trigonometry
algebra
polynomial
function
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real analysis unsolved

Any ideas?

Assume a solution of the form $ u(x, t) \equal{} X(x)T(t)$ to the modified diffusion equation $ ut$ − $ Du_{xx}$ − $ au$ = $ 0$. First show that the equation separates and find the general solution for $ X(x)$ and $ T(t)$.
Next, assuming that $ D > 0$, $ a \geq 0$, $ L > 0$, solve the boundary value problem
$ ut$ − $ Du_{xx}$ − $ au$ = $ 0$, for all $ 0 \leq x \leq L$, $ t \geq 0$
$ u_{x}(0, t) \equal{} u_{x}(L, t) \equal{} 0$, for all $ t \geq 0$
$ u(x, 0) \equal{} cos(\pi x/L) \plus{} cos(2\pi x/L)$.

Thanks

could  you please explain how you came up with this answer?
Thanks

Notation question: should the PDE actually be this?
\[ u_t \minus{} Du_{xx} \minus{} au \equal{} 0
\]
(That is, $ u_t$ rather than $ ut$ - that would match with calling this a variant on the diffusion equation.)

Note that what V.V. said isn't really an answer - it's an intermediate step. He's assumed that you can separate variables and that the $ x$ equation would become $ X''(x) \plus{} \lambda X(x) \equal{} 0,\ X'(0) \equal{} X'(L) \equal{} 0.$

That's a well-known eigenfunction equation, and its solutions are $ X(x) \equal{} \cos(k\pi x/L),$ with $ \lambda \equal{} \frac {k^2\pi^2}{L^2}.$

Your initial condition is a trigonometric polynomial in terms of precisely that form. Hence, provided separation of variables holds, the solution will have just those terms and no others, each multiplied by its own function of $ t.$

What V.V. has not done yet is to look at the equation you get in $ t.$

$ u(x,t) \equal{} T_1(t)\cos(\pi x/L) \plus{} T_2(t)\cos(2\pi x/L)$

$ u_t\minus{}Du_{xx}\minus{}au\equal{}(\dot{T}_1\plus{}(D\frac{\pi^2}{L^2}\minus{}a)T_1(t))\cos(\pi x/L)\plus{}(\dot{T}_2\plus{}(D\frac{4\pi^2}{L^2}\minus{}a)T_2(t))\cos(2\pi x/L)$.

Hence, $ \dot{T}_1\plus{}(D\frac{\pi^2}{L^2}\minus{}a)T_1(t)\equal{}0$, $ \dot{T}_2\plus{}(D\frac{4\pi^2}{L^2}\minus{}a)T_2(t)\equal{}0$.
From initial condition $ T_1(0)\equal{}1$, $ T_2(0)\equal{}1$.

Is this what i should do to prove that the equation separates?

$ u(x,t) \equal{} X(x)T(t)$
$ u_t$ - $ Du_{xx}$ - $ au$ = $ 0$
$ XT' \minus{} DX''T \minus{} aXT \equal{} 0$
$ XT' \minus{} T(DX'' \plus{} aX) \equal{} 0$

$ T'/T \equal{} [DX''/X \plus{} a) \equal{} \minus{}h$

$ DX'' \plus{} (a \plus{} h)X \equal{} 0$

Please correct me.
Thanks

Well, it's nice to have some sort of argument (in English) other than just typing a bunch of symbols.

Essentially yes, you show that you can write it as

$ \frac {T'}T \minus{} D \frac {X''}X \minus{} a \equal{} 0$, so both the parts in $ X$ and $ T$ should be equal to some constant, which you call $ h \in \mathbb{R}$ apparently.

But why is that true?

[quote="georgech"]Any ideas?

Assume a solution of the form $ u(x, t) = X(x)T(t)$ to the modified diffusion equation $ ut$ − $ Du_{xx}$ − $ au$ = $ 0$. First show that the equation separates and find the general solution for $ X(x)$ and $ T(t)$.
Next, assuming that $ D > 0$, $ a \geq 0$, $ L > 0$, solve the boundary value problem
$ ut$ − $ Du_{xx}$ − $ au$ = $ 0$, for all $ 0 \leq x \leq L$, $ t \geq 0$
$ u_{x}(0, t) = u_{x}(L, t) = 0$, for all $ t \geq 0$
$ u(x, 0) = cos(\pi x/L) + cos(2\pi x/L)$.

Thanks[/quote]

$ \forall \varphi \in C^2 ( 0 ;L )$, we have: $ \int _0 ^L u_t \, \varphi\, \text{d} x - D \int _0 ^L u_{xx} \, \varphi \, \text{d} x - a \int _0 ^L u \, \varphi \ \text{d} x = 0$

we use the Green formula:

$ \int _0 ^L u_t \, \varphi\, \text{d} x - D \int _0 ^L u \, \varphi _{xx}\, \text{d} x - \left. \left[ \varphi \, u_x - \varphi _x u \right] \right|_{x = 0}^{x = L} - a \int _0 ^L u \, \varphi \ \text{d} x = 0 \;\; (1)$

Solve the Sturm-Liouville problem : $ \left\{\begin{array}{l} - \varphi '' = \lambda \varphi \\
\varphi_{x} (0) = \varphi_{x} (L) = 0 \end{array}\right. \Rightarrow \left\{\begin{array}{l} \varphi_n(x) = \cos \left( \frac { n \pi x }{ L} \right) \\
\lambda _n = \frac { n^2 \pi ^2 }{ L^2 } \end{array}\right. \;\; (2)$

as we know $ \overline{\left < \left\{\varphi _n \right\} \right > }^{L^2} = L^2 (0;L)$ and $ \left < \varphi _n , \varphi _m \right > = \frac {L}{2} \delta _{m n}$. From $ (1)(2)$, we have:

$ \frac {\text{d}}{\text{d} t } \left[ \int _0 ^L u\, \varphi _n \, \text{d} x \right] + ( \lambda _n D - a) \int _0 ^L u \, \varphi _{n}\, \text{d} x = 0$

ie ,  $ \frac {\text{d}}{\text{d} s } \left\{\mathrm{e}^{ ( \lambda _n D - a) s } \int _0 ^L u (x,s)\, \varphi _n (x) \, \text{d} x \right\} = 0$

integrate $ s \in (0;t)$

 \begin{align} \left < u , \varphi _n \right > & = \int _0 ^L u\, \varphi _n \, \text{d} x = \mathrm{e}^{ - ( \lambda _n D - a) t } \int_0^L u( x , 0) \, \varphi _n ( x ) \text{d} x \nonumber \\
& = \mathrm{e}^{ - ( \lambda _n D - a) t } \int_0^L \left[ \cos \left( \frac {\pi x}{L} \right) + \cos \left( \frac {2\pi x}{L} \right) \right] \, \cos \left( \frac { n \pi x }{ L} \right) \text{d} x \nonumber \end{align}

because $ \left < \varphi _n , \varphi _m \right > = \frac {L}{2} \delta _{m n}$, we have

 $ \left < u , \varphi _n \right > = \left\{\begin{array}{lll} \frac {L}{2} \mathrm{e}^{ - ( \pi ^2 D L^{ - 2} - a) t } & \text{if} & n = 1 \\
\frac {L}{2} \mathrm{e}^{ - (4 \pi ^2 D L^{ - 2} - a) t } & \text{if} & n = 2 \\
0 & \text{if} & n > 2 \end{array} \right.$

expand Fourier $ u(x,t) = \sum\limits_{n = 1}^{\infty } \frac {\left < u , \varphi _n \right > }{\left\| \varphi _n \right\|^2 } \, \varphi _n (x)$

The solution is:

$ u(x,t) = \mathrm{e}^{ - ( \pi ^2 D L^{ - 2} - a) t } \cos \left( \frac {\pi x}{L} \right) + \mathrm{e}^{ - (4 \pi ^2 D L^{ - 2} - a) t } \cos \left( \frac {2\pi x}{L} \right)$

cheer! :lol:

Let me combine together the work of the last three posters in a classical separation of variables argument.

We look for solutions of the form $ u\equal{}X(x)T(t).$ Put that into the original equation and divide by $ XT$ to get
\[ \frac{T'}{T}\minus{}a\equal{}D\frac{X''}{X}\equal{}\lambda\]
The last part , the $ \equal{}\lambda,$ comes from the fact that the only way a function of $ x$ alone could equal a function of $ t$ alone is for both to be constant.

The equation $ \frac{X''}{X}\equal{}\frac{\lambda}{D},\ X'(0)\equal{}X'(L)\equal{}0$ has the well-known solutions $ X(x)\equal{}\cos\left(\frac{k\pi x}{L}\right)$ for $ k\equal{}0,1,2,\dots,$ with $ \frac{\lambda}{D}\equal{}\minus{}\frac{k^2\pi^2}{L^2}$ or $ \lambda\equal{}\minus{}\frac{k^2\pi^2D}{L^2}.$

Now bring this back over to the other equation:

$ \frac{T'}{T}\minus{}a\equal{}\minus{}\frac{k^2\pi^2D}{L^2}$

$ T'\equal{}\left(a\minus{}\frac{k^2\pi^2D}{L^2}\right)T$

This is the equation of exponential growth or decay, and has solutions

$ T\equal{}Ce^{\left(a\minus{}\frac{k^2\pi^2D}{L^2}\right)t}.$

Since the original equation is linear and homogeneous, that means that we can construct solutions from arbitrary superpositions of these functions:

$ u(x,t)\equal{}\sum_{k\equal{}0}^{\infty}a_ke^{\left(a\minus{}\frac{k^2\pi^2D}{L^2}\right)t}\cos\left(\frac{k\pi x}L\right).$

From this $ u(x,0)\equal{}\sum_{k\equal{}0}^{\infty}a_k\cos\left(\frac{k\pi x}L\right).$

But we are given that $ u(x,0)\equal{}\cos\left(\frac{\pi x}L\right)\plus{}\cos\left(\frac{2\pi x}L\right).$ That means that we can match the initial condition by setting $ a_1\equal{}1,a_2\equal{}1,$ and all other $ a_k\equal{}0.$ This gives us a solution of
\[ u(x,t)\equal{}e^{\left(a\minus{}\frac{\pi^2D}{L^2}\right)t}\cos\left(\frac{\pi x}L\right)\plus{}e^{\left(a\minus{}\frac{4\pi^2D}{L^2}\right)t}\cos\left(\frac{2\pi x}L\right)\]

In order to claim that this is the solution, we need some sort of uniqueness theorem. That is, we've shown that we have a solution that matches the initial condition; a uniqueness theorem would say that there is only one such solution.

Note that this solution has certain curious properties, depending on the relationship among $ a,D,$ and $ L.$

If $ a<\frac{\pi^2D}{L^2},$ then the solution tends to zero as $ t\to\infty.$

If $ a>\frac{\pi^2D}{L^2},$ then the solution grows without bound as $ t\to\infty.$

"The" diffusion equation (heat equation) would have $ a\equal{}0$ and the solutions would tend to zero.

hey brethrens,

Solve the problem $ \left\{ \begin{array}{l} \frac{\partial u}{ \partial t}  \minus{} \frac{1}{r} \frac{ \partial }{ \partial r} \left( r \frac{\partial u }{ \partial r}\right) \equal{}  0 \; , \; 0 \le r \le 1 \; , \, t \ge 0 \\\\ u( r , 0) \equal{} \tau _0 \;(\text{const}) \\\\ u ( 1,t) \equal{} 0  \end{array}\right.$

I tried to find such a uniqueness theorem showing that this is the only solution but failed.. :( ...May i ask what do you have in mind?
Thanks

[quote="georgech"]I tried to find such a uniqueness theorem showing that [u][b]this[/b][/u] is the only solution but failed.. :( ...May i ask what do you have in mind?
Thanks[/quote]

what is "this" ?


Let A be a matrix of mxn.What is the algorithm that shows its elements in this order:the elements of the first column upside down then the elements of the second column from the bottom to the top then the elements of the third column from the top to the bottom and so on.

Depends on what you mean by upside down and what you mean by top to bottom. (In terms of array offsets.) But I think you're looking for something like this:
[code]
// The input is two-dimensional array matrix[m][n].

for (int i = 0; i < m; i ++) {
	for (int j = 0; j < n; j ++) {
		print(matrix[i][j]);
	}
	i ++;
	if (i >= m) break;
	for (int j = n - 1; j >= 0; j --) {
		print(matrix[i][j]);
	}
}
[/code]

You may need to change the indexing order depending on which index to the array are rows and which are columns.

I'm sorry for not well expressing myself.I mean the output should give the elements of the first column from its top to its bottom, then the elements of the second column from its bottom to its top and so on.While we move through a column i doesn't change but I can't resume this in a single condition for all columns

[code]// For an array matrix[][]

for (int col = 0; col < matrix[].length; col++) {          // For each column col:
    if (col % 2 == 0) {                                    // If col is even, print the column top to bottom
        for (int row = 0; row < matrix.length; row++) {
            print(matrix[row][col]);
        }
    } else {                                               // If col is odd, print the column bottom to top
        for (int row = matrix.length - 1; row >= 0; row--) {
            print(matrix[row][col]);
        }
    }
}[/code]


number theory
least common multiple
geometry
circumcircle

given three points in a plane , construct a quadrilateral for which these points are mid points of three

Successive equal sides.

Let K, L, M be the 3 given points. Assume that the problem has been solved and K, L, M are the midpoints of the quadrilateral sides AB = BC = CA, respectively. The triangles $\triangle KBL, \triangle LCM$ are both isosceles, hence, the vertices B, C lie on the perpendicular bisectors p, q of the known segments KL, LM. These perpendicular bisectors can be constructed and they intersect in the circumcenter O of the triangle $\triangle KLM$. Since L is the midpoint of BC, OL is the O-median of the triangle $\triangle OBC$. This means that the areas of the triangles $\triangle OBL, \triangle OCL$ are equal. Let $h_p, h_q$ be the L-altitudes of these 2 triangles, respectively, the distances of the given point L from the known lines p, q. We have

$h_p \cdot OB = h_q \cdot OC,\ \ \frac{h_p}{h_q} = \frac{OC}{OB}$

Transfer the segments $h_p, h_q$ to the rays $[OC, [OB,$ of the lines p, q, respectively, so that $C' \in [OC,\ B' \in [OB$ and $OC' = h_p,\ OB' = h_q$. The triangles $\triangle OB'C' \sim \triangle OBC$ with the common angle $\angle B'OC' \equiv \angle BOC$ are then similar by SAS, which implies that their sides $B'C' \parallel BC$ are parallel. Draw a line a through the point L parallel to the line B'C' intersecting the lines p, q at the points B, C, respectively, 2 vertices of the desired quadrilateral ABCD. Complete the quadrilateral by drawing the lines BK, CM and constructing the remaining vertices A, D, such that AK = BK, DM = CM.


geometry
rectangle
LaTeX
number theory
least common multiple
algebra
system of equations

If someone could post the solutions to the problems on the AMC 10B, that'd be really really nice. Here are a few problems:

1:

$What is (-1)^1 + (-1)^2 + ... + (-1)^2006?$

$(A) -2006$
$(B) -1$
$(C) 0$
$(D) 1$
$(E) 2006$

2:

$For real numbers x and y, define x * y = (x+y)(x-y). What is 3 * (4 * 5)?$

$(A) -72$
$(B) -27$
$(C) -24$
$(D) 24$
$(E) 72$

3:

$A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?$

$(A) 10$
$(B) 14$
$(C) 17$
$(D) 20$
$(E) 24$

These are, of course, the easiest ones... here are harder ones  :D

16:

$Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?$

$(A) Tuesday$
$(B) Wednesday$
$(C) Thursday$
$(D) Friday$
$(E) Saturday$

20:

$In rectangle ABCD, we have A = (6, -22), B = (2006, 178), and D = (8, y), for some integer y. What is the area of rectangle ABCD?$

$(A) 4000$
$(B) 4040$
$(C) 4400$
$(D) 40000$
$(E) 40400$

25:

$Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. \"Look, daddy!\" she exclaims. \"That number is evenly divisible by the age of each of us kids!\" \"That's right,\" replies Mr. Jones, \"and the last two digits just happen to be my age.\" Which of the following is [i: 988c6147b8][b: 988c6147b8]not[/b: 988c6147b8][/i: 988c6147b8] the age of one of Mr. Jones's children?$

$(A) 4$
$(B) 5$
$(C) 6$
$(D) 7$
$(E) 8$

Much :coool: er $\text{\LaTeX}$

1:

$\text{What is }(-1)^1 + (-1)^2 + ... + (-1)^{2006}?$
$\text{(A) -2006}$
$\text{(B) -1}$
$\text{(C) 0}$
$\text{(D) 1}$
$\text{(E) 2006}$

2:

$\text{For real numbers }x \text{ and } y\text{, define }x * y = (x+y)(x-y)\text{. What is }3 * (4 * 5)?$

$\text{(A) -72}$
$\text{(B) -27}$
$\text{(C) -24}$
$\text{(D) 24}$
$\text{(E) 72}$

3:

$\text{A football game was played between two teams, the Cougars and the Panthers.}$ $\text{The two teams scored a total of 34 points, and the Cougars won}$ $\text{by a margin of 14 points. How many points did the Panthers score?}$

$\text{(A) 10}$
$\text{(B) 14}$
$\text{(C) 17}$
$\text{(D) 20}$
$\text{(E) 24}$

16:

$\text{Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?}$

$\text{(A) Tuesday}$
$\text{(B) Wednesday}$
$\text{(C) Thursday}$
$\text{(D) Friday}$
$\text{(E) Saturday}$

20:

In rectangle$ABCD$, we have $A = (6, -22),\ B = (2006, 178)$, and $D = (8, y)$ for some integer $y$. What is the area of rectangle $ABCD?$

$\text{(A) 4000}$
$\text{(B) 4040}$
$\text{(C) 4400}$
$\text{(D) 40000}$
$\text{(E) 40400}$

25:

$\text{Mr. Jones has eight children of different ages. On a family trip his oldest child,}$ $\text{who is 9, spots a license plate with a 4-digit number in which each of two}$ $\text{digits appears two times. \"Look, daddy!\" she exclaims.}$ $\text{\"That number is evenly divisible by the age of each of us kids!\"}$ $\text{\"That's right,\" replies Mr. Jones,}$ $\text{ \"and the last two digits just happen to be my age.\" Which of the following is the age of one of Mr. Jones's children?}$

$\text{(A) 4}$
$\text{(B) 5}$
$\text{(C) 6}$
$\text{(D) 7}$
$\text{(E) 8}$

OK my best solutions:
[hide="numero uno"]
For the odd powers, it equals -1
For the even powers, it equals 1
Thus, we can group the powers into 1003 groups of 1 odd 1 even.
The sum of 1 odd and 1 even is $1+-1=0$
$0\cdot1003=0$, the answer[/hide]
[hide="numero dos"]
$4*5=(4+5)(4-5)=9\cdot-1=-9$
$3*-9=(3+(-9))(3-(-9))=-6\cdot12=-72$[/hide]
[hide="numero tres"]
If $x$ is the Cougar's score and $y$ is the Panther's score, you can set up the system of equations:
$x+y=34$
$x-y=14$
Solving gives $(24,10)$, so the score of the Panthers is 10[/hide]
[hide="numero diez y seis"]
You have to first sum the number of days between the leap years. If you sum correctly you get
A number that is $6\mod{7}$(im too lazy to put the number here), so Sunday+6 days=Saturday[/hide]
[hide="numero veinte"]
I didn't get it  :blush: [/hide]
[hide="numero veinte y cinco"]
Clearly only one of the ages 8-1 is not included.
It can't be 1, obviously
If it is 2, then 4,6,8 can't be there
If it is 3, then 6 can't be there
If it is 4, then 8 can't be there
It can be 5
If it is 6, then 2,3,4,8 can't be there
It can be 7
It can be 8
Let's try 5 first.
The lcm of the remaining numbers is
504
Thus the license plate must be a multiple of 504, but not divisible by 5.
If you keep pressing +504 on your calculator, you will get to 5544.
It fits all of the requirements, and the dad is 44 (makes sense).
Thus the answer is 5[/hide]

to Karth: can you post more problems so I can see if I was right or not (I don't have a copy of the test and am not waiting 3 weeks to get the results; Ill even do solutions to the ones I know) :nhl:  :moose:  :football: :nhl:  :moose:  :football: :nhl:  :moose:


function
calculus
derivative
calculus computations

If $ \alpha$ is a real number and $ \alpha x \minus{} \frac{x^{3}}{1\plus{}x^{2}}$ is an increasing function for all values of $ x$, show that $ \alpha \ge \frac{9}{8}$.

If $ f(x)$ is an increasing function, then $ \forall_x,f'(x)\geq 0$ .. We obtain something like $ \forall_x, Ax^4\plus{}Bx^2\plus{}C \geq 0$ and so on .. :cool:

[quote="AndrewTom"]If $ \alpha$ is a real number and $ \alpha x \minus{} \frac {x^{3}}{1 \plus{} x^{2}}$ is an increasing function for all values of $ x$, show that $ \alpha \ge \frac {9}{8}$.[/quote]
[hide="Solution?"]Since it is an increasing function for all x, this means as x approaches infinity, we obtain larger and larger values, thus this is also concave up. Thus we must have that $ f'(x)\ge0$, and $ f''(x)\ge0$.
$ f'(x)\equal{}\alpha\minus{}\frac{x^{4}\plus{}3x^{2}}{(x^{2}\plus{}1)^{2}}\ge0$
$ f''(x)\equal{}\frac{\minus{}2x(x^{2}\minus{}3)}{(x^{2}\plus{}1)^{3}}\ge0$
From the 2nd equation we get that x must be in the interval $ (\minus{}\infty,\minus{}\sqrt{3}]U[0,\sqrt{3}]$
To find the lowest possible value of $ \alpha$, we plug in the max value of x into the first equation. Plugging in $ x\equal{}\sqrt{3}$ into the first derivative results in $ \alpha\ge\frac{9}{8}$[/hide]

[quote="ckck"]
...
$ f'(x) \equal{} \alpha \minus{} \frac {x^{4} \plus{} 3x^{2}}{(x^{2} \plus{} 1)^{2}}\ge0$
$ f''(x) \equal{} \frac { \minus{} 2x(x^{2} \minus{} 3)}{(x^{2} \plus{} 1)^{3}}\ge0$
...
[/quote]
warning , $ f''(x) \equal{} \frac { 2x(x^{2} \minus{} 3)}{(x^{2} \plus{} 1)^{3}}$ .
$ \minus{}\infty < x < \minus{}\sqrt 3 \Longrightarrow f''(x)<0  \Longrightarrow f'(x) \ decreasing$
$ \minus{}\sqrt 3 < x < 0 \Longrightarrow f''(x)>0  \Longrightarrow f'(x) \ increasing$
$ 0< x < \sqrt 3 \Longrightarrow f''(x)>0  \Longrightarrow f'(x) \ decreasing$
$ \sqrt 3 < x < \minus{} \Longrightarrow <\infty f''(x)<0  \Longrightarrow f'(x) \ increasing$
The minimum,maximum of $ f'$ are reached for $ x\equal{}\pm\sqrt 3, x\equal{}0$ so :  $ 0\leq \alpha\minus{}\frac 98\equal{}f'(\pm\sqrt 3) \leq f'(x) \leq f'(0)\equal{}\alpha$
 :cool:

$ f(x) \equal{} \alpha x \minus{} \frac {x^3}{1 \plus{} x^2} \equal{} (\alpha \minus{} 1)x \plus{} \frac {x}{1 \plus{} x^2}$,

$ f'(x) \equal{} \alpha \minus{} 1 \minus{} \frac {x^2 \minus{} 1}{(1 \plus{} x^2)^2}\geq 0,\ \forall x\in{\mathbb{R}}$

$ \Longleftrightarrow \frac {x^2 \minus{} 1}{(x^2 \plus{} 1)^2}\leq \alpha \minus{} 1, \ \forall x\in{\mathbb{R}}$

Let $ g(x) \equal{} \frac {x^2 \minus{} 1}{(x^2 \plus{} 1)^2} \equal{} \minus{} \left(\frac {1}{x^2 \plus{} 1} \minus{} \frac {1}{4}\right)^2 \plus{} \frac {1}{8}$

Since $ 0 < \frac {1}{x^2 \plus{} 1}\leq 1\ \therefore g_{max} \equal{} g(\pm 3) \equal{} \frac {1}{8}$

$ \therefore \frac 18\leq \alpha \minus{} 1$, yielding $ \alpha \geq \frac 98$

Thanks. I had:

$ f'(x) \equal{} \alpha \minus{} \frac{3x^{2}\plus{}x^{4}}{(1\plus{}x^{2})^{2}}$

$ f'(x) \ge 0$ if $ \alpha \ge y\equal{}\frac{3x^{2}\minus{}x^{4}}{1\plus{}2x^{2}\plus{}x^{4}} \equal{}1 \plus{} \frac{x^{2}\minus{}1}{1\plus{}2x^{2}\plus{}x^{4}} \equal{} 1 \plus{} \frac{1}{1\plus{}x^{2}}\minus{}\frac{2}{(1\plus{}x^{2})^{2}}$

$ \frac{dy}{dx} \equal{} \minus{}\frac{2x}{(1\plus{}x^{2})^{2}} \plus{} \frac{8x}{(1\plus{}x^{2})^{3}}$

$ \frac{dy}{dx} \equal{} 0$ for $ 1\plus{}x^{2} \equal{} 4$, that is, $ x^{2}\equal{}3$.

Differentiating again, we see that this gives a maximum and the maximum value is $ \frac{9}{8}$.


geometry
conics
projective geometry
geometry unsolved

the incircle of triangle ABC is tangenc to BC at Z.AZ intersect incircle at P.BP and PC intersect incircle at Y and X.prove that AZ,CY,BX are concur.

in few minutes later i become crazy because of this question hurry up please please :(

where are the lovers of geometry.i can conclude nobody can solve my problems.

The second post 9 (!!!) minutes after the first one.
Are you stupid or what?

yes i become stupid because of this quastion :(  :(

[b]More generally:[/b]
$ \mbox{Let Z the tangency point of BC an inscribed conic in the triangle ABC, and AZ intersect inconic at P. } \\
\mbox{BP and PC intersect the conic at Y and X. Prove that AZ,CY,BX are concur.}$
(view Figure!)

ok,why?(can you prove the main quastion with triangle geometry)

i realy conufsed.a realy nice and hard problem.but ithink you can solve it.help help help help help :(  :(

Let $ E,F$ be the touch points of the incircle (with center $ I$) with $ AC,AB$
then by using similarities (or directly the well-known theorem) we get $ FY . PZ \equal{} ZY . PF$ and similarly $ EX.PZ \equal{} EP.ZX$
now we get $ FY.PZ.EP.ZX \equal{} ZY.PF.EX.PZ$ or in other words $ FY.EP.ZX \equal{} ZY.PF.EX$ which is in fact the sufficient and needed condition for which $ ZP,EY,FX$ are concurrent(use céva's theorem sinus form for proving this).
let the tangent at point $ P$ to $ (I)$ intersect $ BC$ at $ T$ then by the well-known theorem(of pole and polar) we have the concurrency of $ EF,XY,PP,BC$ at point $ T$.($ PP$ is the tangent)
also we know that $ (TZBC) \equal{} \minus{} 1$ and since $ T,Y,X$ are collinear it implies the concurrency of $ BX,CY,PZ$ and we are done.
also note that we can prove $ FY,EX,AZ$ are concurrent (since we know the concurrency of $ BX,CY,PZ$ we can prove this by Menelaus's theorem).
Let this point be $ S \equal{} (FY,EX,AZ)$ then by pascal's theorem we can prove that $ (SPAZ) \equal{} \minus{} 1$.


geometry
geometric transformation
rotation

A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular to each other. Each sector is to be painted with one of four colors. How many distinct ways can the table be painted? (A color may be used on more than one sector, but paintings that are the same after a rotation are not considered distinct.)

There is orginally $ 4!$ ways but you must divide by 4 to compensate for rotation. So the answer is $ 3!\equal{}6$

The answer is 70 (I know from reviewing) and remember, you can have colors repeated.

Can someone elaborate on how the answer is 70?

We use casework! (4,0,0,0) denotes there are 4 of one color, 0 of another, 0 of 3rd, etc.

(4,0,0,0)=4 ways (one for each color)

(3,1,0,0)=4 ways for the 3 and 3 ways for the 1. 4x3=12 ways.

(2,2,0,0)=4c2=6 ways to pick colors. Either the same colored parts must touch or be opposite, for 2 ways of placing onjce you have selected colors. Thus 6x2=12

(2,1,1,0)=4 ways to pick the 2, 3c2=3 ways to pick the 1s. 12 ways to pick colors. There are normally 6 ways to arrange table, but we must divide by 2 as two colored parts are indistinct. Thus 12x3=36 ways.

(1,1,1,1)=1 way to pick colors, 6 ways to arrange (4!/4) so 6 ways.

Adding, 4+12+12+36+6=70


vector
linear algebra
linear algebra unsolved

I think this has been discussed before, but never actually posted in it's own topic, as it deserves :).

Let $A=(a_{ij})$ be a complex matrix. Prove that:

(a) The eigenvalues of $A$ lie in the region $\bigcup_{i=1}^nD_i$, where $D_i=\{z\in\mathbb C\ |\ |z-a_{ii}|\le\sum_{j\ne i}|a_{ij}|\}$;
(b) If the union of $k$ of the disks $D_i$ is disjoint from the other $n-k$ disks, then these $k$ disks contain exactly $k$ eigenvalues of $A$.

The numerical analysts that I know declare this to be their favorite theorem about matrices.

a)Let $\lambda$ an eigenvalue.Take a vector $X$(with $||X||=1$) such that $AX=\lambda X$.There is $i_0$ such that $x_{i_0}=1$.
$AX=\lambda X \rightarrow$$\sum_{j=1}^n a_{i_0j}x_j=\lambda x_{i_0}\rightarrow$
$\sum_{j\neq i_0} a_{i_0j}x_j=(\lambda-a_{i_0i_0})x_{i_0}$.So $|\lambda-a_{i_0i_0}|\leq \sum_{j\neq i_0} |a_{ij}|$ and $\lambda \in D_{i_0}$

[quote="killer"]... with $||X||=1$ ... There is $i_0$ such that $x_{i_0}=1.$ [/quote]
Just a clarification: killer is clearly using the $\infty$ (or sup) norm:

$||x||_{\infty}=\max_{1\le i\le n} |x_i|$


From a group of boys and girls, 15 girls leave. There are then left two boys for each girl. After 45 boys leave, there are then 5 girls for each boy. The number of girls in the beginning was:

(A)40 (B) 43 (C) 29 (D) 50 (E) 60

HI
the number of girls should be 40
let the no of boys be 'b'
let the no of girls be 'g'

From a group of boys and girls, 15 girls leave.

b/g-15=2
b=2g-30

After 45 boys leave, there are then 5 girls for each boy.
b-45/g-15=1/5
b-45=g-15/5
5b-225=g-15
5b=g-15+225
b=g+210/5

equating the two

2g-30=g+210/5
10g-150=g+210
9g=360
[b]g=40[/b]

sorry for not using latex i am not acquainted with it


7 is prime it generates cyclotomic polynomial

$x^7+y^7=(x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)$

[quote="prototypex24"]One question:

What if the exponent isn't prime? For instance, how would you factor $x^6+y^6$?   :huh:[/quote]

$(x^2)^3+(y^2)^3=(x^2+y^2)(x^4-x^2y^2+y^4)$

[quote="prototypex24"]Oh thanks, so I guess all composite powers can be expressed that way.

EDIT: How about $x^4+y^4$? $(x^2)^2+(y^2)^2$ doesn't help much... :|[/quote]

that one cannot be factorise .

if $x^n+y^n$ where $n\in\mathbb{N}$ . Then if $n$ has odd prime , it can be factorised.

Well, you could factor $x^4+y^4$ this way:

$x^4+y^4=(x^2+y^2)^2-2(xy)^2=(x+y+\sqrt{2}xy)(x+y-\sqrt{2}xy)$

Where I used the difference of squares.

i am pretty sure that a^n+b^n can always be factored if n is odd greater than 1, being prime does not matter

is this taught in Algebra II/trig  or is it taught in algebra I?

i don't know if this is taught, but you can easily show that it is true with algebra 2 concepts,

let $f(x)=x^7+a^7$ clearly $-a$ is a root, therefore $(x+a)$ is a factor of $f(x)$ by the factor theorem

But the hard part is how would you show that $x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6$ has no more polynomial factors with integer coefficients... maybe rational root?


geometry
perimeter
inequalities
incenter
trigonometry
inequalities unsolved

Denote by p the semi-perimeter of a triangle ABC, and by L the sum of the
lengths of the interior angle bisectors of the triangle ABC. Prove that the
following inequality takes place
p√3≥L

[quote="Maverick"]$(l_a+l_b+l_c)^2\leq3(l_a^2+l_b^2+l_c^2)\leq3\sum p(p-a)=3p^2$[/quote]

I don't understand your second assertion. Any help please? ;)

$l_a^2 \leq p(p-a)$ this one is easy to prove

[quote="Iris Aliaj"]Denote by $s$ the semi-perimeter of a triangle $ABC$, and by $w_{a},w_{b},w_{c}$ the lengths of the interior angle bisectors of the triangle $ABC$. Prove that the following inequality takes place $w_{a}+w_{b}+w_{c}\leq s\sqrt{3}$.[/quote]
In fact, we have the following inequality chain :
\[w_{a}+w_{b}+w_{c}\leq\frac{3}{2}(A\Gamma+B\Gamma+C\Gamma)\leq\frac{3}{2}(AI+BI+CI)\leq s\sqrt{3}, \]
where $I$ be the incentre and $\Gamma$ be the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33335]Gergonne Point[/url].

[quote="bibobeo"]I remember having seen this
$l_{a}= \frac{2bc}{b^{2}+c^{2}}\sqrt{p(p-a)}$
Is it true, and how to prove it?  :D[/quote]
Actually
$l_{a}=\frac{2\sqrt{bc}}{b+c}\sqrt{p(p-a)}$
By Stewart it's not hard.

:D Sorry, I remember it is $l_{a}=\frac{2bc}{b^{2}+c^{2}}cos\frac{A}{2}$ (I use Sin law to prove this)
btw. who is Stewart?
I use $AD^{2}=AB.AC-DB.DC$ for the first one

[quote="shalex"]if $P$ is on the side $BC$ then
$AP^{2}=AB^{2}\cdot\frac{PC}{BC}+AC^{2}\cdot\frac{BP}{BC}-BP\cdot CP$[/quote]

you forgot something

I think Shalex is right. I've checked it with $m_{a}$, $l_{a}$  and $h_{a}$ with $a=5, b=4,c=3$

oh,yes. it`s ok. I was thinking at stewart without proportion ("/")

[quote="bibobeo"]but ... how to prove it?  :blush:[/quote]
just denote that $\angle APB=\theta$ and
$\frac{AP^{2}+BP^{2}-AB^{2}}{2AP\cdot BP}=\cos\theta$
$\frac{AP^{2}+CP^{2}-AC^{2}}{2AP\cdot CP}=\cos(\pi-\theta)=-\cos\theta$
Sum them up with some computations we reach to the conclusion :wink: .

I use similar triangles and Ptoleme Theorem. But it's quite long.
You have very nice solution Shalex  :lol:


geometry
circumcircle
cyclic quadrilateral
power of a point

In $\triangle ABC$, let $D$ a point on $BC$ so that $BD:DC=5:4$. Also, $\angle BAD=40^\circ$, $\angle CAD=70^\circ$, and $AD=6$. Find $AB$.

I know the answer, but I don't know how to solve it. I heard it's from a contest for elementary school kids, so you can't use any special formula :(  Any idea?

The angle $\angle CAB$ is equal to

$\angle CAB = \angle CAD + \angle BAD = 70^\circ + 40^\circ = 110^\circ$

Let $\triangle PBC$ be a triangle having the common side BC with the triangle $\triangle ABC$, the vertex P lying on the opposite side of the line BC than the vertex A, such that the angles $\angle PBC = 70^\circ,\ \angle PCB = 40^\circ$. The remaining angle of this triangle is then

$\angle BPC = 180^\circ - (\angle PBC + \angle PCB) = 180^\circ - (70^\circ + 40^\circ) = 180^\circ - 110^\circ = 70^\circ$

Since the angles

$\angle CAB + \angle BPC = 110^\circ + 70^\circ = 180^\circ$

add up to a straight angle, the quadrilateral ABPC is cyclic. Let D' be the intersection of its diagonals BC, AP. Since $BD' \cdot CD' = AD' \cdot PD'$ is the power of the point D' to the circumcircle of this quadrilateral, we get

$\frac{BD'}{PD'} = \frac{AD'}{CD'}\ \Rightarrow \triangle BPD' \sim \triangle ACD'\ \Rightarrow \angle CAD' = \angle PBD' \equiv \angle PBC = 70^\circ$

$\frac{BD'}{AD'} = \frac{PD'}{CD'}\ \Rightarrow \triangle BAD' \sim \triangle PCD'\ \Rightarrow \angle BAD' = \angle PCD' \equiv \angle PCB = 40^\circ$

(This certainly can be done without using the power of a point to a circle.) Consequently, the points $D' \equiv D$ are identical, i.e., the point D is the diagonal intersection of the cyclic quadrilateral ABPD. The internal angles $\angle PBC = \angle BPC = 70^\circ$ of the triangle $\triangle PBC$ are equal, hence, this triangle is isosceles and $BC = PC$. Using one more time similarity of the triangles $\triangle BAD \sim \triangle PCD$, we get

$\frac{AB}{AD} = \frac{CP}{CD} = \frac{CB}{CD} = \frac{BD + CD}{CD} = \frac{BD}{CD} + 1 = \frac 5 4 + 1 = \frac 9 4$

$AB = \frac 9 4\ AD = \frac 9 4 \cdot 6 = \frac{54}{4} = 13. 5$

You can pull out the same trick with any other pair of angles $\beta, \gamma$, such that $\beta + 2\gamma = 180^\circ$. In the posted problem, $\beta = 40^\circ, \gamma = 70^\circ$, $\beta + 2 \gamma = 40^\circ + 2 \cdot 70^\circ = 180^\circ$. Angles with an integer number of degrees are constructible iff the number of degrees is divisible by 3. Hence, we may choose the non-constructible angle $\gamma = 53^\circ$, in order to make the problem look difficult. If we add the angle $\beta = 180^\circ - 2\gamma = 74^\circ$, the problem can be solved in exactly the same way.

I think I got an alternate solution,
[hide="Solution"]
we know that the areas $\triangle{ABD}:\triangle{ADC}=5:2$
we also know that area of $\triangle{ABD}=\frac{1}{2}*AB*AD=3AB*sin40^\circ$
Similarily for the other triangle,
Area of $\triangle{ADC}=3AC*sin70^\circ$
Equating the $2$ areas to $5:2$ we get,
$\frac{5sin70^\circ}{2sin40^\circ}=\frac{AB}{AC}$
We also know that $\text{area of}\triangle{ABD}+\text{area of}\triangle{ADC}=\text{area of}\triangle{ABC}$
which gives,
$\frac{AB*6sin40^\circ}{2}+\frac{AC*6sin70^\circ}{2}=\frac{AB*ACsin110^\circ}{2}$
using the 2 equations we can find the value of $AB$
[/hide]

[quote="gunes"]another solution the short one.[/quote]
Oh gosh! Your solution is very quick :o I knew this property and yet I didn't notice that! :ewpu:

What does bisector theorem state ?  :blush: Anyone mind to tell

yes what does it state, Ive never heard of it :blush:


geometry
parallelogram
geometry unsolved

Let ABCD  a convex quadilateral. a, b, c, d are the angles of A, B, C, D respectively. AC and BD intersect in O.
Prove that ABCD  is a parallelogram if and only if
OA(sin a)= OC(sin c)  and OB(sin b)= OD(sin d)

Im wait your ingenuos solutions!!

I am not a very good geometry solver, nor a good explainer.
OA(sin a)= OC(sin c) and OB(sin b)= OD(sin d)
I don't think a // gram must possess this property in accord with my knowledge.
If I am wrong, please tell me.

If $ABCD$ is a parallelogram, then the result is straightforward. So, let's consider the converse part of the problem.

Assume the condition satisfies, then we have $OA/OC = (DAB)/(DCB) = \frac{\sqrt{AB \cdot AD}}{ \sqrt{CB \cdot CD}}$. Analogously, $OB/OD = \frac{\sqrt{BA \cdot BC}}{ \sqrt{DA \cdot DC}}$. Therefore, $(AOB)/(DOC) = \frac{OA\cdot OB}{OC \cdot OD} = AB/CD$. Let $X,Y,Z,W$ be the perpendicular feet from O to segments $AB, BC, CD, DA$, respectively. From the last relation, then we have that $OX = OZ$ and $OY = OW$. In addition, by the sine law, we have that $WX = AO \cdot sin(a) = OC \cdot sin(c) = YZ$. Analogously, $XY = ZW$. Thus, $XYZW$ is a parallelogram, which implies that $X,O,Z$ and $Y,O,W$ are collinear. Finally, it is easy to conclude that $AD || BC$ and $AB || CD$.


floor function
combinatorics unsolved
combinatorics

On a soccer tournament with $n\ge 3$ teams taking part, several matches are played in such a way that among any three teams, some two play a match.
$(a)$ If $n=7$, find the smallest number of matches that must be played.
$(b)$ Find the smallest number of matches in terms of $n$.

We draw a graph such that any line between vertices A and B we color in blue iff A played with B, and in red iff A didn't play with B. Then there no exits triangle ABC such AB, BC and AC are all red. This is well known problem, the maximum of red edges is $ \left\lfloor \frac{n^{2}}{4} \right\rfloor $. So the answer is $ \frac {n(n-1)}{2}-\left\lfloor \frac{n^{2}}{4} \right\rfloor$.


Each square of the three by three grid is painted so that the whole picture has at least the two lines of symmetry indicated. Each grid square is painted one solid color. What is the maximum number of colors that could have been used?
[asy]size(100);
draw((0,0)--(0,3)--(3,3)--(3,0)--cycle);
draw((1,0)--(1,3));
draw((2,0)--(2,3));
draw((0,1)--(3,1));
draw((0,2)--(3,2));
draw((-0.2,-0.2)--(3.2,3.2),dashed);
dot((0,0));dot((3,3)); dot((1.5,0)); dot((1.5,3));
draw((1.5,-0.3)--(1.5,3.3), dashed);[/asy]

Label the squares 1,2,3,4,5,6,7,8,9 from top left to bottom right.
1,3,7, and 9 have to be the same color.
2,4, 6, and 8 have to be the same color.
5 can be any color.
Thus, the maximum number of colors that can be used is $ \boxed{3}$.

[quote=bluehurricane]What if you do this?[/quote]

No, it isn't diagonally symmetrical :)

[quote=bluehurricane]Oh, OK. Thanks for telling me, mossie![/quote]

No problem! :-)

[hide=Solution]
There are three parts of the square. Corner, middle edge, and center. Therefore, the max is $\boxed{3}.$
[/hide]


calculus
integration
function
logarithms
calculus computations

Define the function $ f(t)\equal{}\int_0^1 (|e^x\minus{}t|\plus{}|e^{2x}\minus{}t|)dx$. Find the minimum value of $ f(t)$ for $ 1\leq t\leq e$.

$ f(t) \equal{} \int_0^1 (|e^x \minus{} t| \plus{} |e^{2x} \minus{} t|)dx \geq \int_0^1 (|e^x \plus{} e^{2x} \minus{} 2t|)dx \geq |\int_0^1 (e^x \plus{} e^{2x} \minus{} 2t)dx| \equal{} |e \minus{} 1 \plus{} \frac {e^{2}}{2} \minus{} \frac {1}{2} \minus{} 2t| \geq \frac {1}{2}|t \minus{} 3\parallel{}t \plus{} 1| \geq 2$.

Let $ u \equal{} e^x$, $ du \equal{} e^x \, dx \equal{} u \, dx$, from which it follows that

$ f(t) \equal{} \int_{x \equal{} 0}^1 |e^x \minus{} t| \plus{} |e^{2x} \minus{} t| \, dx \equal{} \int_{u \equal{} 1}^e \frac {|u \minus{} t| \plus{} |u^2 \minus{} t|}{u} \, du \equal{} \int_{u \equal{} 1}^e g(u;t) \, du$,

where $ g(u;t) \equal{} \left|1 \minus{} \frac {t}{u}\right| \plus{} \left|u \minus{} \frac {t}{u}\right|$.

Then since $ 1 < t < e$, for $ 1 < u < \sqrt {t} < t$,

$ g(u;t) \equal{} \frac {2t}{u} \minus{} 1 \minus{} u$,

and for $ \sqrt {t} < u < t$,

$ g(u;t) \equal{} u \minus{} 1$,

and for $ t < u < e$,

$ g(u;t) \equal{} 1 \plus{} u \minus{} \frac {2t}{u}$.

It follows that we must evaluate the integral over each of these three subintervals:

$ f(t) \equal{} \int_{u \equal{} 1}^{\sqrt {t}} \frac {2t}{u} \minus{} 1 \minus{} u \, du \plus{} \int_{u \equal{} \sqrt {t}}^t u \minus{} 1 \, du \plus{} \int_{u \equal{} t}^1 1 \plus{} u \minus{} \frac {2t}{u} \, du$

$ \equal{} \left[2t \log u \minus{} u \minus{} \frac {u^2}{2}\right]_{u \equal{} 1}^{\sqrt {t}} \plus{} \left[\frac {u^2}{2} \minus{} u\right]_{u \equal{} \sqrt {t}}^t \plus{} \left[u \plus{} \frac {u^2}{2} \minus{} 2t \log u\right]_{u \equal{} t}^e$

$ \equal{} t \log t \minus{} \sqrt {t} \minus{} \frac {t}{2} \plus{} \frac {3}{2} \plus{} \frac {t^2}{2} \minus{} t \minus{} \frac {t}{2} \plus{} \sqrt {t} \plus{} e \plus{} \frac {e^2}{2} \minus{} 3t \minus{} \frac {t^2}{2} \plus{} 2 t \log t$

$ \equal{} 3t \log t \minus{} 5t \plus{} \frac {3 \plus{} 2e \plus{} e^2}{2}$.

This function is minimized at a critical point or endpoint, so we compute

$ f'(t) \equal{} 3 \log t \minus{} 2 \equal{} 0$

from which we have $ t \equal{} e^{2/3}$.  Evaluating,

$ f(1) \equal{} (e^2 \plus{} 2e \minus{} 7)/2$,

$ f(e^{2/3}) \equal{} (e^2 \plus{} 2e \minus{} 6e^{2/3} \plus{} 3)/2$,

$ f(e) \equal{} (e^2 \minus{} 2e \plus{} 3)/2$,

and the minimum is attained for $ t \equal{} e^{3/2}$.  If we choose, we can verify this algebraically by computing the differences

$ f(1) \minus{} f(e^{2/3}) \equal{} 3e^{2/3} \minus{} 5 > 3e^{2/3} \minus{} 2e \equal{} f(e) \minus{} f(e^{2/3})$.

We wish to show that $ e^{\minus{}1/3} > 2/3$.  Expanding as a power series, we find

$ e^{\minus{}1/3} > 1 \minus{} (1/3) \plus{} (1/18)$,

from which it follows that $ 3e^{2/3} \minus{} 2e > 0$, and the minimality of $ f(e^{2/3})$ is proven.

[quote="akech"]$ |e \minus{} 1 \plus{} \frac {e^{2}}{2} \minus{} \frac {1}{2} \minus{} 2t| \geq \frac {1}{2}|t \minus{} 3\parallel{}t \plus{} 1|$.[/quote]

That cannot be correct, since when $ t \equal{} (e^2 \plus{} 2e \minus{} 3)/4 \approx 2.456 < e$, the LHS is zero.


Happy New Year everyone! Try this 'evil' problem:

Problem. A positive integer $m$ is called 'evil' if, for some $t \in \mathbb{N}$, the sum of the first $t$ decimal digits of $m$ equals $666$. Prove or disprove: for every $n \in \mathbb{N}$, there is a $k \in \mathbb{N}$ such that each of the numbers $k,2k,\ldots,nk$ is 'evil.'

On that note, has anyone ever wondered which is the most useless entry in MathWorld? http://mathworld.wolfram.com/EvilNumber.html  :D

Cheers!  :D

Hum at fist when i link to the links you wrote there are just few words like
[b]Access Denied to IP Address 203.160.1.67[/b]
---------
The second plz tell me the sum from the left or right?

--------

The last , happy new year :D

[quote="pluricomplex"]
The second plz tell me the sum from the left or right?
[/quote]

From the left: a positive integer is 'evil' if a block of its decimal digits, from left to right, sums to $666$. If we were summing from the right (the last few digits), the answer would (almost trivially) be 'no.'

[quote="darij grinberg"][quote="vess"]On that note, has anyone ever wondered which is the most useless entry in MathWorld? http://mathworld.wolfram.com/EvilNumber.html  :D [/quote]

I disagree. What about http://mathworld.wolfram.com/LeviathanNumber.html ?

  Darij[/quote]

There are many more of it like Apocalypse Number and similar  :D


Find all continuous functions $f: \mathbb R \to \mathbb R$ such that
\[ f(2x - y) = 2f(x) - f(y), \quad \forall x,y\in \mathbb R.\]

[quote="TRAN THAI HUNG"]Find f(x) continuous in R and
$ {\rm{f(2x \minus{} y) \equal{} 2f(x) \minus{} f(y)}}\forall {\rm{x,y}} \in {\rm{R}}$:blush:[/quote]
Let $ P(x,y)$ be the assertion $ f(2x \minus{} y) \equal{} 2f(x) \minus{} f(y)$

If $ f(x)$ is solution, $ f(x) \minus{} f(0)$ is too. So we'll only look for solutions such that $ f(0) \equal{} 0$

$ P(x,0)$ $ \implies$ $ f(2x) \equal{} 2f(x)$
$ P(0,y)$ $ \implies$ $ f( \minus{} y) \equal{} \minus{} f(y)$

So $ 2f(x) \minus{} f(y) \equal{} f(2x) \plus{} f( \minus{} y)$ and so $ f(2x \plus{} ( \minus{} y)) \equal{} f(2x) \plus{} f( \minus{} y)$ and so $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and so $ f(x) \equal{} ax$ since $ f(x)$ is continuous (Cauchy).

So $ f(x) \equal{} ax \plus{} b$ and it's easy to check back that these functions indeed are solutions.


vector
inequalities
geometry unsolved
geometry

Given a set of $n>2$ planar vectors. A vector from this set is called [i]long[/i],  if its length is not less than the length of the sum of other vectors in this set. Prove that if each vector is long, then the sum of all vectors equals to zero.
[i]N. Agakhanov[/i]

Let $v_{i}=(x_{i}, y_{i})$, $1\leq i \leq n$, the long vectors and $S=(X,Y)$ their sum.
For each $i$ we have: \[|v_{i}| \geq |S-v_{i}| \Longleftrightarrow x_{i}^{2}+y_{i}^{2}\geq (X-x_{i})^{2}+(Y-y_{i})^{2}\] \[\Longleftrightarrow 2Xx_{i}+2Yy_{i}\geq X^{2}+Y^{2}\] Adding these $n$ inequalities: \[2X(x_{1}+x_{2}+\ldots+x_{n})+2Y(y_{1}+y_{2}+\ldots+y_{n})=2X^{2}+2Y^{2}\geq n(X^{2}+Y^{2}).\] If $S$ is not the zero vector then $2\geq n$, a contradiction. So $S$ is the zero vector.

$Tipe$

The result is true in any dimension $ d$, for $ n > 2$ vectors $ v_i$ of sum $ \sigma \equal{} \sum_{i\equal{}1}^n v_i$.
If $ | v_i | \geq | \sigma \minus{} v_i |$ for all $ 1 \leq i \leq n$, then $ \sum_{i\equal{}1}^n | v_i |^2 \geq \sum_{i\equal{}1}^n | \sigma \minus{} v_i |^2 \equal{} \sum_{i\equal{}1}^n < \sigma \minus{} v_i, \sigma \minus{} v_i > \ \equal{} \sum_{i\equal{}1}^n (| \sigma |^2 \minus{} 2 < \sigma, v_i > \plus{} \ | v_i |^2) \equal{}$ $ n| \sigma |^2 \minus{} 2< \sigma, \sum_{i\equal{}1}^n v_i > \plus{} \sum_{i\equal{}1}^n | v_i |^2 \equal{} (n \minus{} 2)| \sigma |^2 \plus{} \sum_{i\equal{}1}^n | v_i |^2$, hence $ (n \minus{} 2)| \sigma |^2 \leq 0$, whence $ \sigma \equal{} 0$.


[quote="siuhochung"]CWMO 03 no.7[/quote]
Yes,I have known it but in CWMO 03 no.7 n=5 and my problem is genaral form of no.7

:D  i think this pro isn't easy

it've proposed for test in HANOI-VMO


Find the minimum value of

f(x,y,z) = x^3 + y^3 + z^3 + 6xyz

for x,y,z >= 0 and x+y+z=1.

there's something that i'm not getting

Find the minimum value of
$ f(x,y,z) \equal{} x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz$ for  $ x,y,z\geq0$ and $ x\plus{}y\plus{}z\equal{}1$

[quote="kritikos"]Find the minimum value of
$ f(x,y,z) \equal{} x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz$ for  $ x,y,z\geq0$ and $ x \plus{} y \plus{} z \equal{} 1$[/quote]
By Schur inequality, we have:
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \ge xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)$
It follows that
$ 4(x^3\plus{}y^3\plus{}z^3)\plus{}13xyz \ge x^3\plus{}y^3\plus{}z^3\plus{}3\left(xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)\right)\plus{}6xyz\equal{}(x\plus{}y\plus{}z)^3\equal{}1$
So $ 4f(x,y,z)\equal{}4(x^3\plus{}y^3\plus{}z^3)\plus{}24xyz \ge 4(x^3\plus{}y^3\plus{}z^3)\plus{}13xyz \ge 1$
Equality holds for $ x\equal{}y\frac{1}{2},z\equal{}0$ and permutation.

$ x^2\plus{}y^2\plus{}z^2\ge \frac{(x\plus{}y\plus{}z)^2}{3}$ I think here u have a mistake because we have that $ x^2\plus{}y^2\plus{}z^2\geq{xy\plus{}yz\plus{}xz}$ but $ \frac{(x\plus{}y\plus{}z)^2}{3}\geq{xy\plus{}yz\plus{}xz}$

I was also trying reducing substituting z=1-(x+y).
Then finding some expresssion that we can put t=x+y ... and finally study f(t) for t in [0,1]

CAN SOMEONE RESOLVE IT?

[quote="exalibur"]if you put x=y=1/2 and z=0 then you have :
f = 1/4 that is smaller than 1/3.!!!!!!

also i found a similiar solution.

pls help[/quote]
Sorry, I have some mistakes. I've edited my post.

we have ${ \{x^3}$+${ \{y^3}$+${ \{z^3}$+6xyz=x²+y²+z²-xy-yz-zx+6xyz
                                                                    =1+6xyz-3(xy+yz+zx);
we have (x+y-z)(y+z-x)(x+z-y)≥xyz;
it follows that (1-2x)(1-2y)(1-2z)≥xyz;
           we attain 1+4(xy+yz+zx)≥9xyz + 2
            nextly, 4(xy+yz+zx - 2xyz)+1≥xyz+2≥2
since we have minimax ...
${ \{(x + y + z)^3}$+9xyz ≥4(x+y+z)(xz+yz+xy) since that we have 1+9xyz≥4(xz+yz+zx)
it follows that 1+4(2xyz-zx-xy-zy)≥-xyz≥$ \frac { - 1}{27}$
since we have minimum!


Here's a fun test I found.  See how you do, and announce your results here.
It wouldn't let me make an extension(quite annoying) so here's the test:

Geography Quiz
US Geography

1. Which US State is home to Hot Springs National Park?
 		A. Louisiana
		B. Arkansas
		C. Wyoming
		D. California

2. The bloodiest site of the Civil War, the Antietam battleground, is located in which US State?
		A. Massachusetts
		B. Kentucky
		C. Maryland
		D. Ohio

3. In terms of area, what is the smallest state in the US?
		A. Delaware
		B. Vermont
		C. Maine
		D. Rhode Island

4. What state was the first to grant women suffrage?
		A. Colorado
		B. Montana
		C. Wyoming
		D. New Mexico

5. Which state is nicknamed the "Sooner State"?
		A. Texas
		B. Oklahoma
		C. Kansas
		D. Alabama

6. In which state was the cereal company Kellogg's founded?
		A. Wisconsin
		B. Minnesota
		C. North Dakota
		D. Michigan

7. Which of the following former Presidents was born in Arkansas?
		A. Ronald Reagan
		B. Jimmy Carter
		C. Bill Clinton
		D. George W. Bush

8. Where did the Klondike Gold Rush happen?
		A. Colorado
		B. Alaska
		C. California
		D. Montana

9. The first U.S. City, St. Augustine, was  founded in 1519 by whom?
		A. The Portuguese
		B. The English
		C. The Spanish
		D. The Italian

10. Which are the only two states where gambling is illegal?
		A. Alaska and Hawaii
		B. Utah and Hawaii
		C. Alaska and Utah
		D. Wyoming and Utah

11. Where is the center of the continental United States located?
		A. Nebraska
		B. Colorado
		C. Oklahoma
		D. Kansas

World Geography
12. Which of the following cities is located in Western Australia?
		A. Melbourne
		B. Sydney
		C. Canberra
		D. Perth

13. Which middle eastern city has a capital of Ashgabat?
		A. Afghanistan
		B. Pakistan
		C. Tajikistan
		D. Turkmenistan

14. What is the most densely populated Asian country?
		A. Brunei
		B. Singapore
		C. Taiwan
		D. Qatar

15. Which of the following countries is located on the Iberian Peninsula?
		A. Italy
		B. Portugal
		C. Ireland
		D. Greece

16. In which African country is English the official spoken language?
		A. Mozambique
		B. Namibia
		C. Egypt
		D. Sudan

17. Which country has the highest per capita GDP of any country?
		A. Luxembourg
		B. Singapore
		C. Qatar
		D. Switzerland

18. Which of the following countries is completely enclosed inside another country?
		A. Tunisia
		B. Sierra Leone
		C. Lesotho
		D. Swaziland

19. In which continent is Mount Kilimanjaro located?
		A. Africa
		B. Europe
		C. Asia
		D. South America

20. Open Ended:  What is the largest country that doesn't border an ocean?

Answers:
1. B
2. C
3. D
4. C
5. B
6. D
7. C
8. B
9. C
10. B
11. D
12. D
13. D
14. B
15. B
16. B
17. C
18. C
19. A
20. Kazakhstan

Here's the grading scale:
20-Geography Expert
19-Genius
16-18 - Smart Alec
13-15 - Geography Whiz
11-12 - Pretty Good
8-10 - Average
5-7 - Not too good
4 - Poor
2-3 - Dreadful
1 - Disgraceful
0 - Know-Nothing

There were probably quite a few people who took it but simply didn't bother to post.

Question 13 says "city" instead of "country"

Question 15 has two answers (both Spain and Portugal are on the Iberian Peninsula, the other country is Andorra)

Also, a lot of it is more of a history quiz than a geography quiz

Sorry about 15, but on number 13, the question said
"Which of the following" meaning there could be more than 1 on the Iberian peninsula, but only one is one of the 4 answer choices.


induction
inequalities
inequalities solved

If $x_{1},x_{2}, \ldots ,x_{n} \geq 1 $ then we have $ \sum_{k=1}^{n} \frac{1}{1+x_{k}} \geq \frac{n}{1+ \sqrt[n]{x_{1}x_{2} \ldots x_{n}}}$.

made by Berkolaiko L.K.

It is Old problem!! :)  :)  :)
We prove by induction on n=2^k
or use Jensen ineq :consider f(x)=1/(1+e^x)


trigonometry
analytic geometry
graphing lines
slope
logarithms
trig identities
calculus

Does $ \frac{e^{\left(\displaystyle\sum_{k\equal{}1}^n \tan^{\minus{}1}\frac{1}{k!}\right)}}{n}$ converge as $ n\rightarrow\infty$? If so, what is its limit?

Let's think about this one little piece at a time -- inside the summation, we have $ \arctan \frac{1}{k!}$.  About how big is $ \frac{1}{k!}$?  So, about how big is $ \arctan \frac{1}{k!}$?  Then about how big is the whole summation?

If the summation in the exponent converges to a finite number (and it does), then the limit will converge to 0. The fun bit is proving that the sum converges.

$ \arctan x<x$ for $ x>0$, so $ \sum_{k\equal{}1}^\infty \arctan \frac{1}{k!}<\sum_{k\equal{}1}^\infty \frac{1}{k!}\equal{}e\minus{}1$.

Um, I guess I typed the wrong problem.  It should be

$ \frac{e^{\sum_{k\equal{}1}^{n} \tan^{\minus{}1} \frac{1}{k}}}{n}$.

Anyway, the numerator forms a line with approximate slope 1.35, and a $ y$ intercept of about .71.  I don't know, I can only go up to about 100 on my Ti-84 before it gets too long to wait.  On Mathematica it went to 100000 and still the slope wasn't 0.

Claim: $ \sum_{k \equal{} 1}^n\tan^{ \minus{} 1}\frac1k \equal{} \ln n \plus{} O(1).$

I make that claim by the limit comparison of this series to the harmonic series.

From that $ e^{\sum_{k \equal{} 1}^n\tan^{ \minus{} 1}\frac1k} \equal{} n\cdot O(1).$

Divide by $ n$ and you get something bounded.

By a slight revision of the argument, I'm willing to claim that $ \sum_{k \equal{} 1}^n\tan^{ \minus{} 1}\frac1k \minus{} \ln n$ is a bounded decreasing (positive) sequence, and hence has a limit, which means that the original problem has a finite limit.

By extrapolation from a modest number of terms, I estimate the limit as $ \approxeq 1.352074826.$ That's the slope in what kidperson3 was saying with his last post, and the slope is what matters.

I don't yet know whether this can be expressed in closed form; there's at least some chance, since there are interesting trig identities to be tried with arctangents.

As an aside: my favorite tool for this sort of thing is not a calculator and it's not Mathmatica/Maple/etc. It's a spreadsheet.


i started alcumus right?
so i started to earn karma one day and then the next day i had even less???
do you have an answer??
if so please post it
also the problems keep repeating them selves

Karma is like money, you earn it then you spend it. For each problem you answer, you lose 5 karma. For each video you watch, you lose 5 karma. For each rating you give a problem, you earn 5 karma. For each video you rate, you earn 5 karma. For each comment you submit on a video, you earn 5 karma. Problems and videos can't be rerated, but you can submit more comments.

but how do you earn karma
i don't know how to rate
and i think sending error reports work right?

sending error reports costs karma (i think it's about 5?), but if your error report is valid, then your receive karma (15 or 20?)

to rate a problem, you go to the bottom of the solutions screen
there will be three categories in which you can rate
rate the categories by choosing the number of stars to give to each category

to rate videos, i believe the rating stars are right underneath the video
just click the number of stars you think the video deserves

It costs 5. You recieve 25.

http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints

 :)

You probably used your karma to answer questions, and never earned any again.


geometry
geometric transformation
rotation
reflection
ratio
3D geometry
sphere

Hey.

You can add two segments/squares/cubes/spheres by putting them together. (same goes to subtraction) Quantification is not required, right?

What does it mean to multiply/divide two segments, w/o assigning a measuring unit?

I really, really hope this makes sense. :ewpu:

Multiplication is really just addition so if you can add two things, you can certainly multiply them.  After all, $ 2\plus{}2\plus{}2$ is the same thing as $ 3\times 2$.

This is assuming you mean the standard multiplication defined on $ \mathbb{R}$.  In abstract groups, you can define some new binary operation and call it multiplication (or addition), and it doesn't necessarily send $ (a,b)\mapsto ab$ (or $ a\plus{}b$).  In this case, you don't need the "quantification" you're thinking of.  For example, in the symmetric group $ S_{n}$, the operation is composition.  In the dihedral group $ D_{2n}$, the operation is rotation and reflection (in a group, there is some binary operation $ \circ$ which is called "multiplication," although as I just pointed out, there may be very little actual multiplication going on).

Holy cow I'm humble, I don't know much math. :surrender:
I tried to understand 2nd paragraph, but I have a hunch it's not what I really meant. :D

I mean, if you don't establish a unit length, you can't really multiply two segments. but you can add them, you know, "sticking" them together.

Let's put it this way:
[img]http://img132.imageshack.us/img132/7519/quantification.jpg[/img]

Multiplying two segments the standard way results in a measurement of area.
Dividing two segments results in a pure number, or a ratio, or whatever.
Multipliying a segment by a sphere results in a four-dimensional measurement of four-dimensional volume.
It's mainly just the problem of the dimensions: $ 2 \text{ km} \times 5 \text{ km} \equal{} 10 \text{ km}^2$.

The second paragraph oversimplified: you can define "multiplication" your way, even if it's something like $ x \times y \equal{} \sin \ln^* (xy)^{5^{(x\plus{}xy\plus{}y\plus{}2010\pi)!!!}}$. However the multiplication should be associative, commutative, have an identity element, and satisfy a bunch of other properties.

Repeat: that is an oversimplification.

[quote="math_explorer"]Multiplying two segments the standard way results in a measurement of area.
Dividing two segments results in a pure number, or a ratio, or whatever.
Multipliying a segment by a sphere results in a four-dimensional measurement of four-dimensional volume.
It's mainly just the problem of the dimensions: $ 2 \text{ km} \times 5 \text{ km} \equal{} 10 \text{ km}^2$.
[/quote]

Thanks. So that means, segment * segment cannot "return" segment, but area?

If so, what about sin(x+y) = sin(x)cos(y) - cos(x)sin(y)?
sin(x+y) is represented by a segment, but RHS is square subtraction, so to speak.

Unfortunately sin(x) does not return a length, but a ratio of lengths, or a pure number. Or you can think of it as a length, but the unit is still implicit.

sin(x) is usually defined in one of three ways:
1. of a right triangle with one angle x, the opposite side (to the x angle) over the hypotenuse. Here it is length / length.
2. in a unit circle, the y-coordinate of the point at angle x. This is a length, but relative to the "unit" of the unit circle is defined;
3. the sum of a certain purely numerical series (which I'm too lazy to look up, as usual).


analytic geometry
graphing lines
slope
ratio
trigonometry
calculus
derivative

A hollow cylinder of radius $R$ and of mass $M$ is rolling down a slope with an incline angle $\varphi$. Inside it there is another but solid cylinder of radius $r$ and mass $m$ rolling so that it maintains the same position with respect to the hollow cylinder and the line connecting the axises of the two cylinders (points $O$ and $C$ on the diagram) is always in angle $\psi$ with the vertical. There is pure rolling everywhere. Find the ratio $\frac{M}{m}$ in terms of the given parameters.

I give up. [b]Thaakisfox[/b], would you please post an answer to this one?

I got $\eta = \frac{M}{m}=\frac{\sin\varphi-((\frac{r}{R})^{2}+\frac{4}{3}(1-\frac{r}{R}))\sin\psi}{\sin\varphi-\frac{4}{3}\sin\psi}$

My result is:

$\frac{M}{m}=\frac{\tan\varphi(1-\cos(\varphi-\psi))+3\sin(\varphi-\psi)}{\tan\varphi(2\cos(\varphi-\psi)-1)-4\sin(\varphi-\psi)}$

This is exactly why I gave up; a tiny mistake could lead to a lot of useless work, and so far either [b]Astronum[/b] or [b]golduck[/b] have squandered their time -- or maybe both :roll:

Yeah, I agree with you [b]Djole[/b] - I tried to solve that problem three times, each time I got a different result :) The one above is the most probable but I still don't believe that it's correct.  :roll:

That's a good way to solve problems, isn't it -- you do it a lot of times and then calculate the mean solution :D You could even estimate the error you made ;)

PS: Sorry for the off-topic...

I checked my solution and now I got a different answer.  :lol:

$\frac{M}{m}=\frac{\sin2\varphi+\cos\psi\sin\varphi-\tan\psi(3(\frac{r}{R})^{2}-4\frac{r}{R}+2-\sin\psi\sin\varphi-2\sin^{2}\varphi)}{\sin2\varphi+\cos\psi\sin\varphi+\tan\psi(\sin\psi\sin\varphi+2\sin^{2}\varphi-2)}$

Hello everyone, my exams are finished so im here again. :D
I see you guys have tried a lot to solve this problem :D
But there is a trick in it.
[hide="hint"]Can this arrangement really take place? :wink: [/hide]

I can't find out why it couldn't take place :?:  :(

Hmm [b]Thaakisfox[/b] could you post a solution to this one :?:

Since we have problem where only rolling is involved, then the problem is of mechanics and we'll try to solve the problem with Lagrangian expressed in terms of generalized coordinates $ \psi$ and $ x$, and their derivatives; $ x$ is the length of the line segment between horizontal ground and the center of hollow cylinder, when we draw parallel to inclined plane:
Firstly, we can easily see that we can divide Lagrangian into two parts corresponding to the Lagrangians of the two cylinders we have:
$ L_{1}= M\dot{x}^{2}-Mgx\sin{\phi}$
The Lagrangian of the second is pretty large with respect to the previous one
$ L_{2}= \frac{1}{2}m((R-r)^{2}\dot{\psi}^{2}+\dot{x}^{2}-2(R-r)\dot{x}\dot{\psi}\cos(\psi-\phi))+$
$ +\frac{1}{4}m(\dot{x}^{2}+(R-r)^{2}\dot{\psi}^{2}-2\dot{x}\dot{\psi}(R-r))-mg(x\sin{\phi}-(R-r)\cos{\psi})$.
Now let's write the Lagrangian for the system in total:
$ L= (M+\frac{3}{4}m)\dot{x}^{2}+\frac{3}{4}m(R-r)^{2}\dot{\psi}^{2}-m(R-r)(\cos{(\phi-\psi)}+\frac{1}{2})\dot{x}\dot{\psi}-$
$ -(M+m)gx\sin{\phi}+mg(R-r)\cos \psi$
Now if we use Lagrange's formula we can easily get the equations of motion:
$ \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})=\frac{\partial L}{\partial x}$
$ \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})=\frac{d}{dt}((2M+\frac{3}{2}m)\dot{x}-m(R-r)(\cos(\phi-\psi)+\frac{1}{2})\dot{\psi})=$
$ = (2M+\frac{3}{2}m)\ddot{x}-m(R-r)\sin(\phi-\psi)\dot{\psi}^{2}-m(R-r)(\cos(\phi-\psi)+\frac{1}{2})\ddot{\psi}$
$ \frac{\partial L}{\partial x}=-(M+m)g\sin{\phi}$
Therfore,
$ (2M+\frac{3}{2}m)\ddot{x}-m(R-r)\sin(\phi-\psi)\dot{\psi}^{2}-m(R-r)(\cos(\phi-\psi)+\frac{1}{2})\ddot{\psi}=$
$ =-(M+m)g\sin{\phi}$
Since, we examine a particular solution for the system $ \dot{\psi}=0$, then our equation reduces to
$ (2M+\frac{3}{2}m)\ddot{x}=-(M+m)g\sin\phi$
Now comes the second Lagrange equation for the system:
$ \frac{d}{dt}(\frac{\partial L}{\partial \dot{\psi}})=\frac{\partial L}{\partial \psi}$
$ \frac{d}{dt}(\frac{\partial L}{\partial \dot{\psi}})=\frac{d}{dt}(\frac{3}{2}m(R-r)^{2}\dot{\psi}-m(R-r)(\cos(\phi-\psi)+\frac{1}{2})\dot{x})=$
$ =\frac{3}{2}m(R-r)^{2}\ddot{\psi}-m(R-r)\sin(\phi-\psi)\dot{\psi}\dot{x}-m(R-r)(\cos(\phi-\psi)+\frac{1}{2})\ddot{x}$
$ \frac{\partial L}{\partial \psi}=-m(R-r)\sin(\phi-\psi)\dot{x}\dot{\psi}-mg(R-r)\sin\psi$
Again, making $ {\dot{\psi}=0}$, we get
$ -(\cos(\phi-\psi)+\frac{1}{2})\ddot{x}=-g\sin\psi$
Combining the two equations we got, we'll achieve
$ \frac{(M+m)g\sin{\phi}}{2M+\frac{3}{2}m}=-\frac{g\sin{\psi}}{\cos(\phi-\psi)+\frac{1}{2}}$
$ \frac{2M+\frac{3}{2}m}{(M+m)}=-\frac{\sin\phi(\cos(\psi-\phi)+\frac{1}{2})}{\sin{\psi}}=$
$ =-\frac{1}{2}(\frac{\sin\psi+\sin(2\phi-\psi)+\sin\phi}{\sin\psi})=-\frac{1}{2}(1+\frac{\sin(2\phi-\psi)+\sin\phi}{\sin\psi})=$
$ =-\frac{1}{2}(1+\frac{2\sin(3\phi-\psi)\cos(\phi-\psi)}{\sin\psi})$
Can anybody check the $ crap$ I've just made here...


Let $n$ be a natural number. Define $t(n)$ as the number of positive divisors of $n$ (including $1$ and $n$) en define $\sigma(n)$ as the sum of these numbers. Show that

\[\sigma(n) \geq \sqrt{n}. t(n)\]

[hide]
Let the divisors be $1 = d_{0}< d_{1}< \cdots < d_{k}= n$. Clearly $d_{0}d_{k}= n$ and in general $d_{i}d_{k-i}= n$. Then we clearly have

$d_{0}d_{1}\cdots d_{k}= n^{\frac{k+1}{2}}$.

By AM-GM, we know

$\frac{d_{0}+d_{1}+\cdots+d_{k}}{k+1}\ge \sqrt[k+1]{d_{0}d_{1}\cdots d_{k}}= \sqrt[k+1]{n^{\frac{k+1}{2}}}= \sqrt{n}$.

But $\sigma(n) = d_{0}+d_{1}+\cdots+d_{k}$ and $\tau(n) = k+1$ so

$\frac{\sigma(n)}{\tau(n)}\ge \sqrt{n}\Rightarrow \sigma(n) \ge \tau(n) \sqrt{n}$.[/hide]


[quote="andrew-mathmaster"]LET a,b,c>0
PROVE THAT:
abc+2(a^2+b^2+c^2)>=8+5(a+b+c) :wink:[/quote]
It is wrong: try $a=b=c\to 0$.


hy all i have some freind who have as email adress
xxxxx@gmail.fr but haw do they make it  because i want to have one  :roll:

I signed up for gmail here:

https://www.google.com/accounts/SmsMailSignup1


As the title say, who can prove Nesbitt's inequality for $7$ numbers? Does an elementary solution exist?

Can anybody show me any information about an elementary solution of Nesbitt inequality for $n=7$?


The sum of three numbers is 20. The first is 4 times the sum of the other two. The second is seven times the third. What is the product of all three?

$\left(A\right) 28$
$\left(B\right) 40$
$\left(C\right) 100$
$\left(D\right) 400$
$\left(E\right) 800$

k i saw a lot of unsolved one's so i guess im done for now cuz i'm bored..

[hide]

1. Let $a+b+c=20$

$a=4(b+c)\implies (b+c)=\frac{a}{4}$ sub into 1

$a+\frac{a}{4}=20\implies a=16$

sub $b=7c$ and $a=16$ into 1

$\implies 8c=4\implies c=\frac{1}{2}\implies b=\frac{7}{2}$

$abc=\frac{1}{2}\cdot\frac{7}{2}\cdot 16=\boxed{28}$

[/hide]


inequalities
logarithms
inequalities unsolved

If $ a,b > 0$ and $ x,y \ge 0$ are non-negative real numbers such that $ x \plus{} y \equal{} 1$ , prove
\[ ax \plus{} by\ge a^xb^y\]

Edited: arqady's right. Inequality is defined for real exponents, and $ a,b >0$ . So how to get through with it ?

[quote="Obel1x"]If $ a,b \ge 0$ and $ x,y$ are non-negative numbers such that $ x \plus{} y \equal{} 1$ , prove
\[ ax \plus{} by\ge a^xb^y\]
[/quote]

 $ ab \equal{} 0$  obvious true

$ ab > 0$

$ ax \plus{} by\ge a^xb^y \Longleftrightarrow x\ln{a} \plus{} y\ln{b}\le \ln{(ax \plus{} by)}$

true due to Jensen。

I wonder we can use Bernoulli inequality.

[quote="hedeng123"][quote="Obel1x"]If $ a,b \ge 0$ and $ x,y$ are non-negative numbers such that $ x \plus{} y \equal{} 1$ , prove
\[ ax \plus{} by\ge a^xb^y\]
[/quote]

 $ ab \equal{} 0$  obvious true

[/quote]
I think, if $ ab\equal{}0$ it's obviously wrong.
Indeed,  $ a^x$ defined for real numbers $ a$ and $ x$ only for $ a>0.$ :wink:


I just have a little question, because I can't picture this in my head.

Say you had three cylinders such that they are each the same size and shape, and the diameter of the cylinders is equal to the height of the cylinders. Say you put the cylinders together at different orientations, such that one is facing front-back, one is facing left-right, and one facing up-down, and they all share the same center. Is the intersection of their three volumes a sphere? If not, what is it?

No, it's not a sphere. Check this out for some more info: [url]http://mathworld.wolfram.com/SteinmetzSolid.html[/url]
It's like two-thirds of the way down the page.


linear algebra
matrix
superior algebra
superior algebra solved

We have $A,B \in M_{4}(R)$ and $A-B=AB.$
If it exists $m \in N^*$ so that:$ A^m-B^m $ is invertible, than prove that A and B are invertible.

Since $A-B-AB+I_n=I_n$ (we don't need the fact that they're in $M_4(\mathbb R)$; any $M_n(\mathbb R)$ works), we have $()()=I_n$, so $()()=I_n$, meaning that $A,B$ commute. This means that $A^m-B^m=(A-B)(A^{m-1}+\ldots+B^{m-1})=AB(A^{m-1}+\ldots+B^{m-1})$, and if this is invertible, then $A,B$ must be invertible.


hey.. does anyone know how to sign up for other olimpiads other than math?
like how do i take the physics and bio equivalent of the AMC or AIME?
I dont think my teachers offer it or know much about it.. which is why i missed out last year... although my skool is very involved in the chem olimpiad..

For Bio:

  http://www.cee.org/usabo/usabo_how_to_enter.shtml

For Physics:

  http://www.aapt.org/Contests/olympiad.cfm

For Computer programming:

  http://www.usaco.org

Wait...let me get this straight...for biology, all you have to do, is just be like, "Hey Miss, nominate me for this USABO!"  and then you're in?  Wow, that's so much more easier the the USAMO. :-P

It's not quite the same.  Your teacher can nominate you for the Open Exam which is kind of like taking the Amc 12/10.  I believe the teacher can nominate as many people as possible.  So it s not the same as taking the USAMO.

What about chemistry?  To me, chem way cooler than biology.

Look on this page for all kinds of competitions [url]http://cty.jhu.edu/imagine/linkB.htm[/url].  It has the olympiads under the math/science section somewhere.

I took the Chemistry this year.  I made it past the first round but did horrible in the  second.  I was happy with my results considering that I had only been in chemistry for less than half of a year.

Chemistry is a bit shady.  You're actually competing regionally rather than nationally for the first exam.  For example, out of 60 questions, in some areas you may need 40ish questions correct to move on to the national exam whereas in other areas only the high fifties will allow you to take the next test.  Read Zumdahl, Chang, or some other chem textbook front to back and you should be set.

oh god.. finally.. i was looking for this post for so long.. forgot where i put it.. thanks

oh man.. jus did a few problems on physics test.. i am getting rusty only after a few month.. guess i'll have to study for it before i take it.. anyways.. anyone know what the curves are for both bio and physics?


A negatively charged particle is caused to move between two electrically charged plates, as illustrated in Figure 2.7.  If the mass of this particle is creased while the speed of the particles remains the same, would you expect the bending to increase, decrease or stay the same, and why?

Thanks in advance!

do you mean increased or decreased?

If increased, then ur increasing the particles kinetic energy, hence it will have less bending as the object will need more work acted upon it to change its course... The opposite for decrease.


limit
induction
inequalities
function
integration
calculus
calculus computations

Solve an equation $\ds\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^{2}+\ldots+\sqrt{x^{n}}}}}=2.$
(A. Kukush)

I have managed to prove the following:
1) There exists precisely one $x$ in $\mathbb{R}^{+}$ that solves the equation (it is also clear that this solution is not in $[0,1]\cup[6,\infty)$ through easy calculations, not in say $[0,3.99]\cup[4.01,\infty)$ by numerical calculations with computer).

2. For any $m$ and $x > 1$, $\ds\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^{2}+\ldots+\sqrt{x^{n}}}}}< \sqrt{1+\sqrt{x+\sqrt{x^{2}+\ldots+\sqrt{x^{m}+\frac{x}{2}\left(1+\sqrt{1+4x^{m-1}}\right) }}}}}$ (and in the proof you also see that the larger $m$ the closer the RHS is to th LHS).

I am also certain that the solution is $x=4$ from numerical evidence, but I can't prove it.

kalle your intuition is good probably $x=4$ is the only solution.
Jean_Noël used mapple he told he found $4$.

Find the proof

Do you know a proof? I have spent a lot of time on this problem, and it seems futile. So I will not think about it any more. The approaches I've tried are:

1) To consider the the equation you get for each $n$ when you remove the limit sign. By intermediate value theorem you have that for each $n$ there is precisely one solution $a_{n}$. It is clear that $a_{n}$ converges... from some reasonings around this you get my statement "1)" in my previous post. I have tried to prove by induction that $4-\frac{1}{2^{n}}< a_{n}< 4+\frac{1}{2^{n}}\quad n \ge 2$, but I can't do it.

2) I have tried to calculate the limit directly for each x, mainly focusing on finding good inequalities (which may also help in 1). I think the one I posted is the best one I got. It comes from that I managed to show that:
$\sqrt{ax^{2}+\sqrt{ax^{4}+\sqrt{ax^{8}+\cdots}}}= \frac{x}{2}\left( 1+\sqrt{1+4a}\right)$.

I am not getting further than that.

How do you get this equality?

$\sqrt{ax^{2}+\sqrt{ax^{4}+\sqrt{ax^{8}+\cdots}}}= \frac{x}{2}\left( 1+\sqrt{1+4a}\right)$.

$\sqrt{ax^{2}+\sqrt{ax^{4}+\sqrt{ax^{8}+\cdots}}}= x\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}$.  Presuming convergence,
$S = \sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}\implies S^{2}= a+S$ whence the result.

I don't think this is going to be helpful here, unfortunately -- there's no obvious way to go from powers of 2 to all the integers in such an expression.

Well, my idea was to take a certain number of terms in the original nested radical and then approximate the rest of the terms with such a series. I don't think the problem can be solved by finding a clever equality, I believe it must be solved with inequalities.

Hey,I think I got it,using numerator-rational.
we have:
$2-\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}=\frac{3-\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}{2+\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}}=\frac{5-\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}{\left(2+\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}\right)\left(3+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}\right)}$
$=\cdots=\frac{2^{n+1}+1}{\left(2+\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}\right)\left(3+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}\right)\cdots \left(2^{n}+1+\sqrt{4^{n}}\right)}$
then it is easy to prove the limit is 0.
So we have proved the limit of the left side is 2,when x=4.
Then by proving the limit of the left side is strickly increasing by x.We sloved the whole problem. :)
Am I right?

The above posts culminating with the alleged proof  I have a complaint with inasmuch as they it is like working on a multiple choice math question and eliminating choices and then plugging in one of the remaining values to see that it appears to be working.

None of the posts have taken a constructivist approach to actually solve the equation outright.  If this value of 4 "weren't
even on the map," what would you solvers do instead?

I have found a different solution, which is unfortunately not nearly as simple as zhaobin's. Actually, my solution doesn't quite work. I am almost certain it is correct up to and including the point where I conclude $\lim_{n\to\infty}a_{n}= 4$. However, the argument following is rubbish. I must've been thinking that the limit is a continuous function of $x$ or something similar. Of course it sucks to realize that you're wrong. Could anyone help me by finding an argument for the last bit?

Consider the equation $\sqrt{1+\sqrt{x+\sqrt{x^{2}+\ldots+\sqrt{x^{n}}}}}=2$ for every integer $n$. The LHS expression is strictly increasing in $[0,\infty)$, assuming all values in $[1,\infty)$, according to the intermediate value theorem. Therefore the equation has precisely one solution $a_{n}$. It is clear that $(a_{n})_{n=1}^\infty$ is a strictly decreasing bounded sequence, and therefore converges to a limit $A$.

Now, I hope you can understand the upcoming notation. It might require you to write the following out yourself. By squaring and rearranging the equation many times we find that it's equivalent to
$\sqrt{x^{n}}= \left( \cdots \left( \left(9-x\right)^{2}-x^{2}\right)^{2}-\cdots-x^{n-2}\right)^{2}-x^{n-1}$, (**)
By an inductive argument you can see that for $x=4$ the RHS in (**) is equal to $2^{n}+1$. From this you see that
 $\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}< 2$
and thus $4 < a_{n}$ for every $n$.

By another inductive reasoning the RHS in (**) is always positive for $x \in [0,a_{n}]$. This statement is truthfully a little bit work to show, but it's not hard. Since it is true for every $n$, it implies that you can't find a bracket "inside" the RHS that is negative in $[0,a_{n}]$ (remember, $a_{n}$ is decreasing). This in turn implies that the RHS is strictly decreasing (with increasing $x$) in $[0,a_{n}]$.

Now introduce $f_{n}(x) = \frac{\sqrt{x^{n}}}{\left( \cdots \left( \left(9-x\right)^{2}-x^{2}\right)^{2}-\cdots-x^{n-2}\right)^{2}-x^{n-1}}$

Our equation is $f_{n}(x) = 1$. By what I said above we have that $f_{n}(4) = \frac{2^{n}}{2^{n}+1}= 1-\frac{1}{2^{n}+1}\to 1$ as $n\to\infty$. From the discussion about monotony, we realize that $f_{n}$ is strictly increasing in $[4,a_{n}]$ for every $n$.

I claim that $A = 4$. We will prove it by contradiction: Assume that $A > 4$. Then we can choose a large $n$ so that $f_{n}(4) \approx 1$. Then since $f_{n}$ is stricly increasing in $[4,a_{n}]$ and $f_{n}(a_{n}) = 1$ it follows that for every $b \in [4,A]$, $\lim_{n\to\infty}f_{n}(b) = 1$. In fact, we have $f_{n}\to 1$ uniformly in $[4,A]$! This would have the consequence that $\lim_{n\to \infty}\int_{4}^{A}f_{n}dx = A-4$. However,
$\int_{4}^{A}f_{n}dx > \int_{4}^{A}\frac{\sqrt{x^{n}}}{2^{n}+1}dx$.
A pure calculation shows that the right hand side tends to $\infty$ as $n$ gets large. So we have found a contradiction.

That is, we have now shown that $\lim_{n\to\infty}a_{n}= 4$.

Finally, we show that
$\ds\lim_{n\to\infty}\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}=2$
Suppose the opposite, then
$\ds\lim_{n\to\infty}\sqrt{1+\sqrt{4+\sqrt{4^{2}+\ldots+\sqrt{4^{n}}}}}<2$
Then, since the value of the limit strictly increases with $x$, we can find a $c>4$ such that
$\ds\lim_{n\to\infty}\sqrt{1+\sqrt{c+\sqrt{c^{2}+\ldots+\sqrt{c^{n}}}}}<2$
But this is absurd because eventually $a_{n}< c$.

Sorry for bumping this thread, but as it is I don't think people have seen that I would like some help/discussion. As it stands I have defined a sequence $a_{n}$ which for each $n$ is value of the solution to the equation:
$\sqrt{1+\sqrt{x+\cdots+\sqrt{x^{n}}}}= 2$.
Furthermore, I have shown that $\lim_{n \to \infty}a_{n}= 4$.

However, although I have included a bogus argument, I don't know how to show that this implies that $x=4$ solves the initial equation. Actually, I don't even know how to show that the solution is unique.

Does anyone have any ideas? I'd love some help.

[quote="Kalle"]

Does anyone have any ideas? I'd love some help.[/quote]
I think we can prove that the limit of the left side is strick increasing by $x$.
let $a(n,x)=\sqrt{1+\sqrt{x+\sqrt{x^{2}+\cdots+\sqrt{x^{n}}}}}$
First step:Prove that $a(n,x)$ has limit when n go infinite for every x.
this is because:$x<1$,obvious.
$x>1$,$a(n,x)<xa(n,1)$,but $\lim_{n\to \infty}a(n,1)$ is exist.
Second step:by first step it is to easy show:$\lim_{n\to \infty}\sqrt{x^{2}+\sqrt{x^{3}+\cdots+\sqrt{x^{n}}}}$ is exist.
then for $x>y$ we have:
$\lim_{n\to \infty}\sqrt{x^{2}+\sqrt{x^{3}+\cdots+\sqrt{x^{n}}}}\ge \lim_{n\to \infty}\sqrt{y^{2}+\sqrt{y^{3}+\cdots+\sqrt{y^{n}}}}$
thus:$\lim_{n\to \infty}a(n,x) >\lim_{n\to \infty}a(n,y)$(do it yourself :) )
Now we proved limit of the left side is strick increasing by $x$.
The result follows.This is my approach.

But a guy in Chinese forum proved:
$\lim_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^{2}+\cdots+\sqrt{x^{n}}}}}=\sqrt{x}+\frac{4-x}{2+\sqrt{x}}$
but his proof is only write:
$\lim_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^{2}+\cdots+\sqrt{x^{n}}}}}-\sqrt{x}=\lim_{n\to \infty}\frac{1+\sqrt{x+\sqrt{x^{2}+\cdots+\sqrt{x^{n}}}}-x}{\sqrt{1+\sqrt{x+\sqrt{x^{2}+\cdots+\sqrt{x^{n}}}}}+\sqrt{x}}=\frac{4-x}{2+\sqrt{x}}$
I can't understand his proof,And I also doubt it...

[quote="Kalle"]Maybe he meant something different?

We have that
$\sqrt{x}+\frac{4-x}{2+\sqrt{x}}= 2$ for every $x$.[/quote]
but he said he is wrong...

[quote="zhaobin"]

$\lim_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^{2}+\cdots+\sqrt{x^{n}}}}}=\sqrt{x}+\frac{4-x}{2+\sqrt{x}}$
[/quote]

this is not correct for $x=1$

$\lim_{n\to \infty}\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}=L$ positive
with $L^{2}=1+L$

$L=\frac{1+\sqrt{5}}{2}$

but when you plug $x=1$ in $\sqrt{x}+\frac{4-x}{2+\sqrt{x}}$
we get 1++1=2

Ok,I also think it is impossible...
But now the whole problem is also solved.
Do you agree?

Which is the official solution for this problem?


calculus
real analysis
real analysis unsolved

This problem is from our schools exams .

Let $ f: [a,b] \rightarrow R$ continuous to $ [a,b]$.$ f''$ is also continuous to $ (a,b)$  .If $ f(a)=f(b)=0$, and exists real numbers $ c,d\in (a,b)$ such as$ f(c)f(d)<0$ proove

i) $ f(x)=0$ has at least one real root to $ [a,b]$

ii) exists$ \xi_{1}, \xi_{2}\in (a,b)$ such as  $ f''(\xi_{1})>0$ $ f''(\xi_{2})<0$

(i)  Since $ f(c)f(d)<0$, we must have $ f(c)$ and $ f(d)$ be of opposite sign.  Intermediate Value theorem guarantees a value between $ c$ and $ d$, say $ x$, such that $ f(x)=0$

(ii)  Note that since $ f(a)=f(x)=f(b)=0$, we can not have $ f''(x)=0 \hspace{1 mm}\forall x$.  But we also can not have $ f''(x)>0 \hspace{1 mm}\forall x$ or $ f''(x)<0 \hspace{1 mm}\forall x$ since $ f$ has two interior critical points which follows from Rolle's theorem. Therefore the result follows.


29. The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divisible by 19 or 31. What is the largest possible last digit in this string?

Gracias.  :D

Huh? It's not possible is it? Cause if you have:
$ 1, 9...$ What's after the 9??????

... Wrong problem, wrong thread, jonathan...

The numbers possible are:
19
38
57
76
95

31
62
93

As you can see, 38 can be eliminated (as there is no two-digit number with 8 to follow up) and 62 (no leading 2).

The numbers possible are:
19
57
76
95

31
93

Now, 57,76,and 95 will be eliminated.

19
31
93

Thus it cycles, 193193193... and the 2002 = 1 (mod 3) digit is [b]1[/b]

Notice that there exists no configurability, and there exists only one string.

Uh.. You might want to see our gmail chat. I edited the names:

mr.k_74: you better want to change ur post now
i edited my post

jonathanchou711: oh..
ok

mr.k_74: me bad

jonathanchou711: that ok

So I was we were both referring to the Mathcounts Marathon place, but we both edited. Apparently, you were in the middle of typing urs when I edited.

The answer isn't $ 1$...after the last $ 3$ shows up, the final digit can be anything that the three allows it to be, disregarding the cycles (since there's nothing after it), and the answer is $ 193193\ldots193\boxed8$.


modular arithmetic
search
algorithm
Euler
geometry
3D geometry
number theory

Mr X chooses an integer x={0,1,2,...,1924992000}. He squares it, divides it by 4000000007 and gives us the remainder, so
(x 2) mod 4000000007 = k

The question is:
If we know k - how to find x?

This is not possible, since a value of $k$ can correspondence with different values of $x$. For example $4\equiv2^2\equiv4000000005^2\pmod{400000007}$.

It has to be possible and I know it is - just can't find the right way to solution....

And btw:
1924992000<4000000005, so we don't need to worry about integers larger then 1924992000....

If someone gave $k=4000000006$  or $k=5$ you would be out of luck, as there are no solutions .. on the other hand $k=2$ would be ok. and you should get your $x$..:)

(For example you can try  $ x = 2529181847$ and you would get  $x^2 \equiv 2  \bmod 4000000007$ )
( Another value of x  (if you wanted it  to be less than 2000000000)   is  $x=  1470818160 $ which will give $x^2 \equiv 2 \bmod 4000000007$)

(The original question is "given $k$ how do  take a "square-root to find $x$" ? (all  calculations are mod 4000000007)
(and  within a reasonable amout of time :)  (Hint: It is  lucky  that 4000000007 is prime and  of the form 4m+3)

Important Note:    Between 0 and 4000000006 there are always two  values of x  which will satisfy any given k you must restrict   0<= x <= 2000000003  to  get an unique value of x)

I understand that we have to take such x that remainder mod 4000000007 will be unique. But x is an integer from set {1,2,3,4,...,1924992000} and 2529181847>1924992000, so if x is from the given set - 1470818160 is ok, but 2529181847 is out of it - it's just too big. And of course - as 2000000003>1924992000, there has to be only one value that satisfy given k - I'm just wondering how to find x, when we are given k (the remainder) and all we know is that x has unique value...

Can you please explain me why it's so important that 4000000007 is prime of the form 4m+3?? Is there any "mathematical" way of getting x when we know k?

First: IMO this problem is probably more suitable for the “Pre-Olympiad?  Forum.

Any way here is one method to do this:

All calculations below are $\bmod 4000000007$  and  $x$ or $k$ etc are not $0$.
We are given  $k=x^2$
But, since $4000000007$ is prime,  FLT ===>  : $x^{4000000006} \equiv 1$
Or $k^{2000000003} \equiv x^{4000000006} \equiv 1$
Let $y \equiv k^{1000000002}$
Then $y^2  \equiv  k^{2000000004} \equiv k$
So  $y \equiv  +x \ or \  y \equiv  -x $  You take the value between  1 and 2000000003.

[hide="Method works when number is prime  and 3 mod 4"]Note if $4000000007$ was not of the form $4n+3$ the above method (try it and see why, use $13$ for example) will not work,  and method would be a little  different.
[/hide]

Also finding $k^{1000000002}$  is not really that hard to compute. [hide]You go  by computing 10th power, eg  say  $a \equiv k^{10}$ then compute $a^{10}$ etc.. so you have to do only 8-9  such computations (and not billions).
[/hide]
Hope this is helpful.

[quote="tytus"]Can you please explain me why it's so important that 4000000007 is prime of the form 4m+3?? Is there any "mathematical" way of getting x when we know k?[/quote]

What you are trying to do is a special (very simple) case of a standard cryptographic algorithm.  [You might search for "Berlekamp's Algorithm"]

Let $m$ be the plaintext (an integer in $[0,1924902000]$).  Let $c$ be the ciphertext (an integer in $[0, 4000000007)$.)

You have told us that $c=m^2 \mod 4000000007$.
Is there a number, $d$, such that $m=c^d \mod 4000000007?$

Combining these equations:
$m=(m^2)^d \mod 4000000007$
$m=m^{2d} \mod 4000000007$

In general:
$m=m^{1+\phi(n)} \mod n$

Where $\phi$ is the Euler Totient Function.  In the special case, where $n$ is a prime, we have $\phi(n)=n-1$.  [This is equivalent to Fermat's Little Theorem, as Gyan previously noted.]

So, all we are left is to find:
$2d=1+j\phi(n)$, where $j$ is a positive integer.

$2d=1+j(n-1)$
$2d=1+j(4000000007-1)$
$2d=1+j(4000000006)$

Oops.  There is no value of $j$, such that $2d$ will be odd.  In a roundabout way, this explains why there is not a unique square root, modulo an odd prime.

If $n$ was not prime, then $\phi(n)$ would have a more complex form, and it would be necessary to factor $n$ to find $\phi(n)$.  [Choosing $n$ to be the product of two very large randomly selected primes is the key to the RSA algorithm's strength.]

If you had cubed the number and asked for the cube-root, modulo 4000000007, then we could have found a unique value of $1\le d<4000000007.$  [In fact, $3\times 2666666671=1+2\times 4000000006$.]

Gyan has shown how to obtain the two square roots, under specified conditions.  Between my explanation and his, does this answer your questions?

[quote="tytus"]I understand that we have to take such x that remainder mod 4000000007 will be unique. But x is an integer from set {1,2,3,4,...,1924992000} and 2529181847>1924992000, so if x is from the given set - 1470818160 is ok, but 2529181847 is out of it - it's just too big. ..
[/quote]

Yes, you are right,  and I was just pointing that out in my first message, of course,  $1470818160$  is nothing but $(4000000007 - 2529181847)$ so once you restrict the range, the value is unique.
In my second  post I posted  one method to find that value.

[quote]
Can you please explain me why it's so important that 4000000007 is prime of the form 4m+3?? Is there any "mathematical" way of getting x when we know k?[/quote]
When the modulo mumber is prime of the form 4k+3 (Or a composite number with all factors of the form 4m+3) Things are a little easier, becuase then you can  write x as a power of k.  (The reason for this a little complex)

For numbers not of the type 4k+3,  one may use a  little different method  but it is  still not  really that difficult .  BTW  Note that [b]not [/b]every number will have a  "square-root".

[quote="tytus"]OK - now I understand little bit more, but can you please show me how does it work?? For example: k=1749870067 and I know that then x=1087143639. But what is the way to get x??[/quote]

[b]That's what was  explained before:[/b]
Using a calculator (Maple)  for example: (Time taken : only a few seconds)

[quote]> n:=4000000007;k:=1749870067;

[color=blue](To avoid overflow we take  only 1000 power  of k at a time .. we have to find k^1000000002)[/color]

> y:=k^1000 mod n;    ====>       y := 3603893674      ([color=blue] Done to avoid overflow) [/color]
> y:=y^1000 mod n;   =====>      y := 1787853990     [color=blue] ( k^1000000)[/color]
> y:=y^1000 mod n;   =====>      y := 3398870423      [color=blue] (k^1000000000)[/color]

> y:=y*k^2 mod n;  ======.         y := 2912856368     [color=blue] (k^1000000002)[/color]

> x:=n-y;         [color=red] [size=150] x := 1087143639[/size]   [/color]      [color=blue](Since 2912856...> 2000000002 we  subtract from n) [/color]

[/quote]

Added later: Just curious, what is your back ground as far as number theory is concerned?  I have seen other questions (Like  how to decompose a prime into sum of two squares, and how to find a prime etc... ) Looks like you are  interested in computing methods? ...


inequalities
search
geometry
inradius
circumcircle
trigonometry
Cauchy Inequality

They may have been solved ...but ...I need your solutions...

[hide="6"]
multiply both sides by $ xyz$,we have to show that:

$ x^4 \plus{} y^4 \plus{} z^4\geq xyz(x \plus{} y \plus{} z)$

now note that by cauchy inequality we have:

$ x^4 \plus{} y^4 \plus{} z^4\geq \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}3\geq \frac {(x \plus{} y \plus{} z)^4}{27}$

so our inequality is equivalent to:

$ \frac {(x \plus{} y \plus{} z)^4}{27}\geq xyz(x \plus{} y \plus{} z)$

$ \iff \frac {x \plus{} y \plus{} z}3\geq \sqrt [3]{xyz}$

which is true by AM-GM...[/hide]

The seventh is incorrect,i think it should be:
$ 9(a^3\plus{}b^3\plus{}c^3)\geq (a\plus{}b\plus{}c)^3$.
For a proof,look here:
http://www.mathlinks.ro/viewtopic.php?t=171643
The third is Iran 1996,
See here:
http://www.mathlinks.ro/viewtopic.php?search_id=209710799&t=3547
And i think that it should be:
$ \sum \frac{a}{b\plus{}c}<2$,
instead of
$ \sum \frac{a}{b\plus{}c}\leq 2$.

17.

$ (a\plus{}b\plus{}c)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\geq 9,$ using Jensen's inequality gives

$ \frac{\left(a\plus{}\frac{1}{a}\right)^2\plus{}\left(b\plus{}\frac{1}{b}\right)^2\plus{}\left(c\plus{}\frac{1}{c}\right)^2}{3}\geq \left(\frac{a\plus{}\frac{1}{a}\plus{}b\plus{}\frac{1}{b}\plus{}c\plus{}\frac{1}{c}}{3}\right)^2$

$ \equal{}\left(\frac{1\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}}{3}\right)^2\geq\left(\frac{10}{3}\right)^2$ yielding $ \left(a\plus{}\frac{1}{a}\right)^2\plus{}\left(b\plus{}\frac{1}{b}\right)^2\plus{}\left(c\plus{}\frac{1}{c}\right)^2\geq \frac{100}{3}$

[quote="kim77"]i'm sending some ques.s more![/quote]
For problem 17.
[hide="Generalisation holds: your case is for n=3."]
[u]Problem:[/u]
$ \left(a_1 \plus{} \frac 1{a_1}\right)^2 \plus{} \left(a_2 \plus{} \frac 1{a_2}\right)^2 \plus{} \ldots \plus{} \left(a_n \plus{} \frac 1{a_n}\right)^2\ge K$
Find the best K

[b][i]Claim:[/i][/b]$ K \equal{} \frac {(n^2 \plus{} 1)^2}{n}$
[b][i][u]Proof:[/u][/i][/b]
Denote LHS of your inequality by I,
AM-HM says:
$ (\sum a_i)(\sum \frac1{a_i}) \geq n^2 \Rightarrow (1 \plus{} \sum \frac1{a_i})^2 \geq (n^2 \plus{} 1)^2$-------------(*)
Applying CS:
$ I \geq \frac {{(a_1 \plus{} \frac 1{a_1} \plus{} a_2 \plus{} \frac 1{a_2} \plus{} \ldots a_n \plus{} \frac 1{a_n}})^2}{1 \plus{} 1 \plus{} 1 \ldots \plus{} 1}$
But $ ({a_1 \plus{} \frac 1{a_1} \plus{} a_2 \plus{} \frac 1{a_2} \plus{} \ldots a_n \plus{} \frac 1{a_n}} \equal{} 1 \plus{} \sum \frac1{a_i}$
So $ I \geq \frac {{(1 \plus{} \sum \frac1{a_i})}^2}{n}$.
But by (*), $ \frac {{(1 \plus{} \sum \frac1{a_i})}^2}{n}\geq \frac {(n^2 \plus{} 1)^2}{n}$
Thus $ I \geq \frac {(n^2 \plus{} 1)^2}{n}$
Equality for all $ a_i$s equal.
K cant get any better than that because we showed equality achieving ;)[/hide]

15 follows from AM-GM, applied to both terms....
[hide]$ (\Sigma{a^2b})(\Sigma{ab^2}) \geq 9{a^6b^6c^6}^{\frac13}$
[/hide]

[hide="Problem 16 By AM_GM,"]
$ abc \leq (\frac {a \plus{} 1}{2})^2 \cdot (\frac {b \plus{} 1}{2})^2 \cdot (\frac {c \plus{} 1}{2})^2 \equal{} \frac {8^2}{4^3} \equal{} 1$[/hide]

2. For $ 0 < x < 1,\ \frac {1}{1 \minus{} \sqrt {x}}\geq 4x \plus{} 1\Longleftrightarrow \sqrt {x}(2\sqrt {x} \minus{} 1)^2\geq 0,$

$ \frac {1 \plus{} \sqrt {a}}{1 \minus{} a} \plus{} \frac {1 \plus{} \sqrt {b}}{1 \minus{} b} \plus{} \frac {1 \plus{} \sqrt {c}}{1 \minus{} c} \plus{} \frac {1 \plus{} \sqrt {d}}{1 \minus{} d} \equal{} \frac {1}{1 \minus{} \sqrt {a}} \plus{} \frac {1}{1 \minus{} \sqrt {b}} \plus{} \frac {1}{1 \minus{} \sqrt {c}} \plus{} \frac {1}{1 \minus{} \sqrt {d}}$

$ \geq 4(a \plus{} b \plus{} c \plus{} d) \plus{} 1\cdot 4 \equal{} 8.$

[hide="For problem 10, "]$ 4(a^3 \plus{} b^3) \geq (a\plus{}b)^3 \Leftrightarrow a^3 \plus{} b^3 \geq a^2b \plus{} b^2a$
It follows from Muirhead, since [3,0] majorises [2,1][/hide]

[quote="Erken"][quote="kunny"][quote="isomorphism"][hide="For problem 10, "]$ 4(a^3 \plus{} b^3) \geq (a \plus{} b)^3 \Leftrightarrow a^3 \plus{} b^3 \geq a^2b \plus{} b^2a$
It follows from Muirhead, since [3,0] majorises [2,1][/hide][/quote]

Just using Jensen's inequality kills in 10 seconds.:D[/quote]
Or using Holder's inequality kills it in 5 second  :wink:[/quote]

Regrettably Holder's inequality beyond my ability. :lol:

12. Let $ 2[ABC]\equal{}ah_a \equal{} bh_b \equal{} ch_c \equal{} abc\cdot k\ (k > 0)\Longrightarrow h_a \equal{} bck,\ h_b \equal{} cak,\ h_c \equal{} abk.$

$ \therefore h_c > \frac {h_ah_b}{h_a \plus{} h_b}\Longleftrightarrow a \plus{} b > c.$

3. $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a},z \equal{} \frac {2c}{a \plus{} b}$
It's remain to prove
\[ \sum_{cyc}\frac {a}{b \plus{} c} \geq \sum_{cyc}\frac {2ab}{(a \plus{} c)(b \plus{} c)}
\]

\[ \leftrightarrow \sum_{cyc}\frac {a}{b \plus{} c} \minus{} \sum_{cyc}\frac {2ab}{(a \plus{} c)(b \plus{} c)} \geq 0
\]

\[ \leftrightarrow \frac {\sum_{cyc}a^3 \plus{} 3abc \minus{} \sum_{sym}a^2b}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq 0
\]
but it's obviously true by schur.

[quote="Erken"][quote="apollo"]4. $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a},z \equal{} \frac {2c}{a \plus{} b}$
 [/quote]
I think this is proof for 3rd problem,not 4th. :)[/quote]
Sorry,I've edited it. :blush:

14. It's equivalent to prove: $ 3(x^3\plus{}y^3\plus{}z^3) \geq (x^2\plus{}y^2\plus{}z^2)(x\plus{}y\plus{}z)$
$ \leftrightarrow 2(x^3\plus{}y^3\plus{}z^3) \geq \sum_{sym}x^2y$
but it's obviously true by Am-Gm.

18. It's obviously true by schur inequality;
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \geq \sum_{cyc}x^2(y\plus{}z)$
$ \rightarrow x^6\plus{}y^6\plus{}z^6\plus{}3x^2y^2z^2 \geq \sum_{cyc}x^4(y^2\plus{}z^2)$

11.
I think it should be:
Let $ x,y,z,t \in R , x\plus{}y\plus{}z\plus{}t\equal{}0 , x^2\plus{}y^2\plus{}z^2\plus{}t^2\equal{}1$
Prove that $ \minus{}1 \leq xy\plus{}yz\plus{}zt\plus{}tx \leq 0$ :D

[hide="20"]
note that $ 3 \minus{} 2a \equal{} b \plus{} c \minus{} a$ and so on,so we have:

$ LHS \equal{} (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)(a^2 \plus{} b^2 \plus{} c^2)$

now note that we have $ a^2 \plus{} b^2 \plus{} c^2 > 0$ and $ RHS \equal{} 3a^2b^2c^2 > 0$ so if $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) < 0$ then the inequality becomes trivial because $ LHS < 0 < RHS$ so assume that $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) > 0$,so we have two possibilities:

1.
two of the numbers $ b \plus{} c \minus{} a,c \plus{} a \minus{} b,a \plus{} b \minus{} c$ were negative,for example $ b \plus{} c \minus{} a < 0$ and $ c \plus{} a \minus{} b < 0$...
adding these two inequalities we get that $ c < 0$ which is a contradiction.
so this case never happens...

2.
all of the numbers $ b \plus{} c \minus{} a,c \plus{} a \minus{} b,a \plus{} b \minus{} c$ are positive i.e. $ a \plus{} b > c , b \plus{} c > a$ and $ c \plus{} a > b$ which means there exists a triangle $ \triangle ABC$ with side lengths $ a,b,c$...
now let $ p,r,R,S$ be the semiperimeter,inradius,circumradius and area of this triangle we have:

$ 3 \minus{} 2a \equal{} b \plus{} c \minus{} a \equal{} 2(p \minus{} a)$ so we have:

$ LHS \equal{} 8(p \minus{} a)(p \minus{} b)(p \minus{} c)(a^2 \plus{} b^2 \plus{} c^2)$

now from a well-known theorem we have:

$ S \equal{} \sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)} \equal{} \frac {abc}{4R} \equal{} pr$

so we'll have:

$ (p \minus{} a)(p \minus{} b)(p \minus{} c) \equal{} pr^2$

and

$ abc \equal{} 4Rrp$

so our inequality reduses to show that:

$ 8pr^2(a^2 \plus{} b^2 \plus{} c^2)\leq 48R^2r^2p^2$

$ \iff a^2 \plus{} b^2 \plus{} c^2\leq 6R^2p$

now note that $ p \equal{} \frac {a \plus{} b \plus{} c}2 \equal{} \frac 32$ so we have to prove that:

$ a^2 \plus{} b^2 \plus{} c^2\leq 9R^2$

now by $ \sin$ law in $ \triangle ABC$ we have:

$ \frac {a}{\sin A} \equal{} \frac {b}{\sin B} \equal{} \frac {c}{\sin C} \equal{} 2R$

so we have:

$ a \equal{} 2R\sin A , b \equal{} 2R\sin B , c \equal{} 2R\sin C$

so our inequality is equivalent to:

$ 4R^2(\sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C)\leq 9R^2$

$ \iff \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C\leq \frac 94$

which is a well-known inequality...[/hide]

8)
Seen it hundreds of times.
Given the inequality between the arithmetical and harmonical mean of numbers (a+b),(b+c),(c+a) we end in a result:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}3\equal{}(a\plus{}b\plus{}c)(\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{a\plus{}b})\equal{}
\frac{1}{2}((a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a))(\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{a\plus{}b})\geq \frac{9}{2}$
Which is equivalent to our first inequality
Second part:
We know, that $ a<b\plus{}c$
So we also know that $ a\plus{}b\plus{}c<2(b\plus{}c)$ and $ \frac{1}{b\plus{}c}<\frac{2}{a\plus{}b\plus{}c}$
So $ \frac{a}{b\plus{}c}<\frac{2a}{a\plus{}b\plus{}c}$
In the same method we can easily prove $ \frac{b}{c\plus{}a}<\frac{2b}{a\plus{}b\plus{}c}$ and $ \frac{c}{a\plus{}b}<\frac{2x}{a\plus{}b\plus{}c}$
Summing all 3 inequalities we have:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}<\frac{2a\plus{}2b\plus{}2c}{a\plus{}b\plus{}c}\equal{}2$ Which is even stronger than the inequality given ;)

[quote="apollo"]11.
I think it should be:
Let $ x,y,z,t \in R , x \plus{} y \plus{} z \plus{} t \equal{} 0 , x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 1$
Prove that $ \minus{} 1 \leq xy \plus{} yz \plus{} zt \plus{} tx \leq 0$ :D[/quote]

11. [hide]$ xy\plus{}yz\plus{}zt\plus{}tx\equal{}(x\plus{}z)(y\plus{}t)$. WLOG let $ x^2\plus{}z^2\leq \frac{1}{2}$, otherwise switch $ x,y$ and $ z,t$. We have $ (x\plus{}z)(y\plus{}t)\equal{}\minus{}(x\plus{}z)^2$, so the right inequality is immediate.

For the left inequality, $ (x\plus{}z)^2\leq (x\plus{}z)^2\plus{}(x\minus{}z)^2\leq 2(x^2\plus{}z^2)\leq 1$.[/hide]

3. [hide]WLOG let $ x\geq y\geq z$, then $ y(x\plus{}z\plus{}xz)\equal{}4\minus{}xz$. Let $ s\equal{}x\plus{}z,p\equal{}xz$. Then
\[ (x\plus{}z\plus{}xz)(x\plus{}y\plus{}z\minus{}xy\minus{}yz\minus{}zx)\equal{}(s\plus{}p)(s\plus{}y\minus{}sy\minus{}p)\]
\[ \equal{}s^2\plus{}sp\plus{}4\minus{}p\minus{}s(4\minus{}p)\minus{}sp\minus{}p^2\equal{}s^2\minus{}4s\plus{}4\minus{}p\plus{}sp\minus{}p^2\]
\[ \equal{}(s\minus{}2)^2\plus{}p(s\minus{}p\minus{}1)\equal{}(x\plus{}z\minus{}2)^2\plus{}xz(x\minus{}1)(1\minus{}z)\]
Finally, $ x\geq y\geq z$ and $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$ must imply $ x\geq 1\geq z$ (since $ x,y,z$ can't be all greater than 1 or all less than 1) so the right hand side is nonnegative.[/hide]


vector
Functional Analysis
advanced fields
advanced fields unsolved

Let $X$ be a normed space and $\{x_1,...,x_n\}$, be a linearly independent subset of $X$,. Use the Hahn-Banach extension theorem to show that there exist bounded linear functionals $f_1,...,f_n$, on $X$, such that
$f_i(x_j) = \delta_{i,j}$,

where $\delta_{i,j}=0$, when $i\neq j$, and $1$, when $i=j$,. Can we think about extending this to an infinite sequence $\left\{x_1,...,x_n,...\right\}$ of vectors?

if $F$ is a closed vectorial subspace and $x$ is not in $F$, then there exists a bounded linear functional$f$ that satisfies $f(x)=1$ and $f(y)=0$ for all $y \in F$. This is an immediate consequence of the following well-known result:
Suppose that $F$ is a closed convex set and $K$ a compact convex set, then there is a bounded functional that strictly separates $F$ and $K$.

Hence the result is true if $x_i$ is not in the closed vectorial subset generated by the $\{x_j\}_{j \neq i}$ for every $i \in I$.
(PS: see also http://www.mathlinks.ro/Forum/viewtopic.php?highlight=functional&t=69217 )