The function $ f :  \mathbb{R} \to \mathbb{R}$ satisfies the equation
\[ f(x\plus{}f(y))\plus{}f(f(x)\plus{}y)\equal{}2x\plus{}2y \quad \forall x,y  \in \mathbb{R}.\]
Prove that $ f(2x) \equal{} 2f(x)$ for all real $x$. Open problem:  find all functions $f$ which satisfy the above functional equation.

[quote="nguyenvuthanhha"][i]The function $ f \ : \ \mathbb{R} \to \mathbb{R}$ satisfies the equation :

   $ f(x \plus{} f(y)) \plus{} f(f(x) \plus{} y) \equal{} 2x \plus{} 2y \ \forall \ \ x;y \ \in \ \mathbb{R} \ \ (*)$

   Prove that : $ f(2x) \ \equal{} \ 2f(x)$

  Open problem :  Find all function satisfy  $ (*)$[/i][/quote]

$ f(x)\equal{}x$, $ \forall x\in \mathbb{R}$ is a solution of $ (*)$.

So the $ (*)$ may be written as $ 2f(x\plus{}y)\equal{}2x\plus{}2y$

$ x\equal{}y$ $ \implies$ $ 2f(2x)\equal{}4x$ $ \implies$ $ f(2x)\equal{}2x\equal{}2f(x)$, $ \forall x\in \mathbb{R}$

Q.E.D.

@prester : you proved the result only for the solution $ f(x)\equal{}x$ (in which case it is obvious) and not for all the solutions.
[quote="nguyenvuthanhha"][i]The function $ f \ : \ \mathbb{R} \to \mathbb{R}$ satisfies the equation :

   $ f(x \plus{} f(y)) \plus{} f(f(x) \plus{} y) \equal{} 2x \plus{} 2y \ \forall \ \ x;y \ \in \ \mathbb{R} \ \ (*)$

   Prove that : $ f(2x) \ \equal{} \ 2f(x)$
[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \plus{} f(f(x) \plus{} y) \equal{} 2x \plus{} 2y$

$ P(0,0)$ $ \implies$ $ f(f(0))\equal{}0$.

$ P(f(0),f(0))$ $ \implies$ $ 0\equal{}f(0)$

$ P(x,x)$ $ \implies$ $ f(x\plus{}f(x))\equal{}2x$ and so $ f(f(x\plus{}f(x))\equal{}f(2x)$

$ P(x\plus{}f(x),0)$ $ \implies$ $ f(x\plus{}f(x))\plus{}f(f(x\plus{}f(x)))\equal{}2x\plus{}2f(x)$ and so $ f(f(x\plus{}f(x))\equal{}2f(x)$

Comparing these two lines, we get $ f(2x)\equal{}2f(x)$
Q.E.D.

[quote="prester"][quote="pco"]@prester : you proved the result only for the solution $ f(x) \equal{} x$ (in which case it is obvious) and not for all the solutions.
[/quote]

Yes, but it should be sufficient to prove that for just one solution...Otherwise it should not valid for any solution. I mean that you proved that indipendenlty on the solution of (*) and this is obviously perfect, but I suppose that if you have a solution of (*) and you prove it, it should be ok too...
...Maybe I am wrong...but sincerely I cannot see why... :oops:[/quote]

I dont understand.

The question is to prove for all solutions, not for just one.

For example, the property $ f(1)\equal{}1$ is true for the solution $ f(x)\equal{}x$ but is not true for all solutions (it's false for solution $ f(x)\equal{}\minus{}2x$, for example).

Dear pco,  you're rigth as usual. I've made another stupid mistake...Sorry


Find all complex polynomial $ P(x)$ such that for any three integers $ a,b,c$ satisfying $ a \plus{} b \plus{} c\not \equal{} 0, \frac{P(a) \plus{} P(b) \plus{} P(c)}{a \plus{} b \plus{} c}$ is an integer.

From the equation we have $ \frac{P(a\plus{}b)\plus{}P(0)\plus{}P(c)}{a\plus{}b\plus{}c}$ is an integer . By the condition of problem we obtain : $ \frac{P(a\plus{}b)\plus{}P(c)\plus{}P(0)}{a\plus{}b\plus{}c}\minus{}\frac{P(a)\plus{}P(b)\plus{}P(c)}{a\plus{}b\plus{}c}$ is an integer for all integer $ a,b,c$ . Thus $ a\plus{}b\plus{}c|P(a\plus{}b)\plus{}P(0)\minus{}P(a)\minus{}P(b) ,\forall a,b,c\in Z$ ,take $ c$ c is large enough ,we must have $ P(a\plus{}b)\plus{}P(0)\minus{}P(a)\minus{}P(b)\equal{}0$ hence $ P(n)\equal{}kn\plus{}c$ forall n .Replace to the condition of problem ,it is easy to show that $ c\equal{}0$ and $ k$ is an integer . Thus all solutions are $ f(n)\equal{}kn,k\in Z$ .

[quote="TTsphn"]From the equation we have $ \frac {P(a \plus{} b) \plus{} P(0) \plus{} P(c)}{a \plus{} b \plus{} c}$ is an integer . By the condition of problem we obtain : $ \frac {P(a \plus{} b) \plus{} P(c) \plus{} P(0)}{a \plus{} b \plus{} c} \minus{} \frac {P(a) \plus{} P(b) \plus{} P(c)}{a \plus{} b \plus{} c}$ is an integer for all integer $ a,b,c$ . Thus $ a \plus{} b \plus{} c|P(a \plus{} b) \plus{} P(0) \minus{} P(a) \minus{} P(b) ,\forall a,b,c\in Z$ ,take $ c$ c is large enough ,we must have $ P(a \plus{} b) \plus{} P(0) \minus{} P(a) \minus{} P(b) \equal{} 0$ hence $ P(n) \equal{} kn \plus{} c$ forall n .Replace to the condition of problem ,it is easy to show that $ c \equal{} 0$ and $ k$ is an integer . Thus all solutions are $ f(n) \equal{} kn,k\in Z$ .[/quote]
You can't have $ P(kn)\equal{}kn$ from $ P(a\plus{}b)\plus{}P(0)\minus{}P(a)\minus{}P(b)\equal{}0$
Because $ P(n)$ uncertainty be real number.

[quote="math10"][quote="TTsphn"]From the equation we have $ \frac {P(a \plus{} b) \plus{} P(0) \plus{} P(c)}{a \plus{} b \plus{} c}$ is an integer . By the condition of problem we obtain : $ \frac {P(a \plus{} b) \plus{} P(c) \plus{} P(0)}{a \plus{} b \plus{} c} \minus{} \frac {P(a) \plus{} P(b) \plus{} P(c)}{a \plus{} b \plus{} c}$ is an integer for all integer $ a,b,c$ . Thus $ a \plus{} b \plus{} c|P(a \plus{} b) \plus{} P(0) \minus{} P(a) \minus{} P(b) ,\forall a,b,c\in Z$ ,take $ c$ c is large enough ,we must have $ P(a \plus{} b) \plus{} P(0) \minus{} P(a) \minus{} P(b) \equal{} 0$ hence $ P(n) \equal{} kn \plus{} c$ forall n .Replace to the condition of problem ,it is easy to show that $ c \equal{} 0$ and $ k$ is an integer . Thus all solutions are $ f(n) \equal{} kn,k\in Z$ .[/quote]
You can't have $ P(kn) \equal{} kn$ from $ P(a \plus{} b) \plus{} P(0) \minus{} P(a) \minus{} P(b) \equal{} 0$
Because $ P(n)$ uncertainty be real number.[/quote]

Yes, he can. $ P(a \plus{} b) \plus{} P(0) \equal{} P(a) \plus{} P(b)\Leftrightarrow P(a \plus{} b) \minus{} P(0) \equal{} (P(a) \minus{} P(0)) \plus{} (P(b) \minus{} P(0)),\forall a,b\in\mathbb{Z}$. Let $ Q(x): \equal{} P(x) \minus{} P(0)$ then $ Q(0) \equal{} 0$ and $ Q(a \plus{} b) \equal{} Q(a) \plus{} Q(b),\forall a,b\in\mathbb{Z}$. From that $ Q(n) \equal{} nQ(1),\forall n\in\mathbb{Z}$. Put $ k: \equal{} Q(1)$ then the two polynomials $ Q(n)$ and $ Q*(n) \equal{} kn$ are equal to each other in infinitely many points. Thus $ Q(n) \equal{} kn$, providing that $ P(n) \equal{} kn \plus{} c$ where $ c: \equal{} P(0)$.

Is it true ?  :blush:
Let $P(a,b,c) : a+b+c | p(a)+p(b)+p(c)$ . ( :thumbup: )
$P(a+b,0,c) : a+b+c | p(a+b)+p(0)+p(c)$ . (:winner_first: )
If (:winner_first: ) - ( :thumbup: ) we get $a+b+c | p(a+b)-p(a)-p(b)+p(0)$ .         ( :clap2: )
In  ( :clap2: ) if $\lim_{b \to +\infty }$ we get $p(a+b)=p(a)+p(b)-p(0)  $
Then $p(x) =kx+c$ . :first:


topology
advanced fields
advanced fields unsolved

I am reading the book Topology by Munkres, and I don't understand why the following map $ \phi$ is well-defined:

Let $ p: E \longrightarrow B$ be a covering map and let $ b \in B$.    Choose $ e \in E$ such that $ p(e) \equal{} b$.  If $ [f] \in \pi_1(B,b)$, let $ \widetilde{f}$ be the lifting of $ f$ to a path in $ E$ that begins at $ e$.   Then, the lifting correspondence derived from the covering  map $ p$ is the map $ \phi: \pi_1(B,b) \longrightarrow p^{ \minus{} 1}(b)$ defined as $ \phi([f]) \equal{} \widetilde{f}(1)$.

FS

[quote="-oo-"]use the homotopy-lifting property for homotopies. this shows that homotopic paths in $ B$ lift to paths with the same endpoint.[/quote]

Excuse me, what is this property homotopy-lifting for homotopies ?

FS

forget it ! In the previous page of the definition, there is a theorem showing what I asked.   Sorry!

the homotopy-lifting property applied to a homotopy.


[code]
Perform the idicated opperation:

5x-2                    3
---------     -   --------
x^2-5x-14        x^2+2x

Simplify:

5 - ./2
--------
./2

Simplify:

5./2(7./8+5./3)

Word prob:

Sam emptied his collection of nikels and dimes on his bed. if his 46 coints were worth (dollers)3.10, how many coints of each did he have.

[/code]

I will not keep deleting your posts. Please post these problems in the appropriate forum, classroom math.

Also, some problems with your problems:

1) These look a lot like you copied them from your homework.
2) You need to learn how to latex. If you are going to cheat, at least look good doing it.
3)You do not cite sources.
4)Why are you posting these? Usually other people accompany their posts with things like "hard problem, help" or "Fun problem." We can find problem sets other places, like kalva or the contests button on the forums. Interesting or unusual problems are better than bland busywork sets.

In light of your lack of sources, copy 'n' paste style, and unclear motive, I'm going to lock this thread. I apoligize if you are really trying to help, but these look very suspicious. If you are in fact trying to do something other than get homework help, post in the appropriate forum. Any other discussion may be conducted by PM, [i]or if you can absolutely not figure out how to PM[/i], a new thread.

Also, spellcheck please.


[b]Problem 3[/b]

Let $ k$ and $ n$ be positive integers such that $ k \leq n \minus{} 1$. Let $ S$ be a set of first $ n$ positive integers, and let $ A_{1}, A_{2}, \cdots A_{k}$ be non-empty subsets of $ S$. Prove that it is possible to colour some elements of $ S$ using two colours, red and blue, such that the following conditions are satisfied:

$ i)$ Each element of $ S$ is either left uncoloured or is coloured red or blue.
$ ii)$ At least one element of $ S$ is coloured.
$ iii)$ Each set $ A_{i} ( i \equal{} 1, 2, \cdots , k )$ is either completely uncoloured or it contains at least one red and at least one blue element.

Let wlog $ \vec{a}\equal{}(a_1,a_2,\ldots,a_n) \in \mathbb{R}^n$ our objects, and define the $ k$ indicator vectors $ \vec{\epsilon_{B_i}}\equal{}(\epsilon_{B_i,1},\epsilon_{B_i,2},\ldots,\epsilon_{B_i,n}) \in \{0,1\}^n$ for all $ 1 \le i \le k$ s.t. $ \epsilon_{B_i,j}\equal{}1$ iff $ a_j \in B_i$. Consider the system of $ k$ equations $ \vec{\epsilon_{B_i}} \cdot \vec{a}\equal{}0$(with infinite solution and rank $ k<n$), color $ a_j$ red iff it is positive, blue iff it is negative, non colored otherwise. And we are done.


[color=blue]Let $ u_n$ be a sequence such that[/color] $ \sum |u_n|$ and $ \sum n|u_n|$ [color=blue]converge[/color]
[color=blue]let [/color]$ v_n\equal{} \sum_{k\equal{}n}^{\plus{}\infty} u_k$

[color=blue]1-Prove that [/color]$ \lim_{n\to\plus{}\infty} nv_n \equal{}0$
[color=blue]2-Prove that [/color]$ \sum_{n\equal{}1}^{\infty}v_n\equal{}\sum_{n\equal{}1}^{\infty} nu_n$

1.

\begin{align*} |n v_n | & = |\sum_{k = n}^{\infty}n u_k | \\
& \leq \sum_{k = n}^{\infty} n | u_k| \\
& \leq \sum_{k = n}^{\infty} k |u_k| \\
\end{align*}

The latter goes to 0 because it is the tail of a convergent series.

2.

The RHS converges absolutely, so we can rearrange the terms however we want. Writing out the LHS,

$ \sum_{n = 1}^{\infty} = \sum_{n = 1}^{\infty}\sum_{k = n}^{\infty} u_k = \sum_{k = 1}^{\infty} u_k + \sum_{k = 2}^{\infty} u_k + \sum_{k = 3}^{\infty} u_k + \cdots$

We see that $ u_i$ occurs $ i$ times on the LHS, making it, indeed a rearrangement of the RHS.


A bead of mass $ m$ is positioned at the top of a frictionless hoop of mass $ M$ and radius $ R$, which stands vertically on the ground. A wall touches the hoop on its left and a short wall of height $ R$ touches the hoop on its right, as shown in the figure to the right. All surfaces are frictionless. A bead is given a tiny, negligible, kick and it slides down the hoop as shown. What is the smallest value for $ m/M$ for which the hoop will rise off the ground at some time during the motion?

Not sure if my approach is corrrect...

Call A the point at the location of the sliding mass, B the point at the center, C the corner.  Then, $ \angle {ABC} \equal{} \alpha$, $ \angle {BAC} \equal{} \angle{ACB} \equal{} \theta$

The speed of the mass as a fn of $ \alpha$ is:
\[ v^2 \equal{} 2gR(1 \minus{} \sin{\alpha})\]
The force of contact between the hoop and the ball is:
\[ F_c \equal{} 2mg(1 \minus{} \sin {\alpha}) \minus{} mg\sin\alpha\]
Let's compute the torque about point B (corner).  If there is a net torque, the hoop will rotate upward, and will rise.

Torque due to F_c.
\[ \tau_c \equal{} rF_c \sin\theta \equal{} R\sqrt {2(1 \minus{} \cos \alpha}) (2mg(1 \minus{} \sin {\alpha}) \minus{} mg\sin\alpha)\sin (\frac {\pi \minus{} \alpha}{2}) \equal{} R\sqrt{2}mg \sqrt{1\minus{}\cos\alpha} (2\minus{}3\sin\alpha)(\sin (\frac{\pi\minus{}\alpha}{2}))\]
We want to maximize $ \tau_c$.

Torque due to gravity:

\[ \tau_g\equal{} Rmg\]

I'll post the rest later, but I think my method gets pretty messy.

My solution without torque (solving with torque next):

First, $ v(h)=\sqrt{2gh}$ from $ U_i+K_i=U_f+K_f$ as energy is conserved because everything's frictionless. $ h=R-R\cos\theta$ for $ 0\leq \theta \leq \pi/2$ which is our bounds anyway. That means $ v(\theta)=\sqrt{2gR(1-\cos\theta)}$, and the force exerted on the bead by the hoop is then $ \dfrac{mv^2}{R}=\dfrac{(m)(2gR)(1-\cos\theta)}{m}=2mg(1-\cos\theta)$. By Newton's Third, there is an equal and opposite force on the hoop by the bead. The magnitude of this force is then $ 2mg(1-\cos\theta)$, and we can see that $ \vec F = 2mg(1-\cos\theta)\sin\theta \uv{i} + 2mg(1-\cos\theta)\cos\theta \uv{j}$. We seek to maximize $ F_y$ because the maximum force required enables us to calculate the minimum value of $ m/M$. Let $ k=\cos\theta, 0\leq k\leq 1$, so we want to maximize $ 2mg(1-k)k$ which we can see is done at $ k=1/2$. This means that the maximum normal force in the $ \uv{j}$ direction is $ mg/2$. So now to counterbalance this, $ Mg\leq mg/2\implies m/M \geq 2$.

Yeah, that seems like the typical solution, but in this case, we have to examine both case, raising due to torque, and raising due to linear upward force, simply because m/M might be smaller with the torque soln.

The surfaces are frictionless...why would they rise due to torque only?


trigonometry
calculus
derivative
real analysis
real analysis unsolved

Study the convergence of the following series:

1) $ \sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}(1\minus{}\sqrt[n]{1\plus{}\sin n})$.

2) $ \sum\limits_{n\equal{}1}^{\infty}(1\minus{}\sqrt[n]{1\plus{}\sin n})$.

3) $ \sum\limits_{n\equal{}1}^{\infty}\frac{1\minus{}\sqrt[n]{1\plus{}\sin n}}{n}$.

$ 1\minus{}\sqrt[n]{1\plus{}\sin n}\sim\frac{\sin n}{n}$
so 1) 3)convergence

[quote="mpcfr"]$ 1 \minus{} \sqrt [n]{1 \plus{} \sin n}\sim\frac {\sin n}{n}$
[/quote]
No, this is false.

--------------------------------------------------------------------------
3) - convergence
2) - divergence
1) -  :maybe:  I'm thinking (it seems to be bounded but..)
later I'll post some arguments

Out of those (2) is the easiest: $ (1\plus{}x)^p\le 1\plus{} px \minus{}cpx^2$ for $ 0< p\le \frac 12$ with some $ c>0$ independent of $ p$ (just look at the second derivative to prove it). Thus, the general term of the series in $ 2$ is at least $ \minus{}\frac{\sin n}n\plus{}c\frac{\sin^2 n}{n}$ and the divergence follows from the convergence of $ \sum_{n}\frac{\sin n}n$ and the divergence of $ \sum_{n}\frac{\sin^2 n}n$.

I leave it to others to consider much more interesting questions (1) and (3).

Of course, the absolute convergence in (3) immediately follows from Hata's theorem and you need to use that $ \pi$ cannot be too well approximated by rationals anyway but it is not immediately clear what bound on the approximation is sufficient to demonstrate convergence. The main question here is whether the "high school bound" $ \frac 1{n!}$ will suffice.

(1) is just too good to be spoiled by any hints of mine. ;)


geometry
analytic geometry
absolute value
geometry solved

we have in the plane points : $A(1,-2)$ and $B(3,2)$ and we have also the
line $d$ of equation $y=x+1$ and $Q \in d$.
find all $Q$ so that $S_{QAB}=7$

now what i tried is to do something like this:
the area of $S_{QAB}$ is
$ S_{QAB} =
 \left| \begin{array}{ccc}
Q_x &Q_y &1\\
A_x &A_y &1\\
B_x &B_y &1\end{array} \right|
$
if i'm right this is a way of calculating the area of triangle depending on
coordinates of all points in it so... we have further :
$ S_{QAB} =
 \left| \begin{array}{ccc}
x &x+1 &1\\
1 &-2 &1\\
3 &2 &1\end{array} \right|=-2x+2+3x+3+6-2x-x-1=-2x+10\\
further:S_{QAB}=-2x+10;we,know:S_{QAB}=7 =>x=-\frac{3}{2}
$
but unfortunately this answer does not coincide with what the people who
wrote the book from wich this problem was take say at the end of the book
at the answers...what am i doing wrong ?
also they say that Q can be in 2 positions both satisfying the conditions of
the problems.
why am i finding only one of them and that one also not a correct one?

First, the formula is actually
\[S_{QAB} = \frac{1}{2} \left|
\begin{array}{ccc}
Q_x & Q_y & 1 \\
A_x & A_y & 1 \\
B_x & B_y & 1
\end{array}
\right|.\]
Second, since area is positive, you should take the absolute value of the determinant.  This will give you two solutions.

$
\displaystyle S_{QAB} =\frac{1}{2}\cdot \left|
\left| \begin{array}{ccc}
x &x+1 &1\\
1 &-2 &1\\
3 &2 &1\end{array} \right|\right|=|5-x|\\
5-x=7 => Q(-2,-1)\\
x-5=7 => Q(12,13);
$
thanks alot for the help


Just a tip on an easy way to plot functions, anywhere.
You can plot any function with a web browser using FooPlot:
  http://fooplot.com/

You can also specify functions right after the URL:
[code]
http://fooplot.com/x^2+2x+1
http://fooplot.com/cosh(x)
http://fooplot.com/1/(1+x^2),exp(-x^2)
[/code]

...you also posted this in The Round Table....

@admins: isn't there some rule where you can't make accounts just to advertise stuff? like for example some dude made a new account and started posting Nike stuff in the Classroom math forum...


calculus
integration
derivative
function
real analysis
real analysis theorems

The text reads "The integral $ \int{F(y,y')dx}$ whose end points are fixed, is $ \textbf{stationary}$ for weak variations if y satisfies the differential equation:

$ F \minus{} y'\frac{\partial{F}} {\partial{y'}} \equal{} c$"

I'm looking for clarification on what "stationary" actually means?

Thanks in advance.

To say the integral is stationary (with respect to y) means that the relative change of the integral with respect to variations, $ \nu$, in y is arbitrarily small.

Stated more precisely: $ \left| \frac{\int f(y\plus{}h\nu,y'\plus{}h{\nu}')dx \minus{} \int f(y,y') dx}{h} \right| \rightarrow 0$ as $ h \rightarrow 0$ for smooth variations $ \nu(x)$.

It is like an extention of the idea of a critical point in one dimensional calculus.

Like a derivative of a function being equal to zero in one dimensional calculus.  The integral is either maximum or minimum.  Correct?

Critical points have either a zero value derivative or the derivative may not be defined in one dim. calculus...  Is the same true here?

Well, close. A derivative being equal to zero in one dimensional calculus does not always mean a maximum or a minimum occurs there. It could be a saddle point. Also a maximum or a minimum can occur where the derivative is undefined.

I'm not 100% sure, but I believe stationary means that the variational of the integral does actually exist and is equal to zero. For a stationary integral, the path y could be a maximum, minimum, or neither for the integral.


Let $ a_{n}$ be the first $ n$ terms of that nester radical. The sequence is clearly increasing, so the limit exists.

Note that $ a_{n}=\sqrt{\frac{1}{1}+\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\ldots+\sqrt{\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{\sqrt 2}}}}}}$. The last radical is $ \leq 1$, so the radical before the last is $ \leq 1$ etc. so $ a_{n}\leq\sqrt 2$. Hence, $ \ell$ is finite.

Let's compute $ \ell^{2}-1 = k$ instead. If we look closely, we'll discover the recurrence relation: ${ b_{n+1}=\sqrt{\frac{1}{2}+\frac{1}{2}b_{n}}}$, where $ b_{n}= a_{n}^{2}-1$. Take the limit to get that $ 2 k^{2}= 1+k$, i.e. $ (2k+1)(k-1) = 0$. But $ k > 0$, so $ k = 1$, i.e. $ \ell =\sqrt 2$.


prove that n^2+11n+2 is not divisible by 12769 for all integer n.

Consider modulo 11, and you get $n^{2}$ is congruent to 7 modulo 11, which is not possible.


probability
geometry
perimeter
induction
combinatorics unsolved
combinatorics

Consider a regular pentagon. The six spots for a rabbit are the five vertices of the regular pentagon and the point where all the diagnols meet, thus there are six spots for the rabbit.A rabbit can move from one spot to any other spot randomly choosing the direction each time. What is the probability that the rabbit is at the meeting point of all diagnols or at the sixth spot after a million jumps.

Sorry if the question was not very clear, I meant that the six spots for the rabbit are the five vertices of the pentagon and the sixth spot is the point of intersection of the diagnols.

The rabbit can start at any of the positions  :blush:

Unfortunately the diagonals of a regular pentagon don't all meet at a point  :roll: I'm assuming you mean the rabbit can be in one of 6 places: one of the vertices, and the center of the pentagon. What is also not quite clear are the transition probabilities: if he can move from each spot to any other spot, then the whole geometry of the pentagon is quite irrelevant, and we're just performing a random walk on $ K_6$.

Assuming this is the case, we can analyze this in many ways but a simple one is to distinguish just two states: being at the center or being at the perimeter. At each step, the probability of moving from the center to the perimeter is 1, the probability of staying at the center is 0, the probability of moving from the perimeter to the center is $ 1/5$, and the probability of staying at the perimeter is $ 4/5$.

Let $ p_n$ be the probability that he is at the perimeter after $ n$ steps, and let $ c_n$ be the probability that he's at the center. Using the transition probabilities we just stated, we have

$ p_{n\plus{}1} \equal{} \frac{4}{5} p_n \plus{} c_n$

$ c_{n\plus{}1} \equal{} \frac{1}{5} p_n$

Since $ p_n\plus{}c_n \equal{} 1$, we have

$ p_{n\plus{}1} \equal{} \frac{4}{5} p_n \plus{} (1\minus{}p_n) \equal{} 1\minus{}\frac{1}{5}p_n$

Let's assume he starts at the perimeter (this does actually matter a little bit). Then $ p_0\equal{}1$, $ p_1 \equal{} \frac{4}{5}$, $ p_2 \equal{} \frac{21}{25}$ and in general it's easy to show by induction that

$ p_n \equal{} \frac{5^{n\plus{}1}\plus{}(\minus{}1)^n}{6 \cdot 5^n}$.

For large $ n$, this is very close to $ \frac{5}{6}$, so the probability of being at the center is about $ \frac{1}{6}$. This isn't surprising: the random walk on any connected, regular graph converges to the uniform one, so you're as likely to be on any vertex.

If the question had been about the graph of the pentagon + 1 vertex connected to all the others, the result would be quite different since the center would have degree 5 while the other vertices would have degree 3.

[quote="randomgraph"]If the question had been about the graph of the pentagon + 1 vertex connected to all the others, the result would be quite different since the center would have degree 5 while the other vertices would have degree 3.[/quote]  Specifically, in this case we would have $ p_{n \plus{} 1} \equal{} \frac{2}{3}p_n \plus{} c_n$ and $ c_{n \plus{} 1} \equal{} \frac{1}{3} p_n \equal{} \frac{1 \minus{} c_n}{3}$.  As this approaches a stable distribution, the probability he's in the center satisfies $ c \equal{} \frac{1 \minus{} c}{3}$ and so $ c \equal{} \frac{1}{4}$, i.e., in the limit he's in the center with probability $ \frac{1}{4}$.

Of course, if you're only interested in the limit probabilities, you can use this approach for the $ K_6$ case as well, and for any other graph, too. Since the question was about $ n\equal{}1000000$, I thought the OP wanted the precise formula  :)


Is there is a way I can put left justified and right justified text on the same line?

Use \hfill as in:
this is left \hfill this is right


Let $a,b,n$ be positive integers. Prove that $n!$ divides \[b^{n-1}a(a+b)(a+2b)...(a+(n-1)b)\]

look at this topic:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=21347[/url]


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let u and v be two different circles. Prove that the inversions with respect to the circles u and v commute with each other if and only if the circles u and v are orthogonal.

Nice, isn't it? Synthetically this is not too trivial.

  Darij

I think the necessity isn't too tough to prove. If $\mathcal C_i(O_i,r_i),\ i=1,2$ are the circles, then we can look at the image of $O_1$ under the two compositions of the inversions (or we can use points progressively "very" close to it, if you don't like infinite points). We will find the square of the distance between the centers to be $r_1^2+r_2^2$, which is the necessary and sufficient condition for orthogonality.

I didn't really think about sufficiency :).

[quote="grobber"]
I didn't really think about sufficiency :).[/quote]

the sufficiency should derive from the following lemma RL05012005
(of which I have yet to find a proof  :))

If ABCD is a cyclic quadrilateral and P = AD^BC and Q = AB^DC then

PQ^2 = PA*AD + QC*QD.

[quote="darij grinberg"][quote="sprmnt21"]the sufficiency should derive from the following lemma RL05012005
(of which I have yet to find a proof  :))

If ABCD is a cyclic quadrilateral and P = AD^BC and Q = AB^DC then

PQ^2 = PA*AD + QC*QD.[/quote]

What, really, you have no proof? Here's a hint: By Miquel's theorem, the point of intersection R of the circumcircles of triangles PAB and QBC different from B lies on the line PQ. Note that $PQ^2 = PQ\cdot\left(PR+QR\right)$. Now...

By the way, this is exactly the same proof of the sufficiency as I gave. The only thing I would remark is that you probably use not the lemma RL05012005 itself, but rather its converse...

Bye (will depart tomorrow early and come back on Sunday late).

  Darij[/quote]

Fantastic, Darij!!! very good proof of the lemma.

Perhaps I am confusing with sufficiency and necessity.
By the way I was reffering to the proof of the fact that if c1 and c2 are the two circles and P a generic point, then if Q = Inv_c2(Inv_c1(P)) =  Inv_c1(Inv_c2(P)) then c1 _|_c2.

Dim.

Let P1 = Inv_c1(P) and P2 = Inv_c2(P). Since O1P1*O1P = O1Q*O1P2=r1^2 and  analoguosly  O2P2*O2P = O2Q*O2P1=r2^2, where Oi and ri are the center and the radius of ci, we have that PP1QP2 is cyclic.

Then thanks to the lemma  DG05012005 it foollows that O1O2^2 = r1^2+r2^2, which is what was requested to prove.

[quote="grobber"]There are some typos, but I think I see what you mean.

If $\mathcal P(X)$ is, in general, the power of $X$ wrt $(ABCD)$, then what you want to show is that the circles centered at $P$ and $Q$, of radi $\sqrt{\mathcal P(P)}$ and $\sqrt{\mathcal P(Q)}$ respectively, are orthogonal.

If $Q_1,Q_2$ and $P_1,P_2$ are the points of tangency betwen the tangents from $Q,P$ respectively and $(ABCD)$, then $Q_1Q_2$ is the polr of $Q$, so $P,Q_1,Q_2$ are collinear, meaning that the inversion of pole $P$ and power $\mathcal P(P)$ fixes the circle $(Q,\sqrt{\mathcal P(Q)})$, Q.E.D.

[Darij answered before I managed to submit my post; you got me, Darij! ;) :D][/quote]

Sorry Grobber. I was writing while you posted your proof and remarks. I corrected, I think, the typos you found.

Your proof is, as usual, inversive/impressive.


Define the functions $f: R\rightarrow R\ ,\ f(x)=x^3$ and $g: R\rightarrow R\ ,\ g(x)=6x^2-12x+8\ .$
The our system becomes: $f(x)=g(z)\ ;\ f(y)=g(x)\ ;\ f(z)=g(y)\ .$
But for any $x\in R$ we have $g(x)=6(x-1)^2+2\ge 2$, i.e. $g(x)\ge 2\ .$
Therefore, $f(x)=g(z)\ge 2$ a.s.o $\Longrightarrow$ $x^3\ge 2$ a.s.o. $\Longrightarrow$ $x\ge \sqrt [3] 2$ a.s.o.
The functions $f,g: \left[\sqrt [3] 2,\infty\right)\rightarrow R$ are increasing strictly.
[u]The first case:[/u] $x=y=z\Longrightarrow x^3-6x^2+12x-8=0\Longrightarrow (x-2)^3=0\Longrightarrow \boxed {x=y=z=2}\ .$
[u]The second case:
[/u] $\boxed {x<y}\Longrightarrow f(x)<f(y)\Longrightarrow g(z)<g(x)\Longrightarrow$
$\boxed {z<x}\Longrightarrow f(z)<f(x)\Longrightarrow g(y)<g(z)\Longrightarrow$
$\boxed {y<z}\Longrightarrow y<z<x<y\Longrightarrow y<y$, what isn't truly.
Thus, the our system has a only solution $x=y=z=2\ .$

[b]Another similar exercises.[/b]

$1.\blacktriangleright x+\frac 1x=2y\ ;\ y+\frac 1y=2z\ ;\ z+\frac 1z=2x\ .$
$2.\blacktriangleright x^3=9z^2-27z+27\ ;\ y^3=9x^2-27x+27\ ;\ z^3=9y^2-27y+27\ .$
$3.\blacktriangleright x^2_k+mx_k+\left(\frac{m-1}{2}\right)^2=x_{k+1}$, where $k\in \overline {1,n}$ and $\ m\in R,\ n\in N,\ n>2\ (n+1: =1)$.
$4.\blacktriangleright x+y=\sqrt{4z-1}\ ;\ y+z=\sqrt{4x-1}\ ;\ z+x=\sqrt{4y-1}\ .$

$\sum \rightarrow s_3-6s_2+12s_1-24=0$ where $s_k=\sum_{cyc} x^k$. This newton's sum corresponds to $r^3-6r^2+12r-8=0 \implies (r-2)^3\implies r=2$ and $(x,y,z)=(2,2,2)$.

To Virgil Nicula's excersizes: 1) (1,1,1). 2) (3,3,3). These can both be proven to be the only answer in a similar fashion as what he used.

[quote="Virgil Nicula"]Altheman, I didn't understand your solution. In my opinion, I think you confuse $\longrightarrow$ with $\longleftarrow$ ![/quote]

true! i did not prove i had the only solution

OWNAGE[hide]Here is a beautiful solution.
Firstly, factor each each equation:
a ^ 3 - 6 * b ^ 2 + 12 * b - 8
= (a - 2)(a^2 +2*a + 4) - 6b * (b - 2).
Add the three equations, get:
(x - 2) ^ 3 + (y - 2) ^ 3 + (x - 2) ^ 3 = 0
If all of x,y,z aren't 0, then because the sum is 0, one of  (x - 2), (y - 2), (z - 2) is negative, the other 2 are positive or vise versa. But then, re-writing the equations in terms of absolute values of (x - 2), (z - 2),(y - 2)
, we get a^3 + b^3 = c^3, which is impossible by Fermat's Last Theorem.  :lol:
So, one of x,y,z is 2. Wolog, let it be x.Then:
(y - 2)(y^2 + 2*y +4) = 0, which gives just one real solution for y: y = 2.z = 2 in the same way.
So x = y = z.[/hide]

[quote="rem"]OWNAGE[hide]which is impossible by Fermat's Last Theorem.  :lol:
[/hide][/quote]

nothing says it's integers  :P

For the exercise posted by virgil,
the answer for question 1 should be
(1,1,1) or (-1,-1,-1).
To apply virgil's method, we need to separate the cases for x > 0 or x < 0 because  the monotonicity of the function $x+\frac{1}{x}$. will change according to x > 0 or x < 0.

[quote="Virgil Nicula"][hide="Here is, I dare to say, a nice solution !"]
[color=darkblue][u]Definition.[/u] $x\ .s.s.\ y\Longleftrightarrow x=y=0\ \vee\ xy>0$, i.e. the real numbers $x,y$ have a $\underline{s}\mathrm {ame}\ \underline {s}\mathrm {ign}$ ([u]sgn x=sgn y[/u]).
If $x=2$ then $\boxed {x=y=z=2}$. Suppose $x\ne 2$, i.e. $2\not \in \{x,y,z\}$. In this case, I take the first equation $x^3-6z^2+12z-8=0$ and I will do three different groupings:
$0.\blacktriangleright x^3=6(z-1)^2+2\ge 2\Longrightarrow x\ge \sqrt [3] 2>1$ a.s.o.
$1.\blacktriangleright x^3-8=6z(z-2)\Longrightarrow (x-2)z(z-2)\ge 0\Longrightarrow (x-2)\ .s.s.\ (z-2)$ a.s.o.
$2.\blacktriangleright \left(x^3-z^3\right)=(2-z)^3\Longrightarrow (x-z)\ .s.s.\ (2-z)$ a.s.o.
Therefore, $(x-z)\ .s.s.\ (2-x)\ .s.s.\ (2-z)$ a.s.o. $\Longrightarrow$
$(x-z)\ .s.s.\ (y-x)\ .s.s.\ (z-y)\ .s.s.\ (2-x)\ .s.s.\ (2-y)\ .s.s.\ (2-z)$, what is false. We came to the conclusion that $x=y=z=2$ is the only solution of the system.[/color] [color=darkred]Am I right when I said that it is a nice solution ?[/color][/hide][/quote]

clever!


Let $ f \in L_1$ such that $ \widehat{f} \in L_2$. Then $ f \in L_2$?

Yes. This is a consequence of the Plancherel theorem.

We can prove that if $ f\in L^1\cap L^2,$ then $ \|f\|_2\equal{}\|\widehat{f}\|_2.$ Furthermore, since the inverse Fourier transform has the same structure as the Fourier transform, we can reverse the roles of $ f$ and $ \widehat{f}$ in this.

The Plancherel theorem then allows us to define a unique operator on $ L^2$ which is an isometric isomorphism from $ L^2$ to $ L^2$ and which coincides with the Fourier transform operator on $ L^1\cap L^2.$


inequalities
function
inequalities proposed
highschoolmath
algebra

I mean the inequality:If $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1,a,b,c \ge 0$
Then:
$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\le \frac {9}{2}$
http://www.mathlinks.ro/Forum/viewtopic.php?t=36873

Proof:
Because the inequality is equal to:
$ \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {ca}{1 \minus{} ca}\le \frac {3}{2}$
But we have:
$ \frac {ab}{1 \minus{} ab} \equal{} \frac {ab}{a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab} \equal{} \frac {2ab}{2a^{2} \plus{} 2b^{2} \plus{} 2c^{2} \minus{} 2ab}\le \frac {2ab}{a^{2} \plus{} b^{2} \plus{} 2c^{2}}\le \frac {1}{2}( \frac {a^{2}}{a^{2} \plus{} c^{2}} \plus{} \frac {b^{2}}{b^{2} \plus{} c^{2}})$(The last one you can apply Cauchy's inequality)
So we have:$ \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {ca}{1 \minus{} ca}\le \frac {3}{2}$Directly.
Do you think it is nice? :)

[quote="pvthuan"]Congratulations, Zhaobin! Very nice and direct proof.

There have been at least 4 solutions to this one![/quote]
It is not from me :D

I have not seen yet the other solutions , but this is a very elegant  and clever solution, I hope the best ! Congrats!

  [u]Babis[/u]

Very nice your solution, arqady.  :)
In Algebraic Inequality there are other two solutions.

[quote="pvthuan"]The trick also fails for 4 variable equivalent I proposed long time ago.
Let $a,b,c,d$ be nonegative numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=1,\] prove that \[\frac1{1-ab}+\frac1{1-bc}+\frac1{1-cd}+\frac1{1-da}+\frac1{1-bd}+\frac1{1-ac}\leq8.\] [/quote]
But we have:

$\frac{ab}{1-ab}\le \frac{2ab}{a^{2}+b^{2}+2c^{2}+2d^{2}}\le \frac{1}{2}(\frac{a^{2}}{a^{2}+c^{2}+d^{2}}+\frac{b^{2}}{b^{2}+c^{2}+d^{2}})$

Ok, this one was solved in my book in the spirit of mixing variable. It's not complicated; it's also nice.

[quote="pvthuan"]The trick also fails for 4 variable
[/quote]
I don't care of how you solve your inequality.
But the trick here won't fails for 4 variable,you must admit it first. :D

May I suggest that we refer to this method as the "[b][i]Puzzle method[/i][/b]" of proving an ineq? The RHS being nicely composed like the parts of a puzzle.  :P  :P

Note that in a similar way, we have $\frac{(a+b)^{2}}{3a^{2}+3b^{2}+4c^{2}-2ab}\le \frac{(a+b)^{2}}{2(a^{2}+b^{2}+2c^{2})}\le \frac{1}{2}(\frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}})$,

which yields a direct puzzle proof of the stronger ineq \[\frac{\bigl(\frac{a+b}2\bigl)^{2}}{1-\bigl(\frac{a+b}2\bigl)^{2}}+\frac{\bigl(\frac{b+c}2\bigl)^{2}}{1-\bigl(\frac{b+c}2\bigl)^{2}}+\frac{\bigl(\frac{c+a}2\bigl)^{2}}{1-\bigl(\frac{c+a}2\bigl)^{2}}\le\frac32,\] equivalent to $\frac1{1-\bigl(\frac{a+b}2\bigl)^{2}}+\frac1{1-\bigl(\frac{b+c}2\bigl)^{2}}+\frac1{1-\bigl(\frac{c+a}2\bigl)^{2}}\le\frac92$ as used
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81709]here[/url].
Maybe after all, the second ineq of that thread can also be proved by using CS in some way. :roll:

This PUZZLE METHOD has just three babies. Please produce more, Spenferkel!  :lol:

[quote="zhaobin"]$\frac{2ab}{a^{2}+b^{2}+2c^{2}}\le \frac{1}{2}( \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}})$(The last one you can apply Cauchy's inequality)
[/quote]

Hi !

I'm a beginner and I do not know how to apply Cauchy's inequality here. Can you give me a hint ? Thanks a lot ! :huh:

[quote="mstoenescu"][quote="zhaobin"]$\frac{2ab}{a^{2}+b^{2}+2c^{2}}\le \frac{1}{2}( \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}})$(The last one you can apply Cauchy's inequality)
[/quote]
[/quote]

Write it as $4ab \leq ( (a^{2}+c^{2})+(b^{2}+c^{2}) ) \cdot \left( \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}}\right)$. Now apply Cauchy on the right.

Multsumesc perfect radio ! Thanks a lot !! :roll:

[quote="Vasc"]Nice solution, zhaobin.
However, I think it doesn't work for the sharper inequalitiy
$\frac 1{5-6ab}+\frac 1{5-6bc}+\frac 1{5-6ca}\leq 1$
(page 442 in Algebraic Inequalities).[/quote]

actually it's on page 422. :D
the second solution provided there shows a stronger inequality. :wink:

[quote="pvthuan"]This PUZZLE METHOD has just three babies. Please produce more, Spanferkel!  :lol:[/quote]Two very nice ones: (Well, actually for me most puzzle proofs are "nice", cuz once you see them, they just look so simple and straightforward, though they can prove some very strong/difficult inequalities :D )

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41931]This one from IMO 01[/url] (I think this is the first puzzle proof I've ever seen :) and I kind of like it):
Let $a,b,c >0$. Prove that :
$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ac}}+\frac{c}{\sqrt{c^{2}+8ba}}\geq 1$
This is proved by showing
$\frac{a}{\sqrt{a^{2}+8bc}}\geq \frac{a^{\frac43}}{a^{\frac43}+b^{\frac43}+c^{\frac43}}$.

Or take [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=62256]this one from the chinese forum:[/url]

[quote="zhaobin"]for $a,b,c>0$
prove
$1 \leq \sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}}+\sqrt{\frac{b^{3}}{b^{3}+(c+a)^{3}}}+\sqrt{\frac{c^{3}}{c^{3}+(a+b)^{3}}}< \sqrt 2$
[/quote]To prove it, [quote="scpajmb"]prove
$\frac{{a^{2}}}{{a^{2}+b^{2}+c^{2}}}\leq \sqrt{\frac{{a^{3}}}{{a^{3}+(b+c)^{3}}}}< \frac{{\sqrt 2 a^{\frac{3}{2}}}}{{a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}}}$[/quote]

[quote="pvthuan"]Here is one more baby from MY BOOK: :lol: \[\frac{a}{\sqrt[3]{a^{3}+63bcd}}\geq \frac{a^{\frac{21}{16}}}{a^{\frac{21}{16}}+b^{\frac{21}{16}}+c^{\frac{21}{16}}+d^{\frac{21}{16}}}.\] [/quote]There you go, and it's easy to show that for $n$ variables, when the $63$ is replaced by $n^{n-1}-1$, it works with the exponent $p=\frac{n^{n-1}-1}{n^{n-1}-n^{n-2}}=1+\frac1n+\cdots+\frac1{n^{n-2}}$ :D  :D

your solution looks too easy, and in fact it is not correct. You do the summing up for each case separately, but $ab,ac,bc$ are mixed between the two cases. :(

zhaobin,thank you for pointing out my fault.I forgot that $\frac{1}{ab}\geq 2.$ :(

Yes, you are right, Beat. We do not need the positivity of the numbers. Even the following inequality is true for REAL numbers.

Let $a,b,c,d$ be real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=1,\] prove that \[\frac1{1-ab}+\frac1{1-bc}+\frac1{1-cd}+\frac1{1-da}+\frac1{1-ac}+\frac1{1-bd}\leq8.\]

[quote="Beat"]If a^{2}+b^{2}+c^{2}=1,$a,b,c \ge 0$
Then:
$\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\le \frac{9}{2}$
[i]

 I don`t see where do we use that $a,b,c\ge0$ !!?!?
Can somebody explain me, please[/i][/quote]
Since, the following inequality
[quote="zhaobin"]
$\frac{2ab}{2a^{2}+2b^{2}+2c^{2}-2ab}\le \frac{2ab}{a^{2}+b^{2}+2c^{2}}$
[/quote]
is wrong for $ab<0.$ :wink:

[quote="arqady"]
[quote="zhaobin"]
$\frac{2ab}{2a^{2}+2b^{2}+2c^{2}-2ab}\le \frac{2ab}{a^{2}+b^{2}+2c^{2}}$
[/quote]
is wrong for $ab<0.$ :wink:[/quote] :D  :D But we don't need that one. We only need
$\frac{2ab}{2a^{2}+2b^{2}+2c^{2}-2ab}\le \frac{{a^{2}+b^{2}}}{a^{2}+b^{2}+2c^{2}}$ which is still valid.
And anyway, if in the original ones some of $a,b,c(,d)$ are replaced by their negative, the LHS only becomes smaller.  :D  :lol: So they indeed hold for all $a,b,c,d$.

Very nice, spanferkel. I agree with you. :lol:
But I only have answered a question which  has been asked. :wink:

OK. oops I meant
$\frac{2ab}{2a^{2}+2b^{2}+2c^{2}-2ab}\le \frac{\frac{(|a|+|b|)^{2}}2}{a^{2}+b^{2}+2c^{2}}$ just in case somebody wondered :wink:  :wink:  :wink:

I have a proof of a nice inequality;
here it is:
Let a,b,c be nonegative numbers such that$a^{2}+b^{2}+c^{2}=1; find max and min of f(a,b,c) = \frac{a^{2}}{1-bc}+\frac{b^{2}}{1-ca}+\frac{c^{2}}{1-ab}.$

Congratulations!!Very short and elegant solution!Just what i like in Maths!

Very good,Spanferlel.
However,I think it doesn't work for the problem:

Let a,b,c satisfy a,b,c>=0;$a^{2}+b^{2}+c^{2}=1$
find max and min of

       f(a,b,c)=$\frac{a}{1-bc}+\frac{b}{1-ca}+\frac{c}{1-ab}$ :)

It is more elegantly proved using the Engel Cauchy form inequality or Titu's Lemma.... :)  :roll:

[quote=zhaobin]I mean the inequality:If $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1,a,b,c \ge 0$
Then:
$$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\le \frac {9}{2}$$
http://www.mathlinks.ro/Forum/viewtopic.php?t=36873

Proof:
Because the inequality is equal to:
$$ \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {ca}{1 \minus{} ca}\le \frac {3}{2}$$
But we have:
$ \frac {ab}{1 \minus{} ab} \equal{} \frac {ab}{a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab} \equal{} \frac {2ab}{2a^{2} \plus{} 2b^{2} \plus{} 2c^{2} \minus{} 2ab}\le \frac {2ab}{a^{2} \plus{} b^{2} \plus{} 2c^{2}}\le \frac {1}{2}( \frac {a^{2}}{a^{2} \plus{} c^{2}} \plus{} \frac {b^{2}}{b^{2} \plus{} c^{2}})$(The last one you can apply Cauchy's inequality)
So we have:$ \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {ca}{1 \minus{} ca}\le \frac {3}{2}$Directly.
Do you think it is nice? :)[/quote]
$$\implies$$
Let $a, b, c$ be the non-negative numbers such that: $a^2+b^2+c^2=1$. Prove that
$$\frac{1+ab}{1-bc}+\frac{1+bc}{1-ca}+\frac{1+ca}{1-ab} \leq 6$$
$$\frac{1+ab}{1-bc}+\frac{1+bc}{1-ca}+\frac{1+ca}{1-ab} \leq \frac{1+ab}{1-ab}+\frac{1+bc}{1-bc}+\frac{1+ca}{1-ca} =3+2\left( \frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ca}\right)\leq 6.$$
[url=https://artofproblemsolving.com/community/c6h36873p367902]here[/url] [url=https://artofproblemsolving.com/community/c6h1085792p4798571]here[/url]

[hide=*][quote=spanferkel][quote="le_vi"]I have a proof of a nice inequality;
here it is:
Let a,b,c be nonegative numbers such that$a^{2}+b^{2}+c^{2}=1$; find max and min of $f(a,b,c) = \frac{a^{2}}{1-bc}+\frac{b^{2}}{1-ca}+\frac{c^{2}}{1-ab}.$[/quote]
Yes! For the minimum, $\sum\frac{c^{2}}{1-ab}=\sum\frac{c^{4}}{c^{2}-abc^{2}}\ge \frac{1}{1-abc\sum a}\ge1$ with equality iff $abc=0$.
For the maximum, $\sum\frac{c^{2}}{1-ab}\le2\sum\frac{c^{2}}{1+c^{2}}\le\frac32$ by Jensen & the convex function $\frac x{1+x}$, with equality iff $a=b=c$.  :)[/quote][/hide]
Let a,b,c be nonegative numbers such that  $a^2+b^2+c^2=1.$  Prove that$$1\leq \frac{a^2}{1-bc}+\frac{b^2}{1-ca}+\frac{c^2}{1-ab}\leq \frac{3}{2}$$
$$\frac{ab}{1-bc}+\frac{bc}{1-ca}+\frac{ca}{1-ab}\leq \frac{3}{2}$$
$$ \frac{a}{1-bc}+\frac{b}{1-ca}+\frac{c}{1-ab}\leq\frac{3\sqrt 3}{2}$$
$$\frac{1+ab}{1-bc}+\frac{1+bc}{1-ca}+\frac{1+ca}{1-ab}\leq 6$$
[url=https://artofproblemsolving.com/community/c6h230910p1277116]h[/url] [url=https://artofproblemsolving.com/community/c6h600016p13979130]h[/url]

Let a,b,c be nonegative numbers such that $a^2+b^2+c^2=1.$ Prove that$$\frac{a^2}{k-bc}+\frac{b^2}{k-ca}+\frac{c^2}{k-ab}\leq \frac{3}{3k-1}$$
Where $k\in N^+.$


calculus
integration
function
algebra
domain
calculus computations

How do I solve this double integral?

 $ \int \int_{S1} xy dA$

where S1 is the circle {$ (x,y)\ |\ x^2\plus{}y^2 \le 9$}

I have been stuck on this one for days, help will be deeply appreciated :)

It's an odd function in $ x$ on a domain symmetric across the $ y$ axis. It's an odd function in $ y$ on a domain symmetric across the $ x$-axis. It's a harmonic function with value zero at the center. Any of those is sufficient to make the integral zero.

I believe such questions, if asked in a textbook or as an assignment question is given after being taught about the Change of Variables theorem.  If you want to compute it, you can use a suitable function, which, in this case, is standard.


Find polynoms f and g (for all real x) such that:

(x^2+x+1)f(x^2-x+1)=(x^2-x+1)g(x^2+x1).

Thanks!! If someone would re-write this using LaTeX i'll be very greateful  :D (i don't have installed LaTeX now on my machine). Thanks again  :P

[quote="Hexaditidom"]$ (x^2+x+1)f(x^2-x+1)=(x^2-x+1)g(x^2+x+1) $[/quote]
Notice that f(x)/x has infinite real roots, which forces f(x)=cx

excuse me. yeah like siuhochung wrote  x1 means x+1  :blush: my mistake.

Hey siuhochung your answer is right ... can I have a fully explained solution  :lol:  :D

Set $\ a=x^2+x+1$ and $\ b=x^2-x+1$ so that $\ a f(b)=bg(a)$ so that $\frac{f(b)}{b}=\frac{g(a)}{a}$.  So setting the last equation equal to some other variable $\ c$ we see that $\ f(x)=cx$ and $\ g(x)=cx$

siuhochung's way of realizing its a constant function sounds better than mine but thats how I see it done.

[quote="probability1.01"]I don't understand how g(b) transformed into f(b).  Why can you say f = g?  Could some please explain?  Thanks.[/quote]

Actually f and g are different. The constants are different.
Replace x by -x and multiply them you find that
$f(x^2-x+1)g(x^2+x+1)=f(x^2+x+1)g(x^2-x+1)$
for any real x. WLOG assume the degree of f is higher, then
$f(x^2-x+1)/g(x^2-x+1) = f(x^2+x+1)/g(x^2+x+1)$ for any x. Infinite real roots -> $f/g$ is constant.
I hope this is correct as I am not raelly sure..

Ohh I'm very sorry I misused the letters in my proof I'll edit it.

I found another solution but i don't understand something.

Let $\epsilon_{1}=- \frac{1}{2}+i\frac{\sqrt{3}}{2}$ a root of $X^2+X+1 \Longleftrightarrow 0=g(0)$

Doing again ($\epsilon_{2}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$) that we obtain $f(0)=0$.

Therefor $\exists f_{1}, f_{2} \in R[X]$ such that $f=X \cdot f_{1}$ and $g=X \cdot f_{2}$ and from the text of the problem results that $f_{1}(X^2-X+1)=g_{1}(X^2+X+1)$.

I can't figure it out :(. Can anyone explain this to me ?

[quote="mna851"]$\frac{f(b)}{b}=\frac{g(a)}{a}$[/quote]

Is this constant?! Is there a proof for this? :blush:

because it's true for ALL $a$ and $b$.  if you keep $a$ fixed and vary $b$, the equation is still true, which can only happen if $f(b)/b=c$, where $c$ is a constant.


conics
ratio
probability
vector
absolute value

I tried to post this in the intermediate topics forum, and have only gotten limited interest, so I want to see if it will do better here.

Find the general second degree equation in two variables for a conic section with eccentricity $e$, focus $(\alpha,\beta)$, and directrix $y=k$, in terms of $e$[edit], $\alpha$, $\beta$ and $k$.

General second degree equation in two variables indicates that it should be in the form $ax^{2}+bxy+c^{2}+dx+ey+f$.  Yes, that does mean that the equation should be in rectangular form, not polar form.

[quote="roadnottaken"]I tried to post this in the intermediate topics forum, and have only gotten limited interest, so I want to see if it will do better here.

Find the general second degree equation in two variables for a conic section with eccentricity $e$, focus $(\alpha,\beta)$, and directrix $y=k$, in terms of $3$, $\alpha$, $\beta$ and $k$.

General second degree equation in two variables indicates that it should be in the form $ax^{2}+bxy+c^{2}+dx+ey+f$.  Yes, that does mean that the equation should be in rectangular form, not polar form.[/quote]

If I remember my conic sections, the directrix does not have to be in the form $y=k$. And there may be two directrices. And do you mean in terms of $e$, instead of $3$?

[quote="cincodemayo5590"]If I remember my conic sections, the directrix does not have to be in the form $y=k$. And there may be two directrices. And do you mean in terms of $e$, instead of $3$?[/quote]
I haven't heard about the possibility of two directrices, but your other observations were both correct.  However, for simplicity's sake, I have limited it to just conic sections with a single, horizontal directrix.

Straigh forward,, Let point (x,y) be on the conic section.

distance from focus = $\sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}$
Distance for directrix = $y-k$
Rest is easy: (By definition of a conic section their ratio is $e$ )

[hide]$e(y-k) = \sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}$
Square both sides, if you want a neater form, the type you wrote)

 [/hide]

[quote="roadnottaken"][hide="There are two problems with that solution."]
     1.  Your definition for the distance to the directrix only works if $y$ is above the directrix.
     2.  The equation isn't in the form I asked for.[/hide][/quote]

1. Just put in absolute value bars and its fine.
2. That's not a problem, all you have to do is square it and rearrange terms. That's just a little tedious.

[quote="roadnottaken"][hide="There are two problems with that solution."]
     1.  Your definition for the distance to the directrix only works if $y$ is above the directrix.
     2.  The equation isn't in the form I asked for.[/hide][/quote]

 hmm ..problems are not with the solution..
1.Doesn't matter, if $y$ is above or below, the distance given by the formula is correct   or absolute value of $y-k$ if you insist that the value has to be positive.   (BTW, depending on the sign of $k$ the directrix it self could be "above"  or "below"  the y axix - but who cares..)..  .. of course  when you square, it won't even matter if the sign is positive or negative.

2. Yes IT IS! (that is if you follow the hint and,  square  both sides and do some simple regrouping, as suggested.)

Hope that helps.

I worked it out about a month and a half ago because I was bored in math class.  The solution is
\[x^{2}+(1-e^{2})y^{2}-2\alpha x-2(\beta-e^{2})y+(\alpha^{2}+\beta^{2}-e^{2}k^{2}) = 0 \]

Just out of curiosity, can anyone find the equation if the directrix is $y = ax+b?$  This one I have not worked out. :(

For  $y=mx+c$ [hide](instead of $y=k$)
 the distance = $\frac{y-mx-c}{\sqrt{1+m^{2}}}$

So again,  rest is easy.. regroup the terms in:

$e^{2}(y-mx+c)^{2}= (1+m^{2})((x-\alpha)^{2}+(y-\beta)^{2}))$ [/hide]

[hide="Open at your own risk."]The general equation for a conic! OK, here goes... :maybe:

$(-e^{2}m^{2}+m^{2}+1)x^{2}+(2e^{2}m)xy+(m^{2}-e^{2}+1)y^{2}+(-2e^{2}mc-2m^{2}\alpha-2\alpha)x+(2e^{2}c-2m^{2}\beta-2\beta)y+(-e^{2}c^{2}+m^{2}\alpha^{2}+\alpha^{2}+m^{2}\beta^{2}+\beta^{2}) = 0$
Does anyone want to check my work? :yankchain:

Or, if the directrix is of the form $x = c$,
$(1-e^{2})x^{2}+y^{2}+(2e^{2}-2\alpha)x+(-2\beta)y+(x^{2}+y^{2}-e^{2}c^{2}) = 0$
A two part equation...go figure.[/hide]
 :huh:  :o  :wallbash:  :wallbash_red:  :wallbash:  :roll:  :!:
Bonus question:  What is the probability that I have no life?

[hide="Open at your own risk."]The general equation for a conic! OK, here goes... :maybe:

$(-e^{2}m^{2}+m^{2}+1)x^{2}+(2e^{2}m)xy+(m^{2}-e^{2}+1)y^{2}+(-2e^{2}mc-2m^{2}\alpha-2\alpha)x+(2e^{2}c-2m^{2}\beta-2\beta)y+(-e^{2}c^{2}+m^{2}\alpha^{2}+\alpha^{2}+m^{2}\beta^{2}+\beta^{2}) = 0$
Does anyone want to check my work? :yankchain:

Or, if the directrix is of the form $x = c$,
$(1-e^{2})x^{2}+y^{2}+(2ke^{2}-2\alpha)x+(-2\beta)y+(\alpha^{2}+\beta^{2}-e^{2}c^{2}) = 0$
There was a minor error in this equation. :oops:
The real question: is that behemoth of an equation even useful for anything?
A two part equation...go figure.[/hide]
 :huh:  :o  :wallbash_red:  :roll:
Bonus question:  What is the probability that I have no life? :yup:

I'm not sure why I bothered with this post.  Everyone seems to have lost interest. :(

Did anyone know that you are limited to ten emoticons per post?

[quote="roadnottaken"]Why is that value for the distance true?  I've never seen it before.[/quote]

The formula for the distance of a point $(x_{0}, y_{0})$ from a line $Ax+By+C = 0$ is

$d = \frac{| Ax_{0}+By_{0}+C |}{ \sqrt{A^{2}+B^{2}}}$

This is easy to prove with vectors.  The proof without vectors is somewhat more tedious.


Find maximal constant $C$ such that for all $a,b,c>0$ satisfying $a^2+b^2+c^2=1$

$\frac{a+b}{a+c}+\frac{b+a}{b+c}+\frac{c+a}{c+b}\ge C$

[quote="Namdung"][quote="conceive"]Find constant $C$ such that for all $a,b,c>0$ satisfying $a^2+b^2+c^2=1$

$\frac{a+b}{a+c}+\frac{b+a}{b+c}+\frac{c+a}{c+b}\ge C$[/quote]

C = 1. Just take a = b = 0 (or -> 0), c = 1.[/quote]
Namdung, I don't know why letting a = b = 0, c = 1 show the question. There exists many values of a, b, c make differrent values of C!

But you can see that our expression = $1 + \frac{a+b}{a+c} + \frac{2a}{b+c} > 1$

Note that two last fraction has the same denominator.


inequalities
function
inequalities unsolved

Prove that
$ \frac{a}{1\minus{}a}\plus{}\frac{b}{1\minus{}b}\plus{}\frac{c}{1\minus{}c}\ge \frac{3\sqrt[3]{abc}}{1\minus{}\sqrt[3]{abc}}$
where $ 0<a,b,c<1$

It can be solved using Cauchy-Schwartz and AM.GM

$ (3\minus{}3\sqrt[3]{abc})\left(\sum{\dfrac{1}{1\minus{}a}}\right)\ge\left(\sum{1\minus{}a}\right)\left(\sum{\dfrac{1}{1\minus{}a}}\right)\ge 9$

Then
$ \sum{\dfrac{1}{1\minus{}a}}\ge\dfrac{3}{1\minus{}\sqrt[3]{abc}}$ which is equivalent to the problem.

Clearly it is true for $ n\ge 2$ variables.
Also it can be solved using smoothing.
:)

:)[/quote]
It can be solved using Chebyshev and AM-GM

For $ f(x)\equal{}\frac{1}{1\minus{}x} \Longrightarrow f''(x)\equal{}\frac{2}{(1\minus{}x)^3}>0$

By the Jensen Inequality; $ \frac{1}{1\minus{}x}\plus{}\frac{1}{1\minus{}y}\plus{}\frac{1}{1\minus{}z} \geq \frac{3}{1\minus{}\frac{x\plus{}y\plus{}z}{3}} \geq \frac{3}{1\minus{}\sqrt[3]{xyz}}$

[quote="crazyfehmy"]For $ f(x) \equal{} \frac {1}{1 \minus{} x} \Longrightarrow f''(x) \equal{} \frac {2}{(1 \minus{} x)^3} > 0$

By the Jensen Inequality; $ \frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z} \geq \frac {3}{1 \minus{} \frac {x \plus{} y \plus{} z}{3}} \geq \frac {3}{1 \minus{} \sqrt [3]{xyz}}$[/quote]
The function should be $ f(x) \equal{} \frac {x}{1 \minus{} x}$ then $ f''(x) \equal{} \frac { \minus{} 2}{(1 \minus{} x)^3}$ ;)
We should actually let $ f(x) \equal{} \frac{e^x}{1\minus{}e^x}$ then $ f''(x) \equal{} \frac{e^x \plus{} e^{2x}}{(1\minus{}e^x)^3}$. Then $ ln(a),ln(b),ln(c) < 0$. Thus $ f''(ln(a)),f''(ln(b)),f''(ln(c)) > 0$. So $ f(ln(a)) \plus{}f(ln(b)) \plus{} f(ln(c)) \ge f \left (\frac{ln(a)\plus{}ln(b)\plus{}ln(c)}{3} \right )$ which is exactly what we are asked to prove.

$ f(x)\equal{}\frac{1}{1\minus{}x} \Longrightarrow f''(x)\equal{}\frac{2}{(1\minus{}x)^3}>0$ so
$ \sum xf(x)\ge \frac{x\plus{}y\plus{}z}{1\minus{}\frac{x^2\plus{}y^2\plus{}z^2}{x\plus{}y\plus{}z}}$ by weight Jensen and
$ x\plus{}y\plus{}z\ge3\sqrt[3]{xyz}$ and
$ x^2\plus{}y^2\plus{}z^2\ge(x\plus{}y\plus{}z)\sqrt[3]{xyz}$ and we are done

Mathias if we add $ 3$ both of two parts; the function is $ \frac {1}{1 \minus{} x}$.  :wink:

By symetry we assume that $a\ge b\ge c$, thus $\frac{1}{1-a}\ge \frac{1}{1-b}\ge \frac{1}{1-c}$. Using Chebyshev inequality and then AM-GM we have $\sum\frac{a}{1-a}\ge \frac{1}{3}\left(\sum a\right)\left(\sum \frac{1}{1-a}\right)\ge \frac{1}{3}\cdot 3\sqrt[3]{abc}\left(\sum \frac{1}{1-a}\right)$, so we are left to prove $\sum\frac{1}{1-a}\ge \frac{3}{1-\sqrt[3]{abc}}$. Now by $T'2$ lemma $\sum \frac{1}{1-a}\ge \frac{3^2}{3-(a+b+c)}\ge \frac{3}{1-\sqrt[3]{abc}}$ because the last inequality is equivalent to $a+b+c\ge 3\sqrt[3]{abc}$, we are done :)

[quote="Vikernes"]By symetry we assume that $a\ge b\ge c$, thus $\frac{1}{1-a}\ge \frac{1}{1-b}\ge \frac{1}{1-c}$. Using Chebyshev inequality and then AM-GM we have $\sum\frac{a}{1-a}\ge \frac{1}{3}\left(\sum a\right)\left(\sum \frac{1}{1-a}\right)\ge \frac{1}{3}\cdot 3\sqrt[3]{abc}\left(\sum \frac{1}{1-a}\right)$, so we are left to prove $\sum\frac{1}{1-a}\ge \frac{3}{1-\sqrt[3]{abc}}$.[/quote]
Actually the problem is equivalent to $\sum\frac{1}{1-a}\ge \frac{3}{1-\sqrt[3]{abc}}.$


Prove that if $x+\frac{1}{x}= 2 cos \alpha$
then

$x^{n}+\frac{1}{x^{n}}= 2 cos n\alpha$

This is hardly Classroom Math.

[hide="Anyway, just..."] Recognize that $x = e^{i \alpha}$. [/hide]

i agree with torajirou, but w/e.

[hide]
let $\epsilon = e^{i\alpha}$ then $\frac{1}{\epsilon}= e^{-i\alpha}$ so we have $\epsilon+\frac{1}{\epsilon}= \cos{\alpha}+\cos{-\alpha}+i\sin{\alpha}+i\sin{-\alpha}= 2\cos{\alpha}$

Hence $x = \epsilon$.  By De Moivre's Formula, $x^{n}= \cos{n\alpha}+i\sin{n\alpha}$ and $\frac{1}{x^{n}}= x^{-n}= \cos{-n\alpha}+i\sin{-n\alpha}= 2\cos{n\alpha}$.

note: $\cos{-\alpha}= \cos{\alpha}$ and $\sin{-\alpha}=-\sin{\alpha}$[/hide]

Just a question, how is $x = e^{i\alpha}$? That was the only part of the proof that I didn't get.

I think its calculus based, but its just exponential form
Its in AOPS book 2


Haha, we already got a chess tournament, why not a scrabble one!  Plus, I'm way better at Scrabble than chess. :-D  Anyhow, http://www.games.com has Scrabble, or we could make it Literati on Yahoo! games.  Whichever.  Anyone interested?

I'm not very good at scrabble.  But I ll be willing to participate in a tournament if enough people are interested.

i'll play, but no timed games cuz then i always run out of my 10 minutes

yea have mercy against some of the people who aren't as good here and don't go putting us against the really good people please.

Haha, um, how can someone's "goodness"?   Well, um, I'll post who is up against who in a couple of days (to get more people) and if you guys got any complaints, just let me know.  (Also, get some http://www.games.com IDs.)

[quote="JS1527"]danny, I'll do it as long as you don't put me against confuted.  I know how to play chess, but I don't think I could beat anyone on AoPS  :? but maybe I can do OK in scrabble[/quote]
What makes you think I'm good at Scrabble?

edit: I don't seem to be able to get games.com to work.

Um, you probably are.  Anyway, Danny wants to play you.

Ugh, I stopped playing there for awhile cuz the server sucked (but I think it's fine now), but it is the only site who has scrabble.  Try again I guess.  The last time I was there it wad fine.  I just went on it now.  You need to click the big icon...the big "Games.com"  or you can just click here:  http://play.games.com/playgames/home.jsp  ...

Also, you guys all probably know, but don't install all their spyware crap (i.e. don't click yes when it says, "Download this, it sychronizes your computer clock!")

Sure, I'll play.

What rules though? I have a friend who CONSTANTLY checked dictionaries while playing Literati, and that really bugged me.

Oh god, that irritates me.  Yeah, you can tell sometimes who those people are because all of a sudden they pop up with a bingo and it's like irfghoh.  It's interesting though.  Haha, three years ago I was one of them!  But now, I guess it's good because you remember all those scary LEGAL words like aalii or torrii, or xu and xi...I'll stop with that, I might give you people good words :-D ...Well, the rules are NO DICTIONARIES OR ANAGRAMERS!   But other than that, games.com doesn't have a timer or anything, so you can take as long as you want.   And if you get disconnected, they even allow a full 7 minutes to get back on, which you can easily do if you're quick on a dial-up modem.  Once you get registered, just pm me and tell me what your s/n on games.com is, so when I do the matches, your opponent will know who to look for.

[quote="JS1527"]Danny wants to play you.[/quote]
Who's Danny?

Games.com doesn't seem to want to let me log in now. Hmm.
What about Internet Scrabble Club? http://www.isc.ro/

I sent Osiris a PM...hopefully we can play today.

Darn, I knew I could not beat someone who knew the cauchy-schwartz inequality  :wink:

JS1527-249
Osiris-302

DANNY!!!! My screen name is <b> JS1527</b>.  My last name starts w/S, not K!

Haha...WOW, sorry.  I guess I was thinking "just joking" instead of "Julie S."  Haha...I get it. :-D

I pm'ed hatsup twice.  He hasn't responded to either.  I can't play if (s) he doesn't set up a time to play.

yif is on 4th of july vacation, so our match hasn't been played yet

rep try reloading wordbiz that worked for me.

http://www.games.com doesn't work for ANYONE here?   That's awfully odd.

[quote="SingTheSorrow"]www.games.com doesn't work for ANYONE here?   That's awfully odd.[/quote]

Works for me but i'm not in the tournament.  :lol:

What about Lord Venom and Cats...?  What's up with you guys?

ive never heard of a scabble tournament on AOPS before... am i too late to sign up? ;)

my mom and dad and i play scrablle all the time i am really good at it and want to play the next time think comes up haha!

Says the person who bumped the long dead thread.


geometry
trigonometry
circumcircle
angle bisector
perpendicular bisector
geometry unsolved

Given an acute triangle $ ABC$. The incircle of triangle $ ABC$ touches $ BC,CA,AB$ respectively at $ D,E,F$. The angle bisector of $ \angle A$ cuts $ DE$ and $ DF$ respectively at $ K$ and $ L$. Suppose $ AA_1$ is one of the altitudes of triangle $ ABC$, and $ M$ be the midpoint of $ BC$.
(a) Prove that $ BK$ and $ CL$ are perpendicular with the angle bisector of $ \angle BAC$.
(b) Show that $ A_1KML$ is a cyclic quadrilateral.

according to http://www.mathlinks.ro/viewtopic.php?p=1525553#1525553  we have $ AK\bot BK$  $ AK\bot CL$
Notice ML=MK=MD=c-b =>  ML//BA
done

How do you get $ MK\equal{}ML\equal{}MD$?

$ MK\parallel{}AC$ $ \Rightarrow$ $ \angle CED\equal{}\angle MKD$ and $ \angle CDE\equal{}\angle MDK$ so $ MD\equal{}MK$

$ ML\parallel{}AB$ $ \Rightarrow$ $ \triangle MLD\sim  \triangle BDF$ $ \Leftrightarrow$ $ MD\equal{}ML$

[quote="wangsacl"]How do you get $ MK \equal{} ML \equal{} MD$?[/quote]
AC'=AC  =>  ML=c-b
AB=AB'  =>  MK=c-b

[quote="babylon"]$ MK\parallel{}AC$ $ \Rightarrow$ $ \angle CED \equal{} \angle MKD$ and $ \angle CDE \equal{} \angle MDK$ so $ MD \equal{} MK$

$ ML\parallel{}AB$ $ \Rightarrow$ $ \triangle MLD\sim \triangle BDF$ $ \Leftrightarrow$ $ MD \equal{} ML$[/quote]
No, in this step we have not proved AB//ML  because we want use ML=MK=c-b to prove  AB//ML  MK//AC

Wow, nice.

I use trigonometry to prove $ MK\equal{}ML$ (which is quite long and complicated)...

Dear Mathlinkers,
1. for (a) see : http://perso.orange.fr/jl.ayme/  vol. 4, An unlikely concurrence p. 5-6
2. for (b) see : http://perso.orange.fr/jl.ayme/  vol. 1, Feuerbach's theorem p. 3
this is the "Calabre's circle".
Sincerely
Jean-Louis

[hide="Another proof of b"] Extend $ LM$ until it intersects $ BK$ at $ X$. Triangle $ BMX$ is congruent to triangle $ CML$. Note that $ KM$ is a median of right triangle $ KLX$, thus $ KM\equal{}LM$. With some angle chasing, we get $ \angle LA_1M\equal{}\angle MA_1K$. So $ M$ is the intersection of the angle bisector of $ \angle LA_1K$ and the perpendicular bisector of $ KL$. Clearly $ M$ is on the circumcircle of triangle $ LA_1K$, as desired.[/hide]

[b]An easy hint for b):[/b]

We have

  a) $ \angle MLK  \equal{} \angle A / 2$, since $ LM // AB$

  b) $ \angle MA_1K \equal{} \angle BA_1K \equal{} \angle BAK \equal{} \angle A /2$, since $ BKA_1A$ is insribed($ \angle AA_1B \equal{} 90 \equal{} \angle AKB$).

 Hence $ \angle MLK \equal{} \angle MA_1K$ .

 Babis


Putnam
logarithms
limit
college contests

That's an incredibly beautiful solution, benorin! I just did the question and here is my solution:

$\begin{aligned}\prod_{2 \le n \le k}\frac{n^3-1}{n^3+1}  = \exp \bigg(\ln{\prod_{2 \le n \le k}\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}} \bigg)  = \exp {\sum_{2 \le n \le k}\ln\bigg[\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\bigg]} = \exp(p_{k}).\end{aligned}$

${\begin{aligned}p_{k} & = \sum_{2 \le n \le k}\ln(n-1)-\sum_{2 \le n \le k}\ln(n+1)+\sum_{2 \le n \le k}(n^2+n+1)-\sum_{2 \le n \le k}(n^2-n+1) \\& = \sum_{2 \le n+2 \le k}\ln(n+1)-\sum_{2 \le n \le k}\ln(n+1)+\sum_{2 \le n \le k}(n^2+n+1)-\sum_{2 \le n+1 \le k}(n^2+n+1) \\& = \sum_{0 \le n \le k-2}\ln(n+1)-\sum_{2 \le n \le k}\ln(n+1)+\sum_{2 \le n \le k}(n^2+n+1)-\sum_{1 \le n \le k-1}(n^2+n+1) \\& = \ln(2)-\ln(k^2+k)-\ln(3)+\ln(k^2+k+1) = \ln\bigg(\frac{2}{3}}\bigg)+\ln\bigg(\frac{1+1/k+1/k^2}{1+1/k}\bigg) \to \ln\bigg(\dfrac{2}{3}\bigg), ~~ ~ k \to \infty.\end{aligned}$

${\displaystyle \prod_{n \ge 2 }\frac{n^3-1}{n^3+1} = \lim_{k \to +\infty}\prod_{2 \le n \le k}\frac{n^3-1}{n^3+1} = \exp\bigg[\ln\bigg(\frac{2}{3}}\bigg)\bigg] = \frac{2}{3}. $

[quote="benorin"][hide="My Solution"]Recall that $a^{3}\pm b^{3}=(a\pm b)(a^{2}\mp ab+b^{2})$ and quickly convince yourself that $\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n}}=\frac{1}{b_{2}}\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n+1}}$ and hence that

\[\prod_{n=2}^{\infty}\frac{n^{3}-1}{n^{3}+1}= \frac{1}{9}\prod_{n=2}^{\infty}\frac{n^{3}-1}{(n+1)^{3}+1}= \frac{1}{9}\prod_{n=2}^{\infty}\frac{(n-1)(n^{2}+n+1)}{(n+2)((n+1)^{2}-(n+1)+1)}\]
\[=\frac{1}{9}\prod_{n=2}^{\infty}\frac{(n-1)(n^{2}+n+1)}{(n+2)(n^{2}+n+1)}=\frac{1}{9}\prod_{n=2}^{\infty}\frac{n-1}{n+2}= \frac{1\cdot 2\cdot 3}{9}\prod_{n=2}^{\infty}\frac{n+2}{n+2}=\frac{2}{3}\][/hide][/quote]
The statement: "$ \prod_{n=2}^{\infty}\frac{a_{n}}{b_{n}}=\frac{1}{b_{2}}\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n+1}} $" is NOT true for general $a_n$ and $b_n$.

Remember that the value of an infinite sum/product is the limit of the partial sums/products as the number of terms goes to infinity.

For example, consider $a_n = b_n = n$ for all integers $n \ge 2$.

LHS: $\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n}} =$ $\lim_{N \to \infty}\prod_{n=2}^{N}\frac{a_{n}}{b_{n}} =$ $\lim_{N \to \infty} \prod_{n=2}^{N}\frac{n}{n} =$ $\lim_{N \to \infty} \prod_{n=2}^{N}1 =$ $\lim_{N \to \infty} 1 = 1$

RHS: $\frac{1}{b_{2}}\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n+1}} =$ $\frac{1}{b_{2}} \lim_{N \to \infty}\prod_{n=2}^{N}\frac{a_{n}}{b_{n+1}} =$ $\frac{1}{2} \lim_{N \to \infty}\prod_{n=2}^{N}\frac{n}{n+1} =$ $\frac{1}{2} \lim_{N \to \infty}\frac{2}{N+1} =$ $\frac{1}{2} \cdot 0 = 0$

Also, for $a_n = n^3-1$ and $b_n = n^3+1$, it can be shown that:

$\prod_{n=2}^{N}\frac{a_{n}}{b_{n}} = \dfrac{2N^2+2N+2}{3N^2+3N}$ and  $\frac{1}{b_{2}}\prod_{n=2}^{N}\frac{a_{n}}{b_{n+1}} = \dfrac{2}{3N^3+9N^2+6N}$

Taking the limit as $N \to \infty$ gives:

$\prod_{n=2}^{N}\frac{a_{n}}{b_{n}} = \dfrac{2}{3}$ and  $\frac{1}{b_{2}}\prod_{n=2}^{N}\frac{a_{n}}{b_{n+1}} = 0$

Clearly, "$\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n}} = \frac{1}{b_{2}}\prod_{n=2}^{\infty}\frac{a_{n}}{b_{n+1}}$" is NOT true for all sequences $a_n$ and $b_n$.

Also, please note that since $\dfrac{n-1}{n+2} < 1$ for all $n \ge 2$, we must have $\frac{1}{9}\prod_{n=2}^{\infty}\frac{n-1}{n+2} \le \dfrac{1}{9}$, and thus, $\frac{1}{9}\prod_{n=2}^{\infty}\frac{n-1}{n+2} \neq \dfrac{2}{3}$.

Conveniently, the two mistakes canceled each other out, yielding the correct answer in your solution.

Don't you mean
[quote="ArtOfPrecision"]That's an incredibly beautiful solution, benorin! I just did the question and here is my solution:

$\begin{aligned}\prod_{2 \le n \le k}\frac{n^3-1}{n^3+1}  = \exp \bigg(\ln{\prod_{2 \le n \le k}\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}} \bigg)  = \exp {\sum_{2 \le n \le k}\ln\bigg[\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\bigg]} = \exp(p_{k}).\end{aligned}$

${\begin{aligned}p_{k} & = \sum_{2 \le n \le k}\ln(n-1)-\sum_{2 \le n \le k}\ln(n+1)+\sum_{2 \le n \le k}\ln(n^2+n+1)-\sum_{2 \le n \le k}\ln(n^2-n+1) \\& = \sum_{2 \le n+2 \le k}\ln(n+1)-\sum_{2 \le n \le k}\ln(n+1)+\sum_{2 \le n \le k}\ln(n^2+n+1)-\sum_{2 \le n+1 \le k}\ln(n^2+n+1) \\& = \sum_{0 \le n \le k-2}\ln(n+1)-\sum_{2 \le n \le k}\ln(n+1)+\sum_{2 \le n \le k}\ln(n^2+n+1)-\sum_{1 \le n \le k-1}\ln(n^2+n+1) \\& = \ln(2)-\ln(k^2+k)-\ln(3)+\ln(k^2+k+1) = \ln\bigg(\frac{2}{3}}\bigg)+\ln\bigg(\frac{1+1/k+1/k^2}{1+1/k}\bigg) \to \ln\bigg(\dfrac{2}{3}\bigg), ~~ ~ k \to \infty.\end{aligned}$

${\displaystyle \prod_{n \ge 2 }\frac{n^3-1}{n^3+1} = \lim_{k \to +\infty}\prod_{2 \le n \le k}\frac{n^3-1}{n^3+1} = \exp\bigg[\ln\bigg(\frac{2}{3}}\bigg)\bigg] = \frac{2}{3}. $[/quote]

Note that $\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1} = \prod_{n=2}^{\infty}\frac{n-1}{n+1}\prod_{n=2}^{\infty}\frac{n^2+n+1}{n^2-n+1}$.

We have that $\lim_{k \rightarrow \infty} \frac{n-1}{n+1} = 1, \lim_{k \rightarrow \infty}\frac{n^2+n+1}{n^2-n+1} = 1$, so we may telescope the series:

$\prod_{i=2}^{n}\frac{i-1}{i+1} = \frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}...\cdot\frac{n-2}{n}\cdot\frac{n-1}{n+1} = \frac{2}{n(n+1)}$.

$\prod_{i=2}^{n}\frac{i^2+i+1}{i^2-i+1} = \frac{7}{3}\cdot\frac{13}{7}\cdot\frac{21}{13}...\cdot\frac{(i-1)^2+(i-1)+1}{i^2-i+1}\cdot...\cdot\frac{n^2+n+1}{n^2-n+1}$. Now note that $(n+1)^2-(n+1)+1 = n^2+n+1$. Therefore the numerator in a term cancels out the denominator in the next term. So the product is just $\frac{n^2+n+1}{3}$.

Therefore $\prod_{i=2}^{n}\frac{i-1}{i+1}\prod_{i=2}^{n}\frac{i^2+i+1}{i^2-i+1} = \frac{2}{3}\cdot\frac{n^2+n+1}{n^2+n}$.
Since $\lim_{n \rightarrow \infty} \frac{n^2+n+1}{n^2+n} = 1$, we have that $\lim_{n \rightarrow \infty} \prod_{i=2}^{n}\frac{n^3+1}{n^3-1} = \boxed{\frac{2}{3}}$.

[quote="me@home"]Don't you mean
[/quote]Yes, that's what I meant. Thanks.


Lets consider a=b=c and then we have 4/4+4/4-8/2 which is equal to 1+1-2 and the answer is 0[/img][/list][/code]

I gave them the value of 1, but it they can take any value and the result is the same

By computer we can compute the minimum value, and that is incorrect...

I really have no good way to approach this inequality (the min occurs at $ (a,b,c)\equal{}(3\minus{}2\sqrt{2}, \sqrt{2}\minus{}1,\sqrt{2})$, for example)

I think that the easiest solution is set $ a\plus{}b\plus{}c\equal{}1$ and use lagrange multipliers (clearly that makes it such that all hypotheses of that technique are satisfied).

\bump It came in NMTC Junior final level this year. It was copied from this contest!!

[quote=April]Let $ a, b, c$ be positive reals.  Find the smallest value of
\[ \frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \minus{} \frac {8c}{a \plus{} b \plus{} 3c}.
\][/quote]
[url=https://artofproblemsolving.com/community/c6h1666660p10591747]Find all triples [/url]$(a, b, c)$ of positive real numbers for which the expression
$$K =\frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \minus{} \frac {8c}{a \plus{} b \plus{} 3c}$$obtains its minimum value.

Here you are
$(a,b,c)=\left(\frac{5\sqrt2-7}{2-\sqrt2}c,\ \frac{\sqrt2-1}{\sqrt2}c,\ c\right)=\left(\frac{(\sqrt2-1)^2}{\sqrt2}c,\ \frac{\sqrt2-1}{\sqrt2}c,\ c\right)$ for any $c>0$

[quote=arqady]Approach and comment by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=581]Anh Cuong:[/url]

Take $ x \equal{} a \plus{} 2b \plus{} c;y \equal{} a \plus{} b \plus{} 2c;z \equal{} a \plus{} b \plus{} 3c$. Then $ A \equal{} \frac {2y}{x} \plus{} \frac {4x}{y} \plus{} \frac {4z}{y} \plus{} \frac {8y}{z} \minus{} 17 \geq 12\sqrt {2} \minus{} 17$

Hm,China MO is usually hard and interesting but it seems that it is quite heavy in computation,even for the girl  :D[/quote]

Very inteligent solution for this noncyclic inequality(generating more example,generalizations..)!

[hide=Solution]
Let $a+b+c=1$. Then we get $$\frac{1-b+2c}{b+1}+\frac{4b}{c+1}-\frac{8c}{2c+1}=\left(-1+2 \cdot \frac{c+1}{b+1} \right)+2 \cdot \frac{2b}{c+1}-\left(4-2 \cdot \frac{2}{2c+1} \right)$$
$$=-5+2\left(\frac{c+1}{b+1}+2\cdot \frac{b+1}{c+1}+\frac{2}{2c+1}-\frac{2}{c+1} \right).$$ Thus, we just want to minimize $\dfrac{c+1}{b+1}+2\cdot \dfrac{b+1}{c+1}+\dfrac{2}{2c+1}-\dfrac{2}{c+1}=\left(\dfrac{c+1}{b+1}+2 \cdot \frac{b+1}{c+1} \right)+\left(\dfrac{2}{2c+1}-\dfrac{2}{c+1} \right)$ over all $b+c<1$. The first term is minimized when $\dfrac{c+1}{b+1}=\sqrt{2}$ by AM-GM. Let $f(x)=\dfrac{2}{2x+1}-\dfrac{2}{x+1}$. We can take the derivative and set it equal to $0$ to show that the minimum of $f$ occurs when $x=\dfrac{\sqrt{2}}{2}$. Thus, we let $c=\dfrac{\sqrt{2}}{2}$. From $\dfrac{c+1}{b+1}=\sqrt{2}$ we get $b=\dfrac{\sqrt{2}-1}{2}$. Since it just so happens that $b+c<1$, this means that these values gives the minimum for all $b+c<1,b,c>0$.

The minimum occurs at $(3-2\sqrt{2},\sqrt{2}-1,\sqrt{2})$ by solving for $a$ and scaling by $2$. Plugging this in gives $\boxed{12\sqrt{2}-17}$.

[url=https://artofproblemsolving.com/community/c6h2430986p20070008]Determine the minimum value of [/url]$$\dfrac{3a-11b-10c}{6a+12b+8c} + \dfrac{11b+4c}{24a+4b+16c} - \dfrac{a-2c}{6a+9b+3c}$$ where $a,b,c>0$.
[b]ktom 2018[/b]


absolute value
real analysis
real analysis unsolved

Does every rearrangement of a conditionally convergent series converge? If not, what is the counterexample.

I'm looking at the harmonic series but I'm not seeing the counterexample.

It need not converge, and moreover you can rearrange the terms in such a way as to make the partial sums oscillate between any desired extended real values.  PlanetMath has a [url=http://planetmath.org/encyclopedia/ConditionallyConvergentSeriesOfRealNumbersCanBeRearrangedToConvergeToAnyNumber.html]proof[/url] which is straightforward enough to motivate explicit examples.

Doh. You are right. I forgot about that case.

In the case of the alternating harmonic series (or a conditionally convergent series where the absolute value of the terms is strictly decreasing), is there only one rearrangement which tends to a given real number?

No, you can always switch finitely many numbers, and most likely more than that.

Anyway, I was just thinking about this question in a general normed linear space. Is there a rearrangement theorem for such as well?

Yes. The set of possible sums always forms a closed affine subspace and, if it does not degenerate to one point, there exists a divergent rearrangement. To prove it is quite an exercise ;)


For which positive integer values of $ n$ and $ k$ is it possible to divide the numbers $ 1,2,...,n$ into $ k$ groups in such a way that the sum of the numbers in each group is the same?

Clearly, $ k$ should be a divisor of $ \frac{n(n\plus{}1)}{2}$ (the sum of all the elements), not exceeding $ \frac{n\plus{}1}{2}$.
That may be even a sufficient condition, but I didn't check that carefully.


hi all sorry but i couldent find that question although i saw it befor
thanks for help

[quote="Elixir10"]http://www.mathlinks.ro/viewtopic.php?t=298038[/quote]
thank u verry much for your help Elixir10  :thumbup:
mr mohamed


geometry
circumcircle
geometric transformation
rotation
symmetry
topology
combinatorics unsolved

Can a closed disk can be decomposed into a union of two congruent parts having no common point?

No. Suppose for the sake of contradiction that two such unions S and T exist. Let O be the center of the disk and suppose O is in S. Since S and T are congruent, there exists a point X in T corresponding to O is S. Since X is not the center, there exists diameter AB of the disk not through X.

Let the union of points on boundary of disk be P. Let the union of points in P that belong to S be Q, the rest of points in P be R. If Q contains at least half of points in P, then Q are the set of points that have distance r(radius of disk) from O in S. Thus, because of congruency, in T there are more than half of points on the boundary of the circle $ O_1$ centers at X with radius r. This is a contradiction since more than half of the points on boundary of circle $ O_1$ are outside of the disk. So Q contains less than half of points in P. Now, more than half of points in P belongs to R. Thus, we can find a point Y in R(i.e. in P that is on the different side of AB from X) has distance larger than r from X. Another contradiction since this implies there exists a point on disk that has distance more than r from O.

A bit different approach is the following:

The boundary of the disk is divided also into two congruent parts $A,B$. Suppose that the parts of the disk are labeled $S\supset A,T\supset B$. Then only one of the parts $S,T$ can contain the center $O$ of the disk, wlog, suppose $O \in S$. Then, because $S,T$ are congruent, there exists a point similar to $O$ in $T$ labeled $O_1$. Since the distance from $O$ to any point in $A$ is $R$, the radius of the circle, it follows that the distance from $O_1$ to any point in $B$ is also equal to $R$.

Suppose that we glue $S,T$ back together to get the original circle. Since $B$ contains infinitely many points, $O_1$ is the circumcircle of three of them. But the circle which contains those three points is the boundary of the original disk. Finally, it follows that $O_1=O$ which contradicts the fact that the congruent parts have no point in common.

Another solution can be given using geometric transformations.

Two congruent figures can be obtained one from another by a rigid motion. Denote the figures with $S,T$, and pick $A,B,C$ in $S$ which lie on the boundary of the disk. Their images uniquely determine the rigid motion. Since the figures are congruent, the images of $A,B,C$ are also on the boundary of the circle, and they form a triangle congruent with $ABC$, which can be obtained from $ABC$ by a rotation of center $O$, the center of the disk, or a line symmetry about a diameter of the circle. Therefore the rigid motion is a rotation of center $O$ or a symmetry about a diameter, both which fix $O$, a point which should be in their intersection. Contradiction.

[ why should a boundary point have a corresponding point on the boundary? suppose $A$ is on the boundary of the circle, and consider its corresponding point $B$. If $B$ is not on the boundary of the circle, then a small neighbourhood of $B$ is included in $S$ or intersects both $S,T$. By the inverse mapping, which is a homeomorphism, the small neighbourhood should be mapped inside the disk, which is impossible, since $A$ is a boundary point. ]


[quote="Victory.US"]The Sequence $ (a_n)$ such that $ a_{n + 2} \le \frac {1}{3}.a_{n + 1} + \frac {3}{2}{a_n}$, $ \forall n \ge 1$

and exist $ \alpha ,\beta$ such that :$ \alpha \le a_n \le \beta$ , $ \forall n \ge1$

Prove that $ (a_n)$ has limit[/quote]

Unfortunately, that's wrong : take $ a_{2p}=0$ and $ a_{2p+1}=1$, for example

[quote="pco"][quote="Victory.US"]The Sequence $ (a_n)$ such that $ a_{n + 2} \le \frac {1}{3}.a_{n + 1} + \frac {3}{2}{a_n}$, $ \forall n \ge 1$

and exist $ \alpha ,\beta$ such that :$ \alpha \le a_n \le \beta$ , $ \forall n \ge1$

Prove that $ (a_n)$ has limit[/quote]

Unfortunately, that's wrong : take $ a_{2p} = 0$ and $ a_{2p + 1} = 1$, for example[/quote]

i made mistake. The Sequence $ (a_n)$ such that $ a_{n + 2} \le \frac {1}{3}.a_{n + 1} + \frac {2}{3}{a_n}$, $ \forall n \ge 1$.
But this problem is so well know and easy. sorry all  :blush:


function
number theory unsolved
number theory

Find all function $ f: \mathbb{N}\longrightarrow \mathbb{N}$, such that for all positives integers $ m$ and $ n$  we have :$ f(n)\plus{}f^2(m) \mid (n\plus{}m^2)^2$

[quote="Mashimaru"]I think it is a hard one. The following solution observed from the suggestion of my brother [b]nguyenvuthanhha[/b]  :P

Put $ n \equal{} 1$ then $ f(1)(f(1) \plus{} 1) | 4$, thus $ f(1) \in \{1,2,4\}$, but the directly test shows that only $ f(1) \equal{} 1$ satisfies. Thus $ f(1) \equal{} 1$.
Now let $ m \equal{} 1$ and $ n \equal{} p \minus{} 1$ where $ p$ is a prime we have
\[ f(p \minus{} 1) \plus{} 1 | p^2\]
Thus $ f(p \minus{} 1) \in \{p \minus{} 1, p^2 \minus{} 1\}$. Suppose that for some prime $ p$ we have
\[ f(p \minus{} 1) \equal{} p^2 \minus{} 1\]
then let $ m \equal{} n \equal{} p \minus{} 1$ we obtain
\[ p^2 \minus{} 1 \plus{} (p^2 \minus{} 1)^2 | (p \minus{} 1 \plus{} (p \minus{} 1)^2)^2 \\
\Leftrightarrow p^4 \minus{} p^2 | (p^2 \minus{} p)^2 \\
\Leftrightarrow p^2(p \minus{} 1)(p \plus{} 1) | p^2(p \minus{} 1)^2 \\
\Leftrightarrow p \plus{} 1 | p \minus{} 1\]
which is a contradiction. Therefore, $ f(p \minus{} 1) \equal{} p \minus{} 1$ for every prime number $ p$.
Now let $ n$ be an arbitrary positive integer and $ p$ is a prime number, we have
\[ f(n) \plus{} (p \minus{} 1)^2 | ((p \minus{} 1)^{2} \plus{} n)^{2} \\
\Leftrightarrow f(n) \plus{} (p \minus{} 1)^2 | ((p \minus{} 1)^{2} \plus{} f(n))((p \minus{} 1)^{2} \plus{} 2n \minus{} f(n)) \plus{} (f(n) \minus{} n)^{2}\]
Since then, for every positive integer $ n$ and prime number $ p$ we have
\[ f(n) \plus{} (p \minus{} 1)^2 | (f(n) \minus{} n)^2\]
Now choose $ p$ large enough we must have $ f(n) \equal{} n$ for every integer $ n$.
In conclusion, the only function that satisfies the problem is $ f(n) \equal{} n$.[/quote]

Nice solution  :)


Prove that $ 1\minus{}\cot 23^\circ\equal{}\frac{2}{1\minus{}\cot 22^\circ}$

Also, would you be able to derive an identity from this fact?

Source: 103 Trigonometry Problems

I'm sorry if this problem has been already posted.

Rewrite $ \cot 23^\circ \equal{}\cot(45^\circ\minus{}22^\circ)$.


inequalities
trigonometry
geometry
perimeter
trig identities
Law of Cosines
triangle inequality

Ofcourse there already is an inequalities marathon succesfully running but I believe that it was dominated by Algebraic Inequalities which I think isnt finite and also geometric inequalities have a special place in inequalities.
So i start this marathon (with hope that the above is accepted by other users, IF not then mods can act suitably).
Rules are the same :
1) Post a new problem when solution[b]S[/b] (yes atleast 2) have been posted to the previous problem by the user who posts the first solution to the problem
2) No links , No statements such as "obvious"
3) Geometric solutions are preferred as that is the higlight of this topic! so solutions that convert the inequality into entirely algebraic ones lose the touch of geometry. Trigo solutions are accepted (but preferable avoided)

I hope me being a new member to the forum isnt considered when reading the first lines..

Let us start with an easy one.
[b]Problem 1[/b]
For $ a,b,c$ sides of a triangle , Prove that the following inequality holds -
$ \sqrt{a} \plus{} \sqrt{b} \plus{} \sqrt{c} \ge \sqrt{a\plus{}b\minus{}c} \plus{} \sqrt{b\plus{}c\minus{}a} \plus{} \sqrt{c\plus{}a\minus{}b}$

[hide="solution to problem 1"]

From [b]QM-AM[/b]
\[ \frac {x \plus{} y}{2} \leq \sqrt {\frac {x^2 \plus{} y^2}{2}}\]

\[ \frac {z \plus{} y}{2} \leq \sqrt {\frac {z^2 \plus{} y^2}{2}}\]

\[ \frac {x \plus{} z}{2} \leq \sqrt {\frac {x^2 \plus{} z^2}{2}}\]
adding together we obtain:
\[ \frac {x \plus{} z}{2} \plus{} \frac {z \plus{} y}{2} \plus{} \frac {x \plus{} y}{2} \leq \frac {1}{\sqrt {2}} \cdot (\sqrt {x^2 \plus{} y^2} \plus{} \sqrt {z^2 \plus{} y^2} \plus{} \sqrt {x^2 \plus{} z^2}) \ ... (*)\]
setting
\[ x \equal{} \sqrt {a \plus{} b \minus{} c}\]

\[ y \equal{} \sqrt {a \plus{} c \minus{} b}\]

\[ z \equal{} \sqrt {c \plus{} b \minus{} a}\]

\[ \sqrt {a \plus{} b \minus{} c} \plus{} \sqrt {a \plus{} c \minus{} b} \plus{} \sqrt {c \plus{} b \minus{} a} \leq \frac {1}{\sqrt {2}} \cdot (\sqrt {2a} \plus{} \sqrt {2b} \plus{} \sqrt {2c}) \equal{} \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\]
[/hide]

[b]Problem 2[/b]. Prove that in a triangle $ ABC$ this inequality holds:
\[ \sin^2\alpha \plus{} \sin^2 \beta \plus{} \sin^2 \gamma \leq \frac {9}{4}\]

Use Ravi Substitution,
$ a = x + y , b = y + z , c = z + x$
the inequality becomes -
$ \sqrt {x + y} + \sqrt {y + z} + \sqrt {z + x} \ge \sqrt {2}( \sqrt {x} + \sqrt {y} + \sqrt {z}$
for convenience sake let -
$ x = a^2 , y = b^2 ,z = c^2$
the inequality becomes -
${ \sqrt {a^2 + b^2} + \sqrt {b^2 + c^2} + \sqrt {c^2 + a^2} \ge \sqrt {2}(a + b + c})$
which can concluded from by applying the pythagoras theorem and polygonal inequality in the following figure. [geogebra]3f4911ee5dbc870194db3ea554c8be143fec6e9a[/geogebra]

Where,
$ ZY = YX = (a + b + c)$
$ ZR = XP = a$
$ RS = YQ = b$
$ SY = PQ = c$
$ ZT = \sqrt {a^2 + b^2}$
$ TU = \sqrt {b^2 + c^2}$
$ UX = \sqrt {c^2 + a^2}$
$ ZX = \sqrt {2}(a + b + c)$

Where,
angle R = angle S = angle Y = angle V = anle Q = angle P = 90 degrees

[hide="solution (2)"]

asume  $ C<\frac{\pi}{2}$

$ \sin^2 A \plus{}\sin^2 B \plus{} \sin^2 C \equal{} \frac{1\minus{}\cos 2A}{2}\plus{}\frac{1\minus{}\cos 2B}{2}\plus{}\frac{1\minus{}\cos 2C}{2}$

$ \frac{3}{2}\minus{}\frac{1}{2}(2\cos(A\plus{}B) \cos(A\minus{}B)\plus{}2\cos^2 C \minus{}1)$

$ 2\plus{} \cos C(1\minus{}\cos C) \leq \frac{1}{4}\plus{}2\equal{}\frac{9}{4}$

and the problem is solved  :wink:  :wink:
[/hide]

[b]Problem 3[/b]. Prove that if the triangle is right-angled in point $ A$ than the inequality holds:

\[ a\plus{}b\plus{}c < a \sqrt{6}\]

[quote="bundler"]
[b]Problem 3[/b]. Prove that if the triangle is right-angled in point $ A$ than the inequality holds:
\[ a \plus{} b \plus{} c < a \sqrt {6}\]
[/quote]
[hide="my solution to Problem 3"]
it is a weak result i think ,
First i will prove that for $ a,b,c$ sides of a right triangle with hypotenuse - $ a$,
we have ,
$ b \plus{} c \le a\sqrt {2}$
It can be seen from the following diagram.-
 [geogebra]c8470310e35a961ff4e2b65ad570b3c2a32b4786[/geogebra]
In the figure
$ EB \equal{} c$
$ BA \equal{} b$
$ AD \equal{} b$
$ DC \equal{} c$
$ BC \equal{} \sqrt {2}(c \plus{} b)$
$ BF \equal{} \sqrt {c^2 \plus{} b^2}$
$ CF \equal{} \sqrt {c^2 \plus{} b^2}$
and we know that,
$ BF \plus{} CF \ge BC$
or,
$ \sqrt{2}(c\plus{}b) \le 2\sqrt{c^2\plus{}b^2} \equal{} 2a \implies \minus{} c\plus{}b \le \sqrt{2}a$
Therefore here we have -
$ b \plus{} c \le \sqrt {2}a$
therefore -
$ a \plus{} b \plus{} c \le a(\sqrt {2} \plus{} 1) < 6a$
done!
[/hide]
[b]Problem 4[/b]
For positive $ a,b,c$ prove that -
$ \sqrt {a^2 \minus{} ab \plus{} b^2} \plus{} \sqrt {b^2 \minus{} bc \plus{} c^2} \ge \sqrt {a^2 \plus{} ac \plus{} c^2}$

[quote="The Enigma"]
[b]Problem 4[/b]
For positive $ a,b,c$ prove that -
$ \sqrt {a^2 - ab + b^2} + \sqrt {b^2 - bc + c^2} \ge \sqrt {a^2 + ac + c^2}$[/quote]
[hide="hopefully correct solution to Problem 4"]
$ \sqrt {b^2 - bc + c^2} \ge \frac {1}{2}(\sqrt {b^2 - 2bc + c^2})$(can be checked easily)
therefore,
Factor the terms in the LHS as -
${ \sqrt {\left(a - \frac {b}{2}\right)^2 + \left(\frac {\sqrt {3}b}{2}\right)^2} + \sqrt {\left(\frac {c-b}{2}\right)^2 + \left(\frac {\sqrt {3}(b - c)}{2}\right)^2}}$
$ \ge \sqrt {\left(a + \frac {c}{2}\right)^2 + \left(\frac {\sqrt {3}c}{2}\right)^2}$
This can be deduced by applying pythagoras theorem and triabgle inequality from the following diagram -
 [geogebra]2793c4f2ee787aa3026a4e9461c992ba910da144[/geogebra]

look at the diagram i have attached -
In it,
Let -
\begin{eqnarray*} PQ = a + \frac {c}{2} \\
PT = \frac {c - b}{2} \\
TQ = a - \frac {b}{2} \\
\end{eqnarray*}

\begin{eqnarray*} QR = \frac {\sqrt {3}}{2}\cdot c \\
QS = \frac {\sqrt {3}}{2}\cdot(c - b) \\
SR = \frac {\sqrt {3}}{2}\cdot b \end{eqnarray*}

Therefore by pythagoras theorem -
$ UR = \sqrt {\left(a - \frac {b}{2}\right)^2 + \left(\frac {\sqrt {3}b}{2}\right)^2}$
$ PU = \frac {1}{2}(\sqrt {b^2 - 2bc + c^2})$
$ PR = \sqrt {\left(a + \frac {c}{2}\right)^2 + \left(\frac {\sqrt {3}c}{2}\right)^2}$

Thus we get the desired.
[/hide]

[quote="The Enigma"]
$ a \plus{} b \plus{} c \le a(\sqrt {2} \plus{} 1) < 6a$
[/quote]

it's $ a \plus{} b \plus{} c \le a(\sqrt {2} \plus{} 1) < \sqrt{6}a$  :wink:

[hide="solution to problem 3"]
Since $ a$ is the hypotenuse we have
 \[ a^2\equal{}b^2\plus{}c^2 \implies 2a^2\equal{}a^2\plus{}b^2\plus{}c^2\]
by Cauchy Shwarts
\[ a^2\plus{}b^2\plus{}c^2 \ge \frac{(a\plus{}b\plus{}c)^2}{3}\]
therefore:
\[ 2a^2 \ge \frac{(a\plus{}b\plus{}c)^2}{3} \implies a\cdot \sqrt{6} \ge a\plus{}b\plus{}c\]
Hence we're done  :) [/hide]

[hide="Solution to problem 4"]
We take Quadrilateral ABCD.
DA=a
DB=b
CD=c
and: $ \sphericalangle ADB \equal{} \sphericalangle \equal{} BDC \equal{} \frac {\pi}{3}$
From law of cosines we gets:
$ AB \equal{} \sqrt {a^2 \minus{} ab \plus{} b^2}$
$ BC \equal{} \sqrt {b^2 \minus{} bc \plus{} c^2}$
$ AC \equal{} \sqrt {a^2 \plus{} ac \plus{} c^2}$
from traingle ineqauilty:
AB+BC>=AC
q.e.d. [/hide]

[b]Problem 5[/b]
If $ a,b,c$ are the sides of a triangle then prove that -
$ \frac{a}{b\plus{}c} \plus{} \frac{b}{c\plus{}a} \plus{}\frac{c}{a\plus{}b} < 2$

By the triangle inequality, we have

$ \frac {a \plus{} b \plus{} c}{2} < b \plus{} c$

So,
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}$
is less than
$ \frac {a}{1/2(a \plus{} b \plus{} c)} \plus{} \frac {b}{1/2(a \plus{} b \plus{} c)} \plus{} \frac {c}{1/2(a \plus{} b \plus{} c)} \equal{} 2$

NEW PROBLEM:
Let S be a square of unit area. Consider any quadrilateral which has one vertex on each side of S. If a,b,c and d are the lengths of the quadrilateral, prove that :

$ 2 \leq a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \leq 4$

From the firgure we deduce that -

[geogebra]571effb3b5a3767aad3cb0145a5bd2a700f63b37[/geogebra]
name
From triangle inequality we have -
$ a^2 \le (1\minus{}y\plus{}x)^2$
construct similar inequalities and add up.

[b]Problem 7[/b]
Let $ a,b,c$ form the sides of a triangle with semi-perimeter - $ s$.Prove that -
$ abcs \ge a^3(s\minus{}a) \plus{} b^3(s\minus{}b) \plus{} c^3(s\minus{}c)$

[quote="geniusbliss"]
[b]Problem 7[/b]
Let $ a,b,c$ form the sides of a triangle with semi-perimeter - $ s$.Prove that -
$ abcs \ge a^3(s \minus{} a) \plus{} b^3(s \minus{} b) \plus{} c^3(s \minus{} c)$[/quote]

[hide="solution to problem 7"]
$ s \minus{} a \equal{} \frac {b \plus{} c \minus{} a}{2}$

$ s \minus{} c \equal{} \frac {a \plus{} b \minus{} c}{2}$

$ s \minus{} b \equal{} \frac {a \plus{} c \minus{} b}{2}$

substituting we get:
\[ 2 \cdot abc \cdot \frac {a \plus{} b \plus{} c}{2}\ge a^3(b \plus{} c \minus{} a) \plus{} b^3(a \plus{} c \minus{} b) \plus{} c^3(a \plus{} b \minus{} c)\]
that is equivalent with:
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c) \geq a^3 b \plus{} a^3 c \plus{} b^3 a \plus{} b^3 c \plus{} c^3 a \plus{} c^3 b\]
which is true since it is equivalent with:
\[ a^2(a \minus{} b)(a \minus{} c) \plus{} b^2(b \minus{} a)(b \minus{} c) \plus{} c^2(c \minus{} a)(c \minus{} b) \geq 0\]
or case $ r \equal{} 2$ of schur's inequality  :)

[/hide]
[hide="another solution to problem 2"]
Since:
\[ \sin^{2}\alpha \plus{} \sin^{2}\beta \plus{} \sin^{2}\gamma \equal{} 2 \plus{} 2\cos\alpha\cdot \cos \beta \cdot \cos \gamma\]
the inequality is equivalent with
\[ \cos \alpha \cdot \cos \beta \cdot \cos \gamma \leq \frac {1}{8}\]
which is true, therefore the solution is correct :)
[/hide]

[b]Problem 8.[/b] If the perimeter of the triangle $ ABC$ is $ 1$ , prove that this inequality holds:
\[ \frac {1 \plus{} a}{1 \minus{} 2a} \plus{} \frac {1 \plus{} b}{1 \minus{} 2b} \plus{} \frac {1 \plus{} c}{1 \minus{} 2c} \ge 12\]

[quote="Obel1x"]
[b]Problem 8.[/b] If the semi-perimeter of the triangle $ ABC$ is $ 1$ , prove that this inequality holds:
\[ \frac {1 + a}{1 - 2a} + \frac {1 + b}{1 - 2b} + \frac {1 + c}{1 - 2c} \ge 12\]
[/quote]
[hide="Solution to Problem 8"]
$ LHS = \sum_{cyclic} \(\frac {1 - 2a}{1 - 2a} + \frac {3a}{1 - 2a}\) \ge 12$
$ \implies \sum_{cyclic} \frac {a}{1 - 2a} \ge 3$
using the condition -
$ 1-2a = b+c-a$ and similarly construct and substitute.
Applying AM-GM,
$ abc \ge (a+b-c)(b+c-a)(c+a-b)$
this the famous Padoa's Inequality
Hence proved
[/hide]

[b]Problem 9[/b]
Let $ a,b,c,d$ be sides of a triangle with perimeter 3 . Prove that -
$ \frac {a}{\sqrt {a + b - c}} + \frac {b}{\sqrt {b + c - a}} + \frac {c}{\sqrt {c + a - b}} \ge \frac {9}{ab + bc + ca}$

EDIT : Initially i read the question wrong and proved the wrong inequality now i have posted the correct solution. Thanks to Obel1x for spotting the mistake.
and Bharat you are wrong i never told that the sidelenghts were integer and AIME15 gave a counter example too

Simple Solution:

The only triangle with perimeter $ 3$ is the equilateral 1-1-1 triangle......
So i dont think the greater than symbol is valid.
I wont post a problem until i know this is correct

P.s. ( A triangle has 3 sides not 4  :D  lolz)

I beg to differ.  Consider a triangle with side lengths $ 0.95, 1, 1.05$ :wink:

Aww fail ...... sorry to all. ( assumed it was integers ) :oops:


Prove the following limit
\[ \lim_{n\to\infty}{\frac{\binom{nk-k}{n-1}}{\binom{nk}{n-1}}=\left(\frac{k-1}{k}\right)^k
}\]

Expand the LHS to get

$ {{(nk\minus{}k\minus{}n\plus{}2)\cdots(nk\minus{}n\plus{}1)}\over{(nk\minus{}k\plus{}1)\cdots(nk)}}$

and split into the fractions of the form

$ {nk\minus{}k\minus{}n\plus{}1\plus{}i\over nk\minus{}k\plus{}i}$ for $ i\equal{}\overline{1,k}$

The above fraction tends to $ \frac{k\minus{}1\minus{}{k\minus{}i\minus{}1\over n}}{k\minus{}{k\minus{}i\over n}}\stackrel{n\to\infty}{\longrightarrow}{k\minus{}1\over k}$

Since there are $ k$ such fractions, the result follows.


[quote="Beat"]$(a)\ge (b)$ and $(c)\ge (d)$,[/quote]
how? :blush:

It is not true. Sorry i have made a mistake.

[quote="Beat"]$\frac{-a^2+a}{b+1}+\frac{-b^2+b}{c+1}+\frac{-c^2+c}{a+1}\le -a^2-b^2-c^2+a+b+c\le 0$, [/quote]
I think is is not ture.because we may $a<1$

[quote="ityaa"][quote="Beat"]It is easy to show that for non-negative real numbers satisfying $a+b+c=3$ we have:
(a) $\frac{a^2}{b+1}+\frac{b^2}{c+1}+\frac{c^2}{a+1}\ge \frac{3}{2}$
(b) $\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\ge \frac{3}{2}$
(c) $\frac{a^2}{b^2+1}+\frac{b^2}{c^2+1}+\frac{c^2}{a^2+1}\ge \frac{3}{2}$
(d) $\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge \frac{3}{2}$
Problem: How do the left sides of (b) and (c) compare ?
It is easy to show that $(a)\ge (b)$ and $(c)\ge (d)$, but i want to arrange them all.
Any ideas?[/quote]
I think $(c) \geq (b)$
may be I remenber that:
$\frac{a^2}{b^2+1^2}+\frac{b^2}{c^2+1^2}+\frac{c^2}{a^2+1^2} \ge \frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}$
general:
$\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2} \ge \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$[/quote]
Is it a general?

Btw:I think there is a more general inequality appears in crux of your second inequality.

Yes i susspected that there is a mistake. Can somebody rearrange them all?
BTW: i don`t think that this:$\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2} \ge \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ will help. How will input 1?


calculus
integration
function
limit
algebra
polynomial
calculus computations

Can u plz help me out with this problem. $ \ lim {n\minus{}>\infty} \int_{0}^{1}x^{n}f(x).dx$

sorry... i also missed that the function is continuous on [0 1]

Set $ f_{n}(x) \equal{} x^n f(x)$. Since f is continuous on a closed and bounded set, [0,1], f must be bounded by some M. This together with the fact that $ |x| \leq 1$ imply that the sequences $ \{f_{n}\}$ are bounded. Also, $ \lim_{n \to \infty} f_{n}(x) \equal{} 0$. What you need is a theorem that allows one to commute integrals with limits but make sure the hypotheses are satisfied.

Even i think the ans should be zero as n $ \rightarrow \infty$ then $ x^{n}$ tends to zero so i think it should be zero..

[quote="akech"]What you need is a theorem that allows one to commute integrals with limits but make sure the hypotheses are satisfied.[/quote]
Since we do not have uniform converge in this instance, akech appears to be hinting at the Dominated Convergence Theorem or one of its relatives. That would certainly work, but ...

I'd assume this problem comes from a lower level course than one in which that theorem is familar, so why not be as concrete as possible? You've already quoted the Extreme Value Theorem (that a continuous function on a closed bounded set is bounded and assumes its maximum) to claim that $ |f(x)|\le M$ on $ [0,1].$ Then
\[ \left|\int_0^1x^nf(x)\,dx\right|\le M\int_0^1x^n\,dx\equal{}\frac{M}{n\plus{}1}\]
That should do it.

@ kent

[quote]
\[ \left|\int_{0}^{1}x^{n}f(x)\,dx\right|\le M\int_{0}^{1}x^{n}\,dx \equal{} \frac {M}{n \plus{} 1}
\]
[/quote]
Actually i don understand how that happens . i mean how can u write it that way , can u please explain how u can do that please.

Lemmas:

1) $ \left|\int_Ig(x)\,dx\right|\le\int_I|g(x)|\,dx.$

2) If $ g(x)\le h(x)$ for all $ x\in I,$ then $ \int_Ig(x)\,dx\le\int_Ih(x)\,dx.$

3) $ \int_Iag(x)\,dx\equal{}a\int_Ig(x)\,dx.$

Note that for $ x\in[0,1],$ $ x^n\ge 0$ and thus $ |x^nf(x)|\le x^n|f(x)|\le Mx^n.$

Can you put all of that together?

Thank you all for your discussion. I have been knowledged that the integral value represents the moments of the function and we say that two functions f(x) and g(x) are equal if their moment values( Sequences formed after evaluating the integral with various values of n) are the same. What is the necessity for $ \ x^{n} $ in the definition.???

Yes, I was hinting at the Dominated Convergence Theorem [the level this theorem is covered varies from country to country].

The solution, which relies on the Extreme Value Theorem, that Kent gave is perhaps the nicest. Alternatively, the continuity of f on [0,1] allows us to approximate f by a polynomial and the result will integrate easily. But is Weierstrass Approximation Theorem appropriate for the level of the person asking the question?

For me, it is not easy to tell which theorem to use when one does not know the background of the person posting the question. And yes I agree that applying big theorems such as the Dominated Convergence Theorem, Weierstrass Approximation Theorem, etc, is an overkill when the Extreme Value Theorem [which is well known in elementary calculus courses] can do.


inequalities
algebra
polynomial
inequalities proposed

The same: For all $ a,b\geq0$
$ (a\plus{}b)^{n^2}\geq  2^{n^2\minus{}1}(ab)^{\frac{n^2\minus{}n}{2}}(a^n\plus{}b^n)$

[quote="Edward_Tur"]The same: For all $ a,b\geq0$
$ (a \plus{} b)^{n^2}\geq 2^{n^2 \minus{} 1}(ab)^{\frac {n^2 \minus{} n}{2}}(a^n \plus{} b^n)$[/quote]
It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88270
 :lol:


topology
vector
advanced fields
advanced fields unsolved

Let $\phi$ be embedding of $X$ into $X^{**}$. Let $\tau$ be weak topology in $X$ and $\delta$-weak* topology in $X^{**}$(i.e. the $X^*$-topology in $X^{**}$). Prove that $\phi$ is homomorphism of $(X,\tau)$ onto everywhere dense in $(X^{**},\delta)$ space.

All we need to prove is that given functionals $f_1,\ldots,f_n\in X^*$ and $\varepsilon>0$, and given $\Lambda\in X^{**}$, we can fin $x\in X$ such that $|\Lambda(f_i)-f_i(x)|<\varepsilon,\ \forall i\in\overline{1,n}$. In fact, we can do more: we can find $x$ such that $\forall i,\ \Lambda(f_i)=f_i(x)\ (*)$.

By linearity of $f\mapsto f(x),\ f\mapsto\Lambda(f)$, it suffices to prove $(*)$ for the case when $f_i$ are linearly independent. But in this case the result follows from the fact that the space spanned by $f_i$ can be seen as the dual of $X/\bigcap_i\ker f_i$, which is finite-dimensional, and finite-dimensional vector spaces are reflexive. We take $\bar x\in X/\bigcap_i\ker f_i$ such that $\Lambda(f_i)=f_i(\bar x),\ \forall i$, and then take $x\in X$ to be any vector in the preimage of $\bar x$ through the cannonical projection $X\to X/\bigcap_i\ker f_i$.

We didn't even need to mention any topological facts here. We would have had to play around with norms a bit if we had tried to prove a slightly stronger assertion, namely that the image through $\phi$ of the unit ball of $X$ is dense in the unit ball of $X^{**}$.


If $n \in N $ has exactly $2005$ divisors,  find the product of all divisors.

Easy!!! :)
 As the no has an odd no of factors It must be a perfect square as all the other factors have their complementary ones which when multiplied result  as $n$.
So product of 2004 pairs is $n{}^2{}^0{}^0{}^4$.
The remaining no is $sqrtn$
So product of all 2005 factors is $n{}^2{}^0{}^0{}^4sqrtn$
Ashwath

correction.There should be $2{}^1{}^0{}^0{}^2$ wherever I put $2{}^2{}^0{}^0{}^4$
Sorry. :blush: