calculus algebra function domain symmetry

well think of this, if you have a function and its inverse, and you compound them, you get x, so what function is the inverse of itself

for 2, i am not that sure, i does not state what the inital function is, but there is no restriction on the domain of the later function so the domain is (a,b)

[quote="rifka53"]I have a pre-calculus test that carries over into tomorrow and there are two problems that i am completely lost on.

1.)  if f(f(x)) = x, represents x for the domain, what is the function? What kind of symmetry?

2.)  if (a,b) and (c,d) are the domain and range of a function respectively, and a < b and c < d, what is the domain and range of function f(x) = k(-1/3x) -1

I know these are vague, but I don't know what to do!!![/quote]Number 2 doesn't make sense to me. The domain of $f(x)=k\left(-\frac13x\right)-1$ is whatever you want it to be. Is that exactly how it's written?

Yeah, 2 is weird.  The function is linear for all real k, so the domain and range should both be $\mathbb R$.  I'm confused why the first part is even there, unless you want to find a, b, c, and d...


analytic geometry
inequalities
polynomial

Let $a,b,c$ be real numbers satisfying $a<b<c,a+b+c=6,ab+bc+ac=9$. Prove that $0<a<1<b<3<c<4$

[hide="Solution"]
Let $abc=k$, then $a,b,c\ (a<b<c)$ are the roots of cubic equation $x^3-6x^2+9x-k=0\Longleftrightarrow x(x-3)^2=k$

that is to say, $a,b,c\ (a<b<c)$ are the $x$-coordinates of the interception of points between $y=x(x-3)^2$ and

$y=k$.

$y=x(x-3)^2$ have local maximuml value of $4$ at $x=1$ and  local minimum value of $0$ at $x=3$.

Since the $x$-coordinate of  the interception point between $y=x(x-3)^2$ and $y=4$ which is the tangent line at

local maximum point $(1,4)$ is a point $(4,4)$,Moving the line $y=k$ so that the two graphs $y=x(x-3)^2$ and

$y=k$ have the distinct three interception points,we can find that the range of $a,b,c$ are

$0<a<1,1<b<3,3<c<4
$,we are done.[/hide]

This was also on round 2 of the 1995 BMO: [url=http://www.bmoc.maths.org/home/bmolot.pdf]www.bmoc.maths.org/home/bmolot.pdf[/url]

Here is my solution with attached figure.


In an automobile factory, 42.5 percent of the cars produced had power windows. What is an equivalent way of describing the proportion of cars with power windows?

A. 85 out of every 200 cars

B. 50 out of every 110 cars

C. 45 out of every 100 cars

D. 40 out of every 100 cars

E. 40 out of every 85 cars

42.5 percent is equivalent to 42.5 cars out of 100 cars
if we double both of them, we get 85 out of 200 cars
therefore, our answer is A


search
function
number theory unsolved
number theory

Let $p\ge 5$ be a prime number and let $r$ be the number of ways of placing $p$ checkers on a $p\times p$ chessboard so that not all checkers are in the same row(but they may be all in the same column). Show that $p^{5}|r$. here we may assume that all the checkers are identical.

i've been able to prove that $p^{3}|r$ but can't do more. so help me please. :)

see http://www.mathlinks.ro/Forum/viewtopic.php?p=462321#p462321

P.S: Try use search function!  :wink:


vector
linear algebra
matrix
superior algebra
superior algebra solved

$V$ is a $\mathbb{R}$eal vector space, $\mathrm{dim}(V)=n$, and let $H$ be a subspace of $V$, $\mathrm{dim}(H)=k$. Let $U$ a subset of $L(V,V)$ such that  $T\in U\Longrightarrow T(h)=0$ for all $h\in H$.
Find $\mathrm{dim}(U)$.

[quote="JBL"]This should be bumped into one of the superior algebra subfora.  The answer is $n^2-nk$, though, I believe.[/quote]

Meant to bump it up earlier...but it's now done.

If $\{v_1,\ldots v_k\}$ is a basis for $H$ and $\{v_1,\ldots v_k, v_{k+1},\ldots v_n\}$ is a basis for $V$ then to each linear operator $T$ as in the problem text we associate it's matrix $(a_{ij})_{i,j=\overline{1,n}}$  such that $T(v_i)=\sum_{j=1}^n a_{ij}v_j$. The correspondence $operator\rightarrow matrix$ is injective.
$T(H)=0$ implies $a_{ij}=0\ \forall i=\overline{1,k},j=\overline{1,n}$. The matrices with this property are easily seen to form a vector space of dimension $n(n-k)=n^2-nk$.


calculus
integration
trigonometry
calculus computations

$ I_1 \equal{} \int_0^ \frac {\pi}{2} \cos \theta \cdot f(\sin\theta \plus{} \cos^2\theta) \cdot d\theta$

$ I_2 \equal{} \int_0^ \frac {\pi}{2} \sin 2\theta \cdot f(\sin \theta \plus{} \cos^2 \theta) \cdot d\theta \cdot$

Then, which of the following is true?

$ A)$ $ I_1\equal{}I_2$

$ B)$ $ \minus{}I_1\equal{}2I_2$

$ C)$ $ \frac {I_1}{I_2} \equal{} 2$

$ D)$ $ None$ $ of$ $ these$.

Let $ \sin \theta \equal{} t$, $ I_1 \equal{} \int_0^1 f(1 \plus{} t \minus{} t^2)\ dt$ and $ I_2 \equal{} 2\int_0^1 tf(1 \plus{} t \minus{} t^2)\ dt$ setting $ 1 \minus{} t \equal{} u$ gives

$ I_2 \equal{} 2\int_0^1 (1 \minus{} u)f(1 \plus{} u\minus{}u^2)\ du \equal{} 2I_1 \minus{} I_2$, yielding $ I_1 \equal{} I_2$.

[quote="Rajiv"]$ I_1 = \int_0^ \frac {\pi}{2} \cos \theta \cdot f(\sin\theta + \cos^2\theta) \cdot d\theta$

$ I_2 = \int_0^ \frac {\pi}{2} \sin 2\theta \cdot f(\sin \theta + \cos^2 \theta) \cdot d\theta \cdot$

Then, which of the following is true?.[/quote]

$ I_1-I_2 = \int_0^ \frac {\pi}{2} (\cos \theta -\sin 2\theta )\cdot f(\sin\theta + \cos^2\theta) \cdot d\theta$

Let $ u=\sin\theta + \cos^2\theta\implies du=(\cos\theta-2\cos\theta\sin\theta )d\theta \implies du=(\cos\theta-\sin2\theta )d\theta$
when $ \theta=\frac {\pi}{2}, u=1$

when ${ \theta=0, u=1}$

$ I_1-I_2 = \int_1^1 f(u)du=0 \implies  I_1=I_2$

:roll:

[quote="kunny"]Yes, 　I also noticed your solution after getting my first solution. :wink:[/quote]

Besides these two ways , have u any idea to solve?


x(n) bound, and a(n) convergent sequence s.t.
x(n+1)+a(n) \geq x(n)+a(n+1) for any n>0
Prove that x(n) converge

Lets denote y_n=x_n-a_n
from the fact that:
x_(n+1)-a_(n+1)>=x_n-a_n we get that y_(n+1)>=y_n for any n>=1 so y_n is incresing. we know that a_n and x_n are bounded so also y_n is bounded so y_n converges! and also because a_n converges we get that (y_n+a_n){n>=1}=(x_n){n>=1} converges so x_n converges.

cheers! :D :D


geometry
angle bisector
geometry proposed

Let ABC be a triangle such that <B=2<C. Let C1 be a circle with radii AB centered in A.

a)Show that the external bisector of <B is tangent to C1, except in one case.
b)Find the case where the external angle bisector of <B cuts the circle.

i drew this out and it wasn't correct. are you sure you've got the question written correctly? :?:


Find the minimum of

$ f(x,y,z)\equal{}\sqrt{\lambda\minus{}xy}\plus{} \sqrt{\lambda\minus{}yz}\plus{} \sqrt{\lambda\minus{}zx}$

As a generalization of

http://www.mathlinks.ro/viewtopic.php?t=64739

I post the solution here.

[hide]Actually, fix $ z$ we find

$ \quad f(x,y,z)\\
\equal{}\sqrt{\lambda\minus{}xz\plus{}\lambda\minus{}yz\plus{}2\sqrt{(\lambda\minus{}xz)(\lambda\minus{}yz)}}\plus{}\sqrt{\lambda\minus{}xy}\\
\equal{}\sqrt{2\lambda\minus{}z(1\minus{}z)\plus{}2t}\plus{}\sqrt{\lambda\minus{}\frac{t^2\plus{}\lambda z(1\minus{}z)\minus{}\lambda^2}{z^2}}\\
: \equal{}g(t)$

Where

$ t\equal{}\sqrt{\lambda^2\minus{}\lambda z(1\minus{}z)\plus{}xyz^2}$

It's easy to show

$ g_1(t)\equal{}\sqrt{2\lambda\minus{}z(1\minus{}z)\plus{}2t},\ g_2(t)\equal{}\sqrt{\lambda\minus{}\frac{t^2\plus{}\lambda z(1\minus{}z)\minus{}\lambda^2}{z^2}}$

are both concave. Hence, $ g(t)$ is concave, and it reach its minimum when $ xy$ reaches its minimum or maximum, which means $ xy\equal{}0$ or $ x\equal{}y$. The rest is not hard. Result:

$ f_{\min}\equal{}\min \left( \sqrt{9\lambda\minus{}1}, 2\sqrt{\lambda}\plus{}\sqrt{\lambda\minus{}\frac{1}{4}}\right)$

[/hide]

Another approach:
W.L.O.G, assume that z=min(x,y,z)>0
Now we will prove the following inequality: f(x,y,z)>=f(x-z,y-z,0)
It's sufficient to prove: 3k-(xy+yz+xz)+2sigma cyclic square root(k-xy)(k-yz)>=
k-(x-z)(y-z)
It's equivalent to: 2k-2(xz+yz)+z^2+2sigma cyclic square root(k-xy)(k-yz)>=0(it's true since 2k>=(x+y+z)^2/2>=2(xz+yz), followed by AM-GM and z=min(x,y,z)>0)
Therefore, we just need to find the minimum in case there is one variable equal to 0. Assume that z=0
Now we find the minimum of: square root(k-xy)(x+y=1,x,y>=0)
It's easy to see that the minimum will reach when x=y, which means square root(k-1/4)
So, the minimum is: square root(k-1/4), which differs from shalex's result :(


One number is "sinister" if the sum of the factors prime and the sum of the powers of the factors primes is equal. Ex: 2^3*5^2*11^13 is "sinister"
How much sinisters exist with Four digits?! And find all...

I like this kind of problem, not difficult but fun :-)

* Since 2^(15)*3*5*7 is the smallest possible Sinister with 4 prime factors and is > 10000 there are no solution with 4 prime factors or more.

* Solution with only one prime factor : 5^5=3125 (4^4 < 1000, 6^6 > 10000)

* With 2 prime factors :

3^7 * 5 > 10000 so smallest prime factor has to be 2.
2^12 * 11 > 10000 and 2^2*3 < 1000 so only possible numbers are 2^4 * 5^3, 2^3 * 5^4, 2^8 * 7^1, 2^7 * 7^2

* With 3 prime factors :

same reasonning, smallest prime has to be 2, biggest has to be 5 (2^10*3*7 > 10000),
only possible numbers are so : 2^6 * 3^3 * 5^1, 2^7 * 3^2 * 5^1, 2^7 * 3^1 * 5^2, 2^8 * 3^1 * 5^1

To sum up, I found 9 solutions :

5^5 = 3125
2^4 * 5^3 = 2000
2^3 * 5^4 = 5000
2^8 * 7^1 = 1792
2^7 * 7^2 = 6272
2^6 * 3^3 * 5^1 = 8640
2^7 * 3^2 * 5^1 = 5760
2^7 * 3^1 * 5^2 = 9600
2^8 * 3^1 * 5^1 = 3840


limit
function
calculus
calculus computations

let $f(x)=$
..................$0$ if $x$ is rational
..................$1$ if $x$ is irrational

Prove that $\lim_{x\rightarrow 0}$ does not exist.

[I believe the book wants a $\delta$,$\epsilon$ proof]

Take a rational sequence $a_{n}$ and an irrational one $b_{n}$ that converge to $0$ .Then

$\lim_{n \to \infty }f(a_{n}) = 0$ and $\lim_{n \to \infty }f(b_{n}) = 1$

so the limit does not exist .

thats what i was thinking, but how do you state that rigirously? i.e. using the definition of a limit

[quote="Ottem"]That isn't rigourious enough for you?  - All we need is to come up with two sequences showing the ambiguity of $\lim f$..[/quote]

maybe...i am not quite sure how one develops limits rigirously, i am working through a book right now on limits, i have not really seen this as a theorem, it makes sense of course, but the book proves everything using delta epsilon (i.e. definition of a limit),

but just to clarify, can we always say this is true:

Let given a function $f(x)$, have sequences $a_{n}$ and $b_{n}$ such that:
1) $\lim_{n\rightarrow \infty}a_{n}=\lim_{n\rightarrow \infty}b_{n}=c$,

2) $\lim_{n\rightarrow \infty}f(a_{n})=a$......$\lim_{n\rightarrow \infty}f(b_{n})=b$

where $a\ne b$,

then $\lim_{x\rightarrow c}f(x)$ does not exist.

this is true...
in fact, you're using what our analysis teacher called "bridge theorem", which links sequences to continuous functions and limits..

$f$ is continuous at point $x_{0}$ if and only if for any sequence $(x_{n})$ converging to $x_{0}$, $f(x_{n})$ tends to $f(x_{0})$.

it's not difficult to prove, using $\epsilon$s and $\delta$s.. i'm sure you can manage it on your own :wink:


inequalities
search
inequalities unsolved

A quick search for the string *\frac{9}{4(x+y+z)}* reveals the following two threads:

http://www.mathlinks.ro/Forum/viewtopic.php?t=4908
http://www.mathlinks.ro/Forum/viewtopic.php?t=30051

  Darij


trigonometry
real analysis
real analysis unsolved

Prove that the following differential equation \[ x'\sin 2t=2(x+\cos t) \] has only one solution $\psi_{0}(.)$ defined on interval $\left(-\frac{\pi}{2},\frac{3\pi}{2}\right)$.

hello, i have found
$ x(t)\equal{}\minus{}\sec(t)\plus{}C\cdot\tan(t)$ as the solution.
Sonnhard.


induction
geometry
number theory
number theory proposed

what is the sum of the digits of the number 33...3*33...3
where 3 is repeated n times in each factor.
My real problem:
according to some didactitian, if the student cannot see the relationship between two consecutive steps he cant finish the proof by math induction( induction seems to work if we could describe how we can move from one step to the following one).But in this problem even if we can find the relationship, it is hard to finish it; so we need to change the vision and treat the problem otherwise.
Do you know some problem in number theory or in geometry not extremly hard where math induction seems to work but in fact it cannot.
Thanks

[quote="senouy"]what is the sum of the digits of the number 33...3*33...3
where 3 is repeated n times in each factor.[/quote]
It's easy to establish in direct manner that $ \overline{333.. n\text{ times } ..33}^2$ $ \equal{}\left(\frac{10^n\minus{}1}3\right)^2$ $ \equal{}10^{n\plus{}1}\frac{10^{n\minus{}1}\minus{}1}9\plus{}8\frac{10^n\minus{}1}9\plus{}1$ And so :

$ \overline{333.. n time ..33}^2$ $ \equal{}\overline{111... (n\minus{}1)\text{ times }...1110888... (n\minus{}1) \text{ times }... 8889}$ whose sum of digits is $ S\equal{}(n\minus{}1)1\plus{}(n\minus{}1)8\plus{}9\equal{}9n$

[quote="senouy"]My real problem:...Do you know some problem in number theory or in geometry not extremly hard where math induction seems to work but in fact it cannot.
Thanks[/quote]
Likely a lot, but I've none in mind immediatly :)

If you have time, please post some of them.
thank you.[/hide]


For any positive real numbers $a,b$ and $c$,
\[ \frac{ab^2}{c(c+a)}+\frac{bc^2}{a(a+b)}+\frac{ca^2}{b(b+c)}\ge\frac{3}{2}\cdot\frac{a^2+b^2+c^2}{a+b+c}  \]

[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[\frac{ab^{2}}{c(c+a)}+\frac{bc^{2}}{a(a+b)}+\frac{ca^{2}}{b(b+c)}\ge\frac{3}{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\]
[/quote]
This inequality, we can prove just use the followings
\[\sum a^{2}+6 \ge \frac{3}{2}(a+b+c+ab+bc+ca) \qquad \forall a,b,c>0,abc=1 \]

\[(a+b+c)^{2}\ge 3(ab+bc+ca) \qquad \forall a,b,c>0 \]
:)

\[ \sum a^{2}+6 \ge \frac{3}{2}(a+b+c+ab+bc+ca) \qquad \forall a,b,c>0,abc=1 \]

Dear toanhocmuonmau, did you prove this one by mixing variable method or have you an elementary solution?

Thank you very much.

Has anyone tried to prove this nice inequality of Ductrung or Toanhocmuonmau hint which implies it and  I suspect can be solved only by mixed variables method? I cannot solve in a direct manner neither the inequality :blush: , nor toanhocmuonmau hint  :oops: ?

Thank you very much.

[quote="manlio"]
\[ \sum a^{2}\plus{}6\ge\frac{3}{2}(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)\qquad\forall a,b,c>0,abc\equal{}1\]
[/quote]

Suppose $ a\le b\le c$

Let $ f(a,b,c)\equal{}\sum a^{2}\minus{}\frac{3}{2}(\sum a\plus{}\sum ab)$.

Prove : $ f(a,b,c)\ge f(a,\sqrt{bc},\sqrt{bc})\Leftrightarrow (\sqrt{b}\minus{}\sqrt{c})^{2})((\sqrt{b}\plus{}\sqrt{c})^{2}\minus{}a\minus{}1)\ge 0$  :wink:

[quote="manlio"][quote="Soarer"]manlio, recall the old inequality $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2abc\plus{}1\ge 2(ab\plus{}bc\plus{}ca)$[/quote]

Soarer, I don't see how to use this old ineq to prove

$ 2\sum a^{2}\plus{}12\geq 3(\sum a\plus{}\sum ab)$

Thank you very much[/quote]

We have a famous inequality :

$ 2(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2abc\plus{}1)\ge 4(ab\plus{}bc\plus{}ca)$

And :

$ ab\plus{}bc\plus{}ca\plus{}6\ge 6(a\plus{}b\plus{}c)(1)$

But $ (1)$ is not true when we put $ a\to\infty,b\equal{}c\to 0$  :P

I think this is a very nice and very hard inequality. :)

[quote="toanhocmuonmau"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{ab^{2}}{c(c\plus{}a)}\plus{}\frac{bc^{2}}{a(a\plus{}b)}\plus{}\frac{ca^{2}}{b(b\plus{}c)}\ge\frac{3}{2}\cdot\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{a\plus{}b\plus{}c}\]
[/quote]
This inequality, we can prove just use the followings
\[ \sum a^{2}\plus{}6\ge\frac{3}{2}(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)\qquad\forall a,b,c>0,abc\equal{}1\]

\[ (a\plus{}b\plus{}c)^{2}\ge 3(ab\plus{}bc\plus{}ca)\qquad\forall a,b,c>0\]
:)[/quote]

Can anyone, please, explain to me how toanhocmuonmau's hint can imply ductrung's inequality. I cannot see that :(

Thank you very much.

[quote="manlio"][quote="Soarer"]manlio, recall the old inequality $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2abc\plus{}1\ge 2(ab\plus{}bc\plus{}ca)$[/quote]

Soarer, I don't see how to use this old ineq to prove

$ 2\sum a^{2}\plus{}12\geq 3(\sum a\plus{}\sum ab)$

Thank you very much[/quote]
I'm sorry manlio, I was wrong. :blush:

Dear Manlio,

My apology for having to postpone posting my solution(s). I promise that when I reveal the solution(s) there will be more answers than was asked for (up to 4 pages  :wink:).

Dear Ductrung,

thank you very much for your nice inequalities. :)

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {ab^2}{c(c + a)} + \frac {bc^2}{a(a + b)} + \frac {ca^2}{b(b + c)}\ge\frac {3}{2}\cdot\frac {a^2 + b^2 + c^2}{a + b + c}
\]
[/quote]
Now, Let I post my solution for it
Lemma. If $ x,y,z \ge 0$ such that $ xyz = 1$ then
\[ x^2 + y^2 + z^2 + 6 \ge \frac {3}{2}(x + y + z + xy + yz + zx)
\]
We can prove it by mixing variable method. :)
Back to the original problem
Setting $ a = \frac {1}{x},b = \frac {1}{y},z = \frac {1}{z}$ then the inequality becomes
\[ \sum \frac {x^2}{y^2(z + x)} \ge \frac {3\sum x^2y^2}{2xyz\sum xy}
\]

\[ \Leftrightarrow \sum \frac {x^2}{y^2} + \sum\frac {x^2}{y(z + x)} \ge \frac {3(\sum x^2y^2)(\sum x)}{2xyz\sum xy}
\]
Using the above lemma, we get
\[ \sum \frac {x^2}{y^2} \ge \frac {3}{2}\sum \frac {x^2 + y^2}{xy} - 6
\]
and the Cauchy Schwarz Inequality gives us
\[ \sum\frac {x^2}{y(z + x)} \ge \frac {(\sum x)^2}{2\sum xy}
\]
It suffices to show that
\[{ \frac {3}{2}\sum \frac {x^2 + y^2}{xy} - 6} + \frac {(\sum x)^2}{2\sum xy}\ge \frac {3(\sum x^2y^2)(\sum x)}{2xyz\sum xy}
\]

\[ \Leftrightarrow \frac {7(\sum x^2 - \sum xy)}{2\sum xy} \ge 0
\]
which is true. :)

PS: We can try by a similar way to prove the stronger of ductrung. :)

[quote="ductrung"]Actually, a stronger inequality can be proved: For any positive real numbers $ a,b$ and $ c$,
\[ \frac{ab^{2}}{c(c+a)}+\frac{bc^{2}}{a(a+b)}+\frac{ca^{2}}{b(b+c)}\ge\frac{3}{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{a+b+c}+\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{9(a+b+c)}\][/quote]

Very sharp inequality. With  new technique, I have improve your bound from $\frac19$ to $\frac14$. Indeed, the following statement is sharper
\[ \frac{ab^{2}}{c(c+a)}+\frac{bc^{2}}{a(a+b)}+\frac{ca^{2}}{b(b+c)}\ge\frac{3}{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{a+b+c}+\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{4(a+b+c)}\]


The numbers 1, 2, 3,. . . , 20 are written on the blackboard. Then one will erase two of the numbers a and b and write a + b − 1 on the board instead. After repeating this step 19 times, there will be only one number on the board. What is that number?

[hide]I just paired any two random numbers together and kept doing it until I ended up with one #, and I got 191 this way.[/hide]

Hide your answer

[hide]You're right![/hide]

Monkey, could you do that in less than 80 seconds (assuming this is a sprint problem)?

[hide]Basically, this is just the sum of 1+2+3+...+20 minus 19, which is 20(21)/2=210, 210-19=191.[/hide]

let's see if pairing can be done in 80 seconds:
there willl be about 20 pairings, and each will have to be done in 4 seconds. can be done if you add and subtract fast.

I don't like the 80 seconds thing... Finishing JUST when time is called is kinda bad.

Try it in 20 seconds.  Then you can do 20 operations in just 1 second a piece. ;)

Gets kinda hard on the last couple operations though...

[hide="10sec!"]How to do in 10 sec:
21*20/2 (1st +last *how many/2)
21*10
210
minus 19 (-1 vevery time you do an operation adn you do 19 operations)
>>191<<[/hide]

[quote="mathgeek2006"]Good job dwx.  First place for less than 10 seconds (Out of one person that participated!) If anyone can do it in 5 or less, then they get  :first:[/quote]

I got it in 3. 14159 seconds :D

I feel stupid.  When I read that last response, I thought it said the number of seconds you got it in, then the answer you got. I was really confused :? .  Anyway, You realize that there is an emoticon for pi? (  :pi:  )

yes i know there's a  :pi: thing....but that would be too obvious :lol:

True, and then you could make my self-esteem go way down! :?  :lol:  :P  :P  :P


Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ \frac {ab}{9\minus{}4bc}\plus{}\frac{bc}{9\minus{}4ca}\plus{}\frac {ca}{9\minus{}4ab} \le \frac 3{5}$.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac {ab}{9 \minus{} 4bc} \plus{} \frac {bc}{9 \minus{} 4ca} \plus{} \frac {ca}{9 \minus{} 4ab} \le \frac 3{5}$.[/quote]
$ \sum \frac {ab}{9 \minus{} 4bc} \equal{} \sum \frac {ab}{(a\plus{}b\plus{}c)^2 \minus{} 4bc} \le \sum \frac {ab}{a^2\plus{}2ac\plus{}2ab} \equal{}\sum \frac {b}{a\plus{}2b\plus{}2c} \le \frac{3}{5}$.
There is no case $ a\equal{}0$ and $ b\equal{}c$.

[quote="ehku"][quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac {ab}{9 \minus{} 4bc} \plus{} \frac {bc}{9 \minus{} 4ca} \plus{} \frac {ca}{9 \minus{} 4ab} \le \frac 3{5}$.[/quote]
$ \sum \frac {ab}{9 \minus{} 4bc} \equal{} \sum \frac {ab}{(a \plus{} b \plus{} c)^2 \minus{} 4bc} \le \sum \frac {ab}{a^2 \plus{} 2ac \plus{} 2ab} \equal{} \sum \frac {b}{a \plus{} 2b \plus{} 2c} \le \frac {3}{5}$.
There is no case $ a \equal{} 0$ and $ b \equal{} c$.[/quote]
Yes. It is. :lol:
There is the case $ a\rightarrow 0$ and $ b\equal{}c$, of course. :wink:

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac {ab}{9 \minus{} 4bc} \plus{} \frac {bc}{9 \minus{} 4ca} \plus{} \frac {ca}{9 \minus{} 4ab} \le \frac 3{5}$.[/quote]
We also have a "similar" result
\[ \sum \frac{ab}{3\plus{}bc} \le \frac{3}{4}
\]
:)
Equality holds for $ a\equal{}b\equal{}c\equal{}1$ and $ a\equal{}b\equal{}\frac{3}{2},c\equal{}0.$

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac {ab}{9 \minus{} 4bc} \plus{} \frac {bc}{9 \minus{} 4ca} \plus{} \frac {ca}{9 \minus{} 4ab} \le \frac 3{5}$.[/quote]

See: http://www.mathlinks.ro/viewtopic.php?t=189307

[quote="manlio"]Sorry, I don't understand how

$ \sum \frac {b}{a \plus{} 2b \plus{} 2c} \le \frac {3}{5}$

is connected with your proof.  :blush:  at the link you have pointed.[/quote]
No, I don't want to tell about this one. I just post the link that includes my solution for the original problem.

[quote="manlio"]Sorry , I cannot find in the forum the proof for

$ \sum \frac {b}{a \plus{} 2b \plus{} 2c} \le \frac {3}{5}$
Where can I find it?  :wink:

Thank you very much[/quote]
Manlio, you can prove it by this way :)
Rewrite the inequality as
\[ \sum \frac{a\plus{}2c}{a\plus{}2b\plus{}2c} \ge \frac{9}{5}
\]
By the cauchy Schwarz, we have
\[ \sum \frac{a\plus{}2c}{a\plus{}2b\plus{}2c} \ge \frac{9(a\plus{}b\plus{}c)^2}{\sum (a\plus{}2c)(a\plus{}2b\plus{}2c)} \equal{}\frac{9(a\plus{}b\plus{}c)^2}{5\sum a^2 \plus{}10\sum  ab} \equal{}\frac{9}{5}
\]
:)

Second proof for :
\[ \frac {a}{2a \plus{} 2b \plus{} c} \plus{} \frac {b}{a \plus{} 2b \plus{} 2c} \plus{} \frac {c}{2a \plus{} b \plus{} 2c}\le \frac {3}{5}
\]

\[ \iff \frac {5}{2}[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \plus{} [(a \minus{} c)^2 \plus{} (b \minus{} c)^2]\left(\frac {2b\plus{}c}{a}  \minus{} 3\right)\ge 0
\]
It is alway true when $ a \equal{} min(a,b,c)\ge 0$ :wink:

The general one also holds. It is one of tanpham's problems  :wink:

Let $ a,b,c\ge 0$ and $ r\ge 0, p\equal{}3r\minus{}3\sqrt{r}\plus{}2$. Prove that that:

\[ \frac {a}{pa \plus{} rb \plus{} c} \plus{} \frac {b}{a \plus{} pb \plus{} rc} \plus{} \frac {c}{ra \plus{} b \plus{} pc} \le \frac {1}{1\minus{}\sqrt{r}\plus{}r}\]

[quote="can_hang2007"] We also have a "similar" result
\[ \sum \frac {ab}{3 \plus{} bc} \le \frac {3}{4}
\]
:)
Equality holds for $ a \equal{} b \equal{} c \equal{} 1$ and $ a \equal{} b \equal{} \frac {3}{2},c \equal{} 0.$[/quote]
I have also in my computer this inequality. My proof is by means of Schur's inequality.


Consider the situation shown in the figure below.
The plates of the capacitor have a plate area $ A$ and are clamped. The dielectric slab is released from rest with a length $ a$ inside the capacitor. Neglecting friction or gravity, show that the slab executes periodic motion. Prove that time period of periodic motion is $ 8\sqrt {\frac {(l \minus{} a)lmd}{\varepsilon_0AV^2(K \minus{} 1)}}$

[url=http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=19456&thumb=1]Image[/url]

Try using virtual work principle.
By the way, in your expression either V is squared or out of the square root.

[quote="Immanuel Bonfils"]V is squared or out of the square root.[/quote]

Yeah, right! I made a mistake while typing it. :oops:

[quote="Immanuel Bonfils"]Try using virtual work principle[/quote]

What is that? (either it is something I don't know or know it under a different name)

He means: find the potential energy of the system at maximum displacement and at equilibrium, and then take their difference to find the work you would have to do.  Then you can find the force needed from the work and hence the force as a function of displacement.

May be you could find some help in a Brazilian book: Problemas de Física - Eletricidade - Eletrostática , by Eduardo Wilner, that explains very well this matter.
  BTW, genericme, in this case the force is of constant magnitude.

[quote="Rajiv"][quote="b555"]consider the force  acting  on the capacitor  at  an arbitrary moment  when  a length  x of it is placed  iside the plates.  now u should get it :wink:[/quote]

Could you suggest a few steps? I usually try to compare the situation with SHM of a block under the influence of a spring and get the answers perfect but here.... :wallbash:[/quote]

Find the potential energy $ U$ of the capacitor. Guess what $ \dfrac{\partial U}{\partial x}$ is. Do you really have SHM?

After this the problem should be trivial.

[quote="Immanuel Bonfils"]BTW, do you know what happens to the energy during oscillation?[/quote]

I suppose it is dissipated as heat, is it?

No friction or other mechanical resistant force...

Unless there is some damping it won't be dissipated as heat...

Mechanical energy+ electric field energy= constant.  So calculate U for electrical energy, and compute dU/dx, to find the force.

Now, my problem is I don't know how to compute U for when you insert a dielectric in??
My second problem is I don't know how to compute the period for a period oscillation that is not SHM?  Can sb help me?

[quote="integration"]
Now, my problem is I don't know how to compute U for when you insert a dielectric in??

[/quote]

Isn't it $ \frac {q^2}{2C}$, where $ C \equal{} \frac {A\varepsilon_0 K}{d}$?

[quote="integration"]
My second problem is I don't know how to compute the period for a period oscillation that is not SHM?  Can sb help me?

[/quote]

$ s \equal{} ut \plus{} \frac {1}{2}at^2 \\
 \\
u \equal{} 0 \\
 \\
\therefore t \equal{} \sqrt {\frac {2s}{a}}$

where $ s$ is the amplitude.

[quote="integration"]
Unless there is some damping it won't be dissipated as heat...

[/quote]

You are right, there is no damping.

There is no dumping because the source continually supplies energy (many people, here, are forgetting that the source is connected to the capacitor during the process).

[quote="Immanuel Bonfils"]There is no [b]damping[/b] because the source continually supplies energy (many people, here, are forgetting that the source is connected to the capacitor during the process).[/quote]

So there will be loss of energy through radiation of heat... :|


Do you think that the US is obliged to do no more than its fair share of eliminating global poverty or do you think it should do more.

I personally think that every country should be doing more, though doing so would be difficult.
However, if it was just the United States giving, I think it sufficient outrage against the government for doing so would be enough to persuade congress to not do more (after all, citizen representation in government is a foundation of the United States government), and thus we shouldn't.

Under the event that a majority of citizens in the US does believe that we should give more, then I think that we should do so. The reasons are the same as the reasons I have listed in the Public Education Should We Have It? thread. (Although I admit I honestly have no idea if giving more would create an impact).

[quote="Scrambled"]However, if it was just the United States giving, I think it sufficient outrage against the government for doing so would be enough to persuade congress to not do more (after all, citizen representation in government is a foundation of the United States government), and thus we shouldn't.[/quote]

Americans are fairly evenly divided on whether to reduce international aid or not.  Notably, however, this is based on gross misperceptions about how much we actually spend on it: the median American thinks that the U.S. spends 25% of its budget on foreign aid, and believes that we should "only" spend 10% of our budget on it.  However, since we actually spend only 1% of our budget on it, this means the median American view is that we should increase our international aid by a factor of 10.  (Well, okay, the second half of the previous sentence is clearly misleading.  It does, however, suggest that Americans could easily be convinced to increase our international aid budget, if they were informed of just how small it is.)

These numbers are from a 2002 survey which I found [url=http://www.worldviews.org/detailreports/usreport/html/ch5s8.html]here[/url].  It also says on that page that when able to distinguish between humanitarian and other (presumably military) forms of aid, everyone basically agrees that humanitarian aid is good and important and we should do more of it.

I think the U.S. should work harder to promote poverty around the world and widen the rich-poor gap in today's society. Poverty is crucial in developing a higher sense of self-esteem for the majority of the world's population, and thus is good thing despite all the criticisms.

[quote="fireemblem13"]Do countries like the US really give to the poorer countries and receive nothing in return? When the paper says that a country has given a large amount of money to another needy country, they are not covering the whole truth. Due to whatever was decided between these two countries, the poorer country is forced to pay back the same or even more amount of money to the richer country. This may be done over a period of time and the country may be even allowed to pay at a later time but ends up paying more than it received. I'm not sure of the credibility of this, but it would be sad if so.[/quote]\

Some international aid comes in the form of grants (i.e. free money, often for particular purposes or only on the condition of certain economic changes, but not to be paid back with interest, or the money spent on humanitarian rescue missions (nobody wants Bosnia to pay NATO or the EU for policing that they've done there)) and some comes in the form of loans (which have to be paid back, with interest).  There is also a non-zero history of countries defaulting on such loans, and of "loan forgiveness" (when the lenders stop asking the lendee to pay them back).  There are people who know a lot about the dynamics and history of such things, but not me.

On the second half of your statement: "paying back more than they received" is often worthwhile, especially in the case of loans for economic growth.  On factor is that money tends to get less valuable with time.  The other (more important) factor is that having a billion dollars right now is much more useful than having a million dollars for each of the next thousand years: I can enact far greater changes with a lump sum all at once.  Assuming (and this is a big assumption, for a variety of reasons) that the initial loans are well-spent, the result should be an increase in the economic production of the receiving country that's larger than the interest payments and so results in both short-term and long-term economic growth.  That's the theory, anyhow.

[quote="Ignite168"]Free handouts are never a good thing, they bind us to a commitment (see North Korea) and make people start relying on us rather than looking for solutions.[/quote]

Yeah, that's right, North Korea would be fine if it hadn't been for all those hand-outs.   :roll:

[quote="fredbel6"]I am complete newbie when it comes to economics, but I would really love to hear some expert's view on this.  Are there many countries whose wealth (for instance copper mines, oil, uranium,...) is in the hands of certain (foreign) companies, who in turn got it because of colonialism.
For instance : European country A has a colony B in Africa.  Company X buys a copper mine from country A and exploits it, they sell it to company Y.... and as of now, in 2007, it is still in the hands of a foreign company.

IF so, that's food for a pretty lengthy discussion. :ninja:[/quote]

Well,  I'm no economics expert, so maybe JBL or some other ahs something more to say about this from a theoretical point of view. However I mantain a blog about multinational corporations and their impact, only it is in spanish. To the first issue what I can say is that multinationals are rather recent stuff. Underdeveloped countries in latinamerica achieved independence before the turning point when explottaion of resources came to be an international business. What has happened there is that countries have sold their companies, sometimes for lack of efficiency, others for convenience and corruption. This of course was enhanced even more in the golden years of neoliberalism. You see, there you have one of the problems with democracy as it is understood  in the wes when applied to every country. The thing is that the economic power of multinationals is many times comparable or greater that the countries themselves. So they effectively have the power to put or remove presidents according to their convenience. One of the things in which corporations spend the most along with advertising is in lobbying. So say there are two candidates for presidency in country C and one is willing to sell to some company one big state-run corporation while the other isn't. Then company X will support his candidature with many millions. As you can imagine this is usually crucial. There are examples, for instance, before Evo Morales triumph in Bolivia and subsequent "nationalization" the oil and gas multinationals contracts in which they gave 18% of the profit to the bolivian state and took 82 percent. You understand that with 82% of the money from oil leaving your country there's little hope for progress. That is without taking into account the thousands of people displaced in Bolivia because of the oil explotation, environmental damage, and so on
One important thing to realize is that you can't really say that the presence of corruption in the governments nowadays is the fault of the people of Bolivia. How come that they have so much corruption in the developing countries and not so much in the developed ones? Are they dumb or something? The fact is that there always will be corrupt people ready to handle power. If they were to  compete with fair people with equal opportunities then probably the developed countries would be able to select the fair politicians the same way we are. But usually corrupts are the ones that fit the big multinational interests the most, therefore they get elected frequently.

In Africa sometimes happen other  things. There you have sometimes strong dictatorial governments or there were in the near past. These tend to exert control over the national resources however the money goes to very few hands. This is what happens for instance in Nigeria, where 85% percent of the profit from the oil extraction goes to 1% percent of the population. This data is from a report by Michael Watt, director of the Center for African Studies at Berkeley. There is also a somewhat outdated but very comprehensive [url=http://www.hrw.org/reports/1999/nigeria/index.htm]report[/url] from human rights watch about this issue. Remember Nigeria provides 10% of the oil consumed in the US and it is thought that the Gulf of Guinea will provide 25% by 2015.

So, concluding, multinationals are a problem for developing countries whether they were there before or not, they have way too much power and control much of the government to their own convenience which empirically it is safe to say it is not the good of the people. The conditions in which people work for those companies are usually - I'm just taking care of not absolutizing - conditions not only of explotation but many times inhuman, and as most reports by independent sources confirm, multinationals have no regard for things such as the environment. All of them are things that would be absolutely unacceptable in our countries, but our companies have them everywhere

Ursula


The fixed point of $f(x) = (3-x)^{ \frac{3}{5}}$ doesn't seem to have a nice form.

Note,
$g_{n+1}=(3-g_{n})^{3/5}$
So, if it exists (probably does lets ignore that for now) then,
$g_{n}\to L$ and $g_{n+1}\to L$ so,
$L=(3-L)^{3/5}$
So,
$L^{5}= (3-L)^{3}$
You need to solve this equation.


inequalities
algebra
polynomial
inequalities open

We have already the following famous one

\[\sum a^{4}(b+c) \le \frac{1}{12}(a+b+c)^{5}.\]

From one post of Silouan, I create the following one

$\bigstar$
For what value of $k$ then the following inequality holds for all $a,b,c\ge 0$
\[k(a+b+c)^{3}(a^{2}+b^{2}+c^{2}) \le a^{4}(b+c)+b^{4}(c+a)+c^{4}(a+b).\]

And I think $k=1-\frac{\sqrt{3}}{2}$. I spend more than two hours to solve it but false to construct a complete solution. However, I strong believe that it is solvable. Yet I need your help.

Sorry. Because I can not edit the post any more, so .... I don't know why in these days, I can not edit many posts  :( Is there any mistake in the site?

Actually, the problem is
\[k(a+b+c)^{3}(a^{2}+b^{2}+c^{2}) \ge a^{4}(b+c)+b^{4}(c+a)+c^{4}(a+b).\]

For $k=1-\frac{\sqrt{3}}{2}$, the equality can hold for $(a,b,c)~(3+\sqrt{3},3-\sqrt{3},0)$. However, I just think 30% it is true.


calculus
integration
modular arithmetic
quadratics
number theory unsolved
number theory

Hello..
Solve this equations for $ x,y \in \mathbb{Z}$:
$ x^2\equal{}y^3\plus{}13$.

It is an elliptic curve with rank 0 and torsion group isomorphic to trivial group. Maybe this helps but I'm still learning how to determine the integer points.

Rank 0 implies that all rational points are torsion points. Every torsion point is integral, thus in this case, the solutions are the torsion points. As there are no (finite) torsion points, there is no solution.

Utku, y could also be 4k+2. Can you prove this case?

[quote="badida"]Hello..
Solve this equations for $ x,y \in \mathbb{Z}$:
$ x^2 \equal{} y^3 \plus{} 13$.[/quote]
If $ y$ is even, then $ x^2 \equiv 5 \pmod 8$ which is impossible.
If $ y \equiv 1 \pmod 4$, then $ x^2 \equiv 2 \pmod 4$ which is also impossible.

For the remaining case: $ y \equiv 3 \pmod 4$, note that $ 13 \equal{} 17^3 \minus{} 70^2$.
So $ x^2 \equal{} (y \plus{} 17) (y^2 \minus{} 17y \plus{} 17^2) \minus{} 70^2$.
Since $ y \equiv 3 \pmod 4 \Rightarrow y^2 \minus{} 17y \plus{} 17^2 \equiv 3 \pmod 4$, it follows that $ y^2 \minus{} 17y \plus{} 17^2$ must have a prime factor $ p$ of the form $ 4n \plus{} 3$.
So we have $ x^2 \equiv \minus{} 70^2 \pmod p$.
This is impossible if $ p \ne 7$ because we would have $ (70^{ \minus{} 1}x)^2 \equiv \minus{} 1 \pmod p$, but $ \minus{} 1$ is not a quadratic residue modulo any prime of the form $ 4n \plus{} 3$.

If $ p \equal{} 7$, then $ 7 \mid x$, and hence $ 7^2 \mid (y \plus{} 17) (y^2 \minus{} 17y \plus{} 17^2)$.
Note that $ 7 \nmid \gcd(y \plus{} 17, y^2 \minus{} 17y \plus{} 17^2)$ since $ (y^2 \minus{} 17y \plus{} 17^2) \minus{} (y \plus{} 17) (y \minus{} 34) \equal{} 3 \cdot 17^2$.
So $ 7^2 \mid (y^2 \minus{} 17y \plus{} 17^2)$, and we have
\[ \left( \frac {x}{7} \right)^2 \equal{} (y \plus{} 17) \left( \frac {y^2 \minus{} 17y \plus{} 17^2}{7^2} \right) \minus{} 10^2.\]
Since $ 7^2 \equiv 1 \pmod 4$ and $ y^2 \minus{} 17y \plus{} 17^2 \equiv 3 \pmod 4$, it follows that
\[ \frac {y^2 \minus{} 17y \plus{} 17^2}{7^2} \equiv 3 \pmod 4,\]
and hence $ (y^2 \minus{} 17y \plus{} 17^2)/7^2$ must have a prime factor $ q$ of the form $ 4n \plus{} 3$.
So we have $ \left( \frac {x}{7} \right)^2 \equiv \minus{} 10^2 \pmod q$ which is also impossible because $ \minus{} 1$ is not a quadratic residue modulo $ q$.

Therefore, there is no integral solution to $ x^2 \equal{} y^3 \plus{} 13$.


How many different solutions $ (p,q)$ positive integers are for the equation $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}\frac{1}{2}$

From the given equation we may assume $ p\equal{}2\plus{}a; q\equal{}2\plus{}b$ where $ a,b$ are integers.
Then the equation $ \implies \frac{4\plus{}a\plus{}b}{4\plus{}2a\plus{}2b\plus{}ab}\equal{}\frac{1}{2}$
$ \implies 8\plus{}2a\plus{}2b\equal{}4\plus{}2a\plus{}2b\plus{}ab\implies ab\equal{}4$
Hence all possible solutions to $ (a,b)\equal{}(2,2), (1,4), (4,1)$
So this equation has 3 solutions, $ (6,3), (4,4), (3,6)$
In particular , If $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}\frac{1}{x}\forall x\in\mathbb{P}$, then it has only 3 possible distinct solutions. ;)

$ \frac{1}{p}\plus{}\frac{1}{q}\equal{}\frac{p\plus{}q}{pq}\equal{}\frac{1}{2}$
$ pq\equal{}2(p\plus{}q)$
$ pq\minus{}2p\minus{}2q\equal{}0$
$ pq\minus{}2p\minus{}2q\plus{}4\equal{}4$
$ (p\minus{}2)(q\minus{}2)\equal{}4$
Since $ p,q>2$, $ p\minus{}q$ and $ q\minus{}2$ are positive. Since there are 3 positive integral solutions to $ ab\equal{}4$, there are 3 positive integral solutions to $ (p\minus{}2)(q\minus{}2)\equal{}4$, which means that there are 3 positive integral solutions to $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}\frac{1}{2}$.


Each distinct letter below represents a unique digit. Find the value of the four-digit number $ papa$.

Yeah, I think Ernie's right. I've seen this problem and there's no figure or whatever there's supposed to be. Every time you see it, report it.

P has to be 1 because the sum of two three digit integers cannot be a four digit number greater than 1998.  Since P + P = A, A has to be 2.  Thus, PAPA is 1212.


modular arithmetic
number theory unsolved
number theory

Prove that for every integer $ a$ greater than 2 there exist infinitely many positive integers $ n$ such that that $ a^{n}\minus{}1$ is divisible by $ n$

if $ (a,p)\equal{}1$ where $ p$ is a prime then choose $ n\equal{}p^k$ , $ k\in\mathbb{N}$.

$ a^{p^k}\equiv 1\mod p^k$

you forgot to say that $ p|a\minus{}1$ that's where you use $ a>2$  :wink:

I have this incomplete solution.

Let $ n$ be such that $ (a,n)\equal{}1$. We have $ a^{\phi(n)}\equiv 1\pmod n$. So we'll be done if we can show that $ \phi(n)\mid n$ for infinitely many positive integers $ n$. Let $ n\equal{}p_1^{k_1}p_2^{k_2}\cdots p_N^{k_N}$ be the prime decomposition of $ n$. Then $ \phi(n)\equal{}p_1^{k_1\minus{}1}p_2^{k_2\minus{}1}\cdots p_N^{k_N\minus{}1}(p_1\minus{}1)(p_2\minus{}1)\cdots(p_N\minus{}1)$. So if $ \phi(n)\mid n$ we must have $ (p_1\minus{}1)(p_2\minus{}1)\cdots (p_N\minus{}1)\mid p_1p_2\cdots p_N$. This is possible if and only if $ p_1\equal{}2,p_2\equal{}3$, because $ p_i\minus{}1\mid p_j\implies p_i\minus{}1\equal{}1$ or $ p_j\implies \{p_i,p_j\}\equal{}\{2,3\}$. Thus $ n\equal{}2^x3^y$, where $ x\ge 1, y\ge 0$. If $ y\ge 1$ we have $ \phi(n)\equal{}2^x3^{y\minus{}1}\mid n$. If $ y\equal{}0$ then $ n\equal{}2^x$ and $ \phi(n)\equal{}2^{x\minus{}1}\mid n$.

Now we need to check if $ (a,n)\equal{}1$ holds for $ n\equal{}2^x3^y$. Suppose that $ a\equal{}2^s3^t$. If $ s\equal{}0$ i.e. $ a\equal{}3^t$, then we can take $ n\equal{}2^x$. I cannot find a suitable $ n$ when $ t\equal{}0$ or $ s,t>0$. Can anyone help? :blush:

If n is solution then $ a^{n}\minus{}1$ is also solution since $ a^{n}\minus{}1$ divides $ a^{a^{n}\minus{}1}\minus{}1$. And since a-1 is a solution (different from 1) then
we are done.
BTW there are no soluttions for a=2.


Prove that the following two integral equalities hold:

1) $\int_{0}^{\infty}\frac{xe^{-x^{2}}}{\sqrt{a^{2}+x^{2}}}dx=\frac{a}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-x^{2}}}{x^{2}+a^{2}}dx$, $a>0$.

2) $\int_{0}^{\infty}\frac{e^{-kx}}{1+x^{2}}dx=\int_{k}^{\infty}\frac{\sin(x-k)}{x}dx$, $k>0$.

OK, i did [u]both [/u]the integrals, but the first one is fairly
easy to show, so i will leave it to someone to
show it (and if no one else does, i will post it).  :D

well, below is the solution for the 2nd one, i used complex variables.

let $\textbf f(z)=\frac{e^{\imath z}}{z+k}$
i chose a [b]square [/b]contour with vertices at $\textbf O(0,0),\textbf A(R,0),\textbf B(R,\imath R),\textbf C(0,\imath R)$

since the pole $z=-k$ lies [b]outside [/b]the contour, by Cauchy,
$\oint_\gamma\;\textbf f(z)\;\textbf dz\;=\;2\pi\imath \sum_{z\in\gamma}\textbf Res\;\textbf f(z) \;=\;0$

so, $\textbf I_{1}+\textbf I_{2}+\textbf I_{3}+\textbf I_{4}=0$

i.e. $\left[ \int_{OA}+\int_{AB}+\int_{BC}+\int_{CO}\right] \;\textbf f(z)\;\textbf dz\;=\;0$

   $\textbf y$
   |
   +---------------+ $B(R,\imath R)$
   |[color=white]..............[/color].[color=white].....[/color]|
   |[color=white].........[/color].[color=white]..........[/color]|
   |[color=white]....[/color].[color=white]...............[/color]|
   +---------------+---[color=white].....[/color]$\textbf x$
   $O(0,0)$[color=white].......[/color]$A(R,0)$

now,
$\textbf I_{1}=\int_{0}^{R}\;\frac{e^{\imath x}}{x+k}\;\textbf dx$

$\textbf I_{4}=\int_{R}^{0}\;\frac{e^{\imath (\imath y)}}{k+\imath y}\;\imath \;\textbf dy=-\imath \int_{0}^{R}\;\frac{e^{-kx}}{k(1+\imath x)}\;k \;\textbf dx$

(using the substitution $y=kx$)

$=-\int_{0}^{R}\;\frac{(x+\imath)\;e^{-kx}}{1+x^{2}}\;\textbf dx$

(multiplying both numerator and denominator by $(1-\imath x)$)

$\textbf I_{2}=\int_{AB}\;\textbf f(z)\;\textbf dz$

now $|\int_{AB}\;\textbf f(z)\;\textbf dz\;|\leq \int_{AB}\;|\textbf f(z)|\;|\textbf dz|\;=\;\int_{0}^{R}\frac{|e^{\imath (R+\imath y)}|}{|(R+\imath y)+k|}\;|\;\imath \;\textbf dy\;|$

$\leq \int_{0}^{R}\;\frac{e^{-y}}{R-k-y}\;\textbf dy$

put $u=R-k-y$, and you get

$\frac{e^{k}}{e^{R}}\;\;\int_{-k}^{R-k}\;\frac{e^{u}}{u}\;\textbf du$

since $\int_{-k}^{R-k}\;\frac{e^{u}}{u}\;\textbf du$ is [b]finite[/b], and $e^{R}$ appears in the denominator,
then as $R\to\infty,\;\;\;\left\{ \frac{e^{k}}{e^{R}}\;\;\int_{-k}^{R-k}\;\frac{e^{u}}{u}\;\textbf du\right\}\to0$

hence, it follows,
$|\int_{AB}\;\textbf f(z)\;\textbf dz\;|\leq \int_{AB}\;|\textbf f(z)|\;|\textbf dz|\;\to 0$

which implies $\textbf I_{2}=0$.
and in a very similar manner, it can be shown $\textbf I_{3}=0$.

and so, $\textbf I_{1}+\textbf I_{4}=0$

$\int_{0}^{R}\;\frac{e^{\imath x}}{x+k}\;\textbf dx-\int_{0}^{R}\;\frac{(x+\imath)\;e^{-kx}}{1+x^{2}}\;\textbf dx=0$

in the limit $R\to\infty$, equate [b] [u]imaginary [/u][/b]parts to get

$\int_{0}^\infty\;\frac{e^{-kx}}{1+x^{2}}\;\textbf dx=\int_{0}^\infty\;\frac{\sin x}{x+k}\;\textbf dx$

now put $x=y-k$ in the [i]RHS [/i]integral to get

$\int_{0}^\infty\;\frac{e^{-kx}}{1+x^{2}}\;\textbf dx=\int_{k}^\infty\;\frac{\sin (y-k)}{y}\;\textbf dy$

thank you and goodnight ! :D


Let $a,b$ be natural number. Prove that if
$a^2+{(a+1)}^2=b^4+{(b+1)}^4$ we get a contradiction!


Given $ a, b, c \geq\ 0$.
Prove that: $ (a^2\plus{}\frac{b^2c}{a})(b^2\plus{}\frac{c^2a}{b})(c^2\plus{}\frac{a^2b}{c}) \geq\ (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)$

First , easy to prove :
$ RHS\equal{}\prod (a^2\plus{}bc) \ge abc \prod (a\plus{}b)$
Second :
$ LHS \equal{} \frac{LHS. \prod (a^2\plus{}ac) }{\prod (a^2\plus{}ac) } \ge \frac{RHS^2}{abc\prod(a\plus{}b )} \ge RHS$


trigonometry limit calculus derivative function

Prove that:

$\displaystyle\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi ^2}{6}$

Check out equations (4) and (5) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=47967]here[/url]. If you expand the product and equate coefficients, you'll get the result. If I remember correctly, equate the coefficients of $x$.

(4) is the well-known taylor series for $\sin{x}$. I derived (5) very nonrigorously by observing that the zeros of $\sin{x}$ are $k\pi$ for all integers $k$, so I figured $\sin{x}=\lim{x}\prod_{i=1}^na_n(x^2-n^2\pi^2)$, and I figured that $a_n$ must be the unique real number such that the product to $n$ has a derivative of $1$ at $x=0$. It's easy to show that $a_n=-\frac{1}{n^2\pi^2}$ for this to be true. Hence (5). It may not be rigorous, but it worked. :D

Anyway, I'm glad this topic was made since I've always wondered (since my little adventure with this last year) if there's a more straight-forward way to find $\sum\frac{1}{k^2}$.

[Sidenote:

In Euler's famous calculation that I mentioned at the top, the coefficients of $x$ are equated, but of course this can be done for the coefficients of $x^n$ for any $n\ge1$. I tried to generalize this, and got a pretty formula, but I'm not sure if it even has any meaning the way I wrote it, let alone if it's correct. I posted it in the "beautiful formulas" thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=47967]here[/url]. Please let me know if it makes sense or if it's complete nonsense. :P]

wow that last thing u posted was INSANE

but yea http://www.matharticles.com/ma_calc8e.html

go there and click on the article "who gave you epsilon? couchy and the origins of ..."

and within that article.. on the 2nd page or something you have the proof of Euler's famous calculation

im still trying to understand all of it though


Determine the number of odd binomial coefficients in the expansion of $ (x\plus{}y)^{1000}$.


function
analytic geometry
trigonometry
symmetry
complex analysis
complex analysis unsolved

u is a harmonic real fucntion on (x,y) plane. The conjugate differential of u is defined to be:
*du=-(du/dy)dx+(du/dx)dy

(du/dy should be understood as partial derevative).

Now consider a circle |z|=r, how to show that on this circle, using polar coordinate (r, θ):
*du=r(du/dr)dθ.
(I suspect, this don't need harmonic condition, just replace dx, dy with polar direvative, but I can't work out the detail) thanks.

$\frac{\partial u}{\partial r}= \frac{\partial u}{\partial x}* \frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}* \frac{\partial y}{\partial r}$
$= \frac{\partial u}{\partial x}* \cos \theta+\frac{\partial u}{\partial y}* \sin \theta$.

$d \theta = (\frac{\partial \theta}{\partial x}) dx+(\frac{\partial \theta}{\partial y}) dy = (\frac{-y}{x^{2}+y^{2}}) dx+(\frac{x}{x^{2}+y^{2}}) dy$
$= (\frac{-r \sin \theta}{r^{2}}) dx+(\frac{r \cos \theta}{r^{2}}) dy = (\frac{-\sin \theta}{r}) dx+(\frac{\cos \theta}{r}) dy$.

Therefore, $r(\frac{\partial u}{\partial r}) d \theta = (\frac{\partial u}{\partial x}* \cos \theta+\frac{\partial u}{\partial y}* \sin \theta) * (-\sin \theta dx+\cos \theta dy)$.

Expanding gives:
$-(\frac{\partial u}{\partial x}* \sin \theta \cos \theta+\frac{\partial u}{\partial y}* \sin^{2}\theta) dx+(\frac{\partial u}{\partial x}* \cos^{2}\theta+\frac{\partial u}{\partial y}* \sin \theta \cos \theta) dy$.

Now I guess you just need to show that $\frac{\partial u}{\partial x}* \sin \theta \cos \theta+\frac{\partial u}{\partial y}* \sin^{2}\theta = \frac{\partial u}{\partial y}$,
and $\frac{\partial u}{\partial x}* \cos^{2}\theta+\frac{\partial u}{\partial y}* \sin \theta \cos \theta = \frac{\partial u}{\partial x}$.

[quote="PTynan89"]

Now I guess you just need to show that $\frac{\partial u}{\partial x}* \sin \theta \cos \theta+\frac{\partial u}{\partial y}* \sin^{2}\theta = \frac{\partial u}{\partial y}$,
and $\frac{\partial u}{\partial x}* \cos^{2}\theta+\frac{\partial u}{\partial y}* \sin \theta \cos \theta = \frac{\partial u}{\partial x}$.[/quote]

This last step, you didn't finish. Is it because it is obvious? Can you give details? thanks.

Well, if you can somehow show that $\frac{\partial u}{\partial x}\sin \theta = \frac{\partial u}{\partial y}\cos \theta$.

Then, you can multiply by $\cos \theta$, then add $\frac{\partial u}{\partial y}\sin^{2}\theta$ to both sides to get the first part.

And, multiply by $\sin \theta$, then add $\frac{\partial u}{\partial x}\cos^{2}\theta$ to both sides to get the second condition.

So, proving the first statement involves $\frac{\partial u}{\partial x}= \frac{\partial u}{\partial r}* \frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}* \frac{\partial \theta}{\partial x}$
$= \frac{\partial u}{\partial r}* \cos \theta+\frac{\partial u}{\partial \theta}* \frac{-\sin \theta}{r}$.

And also $\frac{\partial u}{\partial y}= \frac{\partial u}{\partial r}* \frac{\partial r}{\partial y}+\frac{\partial u}{\partial \theta}* \frac{\partial \theta}{\partial y}$
$= \frac{\partial u}{\partial r}* \sin \theta+\frac{\partial u}{\partial \theta}* \frac{\cos \theta}{r}$.

So then, $\frac{\partial u}{\partial y}\cos \theta-\frac{\partial u}{\partial x}\sin \theta = (\frac{\partial u}{\partial \theta})* (\frac{\cos^{2}\theta}{r}+\frac{\sin^{2}\theta}{r}) = \frac{1}{r}(\frac{\partial u}{\partial \theta})$.

Proving that this is equal to $0$ solves it.

Alternatively, the first statement would imply that $\frac{\partial u}{\partial x}\tan \theta = \frac{\partial u}{\partial y}= \frac{\partial u}{\partial x}* \frac{y}{x}$.

$\frac{\partial}{\partial y}(\frac{\partial u}{\partial x}* \frac{y}{x}) = (\frac{\partial^{2}u}{\partial y \partial x}* \frac{y}{x})+(\frac{\partial u}{\partial x}* \frac{1}{x})$.

Then, $\frac{\partial}{\partial y}(\frac{\partial u}{\partial x}* \frac{y}{x})+\frac{\partial^{2}u}{\partial x^{2}}$
$= (\frac{\partial^{2}u}{\partial y \partial x}* \frac{y}{x})+(\frac{\partial u}{\partial x}* \frac{1}{x})+\frac{\partial^{2}u}{\partial x^{2}}$
$= ( (\frac{y}{x}) \frac{\partial}{\partial y}+\frac{1}{x}+\frac{\partial}{\partial x}) \frac{\partial u}{\partial x}$.

Of course, it is equivalent to showing $(y \frac{\partial}{\partial y}+1+x \frac{\partial}{\partial x}) \frac{\partial u}{\partial x}= 0$,
which appears somewhat easier to work with due to symmetry and would also complete the proof.

Unfortunately, I don't see right now how to finish either of those. :-/

This is actually from Ahlfors' book (1953 edition).

It is in Chap V "The Dirichlet Problem".

At very begining of this Chap, he defined conjugate deffrential, then in proving "mean value property",
he mentioned (wihout proof):

if u is harmonic, then on |z|=r, *du=r(du/dr)dθ.
And he used this ro prove mean value property.

Can someone take a look? Thanks.

I was told that:
*du=r(du/dr)dθ-(1/r)(du/dθ)dr, and since on the circle, dr=0, so last term is zero.

how to show *du=r(du/dr)dθ-(1/r)(du/dθ)dr, though?

Since, $r(\frac{\partial u}{\partial r}) d \theta =-(\frac{\partial u}{\partial x}* \sin \theta \cos \theta+\frac{\partial u}{\partial y}* \sin^{2}\theta) dx+(\frac{\partial u}{\partial x}* \cos^{2}\theta+\frac{\partial u}{\partial y}* \sin \theta \cos \theta) dy$,
you just need to find $-\frac{1}{r}(\frac{\partial u}{\partial \theta}) dr$, in a similar expression.

Combining them should yield the desired result.


Let circles $ (C_1)$ and $ (C_2)$ with the common external tangents $ ABCD$ and $ EFGH$. Circle $ (C_1)$ is tangent to $ BF$ at $ M$ and circle $ (C_2)$ is tangent to $ CG$ at $ N$. Lines $ AM$ and $ DN$ meet at $ P$ and lines $ EM$ and $ HN$ meet at $ Q$. Prove that the points $ M$, $ P$, $ N$, and $ Q$ are concyclic (lie on a same circle).

$ \angle QND\equal{}180^o\minus{}\angle HND\equal{}\angle EMA\equal{}\angle PMQ$


I [url=http://matt.mitblogs.com/archives/2005/12/upcoming_non-mi.html#comments]hear[/url] that the program is not being held this year. Has anyone else heard something about this?

[edit] Never mind. I just called them and found out that it's not being held this year but will be held next year. :( [/edit]

That every-other-year schedule is kind of a drag.

http://www.claymath.org/programs/outreach/academy/about.php

Why not every year?

It is a drag. It (both acceptance and/or rejection) was the one thing I was looking forward to senior year. Now not only do I have nothing to break the monotony of exams, but I lost a great opportunity because I didn't know about this every other year deal and chickened out of applying last year.  :(

I'm sure they would like to do it every year. They probably don't have enough funding. It's a great program for people who might apply next year.

I wanted to apply too :(.  Although I hear from some alumni from the program that it never accomplishes its true goal.  Basically the people who get in are really smart but spend the entire time actually learning and getting into the problem and the program is simply too short for that.  That's just some of what I've heard.


number theory
relatively prime
algebra
difference of squares
special factorizations
number theory proposed

Suppose that $a_{1}, a_{2},..., a_{\phi (n)}$ are the $\phi (n)$ numbers that are smaller than $n$ and relatively prime to it. Prove that $\prod_{i=1}^{\phi (n)}a_{i}< \left (\frac{n}{2}\right )^{\phi (n)}$.

It's nearly trivial.  If $gcd(a, n) = 1$ then $gcd(n-a, n) = 1$ as well so we can pair up $a_{i}, a_{j}$ that add up to $n$ and use difference of squares:

$\left( \frac{n}{2}-k \right)\left( \frac{n}{2}+k \right) < \left( \frac{n}{2}\right)^{2}$


function
inequalities
induction
algebra unsolved
algebra

Function $ f: \mathbb{R^\plus{}}\rightarrow\mathbb{R^\plus{}}$ is such that $ f(3x)\geq f(\frac{1}{2}f(2x))\plus{}2x$ for every $ x>0$.

Prove that $ f(x)\geq x$ for all $ x>0$.

[hide]
Define the sequence $ a_i$ by a_0 = 2/3, $ a_k \equal{} \frac{k^2}{3}\plus{}\frac{2}{3}.$

We claim that $ f(x) > a_ix$ for all positive integers i and all positive x. We prove this by induction on i.

Its true when i = 0 because since f(x) is always positive, we get in the function equaiton that $ f(3x) > 2x,$ or $ f(x) > \frac{2x}{3}.$

Assume its true for i = n. Then, plug in x = y/3 to get $ f(y) \ge \frac{2x}{3} \plus{} f(\frac{f(\frac{2x}{3})}{2})$ and we use the inductive hypothesis twice to get the desired result.

Note that as i approaches infinity, a_i approaches 1, so we have f(x) >= x for all x.[/hide]

[quote="pythag011"][hide]
Define the sequence $ a_i$ by a_0 = 2/3, $ a_k \equal{} \frac {k^2}{3} \plus{} \frac {2}{3}.$

[/quote]

I think you meant something like $ a_k \equal{} \frac {a_{k\minus{}1}^2}{3} \plus{} \frac {2}{3}.$

The part that this sequence converges to $ 1$ seems unclear to me. It is true that $ a_k>a_{k\minus{}1}$ for all $ k$, but then why does the sequence converge to $ 1$?


The distance from the center of a circle to its 16-cm chord equals 15 cm. Find the area of the circumscribed triangle if its perimeter equals 200 cm.

i somehow feel bigknick knickmid and 8isgreat are one and the same person wanting to get his homework done


linear algebra
matrix
linear algebra unsolved

$ A, B\in M_{3*5}(R)$:

$ A\equal{}\left[\begin{array}{ccccc}1&a&b&\minus{}1&c\\2&1&0&2&\minus{}1\\1&3&d&1&e\end{array}\right]$

$ B\equal{}\left[\begin{array}{ccccc}f&g&3&h&1\\i&3&3&j&k\\l&m&2&n&0\end{array}\right]$

$ B$ is matrix forthcoming from matrix $ A$ (using only elementary operations).
Find: $ a, b, c, d, e, f, g, h, i, j, k, l, m, n$.

I doubt that if the question is right that to find one solution, but add up two first rows of $ A$ and replace it by second row of A, then the answer will be clear:
$ a\equal{}d\equal{}g\equal{}2, b\equal{}m\equal{}i\equal{}3, c\equal{}l\equal{}n\equal{}f\equal{}j\equal{}1, e\equal{}k\equal{}0$ and $ h\equal{}\minus{}1$


There's a real number written on every field of a $n \times n$ chess board. The sum of all numbers of a "cross" (union of a line and a column) is $\geq a$. What's the smallest possible sum of all numbers on the board¿

$S=$sum of all numbers on the board.
$S_{i,j}=$sum union of a line $i$ and a column $j$.

$(2n-1)S=\sum_{1\le i,j\le n}S_{i,j}\ge n^{2}a$
then $S\ge \frac{n^{2}a}{2n-1}$

and if we take all numbres on the board $=\frac{a}{2n-1}$.
we have $\forall (i,j),\ S_{i,j}=\frac{(2n-1)a}{2n-1}=a\ge a$
and $S=\frac{a}{2n-1}$

so, $S_{min}=\frac{n^{2}a}{2n-1}$
and $S_{max}=+\infty$ :lol:


geometry
3D geometry
analytic geometry
trigonometry
function
calculus
derivative

Given a cube $ ABCD\minus{}EFGH$ with side length 1. Let $ P$ be a point on the edge $ AB$. If $ \angle{EPG}$ is maximal, then prove that $ 0.5<AP<0.6.$

A point of clarification: we're told that $ AB$ is an edge. I would infer that $ EG$ is a diagonal of a face of the cube - is that correct? And $ AE$ is an edge of the cube?

I'll do this with coordinates and vectors. And it will get a little messy.

Label these points:
\[ A(0,0,0)\\B(1,0,0)\\E(0,1,0)\\G(1,1,1)\\P(x,0,0)\]
Then $ \vec{PE}\equal{}\langle \minus{}x,1,0\rangle,\ \vec{PG}\equal{}\langle1\minus{}x,1,1\rangle.$

$ \vec{PE}\cdot\vec{PG}\equal{}x^2\minus{}x\plus{}1,\ \|\vec{PE}\|\equal{}\sqrt{x^2\plus{}1},\ \|\vec{PG}\|\equal{}\sqrt{x^2\minus{}2x\plus{}3}.$ Here, $ x\equal{}AP$ is the quantity we seek.

Note that $ \vec{PE}\cdot\vec{PG}>0$ always so $ \angle EPG$ is always acute.

$ \cos\angle EPG\equal{}\frac{x^2\minus{}x\plus{}1}{\sqrt{x^2\plus{}1}\sqrt{x^2\minus{}2x\plus{}3}}.$

We maximize the angle by minimizing its cosine. Since that cosine is positive, we minimize it by minimizing its square. Hence, we wish to minimize the following function:

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

$ \equal{}1\minus{}\frac{x^2\plus{}2}{x^4\minus{}2x^3\plus{}4x^2\minus{}2x\plus{}3}.$

Differentiate that and set the derivative equal to zero, which means setting the numerator of the derivative equal to zero, which means solving this polynomial equation:
\[ x^5\minus{}x^4\plus{}4x^3\minus{}5x^2\plus{}5x\minus{}2\equal{}0\]
Newton's method works particularly well for polynomial equations, and we see that the only solution to this in $ [0,1]$ is $ x\approxeq 0.596071638.$


algebra
polynomial
topology
function
real analysis
real analysis unsolved

For which values of $a,b,c$ the function $y= x^{2n+3}+ax^{2n+2}+bx^{2n+1}+c$ is a homeomorphism of $\mathbb{R}$ for all $n=1,2,3,...$

"Homeomorphism" = "increasing function" in this context. $y'=x^{2n}((2n+3)x^{2}+(2n+2)ax+(2n+1)b)$ must be nonnegative, which is equivalent to the discriminant of $(2n+3)x^{2}+(2n+2)ax+(2n+1)b$ being nonpositive.


[color=blue] [b] Problem. [/b]

    Let $a, b, c$ three positive real numbers. Prove the inequality

                         \[ \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\geq\frac{bc}{a^{3}(b+c)}+\frac{ca}{b^{3}(c+a)}+\frac{ab}{c^{3}(a+b)}.  \]

  [/color]

 EDIT: I must have been mistaken, when I reduced some terms. :)

can we use:

$\frac{b+c}{2a^2} \ge \frac{bc}{a^2(b+c)}$

The inequalities is vice versa I think.

  [I have edited it according to the comment of Nameless at this post below. Indeed, we don't need to use S.O.S here I think]
Wolg $a \geq b \geq c$
We have:
$\frac{1}{a(b+c)} \leq \frac{1}{b(a+c)} \leq \frac{1}{c(a+b)}$
$\frac{bc}{a^2}\leq \frac{ac}{b^2}\leq \frac{ab}{c^2}$
Let $U=\frac{1}{a(b+c)}+\frac{1}{b(a+c)}+\frac{1}{c(a+b)}$
     $V= \frac{bc}{a^3(b+c)}+\frac{ac}{b^3(a+c)}+\frac{ab}{c^3(a+b)}$
By Chebysep:
$V \geq U \frac{ \frac{bc}{a^2}+ \frac{ac}{b^2}+ \frac{ab}{c^2}}{3} \geq U$
Done!

[quote="Nameless"][quote="Cezar Lupu"][color=blue] [b] Problem. [/b]

    Let $a, b, c$ three positive real numbers. Prove the inequality

                         \[ \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\geq\frac{bc}{a^{2}(b+c)}+\frac{ca}{b^{2}(c+a)}+\frac{ab}{c^{2}(a+b)}.  \]

  [/color][/quote]
This ineq is wrong: $a=1,b=2,c=3$
I think it can be:
 \[ \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\le\frac{bc}{a^{3}(b+c)}+\frac{ca}{b^{3}(c+a)}+\frac{ab}{c^{3}(a+b)}.  \]
It can be solved easily by SOS[/quote]
I think arrangement inequality will be easier than SOS ;)


probability
number theory
divisibility tests

The sum of the digits of a number is 9. What is the probability that the number is prime?

$ \boxed{0}$.  If the sum of digits is $ 9$ it is divisible by $ 9$.  :)

test of divisibility by 9: digits add up to a multiple of 9.

therefore, thsi number is div by 9 and 3, so the answer is $ \boxed{0}$


I'm a mathematics enthusiast with so many field of
interest.One of these is '[b]Combinatorical [/b]Analysis[b].'Can anyone help to locate combinatorical[u] journals[/u] free[/b] of[b] charge[/b] in any internet websites?

                                                      Respecrfully,
                                                      A.Eddington

http://www.combinatorics.org/ is the website of the Electronic Journal of Combinatorics.


Let $ a;b;c$ be positive real numbers saisfying :
$ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\equal{}11$
Find min of:$ A\equal{}(a^{2}\plus{}b^{2}\plus{}c^{2})(\frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}})$


Let a,b,n>1. Prove the following inequality:
$\frac{1}{(a+n)^{2}}+\frac{1}{(b+n)^{2}}\geq\frac{1}{ab+n^{2}}$
i think it's easy but i have a nice solution. :lol:

Well I told it to you today[and I also told you that we don`t need a,b,c>1]
[b]Problem:[/b]Let a,b,n be positive real numbers.Prove that the following inequality occurs: \[\frac{1}{(a+n)^{2}}+\frac{1}{(b+n)^{2}}\geq\frac{1}{ab+n^{2}}\]

This is variant of very old easy problem, $\sum \frac{1}{(x+1)^{2}}\geq \frac{1}{xy+1}$

Anyway here's solution: \[(a+b)(ab+n^{2})\geq (a\sqrt{b}+n\sqrt{b})^{2}=b(a+n)^{2}\\ (b+a)(ab+n^{2})\geq (b\sqrt{a}+n\sqrt{a})^{2}=a(b+n)^{2}\]  \[\Longleftrightarrow \frac{1}{(a+n)^{2}}\geq \frac{b}{a+b}\cdot \frac{1}{ab+n^{2}}\\ \frac{1}{(b+n)^{2}}\geq \frac{a}{a+b}\cdot \frac{1}{ab+n^{2}}\] Add them, we get the desired inequality$\square$

if we take our left part $=LHS$ and our right part of the ineq $=RHS$, we can easily prove that $LHS-RHS \geq  0$ by a not very difficult expanding.(sorry but i`m in a rush right now, i can`t post all the details)


Just wondering, cuz I saw this a few times in AOPS, but what is "Prasolov's book". From what i read, it has something to do with....geometry?

I am not asking for any illegal downloads, but just wondering how to obtain it. Is it on Amazon? Is it in English or another language? What does the book contain...description, purpose, etc.

Thanks!

Prasalov's book is available on the net for free.
It contains Geometry Problems and Answers.

The link to the book (English translation) is http://students.imsa.edu/%7Etliu/Math/planegeo.pdf


inequalities
pigeonhole principle
Ring Theory
number theory
number theory proposed

If $p$ is a prime such that $p \equiv 1(\mod 4)$, then $p=a^2+b^2$ for some integer $a$, $b$.

There is a proof  in Niven's and  Zuckerman's number theory book...

...about five pages
:ewpu: :ewpu: :ewpu: :ewpu: :ewpu:

Please let me know if you have a shorter one :)

Erdos gave a great proof, but I can't recall it now. I think it involves one of Dirichlet's approximation theorems, and it's definitely less than 5 pages. If no one gives a proof soon, I can look it up from my notes.

"Proofs from the book" definately has something more around 1-page.

Well, I think it can easily be done in less then five pages (and was postet before I think):

Let be $p$ a prime $\equiv 1 \mod 4$.

Lemma 1: There is an $z$, such that $z^2 \equiv -1 \mod p$
prove: set for example $z= \left( \frac{p-1}{2} \right) !$. Then $z^2 \equiv  \left( \frac{p-1}{2} \right) ! ^2 \equiv  (-1)^{\frac{p-1}{2}}  (p-1)  ! \equiv (p-1)! \equiv -1 \mod p$ by Wilson's theorem.

Lemma 2 (Thue): Let be $a$ some given integer. then there exist integers $x,y$ with $|x| , |y| < \sqrt{p} , y \neq 0$ and $x \equiv ay \mod p$.
prove: Let be $k$ the biggest integer smaller than $\sqrt p$. Then there are $(k+1)^2 >p$ pairs (x,y) such that the inequality $0 \leq x,y \leq k$ is fulfilled. By the pidgeonhole principle, there are two different ones $(x_1,y_1) , (x_2,y_2)$ with $ay_1-x_1 \equiv ay_2-x_2 \mod p \iff (x_2 - x_1) \equiv a(y_2 - y_1) \mod p$. But now the pair $( x_2-x_1 , y_2-y_1)$ solves our problem ($y_1 \equiv y_2 \mod p$ would imply that $x_1 \equiv x_2 \mod p$ which ios not possible because the pars are different).

prove of the theorem:
Let be $z$ that one from lemma 1.
Choose $a=z$ in lemma 2 and take some working pair $(x,y)$. Then $x=zy \mod p \implies x^2+y^2\equiv 0 \mod p$, but also $0< x^2+y^2 < p + p =2p$. This immediatly gives $x^2+y^2 =p$.

(You can easily get similar results for primes of the type $a^2+2b^2$ or $a^2+3b^2$)

Btw, check out Don Zagier's proof of this fact...awesome...

Any link to Don Zagier's proof? Or if it's not too inconvenient for you Rafal, could you write out the proof? I'm interested...

It uses involutions, tommorow I'll give a link or write that proof.

By ZetaX's lemma 1, there exists z such that p divides z^2+1=(z+i)(z-i).  We have that Z[i] is a ufd, so therefore, if p is prime in Z[i] as well, then either p divides z+i or p divides z-i.  But both of these are clearly absurd.  Therefore, we must have that p is not prime in Z[i], that is, p=(a+bi)(a-bi)=a^2+b^2 for some a and b.


Solve in natural numbers the following equation

$3^x=2^y\cdot 5^z+1$

Let $3^x-1=2^y5^z$.
1.If x=2k is even or $3^k+1=2*5^z,3^k-1=2^{y-1}, \ or \ 3^k-1=2*5^z, 3^k+1=2^{y-1}$ This equatin have only one solution $3^2-1=2^3$, therefore z=1,x=4,y=4.
2. x is odd, then y=2. $3^x-1=2*5^z %Error. "Longarrow" is a bad command.
\ if \ z>0 \ 4*5^{z-1}|x$ It give contradition, therefore z=0,x=1,y=1.


Hi!
I have a question:
I i want a special heading on the \section , \subsection like arial, is there a way to do that?

How to i change the font on the usual text?

 / Johan

[quote="anirudh"]Look on the toolbar. There is a font thingie there.[/quote]If you're talking about TeXnic Center, then all the "Format" menu does is change between different families of the same font, presumably the default Computer Modern font.  It can't change to a completely different font like Arial.  To do that one normally uses the \usefont command.  For example, to switch to Helvetica (which is pretty much the same as Arial), you go
\usefont{T1}{phv}{m}{n}

An easier option if you want Helvetica is to use the helvet package (put a \usepackage{helvet} sometime near the top of your document).  Then \textsf will switch to the Helvetica font.

To change \section, \subsection, etc., you need to redefine those commands using the \@startsection command.  It's kind of complicated, and it's been discussed a bit [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30107]here[/url].


geometry
incenter
inradius
geometry solved

Given a right triangle ABC with C = 90. A point D distinct from A and C is chosen
on the ray AC so that the line through the incenter of ABC parallel to
the bisector of ADB is tangebt to the incircle of BCD

prove that AD = BD

can anyone solve this problem ??

Posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=27798 .

  darij

The principal  reason for the result is Poncelet theorem

  for right ( not wrong ! ) at C  triangle:

     Rc = Ra + r + Rb       where    Rx -  ex-circles radii,  r - inradius

  Thank you.

  M. T.

[i]Dear Armpist[/i], allow me, please, prove (using the standard notations) the mentioned consequence of the [u]Poncelet's theorem:[/u] $\triangle ABC: \ r_a=r_b+r_c+r$ $\Longleftrightarrow$ $A=90^{\circ}.$

[u]A solution.[/u] From the wellknown relations $S=\sqrt{s(s-a)(s-b)(s-c)}$, $S=sr=(s-a)r_a=(s-b)r_b=(s-c)r_c$, $r_a+r_b+r_c=4R+r$, $4Rrs=abc$, $s(r_a-r)=ar_a$ results:

$r_a=r_b+r_c+r$ $\Longleftrightarrow$ $r_a=(4R+r-r_a)+r$ $\Longleftrightarrow$ $r_a-r=2R$ $\Longleftrightarrow$ $ar_a=2Rs$ $\Longleftrightarrow$ $2rr_a=bc$ $\Longleftrightarrow$ $2S^2=bcs(s-a)$ $4(s-b)(s-c)=2bc$ $\Longleftrightarrow$ $a^2-(b-c)^2=2bc$ $\Longleftrightarrow$ $a^2=b^2+c^2$ $\Longleftrightarrow$ $A=90^{\circ}.$


Need help with this problem

Question:

Find all incongruent solutions modulo 35 of the following system
7x + 5y = 13 (mod 35)
11x + 13y = 4 (mod 35)

Not sure what to do or how to get the solutions

Thanks

1st possibility:
Consider these congruencies separately modulo 5 and modulo 7. After solving apply Chineese Remainder Theorem.

2st possibility:
Observe that $11$ and $35$ are coprime. So you can find integers $u$ and $v$ such that $11u+35v= 1$ and hence  $11u \equiv 1 (mod 35)$
Multiply the second congruence by $u$ and receive $x$ through $y$. Use this in the 1st congruence.

Good luck.


Let $ x_1,x_2,...,x_n$ be non-negative real numbers such that $ \sum_{i\equal{}1}^n x_i \equal{}1$.
 Prove the inequality:
$ (\sum_{i\equal{}1}^n \sqrt{x_i}) (\sum_{i\equal{}1}^n \frac{1}{\sqrt{1\plus{}x_i}})\leq\frac{n^2}{\sqrt{n\plus{}1}}$

[quote="mr.danh"]Let $ x_1,x_2,...,x_n$ be non-negative real numbers such that $ \sum_{i \equal{} 1}^n x_i \equal{} 1$.
 Prove the inequality:
$ (\sum_{i \equal{} 1}^n \sqrt {x_i}) (\sum_{i \equal{} 1}^n \frac {1}{\sqrt {1 \plus{} x_i}})\leq\frac {n^2}{\sqrt {n \plus{} 1}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88299


hey if anyone is hosting a competition, can you please tell me about it? I would like to join.. :)

There are many going on right now! :P
i.e., the MC problem solving Marathon.

Keep an eye on the AMC forum. There are many Mock AMC/AIME/USAMO contests, though you don't really compete against other people.
A Mock USAMO just ended, but a Mock AIME just began. :D

[quote="perfect628"]I was going to start one, but I'm too busy to write things in LaTeX. If people would suck it up and let me just type it, I could do one.[/quote]

I could help you write it in $\LaTeX$

[quote="perfect628"]I'm ABLE to write it in LaTeX, I just don't feel like it. If you WANT to, then sure, you can. But, now that I think of it, most of the questions I've written lately don't require LaTeX, only the solutions. I'll think about it.[/quote]

sure, you want to pm me the problems/solutions?


It says I'm missing \begin{document}, but I'm not. What happened?

It's possible that you have a line before \begin{document} that should go after \begin{document}. You'll need to post your document here for us to find out more - if it's long then cut out the body of the document but so that you still get the same error.

Alternately, you may have prematurely ended the document somehow, which could either happen if you typed something after end document OR perhaps you ended something else that wasn't actually ever started (i.e. perhaps you ended an enumerate that you forgot to begin, or ended it twice, or something).


projective geometry
geometry proposed
geometry

Let $ X$, $ Y$, $ Z$ three arbitrary points inside or on the boundary of a given circle $ \rho$. Consider the following intersections: $ \left\{D,E \right\}\in XY \cap \rho$, $ \left\{M,N \right\}\in YZ \cap \rho$ and $ \left\{S, T \right\}\in ZX \cap \rho$. Now denote $ X_{1}\in MS \cap NE$, $ Y_{1}\in NT \cap DS$ and $ Z_{1}\in TE \cap MD$. Prove that the lines  $ XX_{1}, YY_{1}, ZZ_{1}$ are concurrent.

You are talking about a concurrency of three pascal lines, and in particular, this result is known as a special case of the Kirkman's theorem, so I'm afraid it's not at all new, since Kirkman proved a generalisation of this theorem concerning many related concurrencies in 1848.  :wink:

The theorem that is known as Kirkman's theorem states that:
[quote]The Pascal lines of the hexagons ABFDCE, AEFBDC, and ABDFEC are concurrent.[/quote]

See here for a more general approach: http://mathworld.wolfram.com/PascalLines.html

[b][size=100][color=DarkBlue]EQUIVALENT  PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ D,$ $ D'$ and $ E,$ $ E'$ and $ F,$ $ F'$ be, the pairs of points at which are intersected its sidelines $ BC,$ $ AC,$ $ AB$ respectively, from an arbitrary circle $ (O)$ $ ($ the points $ D,$ $ E,$ $ F,$ adjoining to $ B,$ $ C,$ $ A,$ respectively $ ).$

It is formed the convex hexagon $ A'B''C'A''B'C''$ by the intersection points of the sidelines of the triangles $ \bigtriangleup DEF$ and $ \bigtriangleup D'E'F'.$

Prove that the diagonals $ A'A'',$ $ B'B'',$ $ C'C'',$ are concurrent at one point, so be it $ P.$[/color][/size][/b]

[b][size=100]EXTENSION RESULTS.[/size][/b] - Prove that (see the figure):

[b][size=100]1).[/size][/b] – The line segments $ AA',$ $ BB',$ $ CC',$ are concurrent at one point, so be it, $ M.$

[b][size=100]2).[/size][/b] – The line segments $ AA'',$ $ BB'',$ $ CC'',$ are concurrent at one point, so be it, $ N.$

Kostas Vittas.

PS. I think that the equivalent problem is well known. Probably the extension results are not so well known.

[quote="pohoatza"][color=darkred]Let $ X,Y,Z$ be three arbitrary points inside of a circle $ w$. Denote the intersections $ \{\begin{array}{ccc}\{D,E \}& = & XY \cap w\\\ \{M,N \}& = & YZ \cap w\\\ \{S, T \}& = & ZX \cap w\end{array}$ and $ \{\begin{array}{ccc}X_{1}& \in & MS \cap NE\\\ Y_{1}& \in & NT \cap DS\\\ Z_{1}& \in & TE \cap MD\end{array}$.
Prove that the lines  $ XX_{1}, YY_{1}, ZZ_{1}$ are concurrently[/color].[/quote]
[color=darkblue][b]Proof.[/b] Define the intersections $ \{\begin{array}{ccc}U & \in & DM\cap TN\\\ V & \in & ET\cap MS\\\ W & \in & EN\cap DS\end{array}$.

Apply the [b]Pascal's theorem[/b] to the cyclic hexagons : $ \{\begin{array}{cccc}NEDMST\ : & \{\begin{array}{c}X_{1}\in NE\cap MS\\\ X\in ED\cap ST\\\ U\in DM\cap TN\end{array}\| & \implies & U\in XX_{1}\ .\\\\ SDETNM\ : & \{\begin{array}{c}Y_{1}\in SD\cap TN\\\ Y\in DE\cap NM\\\ V\in ET\cap MS\end{array}\| & \implies & V\in YY_{1}\ .\\\\ TENMDS : & \{\begin{array}{c}Z_{1}\in TE\cap MD\\\ W\in EN\cap DS\\\ Z\in NM\cap ST\end{array}\| & \implies & W\in ZZ_{1}\ .\end{array}$[/color]

I"ll come back soon here .

Dear Mathlinkers,
the point of concur is a Kirkman point...
You can see a proof on
http://perso.orange.fr/jl.ayme  vol; 12 Hexagramma... p. 56.
Sincerely
Jean-Louis


Find the smallest positive number $ x$ such that $ 0.8\bar{3}\cdot x$ is an integer.

The answer is [hide]$ 1.2$[/hide], and that works, but I don't know how find it.

[size=75][color=darkred]Edited LaTeX.
-nebula42[/color][/size]

[hide]well, we can use a formula to find that 0.8333333333333... is equal to 5/6. the smallest # that would make 5/6 a # is the reciprocal, 6/5, which is equal to 1.2[/hide]

Who cares about formulas. (Or for Mathcounts, my coach had me memorize a bunch of fractions and their decimals). Heres the other way
[hide]Let $ x\equal{}0.8\bar{3}$.
$ 100x\equal{}83.\bar{3}$
$ 10x\equal{}8.\bar{3}$.
Subtracing gives $ 90x\equal{}75$ or $ x\equal{}\frac{5}{6}$.
The smallest one we want is the reciprocal, or $ \frac{6}{5}\equal{}1.2$[/hide]

[quote="xpmath"]Who cares about formulas. (Or for Mathcounts, my coach had me memorize a bunch of fractions and their decimals). Heres the other way
[hide]Let $ x \equal{} 0.8\bar{3}$.
$ 100x \equal{} 83.\bar{3}$
$ 10x \equal{} 8.\bar{3}$.
Subtracing gives $ 90x \equal{} 75$ or $ x \equal{}\frac{5}{6}$.
The smallest one we want is the reciprocal, or $ \frac{6}{5}\equal{} 1.2$[/hide][/quote]

yeah, thats the formula i was talking abt

i misunderstood the problem. do u mean what number added to .8333.. makes an integer?

mz94 is correct. $ .8\bar{3}* x$

It's easier to see the muliplication symbol in tex if it is a *, even if that is incorrect or uncommon.


Prove that $ \sum_{i=0}^{n}\left[\binom{n}{i}\sum_{j=0}^{i}\binom{i}{j}\right]=3^{n}\forall n\in\mathbb N$.

Because $ 2^{r}=(1+1)^{r}=\sum_{k=0}^{n}C_{r}^{k},\forall r \in \mathbb N$ our identity is equivalent to:
$ \sum_{i=0}^{n}2^{k}\cdot C_{n}^{k}=(1+2)^{n}=3^{n}$

The original sum is the number of ways to choose two nested subsets of an $ n$-set.  This is clearly $ 3^{n}$, since for each element it can be in neither subset, in the outer but not inner subset, or in both subsets.


integration
logarithms
calculus
real analysis
real analysis theorems

I was wondering, how does one prove that something has no closed form? Certainly "everyone knows" that certain things like $\int x^x dx$ and $\int e^{-x^2}dx$ have no antiderivatives that are finite linear combinations of elementary functions. But how do we prove it? Also, why do we care? I mean, no one seems concerned about the value of $\int_0^\infty (\sin(x))^x dx$. But people are interested in $\int_0^{\frac{\pi}{4}} x\ln(\sin(x))dx$ (if they know that this is related to $\zeta(3)$) (I hope I recall that integral correctly). After like 270 years, a fifth of which included Euler's best efforts, if nobody has managed to find a closed form, it doesn't seem unreasonable to expect that there is none. But how would one go about trying to prove it? Or is it even worth our time?

[quote="Xevarion"] Certainly "everyone knows" that certain things like $\int x^x dx$ and $\int e^{-x^2}dx$ have no antiderivatives that are finite linear combinations of elementary functions. But how do we prove it?
[/quote]
See [url]http://math.hunter.cuny.edu/ksda/papers/churchill.pdf[/url]


A $ \textit{palindrome}$ is a number that reads the same forward as backward. How many 3-digit palindromes are multiples of 3?

[quote="FantasyLover"]if the sum of the digits of a number is multiple of 3, the number is multiple of 3.

since xyyx=multiple of 3, xy also is multiple of 3.

xy can range from 12 to 99.

12=3*4
99=3*33

33-4+1=30

answer: 30[/quote]
they're 3 digit numbers, not four digit #'s, as FantasyLover said

the number is xyx

If the x is not divisible by three, there will be 3 choices for y. there are 6*3=18 of these.

if the x is divisible by three, y has 4 possibillities. (0,3,6,9) x has 4 choices, but if the leading digit is 0, it will be 2-digit, so there are 3*4=12 of these.

12+18=\boxed{30}

How does $x$ not being divisible by 3 mean that there are 3 choices for y?


7 distinct primes can be written as a, b, c, a+b+c, a+b-c, a-b+c, -a+b+c, where two of a, b, c sum to 800. Let d be the difference between the largest and the smallest of the 7 primes. What is the largest possible value of d?

Obviously none of these primes can be 2, then we have far too many even numbers.

Suppose none of them were 3. Then all are 1 or -1 mod 6, but if a, b, c are all 1 or -1 mod 6 then one of the last 4 cannot be 1 or -1. So one of these primes must be 3.

If it was a (or b or c similarly), then again b and c must be 1 or -1 mod 6, but this doesn't work as one of 3+b-c, 3+b+c would turn out to be 3 mod 6. a+b+c also can't be 3 as a, b, and c are distinct primes. So assume a+b-c = 3.

Now our primes are a, b, a+b-3, 2a+2b-3, 3, 2a-3, 2b-3.
Firstly suppose a+c (or b+c similarly) was 800. Then a=797, but 2a-3 = 1591 = 37*43.

So we have a+b = 800. Now those primes are a, 800-a 797, 1597, 3, 2a-3, 1597-2a. But for 1597-2a to be a prime, 2a<1597, which means 2a-3 < 1594 and all these primes are less than 1597 (as a<800 also). So d is 1594.

I've probably missed one case somewhere since it says biggest value of d.. but I'm sure that'll be easily fixed.


Solve for y in terms of x $x=\frac{\frac{4}{9}y+2}{\frac{1}{6}y-1}$

[quote="anirudh"][hide="Hint :wink: "]You multiply both sides by $\frac{1}{6}y-1$[/hide][/quote]

Don't provide hints like that please.


[u][i]   Problem[/i][/u]

[i]   Let $ a , b , c$ be positive real numbers such that $ abc \leq 8$

  Prove that

  $ \frac{1}{ a^{2} \minus{} a \plus{}1 } \plus{}  \frac{1}{ b^{2} \minus{} b \plus{}1 } \plus{} \frac{1}{ c^{2} \minus{} c \plus{}1 }  \geq 1$[/i]

[i]
  Love An Forever

 Huỳnh Võ Phương An[/i]

[quote="nkht-tk14"][u][i]   Problem[/i][/u]

[i]   Let $ a , b , c$ be positive real numbers such that $ abc \leq 8$

  Prove that

  $ \frac {1}{ a^{2} \minus{} a \plus{} 1 } \plus{} \frac {1}{ b^{2} \minus{} b \plus{} 1 } \plus{} \frac {1}{ c^{2} \minus{} c \plus{} 1 } \geq 1$[/i]

[i]
  Love An Forever

 Huỳnh Võ Phương An[/i][/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35975


function
limit
vector
real analysis
real analysis unsolved

Let $ C^1[ \minus{} 1,1]$ be the set of all continuous diffential functions on $ [ \minus{} 1,1]$ with norm:$ \parallel{}x\parallel{} \equal{} \sup_{t\in[ \minus{} 1,1]}|x(t)| \plus{} \sup_{t\in[ \minus{} 1,1]}|x'(t)|$.
Put $ f_\epsilon: C^1[ \minus{} 1,1]\rightarrow\mathbb{R}$ where $ f_\epsilon(x) \equal{} \frac {1}{2\epsilon}[x(\epsilon) \minus{} x( \minus{} \epsilon)]$, $ x\in C^1[ \minus{} 1,1]$.
Calculate $ \parallel{}f_\epsilon\parallel{}$

I'm going to edit some absolute values into your definition of norm - you really need them.

The Mean Value Theorem is your friend.

$ f_{\epsilon}(x) \equal{} x'(\xi)$ where $ \xi\in( \minus{} \epsilon,\epsilon).$

That's bounded by $ \sup |x'(t)|,$ which is in turn bounded by $ \|x\|,$ so the norm of the functional $ f_{\epsilon}$ is no greater than $ 1.$

On the other side, we could create a function $ x$ where $ x(t) \equal{} t$ on a very short interval around zero and $ x$ is flat outside that interval, so $ \|x\|$ is only slightly larger than $ 1.$ From that I see that $ \lim_{\epsilon\to0}\|f_{\epsilon}\| \equal{} 1.$ The exact computation for any specific $ \epsilon$ may be a bit messier.

======

Added comment:

Making that example a little more quantitative, I can see that $ \|f_{\epsilon}\|\ge\frac1{1\plus{}\epsilon}.$ I suspect, but haven't proved, that that is an equality.

[quote]I'm going to edit some absolute values into your definition of norm - you really need them. [/quote]
I don't undersatand what's your means?

[quote]On the other side, we could create a function $ x(t) \equal{} t$ where  on a very short interval around zero and  is flat outside that interval[/quote]
with the function $ x$ above, i think it's not in $ C^1[\minus{}1,1]$

You actually wrote this:

$ \|x\| \equal{} \sup_{t\in[ \minus{} 1,1]}x(t) \plus{} \sup_{t\in[ \minus{} 1,1]}x'(t)$.

The problem is, that's not a norm. A norm on a (real or complex) vector space has to satisfy these properties:

1. $ \|x\|\ge 0,$ and $ \|x\|>0$ for all $ x\ne 0.$

2. $ \|\alpha x\| \equal{}|\alpha|\|x\|$ for all $ x$ and for all scalars $ \alpha.$

3. $ \|x\plus{}y\|\le \|x\|\plus{}\|y\|$ for all $ x$ and $ y.$

The problem with what you wrote, without the absolute values in it, is that it's not a norm at all. It violates these conditions. Notably, it can be negative, and it doesn't scale right when you factor out negative scalars. Hence what you originally wrote makes no sense as a problem. With the absolute values signs that I edited in, it is a norm and the problem does make sense.

[quote]with the function $ x$ above, i think it's not in $ C^1[\minus{}1,1]$[/quote]
Eh, don't be so literal-minded and start thinking like an analyst. Sure, a nontrivial piecewise-linear continuous function isn't $ C^1$ - but all you have to do is approximate it very, very slightly by rounding off the corners very slightly and it will be. Focus on what you can get with an example like this.

[quote="Kent Merryfield"]You actually wrote this:

$ \|x\| \equal{} \sup_{t\in[ \minus{} 1,1]}x(t) \plus{} \sup_{t\in[ \minus{} 1,1]}x'(t)$.[/quote]
Thans for point out my mistake.

[quote]with the function $ x$ above, i think it's not in $ C^1[ \minus{} 1,1]$
Eh, don't be so literal-minded and start thinking like an analyst. Sure, a nontrivial piecewise-linear continuous function isn't $ C^1$ - but all you have to do is approximate it very, very slightly by rounding off the corners very slightly and it will be. Focus on what you can get with an example like this.[/quote]

Yes, i see, thanks so much

[quote="Kent Merryfield"]

Making that example a little more quantitative, I can see that $ \|f_{\epsilon}\|\ge\frac1{1 \plus{} \epsilon}.$ I suspect, but haven't proved, that that is an equality.[/quote]

What exactly $ x$ did you use to get this estimate, Kent?  :)

I let $ x$ be the pieceswise linear function that is equal to $ t$ on $ (\minus{}\epsilon,\epsilon),$ $ \plus{}\epsilon$ to the right of that and $ \minus{}\epsilon$ to the left of that. Yes, this $ x$ isn't $ C^1$ but we can approximate it arbitrarily closely by functions that are $ C^1.$ We have that $ f_{\epsilon}(x)\equal{}1$ while $ \|x\|\equal{}\|x\|_{\infty}\plus{}\|x'\|_{\infty}\equal{}\epsilon\plus{}1.$ Then $ \frac{|f_{\epsilon}(x)|}{\|x\|}\equal{}\frac1{1\plus{}\epsilon}.$ Again, we can't actually have this as an equality, but be can approximate it arbitrarily closely.

I suspect that this is the worst-case scenario, which is why I think that is the operator norm, but I haven't actually proved that it's the the worst-case scenario.


can someone prove the following equation for polygonal number:

$P^n_r=\frac{1}{2}r((n-2)r-(n-4))$

seen here:
http://mathworld.wolfram.com/PolygonalNumber.html

Count the points lying on the i-th n-gon not lying on the bottom common edge. For the first one there are n-2 such ponts. For each next, there are n-2 more (1 is added on each non-comon edge), so for all the n-gons there are
(n-2)+2(n-2)+...+(r-1)(n-2)=(n-2)(1+2+...+r-1)=(n-2)r(r-1)/2.
Now we have to add the points lying on the common bottom edge - there are r such points. So the number of the points is (n-2)r(r-1)/2+r, which is easy to get into the form we want.


ARML AMC USA(J)MO USAMO geometry \/closed

I'm going to study for high school in US but I haven't decided where to go yet.

I wonder if there are differences among high schools in different states.
I mean may the high schools in some states be better (especially in Math) than those in some other states.

I am considering high schools in San Jose (California) and Austin (Texas) but haven't  know which is better.

Thanks for any help.

I am not knowledgeable enough to make any comparisons.

If you are interested in math, I know LBJ High School in Austin, TX is pretty good. A lot of people on the Texas ARML team are from that school, and they usually have several USAMO qualifiers. Their coach is really cool. He's a grad student in math at UT Austin and he did pretty well in contests in high school and college. He also used to run his own math contest.

Also, my mom teaches at Pflugerville high school (also an Austin suburb), so I feel compelled to put in a good word for them, even though I don't think they have a particularly good reputation for math excellence. They just got a great Latin teacher, though! ;)

There were about 40 USAMO qualifiers last year from the southern portion of the San Francisco Bay area, which is to say from communities within 20 miles of San Jose. These 40 or so were spread over a number of different high schools (and middle schools). Some of them went to private schools but most were from public schools. I know a few of the names: San Jose Lynbrook, Cupertino Monta Vista, Fremont Mission San Jose, Palo Alto Gunn, but there are others. Maybe someone who lives there could fill in the details.

Of course, housing is ridiculously expensive around San Jose.

Montgomery blair in MD
thomas jefferson high
exeter acadamy

all very good math schools...

Thanks for the help from you all.

[quote="Xevarion"]
If you are interested in math, I know LBJ High School in Austin, TX is pretty good. A lot of people on the Texas ARML team are from that school, and they usually have several USAMO qualifiers. Their coach is really cool. He's a grad student in math at UT Austin and he did pretty well in contests in high school and college. He also used to run his own math contest.
[/quote]

I have surfed the web for more information about the LBJ HS.
I can see their achievements in math, but I don't know why the school wasn't ranked very well by the [i]greatschools.net[/i] and some other Web pages.

I wonder if the school has any shortcomings.


inequalities geometry vector function trigonometry conics parabola

can anyone please explain in detail or give links for..........
1)power mean theory.........
2)cauchy schwarz inequality
3)bernoulli's inequality
4)convexity

Q)proove
(1)  a+b+c<=a*a+b*b+c*c    abc are real

(2)  (IMO 95/2) a; b; c are positive reals with abc = 1. Prove that
       1/a^3(b + c) + 1/b^3(c + a) + 1/c^3(a + b)>=3/2

please give detailed solutions...........

i think i've heard abt cauchy-schwarz inequality...its related to number theory :(  :)

hmm...cauchy is not necessarily related to number theory.....it has wide applications-even in geometry,combinatorics netc.....

it has
but i think u do read it first in Number theory...... :)  :(

The Cauchy–Schwarz inequality states that for all vectors x and y of an inner product space,

    $ |\langle x,y\rangle|^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle,$

where $ \langle\cdot,\cdot\rangle$ is the inner product. Equivalently, by taking the square root of both sides, and referring to the norms of the vectors, the inequality is written as

    $ |\langle x,y\rangle| \leq \|x\| \cdot \|y\|.\,$

Moreover, the two sides are equal if and only if x and y are linearly dependent (or, in a geometrical sense, they are parallel or one of the vectors is equal to zero).

If $ x_1,\cdots, x_n\in\mathbb C and y_1,\cdots, y_n\in\mathbb C$ are the components of x and y with respect to an orthonormal basis of V the inequality may be restated in a more explicit way as follows:

    $ |\overline{x}_1 y_1 \plus{} \cdots \plus{} \overline{x}_n y_n|^2 \leq (|x_1|^2 \plus{} \cdots \plus{} |x_n|^2) (|y_1|^2 \plus{} \cdots \plus{} |y_n|^2).$

Equality holds if and only x and y are linearly dependent, that is, one is a scalar multiple of the other (which includes the case when one or both are zero).
PROOF

As the inequality is trivially true in the case y = 0, we may assume <y, y> is nonzero. Let λ be a complex number. Then,

   $ 0 \leq \left\| x\minus{}\lambda y \right\|^2 \equal{} \langle x\minus{}\lambda y,x\minus{}\lambda y \rangle \equal{} \langle x,x \rangle \minus{} \bar{\lambda} \langle y,x \rangle \minus{} \lambda \langle x,y \rangle \plus{} |\lambda|^2 \langle y,y\rangle.$

The above expression is valid for any complex λ. Choosing

   $ \lambda \equal{} \langle y,x \rangle \cdot \langle y,y \rangle^{\minus{}1}$

we obtain

   $ 0 \leq \langle x,x \rangle \minus{} |\langle x,y \rangle|^2 \cdot \langle y,y \rangle^{\minus{}1}$

which is true if and only if

    $ |\langle x,y \rangle|^2 \leq \langle x,x \rangle \cdot \langle y,y \rangle$

or equivalently:

    $ \big| \langle x,y \rangle \big| \leq \left\|x\right\| \left\|y\right\|,$

which is the Cauchy–Schwarz inequality.

4) Concavity , convexity means the graph of the function is concave or convex, e.g sine curve is convex from 0 to pi whereas the parabola x^2 = y is concave. If you have a couple of points lying on such curves , then you can connect them and draw triangles or quadrilaterals and prove many (in)equalities , more generally called the Jenson's inequality.

[quote="Arnav...!!!"]
4)convexity

Q)proove
(1)  a+b+c<=a*a+b*b+c*c    abc are real

(2)  (IMO 95/2) a; b; c are positive reals with abc = 1. Prove that
       1/a^3(b + c) + 1/b^3(c + a) + 1/c^3(a + b)>=3/2

please give detailed solutions...........[/quote]

A curve is said to be convex in a given interval if it lies below any of its tangent lines and concave if it lies above any of its tangent lines. An exercise could be to prove that the double derivative of a concave function is always positive and that for a convex function is negative.

A very useful inequality in this context is Jensen's inequality, which is useful in innumerable places.

You can wiki it to find the exact form but here is how i know it.

For a [b]concave[/b] function, $ f$ in a given interval if a,b,c be any points in this interval,
$ f(\frac {a \plus{} b \plus{} c}{3}) \le \frac {f(a) \plus{} f(b) \plus{} f(c)}{3}$

For a [b]convex[/b] function, $ f$ in a given interval if a,b,c be any points in this interval,
$ f(\frac {a \plus{} b \plus{} c}{3}) \ge \frac {f(a) \plus{} f(b) \plus{} f(c)}{3}$

I can give you a very elementary proof (almost 9th stanadard) of the above inequalities, which i doubt is available on the net, as it is original(?) due to my teacher

You can try the imo problem with CS (Cauchy-Schwarz) inequality, it is infact an easy problem, once you apply CS inequality.

well thanks to all of u...........everyone explaind very well ..........but keep telling solutions for the probl left.........

@hell_ever............please tell the proof......... :P

well not clear still wid cauchy schwarz........ :huh:  :oops:

the second problem can also be proved using titu's lemma  :roll:

We make the substitution$ , , .$ Then

\begin{align*}
    \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} &= \frac{x^3}{xyz(1/y+1/z)} + \frac{y^3}{xyz(1/z+1/x)} + \frac{z^3}{xyz(1/x+1/z)} \\
    &= \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} .
    \end{align*}

We make the substitution$ , , .$
Then
    $ \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{2} \left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) .$

By the Power Mean Inequality$ ,$
By the Power Mean Inequality,
    $ \frac{1}{2}\left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) \ge \frac{1}{2}(x+y+z) .$
    $ \frac{1}{2}(x+y+z) \ge \frac{3}{2}xyz = \frac{3}{2},$

I can give a hint to the proof, dont you want to try yourself? If you dont get it even then, i will give the whole proof.

[hide="Hint 1"]What does $ \frac {a\plus{}b\plus{}c}{3}$, imply geometrically?[/hide]

[hide="hint 2"]Think co-ordinate geometry![/hide]

It is not very difficult.  :)

hey anyone having some good material on congruences...
I am unable to understand.... Kindly help.... Please write in few words to explain congruences and their basic functions and their application... :?:

@ rt_08
Your soultion can be made more elegant by means of the Cauchy Schwarz inequality
Write x={x(y+z)^1/2}/(y+z)^1/2 and similarly for y and z and then apply cauchy's inequality to get the result.

i know that aravind!
thanks still!
i just wanted to know whether i was getting the correct solution or not