal{} BD$ then $ CM \equal{} AB$

Construct the parallelogram $ ABGC$, and denote by $ H$ the intersection of $ CE$ with $ BG$.
By [b]Thales theorem[/b], we have $ \frac{OH}{OC}\equal{}\frac{BH}{CF}\equal{}\frac{BH}{BC}$, it follows that $ BO$ is the angle bisector of angle $ GBC$. Similarly, $ CO$ is the angle bisector of angle $ GCB$. Hence, we conclude that $ O$ is the incentre of triangle $ GBC$, so $ GO$ is the angle bisector of angle $ BGC$ $ (1)$.

On the other hand, we have the quadrilateral $ ABGC$ is a parallelogram, so the angle bisector of angle $ BAC$ is parallel to the angle bisector of angle $ BGC$ $ (2)$.

From $ (1)$, $ (2)$ and the statement $ OM\parallel AD$, we get $ G$, $ O$ and $ M$ are collinear.
Now, we have $ \angle CGM\equal{}\angle BGM\equal{}\angle GMC$, so $ CM\equal{}CG$, i.e. $ CM\equal{}AB$, as desired.


geometry
geometric transformation
rotation
linear algebra
matrix

If we have M(x,y) then the formula looks thus:
x=Xcosα - Ysinα
y=Xsinα+Ycosα

What will be modified the formula if we have z-axis.
Such as for N(x,y,z)

x=?
y=?
z=?
 :maybe:

See [url=http://en.wikipedia.org/wiki/Rotation_matrix]the Wikipedia article[/url].  The relevant formula occurs a little less than halfway through.


If a charge is placed inside a cavity made in a solid conducting sphere, what governs the magnitude of the charges appearing on the outside surface of the sphere and the ones inside of the cavity.

[hide]I can understand that the charge will repel/attract electrons (depending on the charge-it is positive or negative) of the conductor but how do I find the magnitude of these charges? I'm also aware that the charges migrate to the surface in case of a conducting solid and that the net charge on the conductor is zero.

Also, what will happen if the body is non-conducting?
[/hide]

[i][color=red]PS[/color]: Maybe this question makes me look like a fool but sometimes one understands the tougher aspects of the subject but surprisingly cannot appreciate the simpler ones, sometimes even basics and the foundation of everything that follows.[/i]

[quote="Rajiv"]If a charge is placed inside a cavity made in a solid conducting sphere, what governs the magnitude of the charges appearing on the outside surface of the sphere and the ones inside of the cavity.

[hide]I can understand that the charge will repel/attract electrons (depending on the charge-it is positive or negative) of the conductor but how do I find the magnitude of these charges? I'm also aware that the charges migrate to the surface in case of a conducting solid and that the net charge on the conductor is zero.

[/hide]

[i][color=red]PS[/color]: Maybe this question makes me look like a fool but sometimes one understands the tougher aspects of the subject but surprisingly cannot appreciate the simpler ones, sometimes even basics and the foundation of everything that follows.[/i][/quote]

Because the E-field inside the conductor must be zero, a charge of $ \plus{}q$ inside the cavity induces a charge of $ \minus{}q$ on the cavity's surface. The charge on the outer surface of the sphere is then $ \plus{}q$, by similar reasoning, as well as a straight forward application of Gauss's law to the outer surface of the sphere.

[quote="Rajiv"]Also, what will happen if the body is non-conducting?[/quote]

[hide]Ampere's law is given by: $ \nabla \times \vec{H} \equal{} \vec{J} \plus{}\varepsilon\frac{\partial \vec{E}}{\partial t}$ from Maxwell's equations.

One now observes that $ \vec{J} \equal{}\sigma \vec{E}$, where $ \sigma$ is the electrical conductivity.

Hence $ \nabla \times \vec{H} \equal{} \sigma\vec{E} \plus{}\varepsilon\frac{\partial \vec{E}}{\partial t}$.

If we take the divergence of this equation,

\[ \underbrace{\nabla\cdot (\nabla \times \vec{H})}_{0} \equal{} \sigma\nabla\cdot\vec{E} \plus{}\varepsilon\frac{\partial \nabla\cdot\vec{E}}{\partial t}\]

Since $ \nabla\cdot\vec{E} \equal{}\frac{\rho}{\varepsilon}$, we obtain

\[ 0 \equal{} \frac{\sigma}{\varepsilon}\rho \plus{}\frac{\partial \rho}{\partial t}\]

This equation is known as the continuity equation. For an electrical conductor $ \sigma >> \varepsilon$, and this implies that $ \rho$ decays very rapidly inside a conductor -- this is why charges migrate to the outer surface of a conductor.

For a perfect insulator, there isn't really such a thing but for purposes of answering your question, $ \sigma \equal{} 0$, which leads to $ \frac{\partial \rho}{\partial t} \equal{} 0$. Obviously, in such a case the charge density remains constant; the charges stay in place.[/hide]


geometry
circumcircle
Euler
geometry proposed

Given arbitrary quadrilateral $ABCD$ let $H_{a},G_{a},O_{a}$ be orthocenter, centroid, circumradius of triangle $BCD$, similarly for triangles $CDA,DAB,ABC$ let $G_{a}G_{c}\cap G_{b}G_{c}=\{G\},H_{a}H_{c}\cap H_{b}H_{c}=\{H\},O_{a}O_{c}\cap O_{b}O_{c}=\{O\}$ prove $O,G,H$ are colinear and $2\vec{OH}+3\vec{HG}=\vec{0}$ we will call the line conecting these points be Euler line of $ABCD$

This problem is the [b][size=109]''Ganin's theorem''[/size][/b] $($ discovered by a Russian student named Jaroslav Ganin $)$ and presented by his teacher Alexei Myakishev, in Hyacinthos forum one year ago.

Please see at: http://tech.groups.yahoo.com/group/Hyacinthos/message#12400

Also it has already been posted by [b][size=109]Tiks[/size][/b], another its formulation in this forum.

http://www.mathlinks.ro/Forum/viewtopic.php?t=107395

Kostas Vittas.

It is also discussed in this article "On Two Remarkable Lines Related to a Quadrilateral", Forum Geometricorum:
[url]http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf[/url]

Thanks, I only see accidentaly by Cabri while It was well know before! :oops:


The function $ d(n)$ is defined for non-negative integers $ n$ and is equal to the sum of the squares of the digits of the decimal representation of $ n$. Thus, for example, $ d(23)\equal{}2^{2}\plus{}3^{2}\equal{}13$.

Find all $ n$ such that $ d(n)\equal{}n$.

[quote="AndrewTom"]The function $ d(n)$ is defined for non-negative integers $ n$ and is equal to the sum of the squares of the digits of the decimal representation of $ n$. Thus, for example, $ d(23) \equal{} 2^{2} \plus{} 3^{2} \equal{} 13$.

Find all $ n$ such that $ d(n) \equal{} n$.[/quote]
If $ n$ has $ k\ge 4$ digits, then $ d(n)\le 81k<10^{k\minus{}1}\le n$ and so $ n$ may only have $ 1,2$ or $ 3$ digits

1) solutions with $ 1$ digit : $ a^2\equal{}a$ and so $ n\in\{0,1\}$

2) solutions with $ 2$ digits : $ 10a\plus{}b\equal{}a^2\plus{}b^2$
So $ a(10\minus{}a)\equal{}b(b\minus{}1)$ and so no solution (LHS may only be $ 9,16,21,24$ or $ 25$ and RHS may only be $ 0,2,6,12,20,30,42,56$ or $ 72$.

3) solutions with $ 3$ digits : $ 100a\plus{}10b\plus{}c\equal{}a^2\plus{}b^2\plus{}c^2$
$ RHS\le3\cdot 81\equal{}243$ and so $ a\in\{1,2\}$
3.1) solutions $ 1bc$ : $ 99\plus{}10b\plus{}c\equal{}b^2\plus{}c^2$
$ LHS\ge 99$ and so at least one of the two numbers $ b,c$ is $ >7$, so $ 4$ possibilities :
$ 18c$ $ \implies$ $ 99\plus{}80\plus{}c\equal{}64\plus{}c^2$ $ \implies$ $ c^2\minus{}c\minus{}115\equal{}0$ and no integer solution
$ 19c$ $ \implies$ $ 99\plus{}90\plus{}c\equal{}81\plus{}c^2$ $ \implies$ $ c^2\minus{}c\minus{}108\equal{}0$ and no integer solution
$ 1b8$ $ \implies$ $ 99\plus{}10b\plus{}8\equal{}b^2\plus{}64$ $ \implies$ $ b^2\minus{}10b\minus{}43\equal{}0$ and no integer solution
$ 1b9$ $ \implies$ $ 99\plus{}10b\plus{}9\equal{}b^2\plus{}81$ $ \implies$ $ b^2\minus{}10b\minus{}27\equal{}0$ and no integer solution

3.2) solutions $ 2bc$ : $ 196\plus{}10b\plus{}c\equal{}b^2\plus{}c^2$
$ LHS\ge 196$ while $ RHS\le 9^2\plus{}9^2\equal{}162$ so no solution

Hence the only two solutions :
$ n\equal{}0$
$ n\equal{}1$


Hi everyone,

I would like to inquire where I could find good math materials (problem books, etc) for elementary grades in Seattle area? My son has Math talent (as far as I can tell) but lack of reliable math books and/or Math programs make him loose his Math knowledge consistency.

Thanks for the help,

Dapyx

If you want good mathcount problems go to [url]http://yashar.biz[/url] and click on mathcounts in the side bar.

the Art of Problem Solving text books are a good resource for math. YOu can find them on this website.


What are "Directed Angles"?  How are they different from normal angles?

I Googled "directed angles," and I got [url=http://de.geocities.com/darij_grinberg/]Darij's Web site[/url], a site by a MathLinks-AoPS participant. Cool.

Thanks, of course, for this publicity, but the directed angles article is in German... and anyway, it's not the best way of explaining these concepts... on earth, don't ask me where a better one can be found...

  darij

Moved to intermediate - this is a little tough for Getting Started!

From Kiran Kedlaya's notes:

http://caltizzle.caltech.edu/explications/geom/node6.html


inequalities
geometry
AoPS Books
\/closed

I have asked a couple admins, and they have given me mixed feedback. In addition to WOOT, I might like to take an additional subject course, as was suggested, and as I have not much experience with Olympiad problems or formal inequalities, I was wondering which would be more helpful to reach my goal of MOP? Or, might WOOT alone be enough to make MOP? I'm currently reading the AOPS books, and plan to finish them before I start a course. Any suggestions? Thank you in advance :D

Well..... I would say the problem solving course, for a couple of reasons. First, the inequalities, in general, is a very easy(well i didnt mean easy as in VERY EASY) topic to be learned before USAMO. in another word, inequality is one of the topics that could be learned quickly and effectively. However, I think that problem solving takes time becuase it is the basis for everything about writing proof. so I would recommend Problem SOlving, but that's just me. I just have a thing for getting inequalitys.

[quote="sirorange"]Anyone else have advice? Also, if I don't take inequalities, how else might I learn inequalities?[/quote]

I have asked many peopel about that, and they have said that 1. Inequality section of the forum in the Olympiad Section has one of the most resources in terms of inequalities, and 2. there is a marathon going on in intermediate that is going to help you learn. Getting a book wont hurt you either. Specific inequalities books such as one by Hardy or simply books that have inequalities, such as Engel or Zeitz.

I teach the Olympiad Inequalities course and here is my opinion:

The Olympiad Problem Solving course is material that is most helpful in the broad sense of learning problem solving.  In that sense, I would recommend it before any of our other two Olympiad level subject classes.

But if you want to focus on either Inequalities or Geometry at some point in time, the classes are designed to help you focus on those fields.

Still, in the end, it's all a matter of which subjects you feel the greatest need to work at with instructors and "curriculum managers" -- the latter of which is as much our role in building the classes as the first.

[quote="bubala"]Wait, didn't Mr. Patrick say that if you are taking WOOT then you should not take olympiad problem solving because it covers the same thing?[/quote]

The WOOT curriculum is something different from our course curriculum.  There might be some overlap, but I don't think WOOT covers the material from one Olympiad class much more than another.

However, taking both WOOT and Olympiad Problem Solving at once might be a fairly large amount of work, depending on your schedule.


quadratics conics parabola function inequalities algebra polynomial

Find the value of $ a$ for which   $ ax^{2} \plus{} (a \minus{} 3)x \plus{} 1 < 0$   for at least one positive real $ x$.

Thanks for any hints.

Aren't there many values of $ a$?
$ a\equal{}1$ and $ a\equal{}\minus{}1$ work, for starters.

$ f(x)\equal{}ax^{2} \plus{} (a \minus{} 3)x \plus{} 1$ graphs as a parabola, with $ f(0)\equal{}1$.

For some values of $ a$ that function will have a maximum. Teachers may say that the parabola "opens down" or is "convex." Would that make the function negative for some positive $ x$? What values of $ a$ cause the function to have a maximum? Green Turtle suggests $ a\equal{}\minus{}1$. Does that work? Once you substitute $ a\equal{}\minus{}1$, the function simplifies. What does it become? Is it negative for some positive $ x$, like $ x\equal{}1$?

Were there any other conditions to complicate the problem, such as that it must be $ a>0$, for example?

If the function has a minimum, the problem gets more complicated. The minimum could be such that the function would not be negative for any positive $ x$. For example, for $ a\equal{}1$, the function is
$ f(x)\equal{}x^2\minus{}2x\plus{}1\equal{}(x\minus{}1)^2$, and it has a minimum with $ x\equal{}1, y\equal{}0$. For all other values of $ x$, the function is positive. It is never negative.

I) if $ a > 0$, the parabola opens up and for the inequality will have solutions $ x$ between the roots of the equation (inclusive).  Thus, there only exist such $ x$ if the equation has solutions - i.e. if the discriminant is positive.  $ d \equal{} (a \minus{} 3)^2 \minus{} 4a \equal{} a^2 \minus{} 10a \plus{} 9 > 0$ gives $ (a \minus{} 1)(a \minus{} 9) > 0$ hence $ a > 9$ or $ a < 1$ give the inequality real solutions, but not necessarily positive.

we now need to require that the larger zero of the polynomial is positive, so we have $ \frac {3 \minus{} a \pm \sqrt {d}}{2a} > 0$.  with $ a > 0$, we must then have $ 3 \minus{} a \plus{} \sqrt {d} > 0$ so $ 3 \plus{} \sqrt {d} > a$ which is clearly true for $ a < 3$, and hence all $ 0 < a < 1$ work.

Now we need $ 3 \plus{} \sqrt {(a \minus{} 1)(a \minus{} 9)} > a$ for $ 9 < a$.  So we have $ \sqrt {(a \minus{} 1)(a \minus{} 9)} > a \minus{} 3$.  note that both sides must be positive, so squaring preserves order: $ a^2 \minus{} 10a \plus{} 9 > (a \minus{} 3)^2 \equal{} a^2 \minus{} 6a \plus{} 9$ so $ 0> 4a$ which is true for $ a < 0$ (for $ a < 3$ the squaring isn't valid).  Thus none of $ a > 9$ work.

II) if $ a \equal{} 0$, we have $ \minus{} 3x \plus{} 1 < 0 \implies x > 1/3$.  so $ a \equal{} 0$ works.

III) if $ a < 0$, the parabola opens downward.  Hence $ \lim_{x \to \infty} ax^2 \plus{} (a \minus{} 3)x \plus{} 1 \equal{} \minus{} \infty$ (this means that for every $ e$ there exists $ N$ such that for all $ x > N$, $ e > ax^2 \plus{} (a \minus{} 3)x \plus{} 1$.  take some $ e \equal{} 0$ and then $ x \equal{} N \plus{} 1$ necessarily works) so clearly all $ a < 0$ satisfy the condition.

In summary, our solutions are $ a \in ( \minus{} \infty, 1)$

You can fill pages with algebra looking at what happens for $ a>0$, and I confess I did just that at first. Using my 20:20 hindsight, I see that it is a very easy problem when looked at from the right angle.

The expression can be taken as a linear function on $ a$:
$ g(a)\equal{} ax^2\plus{}(a\minus{}3)x\plus{}1\equal{}a(x^2\plus{}x)\plus{}(1\minus{}3x)$
It is a line with positive slope $ x^2\plus{}x>0$ for every positive $ x$.

For every positive $ x$, $ g(a)$ increases with increasing $ a$.

Since $ g(1)\equal{}x^2\minus{}2x\plus{}1 \ge 0$ for every  positive $ x$, $ a\equal{}1$ is not an answer to the problem.
Also for all $ a>1$, it will be $ g(a)>0$ for every positive $ x$, so no $ a>1$ would work either.
In sum, we know there are no $ a \ge 1$ answers to the problem.

For $ a<1$, with $ x\equal{}1$, $ g(a)<g(1)<0$, so there is one positive real $ x$ $ ($namely $ x\equal{}1)$ that makes the expression negative for not one $ a$, but all $ a<1$.

The answer to the problem is any $ a<1$.
IFF $ a<1$, there is at least one positive real $ x$ that makes the expression negative.

Moral to this story: Walk all around the problem and look at it from several angles.

PS _ And if you see a flaw in my reasoning, let me know. I come here to learn how to think and an occasional humiliating rebuke is very educational.

hmmm ...
 I think the function $ f(x) \equal{} ax^{2} \plus{} bx \plus{} c$ is always negative when $ a < 0$ and $ D \equal{} b^{2} \minus{} 4ac < 0$

Edited Now : thanks KMST

[quote="Patryk22"]Since you see that the equation has +1, then that means it will have two nonreal roots, unless you have a negative in either the ax^2 or bx part.
That should set you on the right path.[/quote]
I'm not sure what you mean   ...
In this case $ D\equal{}(a\minus{}3)^{2} \minus{}4a \equal{}(a\minus{}1)(a\minus{}9)$.And this is positive for every $ a>9$ and for every $ a<1$

[quote="enndb0x"]I think the function $ f(x) \equal{} ax^{2} \plus{} bx \plus{} c$ is always negative when $ a > 0$ and $ D \equal{} b^{2} \minus{} 4ac < 0$[/quote]You must have meant
"$ f(x)$ is always negative when $ a < 0$ and $ D < 0$"
or maybe
"$ f(x)$ is always positive when $ a > 0$ and $ D< 0$"

If $ D<0$ the function stays to one side of the x-axis. Even without invoking calculus and limits, you can see that for an $ x$ with a sufficiently large $ |x|$, the term $ ax^2$ "overwhelms" the other two terms, meaning that
$ |ax^2|>>|bx\plus{}c|$, so the sign of $ a$ determines the sign of the function.

Looking at it from the algebra viewpoint:
$ f(x) \equal{} ax^{2} \plus{} bx \plus{} c\equal{}a\left( x\plus{} \frac{b}{2a} \right)^2\minus{} \frac{b^2\minus{}4ac}{4a}$
or
$ f(x) \equal{} a\left( x\plus{} \frac{b}{2a} \right)^2\minus{} \frac{D}{4a}$

When $ a < 0$ and $ D < 0$, both terms are negative, so $ f(x) <0$.

When $ a > 0$ and $ D < 0$, both terms are positive, so $ f(x) >0$.


Point M is inside the $ ABCD$ quadrangle such that $ \widehat{CMD}$ $ \equal{}$ $ \widehat{MCB}$ $ \plus{}$ $ \widehat{MAB}$ , $ \widehat{AMB}$ $ \equal{}$ $ \widehat{MAD}$ $ \plus{}$ $ \widehat{MCD}$ and $ MA$ $ \equal{}$ $ MC$.
prove that:

$ BM$ $ \times$ $ AD$ $ \equal{}$ $ MA$ $ \times$ $ CD$ , $ AB$ $ \times$ $ CM$ $ \equal{}$ $ BC$ $ \times$ $ MD$


[quote="Jane_Bennet"]Prove that $ \sin{25^\circ}\sin{35^\circ}\sin{60^\circ}\sin{85^\circ} \equal{} \sin{20^\circ}\sin{40^\circ}\sin{75^\circ}\sin{80^\circ}$.[/quote]
First, we will prove that: $ sin a sin (60^o\minus{}a)sin (60^o\plus{}a)\equal{}\frac{1}{4}sin 3a$
Proof:
$ LHS\equal{}sin a (sin 60^o cos a\minus{}sin a cos 60^o)(sin 60^o cos a\minus{}sin a cos 60^o)$
$ \equal{}sin a(\frac{\sqrt{3}}{2})cos a\minus{}\frac{1}{2}sin a)(\frac{\sqrt{3}}{2})cos a\plus{}\frac{1}{2}sin a)$
$ \equal{}sin a(\frac{3}{4}cos^2 a\minus{}\frac{1}{4}sin^2 a)\equal{}\frac{1}{4}sin a[3(1\minus{}sin^2 a)\minus{}sin^2 a]$
$ \equal{}\frac{1}{4}(3sin a\minus{}4sin^3 a)\equal{}\frac{1}{4}sin 3a$
It follows that:
$ \sin{25^\circ}\sin{35^\circ}\sin{60^\circ}\sin{85^\circ}\equal{}\frac{1}{4}\sin{75^\circ}\sin{60^\circ}$
$ \sin{20^\circ}\sin{40^\circ}\sin{75^\circ}\sin{80^\circ}\equal{}\frac{1}{4}\sin{60^\circ}\sin{75^\circ}$
So we are done.

Ooh, brilliant! :clap2: Thank you! :omighty:

[quote="Jane_Bennet"]Prove that $ \sin{25^\circ}\sin{35^\circ}\sin{60^\circ}\sin{85^\circ} = \sin{20^\circ}\sin{40^\circ}\sin{75^\circ}\sin{80^\circ}$.[/quote]

We have
\begin{align*}LHS&=\sin 60\cdot\frac 12\sin 35(\cos 60-\cos 110)\\
&=\sin 60\left(\frac 14\sin 35-\frac 12\cos 110\sin 35\right)\\
&=\sin 60\left(\frac 14\sin 35-\frac 14(\sin 145-\sin 75)\right)\\
&=\sin 60\left(\frac 14\sin 35-\frac 14(\sin 35-\sin75)\right)\\
&=\frac 14\sin 60\sin 75\end{align*}
On the other hand,
\begin{align*}RHS&=\sin 75\cdot\frac 12\sin 40(\cos 60-\cos 100)\\
&=\sin 75\left(\frac 14\sin 40-\frac 12\cos 100\sin 40\right)\\
&=\sin 75\left(\frac 14\sin 40-\frac 14(\sin 140-\sin 60)\right)\\
&=\sin 75\left(\frac 14\sin 40-\frac 14(\sin 40-\sin 60)\right)\\
&=\frac 14\sin 75\sin 60\end{align*}
Thus $ LHS=RHS$.

[quote="NguyenDungTN"]First, we will prove that: $ sin a sin (60^o - a)sin (60^o + a) = \frac {1}{4}sin 3a$.[/quote]

Another proof: We have
\begin{align*}\sin a\sin(60^{\circ} + a)\sin(60^{\circ} - a) & = \frac 12\sin a(\cos 2a - \cos 120^{\circ}) \\
& = \frac 12\sin a\cos 2a + \frac 14\sin a \\
& = \frac 14(\sin 3a - \sin a) + \frac 14\sin a \\
& = \frac 14\sin 3a\end{align*}


calculus
integration
derivative
real analysis
real analysis unsolved

There is something mistaken when I am trying to prove Cauchy Iteration formula as follow?

Find   ${I={d\over {dx}}\int_{t=0}^x(x-t)^{n-1}f(t)dt={d\over {dx}}\int_{t=0}^xF(t)dt}$

Should we get $I=F(x)=(x-x)^{n-1}f(x)=0$ which is rather contradictory? :blush:

You can't just differentiate the left side by the fundamental theorem; the integrand depends on $x$ as well.

What's $F$ here? This doesn't feel right.

That comes up when I was trying to prove Cauchy Interation formula

$J^n(f)(x)=\int_{u_{n-1}=0}^x\int_{u_{n-2}=0}^{u_{n-1}}...\int_{u_{0}=0}^{u_1}f(u_{0})\,du_{0}... \,du_{n-2}\,du_{n-1}={1\over {(n-1)!}}\int_{u_{0}=0}^x(x-u_{0})^{n-1}f(u_{0})\,du_0$

by differentiating the RHS. I could only finish the prove by binomial expansion of $(x-u_0)^{n-1}f(u_0)$ and use backward induction. Could it be possible to apply a direct differentiation on the RHS?


A cross on a chessboard consists of a square together with its neighbors above, below, to the right, and to the left. With proof, find the largest number of crosses that can fit without intersection on a regular chessboard.

(A regular chessboard is and 8x8 square, for anyone who didn't know.)

Is it 7?  I was thinking find the total number of squares(64) and dividing it by 9, since that's kind of the number of squares a cross would take up, but it doesn't really take everything into account (like how the 3x3 areas of the crosses can overlap, just not the 5 squares).  I drew a picture of the board and did it that way, and I still got 7.  Can someone tell me how you're really supposed to do this?

I'm not really sure how to do it but I did find a way to fit 8 on so 7 would definitely be wrong.

How did you find it?  I really want to know how you're supposed to do this without drawing a picture.  It seems so simple, and I feel that I should know it.

Yeah I feel the same way.  I haven't found a way to do it w/out drawing it.  I've been thinking about it and considering the maximum # of squares that can be filled by part of a cross in certain rows and columns but haven't gotten all that far.  I'll try to think about it some more and maybe I'll figure it out.

No one posted in this for a while. So I am not sure if it is at Getting Started Level. Can a moderator help me out here?

In general, if I see a problem that says "with proof," I figure it's an Intermediate level problem. I let this problem stay on Getting Started for a while because the problem statement is accessible to beginners, but just as the conjecture known as Fermat's Last Theorem is easy to understand (as a problem statement) but hard to prove, so too this problem seems to be harder than it looks at first glance. I'll promote it in a few minutes.


Is it possible for a class or seminar to fill up (in terms of enrollment) before the first class session? If so, what is the maximum capacity for each class?

Yes, it is possible.  The cap varies from class to class depending on the level of material.  We usually set it at around 100.  We don't often hit it, but we have in the past.

mr. r, u have 4999 posts! (as of Sunday, January 28, 2007 11:05, 28 seconds pm)

didn't we do this with the mcounts prob series?

i know our user counter got to 92ish the first meeting

-jorian


Prove that any power of 2 cannot end in 4 identical digits.

Couldn't help noticing that at this moment 1234567890 has 1234 posts :D  :D


i want to take amc12B, but i have not registed it (deadline was 13th),  the problem is i can't find any school near to my home (washington state)which is offering amc12b. please help me what should i do?

umm today's the 10th and Mr DPatrick posted this site:

http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml


geometry
ratio
symmetry
analytic geometry
graphing lines
slope
Pythagorean Theorem

[quote="tianle"]ABCD us a square, E,F,G,H are on AB,BC,CD,DA respectively and the divide the respective side in the ratio 1:2. AG, BH, CE, DF forms a smaller square SPRQ in the centre of the big square. find the ratio of the small square to the big square.[/quote]

[hide]
Let the sides of the square be 1.  Draw diagonals $AC$ and $BD$.  You know that these are $\sqrt{2}$, since they are diagonals.  By midline theorem, $AG=BH=CE=DF=\frac{\sqrt{2}}{2}$.  The area of this is $\frac{1}{2}$, so the ratio is $\frac{\frac{1}{2}}{1}=\frac{1}{2}$.
[/hide]

I think that ccy has'nt the good solution: it seems to me that $E$ is not the middle of $AB$ and so on... :?
My construction "tells" me that the ratio would be 1:16.(or something like that...)
I'm workink at... see you soon !

[hide="My Incredibly Long and Non-Elegant Solution"]Let $AE=BF=CG=DH=x$ and $EB=FC=GD=HA=2x$.  Then $AB=BC=CD=DA=3x$.  We have right triangle $\bigtriangleup DAG$ with legs $DA=3x$ and $GD=2x$.  By Pythagorean Theorem, $AG=x\sqrt{13}$.  By symmetry, $AG=BH=CE=DF=x\sqrt{13}$.  The sides of the small square are parts of these four segments.  To find the side length of the small square, we can subtract the lengths of the extra parts of the segments from the length of the whole segment.

If you draw a diagram including segments $AG$, $BH$, $CE$, and $DF$, the large square will be divided into various shapes, including four congruent triangles.  (Draw it to see.)  The two "extra parts" of the segments are the two different legs of the triangles.

To find the lengths of the legs of the four congruent triangles, we can first note that $[ABH]=\left(\frac{1}{2}\right)(3x)(2x)=3x^{2}$.  Then the altitude to the hypotenuse of $\bigtriangleup ABH$, which is also the long leg of one of the four congruent triangles, is equal to (from the formula $A=\frac{1}{2}bh$)
$\frac{2[ABH]}{BH}=\frac{6x^{2}}{x\sqrt{13}}=\frac{6x}{\sqrt{13}}=\frac{6x\sqrt{13}}{13}$

By Pythagorean Theorem, the short leg of the four congruent triangles is equal to
$\sqrt{4x^{2}-\frac{36x^{2}}{13}}=2x\sqrt{1-\frac{9}{13}}=2x\sqrt{\frac{4}{13}}=\frac{4x}{\sqrt{13}}=\frac{4x\sqrt{13}}{13}$

Then the side length of the small square is equal to
$x\sqrt{13}-\frac{6x\sqrt{13}}{13}-\frac{4x\sqrt{13}}{13}=\frac{3x\sqrt{13}}{13}$

The ratio of the side lengths of the small and large squares is $\frac{\frac{3x\sqrt{13}}{13}}{3x}=\frac{\sqrt{13}}{13}$.  The ratio of the areas of the small and large squares is $\left(\frac{\sqrt{13}}{13}\right)^{2}=\boxed{\frac{1}{13}}$.[/hide]

The explanation is a little confusing, but if you draw the diagram it should make plenty of sense.

[hide]

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=548[/img]

Let $AB=3$. Since $\triangle ASH \sim \triangle ADG$, we find that $AH/SH = AG/DG$ or $2/SH = \sqrt{13}/2 \implies SH = \frac{4}{\sqrt{13}}$. This means that $AS = BP = \sqrt{2^2-(4/\sqrt{13})^2}=6/\sqrt{13}$. Finally, since $SH + SP + BP = \sqrt{13}$, we find that \[ BP^2 = (\sqrt{13}-\frac{10}{\sqrt{13}})^2 = \frac{9}{13}, \] so $[SPRQ]/[ABCD]=\boxed {\frac{1}{13}}$.
[/hide]

:P Same idea as mine, but much simpler and easier to understand... nice!

using chess's diagram, say the side is 3s, then the slope of DF, and BH is 2/3, and the y-intercepts are respectively (0,0) and (0,s), using the line line distance formula, $d=\frac{s}{\sqrt{1+\frac{4}{9}}}=\frac{3s}{\sqrt{13}}$ then the area ratio is $\frac{d^2}{3s^2}=\frac{1}{13}$


geometry
trapezoid
logarithms
LaTeX
ARML
ratio
number theory

Hey I was just wondering, is it possible to start a marathon here? Anyway, if so, then I'll post the first problem which is fairly easy:

Problem 1:

Suppose that the area of trapezoid $ABCD$ is $72$. If $AB=8, CD=16$ and angle $ADC=45  degrees$, Find $BC$.

[hide="answer n solution"]answer: $\sqrt{40}$
solution: from the general area formula, A=(1/2)(B1+B2)(h), the height is 6. from there create points E and F directly under A and B, respectfully (to create a 8x6 rectangle). thus EF=8. using basic geometry, DE=6. that leaves FC=2. to find BC, do the pythagorean theorum with FC and the height. [/hide]

Problem 2
simplify:
$\frac{bx(a^2x^2+2a^2y^2+b^2y^2)+ay(a^2x^2+2b^2x^2+b^2y^2)}{ay+bx}$

this problem hasn't gotten solved for two reasons:
1. nobody cares about the carolina threads  :(  OR
2. it's a little difficult and somebody wants a hint.
[hide="hint"]try some expansion and then nifty grouping.[/hide]

amazing how quickly a marathon dies... :?

I would do it, but I've already posted a problem before it, and I just about hate factoring, I guess this thread's gone... :mad:  :mad:

Well, there's only a single digit number of people who frequent this forum...o, well.

[hide="Solution"]
We guess that the numerator will factor into $bx+ay$ times something.  Keeping that in mind, we consider just the numerator

\begin{eqnarray*}
bx((ax)^2+2(ay)^2+(by)^2)+ay((ax)^2+2(bx)^2+(by)^2) &=& bx(ax)^2+2bx(ay)^2+bx(by)^2+ay(ax)^2+2ay(bx)^2+ay(by)^2\\
&=& bx(ax)^2+ay(ax)^2 + 2bx(ay)^2+2ay(bx)^2+bx(by)^2+ay(by)^2\\
&=& (ax)^2(bx+ay)+2abxy(bx+ay)+(by)^2(bx+ay)\\
&=& (bx+ay)((ax)^2+2ab xy+(by)^2)
\end{eqnarray*}

Thus we are left with $(ax)^2+2(ax)(by)+(by)^2$ or $(ax+by)^2$.  [/hide]

Do you want me to post the next problem...or is this thread done for?

Well, I'll post the next problem anyways.

[b]Problem 3[/b]

Find the number of triples $(a,b,c)$ of positive integers with $a\ge b\ge c$ that satisfy the equation

\[\frac 1a +\frac 1b+\frac 1c=1.\]

Hmmm...Is there an organized way to do this? I just kept using trial and error finding restrictions of what each integer can't be and I only found two: $(3,3,3)+(4,4,2)$, but I'm probably missing a few, joe, or anyone else, can you show an organized way of doing this besides trial and error?

You are missing 1 triple.  My solution looks long, but that's because I tried to explain every step.

[hide="Organized Solution"]
We organize our solution by considering what values $c$ can be.  It is important to notice that $c$ is at most 3.  If $c$ were more than $3$ than one of $\frac 1a$ or $\frac 1b$ would be greater than $\frac 13$, let's say that $\frac 1a$ is larger than $\frac 13$.  Then $a$ would be smaller than 3 which contradicts $c$ being the smallest.

So we only have to consider $c=1,2,3.$

[b]Case 1: $c=1$[/b]

This leads to $\frac 1a +\frac 1b=0$ which has no solutions in positive integers.

[b]Case 2: $c=2$[/b]

Now we have

\[\frac 1a+\frac 1b=\frac 12\Rightarrow 2b+2a=ab.\]

Then we use Simon's Favorite Factoring trick (that's what people on AoPS call it, at least ;) ):

\[ab-2a-2b=0\Rightarrow (a-2)(b-2)=4.\]

Thus, we have one solution for $a-2=4$ and $b-2=1$...$(a,b)=(6,3)$.  And one more solution for $a-2=2$ and $b-2=2$...$(a,b)=(4,4).$

Therefore, case 2 gives the solutions $(6,3,2)$ and $(4,4,2).$

[b]Case 3: $a=3$[/b]
We proceed in a similar fashion to [b]Case 2[/b].

\begin{eqnarray*}
\frac 1a+\frac 1b &=& \frac 23\\
\Rightarrow ab-\frac 32a-\frac 32b=0\\
\Rightarrow (a-\frac 32)(b-\frac 32)=\frac 94\\
\Rightarrow (2a-3)(2b-3)=9
\end{eqnarray*}

Thus $2a-3=3$ and $2b-3=3$ or $2a-3=9$ and $2b-3=1$.  The second produces an invalid solution (remember $b\ge c$).  The first produces the solution $(3,3,3).$

Thus the 3 triples are $(6,3,2), \ (4,4,2), \ \text{and} \ \, (3,3,3).$

NOTE:  Solving the equation $\frac 1x+\frac 1y=\frac 1k$ is a pretty common question in math competitions.  It even showed up on USC last year, iirc.  It's important to know how to solve it for certain constants, $k$.  It would be even better to know how to solve the general eqution too.

There were three important concepts in this problem: $(1)$ breaking the problem into cases (only do this if there are a manageable number of cases) $(2)$ Simon's Favorite Factoring Trick and $(3)$ solving the diophantine equation $1/a+1/b=1/k.$[/hide]

Here's a problem from an AMC-12, number 18 I think:

$\log{xy^3}=1$ and $\log{x^2y}=1$, what is $\log{xy}$?

IT'S ALIVE!!!!

[hide="answer_solution"]with properites of logs we have x*y^3=10 and y*x^2=10. solving this, we have y^2=x. plugging that in, log (y^5)=1.
then, 5 log y=1. log y = 1/5. now, we want log xy which equals log y^3 = 3 log y = 3 (1/5) = (3/5).
*probably should have used $LaTeX$  :D [/hide]

[b]Problem 5.[/b]

[size=150]Find the sum of the divisors of 420.[/size]

Too easy :P

[hide="Solution"]
$420=2^2\cdot3\cdot5\cdot7.$

Thus $\text{Sum of Divisors}=(1+2+4)(1+3)(1+5)(1+7)=1344.$
[/hide]

[b]Problem 6[/b]

Compute $S=\frac 15 + \frac 1{25} + \frac 2{125} + \frac 3{625} + \frac 5{3125}+\cdots$, where each numerator is the sum of the two preceeding numerators and, and each denominator is 5 times the preceding one.

[quote="joml88"]Too easy :P[/quote]
why are you solving these joe? there is no problem that iversonfan and I can come up with that you can't solve.

Well, if I didn't solve them the marathon would be dead right now ;)  And I was jk.

I'm not sure about this one, and I have no clue on Latex, so I will just give my answer:

[hide="Problem 6"]
5/19
????
[/hide]

Here's one from the 1995 ARML:

[u][b]Problem 7[/b][/u]

Starting at the same time at Corner M, Allison and Ben run in opposite directions around square track MNPQ, each travelling at a constant rate.  The first time they pass each other at corner P is the tenth time that they meet.  Compute the smallest possible ratio of the faster person's speed to the slower person's.

I love this question!
All words, few numbers!


calculus
integration
function
limit
derivative
absolute value
calculus computations

Let $ f(x,y)$, $ 0\leq x,y\leq 1$, satisfy the following conditions: for each $ x$, $ f(x,y)$ is an integrable function of $ y$, and $ \frac{\partial f(x,y)}{\partial x}$ is a bounded function of $ (x,y)$.  Prove that $ \frac{\partial f(x,y)}{\partial x}$ is a measurable function of $ y$ for each $ x$ and $ \frac{d}{dx}\int_{0}^{1}f(x,y)\,dy\equal{}\int_{0}^{1}\frac{\partial f(x,y)}{\partial x}\,dy$.

For the first part, I defined $ g_{n}\equal{}n\left(f(x\plus{}\frac{1}{n},y)\minus{}f(x,y)\right)$, which is integrable, and hence measurable for each $ n$, $ x$.  Then the limit function, $ \partial f/\partial x$ is measurable as well.

For the second part, I am thinking that we can bound $ \partial f/\partial x$ by $ \sup_{x,y}\partial f/\partial x$, which is integrable, so we can apply DCT.  Then I am thinking that $ \frac{d}{dx}\int_{0}^{1}f(x,y)\,dy\equal{}\lim_{n\rightarrow\infty}\int_{x}^{x\plus{}1/n}\left(\int_{0}^{1}f(x,y)\,dy\right)\,dt$, but can I switch the limits?

To show the differentiability it is not sufficient to consider the sequence $ \let(\frac{1}{n})$, but an arbitrary sequence $ h_n\to 0$. So just replace $ \frac{1}{n}$ by $ h_n$ to have a rigorous argument.

Note that $ g_n$ is integrable over $ [0,1]^2$ and bounded by $ M$, by the mean value theorem, if $ M$ denotes an upper bound of $ \frac{\partial f(x,y)}{\partial x}$. Since $ g_n(x,y)\to  \frac{\partial f(x,y)}{\partial x}$ pointwise, we get \[ \frac{d}{dx}\int_0^1f(x,y)dy=\lim_{n\to\infty}\int^1_0 g_n(x,y)dy=\int_0^1 \lim_{n\to\infty}g_n(x,y)dy=\int_0^1 \frac{\partial f(x,y)}{\partial x}dy\]by DCT.

Makes sense.  What are you choosing as your upper bound through the mean value theorem?  Could we use what I suggested, $ \sup_{x,y}\frac{\partial f(x,y)}{\partial x}$, since we are told that $ f_{x}(x,y)$ is a bounded function, so we just use the constant function at that point?

Right. I was a bit hasty: I interpret the assumption that the partial derivative is bounded as bounded from above and below. So $ M$ is meant to be the supremum of the [u]absolute value[/u] of the partial derivative.

If you apply the MVT w.r.t. $ x$ to the difference quotient $ g_n$ for fixed $ y$, then you get $ \mid g_n\mid\le M$.

Yes, of course I mean the supremum over the absolute value.  Everything else seems to look good. :)


$ f: [a,b]\rightarrow\mathbb{R}$ contineous and such that for all $ x\in]a,b[$ there exists $ t_x > 0$;$ t_x\in]0,inf(x \minus{} a;b \minus{} x)[$;$ f(x) \equal{} \frac {1}{2}(f(x \plus{} t_x) \plus{} f(x \minus{} t_x))$
Prove that $ f$ is convex.

[quote="mlok"]A counterexample: $ f(x) \equal{} \max(0,1 \minus{} |x|)$ on the interval $ [ \minus{} 10,10]$.
Did you mean $ \dots\le \frac {1}{2}\dots$ instead of $ \dots \equal{} \frac {1}{t_x}\dots$?[/quote]
You are right,I corrected.

But then $ f(x)\equal{}px\plus{}q$ for some constants $ p,q$. Let's first consider the case $ f(a)\equal{}f(b)\equal{}0$. If $ f$ is not identically $ 0$, then it attains either maximum or minimum in $ (a,b)$. If $ M$ is its (positive) maximum, then let $ c\equal{}\max\{x\colon f(x)\equal{}M\}$. By the assumption $ M\equal{}f(c)\equal{}\frac12(f(c\plus{}t_c)\plus{}f(c\minus{}t_c))$. But $ f(c\plus{}t_c)< M$ and $ f(c\minus{}t_c)\le M$, a contradiction. This proves $ f\equiv 0$.
The general case is reduced to the above by subtracting an appropriate linear function from $ f$.


I was wondering, how would you relate the function $\sum_{n=0}^{\infty}a_{n}^{2}x^{n}$ to $A(x) = \sum_{n=0}^{\infty}a_{n}x^{n}$?

I've tried to work out other problems with a generating function method, but I always get stumped when I get to something like this.

Thanks.

[hide="Cheating"]
If $a_{n+1}= a_{n}$ then, $\sum_{n=0}^{\infty}a_{n}^{2}x^{n}\rightarrow a_{n}\sum_{n=0}^{\infty}a_{n}x^{n}\rightarrow a_{n}A \left( x \right)$.
[/hide]

[hide="a thought"]

lets say, for simplicity
$a_{n}= n$

then you look at the 2nd differences. Those should be constant.

$f(x) = x+4x^{2}+9^{2}x^{3}+16x^{4}...$

$x f(x) = x^{2}+4x^{3}+9x^{4}+16x^{5}...$

$f(x)-xf(x) = x+3x^{2}+5x^{3}+7x^{4}... = g(x)$

$x g(x) = x^{2}+3x^{3}+5x^{4}...$

$g(x)-xg(x) = x+2x^{2}+2x^{3}+2x^{4}... = x+\frac{2x^{2}}{1-x}$

$f(x)-2xf(x)+x^{2}f(x) = x+\frac{2x^{2}}{1-x}$

solving for f(x) should be trivial from there,
is there a way to substitute a generalized sequence a_n into that logic somehow?
[/hide]

[quote="chickendude"]is there a way to substitute a generalized sequence a_n into that logic somehow?[/quote]

No.  As far as I know, there is no general method to relate those two generating functions.


inequalities
function
algebra
domain
logarithms
inequalities proposed

Let $f$ be a function such that :
$f(x_1)f(x_2)\ge{(f(\frac{x_1+x_2}{2}))}^2$ for all $x_1,x_2$ in its domain.Prove that for all $x_1,...,x_n$ in its domain :
$f(x_1)...f(x_n)\ge{(f(\frac{x_1+...x_n}{n}))}^n$.

let $g(x)=\ln f(x)$
by the condition one we have
$f(x_1)f(x_2)\ge{(f(\frac{x_1+x_2}{2}))}^2$
therefore
$\ln f(x_1)f(x_2)\ge \ln {(f(\frac{x_1+x_2}{2}))}^2$
$\ln f(x_1) + \ln f(x_2) \ge 2\ln {(f(\frac{x_1+x_2}{2}))}$
$\frac{g(x_1)+g(x_2)}{2}\ge g(\frac{x_1+x_2}{2})$
therefore $g(x)$ satisfies jensen, we do the inverse of the first step and we are done  ;)


calculus
integration
function
complex analysis
complex analysis unsolved

hello all,

there's this c.p.v that I was supposed to calculate, but the final solution offered by my teacher doesn't coincide with mine.(#1)
could someone please tell me which is right and why?

Gilad

The second one is correct.  I'm going to omit a lot of steps but I'll give you a sketch and you can fill in the details.

There's a singularity at 0 and 4.  Choose a branch such that the branch cut lies along the non-negative real axis.  Then there will be 2 circular contours, one around each of the poles.  You'll have to close the path with a large circle.  I'll leave you to check that the large circle goes to 0 as the radius goes to infinity, and that the one around 0 goes to 0 as the radius goes to 0.  The function is multi-valued so when you go on one side of the branch, we get $ I$ (what we want), and $ -e^{-2\pi i\lambda}I$ (opposite direction, hence the negative).  On the top, the integral along the semi-circle over 4 is $ -\pi\text{Res}(4)=-i\pi 4^{-\lambda}$.  Along the bottom, it's $ -i\pi 4^{-\lambda}e^{-2\pi i\lambda}$.  There are no singularities inside the contour so the residue theorem yields that the entire contour integral is 0.  Hence
$ \oint_{\Gamma}f(z)\ dz=0=I-e^{-2\pi i\lambda}I-i\pi 4^{-\lambda}e^{-2\pi i\lambda}-i\pi 4^{-\lambda}\Rightarrow$
$ I=i\pi 4^{-\lambda}\frac{1+e^{-2\pi i\lambda}}{1-e^{-2\pi i\lambda}}=i\pi 4^{-\lambda}\frac{e^{i\pi\lambda}+e^{-i\pi\lambda}}{e^{i\pi\lambda}-e^{-i\pi\lambda}}=\pi 4^{-\lambda}\cot(\pi\lambda)$

there's one thing I don't understand, why the integral along the bottom semi-circle around x=4
gets multiplied by the exponent?

suppose we were to calculate the integral of f(z) over a small circle around x=4,
wouldn't we get the following?

No, you're missing the point that you're dealing with a multi-valued function and as a result, there is a branch.  When you cross the branch, you're going onto another branch so you have to multiply by $ e^{2\pi i\alpha}$, where $ \alpha$ is whatever the exponent is.  In this case, it's $ \minus{}\lambda$.  You're not really making a complete circle, you're making two half circles, which are on completely different branches.


Find all real numbers $a,$ $b$ and $c$ from $[-\frac{1}{3},+\infty)$ such that
$a^{2}+b^{2}+c^{2}\geq3(a^{3}+b^{3}+c^{3})$ and $a+b+c\geq1.$

Nice one.
By the condition we have that $(a+b+c)(a^{2}+b^{2}+c^{2})\geq 3(a^{3}+b^{3}+c^{3})$

But from Chebychev we have $(a+b+c)(a^{2}+b^{2}+c^{2})\leq 3(a^{3}+b^{3}+c^{3})$

so the equality holds and holds when a=b=c and the rest is easy....... :)

[quote="silouan"]
But from Chebychev we have $(a+b+c)(a^{2}+b^{2}+c^{2})\leq 3(a^{3}+b^{3}+c^{3})$
[/quote]
But $a,$ $b$ and $c$ can be negative.
Id est, if $a\geq b\geq c$ then not true that $a^{2}\geq b^{2}\geq c^{2}.$ :wink:

You are right Isolved it only for positive numbers....But now I have another solution hope it is right..... :)
By the condition we have that
$a^{2}+b^{2}+c^{2}-3(a^{3}+b^{3}+c^{3})+\frac{a+b+c-1}{3}\geq 0$  or equivalent (by your method  :lol: )

$-\frac{1}{9}(3a-1)^{2}(3a+1)-\frac{1}{9}(3b-1)^{2}(3b+1)-\frac{1}{9}(3b-1)^{2}(3b+1)\geq 0$ which is true only for

$a=b=c=\frac{1}{3}$     :)

[quote="silouan"]
By the condition we have that
$a^{2}+b^{2}+c^{2}-3(a^{3}+b^{3}+c^{3})+a+b+c-1\geq 0$  [/quote]
Only $a^{2}+b^{2}+c^{2}-3(a^{3}+b^{3}+c^{3})+\frac{1}{3}\cdot(a+b+c-1)\geq 0.$ :wink:
My congratulations,  silouan! :lol:

[quote="arqady"]
My congratulations,  silouan! :lol:[/quote]

Thank you Arqady , and thanks for the nice problem...... :D


Working from seamus...

1/3 + 2/9 + 3/27...
=1/3 + 1/9 + 1/27... + 1/9 + 1/27... + 1/27 + ...
=(1/3)/(1-(1/3)) + (1/9)/(1-(1/3) + ...
=1/2 + 1/6 + 1/18...
=(1/2)/(1-(1/3))
=3/4

So the answer is 2^(3/4) (I think!)
If it wasn't clear, the way I evaluated the sequence was that I treated it as an infinite series of infinite series. Sorry if that wasn't clear. I'm really not sure if this is right either...

[quote="The Original Pi Guy"]
So the answer is 2^(3/4) (I think!)
[/quote]

Good job, you're right.

I went through it like that, and indeed, it seems to work. Good job, you two!

[quote="The Original Pi Guy"]If it wasn't clear, the way I evaluated the sequence was that I treated it as an infinite series of infinite series.[/quote]

More formally,
$\sum_{n=1}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty\sum_{k=1}^n\frac1{3^n}$
$=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1{3^n}= \sum_{k=1}^\infty\frac1{3^k}\cdot\frac1{1-\frac13}= \frac13\left(\frac1{1-\frac13}\right)^2=\frac34$

The step that needs justification is the interchanged order of summation. In this case, we can easily see that each term is positive and that the sum of the absolute values is finite. Either of those would be enough to invoke an appropriate theorem and conclude that the two orders of summation produce the same result.
Such theorems belong to calculus- one name to look for is "Fubini's theorem".


In a certain year the price of gasoline rose by $ 20\%$ during January, fell by $ 20\%$ during February, rose by $ 25\%$ during March, and fell by $ x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $ x$?

$ \textbf{(A)}\ 12\qquad
\textbf{(B)}\ 17\qquad
\textbf{(C)}\ 20\qquad
\textbf{(D)}\ 25\qquad
\textbf{(E)}\ 35$

[hide="Answer"]
Let $ c$ denote cost of gas. You are given $ c = c \cdot \frac {6}{5} \cdot \frac {4}{5} \cdot \frac {5}{4} \cdot x$ Since $ \frac {5}{4} \cdot \frac {4}{5}$ cancel out, you still need to cancel out $ \frac {6}{5}$ with a $ \frac {5}{6}$ which is multiplying by roughly $ .83$ or decreasing by $ 17% \implies \boxed{(B)}
$
[/hide]

[hide]
Let p be the price of gas at the beginning of January. Hence, $ p*(1.2)*(.8)*(1.25)*(1\minus{}x)\equal{}p$. We can simplify this to $ 1.2(1\minus{}x)\equal{}1$, so $ 1\minus{}x\sim .83$ and $ x\sim 17 \Rightarrow \boxed{B}$[/hide]


geometry
geometric transformation
projective geometry
geometry proposed
Desargues
Monge theorem

Let $ABC$ be a triangle, and $\omega_{a}$, $\omega_{b}$, $\omega_{c}$ be circles inside $ABC$, that are tangent (externally) one to each other, such that $\omega_{a}$ is tangent to $AB$ and $AC$, $\omega_{b}$ is tangent to $BA$ and $BC$, and $\omega_{c}$ is tangent to $CA$ and $CB$. Let $D$ be the common point of $\omega_{b}$ and $\omega_{c}$, $E$ the  common point of $\omega_{c}$ and $\omega_{a}$, and $F$ the common point of $\omega_{a}$ and $\omega_{b}$. Show that the lines $AD$, $BE$ and $CF$ have a common point.

Let $ U, V$ and $ W$ be the centers of the circles $ \Gamma_{A}, \Gamma_{B}$ and $ \Gamma_{C}$ respectively. The pairs of lines $ EF$ and $ VW$, $ FD$ and $ WU$, and $ DE$ and $ UV$ meet at $ X, Y$ and $ Z$ respectively (one or all three points may be at infinity, but the argument below works in those cases too). By applying Menelaus to triangle $ UVW$ and transversals $ \over arrow{E - F - X}, \over arrow{F - D - Y}$ and $ \over arrow{D - E - Z}$ yields that $ X, Y$ and $ Z$ are the intersection points of the common external tangents to $ \Gamma_{B}$ and $ \Gamma_{C}$, $ \Gamma_{C}$ and $ \Gamma_{A}$, and $ \Gamma_{A}$ and $ \Gamma_{B}$ respectively.
Now by Monge theorem we have that the points $ X, Y$ and $ Z$ are collinear (see here [url]http://mathworld.wolfram.com/MongesCircleTheorem.html[/url] ).
Consequently, the triangles $ ABC$ and $ DEF$ are perspective, therefore the lines $ AD, BE$ and $ CF$ are concurrent (by Desargues' Theorem).

we don`t really have to use monge`s theorem.
the incircle of triangle $\triangle UVW$ touches it`s sides at $D,E,F$.
$X$ is in fact the harmonical conjugate of $D$ wrt $V,W$, and it can be proven using very basic geometry facts that $X$ lies on $EF$ and $BC$. now, the fact that $X,Y,Z$ are collinear is another menelaus, or, speaking from projective geometry`s point of view, note that $X$ is the pole of $UD$ wrt the incircle of $UVW$. since $UD,VE,WF$ are concurrent, it follows that $X,Y,Z$ are collinear.
note : some of the points $X,Y,Z$ may be points at infinity, but the arguments above remain true.

[quote="maky"]note that $ X$ is the pole of $ UD$ wrt the incircle of $ UVW$. since $ UD,VE,WF$ are concurrent, it follows that $ X,Y,Z$ are collinear.[/quote]

why X is the pole of UD wrt the incircle of UVW??

and how menelaus yields that X,Y,Z are the intersections of tangents??

i am really a beginner of projective geom, :(  but i really want to understand the problem. :)

Let $P,Q,R$ be centers of the circles $\omega_a, \omega_b, \omega_c$. $E$ is the insimilicenter of $w_a,w_c$ , $F$ is the insimilicenter of $w_a,w_b$, and $QR\cap BC= P_1$ (say) is the exsimilicenter of $\omega_b,\omega_c$. So by Monge's theorem $E,F,P_1$ are collinear. So $EF \cap BC = P_1$, similarly we have $ED \cap AB = R_1$ and $FD \cap AC = Q_1$, where $P_1, Q_1, R_1$ are exsimilicenters of $\omega_b,\omega_c$ and $\omega_a, \omega_c$ and $\omega_a, \omega_b$ respectively. Again by Monge's theorem  $P_1,Q_1,R_1$ are collinear. So $\triangle ABC$ and $\triangle DEF$ are perspective from a line, so they must be perspective from a point. So $AD,BE,CF$ are concurrent.

I think it's sufficient to use three-center theorem on circles $Wa,Wb,Wc$,then by Desargues (in triangle DEF and ABC) it's trivial.

Let $ X=DE \cap AC$ and define $Y,Z$ similarly. By Monge, the internal centers of $(w_c,w_b),(w_b,w_a)$ and the external center of $(w_a,w_c)$ are collinear. That is, $Y$ is the external center of $(w_a,w_c)$. By Monge, the external centers of $(w_a,w_b),(w_c,w_a),(w_b,w_c)$ are collinear. That is, $XYZ$ are collinear. Thus by Desargues, $AD, BE, CF$ are concurrent.

Let $X$, $Y$, $Z$ be the exsimilicenteres of $\omega _b$ and $\omega _c$, $\omega _a$ and $\omega _c$, $\omega _a$and $\omega _b$.

From Monge's theorem,$X$, $Y$, $Z$ are colliniar,and because the exsimilicenter, insimilicenter and the centers of two circles are colliniar, we get that triangles $ABC$ and $DEF$ are perspective.So,by Desargues' theorem, the conclusion follows.

Notice $X=\overline{AB}\cap\overline{DE}$ is the exsimilicenter of $\omega_a$ and $\omega_b$ by 2-1 Monge. Similarly define $Y$ and $Z.$ By Monge, $X,Y,$ and $Z$ are collinear. Hence, $\triangle ABC$ and $\triangle DEF$ are perspective and Desargues finishes. $\square$