inequalities calculus derivative function ratio geometry

1/Let $ a_1,a_2,a_3,b_1,b_2,b_3$ be non-negative numbers satisfying $ a_1\plus{}b_1\equal{}a_2\plus{}b_2\equal{}a_3\plus{}b_3\equal{}k$
Prove that   $ a_1\cdot b_2\plus{}a_2\cdot b_3\plus{}a_3\cdot b_1\le k^2$
2/Prove that $ x^2\plus{}y^2\plus{}z^2\le xyz\plus{}2$ where $ x,y,z\in [0,1]$

I'm not sure if this is valid

[hide="Question 2"]
Let $ f(x,y,z) \equal{} 2\plus{}xyz \minus{}x^2\minus{}y^2\minus{}z^2$

Since $ f$ is concave in each variable (second derivative strictly negative) the minimum occurs at extreme points. Checking these points yeilds the solution[/hide]

[quote="SimonM"]I'm not sure if this is valid

[hide="Question 2"]
Let $ f(x,y,z) = 2 + xyz - x^2 - y^2 - z^2$

Since $ f$ is concave in each variable (second derivative strictly negative) the minimum occurs at extreme points. Checking these points yeilds the solution[/hide][/quote]

I believe the answer is easier when you refer to normal $ f(x)$ while keeping $ yz$ fixed.
So, if we have $ f: <0,1>\rightarrow \mathbb b{R} f(x)=-x^{2}+x*yz+(2-y^{2}-z^{2})$ and
$ f'(0)=yz>0 \ f''(x)=-2$ thus on the interval $ <0,1>$ the function is rising up to some point (maybe it is rising on the whole interval) and then it is strictly decreasing, so we can work only on $ f(0)$ or $ f(1)$.

Another approach involves some computations (still it is similar):
Assume $ x\geq y$ Then defining $ f(x,y,z)=2+xyz-x^{2}-y^{2}-z^{2}$ observe that $ f(x,y,z)-f(x+k,y-k,z)\geq 0$ thus we can put $ 0$ or $ 1$ for one of the variables.

I believe that the inequalities $ 2-y^{2}-z^{2}\geq 0$ and $ 1+yz-y^{2}-z^{2}\geq 0$ are really obvious :wink:

[quote="Kyoshiro"]
2/Prove that $ x^2 \plus{} y^2 \plus{} z^2\le xyz \plus{} 2$ where $ x,y,z\in [0,1]$[/quote]
[hide]Here's my solution(a very slick solution I think):
From the restriction we have: (1-x^2)(1-y^2)>=0, which is: 1+(xy)^2>=x^2+y^2(1)
On the other hand we also have: 1>=z^2(2)
Adding (1) and (2) we get: 2+(xy)^2>=x^2+y^2+z^2
Therefore, it is sufficient to prove that: 2+xyz>=2+(xy)^2, which is equivalent to: z>=xy (*)
To prove this inequality,W.L.O.G, assume that z=max(x,y,z).
So, z>=x>=xy(since 0=<y=<1)
We can conclude that (*) is true(Q.E.D)
By the way, #1 is a very old inequality. If I recalled correctly, it was a problem in IMO Shortlist(I'm not too sure though).[/hide]

Hmm... I have a geometric proof of 1.

[hide]Consider an equilateral triangle with side length k. Consider points M, N, P on AB, BC, and CA, respecitvely, such that MA = b1, NB = b2, PC = b3. Then, AP = a3, NC = a2 and MB = a1. Consider the ratio of areas [AMP]/[ABC], [MBN]/[ABC], and [CNP]/[ABC]. These ratios are a3b1/k^2, a1b2/k^2, a2b3/k^2. We know that the sum of these ratio of areas is obviously <= 1, since [AMP]+[MBN]+[CNP] is obviously <= [ABC]. Using that, we get the desired inequality.[/hide]

[quote="Kyoshiro"]
2/Prove that $ x^2 \plus{} y^2 \plus{} z^2\le xyz \plus{} 2$ where $ x,y,z\in [0,1]$[/quote]

By Permutation Inequality we have:
$ x^2 \plus{} y^2 \plus{} z^2\leq 1\plus{}x^2y^2\plus{}z^2 \leq 2\plus{}x^2y^2z^2 \leq 2\plus{}xyz$

Another proof of [hide="#1"]
Put $ a_1 \equal{} k \minus{} b_1,a_2 \equal{} k \minus{} b_2,a_3 \equal{} k \minus{} b_3$ in the inequality to obtain the equivalent:
$ k(b_1 \plus{} b_2 \plus{} b_3) \minus{} (b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le k^2$
So we need to prove
$ f(k)\equal{}k^2\minus{}k(b_1\plus{}b_2\plus{}b_3)\plus{}(b_1b_2\plus{}b_2b_3\plus{}b_3b_1)\ge 0$
We have $ \Delta_f \equal{} b_1^2 \plus{} b_2^2 \plus{} b_3^2 \minus{} 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le 0$
so that $ f(k)\ge 0$ as desired [/hide]

[quote="Kyoshiro"]Another proof
We have $ \Delta_f \equal{} b_1^2 \plus{} b_2^2 \plus{} b_3^2 \minus{} 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le 0$
so that $ f(k)\ge 0$ as desired [/hide][/quote]

I don't know how to prove $ \Delta_f \equal{} b_1^2 \plus{} b_2^2 \plus{} b_3^2 \minus{} 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le 0$
What I think is that if $ b_3 \equal{} 0$ or approaching $ 0$ then $ \Delta_f \geq 0$  :)

[quote="Kyoshiro"]Another proof of [hide="#1"]
Put $ a_1 \equal{} k \minus{} b_1,a_2 \equal{} k \minus{} b_2,a_3 \equal{} k \minus{} b_3$ in the inequality to obtain the equivalent:
$ k(b_1 \plus{} b_2 \plus{} b_3) \minus{} (b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le k^2$
So we need to prove
$ f(k) \equal{} k^2 \minus{} k(b_1 \plus{} b_2 \plus{} b_3) \plus{} (b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\ge 0$
We have $ \Delta_f \equal{} b_1^2 \plus{} b_2^2 \plus{} b_3^2 \minus{} 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le 0$
so that $ f(k)\ge 0$ as desired [/hide][/quote]

$ \boxed{\Delta_f \equal{} b_1^2 \plus{} b_2^2 \plus{} b_3^2 \minus{} 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\le 0}$

I also have doubt if your stated inq. is true, when we select

$ b_1\equal{}3 ,b_2\equal{} 1 ,b_3\equal{}0.1$
$ b_1^2 \plus{} b_2^2 \plus{} b_3^2\equal{}9\plus{}1\plus{}0.01\equal{}10.01$
$ 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)\equal{}2\times(3\plus{}0.1\plus{}0.3)\equal{}6.8$

 :oops:  :P

If $ b_1,b_2,b_3$ are the sides of a triangle we have,
$ b_1+b_2> b_3 \Longrightarrow b_1.b_3+b_2.b_3 >b_3^2$......(i)
Similarly  $ b_1.b_2+b_2.b_3 >b_2^2$.........(ii)
And $ b_1.b_2+b_1.b_3 >b_1^2$........(iii)
(i)+(ii)+(iii)$ \Longrightarrow b_1^2+b_2^2+b_3^2<2(b_1b_2 + b_2b_3 + b_3b_1)$
But we do not have it given that $ b_1,b_2,b_3$ are necessarily the sides of a triangle.

Oh pythag011's proof is true and very nice.
 [quote="Kyoshiro"] :oops:   sorry every one.
The inequality is true if and only if  are the sides of a triangle.
I forgot it.  :blush: [/quote]
The inequality here is $ b_1^2 \plus{} b_2^2 \plus{} b_3^2 < 2(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)$
By the way, anyone has another solution to #1?

[quote="ghjk"][quote="pythag011"][quote="Kyoshiro"]:oops:  sorry every one.
The inequality is true if and only if $ b_1,b_2,b_3$ are the sides of a triangle.
I forgot it.   :blush:[/quote]

The inequality is always true; see my proof. It did not assume that b_1, b_2, b_3 are the sides of a triangle.

Also: Let b1 = b2 = 0. b3 = k. Then it is true... but b1, b2, b3 are not sides of a triangle.[/quote]
Your example doesn't work since b_1+b_2=0, which is wrong! I'll check your proof since I think it has some "flaws" in it. :P[/quote]

Isn't it supposed to be a2 + b2 = k...?

[quote="pythag011"][/quote][quote="ghjk"][quote="pythag011"][quote="Kyoshiro"]:oops:  sorry every one.
The inequality is true if and only if $ b_1,b_2,b_3$ are the sides of a triangle.
I forgot it.   :blush:[/quote]

The inequality is always true; see my proof. It did not assume that b_1, b_2, b_3 are the sides of a triangle.

Also: Let b1 = b2 = 0. b3 = k. Then it is true... but b1, b2, b3 are not sides of a triangle.[/quote]
Your example doesn't work since b_1+b_2=0, which is wrong! I'll check your proof since I think it has some "flaws" in it. :P[/quote][quote="pythag011"]

Isn't it supposed to be a2 + b2 = k...?[/quote]
Oopps! My bad! Your counterexample is right!Sorry for my carelessness. :oops:


see the picture blow.the system is static.find T,f(K),f(k'),and disscuss about M1,M2.

Im not sure it's possible unless you can confirm more details.. eg is there any friction, is it a right triangle,

I think that f(k) and f(k') are the friction (coefficients?) we're supposed to find. It would be nice to know the ratio of the sides and/or the angles of the triangle. I alsow wonder what are m(s) and m(s2)...

suppose we know all the angles and any thing else that we need.

It is impossible to find $T$, $f_k$ and $f_k'$ exactly, there's not enough data. However, if you knew at least one of these three you could find the other two...


How would u do this problem?

A sun is cut from construction paper by first cutting a regular octagon, and then cutting eight equilateral triangles from the octagon, each of which has a side which is also a side of the octagon. What is the measure, in degrees, of each of the angles at the tips of the suns rays?

What is the measure of the interior angle?

$180(8-2)/8=135 degrees$

2 equilateral triangles share this angle, and however many degrees are left is our answer. Each equilateral triangle has angle equal to 60 degrees:

$135-2(60)=15 degrees$


geometry
3D geometry
algebra
polynomial
Vieta
quadratics
number theory unsolved

Prove that if x,y and (x^2 + y^2 +6)/xy are natural numbers thefore ((x^2 + y^2 +6)/xy) is cube of natural number.
 :moose:

That integer is actually equal to 8.

Suppose we had $ x^2\plus{}y^2\plus{}6\equal{}kxy$ for $ k\neq 8$. Consider such a solution with least $ x\plus{}y$. Then $ x$ divides $ y^2\plus{}6$. Letting $ a\equal{}\frac{y^2\plus{}6}{x}$, we have $ a$ a positive integer and $ a^2\plus{}y^2\plus{}6\equal{}kay$ as well. So we must have $ a\plus{}y\geq x\plus{}y$, that is, $ a\geq x$. Doing a similar thing for $ y$, we have
\[ y^2\plus{}6\geq x^2,x^2\plus{}6\geq y^2\]
WLOG let $ x\geq y$. As $ y^2\plus{}6$ cannot equal $ x^2$ (fails modulo 4, for example), we must have $ y^2\plus{}6\geq x(x\plus{}1)$. Suppose $ x\neq y$. Then $ y\leq x\minus{}1$ so
\[ x^2\plus{}x\leq x^2\minus{}2x\plus{}7\Rightarrow 3x\leq 7\Rightarrow x\leq 2\]
The only possibility here is $ x\equal{}2,y\equal{}1$ which is not a solution.

So $ x\equal{}y$ giving $ k\equal{}2\plus{}\frac{6}{y^2}$. Since $ y^2$ divides 6, $ y$ must equal 1 and so $ k\equal{}8$, a contradiction.

So the only possible $ k$ is 8.

[quote="scorpius119"]So we must have $ a \plus{} y\geq x \plus{} y$, that is, $ a\geq x$. Doing a similar thing for $ y$, we have
\[ y^2 \plus{} 6\geq x^2,x^2 \plus{} 6\geq y^2
\][/quote]

Could you explain it? Thanks

 :blush:

Because $ \frac{y^2\plus{}6}{x}\equal{}a\geq x$, we get the first inequality. Switching $ x$ and $ y$ gives the second.

If you're asking about how I got $ a\plus{}y\geq x\plus{}y$, that follows from my choice of $ x,y$: they were chosen to have smallest value of $ x\plus{}y$ among all ordered pairs of positive integers $ (x,y)$ such that $ \frac{x^2\plus{}y^2\plus{}6}{xy}$ was an integer other than 8 (assuming one existed).

But $ a$ is a positive integer with $ \frac{a^2\plus{}y^2\plus{}6}{ay}\equal{}\frac{x^2\plus{}y^2\plus{}6}{xy}$, so $ \frac{a^2\plus{}y^2\plus{}6}{ay}$ is an integer other than 8.

The technique scorpius uses is known as "root flipping," "root jumping," or "Vieta jumping."  Given a particular pair of integers $ x, y$ such that $ x^2 \plus{} y^2 \plus{} 6 \equal{} kxy$, we consider the equation as a quadratic polynomial in $ x$ for fixed $ y, k$ and find the other root; that gives us another solution.  This is a useful technique that can be applied to a lot of quadratic Diophantines.  :)

Oh I see  :D
That's cool, thanks a lot  :)


How many ways are there to fill a 4 x 4 chessboard with non negative integers such that the sum of numbers in each row and each column does not exceed 3?

I'm sure that it has been discussed before.

sorry, but i cannot find it. can you give me the link please? :blush:


inequalities
triangle inequality
absolute value

I have a test on inequality tomorrow so helping myself out by doing a short review and also showing what the inequality is to anyone who is not familiar with this.

This is just plain inequality. Not like trivial or any other ones.

Just plain ones.

[b]Inequality[/b]-Expression using any of the followings:
\[>, <, \geq, \leq, \]

Example:
\[x<5\], \[q>8\], \[z \geq {50}\]. \[ p \leq {27}\]

Pretty simple, isn't it?

But inequality, which is a concept that everyone say, "hey, I know this!" was the one of the round one COP question that several people missed. Of course, that was the [b]Triangle Inequality[/b]which isn't exactly same to what I'm talking now but related. I hope I can tell more about it later.

Problems:
1. Find the value of x.
\[2x<5\]
\[x<5/2\]
So, x is any real number less than 5/2.
2. Find the value of z.
\[5z+2>7\]
\[5z>5\]
\[z>1\]
So, z is any real number greater than 1.
3. Find the value of o.
\[7o+2\geq74-2o\]
\[9o\geq72\]
\[o\geq8\]
So, o is any real number greater or equal to 8.
4. Find the value of p.
\[5p+25\leq10p+35\]
\[-5p\leq10\]
\[p\geq-2\]

Please note that when in division or multiplication (which are basically same, remember that dividing 2 is equal to multiplying by 1/2?) using negative number will flip the inequality's direction.

[b]Using Absolute Values[/b]

\[|x| < 5\]
\[x<5\] and \[x>-5\]
So, any number between 5 and -5.

\[|x-10|\geq 3\]
\[x-10\geq 3\] and \[x-10\leq -3\]
\[x\geq 13\] and \[x\leq 7\]
So, any number less than 7 or greater than 13. Notice that the [b] interval [/b] of the solution is not same as the upper example.

[b]Fractions in Inequality[/b]

\[-4<\frac {2x-3}{3}<4\]
\[-4<\frac {2x-3}{3}\] and
\[\frac{2x-3}{3}<4\]
\[-12<2x-3\]
and \[2x-3<12\]
\[-15<2x\]
and \[2x<15\]
\[x<-15/2\]
and \[x<15/2\]

At last, the absolute value with fractions!

\[|\frac {x-3}{2}|\geq{5}\]
\[\frac {x-3}{2}\geq{5}\]
and \[\frac {x-3}{2}\leq{-5}\]
\[x-3\geq{10}\]
and \[x-3\leq{-10}\]
\[x\geq{13}\]
and \[x\leq{-7}\]

That's our final answer.

That's about it. I don't know how it would help anyone but oh well.
Better than nothing, I hope. ;)

thx for the review i remember covering it awhile ago in math class.  I will be sure to remember it for any problem that can be solved by inequalities


function
induction
real analysis
real analysis unsolved

(This problem on metrics is taken from one method developed for the comparison of DNA segments.) Let A be a finite set of objects which we call an alphabet (in practice, the 20-letter amino acid alphabet of proteins). Let d(a, b) be a distance function on A for a, b ∈ A (in practice, the cost of a mutation from a to b). Let g(a) be a positive-valued function on A for a ∈ A (in practice, the positive cost of inserting or deleting the letter a). For two finite sequences a = (a1, a2, · · · , am) and b = (b1, b2, · · · , bn), not necessarily of the same length, define D(a, b) by induction on m and n, as follows.
D(a, b) = 0 if m = n = 0,
D(a, b) = g(b1) + g(b2) + · · · + g(bn) if m = 0 and n > 0,
D(a, b) = g(a1) + g(a2) + · · · + g(am) if m > 0 and n = 0,
D(a, b)
= min (D(a′, b) + g (am) , D(a′, b′) + d (am, bn) , D(a, b′) + g(bn))
if m > 0 and n > 0,
where a′ = (a1, a2, · · · , am-1) and b′ = (b1, b2, · · · , bn-1). Verify that D(·, ·) is a metric on the space of all finite sequences of letters of the alphabet A.

Interesting!!

First clearly that $D(a,b)\geq 0$ and $D(a,b)=D(b,a)$.
If $a=b$ then
\[ D(a,b)\leq D(a',b')+d(a_m,b_m)\leq \cdots \leq \sum_1^m d(a_i,b_i)=0  \]
So $D(a,b)=0$. Conversely let $D(a,b)=0$. Since $g(a_m), g(b_n)$ are positive then
\[ 0=D(a,b)=D(a',b')+d(a_m,b_n)  \]
This implies that $d(a_m,b_n)=0$ and $D(a',b')=0$. Using the similar argument to $D(a',b')$ we have $d(a_{m-1},b_{n-1})=0$. Using this argument repeatedly, we conclude that we must have $m=n$ otherwise if $m<n$ then $D(a,b)=g(1)+\cdots +g(n-m)>0$. Then $d(a_i,b_i)=0$. So $a_i=b_i$ for every $i$, i.e., $a=b$.

I don't have time for the triangle inequality......... :D


Let f=mx+ny, where m,m are fixed positive integers, and x,y are positive numbers such that xy is a fixed constant. Find the minimum value of f.

Let $ xy=k$. $ f=mx+ny\ge 2\sqrt{(mx)(ny)}=\boxed{2\sqrt{mnk}}$ by the AM-GM inequality. Equality occurs when $ mx=ny$, $ xy=k\implies (x,y)=\left( \sqrt{\frac{kn}{m}},\sqrt{\frac{km}{n}}\right)$.


If $a$ and $b$ are digits for which

\begin{tabular}{ccc} & 2 & a\\ $\times$ & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{tabular}

Then $a+b =$

A. 3
B. 4
C. 7
D. 9
E. 12

[hide]
If a*3=9, then a=3
920/23=40, so b=4
3+4=7
So the answer is C[/hide]

If $ a$ and $ b$ are digits for which

\begin{tabular}{ccc} & 2 & a \\
$\times$ & b & 3 \\
\hline & 6 & 9 \\
9 & 2 \\
\hline 9 & 8 & 9\end{tabular}

Then $ a + b =$

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 12$

A has to be 3 because 9/3=3
B has to be 4 because 92/23=4
3+4=7 so therefore the answer is C
 :)

Hmm the $\LaTeX$ here seems to be a bit off.  Can a moderator of this forum adjust it to the following?  Thanks!

[quote=worthawholebean]If $ a$ and $ b$ are digits for which

$$\begin{array}{ccc} & 2 & a \\
\times & b & 3 \\
\hline & 6 & 9 \\
9 & 2 & \\
\hline 9 & 8 & 9 \end{array}$$

Then $ a + b =$

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 12$[/quote]


The first spinner is divided into 4 equal sections, and the second is divided into 3 equal sections. When these 2 fair spinners are each spun once, what is the probability that the product of the 2 numbers to which the arrows point is odd? Express your answer as a common fraction.

[asy]size(5cm,5cm);

draw(Circle((0,0),1));

draw(Circle((-3,0),1));

draw((0,0)--(0,1));

draw((0,0)--(-0.9,-0.47));

draw((0,0)--(0.9,-0.47));

draw((-2,0)--(-4,0));

draw((-3,1)--(-3,-1));

label("$4$",(-0.5,0.3));

label("$8$",(0.5,0.3));

label("$3$",(0,-0.5));

label("$7$",(-2.6,-0.4));

label("$6$",(-2.6,0.4));

label("$1$",(-3.4,0.4));

label("$5$",(-3.4,-0.4));

draw((0,0)--(-0.6,-0.6),Arrow);

draw((-3,0)--(-2.2,0.2),Arrow);[/asy]

Both numbers must be odd for an odd product.

For the first spinner, the probability of spinning an odd number is $ \frac{3}{4}$.

For the second spinner, the probability of spinning an odd number is $ \frac{1}{3}$.

Probability of getting odd product is $ \frac{3}{4}*\frac{1}{3}\equal{}\boxed{\frac{1}{4}}$.


[hide]
if the sequnce is 1, 7, x then 8+x=10, and x=2 so the sequence is 1, 7, 2.  The 99th term is 2 because 99/3=33 and so the answer is $\boxed{2}$[/hide]

let sequence be 1,7,a then 8+a=10 and a=2
therefore the 99th term is 99/3=33 so ans is 2[hide][/hide]

[quote="236factorial"]What is the 99th term of this sequence, in which each three consecutive terms add up to 10:

1, 7, ...[/quote]
[hide="clicky here"]the pattern is 1,7,2,1,7,2......and so on so the 99th term willl be 99/3 = no remainder so it will be 2.

[size=150]2[/size][/hide]

[hide]
It keeps on going 1,7,2,1,7,2,...
So, you do 99/3=33
So, the answer is $\boxed{2}$[/hide]

$\frac{99}{3}=33$

dar natijeh the answer is $2$.


The sum of the areas of all triangles whose vertices are also vertices of a $ 1 * 1 * 1$ cube is $ m+\sqrt{n}+\sqrt{p}$, where $ m, n$, and $ p$ are integers.  Find $ m+n+p$.

Well, what I did was fairly simple.  I just figured out all of the cases, and did this by looking at the dimensions of the possible triangles, noting that triangles with three equal sides are in the same case.

I said that with a $ 1*1*\sqrt{2}$ triangle, you can choose the vertex on where you start in $ 8$ ways, the second vertex in $ 3$ ways (assuming the edge connecting the first and second vertexes that we choose has a length of $ 1$, there are only $ 3$ such paths).  And then, to stay on the same face and so that the created edge will be of length $ \sqrt{2}$, there are $ 2$ ways.  For the last choice, just go back to the original vertex, so only $ 1$ way.  That's a total of $ 8*3*2*1=48$ ways.  But the total number of possible triangles is only $ 56$, and the number 48 cannot be right...what am I counting wrong?

 Is it illogical to say that there are a certain number of ways to start at a given vertex, and then count the number of ways to construct a given triangle with the given dimensions...?  If it's not totally wrong, where am I overcounting?  This happened with triangles of other dimensions too.

Here I think is an example of where you over count.
Imagine one face of the cube,
AB
CD
In your method you count C-D-A and C-A-D as different triangles.

Here's how I would do it:Per face, there are 4 different 1-1-sqrt2 triangles.  Multipying by 6 faces we get 24 triangles.
There's also something here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87256[/url]


modular arithmetic
number theory theorems
number theory

It is not true, for example $ \phi(8)\equal{}\phi(5)\equal{}\phi(12),...$

[quote="Murat Aygen"]Is the following lemma valid?

If $ \varphi(n) \equal{} \varphi(m)$ and $ 2\parallel{}\varphi(n)$, then either $ n \equal{} 7$, $ m \equal{} 9$ or $ n \equal{} m$.[/quote]
$ 2\parallel{}\varphi(n)$ if and only if $ n\equal{}2p^k,p^k$ where p is prime in the form : $ 4a\plus{}3$
Therefore we only need to consider equation :
$ q^{k\minus{}1}(q\minus{}1)\equal{}p^{m\minus{}1}(p\minus{}1)$
Case 1 $ p\equal{}q$ then $ m\equal{}n$ or $ (m,n)\equal{}(p^k,2p^k),(2p^k,p^k)$ where $ p\equiv 3(\mod 4)$
Case 2 : $ p>q$
$ p^{m\minus{}1}|q\minus{}1$ ,it means that exist an integer t such that
$ q\minus{}1\equal{}tp^{m\minus{}1},p\minus{}1\equal{}tq^{k\minus{}1}$
Because $ p>q$ so $ m\equal{}1$
So $ p\minus{}1\equal{}q^{k\minus{}1}(q\minus{}1) \Leftrightarrow p\equal{}q^{k\minus{}1}(q\minus{}1)\plus{}1$
It is an open problem such that :Find $ q\in P,k\in N,k>1$ such that $ q^{k\minus{}1}(q\minus{}1)\plus{}1$ is a prime.

[quote="Rust"]It is not true, for example $ \phi(8) \equal{} \phi(5) \equal{} \phi(12),...$[/quote]

Your counterexample is invalid.  $ 2 \parallel{} a$ means that $ a$ is divisible by $ 2$ but not by $ 4$.  In general, the notation $ p^k \parallel{} a$ means that $ p^k$ divides $ a$ but any larger power of $ p$ does NOT divide $ a$.

I think we should start with $ 2 \parallel{} \phi(n)$.  What does that mean?  If $ p_1,p_2, \ldots$ are the prime factors of $ n$, we have $ \phi(n) \equal{} n \cdot\frac {p_1 \minus{} 1}{p_1} \cdot \frac {p_2 \minus{} 1}{p_2} \cdots$.

If a prime factor of $ n$ is odd, then $ \phi(n)$ has to be even, and if we have $ 2$ odd prime factors, then $ \phi(n)$ is a multiple of $ 4$.  So we have $ n$ is an odd prime (because it factors into only an odd prime) but clearly both $ n$ and $ m$ cannot be odd primes because this does not fulfill $ \phi(n) \equal{} n \minus{} 1 \neq m \minus{} 1 \equal{} \phi(m)$ unless $ n \equal{} m$.

Now lets consider if $ n$ has even prime factors - that is, if $ 2$ is a factor.  If $ n \equal{} 2p$ where $ p\neq 2$ is prime, then the totient is $ \phi(n) \equal{} 2p(1/2)((p \minus{} 1)/p) \equal{} p \minus{} 1$, which fulfills the requirement $ 2 \parallel{} p \minus{} 1$ iff $ p \equal{} 4k \plus{} 3$ for some $ k\in\mathbb{Z*}$.

Now what about $ n \equal{} 2^k p$ for $ k \ge 2$?  Then $ \phi(n) \equal{} 2^{k \minus{} 1} (p \minus{} 1)$ which clearly fails $ 2 \parallel{} \phi(n)$.

Last case: if $ n \equal{} 2^k$, then we have $ \phi(n) \equal{} 2^{k \minus{} 1}$ which satisfies $ 2 \parallel{} \phi (n)$ iff $ \phi(n) \equal{} 2 \implies k \equal{} 2 \implies n \equal{} 4$.

In summary: $ 2 \parallel{} \phi(n)$ if (1) $ n$ is an odd prime of the form $ 4k \plus{} 3$, (2) $ n \equal{} 2p$ for some $ p$ prime such that $ p \equal{} 4k \plus{} 3$ or (3) $ n \equal{} 4$.

And your lemma is invalid since $ \phi(3) \equal{} \phi(4) \equal{} 2$.

EDIT: I just realized I ignored the cases $ n\equal{}p^k$ and $ n\equal{}2p^k$, which are both perfectly possible as long as $ p\equal{}4k\plus{}3$ is prime.

[quote="rofler"]Lol, it is not $ 2p^k$ it is $ 2*m$ for some odd m.[/quote]

If this is an answer to TTsphn's statement that $ n \equal{} p^k,2p^k$, then your response is wrong (the only thing he missed is $ n\equal{}4$, which is really not important).

The case $ n\equal{}4$ is pretty insignificant to the overall problem, but we should also note that $ \phi(p^k) \equal{} \phi(2p^k)$, and if $ p \equiv 3 \pmod 4$ is prime, this satisfies the condition $ 2 \parallel{} \phi(p^k)$.  So the proposed lemma misses the case $ n\equal{}2m$ or $ 2n\equal{}m$.

TTsphn, i think you reversed the divisibility in your case $ p>q$.  Shouldn't we have $ q\minus{}1 | p^{m\minus{}1}$ so $ p^{m\minus{}1}\equal{}t(q\minus{}1)$ and $ p\minus{}1 | q^{k\minus{}1} \implies q^{k\minus{}1} \equal{} t(p\minus{}1)$.  Hence $ t|p^{m\minus{}1}$ and $ t|q^{k\minus{}1}$ which implies that $ t\equal{}p^x\equal{}q^y \implies p\equal{}q, x\equal{}y$ (prime factorizations are unique) if $ k,m \neq 1$.  So we must have that either $ m$ or $ k$ is $ 1$ and thus $ t\equal{}1$.

Two cases:
$ m\equal{}1$: then $ p\equal{}q^{k\minus{}1}(q\minus{}1)\plus{}1$

$ k\equal{}1$: then $ q\equal{}p^{m\minus{}1}(p\minus{}1)\plus{}1$ which is impossible if $ p>q$.

looks like the mistake didn't change your conclusion, TTsphn.

Anyway, with a little calculator fiddling:
$ q\equal{}3, k\equal{}3 \implies p\equal{}19$ -> $ \phi(27) \equal{} \phi(54) \equal{} \phi(19) \equal{} \phi(38)$
$ q\equal{}3, k\equal{}5 \implies p\equal{}163$ -> $ \phi(243) \equal{} \phi(486) \equal{} \phi(163) \equal{} \phi(326)$
$ q\equal{}3, k\equal{}6 \implies p\equal{}487$ -> $ \phi(729) \equal{} \phi(1458) \equal{} \phi(487) \equal{} \phi(974)$
$ q\equal{}3, k\equal{}7 \implies p\equal{}1459$ -> $ \phi(2187) \equal{} \phi(4374) \equal{} \phi(1459) \equal{} \phi(2918)$
$ q\equal{}5, k\equal{}3 \implies p\equal{}101$ -> $ \phi(125) \equal{} \phi(250) \equal{} \phi(101) \equal{} \phi(202)$
etc.  So there appear to be many (perhaps infinitely many?) $ q$ and $ k$ pairs which give $ p$ prime.


I mean prove by the definition of the limit of the sequence, not use any konwn facts.


Find the sum of the even integers from 2 to 220 inclusive.
$ (2\plus{}4\plus{}6\plus{}...\plus{}220)$

Alternatively, the sum of an arithmetic sequence is $ n(a\plus{}z)\div2$, where $ n$ is the number of terms, $ a$ is the first, and $ z$ is the last.

We have $ 110(2\plus{}220) \div 2 \implies 55(222) \implies \boxed{12210}$.


Prove that if $p$ and $p+2$ are prime integers greater than 3, then 6 is a factor of $p+1$.

[b]Solution.[/b]  Clearly, all primes are of the form $6k\pm 1$.  Since $p+2$ is also a prime, $p=6k-1\implies 6|p+1$, as required.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1973Problem3

Vo Duc Dien


trigonometry
real analysis
real analysis unsolved

I have just started studying PDE's.

the problem is a heat equation of the form

u_t = alpha^2 * u_xx

The obejct is insulated ath the lateral boundary so the bc's are both zero.

However, the IC is

u(x,0)=x-x^2

The book asks me to solve the problem using the fourier sine transform.

I just don't quite understand what im supposed to do with teh x-x^2 part, and im not completely comfortable with the fourier sine transform yet either.

thanks for any help

i figured it out, it was actually really easy.

  Multiply both sides of the ic by sin(mnPI) and use orthogonality by the fourier series and its really easy to solve for the
coeffecients  that way.

How do you guys do the notation via tags?


function
integration
inequalities
Functional Analysis
real analysis
real analysis unsolved

Suppose $ r>1$ and $ a_n$ is a nonnegative sequence such that $ \sum_na_n\equal{}\infty.$ Prove that one can choose a nonnegative sequence $ b_n$ such that $ \sum_nb_na_n\equal{}\infty$ but $ \sum_nb_n^ra_n<\infty.$

$ a_n \equal{} n$
$ b_n \equal{} \frac {1}{n^2}$

this only comes to mind so quickly because i was thinking about a similar problem

find $ f\in L^1$ but $ f \notin L^p, p\ne 1$

oh oh i read it better now, let me think something less easy up

There's an implied $ \forall$ in there: given any divergent sequence $ a_n,$ one may choose $ b_n$ (depending on $ a_n$) such that ...

One example isn't enough.

anyone have any ideas? I've solved all but one case.

consider $ [0,\infty ) \equal{} \cup_{n \in N}X_n$

where $ X_n \equal{} [\sum_{k\equal{}1}^{n\minus{}1} a_k , \sum_{k\equal{}1}^n a_k )$

then we are looking for a function $ f(x) \equal{} \sum_{n\in N} \lambda_n 1_{X_n} (x)$

with $ f \in L^r , f \notin L^1$

i keep trying stuff but running into problems with "bad" $ a_n$'s

Ok I solved it in the shower.  :maybe:

I will wait for kenn4000 and KM to show their proofs.

All right, here's [hide="my proof."]
With the convention that $ n_0\equal{}0,$ choose $ n_k$ so that $ \sum_{n\equal{}n_{k\minus{}1}\plus{}1}^{n_k}a_n\ge 1.$ We can do this because $ \sum a_n\equal{}\infty.$

Define $ c_k\equal{}\sum_{n\equal{}n_{k\minus{}1}\plus{}1}^{n_k}a_n.$ We know that $ c_k\ge 1.$

Now, for $ n_{k\minus{}1}<n\le n_k,$ define $ b_n\equal{}\frac1{kc_k}.$

Then $ \sum_{n\equal{}1}^{\infty}b_na_n\equal{}\sum_{k\equal{}1}^{\infty} \frac1{kc_k}\sum_{n\equal{}n_{k\minus{}1}\plus{}1}^{n_k}a_n\equal{}\sum_{k\equal{}1}^{\infty}\frac1k \cdot\frac{c_k}{c_k}\equal{}\sum_{k\equal{}1}^{\infty}\frac1k\equal{}\infty.$

But $ \sum_{n\equal{}1}^{\infty}b_n^ra_n\equal{}\sum_{k\equal{}1}^{\infty} \frac1{k^rc_k^r}\sum_{n\equal{}n_{k\minus{}1}\plus{}1}^{n_k}a_n$

Note that since $ c_k\ge 1$ and $ r>1,$ we have $ \frac1{c_k^r}\le\frac1{c_k}.$

$ \sum_{n\equal{}1}^{\infty}b_n^ra_n\le\sum_{k\equal{}1}^{\infty} \frac1{k^rc_k}\sum_{n\equal{}n_{k\minus{}1}\plus{}1}^{n_k}a_n\equal{} \sum_{k\equal{}1}^{\infty}\frac1{k^r}\frac{c_k}{c_k}\equal{} \sum_{k\equal{}1}^{\infty}\frac1{k^r}<\infty$

Now, how did I come to propose such a problem?

[hide="Motivation."]For a graduate analysis course I'm teaching, I wanted to prove the following. (I think it's appeared elsewhere on this forum in some form.)

Let $ 1<p<\infty$ and $ \frac1p\plus{}\frac1q\equal{}1.$ Suppose $ f$ is a measurable function on some measure space and suppose that

$ \left|\int fg\right|<\infty$ for every $ g\in L^q.$

Prove that $ f\in L^p.$

The problem above amounts to the central part of the argument for the contrapositive of this theorem - if you're going to prove it in a concrete, constructive way. You could also find a way to quote the Uniform Boundedness Principle, but I like the direct construction.

[/hide][/hide]

ah, thank you Kent. thats what i was trying and failing to do.

[quote="Kent Merryfield"]All right, here's [hide="my proof."]
With the convention that $ n_0 \equal{} 0,$ choose $ n_k$ so that $ \sum_{n \equal{} n_{k \minus{} 1} \plus{} 1}^{n_k}a_n\ge 1.$ We can do this because $ \sum a_n \equal{} \infty.$

Define $ c_k \equal{} \sum_{n \equal{} n_{k \minus{} 1} \plus{} 1}^{n_k}a_n.$ We know that $ c_k\ge 1.$

Now, for $ n_{k \minus{} 1} < n\le n_k,$ define $ b_n \equal{} \frac1{kc_k}.$

Then $ \sum_{n \equal{} 1}^{\infty}b_na_n \equal{} \sum_{k \equal{} 1}^{\infty} \frac1{kc_k}\sum_{n \equal{} n_{k \minus{} 1} \plus{} 1}^{n_k}a_n \equal{} \sum_{k \equal{} 1}^{\infty}\frac1k \cdot\frac {c_k}{c_k} \equal{} \sum_{k \equal{} 1}^{\infty}\frac1k \equal{} \infty.$

But $ \sum_{n \equal{} 1}^{\infty}b_n^ra_n \equal{} \sum_{k \equal{} 1}^{\infty} \frac1{k^rc_k^r}\sum_{n \equal{} n_{k \minus{} 1} \plus{} 1}^{n_k}a_n$

Note that since $ c_k\ge 1$ and $ r > 1,$ we have $ \frac1{c_k^r}\le\frac1{c_k}.$

$ \sum_{n \equal{} 1}^{\infty}b_n^ra_n\le\sum_{k \equal{} 1}^{\infty} \frac1{k^rc_k}\sum_{n \equal{} n_{k \minus{} 1} \plus{} 1}^{n_k}a_n \equal{} \sum_{k \equal{} 1}^{\infty}\frac1{k^r}\frac {c_k}{c_k} \equal{} \sum_{k \equal{} 1}^{\infty}\frac1{k^r} < \infty$

Now, how did I come to propose such a problem?

[hide="Motivation."]For a graduate analysis course I'm teaching, I wanted to prove the following. (I think it's appeared elsewhere on this forum in some form.)

Let $ 1 < p < \infty$ and $ \frac1p \plus{} \frac1q \equal{} 1.$ Suppose $ f$ is a measurable function on some measure space and suppose that

$ \left|\int fg\right| < \infty$ for every $ g\in L^q.$

Prove that $ f\in L^p.$

The problem above amounts to the central part of the argument for the contrapositive of this theorem - if you're going to prove it in a concrete, constructive way. You could also find a way to quote the Uniform Boundedness Principle, but I like the direct construction.

[/hide][/hide][/quote]

Oh, dear Kent ,why you sent your solution so soon ? I had this solution  :(

A basically the same, but somewhat cleaner example would be
\[ b_n \equal{} \frac1{a_1 \plus{} \ldots \plus{} a_n} : \equal{} \frac1{s_n} .
\]
This is handled in Rudin's Principles (p.79), for $ r \equal{} 1,2$. The first time around I couldn't prove that $ \sum_{n \geq 1} a_n b_n^r$ converges for $ r > 1$, but it's actually pretty easy to generalize the idea used there. More exactly, the following inequality
\[ \frac {a_n}{s_n^r} \leq K_r \left( \frac1{s_{n \minus{} 1}^{r \minus{} 1}} \minus{} \frac1{s_n^{r \minus{} 1}} \right)
\]
holds for suitable $ K_r > 0$. Applying MVT to $ x \mapsto x^{1 \minus{} r}$ we get our result.

As a side note, a similar statement holds for convergent non-negative series (it's on the same page in Rudin).


probability
function
number theory
relatively prime

you have $ 21$ jars in front of you, each containing $ 21$ balls and balls marked $ 1, 2, 3, 4,\cdots 21$. You are asked to pick out one ball from each container blindfold. Now note down the numbers marked on the $ 21$ balls thus picked out.

What is the probability of the product of numbers of picked out balls being completely divisible by the digit $ 21$ ?

We get a number divisible by 21 if and only if we pick a number divisible by 3 and a number divisible by 7. Consider the probability the event doesn't occur.

Ways to pick balls, none divisible by 3: $ 14^{21}$.
Ways to pick balls, none divisible by 7: $ 18^{21}$.
Ways to pick balls, none divisible by 3 or 7: $ 12^{21}$.

By PIE, the answer is $ 1\minus{}\frac{14^{21}\plus{}18^{21}\minus{}12^{21}}{21^{21}}$. Uh, I'm not supposed to compute that, right?

No, I included that 1 - x part.

That's the actual probability, which makes sense.

The odds are pretty damn close to 100% that you'll get a 21 somewhere in the 21 draws you make, making the number divisible by 21.

Oops!

I posted without thinking, again.

You're completely correct as usual  :lol:

[quote="JBL"][quote="DiscreetFourierTransform"]The odds are pretty - close to 100% that you'll get a 21 somewhere in the 21 draws you make, making the number divisible by 21.[/quote]

This is, in fact, completely wrong.  The odds of drawing a 21 at some point are $ 1\minus{}\left(\frac{20}{21}\right)^{21}\equal{} 1\minus{}\left(1\minus{}\frac{1}{21}\right)^{21}$.  The second term in this expression is very close to $ \frac{1}{e}$, so the odds of drawing a 21 at some point are very close to $ 1\minus{}\frac{1}{e}$, which is about $ 63\%$.  The fact that we get an answer close to 1 has to do with the number of numbers less than 21 which are not relatively prime to it (the [i]totient[/i] of 21), and in particular the fact that it is nonzero.

Generalizations/variations of the original question:

1) Replace 21 with $ p$ for prime $ p$.  Calculate a few values numerically.
2) Replace 21 with $ pq$ for primes $ p$ and $ q$.  Calculate a few values numerically.
3) Replace 21 with $ pqr$ for primes $ p$, $ q$, $ r$.
4) Replace 21 with $ p^{2}$ for prime $ p$.
5) Replace 21 with $ p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ for primes $ p_{i}$ and positive integers $ a_{i}$.
6) Can you describe the limiting behavior (as one or several of the primes involved grow to infinity) in each of these cases?[/quote]

if the digits $ 3$ and $ 7$ are picked in the course, then also the product will be divisible by $ 21$. So what will be the probability in that case ?

MellowMelon's post contains a full solution, so I'm not sure what the purpose of your question is.  I'm sad no one has tried any of mine  :(

[hide="1"]Consider the complement - the probability of getting numbers whose product is not divisible by $ p$. This means all the numbers we pick aren't $ p$. We have to pick $ p$ numbers, and the probability of not getting $ p$ is $ \frac{p\minus{}1}{p}$. So the answer is $ 1\minus{}\left(\frac{p\minus{}1}{p}\right)^{p}$.

As $ p$ goes to infinity, this converges to $ 1\minus{}\frac{1}{e}$.
[/hide]

[hide="2"]
Now we have to use PIE. The number of ways to pick a number not divisible by $ p$ is $ \frac{pq\minus{}q}{pq}\equal{}\frac{p\minus{}1}{p}$, so that doesn't change. We get a similar thing for $ q$. Finally, the number of ways to pick a number not divisible by $ pq$ is also $ \frac{pq\minus{}1}{pq}$. The pattern here suggests a lemma which we will prove:

[b]Lemma[/b]: Let $ S \equal{}\{1,2,\ldots,n\}$. The probability of picking a number from $ S$, if all numbers are equally likely to be selected, that is divisible by $ d$ is $ \frac{1}{d}$ if and only if $ d\mid n$.
[hide="Proof of right-to-left implication"]Assume $ d\mid n$. Now we know there exists an $ m$ such that $ n \equal{} md$. Then the subset of $ S$ containing numbers divisible by $ d$ is $ \{d,2d,3d,\ldots,md\}$, which contains $ m$ elements. So the probability will be $ \frac{m}{n}\equal{}\frac{m}{md}\equal{}\frac{1}{d}$.[/hide][hide="Proof of left-to-right implication"]Now assume the probability is $ \frac{1}{d}$. We know the probability can always be written in the form $ \frac{a}{n}$ for some nonnegative integer $ a\leq n$. So we have $ \frac{1}{d}\equal{}\frac{a}{n}\Rightarrow n \equal{} ad$. Hence $ d\mid n$.[/hide]
Anyways, back to actually using PIE. Since we are picking $ pq$ numbers, we take each fraction to the $ pq$th power. So the final answer is $ 1\minus{}\left(\frac{p\minus{}1}{p}\right)^{pq}\minus{}\left(\frac{q\minus{}1}{q}\right)^{pq }\plus{}\left(\frac{pq\minus{}1}{pq}\right)^{pq}$.

I think if $ p$, $ q$, or both go to infinity, you get $ 1\minus{}\frac{1}{e}$ again in all 3 cases. The only one I'm not absolutely sure about is if both go to infinity.
[/hide]

[hide="3"]
Using similar techniques to the other 2, we get $ 1\minus{}\left(\frac{p\minus{}1}{p}\right)^{pqr}\minus{}\left(\frac{q\minus{}1}{q}\right)^{pqr}\minus{}\left(\frac{r\minus{}1}{r}\right)^{pqr}\plus{}\left(\frac{pq\minus{}1}{pq}\right)^{pqr}\plus{}\left(\frac{pr\minus{}1}{pr}\right)^{pqr}\plus{}\left(\frac{qr\minus{}1}{qr}\right)^{pqr}\minus{}\left(\frac{pqr\minus{}1}{pqr}\right)^{pqr}$.
[/hide]

[hide="4"]
There are two ways to get a product divisible by $ p^{2}$. Either we pick $ p^{2}$, or we pick a two numbers divisible by $ p$. We're still considering the complement, so how not to get a product divisible by $ p^{2}$. Either we pick no numbers divisible by $ p$, or we pick one number divisible by $ p$ (and not by $ p^{2}$). The probability of getting no numbers divisible by $ p$ is $ \left(\frac{p^{2}\minus{}1}{p^{2}}\right)^{p^{2}}$. The probability of getting exactly one number divisible by $ p$ is $ p^{2}\cdot\left(\left(\frac{p^{2}\minus{}1}{p^{2}}\right)^{p^{2}\minus{}1}\cdot\left(\frac{p\minus{}1}{p^{2}}\right)\right)$. The $ p^{2}$ coefficient is there because any of the $ p^{2}$ numbers we pick might be the one divisible by $ p$; it could be written as $ {p^{2}\choose 1}$ to make it more clear why it's there. Anyways, we take 1 minus the sum of these two probabilities to get our answer.
[/hide]

[hide="5?"]
So if you read the answer to 4, you'll realize quickly that the method it uses doesn't generalize well; you would have to use the partition function and large binomial expansions for higher prime powers. So I haven't gotten this problem yet. :(
[/hide]

Sorry for the topic revival. I was wondering if JBL or anyone else had the answer to his 5th question. I still can't get it.


Let cos4A = 1/3   Determine cos^8(A) - sin^8(A)

Try decomposing that eight-power expression as a product of quadratics which you know how to solve.

Fudge sorry I didn't realize that was the cosine of 4A.


The easiest way to explain what I'm looking for is through the code:

[code]2(1 + \sqrt{x}) - \ln(1 + \sqrt{x})^2 + C $ $ $ $ $ $ $ $ $ $ $ $(x \geq 2)$[/code]

Clearly this isn't very efficient. What's the best way to go about doing this?

[url=http://www-h.eng.cam.ac.uk/help/tpl/textprocessing/teTeX/latex/latex2e-html/ltx-241.html]\hspace[/url] is what you're looking for.


LaTeX
algebra
polynomial
function
ratio
factorial
binomial coefficients

[quote="Arvind_sn"]What is the sum of the convergent geometric series $1, 0.5, 0.25, 0.125...$?[/quote]

[hide]$\frac{1}{0.5}=\boxed{2}$

Does this really belong in intermediate?[/hide]

[quote="ehehheehee"](How do you latex summation?)[/quote]

\sum

\sum_{i=0}^{\infty} gives lower and upper bounds on the summation.

[quote="t0rajir0u"][quote="ehehheehee"]Ok, here is another one:

1/2+4/4+9/8+16/16+25/32...

or the summation of K^2/2^k as k ranges from one to infinity (How do you latex summation?)[/quote]

[hide="Solution"] Let $S = \sum_{k=0}^{\infty}\frac{k^{2}}{2^{k}}$.  Then

$2S = \sum_{k=0}^{\infty}\frac{(k+1)^{2}}{2^{k}}$
$2S-S = \sum_{k=0}^{\infty}\frac{2k+1}{2^{k}}$.

Let's sum the linear term.  Let $R = \sum_{k=0}^{\infty}\frac{k}{2^{k}}$.  We can use a similar trick here.

$2R = \sum_{k=0}^{\infty}\frac{k+1}{2^{k}}$
$2R-R = \sum_{k=0}^{\infty}\frac{1}{2^{k}}= 2$

Hence

$2S-S = 2 \cdot 2+2 = \boxed{6}$.  I think.[/hide][/quote]

That's an interesting method! What about $\sum_{k=0}^{\infty}\frac{k^{3}}{3^{k}}$? And $\sum_{k=0}^{\infty}\frac{k^{a}}{a^{k}}$ where $a$ is an arbitrary positive integer?

That method will actually work for $a$ any real number larger than 1.  The "good" polynomials for the top are the binomial coefficients, $k \choose n$ for fixed $n$.  An arbitrary polynomial in $k$ can be expressed as sums of such polynomials, and that makes it all nice.

Hmm, but how would you expand $(k+1)^{a}$ where $a=\sqrt{2}$ for example?  :|

But even with $a\in\mathbb{N}$...ok, the methode works for $a=2,3, 4, ...$ but is there a closed formula? And if so, could we prove it with this method? I find it interesting anyway  :)

[quote="Kurt Gödel"]Hmm, but how would you expand $(k+1)^{a}$ where $a=\sqrt{2}$ for example?  :| [/quote]

[b]Definition:[/b]  ${a \choose k}= \frac{a(a-1)(a-2)...(a-(k-1))}{k!}$ for a non-negative integer $k$ and for any $a$.

We can have $a \in \mathbb{R}$ or even $a \in \mathbb{C}$ if we wish.

[b]Theorem:[/b]  $(x+1)^{a}= \sum_{k=0}^{\infty}{a \choose k}x^{k}$

[quote="JBL"]Even if you didn't happen to know the binomial expansion worked for non-integers (which would be a totally reasonable thing to not know), you could still do a pretty darn good generalization: $\sum_{k = 0}^\infty \frac{P(k)}{a^{k}}$ for any polynomial $P$ and any real $a > 1$.  In fact, any complex number $a$ such that $|a| > 1$.[/quote]

Looks interesting!

But I still don't really get how you could generalize this... If we use your method, t0rajir0u, for $a=3$ we would expand $(k+1)^{3}= k^{3}+3k^{2}+3k+1$ so we would have to examinate $\sum \frac{3k^{2}}{3^{k}}$ and $\sum \frac{3k}{3^{k}}$. Ok, that's fine, but for $a=2006$ for example, this is tricky... :dry: And for $a\in\mathbb{R}$ or $a\in\mathbb{C}$ I'm totally lost  :P Help!

I assume that we could use the regular combinations formula with the Gamma function to achieve the same result.

[quote="t0rajir0u"]We can use your example for $a = \sqrt{2}$.

$(1+k)^{ \sqrt{2}}= k^{0}+\sqrt{2}k^{1}+\frac{ \sqrt{2}( \sqrt{2}-1) }{2!}k^{2}+\frac{ \sqrt{2}( \sqrt{2}-1)( \sqrt{2}-2)}{3!}k^{3}+...$[/quote]

I think I get the binomial stuff. However, I still don't get how to evaluate the sum $\sum \frac{k^{\sqrt{2}}}{\sqrt{2}^{k}}$ for example.

The technique used earlier only works on polynomials, since otherwise the process of repeated subtraction doesn't terminate.  You'd be left with an even harder infinite sum to evaluate  :P

Yes, sorry for misleading you there -- I was looking only at the $a$ in the denominator, not the numerator.

Hehe, no problem  :P

In meantime, I found an interesting document about sums like \[\sum_{k\geq 1}\frac{k^{n}}{m^{k}}\] where $n\in\mathbb{N}\cup\{0\}$ and, indeed, $|m| > 1$. It's right [url=http://www.math.psu.edu/sellersj/p25.pdf]here[/url].

[quote="t0rajir0u"]We can have $a \in \mathbb{R}$ or even $a \in \mathbb{C}$ if we wish.[/quote]

How would you work with it if $a \in \mathbb{C}$?

JBL:

the method i used can be generalized for sum[(p(x)/a^k]

for degree of p, call the summation a n-tuple telescoping summation

then just multiply thru the expanded expression by the common ratio and subtract.  repeat n times.  this gives the answer for any polynomial function over a exponential one for summations.

[quote="t0rajir0u"][quote="lingomaniac88"]How would you work with it if $a \in \mathbb{C}$?[/quote]

It's still just multiplication.  For example,

${3+4i \choose 5}= \frac{ (3+4i)(2+4i)(1+4i)4i(4i-1)}{5!}$[/quote]

Oh.  Didn't realize it was that simple.  Also, I was familiarized with $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, and I just couldn't see doing something like $(3+4i)!$.[url=http://www.google.com/search?hl=en&q=%283%2B4i%29%21]Apparently, you can do something like that.  Who knew?[/url]

Here are the graphs of the factorial function over real and complex. http://mathworld.wolfram.com/Factorial.html
You have to scroll down. :D


In oscillation motion, I know that a=(omega)^2*position. Is it also true that alpha =(omega)^2*angular position?

Define your variables.  $ a\equal{}\alpha r$, $ v\equal{}\omega r$, and $ x\equal{}\theta r$.


Let $ a,$ $ b$ and $ c$ are real numbers such that $ \prod_{cyc}\left(2(a \plus{} b)^2 \plus{} c^2\right)\neq0.$ Prove that
\[ \frac {(a \plus{} b \minus{} c)^2}{2(a \plus{} b)^2 \plus{} c^2} \plus{} \frac {(a \plus{} c \minus{} b)^2}{2(a \plus{} c)^2 \plus{} b^2} \plus{} \frac {(b \plus{} c \minus{} a)^2}{2(b \plus{} c)^2 \plus{} a^2}\geq\frac {1}{3}
\]
This inequality I posted in the pre-olympiad forum but it has been moved into here. I am sorry.

Here is a similar inequality, but still it's very easy.
Let $ x,$ $ y$ and $ z$ are real numbers such that $ xy\plus{}xz\plus{}yz\plus{}2xyz\equal{}1.$ Prove that
\[ (x\minus{}1)^2\plus{}(y\minus{}1)^2\plus{}(z\minus{}1)^2\geq\frac{3}{4}\]

Use your method. [url]http://www.mathlinks.ro/viewtopic.php?t=258462[/url]
As for the second ineq put $ x \equal{} \frac {a}{b \plus{} c}$, $ y \equal{} \frac {b}{c \plus{} a}$, $ z\equal{}\frac {c}{a \plus{} b}$ and again apply your method.


mobile recharge cards of 100(a) and 109(b) give a talktime of 91 and 100 respectively.which one is BETTER?

possible approaches:

1)subtract one ratio from the other and see if the result is positive or not.(NOT GOOD)

2)MY APPROACH: I WILL USE AN EXAMPLE TO ILLUSTRATE IT:

SYSTEM X SYSTEM Z
6/7 ? 11/12. (=(6+5)/(7+5))

observe that N-D is 1 for both the ratios.
so,for system z we get a HIGHER OUTPUT for the COMMON LOSS OF 1.
hence system z is better,HENCE 11/12 > 6/7.

answer to the main problem: for a service charge of 9, offer 'b' is better than offer 'a'.
[b]try to look at the mathematics only and not the business related aspects[/b]

now,
consider a ratio p/q, p<q, p>0,q>0.
addition of a POSITIVE constant,say m, to the numerator(N) and to the denominator(D) gives a ratio greater or lesser than the ratio we started with?(now we should be able to tell the answer right away)

please comment on my article.
THANK YOU FOR READING PATIENTLY...

http://www.mathlinks.ro/Forum/viewtopic.php?p=935065#935065

Don't post one  issue in two topics  :(


Evidently there is some question about whether you actually have room for last minute teams. I think we signed up as quagga (in case this is ambiguous, we are the quagga mostly from NC). Can we have confirmation as to whether there is room for us, so that we can make travel plans? Thanks.

Please direct questions like this to hmmt-request (at mit dot edu).  Those in charge of registration won't necessarily check this forum very often.

how many members does quagga have right now? there's some TX guys looking for a team to merge with.


inequalities
geometry
triangle inequality
algebra unsolved
algebra

Let $ x, y$ be real numbers satisfying:
\[ x^2 \plus{} y^2 \minus{} 4x \minus{} 6y \plus{} 12 \equal{} 0
\]
Find $ x, y$ such that  $ A \equal{} x^2 \plus{} y^2$ be maximal and find the maximal value of $ A$.

Inequalities (or maybe Geometry) Section.

hello, with the method of Lagrange multipliers we get that the searched maximum is $ 2(7\plus{}\sqrt{13})$ for
$ x\equal{}\frac{2}{13}(13\plus{}\sqrt{13})$ and $ y\equal{}\frac{3}{13}(13\plus{}\sqrt{13})$.
Sonnhard.

$ (x \minus{} 2)^2 \plus{} (y \minus{} 3)^2 \equal{} 1$

We want to find the point on this circle that has maximal distance from $ (0,0)$.

Let the point $ C \equal{} (2,3)$, the origin denoted $ O$, and the unknown point lying on the circle as $ B$, we have by the triangle inequality $ BO\le BC \plus{} OC \equal{} 1 \plus{} \sqrt {13}$ with equality only if $ B,O,C$ lie on the same line in that order.

Since clearly such a point $ B$ exists, we let $ B$ be this point, and we have the maximal value for $ A \equal{} (BO)^2 \equal{} (BC \plus{} OC)^2 \equal{} (1 \plus{} \sqrt {13})^2 \equal{} 2(7 \plus{} \sqrt {13})$

Problem can be done by use this inequation: $ |a_1b_1 \plus{} a_2b_2| \le \sqrt {(a_1^2 \plus{} a_2^2)(b_1^2 \plus{} b_2^2)}$

This is my solution:

$ \bullet$  $ x^2 \plus{} y^2 \minus{} 4x \minus{} 6y \plus{} 12 \equal{} 0 \Longleftrightarrow (x \minus{} 2)^2 \plus{} (y \minus{} 3)^2 \equal{} 1$

$ \bullet$  $ x^2 \plus{} y^2 \equal{} 4x \plus{} 6y \minus{} 12 \equal{} 4(x \minus{} 2) \plus{} 6(y \minus{} 3) \plus{} 14$

$ \le \sqrt {(4^2 \plus{} 6^2)[(x \minus{} 2)^2 \plus{} (y \minus{} 3)^2]} \plus{} 14 \equal{} 2\sqrt {13} \plus{} 14 \equal{} 2(\sqrt {13} \plus{} 7)$


modular arithmetic
number theory
number theory unsolved

in 2,we have $LHS=\frac{1^{p-1}}{1}+\frac{2^{p-1}}{2}+\frac{3^{p-1}}{3}+\dots+\frac{(p-1)^{p-1}}{p-1}\equiv 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{p-1}$. by Wolstenholmes Theorem, this sum has the form $\frac{p^{2}m}{n}$ with $(m,n)=1$.
on the other hand, by Fermat's Little Theorem, $2-2^{p}\equiv 0\pmod p$. so if your problem is correct, we have that $\frac{p^{2}m}{n}+\frac{2^{p}-2}{p}\equiv 0\pmod p$ so $p^{3}m+n(2^{p}-2)\equiv 0\pmod{p^{2}}$ hence $n(2^{p}-2)\equiv 0\pmod{p^{2}}$, which can't be true for all $p$ since $(p,n)=1$. for a counterexample of your problem, take $p=5$. then we have $LHS=100,RHS=-6$. :P


it would be helpful if you corrected the initial post and then people wouldn't start doing an incorrect question :)

get an expression for $ x_{k}$ in terms of a continued fraction and then solve... is rather pretty

I would've, except I couldn't edit my post.


Does anybody where to get past ARML Questions and Solutions?

Here is some information about the books:
http://www.arml.com/books.htm

I don't think you'll find better public resources for ARML questions than these:
ARML Problem Solving Marathon: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40531
ARML Practice Series: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264
with solutions here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32273


[quote="linda2005"]Hello all,

Solve this equations :

1) $ [x] \plus{} [2x] \plus{} [4x] \plus{} [8x] \plus{} [16x] \plus{} [32x] \equal{} 12345$

2) $ [x^2] \equal{} [x]^2$

3) $ [\frac {x^2 \minus{} 2x}{3}] \equal{} \frac {x}{2}$[/quote]

1.

A quick check - plugging in $ x\equal{}196$ gets us the result 12348 on the RHS. We're going to need to take off three, which is fairly easy because it's impossible to do so. Letting $ x \equal{} 196 \minus{} \epsilon$ reduces each of the six terms in the LHS by 1. No solution.

[hide="2 and 3"][b]Equation 2.[/b] Put $ x \equal{} n \plus{} \alpha$ where $ n$ is integer and $ 0\leq\alpha < 1$. Then

$ [n^2 \plus{} 2n\alpha \plus{} \alpha^2] \equal{} n^2$

$ n^2 \plus{} [2n\alpha \plus{} \alpha^2] \equal{} n^2$

$ [2n\alpha \plus{} \alpha^2] \equal{} 0$

$ 0\leq 2n\alpha \plus{} \alpha^2 < 1$

Case 1. $ \alpha \equal{} 0$. Then the above inequation system is trivially satisfied. Therefore, [u]any integer[/u] $ x$ is the solution of the initial equation.

Case 2. $ \alpha\neq 0$. Then the lefthand part of the system is satisfied for $ n\geq 0$, and the righthand part gives

$ n < {1 \minus{} \alpha^2\over 2\alpha}$

Therefore, for any arbitrarily chosen $ \alpha\in(0,1)$ we can choose $ n$ out of all the integer values from $ \left[0, {1 \minus{} \alpha^2\over 2\alpha}\right)$, to get $ x \equal{} n \plus{} \alpha$ as a solution.

Example: For $ \alpha \equal{} 0.01$ we get $ n\in[0;49.995)$, hence $ x\in\{0.01;1.01;2.01;\dots;49.01\}$

[b]Equation 3.[/b] It's obvious that $ {x\over 2}$ must be integer, so $ x \equal{} 2n, n\in\mathbb{Z}$. Then

$ \left[{4n^2 \minus{} 4n\over 3}\right] \equal{} n$

$ n\leq{4n^2 \minus{} 4n\over 3} < n \plus{} 1$

Lefthand part:

$ 4n^2 \minus{} 4n\geq 3n$

$ 4n^2 \minus{} 7n\geq 0$

$ n(4n \minus{} 7)\geq 0$

Either $ n\leq 0$ or $ n\geq {7\over 4}\implies n\geq 2$

Righthand part:

$ 4n^2 \minus{} 7n \minus{} 3 < 0$

This is satisfied for $ n\in\left({7 \minus{} \sqrt {97}\over 8},{7 \plus{} \sqrt {97}\over 8}\right)$.

The only integer values in this interval are $ 0,1,2$. Hence $ n \equal{} 0$ or $ n \equal{} 2$, which gives $ x\in\{0,4\}$[/hide]

hello
thank you Farenhajt , and the first ? some one can explain me please!

[quote="Lazarus"][quote="linda2005"]Hello all,

1.

A quick check - plugging in $ x \equal{} 196$ gets us the result 12348 on the RHS. We're going to need to take off three, which is fairly easy because it's impossible to do so. Letting $ x \equal{} 196 \minus{} \epsilon$ reduces each of the six terms in the LHS by 1. No solution.[/quote][/quote]

thanks Lazarus  , but x in IR isn't just integer  , you can explain me more please , thanks

[quote="linda2005"][quote="Lazarus"][quote="linda2005"]Hello all,

1.

A quick check - plugging in $ x \equal{} 196$ gets us the result 12348 on the RHS. We're going to need to take off three, which is fairly easy because it's impossible to do so. Letting $ x \equal{} 196 \minus{} \epsilon$ reduces each of the six terms in the LHS by 1. No solution.[/quote]

thanks Lazarus  , but x in IR isn't just integer  , you can explain me more please , thanks[/quote][/quote]

$ \epsilon$ is traditionally a quantity very slightly greater than 0. Hopefully that clears things a little bit up.

[quote="linda2005"][u]Letting  reduces each of the six terms in the LHS by 1. No solution.[/u]  I don't understand how you are doing  :(  sorry[/quote]

When he plugs $ x\equal{}196$, the result is $ 12348$. Now imagine we decrease that $ 196$ slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, $ [32\cdot 196]\equal{}6272, [32\cdot 195.99]\equal{}6271$), therefore the sum will decrease by AT LEAST six, and that's too much. Do you get it now? He should have said not "by 1", but "by at least 1".

[quote="Farenhajt"][quote="linda2005"][u]Letting  reduces each of the six terms in the LHS by 1. No solution.[/u]  I don't understand how you are doing  :(  sorry[/quote]

When he plugs $ x \equal{} 196$, the result is $ 12348$. Now imagine we decrease that $ 196$ slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, $ [32\cdot 196] \equal{} 6272, [32\cdot 195.99] \equal{} 6271$), therefore the sum will decrease by AT LEAST six, and that's too much. Do you get it now? He should have said not "by 1", but "by at least 1".[/quote]

thanks you Farenhajt now it's ok :10: , please you can give me another solution to question 2) ?

Why do you need another?  :lol: Besides, that's the only one I could ever come up with...

[hide="Appendix to solution 2"]From $ \alpha^2 \plus{} 2n\alpha \minus{} 1 < 0$ we get $ \alpha\in( \minus{} \sqrt {n^2 \plus{} 1} \minus{} n,\sqrt {n^2 \plus{} 1} \minus{} n)$. It's easily shown that $ \sqrt {n^2 \plus{} 1} \minus{} n < 1$ for natural $ n$, hence this boils down to $ \alpha\in[0,\sqrt {n^2 \plus{} 1} \minus{} n)$. Therefore, as $ x \equal{} n \plus{} \alpha$, we get $ x\in\mathbb{Z}^{\minus{}}\cup\left(\bigcup_{n \equal{} 0}^{\infty}[n,\sqrt {n^2 \plus{} 1})\right)$[/hide]

@Farenhajt
Why $ [n^2 \plus{} 2n\alpha \plus{} \alpha^2] \equal{} n^2$
-put $ n \equal{} 1$ and $ \alpha \equal{} \frac {1}{2}$ we have $ [1^2 \plus{} 2\cdot 1 \cdot \frac {1}{2} \plus{} \frac {1}{4}] \equal{} 2 \not \equal{} 1^{2}$

enndbox, what you stated is not an IDENTITY, but an EQUATION to be solved.

And if you confer final formula, you'll see that $ x\equal{}n\plus{}\alpha\equal{}1.5$ falls outside the given intervals, hence is not a solution.


conics
parabola
geometry
rectangle
analytic geometry
geometry proposed

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=13840[/img]

 :(

Hi again, jrrbc! I like your problems!  :)

Let $ v$ be the line perpendicular to the axis of the parabola, passing through $ V$

Suppose that we have constructed the common tangent (the red line). This line touches the circle at $ D$ and intersects the line $ v$ at $ A$

It is tangent to the parabola, so it should be $ FA\perp DA$
(indeed, if $ F'$ is the symmetric of point $ F$ with respect to $ A$ and $ T$ the tangent point, then $ \triangle TF'F$ sould be isosceles)

Now, we construct the rectangle $ ABCD$. It is clear that the points $ A,B,C,D,V$ are concyclic. Then we have

$ FV \cdot FC \equal{} FA \cdot FB \Rightarrow$

$ FV \cdot FC \equal{} FA \cdot (FA \minus{} r) \ \ (1)$, where $ r$ is the radius of the given circle

In $ (1)$ the only unknown is $ FA$

$ (1)$ can be transformed to $ FA^2 \minus{} r\cdot FA \minus{} (FV\cdot FC) \equal{} 0$

whose positive solution is $ \boxed{FA \equal{} \frac r 2 \plus{} \sqrt {FV\cdot FC \plus{} \frac {r^2} 4}}$, which is constructible

So, we find the point $ A\in v$ such that the length $ FA$ is given above, and from this point we bring perpendicular to $ FA$
(*) There are two points $ A$ and $ A'$, symmetric wrt the axis of the parabola

Oh, you're right...

With analytic geometry I find that $ \boxed{x_1 \equal{} \frac {r^2}{p} \minus{} x_o \pm \frac {r}{p}\sqrt {r^2 \plus{} p^2 \minus{} 2px_o}} \ \ (1)$, where:

The parabola has equation $ \boxed{y^2 \equal{} 2px}$

The circle's equation is $ \boxed{(x \minus{} x_o)^2 \plus{} y_o^2 \equal{} r^2}$

$ (x_1,y_1)$ is the touch point on the parabola

Each solution of $ (1)$ gives of course a pair of points $ (x_1,y_1)$ and $ (x_1, \minus{} y_1)$

One pair is for the internal tangents ans the other is for the external

But I didn't find yet some elegant geometric meaning to $ (1)$

[size=150][hide="A complete solution"][/size]
Let $ \boxed{y^2 \equal{} 2px}$ and $ \boxed{(x \minus{} x_0)^2 \plus{} y_0^2 \equal{} r^2}$ be the equation of the parabola and the equation of the circle respectively

This means that $ F\left(\frac {p}{2},0\right)$ is the focus of the parabola, $ C(x_0,y_0)$ is the center of the circle and $ r$ is the radius of the circle

Also, $ O(0,0)$ is the vertice of the parabola

Let $ A(x_A,0)$ be the point where some of the common tangents intersect the x-axis

It's easy to find (using analytic geometry) that $ x^2_A \plus{} 2\left(\frac {r^2}{p} \minus{} x_0\right)x_A \plus{} (x^2_0 \minus{} r^2) \equal{} 0$

or simply $ x^2_A \minus{} 2H\cdot x_A \plus{} P \equal{} 0 \ \ (1)$

where $ H \equal{} \left(x_0 \minus{} \frac {r^2}{p}\right)$ and $ P \equal{} x^2_0 \minus{} r^2$

(both $ H$ and $ P$ are constructible)

Then we have to find two numbers $ x_1$ and $ x_2$ such that $ x_1 \plus{} x_2 \equal{} 2H$ and $ x_1x_2 \equal{} P$

[color=brown]$ \star$ Notice that since $ x_0$ is negative, $ H$ will be also negative and consequently the two solutions of $ (1)$ will be negative.
$ P$ is the power of the point $ O(0,0)$ wrt the circle $ (C,r)$, and it is positive because $ O$ is outside of the circle[/color]

$ H \equal{} \left( \minus{} |x_0| \minus{} \frac {r^2}{p}\right) \equal{} \left( \minus{} OC \minus{} \frac {r^2}{p}\right) \equal{} \minus{} \left(OC \plus{} \frac {r^2}{p}\right)$

Let $ T$ be the point $ T(H,0)$

Draw the circle $ (T)$ with radius $ H^2 \minus{} P$

The points where this circle intersects the x-axis are the solutions of $ (1)$, that is, the desired points of the common tangents [/hide]
------------------------------------------------------------------------------------------------------------------------------------------------------
Now:

[color=blue][i]Just an idea how to make the construction cute:[/i]

1) Let $ L$ be the point such that $ \vec{CL} \equal{} 2\vec{FO}$ (so, [b]L[/b] must be on the 'left' of [b]C[/b])

2) Let $ M$ be the inverse of $ L$ w.r.t. $ (C)$

3) Let $ w$ be the circle centered at $ O$, which is perpendicular to $ (C)$

4) Let $ m$ be the circle centered at $ M$, which is perpendicular to $ w$

5) Let $ A_1,A_2$ be the points where the circle $ m$ intersects the x-axis

6) Bring the tangents from $ A_1$ and from $ A_2$ to $ (C)$[/color]


trigonometry
function
geometry
vector
algebra
polynomial
geometric transformation

I attended a lecture last month at Trent University about a new theory called rational trigonometry that was simply fascinating. It claims to be revolutionary in that it dispenses with all the usual complicated trig functions and relations and establishes a trig based entirely on elemental arimetric and algebra.  It seemed like such an elegant approach to trig... in the span of an hour, the prof not only proved his five new trig "laws", but also solved three complex trig problems, using his new theory. I have since visited the lecturing professor's website and downloaded the first few chapters of his new book on the subject and am still as intrigued as ever. (website http://wildegg.com)  I also just re-visited his site, and discovered a special pre-publication price for the book... so now it's even more attractive. I am interested in any feedback on this new theory, so please contact me and let me know your comments. Thanks,

Krista

The book appears to be correct mathematically, and very interesting to compare with the usual development.
As the author points out, it is well-known that it is more "correct" to develop geometry using the quantities that work better algebraically (squared distance, dot product of vectors rather than angles, quantities that are polynomial functions of the coordinates).  I think he oversimplifies the issues, though, in claiming un-naturality of the standard approach; angle measure in its usual form is forced on us by the structure of the rotation group, and so is unavoidable.

By the way, breathless enthusiastic descriptions give the impression of promoting a book.  I think there is enough in this book to make it interesting on its own merits, even if not revolutionary.  Certainly it's thought-provoking and maybe even controversial, in ways that are unusual for a large-audience mathematical book.


Is there any strictly increasing and infinite sequence which is defined on positive integers $ a_1,a_2...$ such that for every positive integer $ i$ ,    $ a_i.a_{i \plus{} 1} \plus{} 1$, $ a_i.a_{i \plus{} 2} \plus{} 1$, $ a_i.a_{i \plus{} 3} \plus{} 1$ are all perfect squares?

I have been trying to solve for months, but haven't solved yet. I also think that it is a nice and hard number theory problem.  :(


Let delta ABC and M is the point in delta ABC . Let A' , B' , C' on BC , CA , AB with A'B' // AM , B'C' // BM , C'A' // CM. Prove that : 3.S(A'B'C') =< S(ABC)
 :roll:  :roll:

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258369[/url]


given $ k\in\mathbb{Z^{+}}$, show that there exists $ x,y\in\mathbb{Z}$ such that $ x^{2}-ky^{2}=1$.
($ k$ is not a square).

$ x=1,y=0$ work. :P
for more details, see here: http://en.wikipedia.org/wiki/Pell's_equation


[quote="cezar lupu"]if x,y,z>0 prove the following inequality:

         (1/x+y+1/y+z+1/z+x)^2>=27/4(xy+yz+zx).[/quote]
can you use latex to write it? I can't understand you.

[quote="cezar lupu"]I'm sorry. The inequality is:  (1/(x+y)+1/(y+z)+1/(z+x))^2>=27/4(xy+yz+zx).[/quote]
I still think it is wrong. I think you mean $\frac{25}{4}$. Anyway, this inequality was posted many times already.

Anyway can you post it with a solution,please?


function
inequalities
calculus
derivative
triangle inequality
topology

Let function $ d: \mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ determinated by $ d(x,y) \equal{} \frac {|x \minus{} y|}{\sqrt {1 \plus{} x^2}\sqrt {1 \plus{} y^2}}$.
 Show that $ (\mathbb{R},d)$ is an incomplete metric space.

There are two parts.

Part 1: Show it's a metric. The triangle inequality is the only nontrivial part here.
That's in inequality calculation; given $ x < y < z$, we want to show $ \frac {y \minus{} x}{\sqrt {1 \plus{} x^2}\sqrt {1 \plus{} y^2}} \plus{} \frac {z \minus{} y}{\sqrt {1 \plus{} y^2}\sqrt {1 \plus{} z^2}}\ge \frac {z \minus{} x}{\sqrt {1 \plus{} x^2}\sqrt {1 \plus{} z^2}}$
Clearing the denominators, that's $ (y \minus{} x)\sqrt {1 \plus{} z^2} \plus{} (z \minus{} y)\sqrt {1 \plus{} x^2}\ge (z \minus{} x)\sqrt {1 \plus{} y^2}$. Let $ y \equal{} tz \plus{} (1 \minus{} t)x$; $ 0 < t < 1$ by the ordering hypothesis. This becomes
$ t(z \minus{} x)\sqrt {1 \plus{} z^2} \plus{} (1 \minus{} t)(z \minus{} x)\sqrt {1 \plus{} x^2}\ge (z \minus{} x)\sqrt {1 \plus{} (tz \plus{} (1 \minus{} t)x)^2}$
Clear out the common factor of $ z \minus{} x$, and that's convexity of $ \sqrt {1 \plus{} x^2}$. The second derivative of $ \sqrt {1 \plus{} x^2}$ is $ \frac {1}{\sqrt {1 \plus{} x^2}} \minus{} \frac {x^2}{(1 \plus{} x^2)^{\frac32}} \equal{} \frac {1}{(1 \plus{} x^2)^{\frac32}} > 0$. It's convex everywhere, and the triangle inequality is always satisfied.

Note that this also shows that $ d(x,y) \equal{} \frac {|x \minus{} y|}{f(x)f(y)}$ is a metric whenever $ f$ is a positive convex function.

Part 2: Construct a Cauchy sequence that doesn't converge to any real value. Looking at the denominator, it looks like a sequence going to $ \infty$ would work. Try the natural numbers; $ d(m,n) \equal{} \frac {|m \minus{} n|}{\sqrt {1 \plus{} m^2}\sqrt {1 \plus{} n^2}}\le\frac {|m \minus{} n|}{mn}\le \frac {\max(m,n)}{mn} \equal{} \frac1{\min(m,n)}$. That's a Cauchy sequence, and it can't possibly converge to anything finite.

[quote="jmerry"]There are two parts.

Part 1: Show it's a metric. The triangle inequality is the only nontrivial part here.
That's in inequality calculation; given $ x < y < z$, we want to show $ \frac {y - x}{\sqrt {1 + x^2}\sqrt {1 + y^2}} + \frac {z - y}{\sqrt {1 + y^2}\sqrt {1 + z^2}}\ge \frac {z - x}{\sqrt {1 + x^2}\sqrt {1 + z^2}}$
Clearing the denominators, that's $ (y - x)\sqrt {1 + z^2} + (z - y)\sqrt {1 + x^2}\ge (z - x)\sqrt {1 + y^2}$. Let $ y = tz + (1 - t)x$; $ 0 < t < 1$ by the ordering hypothesis. This becomes
$ t(z - x)\sqrt {1 + z^2} + (1 - t)(z - x)\sqrt {1 + x^2}\e (z - x)\sqrt {1 + (tz + (1 - t)x)^2}$
Clear out the common factor of $ z - x$, and that's convexity of $ \sqrt {1 + x^2}$. The second derivative of $ \sqrt {1 + x^2}$ is $ \frac {1}{\sqrt {1 + x^2}} - \frac {x^2}{(1 + x^2)^{\frac32}} = \frac {1}{(1 + x^2)^{\frac32}} > 0$. It's convex everywhere, and the triangle inequality is always satisfied.

Note that this also shows that $ d(x,y) = \frac {|x - y|}{f(x)f(y)}$ is a metric whenever $ f$ is a positive convex function.

Part 2: Construct a Cauchy sequence that doesn't converge to any real value. Looking at the denominator, it looks like a sequence going to $ \infty$ would work. Try the natural numbers; $ d(m,n) = \frac {|m - n|}{\sqrt {1 + m^2}\sqrt {1 + n^2}}\le\frac {|m - n|}{mn}\le \frac {\max(m,n)}{mn} = \frac1{\min(m,n)}$. That's a Cauchy sequence, and it can't possibly converge to anything finite.[/quote]

that right!
but $ t(z - x)\sqrt {1 + z^2} + (1 - t)(z - x)\sqrt {1 + x^2}\geq (z - x)\sqrt {1 + (tz + (1 - t)x)^2}$

Fixed. I had \e, which is not a valid code.


[quote="pvthuan"]Prove that if $a,b,c \geq0$, then
\[ \frac a{a^2+bc}+\frac b{b^2+ca}+\frac c{c^2+ab}\geq\frac4{a+b+c}.  \][/quote]

Let $a=b=1$. Then
\[ \frac a{a^2+bc}+\frac b{b^2+ca}+\frac c{c^2+ab}-\frac4{a+b+c}=\frac{2}{1+c}+\frac{c}{c^2+1}-\frac{4}{2+c}=\frac{c^2(3-c)}{(c+1)(c+2)(c^2+1)}  \]
which is negative for $c > 3$.

Very good counterexample, ductrung. Thank you.

But the following ineq is TRUE for $a,b,c\geq0$
\[ \frac a{b^2+ca}+\frac b{c^2+ab}+\frac c{a^2+bc}\geq\frac4{a+b+c}. \]

No it's not.take a=0;b=c/2Is your book a colection of conjecture

[quote="pvthuan"]Prove that if $a,b,c \geq0$, then
\[ \frac a{a^2+bc}+\frac b{b^2+ca}+\frac c{c^2+ab}\geq\frac4{a+b+c}.  \][/quote]Let $a \rightarrow +\infty;b=c=1$,your ineq is wrong.
[u][b]The true inequality:[/b][/u]
$\sum \frac{a}{a^2+bc}\ge \frac{3}{a+b+c}$  :)  and $3$ is the best constant.  :)

Surely the inequality is wrong ...

Davron

No it shouldn't... if it fails for 4 why would it hold for 9/2 > 4?

[quote="pvthuan"]Prove that if $a,b,c \geq0$, then
\[ \frac a{a^2+bc}+\frac b{b^2+ca}+\frac c{c^2+ab}\geq\frac4{a+b+c}.  \][/quote]
IT's an old ineq.Let$a=xy,b=yz,c=zx$ we have an ineq
$\sum\frac{1}{x^2+yz}\ge \frac{3}{xy+yz+zx}$ :)
Click [url=http://www.mathlinks.ro/Forum/topic-55584.html]here[/url]


A certain physicist , who is always in hurry, walks up an upgoing escalator at the rate of one step per second. Twenty steps bring him to the top. Next day he goes up at two steps per second, reaching the top in 32 steps. How many steps are there in the escalator ?

Not sure about this but I think it might be $ 80$ steps.

[hide=" click to reveal hidden content"]

The escalator travels at $ x$ steps per second.

On the first day, it takes him $ 20$ seconds to reach the top. During that time, the escalator moves $ 20x$ steps, so if he'd been standing still, he'd have another $ 20$ steps to go.

On the next day, it takes him $ 16$ seconds to reach the top.

So $ 20x \plus{} 20 \equal{} 16x \plus{} 32$

$ x\equal{}3$.

$ 20x\plus{}20\equal{}16x\plus{}32\equal{}80$

[/hide]


Put $a_{n}=\frac{(n+1)^{n}}{n^{n-1}}$ for all $n \geq 1,n \in N$
Prove that:
                   $lim_{n\to  \infty}(\frac{a_{n+1}}{a_{n}})^{n}=e$

After some computations we get: \[\frac{a_{n+1}}{a_{n}}=\left( 1+\frac{1}{n+1}\right)^{2}\left( 1-\frac{1}{(n+1)^{2}}\right)^{n-1}\] But $\left( 1+\frac{1}{n+1}\right)^{2n}\rightarrow e^{2}$ because $\frac{2n}{n+1}\rightarrow 2$ and  $\left( 1-\frac{1}{(n+1)^{2}}\right)^{n(n-1)}\rightarrow e^{-1}$ because $-\frac{n(n-1)}{(n+1)^{2}}\rightarrow-1$.

So $\lim_{n \to \infty}\left( \frac{a_{n+1}}{a_{n}}\right)^{n}=e$.


induction
combinatorial geometry
combinatorics proposed
combinatorics

Do there exist two disjoint infinite sets $A,B$ of points in the plane satisfying the following properties?

(a) No three points in $A\cup B$ are collinear;

(b)The distance between any two points in $A\cup B$ is at least $1$;

(c) For every three points in $A$ we can find a point of $B$ inside the triangle determined by them, and the same holds if we reverse the roles of $A$ and $B$.

Lemma 1: by condition b, any finite polygon contains a finite number of points of the set $A \cup B$ (to prove it, just tile the plane with squares of side $\frac{\sqrt{2}}{2}$; then each square can contain at most one point, and a finite number of squares cover the polygon).

Lemma 2: if given a set of enough points, we can find a subset of $5$ of them that form a convex pentagon (it is known that $9$ are enough, but we can prove a weaker statement by Ramsey, which is enough for the problem, since we're given infinite sets). It is easy to prove that among any $5$ points there are $4$ that form a convex quadrilateral, and it is also easy to prove that there's a number $R(4, n)$ such that if we have a set of at least $R(4, n)$ elements and we color each 4-element subset white or black, there will be a subset of $n$ elements with all its 4-subsets of the same color. So take a set of $R(4, 5)$ points in the plane (with each quadrilateral colored white if it is convex and black otherwise), and among them take a 5-subset with all its 4-subsets of the same color. Since among any 5 points there are 4 that form a convex quadrilateral, at least one 4-subset is white, so all of them are white. It is easy to see this implies that the pentagon is convex.

Lemma 3: Given a (not necessarily convex) $n$-gon in the plane, we can find a set of at least $n-2$ triangles with vertices among those of the $n$-gon and not sharing any interior points. This is easy to prove by induction, looking at a couple of cases.

Now take set $A$ and find a convex pentagon with vertices among the points in $A$ (we can do this by lemma 2). Suppose this pentagon has $k$ interior points in $A$ ($k$ is a finite number by lemma 1). If we triangulate the subset of $k+5$ points with at least $k+3$ disjoint triangles (this is lemma 3), there must be at least $k+3$ $B$ points inside the pentagon (since there's one inside each triangle and the triangles are disjoint). Now triangulate these $k+3$ points with $k+1$ triangles. Each triangle must contain an $A$ point, but the only points that can be inside these triangles are the $A$ points interior to the pentagon, and we have just $k$ of them. Contradiction.

Thanks for the interest, guys. I had completely forgotten my proof, so I had to reconstruct it. Here it is:

Suppose we can find such an arrangement in which $|A|\ge 5$. As in Severius's first observation, every polygon contains finitely many points, so we can find five $A$-points s.t. their convex hull $\mathcal H$ contains no other $A$-points. There are three possibilities:

[b](1)[/b] $\mathcal H$ is a pentagon.

In this situation, we can find three disjoint triangles formed by the five points, and each one must contain at least one $B$-point. Any triangle formed by three of these $B$-points will contain no $A$-point inside, so we get a contradiction.

[b](2)[/b] $\mathcal H$ is a quadrilateral.

We can now form $4$ disjoint triangles with the five $A$-points (connect the point inside $\mathcal H$ to all the vertices of $\mathcal H$), and each of these contains at least a $B$-point. With these $4\ B$-points, we can create two disjoint triangles, and they can't both contain the $A$-point inside $\mathcal H$, so one of the triangles will have no $A$-point inside - contradiction again.

[b](3)[/b] $\mathcal H$ is a triangle.

We can now create $5$ disjoint triangles with vertices among the five points. We will thus have $\ge 5\ B$-points inside $\mathcal H$, with which we will be able to form at least three triangles. However, we only have two $A$-points inside $\mathcal H$, which can't fill up the $\ge 3\ B$-triangles, so, once more, we get a contradiction.

P.S.

I didn't read Severius's proof carefully, but I think it's pretty much the same idea.

nice, direct, and simple proof grobber !  :)


Suppose there are $n$ distinct points on plane. There is circle with radius $r$ and center $O$ on the plane. At least one of the points are in the circle. We do the following instructions. At each step we move $O$ to the baricenter of the point in the circle. Prove that location of $O$ is constant after some steps.

assume that the sequence of circles is $C_i$ then we want to say that $C_i$ is constant eventually,so suppose that $A_{1,C_i},\dots,A_{k_i,C_i}$ are in $C_i$ then define such a function $F(C_i)=\sum_{j=1}^{i_k} (R^2-|O_iA_{j,C_i}|^2)$ then this function if $O_i\neq O_{i+1}$ increase .(this part is cmputational) defining such afunction is not that hard i think that if you define $F(C_i)=\sum (|O_iA_{j,C_i}|^2)+(n-i_k)R^2$ this decrease so it will stop.any way no one could'nt solve it in exam.

Cool problem! I have the same idea, but I wanted to express it via moment of inertia only, and I, obviously, failed to get monotonocy.

[quote="sam-n"]define such a function $F(C_i)=\sum_{j=1}^{i_k} (R^2-|O_iA_{j,C_i}|^2)$ then this function if $O_i\neq O_{i+1}$ increase .(this part is cmputational)[/quote] Indeed, $\sum_{j=1}^{i_k} |O_iA_{j,C_i}|^2$ is the moment of inertia of $\{A_{j,C_i}\}$ wrt $O_i$. Due to Steiner's theorem when we move the point $O_i$ to the center of gravity $O_{i+1}$ this sum decreases, moreover we lost some negative summands $R^2-|O_iA_{j,C_i}|^2$, if $A_{j,C_i}$ becomes out of circle, and we gain new positive summands from the new incoming points. Therefore, $f(C_i)$ increases!

I feel myself stupid. Is the circle the set of $X$ for which $OX=R$ or the set of $X$ for which $OX\le R$? I was sure that the first case is correct, than there is a counterexample: take a rectangular $ABCD$, put $O$ in a midpoint of $AB$, and choose $r=OC=OD$. Then $O$ moves from midpoint of $AB$ to midpoint of $CD$ and so on.

Indeed, we consider a disk instead of a circle in this problem.

This problem was the most difficult problem. I think nobody solved it.

hey people,
i need some help here understanding the proof.
how do you prove that  $f(C_i)$ keeps increasing?
and myth, what's Steiner's theorom and what does moment of inertia refer to ?

Here's a solution by [b]geoishard[/b] and me.

There are finitely many centroids, so it suffices to show that no nontrivial cycles exist.

Suppose that centers $O_1,O_2,\ldots,O_n$ of circles $C_1,C_2,\ldots,C_n$ ($n>1$; indices modulo $n$) form a cycle so that for all $k$, $O_{k+1}$ is the centroid of the points $P\in C_k$.

The key idea is that the centroid (uniquely) minimizes the sum of squares of distances to vertices (while centers of circles essentially minimize maximal distances), so $\sum_{P\in C_k}PO_k^2-PO_{k+1}^2>0$ for all $k$ since $O_k\ne O_{k+1}$. On the other hand,
\begin{align*}
\sum_{k=1}^{n}\sum_{P\in C_k}PO_k^2-PO_{k+1}^2
&=\sum_{k=1}^{n}\sum_{P\in C_k}PO_k^2-\sum_{P\in C_{k-1}}PO_k^2 \\
&=\sum_{k=1}^{n}\sum_{P\in C_k\setminus C_{k-1}}PO_k^2-\sum_{P\in C_{k-1}\setminus C_k}PO_k^2 \\
&\le r^2\sum_{k=1}^{n}|C_k\setminus C_{k-1}|-|C_{k-1}\setminus C_k|=r^2\sum_{k=1}^{n}|C_k|-|C_{k-1}|=0,
\end{align*}contradiction.

This problem is closely related to the [b]K-means algorithm[/b] from signal processing and unsupervised machine learning. You can learn more about this topic on the corresponding [url=https://en.wikipedia.org/wiki/K-means_clustering#Standard_algorithm_(naive_k-means)]Wikipedia article[/url] or [url=https://www.coursera.org/lecture/machine-learning/k-means-algorithm-93VPG]Coursera video[/url]. First I will introduce the algorithm and show why it terminates, and then I will apply it to solve the problem.

[b]Introducing the K-means Algorithm:[/b]

Suppose we have $m$ points $P_1, P_2, \dots, P_m$ in some Euclidean space and we want to group them into $k$ clusters of "nearby" points. We accomplish this with an iterative process as follows: At the initial iteration $t=1$ start with $k$ random points $Q_1^0, Q_2^0, \dots, Q_k^0$ in the space which represent the centers of the clusters. Now at each iteration $t$ of the algorithm, we will group the points $P_1, P_2, \dots, P_m$ based on which of the $k$ points $Q_1^{t-1}, Q_2^{t-1}, \dots, Q_k^{t-1}$ they are closest to; this partitions the $P_i$ into $k$ clusters $C_1,C_2, \dots, C_k$ such that all points in $C_i$ are closer to $Q_i^{t-1}$ than $Q_j^{t-1}$ for any $j\neq i$ (ties are broken arbitrarily). Now for each $1\le i\le k$ we will define $Q_i^t$ as the centroid of the points in $C_i$ (with no change if $S_i$ is empty), and then move on to the next iteration of the algorithm. There are nice pictures illustrating this in the links I wrote above. We terminate when the $Q_i$ stop moving (ie. when $Q_i^{t-1} = Q_i^t$ for each $i$).

[b]Proof the algorithm terminates:[/b]

Clearly there are only finitely many possible positions for the $Q_i^t$, so if we can find a monovariant then the algorithm must terminate. At any given iteration $t$ of the algorithm, define $f_t(i)=j$ if $P_i\in C_j$, so that $f_t(i)$ represents the point $Q_j^t$ closest to $P_i$ at step $t$. Then it's easy to see that the cost $\displaystyle\sum_{1\le i\le m} |P_iQ_{f_t(i)}^{t-1}|^2$ decreases whenever we go from $t\to t+1$ until the algorithm terminates: This is because the step where we cluster the $P_i$ based on which of the $Q_j^t$ they are closest to necessarily minimizes each individual term $|P_iQ_{f(i)}^{t-1}|^2$ over all possible index assignment functions $f$, and the step where we reassign $Q_j$ to be the average of the points $C_j$ necessarily minimizes $\sum_{f_{t+1}(i) = j} |P_iQ_j^t|^2$ over all possible choices of $Q_j^t$ for each $j$, since it's well-known the centroid minimizes the sum of squares of distances. (Basically, the index reassignment and $Q_j$ reassignment steps both act to greedily minimize the cost).

[b]Solving the Iran problem:[/b]

WLOG suppose we are working in the plane $z=0$ and denote $O  = (x_0,y_0,0)$ the initial position of $O$, the center of the disk. For each point $P_i = (x_i,y_i,0)$ for $1\le i\le n$ we will introduce a duplicate point $P_i' = (x_i,y_i,2r)$ in an upper plane so there are $2n$ total $P$'s. We will run the $K$-means algorithm on these $2n$ points with $n+2$ clusters. For each pair $P_i = (x_i,y_i,0), P_i' = (x_i,y_i,2r)$ we introduce a cluster center $Q_i^0 = (x_i,y_i,r)$; we also introduce two new clusters at positions $Q_{n+1}^0 = (x_0,y_0, 0), Q_{n+2}^0 = (x_0, y_0, 2r)$. First note that as the algorithm progresses, the movements of $Q_{n+1}, Q_{n+2} = O, O'$ will mirror each other, on parallel planes. This means by induction on $t$ that the positions of the points $Q_1^t, Q_2^t, \dots, Q_n^t$ will always remain fixed, because for each $1\le i\le n$ the only points which could possibly lie in $C_i$ at any given moment in time are $P_i, P_i'$ (since $Q_i^t$ will be $r$ units away from $P_i, P_i'$ and $>r$ units from everything else), and the average of $P_i, P_i'$ is precisely $Q_i^0$. Therefore the movements of $Q_{n+1}, Q_{n+2}$ in the $K$-means algorithm precisely mirror the movement of $O, O'$ in the original problem, since the only points in $C_{n+1}, C_{n+2}$ will be the ones $\le r$ units away from $Q_{n+1}^t, Q_{n+2}^t$, and we keep updating $Q_{n+1}, Q_{n+2}$ to be the centroids of the points in $C_{n+1}, C_{n+2}$! Thus since the $K$-means algorithm terminates, $O$ and $O'$ will eventually stop moving as well $\blacksquare$

(Obviously the cost function we use to prove the $K$-means algorithm terminates is identical to the monovariant used in the above solutions so there is nothing new here, but I thought it would be instructive to point out the connection here, and besides, this is a nice algorithm to learn about in its own right.)


Solve this equation in $ Z$:
$ (4y\minus{}1)x^3\plus{}(6y^2\minus{}3z\minus{}58)x^2\plus{}(4y^3\minus{}3z^2\minus{}90z\minus{}2)x\plus{}(y^4\minus{}z^3\minus{}45z^2\minus{}z\minus{}56)\equal{}0$

This is a rather strange problem.  It can be rewritten as $ (x \plus{} y)^4 \equal{} x^4 \plus{} (x \plus{} z)^3 \plus{} 45(x \plus{} z)^2 \plus{} (x \plus{} z) \plus{} 13x^2 \plus{} x \plus{} 56$, but that's just a few variables too many for me.

It isn't strange like you think  :blush:


Given the numbers x,y,z verifying the equation max(x,y)+max(z,2004)=min(x,z)+min(y,2005). Prove if y>=z.

very simple,
assume by contrary that y<z.
one can easily see that we can't have y>=2005.
so assume y<2005.
this is very simple to consider.


geometry
geometric transformation
rotation
combinatorics proposed
combinatorics

There are $n$ cars waiting at distinct points of a circular race track. At the starting signal each car starts. Each car may choose arbitrarily which of the two possible directions to go. Each car has the same constant speed. Whenever two cars meet they both change direction (but not speed). Show that at some time each car is back at its starting point.

WLOG Assume 1 car takes 1 hour to complete the track.
We will Show that after $ n$ hours, each car will be back at starting pos. Consider another acr race such that if two cars meet they [i]pass through each other[/i] instead of changing sirection.Although two races are different,at any given time the points on the racecourse occupied by the cars are same.after 1 hour of the "changing direction" car race,there will be a car at each starting point and that car will be going in the same direction as the first one.In addition, although we may not know which car is at starting oint in one hour, we know that the cars in the "changing direction" race maintain same relationship(i.e. clockwise or counterclockwise w.r.t. others) with others.therefore after 1 hour of the 'changing direcn' race,the cars at the starting pt. on the racecourse are some rotation of the original cars on course and going in same direction as those.Therefore after n hours,the cars in the "changing direction " car race will be back at starting point.


some compounds eh!
http://en.wikipedia.org/wiki/Olympiadane
http://en.wikipedia.org/wiki/Catenane

olympiadane is shaped like the 5 rings of the olympics, heh.

[quote="pardesi"]thanks for the info  :D
now propose a synthesis for catenane from a cyclic compoun containing 34 carbon atoms and a double bond

this may be absolutely no syntheic use but just for the sake of it  :)[/quote]
acyloin condensation???? :huh:  but it req a dicarboxyllic acid of 34 carbons not a double bond. but ur rite the method sux, as it is in fact a chance product depending upon the threading of diester molecule through the acyloin ring before it closed. the yield is around 1-2%
i already posted this for madness.
[quote="gregx007"]Unfortunately, Wikipedia is blocked in this country so I can't see what you're referring to.  What makes these compounds so cool?[/quote]
why the hell is wikipedia blocked in china? :o

neways here's catenane

[quote="VARUNARASU"]i already posted this for madness.
[/quote]

I didnt get what you meant here  :maybe:   :huh:

i had already posted this question for u, in the iit jee thread. :D

why the hell is wikipedia blocked in china? :o
that's because of some communist troubles :rotfl:

Probably because of the problems that Tibet have with China.

No, Wikipedia (as well as sites such as Blogspot, Xanga, and the political ticker on CNN.com) seem to be permanently blocked.  I've heard different reasons for it, but I'm sure you can guess as well as I can: maybe these sites have badmouthed China at one point or another, or perhaps China wants to be able to more closely control what its citizens view.  In any case, its their country and therefore their prerogative (although it does make things inconvenient!).

Regarding Tibet, China temporarily blocked YouTube when that situation fired up started, but it's working again.


hi brother i m morrocan to :)
so
(2+a^3)=1+1+a^3 >= 3a   (am >= gm)
so the left is greater than 3(a/b+b/c+c/a) which is always greater than 3 :)

khouk ana ;)
thella

you are from witch city and did you participate at imo this year?

[quote="kam"]here is an other problem
a,b,c>o /
prove that
a/b+b/c+c/a>=a+b+c[/quote]

Hmm, this is not true.

if a=b=c=2
that's absolutly not true brother
i m from essaouira
i participate in imo 01 and imo 02
what about you ,you were in greece this year right ! congratulations from moroccan team

there is a condition to the inequality

yes you are absolutly right : abc=1
so if a,b,c >0 and abc=1
then a/b+b/c+c/a >= a+b+c

x->1/x is convex
so
let's note a/a+b+c=x,b/a+b+c=y,and c/a+b+c=z
the inequality becomes
A=x/b+y/c+z/a>=1
 or
*****by jensen we have
A>=1/(xb+yc+za)
******

in the other hand using the abc=1
the ineq becomes
A=az+bx+ca>=1

x->x is also convex

soA>=(az+bx+yc)

so A^3=A*A>=(az+bx+yc)/(az+bx+cy)=1
and we have the ineq prooved
:))

your solution is too nice but I have found an other solution:
we have by AM-GM a/b+a/b+b/c>=3(a^2/bc)^(1/3)=3a
and b/c+b/c+c/a>=3b; c/a+c/a+a/b>=3c
and we have the result by suming


algebra
polynomial
modular arithmetic
Ring Theory
number theory theorems
number theory

Please, give me a proof of Hensel´s Lema in analisis.

I will write $p^s||a$ to express that $s$ is the maximal exponent so that $p^s|a$.

Proof:

Lemma:
$p^x||a-1 \Rightarrow p^{x+1}||a^p-1$:
write
$a-1=mp^x,\gcd(m,p)=1 \Rightarrow a^p=(1+mp^x)^p=\sum_{i=0}^p \binom{p}{i} m^i p^{ix} = 1+p^{x+1}(m+p^x(...) )$

Now use the lemma $y$ times to get $p^{x+y}||a^{p^y}-1$, and by writing $n=lp^y$ and doing the same as above you get $p^{x+y}||a^n-1$

Maybe someone will write what actually is proved in this topic?

I found at least two Hensel's lemmas. And they both related to polynomials.

Myth! you are right, I dont know the name of this lemma, but it is not HENSEL'S lemma! I will try to find the name of this very nice result.

WHAT result?

Can you, please, state it here?

Hensel's lemma states the following: if $p$ is a prime, $r$ is a positive integer, and $f(x) \in \mathbb{Z}[x]$ is a polynomial such that there exists an $a_r$ for which (i) $p^{2r-1} \mid f(a_r)$; and (ii) $p^r \nmid f'(a_r)$, then there is a (unique modulo $p^r$) integer $a_{r+1} \equiv a_r \pmod{p^r}$ such that $p^{2r+1} \mid f(a_{r+1})$ and $p^{r+1} \nmid f'(a_{r+1})$. Hence, there is a unique $p$-adic number $\alpha$ such that $f(\alpha) = 0$ and $\alpha \equiv a_r \pmod{p^r}$.

It seems that is one particular variant of what I see about Hensel lemma.  :P

Sorry, but when I posted, the lemma was writen in the first post.

So I am repeating my question: Please, state what have you proved.

I think it was the following:
If you have $p^x||a-1,x \geq 1$ and $p^y||n$, then $p^{x+y}||a^n-1$.
And I see now that this can be proved more directly.

You can actually prove the stronger result that if $p^{\alpha}\|a-b$ where $(a,p)=(b,p)=1$ and $p^{\beta}\|n$ then
$p^{\alpha+\beta}\|a^n-b^n$

This has been proved in this forum before somewhere. I made a proof of it and have it on pdf, but is in spanish. If anybody is interested let me know.

Your theorem follows immediatly from the theorem I proved above. Only take $c:=ab^{-1} \mod p^m$ for big enough $m$ or substitute every '$1$' by $b$ in the proof above.

1. Please don't delete your given statement/change it totally again!
2. Do you mean the following theorem¿:

Let be $K$ some ultrametric and complete field.
Let be $R$ the maximal (by inclusion, so exactly the subset $\{x \in K | |x|\leq 1 \}$ ) subring of $K$ and $p$ a polynomial with coefficients in $R$.
If you now have some $x \in R$ such that $|p(x)| < |p'(x)|^2$ then there exists exactly one $\xi \in R$ such that $p(\xi ) =0$ and $|x-\xi | < |p'(\xi )|$


I don't know about you guys, but Nats was a reallly big disappointment for me...., Washington had a really good team, but we did soo bad, we got scores of 24, 29, 32, and 34, and a team round of 8....we took 22nd, oh well... (i'm the one that got a 24, oops)

wow, that's almost the exact same scores as my team, except we had two people get in the 100s. We had a 34 and a 32 and two other scores i don't know. we got a 9 on the team, so we wound up pretty well, 14th. i'm from connecticut by the way

This year's competition was obviously harder.  IL: 44 (Greg), 40 (Me :-)), 36, and 32.  Team round: 10.  Last year CA had like 3 people in the 40s.

it was pretty interesting. of course, everything else was total  boredom to me, so it could've just been relative. lets just say the best speech given that day

[quote="mr_um2312"]what did you think of the chicks speech?[/quote]

LOL, anyone who says it's OK to stay up the entire night's fine by me.

I scored something like a 36, and ranked 20th. I'm OK with that, I guess. :)

The last MC night was interesting for some of us.  Fermatprime and captcha can second that.

[quote="justinian"]My coach is so cool though, he just let us stay up all night, so i got pretty much no sleep the last 2 (3?) days.  We got in some trouble i think, but my coach didn't care.  It was great, i had fun even though i did bad at the competition.[/quote]
o man, my roommate and i slept so late. the night before the competition, we slept at a reasonable time, 10:45. On friday night, we went to sleep at 11:30 and then we slept at 1:15 on saturday night, mainly because of saturday night live.

Heh.  Thursday night I went to bed around 10:30.  I woke up around 1:00 competition day and couldn't fall asleep because of the annoying air conditioning that I couldn't turn off.  I read Countdown for the rest of the time and somehow didn't fall asleep during the competition.  What wonderful rest I had!  Greg didn't notice a thing...

i bet my room was the noisiest room in the hotel. good thing the only people next to us were our teammates and our coach. i brought a tennis ball and kept chucking it against the wall at around 11:00 at night. also, we'd play the super mario theme song really loud just to get people mad.

Seven hours straight?? you guys left the room like two hours before I did...i got like no sleep wutsoever...im still catching up on sleep...my coach was right next door to me so I couldnt really make any noise so we had to play cards and talk and stuff in the other people's rooms...btw i got a 36...placed 18th

[quote="probability1.01"]...couldn't fall asleep because of the annoying air conditioning that I couldn't turn off...[/quote]

You're lucky that you at least [i]had [/i]air conditioning. The last two years when I went to nationals, they didn't have the air conditioning and the room got too hot and stopped me from sleeping. Particularly in 2002 (Although the next night I stayed up all night voluntarily - I'll leave that story for another day - it's quite an amusing story though)

ha...i got a cold because my roommate refused to turn off the AC

I almost gave two of my teammates a cold because I lowered their air conditioning to about 45 degrees, but it didn't work.
Maybe I forgot to push on or something.

i stayed up to like 3AM on saturday..
now i have a cold

what really annoyed me was that their TV didnt have an input cable link so we couldn't play the two gamecubes our team brought. And their "TV internet" didnt have java lol.

haha o well as wutsherface said one nights worth of sleep lost is gonna have no effect on u in the long run

i thought the team activity was pretty cool, especially since we got to make something. it was just different from like going to the zoo, because you're doing something and building.

yea the adults made our team  go to the zoo on saturday anyway...nothing really special there i guess i mean ATL zoo has pandas and everything too so...but the team activity was fun even if we didnt win...

on sunday, i went to the space & flight museum. didn't get to go on the flight simulator though...

We left the hotel at 7 am sunday morning we did nothing on sunday although on saturday we took a tour on one of the little tram thingys and went to the zoo

who knows anything about california team scores? :D

two of them were higher than 36 and two were lower

[quote]The adults with our team made us go see everything, so we ended up walking around for about 6 hours.[/quote]
We didn't walk the entire 6 hours. You and Greg spent a lot of time making that card game and we sat in front of the international spy museum for at least a half hour.

Our team started walking on Sat. about 8ish and stopped walking about 5 in the afternoon... our coordinator left us in the middle cause she was too tired

judging by the way he made his post i guess JMI is nahlik?? or jenny??

I'd guess Jenny because [b]J[/b]enny [something (M?)] [b]I[/b]glesias

[quote="probability1.01"][quote="captcha000"]i take back the thing i said about how did you become quarter or semi finalist in the other thread[/quote]

?????[/quote]

I beleated it.

[quote="Fermatprime"][quote="justinian"]what did you do?[/quote]

played games for about 7 hours straight...[/quote]

Ya well that is about the same thing we did, including poker etc.  We were sitting next to the people from NMI (northern mariana islands) for the awards ceremony, and they have a 10 hout time difference, so they were like falling asleep at the table.  It was pretty funny.  (sorry if either of the two people I am mentioning are actually looking at this forum)

The Nationals were pretty good... the entire NJ team got in the top 100s :D . 17th, 36th, 64th, and 81st, if I remember correctly... top ten teams, too. I got in 36th place :D

I played Hearts with the people from Pennsylvania and my teammates until about 11 PM...