Prove that for any prime $ p > 2$ that

 $ \displaystyle\frac{2 \minus{} 2^p}{p} \equiv \minus{}\displaystyle\sum_{j\equal{}1}^{p \minus{} 1}\frac{2^j}{j}$ $ {mod \,p}$

More generally for any natural number $ n$, and an odd prime $ p$, prove that

 $ \displaystyle\frac{n \minus{} n^p}{p} \equiv \minus{}\displaystyle\sum_{r\equal{}1}^{p \minus{} 1}\frac{1^r \plus{} 2^r \plus{}.. \plus{} n^r}{r}$ $ {mod \,p}$

Am I missing something?  I checked the first equivalence for the first 150 odd primes (with Maple) and found that it is false for all of them.  Perhaps, you meant something else?


I appologize for the severely alliterated title :).

Find all solutions x for:

$x\equiv 1 \pmod{3}$
$x\equiv 2 \pmod{5}$
$x\equiv 1 \pmod{7}$
$x\equiv 1 \pmod{11}$
$x\equiv 1 \pmod{13}$

Obviously there are an infinite number, so express your answer in the form $x \pmod{n}$.

[hide]Look at $x-1$.  It is:
$\equiv0 \pmod{3}$
$\equiv0 \pmod{7}$
$\equiv0 \pmod{11}$
$\equiv0 \pmod{13}$
So it's $x-1\equiv0 \pmod{3\cdot7\cdot11\cdot13}$, or $x-1\equiv0 \pmod{3003}$.  Look at 6006.  It satisfies that as a value of $x-1$, and also, $6006\equiv1 \pmod{5}$.  Clearly, $x\equiv6007 \pmod{3\cdot5\cdot7\cdot11\cdot13}$, or $\boxed{6007 \pmod{15015}}$.[/hide]

I have a decent solution to this that doesn't necessarily require a calculator.  If you'd rather see a solution than find one, look it over.

[hide]Going down the line, top to bottom:
(I will label each important equation with a symbol).

$\clubsuit x\equiv 3a + 1 \Rightarrow 3a+1\equiv 2 \pmod{5}$
Logically, then:
$3a\equiv 1 \pmod{5} \Rightarrow a\equiv 2 \pmod{7}$

Substitute this back into equation $\clubsuit$, and:

$\spadesuit x\equiv 3(5b +2) + 1 \Rightarrow x\equiv 15b + 7$

$15b + 6\equiv 1 \pmod{7} \Rightarrow 15b\equiv 1 \pmod{7}$
Just by trying a few numbers, you soon realize that:
$b\equiv 1 \pmod{7} \Rightarrow b\equiv 7c + 1$

Plug this into $\spadesuit$ and:
$x\equiv 15(7c+1)+7 \Rightarrow \diamondsuit x\equiv 105c + 22$
Put this into mod 11, and:
$105c + 22\equiv 1 \pmod{11} \Rightarrow 105c\equiv 1 \pmod{11}$

Think of this as $99c+6c\equiv 1 \pmod{11} \Rightarrow 6c \equiv 1 \pmod{11}$
So logically, $c\equiv 2 \pmod{11} \Rightarrow c\equiv 11d + 2$

Put this into $\diamondsuit$ and:

$x\equiv 105(11d+2) + 22 \Rightarrow \heartsuit x\equiv 1155d +232$

We know with the given then that:

$1155d+232\equiv 1 \pmod{13} \Rightarrow 1155d\equiv 3 \pmod{13}$

The closest multiple of 13 is 1144, so think of 1155d as:

$1144d + 11d$ so $11d\equiv 3 \pmod{13} \Rightarrow d\equiv 5 \pmod{13}$

Put this in terms of e, and:

$d\equiv 13e + 5$  Now plug this into $\heartsuit$.

$x\equiv 1155(13e+5) + 232 \Rightarrow x\equiv 15015e + 6007$

Hurray!  $x\equiv 6007 \pmod{15015}$[/hide]

I realize this is extremely long, but it definitely proves x to be, well, you know.  Lingo, I like your reasoning (and that was my first impulse, just to say 6007) but I was never really quite sure that I could prove it :).


projective geometry
geometry
trigonometry
geometry proposed

Two circles $ k_1,k_2$ with centers $ O_1,O_2$ respectively are externally tangent at $ P$. A circle $ k_3$ is tangent to $ k_1$ at $ Q$ and to $ k_2$ at $ R$. The lines $ PQ,PR$ meet $ k_3$ again at $ A,B$ respectively. If $ AO_2$ and $ BO_1$ meet at $ S$, show that $ SP$ is perpendicular to $ O_1O_2$

Here's a sketch of a solution:
Let centre of third circle be $ O_{3}$. An angle chase gives us that $ AO_{3}$ is parallel to $ O_{1}P$ (indeed, let the tangent at $ Q$ to 1st, 3rd circles be $ l$. It's easy to see $ \angle QO_{1}P \equal{} 2 \angle (l, QP) \equal{} \angle AO_{3}Q$). Analogously, $ BO_{3}$ is parallel to $ O_{1}O_{2}$, so $ A,B,O_{3}$ are collinear on a line parallel to $ O_{1}O_{2}$. Applying Desargues on the sets of points, lying on parallel lines $ A, O_{3}, B$ and $ O_{1}, P, O_{2}$, tells us that $ S$ lies on $ QR$. We wish to prove that $ SP$ is perpendicular to $ O_{1}P$. For simplicity, let's convert this problem to a triangle geometry problem:

Problem: In triangle $ ABC$ with angles $ 2A, 2B, 2C$, let the incircle touch the sides at $ A_{1}, B_{1}, C_{1}$. Choose $ S$ on $ B_{1}C_{1}$, with $ A_{1}S$ perpendicular to $ BC$. let $ C_{1}A_{1}$ intersect the line through $ A$, parallel to $ BC$, at a point $ K$. then $ K, S, C$ are collinear.

One way to nuke this is to prove that,
$ \frac {sin \angle SCB_{1}} {sin \angle SCA_{1}} \equal{} \frac {sin \angle KCA} {sin \angle KCA_{1}} ... (*)$
After proving this, clearly the conclusion will follow. Since we can angle chase all of the necessary angles in quadrilateral $ SB_{1}A_{1}C$, we can use Sin Rule to find the LHS of equation $ (*)$. A bit more trig allows us to find the lengths $ KA, KA_{1}$, and we can angle-chase some angles in quadrilateral $ KAA_{1}C$, and this allows us to find the RHS of $ (*)$, so in that way we can prove $ (*)$.


Mike drives to work traveling 40 mph.  Half way there he realizes he forgot his wallet at home and he travels 50 mph on the way home to get it.  It takes him awhile to find his wallet, so when he heads out to work again he travels 60 mph all the way there.  What is Mike's average speed during this morning adventure expressed as a mixed number?

I think you need to know how long it takes for him to find his wallet.

[hide]Assuming that you don't count the time when he is looking for his wallet, I just pretended that the distance was 1200 miles (although unreasonable), so he would travel 2400 miles total and take 600/40+600/50+1200/60 = 47, and 2400/47 = [b]51 3/47.[/b][/hide]

[hide]Assume the distance is 1200.

R___T___D
40__15__600
50__12__600
60__20__1200

Adding the time column, we get a total of 47 hrs.
Adding the distance column, we get a total of 2400 miles.
Dividing 2400/47, we get $\boxed{51\frac{3}{47}}$.[/hide]

A more eloquent way (no assumptions necessary, thus it clearly works for all cases):
[hide]
$rt=d$
$40t=d$
$50(4t/5)=d$
$60(2t/3)(2)=2d$

Avg r= total d/total t

$=(40t+50(4/5)t+60(4/3)t) / (t+(4/5)t+(4/3)t)$
$=2400/7$[/hide]

Harmonic mean doesn't work here? I tried that and got $\frac{1800}{37}$.

I think harmonic mean is a bit tedious here, but here is how you would do it.
[hide]
To use harmonic mean, the point is that each leg must be the same distance. So break down the trip into 4 legs of half the distance to work. He takes 4 of these, 40mph,50 mph, 60mph, and another at 60mph.

So for harmonic mean you want $\frac{1}{\frac{\frac{1}{40}+\frac{1}{50}+\frac{1}{60}+\frac{1}{60}}{4}}$, which is equal to $51\frac{3}{47}$[/hide]

[quote="ajai"]what is harmonic mean? :?  :huh:[/quote]

Harmonic mean of $a_1, a_2, a_3...,a_n$ is $\frac{1}{\frac{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}}}{n}$.

[quote="ajai"]what is harmonic mean? :?  :huh:[/quote]
basically the inverse of the average of the inverses.


function
calculus
integration
factorial
calculus computations

I know the very *basics* of how the Gamma function works, but how would I solve an equation like
$\Gamma(x) = 110$?

As far as I know, $\Gamma^{-1}(x)$ doesn't have a closed form, or even a series or product form.

This isn't the sort of thing you use the Gamma function for, anyway.

Really? I'm new to this, but I read somewhere that gamma can be defined as a definite integral. Are you sure that the inverse function isn't defined similarly?

Can you find a way to invert an integral to form another integral?  Because I can't.  :P

Fair enough. I'm new to calculus, I've only been introduced to differentials, and integrals are completely beyond me now. I've been doing contest math for a while, though, and I've never seen any calculus on any competitions, so I don't think I'll need to use the Gamma function much.

It's the extension of factorial to real and complex numbers, not just natural numbers.


modular arithmetic
number theory proposed
number theory

[quote="Jumbler"]Prove that for any number $ l$ such that $ 0\leq l\leq 9$, one can find $ k$ such that the $ n$th digit (counting from the right) of $ 2^k$ is $ l$.
[/quote]

Wrong : Take $ n\equal{}1$ and $ l\equal{}3$ and no such $ k$ exists.

[quote="pco"][quote="Jumbler"]Prove that for any number $ l$ such that $ 0\leq l\leq 9$, one can find $ k$ such that the $ n$th digit (counting from the right) of $ 2^k$ is $ l$.
[/quote]

Wrong : Take $ n \equal{} 1$ and $ l \equal{} 3$ and no such $ k$ exists.[/quote]
k=5 solution, because $ 2^5\equal{}32$ and first (n=1) (counting from the right ) is l=3.
It is followed from fact, that $ log_{10}2$ is irrational.

[quote="Rust"][quote="pco"][quote="Jumbler"]Prove that for any number $ l$ such that $ 0\leq l\leq 9$, one can find $ k$ such that the $ n$th digit (counting from the right) of $ 2^k$ is $ l$.
[/quote]

Wrong : Take $ n \equal{} 1$ and $ l \equal{} 3$ and no such $ k$ exists.[/quote]
k=5 solution, because $ 2^5 \equal{} 32$ and first (n=1) (counting from the right ) is l=3.
It is followed from fact, that $ log_{10}2$ is irrational.[/quote]

Sorry, Rust but first digit counting from the right in $ 32$ is $ 2$.  :P

$ \longrightarrow$ left $ 3 2$ right $ \longleftarrow$

For any $ n>1$, because $ 2^n\equal{}\frac{10^n}{5^n}<\frac{10^{n\minus{}1}}{2}$, there exist two consecutive positive multiples of $ 2^n$ with $ n$ digits and a leading digit of $ l$, so at least one of them is not a multiple of 5. Let $ m$ be an $ n$-digit multiple of $ 2^n$ with leading digit $ l$ which is not divisible by 5. Then if we can show 2 is a primitive root modulo $ 5^n$, then there exists $ k$ such that $ 2^k\equiv m\pmod{5^n}$ and, by adding a multiple of $ 4\cdot 5^{n\minus{}1}$ if necessary, have $ k\geq n$, obtaining the result.

We will now prove this assertion by induction. For $ n\equal{}1$, it's easy to check. For $ n\equal{}2$, the order is divisible by 4 (from the $ n\equal{}1$ case) and a divisor of 20. We can easily see it's not 4, so must equal 20.

Now given true for $ n\geq 2$, we'll prove for $ n\plus{}1$. So we have that $ 2^{4\cdot 5^{n\minus{}2}}$ is congruent to 1 modulo $ 5^{n\minus{}1}$ but not modulo $ 5^n$. Write $ 2^{4\cdot 5^{n\minus{}2}}\equal{}1\plus{}a\cdot 5^{n\minus{}1}$ where $ a$ is not divisible by 5. Then
\[ 2^{4\cdot 5^{n\minus{}1}}\equal{}(1\plus{}a\cdot 5^{n\minus{}1})^5\equiv 1\plus{}a\cdot 5^n\pmod{5^{n\plus{}1}}\]
since in the binomial expansion, the second term is divisible by $ 5^{2n\minus{}1}$ and the remaining terms are divisible by $ 5^{3n\minus{}3}$. As $ n\geq 2, 3n\minus{}3\geq 2n\minus{}1\geq n\plus{}1$. So the order of 2 modulo $ 5^{n\plus{}1}$ is divisible by, but not equal to, $ 4\cdot 5^{n\minus{}1}$, and a divisor of $ 4\cdot 5^n$. So the order must equal $ 4\cdot 5^n$ making 2 a primitive root as desired.

[quote="scorpius119"] by adding a multiple of $ 4\cdot 5^{n \minus{} 1}$ if necessary, have $ k\geq n$, obtaining the result.

[/quote]

What do you mean here?  :maybe:

If $ 2^k\equiv m\pmod{5^n}$ but $ k<n$, then $ 2^{k\plus{}4\cdot 5^{n\minus{}1}}\equiv m\pmod{5^n}$ and $ k\plus{}4\cdot 5^{n\minus{}1}\geq n$.


geometry
analytic geometry
geometric transformation
reflection
rotation
rectangle
ratio

A train traveling at a speed of 30 miles per hour passes point A on its way to point B. At the same time, on a parallel track, another train traveling at a speed of 70 miles per hour passes point B on its way to Point A. If point A and B are 300 miles apart, how from point B will the trains meet?

2.) There are 10,000 fish in a pond. Of the 50 fish caught, 35 are female and 15 are male. What is a good estimate of male fish before the 50 fish were caught?

3.) A cylindrical soup can is 4 inches tall and has a radius of 1.5 inches. What is the area of a label that will completely cover the side of the can, not including the top and bottom.

4. One diagnal of a square lies on the y axis of a coordinate system. The coordinates of one corner are (3,7). What are the coordinates of the opposite corner?

5.) What do you do to the coordinates if you are rotating it 90 degrees. Like (x,y) => (?, ?)

Let P= (2,3). First translate P one unit right and call it R. Then reflect R over the y axis and call this image S. Finally rotate S 90 degrees and call it T, what are the coordinates of T?

6.) IN a scale diagram 1 inch= 100 ft, how many square inches represent 1 square foot?

7. When N is divided by 5 the remainder is 2, what is the remainder when n+4 is divided by 5?

-Show all work so maybe I'll be able to do it like you guys!  :D

[hide="#1"]
The trains will meet in $ \frac {300}{70 \plus{} 30} \equal{} 3$ hours. The train starting from point B has a speed of $ 70$ mph, so, in that time train will move $ 210$ miles. Since the trains meet at that point, our answer is $ \boxed{210}$ miles.
[/hide]

[hide="#2"]
$ \frac {15}{50}$ is male, so, out of 10,000 fish, $ 10000\cdot \frac {15}{50}$ are male, or $ \boxed{3000}$ are male.
[/hide]

[hide="#3"]
The area is a rectangle, the length of which is the circumference and the width is the height. The circumference is $ \pi\cdot 1.5\cdot 2 \equal{} 3 \pi$.
$ 4\cdot 3pi \equal{} \boxed{12\pi} \approx \boxed{38}$ square inches.
[/hide]

[hide="#4"]
The square is rotated 45 degrees (diamond shape).
Squares are rhombi. Thus, we find the shortest distance from the corner to the y-axis (the perpendicular diagonal), and double it to get to the other corner. So, The distance from $ (3,7)$ to the y-axis is $ 3$. From $ 0,7$, subtract another 3 from the x-coordinate to get $ \boxed{( \minus{} 3,7)}$.
[/hide]

[hide="#5"]
First off, it depends what direction it is rotated.
By logic, switch the x and y coordinates. Then adjust the positive/negative signs accordingly to the quadrant the final point ends up on.
Another way is rotation formulas, which do not belong here, but look on Wikipedia for more info.

"http://en.wikipedia.org/wiki/Rotation_(mathematics)"

Also, the direction rotated [b]must[/b] be specified.

For the actual problem:
P=(2,3)
R=(3,3)
S=(-3,3)
T=(-3,-3)
For T, If you draw it, the answer is obvious without knowing any methods.
[/hide]

[hide="#6"]
Wow. huge numbers.
Ratio=$ 1: 100$
Ratio of squares= $ 1^2: 100^2$, or $ 1: 10,000$
And then divide accordingly. $ \frac {1}{10000}: 1$
So, the answer is $ \frac{1}{10000}$ square inches.
[/hide]

[hide="#7"]
Modular arithmetic.
Basically, $ n \equiv 2 \mod 5$, which means that $ n$ has remainder $ 2$ when divided by $ 5$. We can add $ 4$ to both sides of the modular equation.$ n \plus{} 4 \equiv 6 \mod 5$. In other words, $ n \plus{} 4$ divided by $ 5$ is the same as $ 6$ divided by $ 5$, and thus the answer is $ \boxed{1}$.
[/hide]


Is it true?
for all $x,y,z \in R$
$8 (x^{3}+y^{3}+z^{3}) + 30xyz \geq 9 \Sigma xy(x+y)$
I have proved it right. But I'm not sure if it is correct.

Look for $x=y=0,z=-1$   :?: (or $1,1,5$ if we consider only $> 0$ numbers)

I proved for positive x,y,z
Sorry, I found that my solution wrong.
But for positive x,y,z, the inequality is also wrong.
Ex: x=1.7,y=0.01,z=1.29.


probability
function
vector
complex numbers
abstract algebra
Functional Analysis
real analysis

In physics it is often interesting to look at two subsystems “living” in two Hilbert spaces. To make my question a bit more simple, let’s say we have two 2-dimensional Hilbert spaces H1 and H2.

In quantum mechanics we often look at the probability of a given system q1(x), is in a specific state h1(x), where q1,h1 in H1. This probability is given by the inner product <q1 | h1>  (thinks it’s defined from the L2-norm). So if h is fixed < | > define a map from H1 to the complex numbers. The same goes for the inner product on H2.

If I then have a system consisting of the two subsystems I can calculate the probability of that system being in the state (q1,q2) given by    P(q1,q2) = <q1 | h1><q2 | h2>, the map P from H1 x H2 (x means the direct product) to the complex numbers, I guess is a bilinear map.

My understanding of the tensor product is that every bilinear map on H1 x H2 can be calculated in H1 t H2 (t means the tensor product), so I can define P(q1 t q2) = <q1 t q2 | h1 t h2> = <q1 | h1><q2 | h2>, so now P is a mapping form H1 t H2 to the complex numbers.

My problem is that this is very theoretical, some of the theorems I have read give me very useful information’s like H1 t H2 has dimension 2 * 2 = 4, so my complete system is 4-dimensional, but I can’t understand what the elements q1 t q2 of H1 t H2 are. Can you give me an explanation of this maybe some general an maybe an example with q1 = x^2 + 1 q2= 2x + 4 and so on, not necessarily these functions but the ones who illustrates it the best.

Thanks in advance, Anders Nørløv Berthelsen

If $H_{1}$ is a space of linear functionals on a vector space $V_{1}$ (over a field $F$) and $H_{2}$ is a space of linear functionals on a vector space $V_{2}$, there is a canonical identification of $H_{1}\otimes H_{2}$ with a space of bilinear functionals on $V_{1}\times V_{2}$.

If $H_{1}$ and $H_{2}$ aren't dual to spaces we know, it's not really worth trying to understand what the elements of $H_{1}\otimes H_{2}$ are. It's just too confusing.

It is important to note that elements of the form $a\otimes b$ aren't the whole tensor product; you need linear combinations of these.

Ok thanks, i know I need linear combinations I i just asked for a single element because it was the easiest. So wht use the tensor product is the only reason (in my case) that i get a theoretical new space for my whole system, I mean I can only use it to describe general things of my system space like it is 4-dimensional, and then I do all caculations in the two hilbertspaces and then calculate <q1 t q2 | h1 t h2> by <q1 | h1><q2 | h2>?

I have made some theoretical theorems in my new system space where I use it is 4-dimensional, and use that I can choose a basis for it, but i didn't use any particular basis, so when I do my real pratical calculations I would never try to find a basis for my new space and calculate in this space, if I am understanding what your are saying?

sorry for the late reply, I had forgot to check "Notify me when a reply is posted", so I didn't think anybody had replied.

ps. how do I use tex-code?

[quote="mrAndersDk"]how do I use tex-code?[/quote]
If you know the code for what you want, just put it between dollar signs.

[quote="mrAndersDk"]... and then calculate <q1 t q2 | h1 t h2> by <q1 | h1><q2 | h2>?[/quote]
That looks completely different than what I was thinking of. You've just defined an inner product on $H_{1}\otimes H_{2}$, using the inner products on $H_{1}$ and $H_{2}$. This has nothing to do with the interpretation of tensor products as bilinear maps.
My best attempt at interpretation:
$H_{1}$ represents the distribution of one system across some states. $H_{2}$ represents the distribution of another system across some other states. $H_{1}\otimes H_{2}$ represents the joint distribution of the combined system. An element of the form $q_{1}\otimes q_{2}$ represents a joint state in which the two systems are probabilistically independent; such a state is unlikely if the systems interact.

yes $q_{1}\otimes q_{2}$ is an element in the joint system, but my problem is that I don't know what the element is if I have some concrete elements $q_{1}$ and $q_{2}$, so my question is that rather than calculating on $q_{1}\otimes q_{2}$ (which I odn't know what is) I calculate on $q_{1}$ and $q_{2}$ individual.

fx. I know what to do with the regular times or $\circ$ for two functions:

$f(x) \cdot g(x) = f(x)g(x)$     (not a very good illustration)
$f(x) \circ g(x) = f(g(x))$

but

$f(x) \otimes g(x) = ???$

You see, I don't know what the new elements are in concrete cases so i can't do calculations in my new space $H_{1}\otimes H_{2}$, so I can only use it for theoretical things where I don't need to know what the elements are.

So my question is do you ever calculate $f(x) \otimes g(x) = ???$ or is it only for theoretical things?

I've seen $A \otimes B$ defined for matrices by the kroenecker product but never for functions that are not linear

If $f$ and $g$ belong to spaces of functions on $U$ and $V$ respectively, I'd say that $f\otimes g$ is a function on $U\times V$ given by $(f\otimes g)(x,y)=f(x)g(y)$. This doesn't really need linearity anywhere, although making some things linear does help.

I think I'm a bit confused.

If you define $g \otimes f = g(x) \cdot f(y)$ then if you take the hilbert space with basis $\{1,x\}$ lets call it $H$, then $H$ is 2-dimenseional, so $H \otimes H$ is 4 dimensional, but isn't a basis for $H \otimes H$ given by $\{1 \otimes 1,1 \otimes x, x \otimes 1,x \otimes x \}$ with the definition of $\otimes$ this must equal $\{1,x,x^{2}\}$ but this is only 3-dimensional.

Maybe I just read to many different versions of the tensor product and I got too confused, there are so many different ways to explaine it and not many examples on how to use it.

Maybe you can recomment a good book, there introduce the tensor product very clear and give some good examples ?

I've found alot of books on the net that are free, and most of them introduce it as a way to make a multilinear map linear by factoring through the tensor product, but they never give any examples, so you can see how to use it in practice.

If you know some free books or websites that got what i need it would be great, if you only know books that are not free, I must try to get them somehow.

Thanks for trying to help me.

Your example is wrong- the basis of $H\otimes H$ should be $1,x,y,xy$. That "x" in your definition of $H$ is really meant to be the identity function, so $x(y)=y$. If we're treating these as functions on some space like $\mathbb{R}$, the tensor product is a space of functions on $\mathbb{R}^{2}$, not a space of functions on $\mathbb{R}$.

[quote]Maybe you can recomment a good book, there introduce the tensor product very clear and give some good examples ?[/quote]
Unfortunately, I can't. I've never seen one.
The first time I saw tensor products was in a graduate algebra context, in the form of tensor products of modules. This can get really weird- the tensor product $\mathbb{Z}/n\mathbb{Z}\otimes \mathbb{Z}/m\mathbb{Z}$ over $\mathbb{Z}$ is trivial if $m$ and $n$ are relatively prime.
General tensor products are always treated at arm's length, with abstract nonsense and universal properties used for everything. The only situation that can be made concrete is a tensor product of function spaces.

Okay, thanks for your help. I also read about tensor products in algebra, but I find it hard to make sense of, when I can't se any really practical use of it, the proofs with the tensor product (i've seen) are fairly simple, but for me it's hard to figure out what makes the tensor product so smart.

For me it's a bit like introducing matrices and don't know that you can solve linear equations and represent linear maps very easy.

One last question before my quest goes on. Do you define a tensor product of a given space, like you say the tensor product on a function space is given by $f \otimes g = f(x)g(y)$, could there be another tensor product on the space, or is it something that you can proof that the tensor product is given by $f \otimes g = f(x)g(y)$.

There is only one tensor product, and it's abstractly defined. We have a canonical isomorphism of this product with a space (not necessarily all) of functions on the Cartesian product, so there's usually no harm in identifying the two.

An example: Let $X=C[0,1]$, the space of real-valued continuous functions on $[0,1]$. $X\otimes X$ can be identified with a space of continuous functions on the unit square $S=[0,1]\times [0,1]$, but it's not the full space $C(S)$. $X\otimes X$ is dense in $C(S)$, but $X\otimes X$ does not contain the function $f(x,y)=\frac1{1+xy}$.

The drawback of this identification is that it can be hard to tell exactly what's in the space. The advantages of having something concrete to work with usually outweigh that.

Thanks for your help. I don't understand it all but it helped a great deal. I think I really got blinded by wanting to know what the space $H_{1}\otimes H_{2}$ looked like instead of just using the properties of the space.


Write $2\log_{2}40-\log_{4}125+\frac{1}{2}\log_{8}5$ in the form $p+q\log_{2}r$, where $p$ and $r$ are integers, and $q\in\mathbb{R}$.


limit
trigonometry
function
calculus
calculus computations

Here it goes:

${ \lim_{x\rightarrow 0}{\frac{\sqrt{1-e^{-x}}-\sqrt{1-\cos x}}{\sqrt{\sin x}}}=\lim_{x\rightarrow 0}{\frac{\sqrt{1-e^{-x}}-\sqrt{1-\cos x}}{\sqrt{x}}}\cdot\frac{\sqrt{x}}{\sqrt{\sin x}}}$, which obviously equals $ \lim_{x\rightarrow 0}{(\frac{\sqrt{1-e^{-x}}}{\sqrt{x}}-\frac{\sqrt{1-\cos x}}{\sqrt{x}})\cdot\frac{\sqrt{x}}{\sqrt{\sin x}}}$.
Since we now have three well known limits: $ \lim_{x\rightarrow 0}{\frac{1-e^{-x}}{x}}=1$, $ \lim_{x\rightarrow 0}{\frac{1-\cos x}{x}}=0$ and $ \lim_{x\rightarrow 0}{\frac{x}{\sin x}}=1$ and $ \sqrt{\cdot}$ is a continuous function, we get $ \lim_{x\rightarrow 0}{\frac{\sqrt{1-e^{-x}}-\sqrt{1-\cos x}}{\sqrt{\sin x}}}=(\sqrt{1}-\sqrt{0})\cdot \sqrt{1}=1$

there is no limit as x approaches 0,

but there is a limit as x approaches 0 from the right.  ( the value is indeed 1 )
The error in the solution is when you multiplied and divided by the square root of x ( which is undefined on the left of 0 ).

By the way, A quick  look at the problem clearly reveals that the limit at x approaches 0  CANNOT exist since the square root of the sine function is undefined on the left of 0.  And how about the square root of (1 - e^-x) ....??  well, is it defined on BOTH sides of 0 ??

  :)


In nr/septembrie 2007 al revistei maghiare "$ KOMAL$" a aparut urmatoarea problema data la olimpiada din CHINA-2007:
"$ {A. 431.}$ A nem egyenlő szárú ABC háromszög körülírt és beírt körének középpontja O, illetve I. A beírt kör a BC, CA, AB oldalakat rendre a D, E, F pontokban érinti. Az FD és AC egyenesek metszéspontja P, a DE és AB egyenesek metszéspontja Q. Az EP és FQ szakaszok felezőpontja M, illetve N. Bizonyítsuk be, hogy MN merőleges OI-re.  (Kínai Matematikai Olimpia, 2007)"

In traducere:
"Fie $ O$ şi $ I$ centrul cercului circumscris şi înscris în triunghiul neisoscel $ ABC$. Notăm cu $ D$, $ E$ şi $ F$ punctele de tangenţă ale laturilor $ [BC]$, $ [CA]$ şi respective $ [AB]$ cu cercul înscris, cu {$ P$} = $ {AC}\cap{DF}$ şi {$ Q$} = $ AB\cap{DE}$ şi fie $ M$ şi $ N$ mijloacele segmentelor [$ PE$] şi [$ QF$]. Arătaţi că dreptele $ OI$ si $ MN$ sunt perpendiculare."

Iata solutia data de mine acestei probleme:
1). [$ ME$] fiind tangenta la cercul ($ I$; $ r$) $ \Rightarrow$ cercul ($ M$; |$ ME$|) este ortogonal cercului ($ I$; $ r$).
2). Dreptele $ AD$, $ BE$ si $ CF$ sunt concurente in punctul lui $ {Gergonne}$ al triunghiului $ ABC$, iar dreapta $ PQ$ este polara triliniara a punctului lui
$ {Gergonne}$ $ \Rightarrow$ punctele $ E$ si $ P$ sunt conjugate armonic fata de punctele $ A$ si $ C$ $ \Rightarrow$ cercdul ($ M$; |$ ME$|) este ortogonal la
cercul ($ O$, $ R$), circumscris triunghiului $ ABC$.
3). In mod analog se arata ca cercul ($ N$; |$ NF$|) este ortogonal la fiecare dintre cercurile: ($ I$; $ r$) si ($ O$, $ R$).
4). In fine cercurile ($ M$; |$ ME$|) si ($ N$; |$ NF$|) fiind ortogonale la cercurile ($ I$; $ r$) si ($ O$, $ R$) $ \Rightarrow$ $ {MN}\perp{OI}$.

[color=red]Domnul profesor [b]Mihai Miculita[/b] ne ofera din nou o frumoasa problema. Va multumesc.
Va rog, in viitor sa nu mai atasati imediat si propria solutie, ci doar dupa "chinul" nostru ...[/color]
[quote="mihai miculita"][color=purple]"Fie cercul circumscris $ w = C(O,R)$ şi cercul înscris  $ C(I,r)$ pentru triunghiul neisoscel $ ABC$. Notăm cu $ D$, $ E$ şi $ F$ punctele de tangenţă ale laturilor $ [BC]$, $ [CA]$ , $ [AB]$ respectiv cu cercul înscris, intersectiile $ P\in AC\cap DF$ şi $ Q\in AB\cap DE$ şi mijloacele $ M$ , $ N$ ale segmentelor $ [PE]$ , $ [QF]$ respectiv. Sa se arate că dreptele $ OI$ si $ MN$ sunt perpendiculare."[/color][/quote]

[color=darkred][b]Lemma.[/b] Daca punctele $ C$ , $ D$ sunt conjugate armonic in raport cu punctele $ A$ , $ B$
si punctul $ M$ este mijlocul segmentului $ [CD]$ , atunci exista relatia $ MC^2 = MA\cdot MB\ .$[/color]

[color=darkblue][b]Demonstratia problemei propuse.[/b] Notam $ p_w(X)$- puterea punctului $ X$ fata de cercul $ w$. Asadar, $ \{\begin{array}{c} p_w(M) = MA\cdot MC = MO^2 - R^2 \\
\ p_w(N) = NA\cdot NB = NO^2 - R^2\end{array}$.  Se stie ca punctele $ P$ , $ E$ sunt conjugate armonic in raport cu punctele $ A$ , $ C$ si punctele $ Q$ , $ F$ sunt conjugate armonic in raport cu punctele $ A$ , $ B$. Folosind lema mentionata mai sus obtinem relatiile $ \{\begin{array}{c} ME^2 = MA\cdot MC \\
\ NF^2 = NA\cdot NB\end{array}\|$ $ \implies$ $ \{\begin{array}{c} ME^2 = MO^2 - R^2 \\
\ NF^2 = NO^2 - R^2\end{array}$ . Dar $ \{\begin{array}{c} ME^2 = MI^2 - r^2 \\
\ NF^2 = NI^2 - r^2\end{array}$.
In concluzie, $ MO^2 - MI^2 = NO^2 - NI^2 = R^2 - r^2$, adica $ MN\perp OI\ .$

[b]Remarca.[/b] Notam $ S\in MN\cap OI$. Atunci $ I\in (OS)$ si $ \frac {SO}{SI} = 1 + \frac {2R(R - 2r)}{r(2R - r)}\ge 1\ .$[/color]


calculus
integration
trigonometry
calculus computations

Let $ a,\ b$ be real constants. Find the minimum value of the definite integral:

$ I(a,\ b)\equal{}\int_0^{\pi} (1\minus{}a\sin x \minus{}b\sin 2x)^2 dx.$

$ I(a,\ b)\equal{}\left( \sqrt{\frac{\pi }{2}}a\minus{}2\sqrt{\frac{2}{\pi }}\right) ^{2}\plus{}\frac{\pi }{2}b^{2}\plus{}\pi \minus{}\frac{8}{\pi }$

$ \vspace{1pt}I_{\min }\equal{}\pi \minus{}\frac{8}{\pi },$ occurs at $ a\equal{}\frac{4}{\pi },b\equal{}0$


how many four digit number are possible if at most 2 digit are to be  used

Do you mean 4-digit numbers or 4-digit combinations?  In other words, is a number with leading zeros (like $ 0220$) acceptable?

if 4 digit combinations(0 can be 1st entry), then there are :

$ 10\times9\times{4\choose2} \equal{} 540$

if 4 digit numbers(0 cant be 1st entry), then there are:

$ 10\times9\times{4\choose2} \minus{} 9\times{3\choose2} \equal{} 513$

hmm i got different answer

2 a's and 2 b's

9*9*6=486

all a

1*9=9

486+9=495

answer : 495

Almost right. You forgot to count 3 of one and one of the other. Also, you have to count that 2 digits that are the same cannot be interchanged to get a new combination.


geometry
power of a point
radical axis
geometry unsolved

(F.Nilov) Two arcs with equal angular measure are constructed on the medians $ AA'$ and $ BB'$ of triangle $ ABC$ towards vertex $ C$. Prove that the common chord of the respective circles passes through $ C$.

Denote these circles as $\omega_{1}$ and $\omega_{2}$ for medians $AA'$ and $BB'$ respectively. Assume that $\omega_{1}$,  $\omega_{2}$ interesect at $P$ and $Q$, line $AC$ intersect $\omega _{1}$ at point $K$ and line $BC$ intersect $\omega_{2}$ at point $L$.  We have $\angle B'LB = \angle A'KA',  $ so triangles $B'LC$ and $KCA'$ are similar. Thus, because points $A'$ and $B'$ are midpoints of sides $BC$ and $AC$ respectively, we have that triangles $LCA$ and $KCB$ are similar. So quadrangle $ABLK$ is cyclic.  Thus using radical axis thorem,  lines $PQ$, $BL$ and $AK$ are concurrent at point $C$.

Well, little bit faster:
We need to show that $C$ has equal powers w.r.t. $\omega_1,\omega_2$; keep the same meaning for $K,L$ from first solution.
As shown, $\angle AKA'=\angle BLB'$, consequently $K,L,B',A'$ are concyclic and $LC\cdot CA'=CK\cdot CB'\iff CK\cdot CA=CL\cdot BC$, done.

Best regards,
sunken rock

Strictly speaking, it should be $\measuredangle AKA'=\measuredangle B'LB$, since $K,L$ may lie on the extension of $CA$ or $CB$.


calculus
integration
calculus computations

Find
$ \sum_{n\equal{}0}^{\plus{}\infty}{2n\choose n}\frac{1}{7^{n}}$.

[quote="Spasty"]Find
$ \sum_{n \equal{} 0}^{\plus{}\infty}{2n\choose n}\frac{1}{7^{n}}$.[/quote]

[hide="Hint"]
$ {{2n}\choose n}\equal{}\frac{1}{2\pi i}\oint_{\gamma}\frac{(1\plus{}z)^{2n}}{z^{n\plus{}1}}dz$ where $ 0\in I(\gamma)$.

So,
$ \sum_{n\equal{}0}^{\infty}{{2n}\choose n}\frac{1}{7^{n}}\equal{}\frac{1}{2\pi i}\sum_{n\equal{}0}^{\infty}\oint_{\gamma}\frac{(1\plus{}z)^{2n}}{(7z)^{n}}\cdot\frac{dz}{z}$

Now change interchange summation and integral ... kinda messy ... answer.
[/hide]

All you need is that the summation $ \sum_{n\geq 0}\binom{2n}{n}x^{n}$ has the closed form $ \frac{1}{\sqrt{1\minus{}4x}}$.  (This one seems to pop up reasonably frequently as a power series that people don't realize has a closed form, as in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160578[/url].  Maybe it is known better by students of combinatorics than by students of calculus :).)


By AM-GM $x^2 - xy + y^2 \ge 2|xy| - xy$

Case 1
$xy < 0 \Rightarrow -xy = |xy|$
$\Rightarrow 2005 \ge 3|xy| \Rightarrow 668 + 1/3 \ge |xy|$
Since $x$ and $y$ are integers, their product is also, which gives:
$668 \ge |xy| \Rightarrow -668 \le xy \le 668$
$\Rightarrow -668 \le xy < 0$

Case 2
$xy > 0 \Rightarrow -xy = -|xy|$
$\Rightarrow 2005 \ge |xy| \Rightarrow -2005 \le xy \le 2005$
$\Rightarrow 0 < xy \le 2005$

Case 3
$xy = 0 \Rightarrow \text{WLOG } x = 0$
$\Rightarrow y^2 - 0 \cdot y + 0^2 = 2005 \Rightarrow y^2 = 2005$
Therefore, $y$ can't be an integer, since 2005 is not a perfect square and this case yields no solutions.

Since our cases cover all possibilites, we know there are only a finite combination of integers that give products between -668 and 2005.

----------
Solution 2: Oops Duh!
Let $(a,b)$ be one of the 2 permutations of $(x,y)$. We know we have $a^2 - ab + b^2 = 2005$
$\Rightarrow 4a^2 - 4ab = 4(2005) - 4b^2$
$\Rightarrow (2a - b)^2 = 4a^2 - 4ab + b^2 = 8020 - 3b^2$
By trivial, we have $0 \le (2a - b)^2 = 8020 - 3b^2$
$\Rightarrow 3b^2 \le 8020 \Rightarrow b^2 \le 2673 + 1/3$
Since $b$ must be an integer, we know $b^2 \le 51^2 = 2601$
because $51^2 = 2601 < 2673 + 1/3 < 2704 = 52^2$. From this we have $-51 \le b \le 51$. Since $b$ was arbitrarily chosen, it could be $x$ or $y$, which tells us that the bounds hold for both:
$-51 \le x,y \le 51$. As our Case 3 from the first argument said, neither can be 0, which leaves us with 102 choices for $x$ and 102 choices for $y$. Which means there are no more than 102^2 solutions, which just so happens to be finite.

$x=\frac{-y\pm\sqrt{y^2-4(y^2-2005)}}{2}=\frac{-y\pm\sqrt{4*2005-3y^2} }{2}$

so $-52<y<52$, by symmetry $x$ is also in that range, so x,y must be finite, giving a finite number of possible solutions

A minor note, Pork_chop: your first proof isn't complete until you show that there are only finitely many solutions so $0 < |xy| < 688$ in integers.

[quote="quantum leap"]Show that there are only finitely many integers $x,y$ such that $x^2-xy+y^2=2005$.[/quote]

Supose :

  (1) first case :$\large{\,\,x.y\,>\,0}$

We have: $\large{\,\,x^2-xy+y^2\,=\,(x-y)^2+xy}$

  How $\large{(x-y)^2\,\geq\,0\,\,\,\textrm{and}\,\,\,xy\,>\,0}$

  We have that $\large{(x-y)^2+xy\,=\,2005}$ have finite number, bescause , doing $\large{k^2\,=\,(x-y)^2\,\,\,,j^2\,=\,xy\,\,|\,k\,\in\,\mathbb{Z}\,\,\,\textrm{and}\,\,\j\,\in\,\,\mathbb{R}}$ ,

  $\large{k^2+j^2\,=2005\,}$ , that, obviousily, has finite solutions.

  (2) second case : $\,\,\,\large{x.y<0}$

  We have $\large{x^2-xy+y^2\,=\,(x+y)^2-3xy\,\,\,}$.

   How $\large{xy<0}$, we do $\large{a^2\,=\,(x+y)^2\,\,\,\textrm{and}\,\,\,b^2\,=\,-3xy\,\,\,|\,\,a\,\in\,\,\mathbb{Z}\,\,\,\textrm{and}\,\,\,b\,\,\in\,\,\mathbb{R}\,\,}$ , then :

   $\large{a^2+b^2\,=\,2005\,\,}$, that has finite solutions.

   (3) 3 case : $\large{xy=0}$

     It is easy see that $\large{x^2-xy+y^2=2005}$ doesn`t have a solution.


trigonometry
complex numbers
algebra unsolved
algebra

Suppose that real numbers $ x,y,z$ satisfy
\[ \dfrac{\cos{x}+\cos{y}+\cos{z}}{\cos{(x+y+z)}}=\dfrac{\sin{x}+\sin{y}+\sin{z}}{\sin{(x+y+z)}}=p\]
Prove that \[ \cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}=p.\]

$ \frac{cos(x)\plus{}cos(y)\plus{}cos(z)}{cos(x\plus{}y\plus{}z)}\equal{}\frac{sin(x)\plus{}sin(y)\plus{}sin(z)}{sin(x\plus{}y\plus{}z)}$

$ \iff 2\frac{sin(x\plus{}y)\plus{}sin(y\plus{}z)\plus{}sin(z\plus{}x)}{sin(2(x\plus{}y\plus{}z))}\equal{}0$

so $ sin(x\plus{}y)\plus{}sin(y\plus{}z)\plus{}sin(z\plus{}x)\equal{}0$

therefore $ e^{i(x\plus{}y)}\plus{}e^{i(y\plus{}z)}\plus{}e^{i(z\plus{}x)}\equal{}e^{\minus{}i(x\plus{}y)}\plus{}e^{\minus{}i(y\plus{}z)}\plus{}e^{\minus{}i(z\plus{}x)}$

and we have : $ e^{ix}\plus{}e^{iy}\plus{}e^{iz}\plus{}e^{\minus{}ix}\plus{}e^{\minus{}iy}\plus{}e^{\minus{}iz}\equal{}(e^{i(x\plus{}y\plus{}z)}\plus{}e^{\minus{}i(x\plus{}y\plus{}z)})p$

and : $ e^{ix}\plus{}e^{iy}\plus{}e^{iz}\minus{}e^{\minus{}ix}\minus{}e^{\minus{}iy}\minus{}e^{\minus{}iz}\equal{}(e^{i(x\plus{}y\plus{}z)}\minus{}e^{\minus{}i(x\plus{}y\plus{}z)})p$

so $ 2(e^{ix}\plus{}e^{iy}\plus{}e^{iz})\equal{}2pe^{i(x\plus{}y\plus{}z)}$

$ \iff e^{\minus{}i(x\plus{}y)}\plus{}e^{\minus{}i(y\plus{}z)}\plus{}e^{\minus{}i(z\plus{}x)}\equal{}p$

Q.E.D

[hide="Solution without complex numbers"]$ \cos{(x \plus{} y)} \plus{} \cos{(y \plus{} z)} \plus{} \cos{(x \plus{} z)} \equal{} \cos{(x \plus{} y\plus{}z\minus{}x)} \plus{} \cos{(x\plus{}y \plus{} z\minus{}y)} \plus{} \cos{(x\plus{}y\plus{}z \minus{}z)}\equal{}$

$ \equal{}\cos{(x\plus{}y\plus{}z)}[\cos{x}\plus{}\cos{y}\plus{}\cos{z}]\plus{}\sin{(x\plus{}y\plus{}z)}[\sin{x}\plus{}\sin{y}\plus{}\sin{z}]\equal{}p[\cos^2{(x\plus{}y\plus{}z)}\plus{}\sin^2{(x\plus{}y\plus{}z)}]\equal{}p$.[/hide]

The given conditions imply $ e^{ix} \plus{} e^{iy} \plus{} e^{iz} \equal{} pe^{i(x\plus{}y\plus{}z)}$

Hence $ e^{\minus{}i(x\plus{}y)} \plus{} e^{\minus{}i(y\plus{}z)} \plus{} e^{\minus{}i(z\plus{}x)} \equal{} p$

and so $ \sum \cos (x\plus{}y) \equal{} p$ and $ \sum \sin (x\plus{}y) \equal{} 0$

This problem was proposed in 2000 at Moldova NMO for 10th grade.


Circles $ \omega_1$ and $ \omega_2$ meet at $ P$ and $ Q$. Segments $ AC$ and $ BD$ are chords of $ \omega_1$ and $ \omega_2$ respectively, such that segment $ AB$ and ray $ CD$ meet at $ P$. Ray $ BD$ and segment $ AC$ meet at $ X$. Point $ Y$ lies on $ \omega_1$ such that $ PY \parallel BD$. Point $ Z$ lies on $ \omega_2$ such that $ PZ \parallel AC$. Prove that points $ Q,X,Y,Z$ are collinear.

This is an old problem.See [hide="here"]http://www.mathlinks.ro/Forum/viewtopic.php?p=982011#p982011[/hide]


trigonometry
function
algebra
polynomial
limit
geometry
geometric transformation

Prove that
$ \sin x\equal{}x\prod_{n\equal{}1}^{\infty}\left(1\minus{}\frac{x^2}{\pi^2n^2}\right)$

I am quite tired and going to bed but the outline is changing it to $ \frac{\sin(x)}{x} \equal{} something$
Now since when $ x\equal{}n\cdot \pi$ for integers $ n$, the function will be $ 0$. That is $ ...\minus{}3,\minus{}2,\minus{}1,1,2,3,...$ are roots of the equation.
So write the function out in terms of its roots.
$ \frac{\sin(x)}{x} \equal{} (1\plus{}\frac{x}{\pi})(1\minus{}\frac{x}{\pi})(1\plus{}\frac{x}{2\pi})(1\minus{}\frac{x}{2\pi})...$ and it follows.

It really doesn't; it's not at all obvious that you can factor arbitrary functions the same way you can factor polynomials.  (What's the factorization of $ e^x$?)  You'll want to look at the [url=http://mathworld.wolfram.com/WeierstrassProductTheorem.html]Weierstrass product theorem[/url] for the necessary rigor.

Aside from it being not at all obvious, it is also not at all true. Consider $ (e^x \plus{} e^{\minus{}x}) P(x)$ for polynomial $ P(x)$... if you tried to construct that in terms of its roots, youd end up with some scalar multiple of $ P(x)$, which is wrong.

For a non-integer $ z\in\Bbb{C}$

$ \lim_{n\to\infty} \frac{n!\, n^z}{z\, (z\plus{}1)\cdots (z\plus{}n)}\equal{}\Gamma(z)$

$ \Rightarrow \, \, \lim_{n\to\infty} \frac{n^z}{\prod\limits_{k\equal{}1}^n \left(1\plus{}\frac{z}{k}\right)}\equal{}z\,\Gamma(z)\equal{}\Gamma(1\plus{}z)$ .

Similarly $ \lim_{n\to\infty} \frac{n^{\minus{}z}}{\prod\limits_{k\equal{}1}^n \left(1\minus{}\frac{z}{k}\right)}\equal{}\Gamma(1\minus{}z)$ .

Hence $ \lim_{n\to\infty} \frac{1}{\prod\limits_{k\equal{}1}^n \left(1\minus{}\frac{z^2}{k^2}\right)}\equal{}\Gamma(1\plus{}z)\, \Gamma(1\minus{}z)$

which is $ \frac{\pi z}{\sin \pi z}$ after Euler's reflection formula.

If we focus on only sine function instead of general holomorphic functions, we may prove this identity only with elementary analysis knowledges.

For $ n \geq 0$, define $ I_n(\alpha) = \int_{-\pi/2}^{\pi/2} \cos^n t \, e^{\alpha t} \, dt$ and $ K_n (x) = \frac{\cos^n x}{I_n(0)}$.

[b]Lemma 1.[/b] For all $ n \geq 0$ and complex $ \alpha$, we have $ I_n(\alpha) = \frac{(n+2)^2 + \alpha^2}{(n+1)(n+2)} \, I_{n+2}(\alpha)$.

[i]pf.[/i] This is an easy consequence of integration by parts.

[b]Lemma 2.[/b] For all nonnegative integer $ n$, we have $ I_n (0) \geq (n + \tfrac{1}{2})^{-1/2}$.

[i]pf.[/i] Use induction on $ n$, with Lemma 1.

[b]Lemma 3.[/b] If $ f(x)$ is bounded on $ [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ and continuous at $ x = 0$, then $ \int_{-\pi/2}^{\pi/2} f(x) K_n(x) \, dx \to f(0)$ as $ n \to \infty$.

[i]pf.[/i] For $ \epsilon > 0$, choose $ 0 < \delta < \pi/2$ such that $ |f(x) - f(0)| < \epsilon$ whenever $ |x|<\delta$. Let $ M \geq 0$ be a bound for $ f$. Then

\begin{eqnarray*}
\left| \int_{-\pi/2}^{\pi/2} f(x) K_n(x) \, dx - f(0) \right|
& \leq & \int_{-\pi/2}^{\pi/2} |f(x) - f(0)| K_n(x) \, dx\\
& \leq & \int_{|x|<\delta} \epsilon K_n(x) \, dx + \int_{\delta \leq |x| \leq \pi/2} M K_n (x) \, dx\\
& \leq & \epsilon + \pi M \sqrt{n + \tfrac{1}{2}} \cos^n \delta,
\end{eqnarray*}
which completes the proof.

Inductive application of Lemma 1 shows that

$ \frac{I_n(\alpha)}{I_n(0)} = \frac{I_{n+2m}(\alpha)}{I_{n+2m}(0)} \prod_{k=1}^{m} \left(1 + \frac{\alpha^2}{(n+2k)^2} \right).$

Now taking $ m\to\infty$, by Lemma 3,

$ \lim_{m\to\infty} \frac{I_{n+2m}(\alpha)}{I_{n+2m}(0)} = \lim_{m\to\infty} \int_{-\pi/2}^{\pi/2} e^{\alpha t} K_{n+2m}(t) \, dt = 1.$

Thus for all $ \alpha \in \mathbb{C}$, we have

$ \frac{I_n(\alpha)}{I_n(0)} = \prod_{k=1}^{\infty} \left(1 + \frac{\alpha^2}{(n+2k)^2} \right).$

Especially, plugging $ n = 0$ and $ \alpha = \tfrac{2 i \omega}{\pi}$ gives the desired result. If we plug $ n = 1$ instead, we obtain an infinite product formula for $ \cos \omega$.

[quote="Doe John"]For a non-integer $ z\in\Bbb{C}$

$ \lim_{n\to\infty} \frac {n!\, n^z}{z\, (z \plus{} 1)\cdots (z \plus{} n)} \equal{} \Gamma(z)$ [/quote]
Is there a quick proof for this identity?

Let $ F: \Bbb{R}^{ > 0}\to\Bbb{R}^{ > 0}$ be a function with

$ F(1) \equal{} 1$ and $ F(x \plus{} 1) \equal{} x\, F(x)$

then $ F(n) \equal{} (n \minus{} 1)!$ for all $ n\in\Bbb{Z}^{ > 0}$ .

Then additionally let $ F$ be log-convex, i.e. $ \ln F\Big((1 \minus{} \vartheta)x_1 \plus{} \vartheta x_2\Big)\le (1 \minus{} \vartheta) \ln F(x_1) \plus{} \vartheta \ln F(x_2)$ .

When we choose $ \vartheta \equal{} x \, , \, x_1 \equal{} n \, , \, x_2 \equal{} n \plus{} 1$ and $ \vartheta \equal{} 1 \minus{} x \, , \, x_1 \equal{} n \plus{} x \, , \, x_2 \equal{} n \plus{} 1 \plus{} x$

we obtain the estimate $ n!\, (n \plus{} x)^{x \minus{} 1}\le F(n \plus{} x)\le (n \minus{} 1)!\, n^x$ .

Hence $ a_n(x): \equal{} \frac {n!\, (n \plus{} x)^{x \minus{} 1}}{x\, (x \plus{} 1)\cdots (x \plus{} n \minus{} 1)}\le F(x)\le \frac {(n \minus{} 1)!\, n^x}{x\, (x \plus{} 1)\cdots (x \plus{} n \minus{} 1)} \equal{} : b_n(x)$ .

Because of $ \frac {b_n(x)}{a_n(x)} \equal{} \left(\frac {n}{n \plus{} x}\right)^{x \minus{} 1}\to 1$ for $ n\to\infty$

$ F$ is unique and equals $ \lim_{n\to\infty} \frac {n!\, n^x}{x\, (x \plus{} 1)\cdots (x \plus{} n)}$ .

The $ \Gamma$ -function is also log-convex ([url=http://planetmath.org/encyclopedia/GausssProduct.html]proof[/url]), therefore $ F(x) \equal{} \Gamma(x)$ for all $ x > 0$ .

Moreover the equation $ F(z) \equal{} \Gamma(z)$ is true for all $ z\in\Bbb{C}\setminus \Bbb{Z}^{\le 0}$ (analytic continuation).


$a,b,c >0$. find the Minimal value of

$\frac{a+b}{a+2b+c}+\frac{4b}{a+b+2c}+\frac{8c}{a+b+3c}$

I'm waiting for a beautiful solution !

is it just me or is the inequality silly?
[hide]
\[\frac{a+b}{a+2b+c}+\frac{4b}{a+b+2c}+\frac{8c}{a+b+3c}\geq \frac{a+b}{a+2b+3c}+\frac{4b}{a+2b+3c}+\frac{8c}{a+2b+3c}\]
\[=\frac{a+5b+8c}{a+2b+3c}\geq 1\]
Therefore, minimum is $\boxed{1}$. It can attained when $a=1,b=c=0$.
[/hide]

[quote="nayel"][quote="vietnam2007"]$a,b,c> 0$ so we can't have $b=c=0$[/quote]

but you said a,b,c are nonnegative![/quote]

I have edited it.I have really a bad English because I wanted to say ( strictement positifs)

The following question is similar ( I did not find the link ):
Let $a,$ $b$ and $c$ are positive  numbers. Find the minimal value of

$\frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}.$
For you, scorpius119. :)

$a+2b+c=x$ , $a+b+2c=y$ and $a+b+3c=z$

the solution comes after that

Does making $a, b, c$ positive instead of nonnegative change anything? I think that the only difference is that $1$ is now an infimum (i.e. value of our expression is always greater than $1$ and it can be arbitrary close to $1$).

But I don't think that $1$ is attained for $a,b,c>0$ so we can't say that $1$ is its minimal value.

Simply forcing $a,b,c>0$ doesn't do very much: you change the answer from "minimum of 1" to "technically no minimum, but 1 is best lower bound". This is even true if you required $a,b,c\geq 1$ since we can choose $a=N,b=c=1$ where $N$ can be as large as you want.

Now, if you required a closed interval for $a,b,c$, that would make the problem completely different.

[hide="for the one given by arqady"]
This seemed pretty much impossible after simply multiplying it out and trying to come up with something, but then I thought, "wait, there has to be some logical way to come up with the minimum!" and then trying substituting variables for the denominators. Letting
\[a+2b+c=u,\ a+b+2c=v,\ a+b+3c=w \]
we have a system of three equations on three variables. From the second two, $c=w-v$. From the first two, $c-b=v-u$ so $b=u-2v+w$. Finally, substitute into any equation to get $a=-u+5v-3w$.

Now we have
\[\frac{-u+2v}{u}+\frac{4(u-2v+w)}{v}-\frac{8(w-v)}{w}\]

\[=2\cdot \frac{v}{u}+4\cdot \frac{u}{v}+4\cdot \frac{w}{v}+8\cdot \frac{v}{w}-17 \]
Yay, no $\frac{u}{w}$ or $\frac{w}{u}$ terms to worry about! Now just apply AM-GM:
\[\frac{v}{u}+2\cdot \frac{u}{v}\geq 2\sqrt{2},\ \frac{w}{v}+2\cdot \frac{v}{w}\geq 2\sqrt{2}\]
This gives a lower bound of $\boxed{12\sqrt{2}-17}$, but we need to show that this minimum works! (Had we applied AM-GM on $2\cdot \frac{w}{v}+\frac{v}{w}$, we would force one of $a,b,c$ to be negative; the best lower bound of this expression is actually 3). To do this, observe that equality holds when
\[u=\frac{v}{\sqrt{2}}=\frac{w}{2}\]
Let's see what happens when $u=1,\ v=\sqrt{2},\ w=2$. This gives
\[a=5\sqrt{2}-7,\ b=3-2\sqrt{2},\ c=2-\sqrt{2}\]
Fortunately, all are positive, so $12\sqrt{2}-17$ is indeed the minimum.
[/hide]

In my book, I proposed the following similar inequality

Given positive real numbers $x,y,z$, then

\[\frac{4}{3}\ge \frac{x+y}{x+4y+z}+\frac{y+z}{y+4z+x}+\frac{z+x}{z+4x+y}\ge 1.\]

I am waiting for a nice solution to it.


real analysis
search
real analysis unsolved

Does every subset of the real line with positive external measure contain a measurable subset with positive (Lebesgue) measure?

*bump*

(I used the search button.)

"external" ought to be "outer"

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156171
(I also used the search button.)


I'm stuck on the equation
$ 4 \sin^2{x} \equal{} 2 \cos{x} \plus{} 1$
Solve for x in the interval [0, 2$ \pi$]
I've tried squaring both sides but it doesn't work too well. I've also tried drawing a triangle with sides $ \sqrt{a^2 \plus{} b^2}$, $ a$, and $ a tan{x}$. Eventually, this approach just led back to the original equation.
I was wondering if anyone could give me a hint on how to start solving this?

[hide="hint"]
Substitute $ \sin^2 x\equal{}1\minus{}\cos^2 x$ to get \[ 4\cos^2x\plus{}2\cos x\minus{}3\equal{}0\]
[/hide]

from what the person above me said, let cos x=y, then solve for y, then solve for x.

since sin^2x+cos^x=1, then 1-cos^2x-sin^2x.  We can transform sin^2x to 1 -cos^2x.  Solve the parenthesis to get
4-4cos^2x=2cos+1
solve the equation in a way to construct a quadratic equation of -4cosx^2-2cosx+3 then solve


Find $ x;y;z \in Z$ such that
$ \frac {13}{x^2} \plus{} \frac {1996}{y^2} \equal{} \frac {z}{1997}$


vector
linear algebra
matrix
induction
linear algebra unsolved

In R^3 let "o" be such that:
(x, y, z) o (u, v, t) = (xu + yt + zv, xv + yu + zt, xt + yv + zu).
Prove that if K is a positive integer and (x,y,z)^k = (0,0,0) we have
(x,y,z) = (0,0,0).
(x,y,z)^k = (x,y,z)^(k-1) o (x,y,z) for k>1 and (x,y,z)^1 = (x,y,z).

[quote="campos"]let $ \omega \equal{} \frac { \minus{} 1 \plus{} i\sqrt 3}{2}$... note that

$ (1,\omega, \overline{\omega})\circ (x,y,z) \equal{} \dfrac{(2x \minus{} y \minus{} z) \plus{} i(y \minus{} z)\sqrt3}{2}\cdot (1,\omega, \overline{\omega})$,

so, if $ (x,y,z)^k \equal{} 0$, then,

$ 0 \equal{} (1,\omega, \overline{\omega})\circ (x,y,z)^k \equal{} \left(\dfrac{(2x \minus{} y \minus{} z) \plus{} i(y \minus{} z)\sqrt3}{2}\right)^k\cdot (1,\omega, \overline{\omega})$

this implies that $ 2x \minus{} y \minus{} z \equal{} 0$ and $ y \minus{} z \equal{} 0$, this is, $ x \equal{} y \equal{} z$... from this it's easy to conclude that $ (x,y,z) \equal{} (0,0,0)$ as we wanted to prove...  :D[/quote]
 Hi[b] campos [/b]It is a solution in $ C^3$ but here [b]joao1[/b] talk in $ R^3$

You can check my solution ! Thanks :  :P
 Put $ J \equal{} \begin{bmatrix}0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0\end{bmatrix}$ Then eassy that $ J$ have eigenvalues are $ \lambda_1 \equal{} 1,\lambda_2 \equal{} \frac { \minus{} 1 \plus{} i\sqrt {3}}{2},\lambda_3 \equal{} \frac { \minus{} 1 \minus{} \sqrt {3}}{2}$
 We have : $ (x,y,z)o(u,v,t) \equal{} (u,v,t)\begin{bmatrix}x & y & z \\
z & x & y \\
y & z & x\end{bmatrix} \equal{} (u,v,t)(xI_3 \plus{} yJ \plus{} zJ^2)$
By induction we can prove that  : $ (x,y,z)^ko(u,v,t) \equal{} (u,v,t)(xI_3 \plus{} yJ \plus{} zJ^2)^k$
By suppose we have :$ (u,v,t)(xI_3 \plus{} yJ \plus{} zJ^2)^k \equal{} O$ ,$ \forall (u,v,t) \in \mathbb{R}^3$
eassy  $ (xI_3 \plus{} yJ \plus{} zJ^2)^k \equal{} O$ Implies $ (x \plus{} y \plus{} z)^k \equal{} 0,(x \plus{} \lambda_2y \plus{} \lambda_2^2z)^k \equal{} 0,(x \plus{} \lambda_3y \plus{} \lambda_3^2z)^k \equal{} 0$ Implies $ x \plus{} y \plus{} z \equal{} 0,x \plus{} \lambda_2y \plus{} \lambda_2^2z \equal{} 0,x \plus{} \lambda_3y \plus{} \lambda_3^2z \equal{} 0$
From here $ x \equal{} y \equal{} z \equal{} 0$
 Sorry if i am wrong !  :blush:


Let the sum of positive reals $a,b,c$ be equal to 1. Prove an inequality \[
 \sqrt{{ab}\over {c+ab}}+\sqrt{{bc}\over {a+bc}}+\sqrt{{ac}\over {b+ac}}\le 3/2
 \].

[b]proposed by Fedor Petrov[/b]

We write $a=1-b-c,b=1-c-a,c=1-a-b$ and replace the $a,b,c$ in the denominators. We get $\sum\sqrt{\frac{ab}{(1-a)(1-b)}}\le\frac 32$. That's equivalent to $\sum\sqrt{\frac{ab}{(a+c)(b+c)}}\le\frac 32$.

We now apply AM-GM to each of the square roots in the form $2\sqrt{\frac{ab}{(a+c)(b+c)}}\le\frac a{a+c}+\frac b{b+c}$ and add up the three inequalities. The LHS will be twice our expression, and the RHS will be exactly $3$.


inequalities
calculus
derivative
inequalities unsolved

[quote="Vasc"]If $a,b,c$ are non-negative numbers, then
$\sqrt{\frac{a}{4a+5b}}+\sqrt{\frac{b}{4b+5c}}+\sqrt{\frac{c}{4c+5a}}\leq 1$.[/quote]

Magnificient inequality!!!

Put $\frac {b}{a}=x_1 , \frac {c}{b}=x_2 , \frac {a}{c}=x_3$, and use a more general inequality :
$\displaystyle  \sum_{k=1}^{n}(u + x_k)^\alpha \leq  n*(u + (\prod_{k=1}^{n}x_k)^{\frac {1}{n}})^\alpha$ , for $u > 0, x_k > 0, \alpha < 0$ .

:cool:  Do you  know this one ?!  :cool:

But this general inequality is not true for any u>0. :)

Can you please post your solution for this one , Vasc?

Thank you very much.

Dear Vasc, can you post your solution in the forum or you must deserve it for a magazine in which you published your inequality?

If so, can you please tell me the name of the review as I am very interested in your solution?

When will you write a book with all your very nice inequalities and solutions?

Thank you very much, Vasc.

Sorry, if I repeat myself.

Have you a direct proof for this one, Vasc?

[quote="manlio"]Dear Vasc, can you post your solution in the forum or you must deserve it for a magazine in which you published your inequality?
If so, can you please tell me the name of the review as I am very interested in your solution?
Have you a direct proof for this one, Vasc? [/quote]

Yes Manlio. I have a direct proof for this inequality.
But I cannot now post it because this inequality was published in
Gazeta Matematica A, no. 1, 2004, in a generalized form, and this
direct proof appear in my solution.
The solution will appear in GM-A, in 2005.

Well, I suppose method called sum of derivatives [in Poland at least] could be useful here. It states:

If $f(x_1,...,x_n)$ is on interval [0,$\infty$) then if we prove $f(0,x_2,...,x_n)\geq0$ [one of the variable is zero-well, we have to check every variable, but one is obviously enough if inequality is symmetric] and $f^{'} _{x_1}+...f^{'} _{x_n}\geq0$ we have $f(x_1,...,x_n)\geq0$.

 Let $f(a,b,c)=1- (\sqrt\frac{a}{4a+5b}+\sqrt\frac{b}{4b+5c}+\sqrt\frac{c}{4c+5a})$ we have to prove $f(a,b,c)\geq0$

I. Consider $f(0,b,c)\geq0$ we have:

$1-\sqrt\frac{b}{4b+5c}-\frac{1}{2}\geq0$
$\frac{1}{4}\geq \frac{b}{4b+5c}$
$5c\geq0$
which is true [from given conditions]

II. Now we should apply sum of partial derivatives [I don't know if it is so called in English] but I have problems with it [just don't know what will derivative of $\sqrt\frac{a}{4a+5b} \ da$ be like]. Maybe it's silly but I suppose we could get some results with it  :maybe:

Megus, in your post you refer to a very interesting method of proving certain inequalities. Where did you learn it? Did you find it in a book?

Yeah, I suppose it's in a book called "Powrot do krainy nierownosci" written by Lev Kourliatchik.


let mis integer >=2.find n is smallest such that every partition set{m,m+1,...n} into two subsets ,there exist one subset include three elements a,b,c st c=a^b


What is the radius of a hydrogen atom in its fourth excited state?

I'll see if I can find it somewhere, but what do you need it for?

you don't have to look it up, just use the formula

well the question is a bit unclear .
first what does the radius have to do with the excited state of an electron.even if there weren't any electrons in them the shells(at least theroitically) have their existance.so basically if u define radius as teh average distance betwen the 'last' electron and the nucleus then this seems right.

in general for $ H$ like species the radius in the the $ (n-1)^{th}$ exciited state is given by

$ r_{n}=0.529 \frac{n^{2}}{Z}\times 10^{-10}m$


inequalities
linear algebra
matrix
inequalities proposed

[quote="alex2008"]Let be $ k\geq 0$ . Find the maximum value of $ k$ such that for every positive real $ a,b,c$, satisfies :

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2)\geq k(a^2 \plus{} b^2 \plus{} c^2) \plus{} (9 \minus{} k)(ab \plus{} bc \plus{} ca)$.

When $ k \equal{} 0$, it becomes the APMO 2004 inequality .[/quote]

By letting $ a \to \inf, b \to 0, c \to 0$ we have $ k \geq 4$. Now we only need to prove that $ k \equal{} 4$ indeed satisfies, or find a sharper limit.

[quote="alex2008"]Let be $ k\geq 0$ . Find the maximum value of $ k$ such that for every positive real $ a,b,c$, satisfies :

$ (a^2 + 2)(b^2 + 2)(c^2 + 2)\geq k(a^2 + b^2 + c^2) + (9 - k)( bc + ca + ab ).$[/quote][quote="hendrata01"]By letting $ a \to \infty, b \to 0, c \to 0$ we have $ k \geq 4$. Now we only need to prove that $ k = 4$ indeed satisfies, or find a sharper limit.[/quote]$ (a^2 + 2)(b^2 + 2)(c^2 + 2) - 4(a^2 + b^2 + c^2) - 5( bc + ca + ab)$

$ = a^2b^2c^2 + 2(b^2c^2 + c^2a^2 + a^2b^2) - 5(bc + ca + ab) + 8$

$ \equiv xyz + 2(x^2 + y^2 + z^2) - 5(x + y + z) + 8$

$ = \frac {15 (y - z)^2 + (4x + y + z - 6)^2 + 4(2x + 7)(y - 1)(z - 1)}{8}\geq0,$

which is clearly true for $ (y - 1)(z - 1) = \max\{(y - 1)(z - 1),(z - 1)(x - 1),(x - 1)(y - 1)\}.$

This problem is introduced as Comment 2 for Example 2.4.18 on page 80 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.bjbb.com/product/detail.php?catalog=1&id=2016939]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

See also : http://www.mathlinks.ro/viewtopic.php?t=76508

[quote="sqing"]$  a~,~b~,~c\geq 0 $,prove that\[(a^2+2)(b^2+2)(c^2+2)\geq 4(a^2+b^2+c^2)+5(ab+bc+ca)\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538619&p3101952#p3101952[/quote]
Ine $\Leftrightarrow (abc)^2+8+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2+b^2+c^2) \geq 4(a^2+b^2+c^2)+5(ab+bc+ac)$
$\Leftrightarrow (abc)^2+8+2(a^2b^2+b^2c^2+c^2a^2)\geq 5(ab+bc+ac)$
Set $ab=x,bc=y,ca=z$
We must prove that
              $xyz+8+2(x^2+y^2+z^2) \geq 5(x+y+z)$
Use $x^2+y^2+z^2+2xyz+1 \geq 2(xy+yz+xz)$
$\Rightarrow xyz \geq (xy+yz+xz)-\frac{x^2+y^2+z^2+1}{2}$
$\Rightarrow xyz+8+x^2+y^2+z^2 \geq (xy+yz+xz)+\frac{3(x^2+y^2+z^2)}{2}+\frac{15}{2}$
So we will prove that
              $(xy+yz+xz)+\frac{3(x^2+y^2+z^2)}{2}+\frac{15}{2} \geq 5(x+y+z)$
    $\Leftrightarrow (x+y+z)^2+15+2(x^2+y^2+z^2) \geq 10(x+y+z)$
But it's true for AM-GM
            $\left\{\begin{matrix}
(x+y+z)^2+9 \geq 6(x+y+z)\\2(x^2+y^2+z^2)+6 \geq 4(x+y+z)

\end{matrix}\right.$
$\Rightarrow$ Q.e.D


Euler
function
inequalities
logarithms
real analysis
real analysis unsolved

Nature of the serie

$ \sum_{k >0} \frac{(\minus{}1)^k}{\phi(k)}$

$ \phi(k)$ Totient Euler function

One thought that ultimately doesn't prove anything:

Divide $ \mathbb{N}$ into three sets: The odd numbers, the numbers that are equivalent to $ 2$ mod $ 4,$ and the numbers that are divisible $ 4.$

For each term $ \frac{\minus{}1}{\phi(k)}$ from the set of odd numbers, there's a matching term $ \frac{1}{\phi(2k)}\equal{}\frac{1}{\phi(k)}$ from the set of numbers equivalent to $ 2$ mod $ 4;$ in fact, this is a one-to-one correspondence. That leaves the set of numbers divisible by $ 4$ and there,
\[ \sum_{k\equal{}1}^{\infty}\frac{1}{\phi(4k)}\ge \sum_{k\equal{}1}^{\infty}\frac{1}{4k}\equal{}\infty.\]
But, as I said, I haven't really proved anything yet. I've shown that some rearrangement of this series diverges to $ \plus{}\infty.$ However, having such a rearrangement still does not preclude the possibility that the original series converges conditionally.

[quote="Kent Merryfield"]One thought that ultimately doesn't prove anything:
[/quote]
On the contrary, Kent, your idea takes care of the problem rather neatly. All we need to show is that $ \minus{}\sum_{2n< k\le 4n, k\text{ odd}}\frac{1}{\varphi(k)}\plus{}\sum_{1\le k\le n, k\text{ odd}}\frac{1}{\varphi(4k)}$ is unbounded from above (the first sum lists all remaining terms that push the sum up to $ 4n$ down and the second sum includes just some terms that push it up).

Now, let $ a_k\equal{}\frac{1}{\varphi(k)}$ for odd $ k$ and $ 0$ for even $ k$. Since $ \varphi(4k)\equal{}2\varphi(k)$ for odd $ k$, we see that assuming the opposite leads to the inequalities
\[ \sum_{2n\le k\le 4n}a_k\ge \frac 12\sum_{1\le k\le n}a_k\minus{}C\]
Since the series $ \sum_k a_k$ diverges, this implies that $ \sum_{k\le 4n}a_k\ge A\sum_{k\le n}a_k$ for sufficiently big $ n$ with any $ A<\frac 32$. This allows us to conclude that $ \sum_{k\le n}a_k\ge n^\gamma$ for large $ n$ with any $ \gamma<\log_4\frac 32$ but this is certainly not the case because $ \varphi(k)>k^{1\minus{}\delta}$ for large $ k$ with any $ \delta>0$.

Of course, this argument makes it plausible (but not certain) that the partial sums tend to $ \plus{}\infty$ (we just proved that they are unbounded from above). I leave it to somebody else to prove this refined statement. If it seems too easy now, try to find the exact order of growth of the partial sums ;).

You know, that $ (\minus{}1)^k$ is really inconvenient; it makes this not a multiplicative function. Fortunately, we can reverse the signs to $ (\minus{}1)^{k\plus{}1}$ and have it actually be multiplicative.

OK, now we bring in a generating function technique (Dirichlet series, in this case). Define $ f(s)\equal{}\sum_{n\equal{}1}^{\infty} \frac{(\minus{}1)^{n\plus{}1}}{\phi(n)}\cdot n^{\minus{}s}$. We want to look at its behavior as $ s\to 1^\plus{}$.
Since those coefficients are a multiplicative function, we can split the sum into a product of sums:
$ f(s)\equal{}\left(1\minus{}\frac{2^{\minus{}s}}{1}\minus{}\frac{2^{\minus{}2s}}{2}\minus{}\frac{2^{\minus{}3s}}{4}\minus{}\cdots\right)\cdot\prod_{p\text{ odd prime}}\left(1\plus{}\frac{p^{\minus{}s}}{p\minus{}1}\plus{}\frac{p^{\minus{}2s}}{p(p\minus{}1)}\plus{}\cdots\right)$

Aside from the initial $ 1$ terms, each of those inner sums is a geometric series. The sums are $ 1\minus{}\frac{2}{2^{s\plus{}1}\minus{}1}$ and $ 1\plus{}\frac{p}{(p\minus{}1)(p^{s\plus{}1}\minus{}1)}$.

For $ s$ near zero, we have one term negative and all others positive, making the whole product negative. The product does converge for all $ s>0$, since $ \sum_p p^{\minus{}s\minus{}1}$ converges. It grows like a constant multiple of $ \frac1s$; the logarithms of the terms boil down to $ \frac1{p^{s\plus{}1}}$ and other terms that converge independent of $ s$.

For the rate of growth at $ s\equal{}0$, this form is less obviously useful.


inequalities
logarithms
real analysis
real analysis unsolved

I believe that the inequality below should be true. However I feel difficult to
construct a formal proof.

With $n=2,3$ where I can easily manipulate it,
I found that the LHS of the inequality
is convex w.r.t. $\{h_1,\ldots,h_{n-1}\}$ where $h_n = 1-\sum_{i\neq n} h_i$ and the min of the LHS is reached
for $h_i = 1/n$.

For the proof of the general case, can anyone give me some ideas ? Thanks.

The inequality is:
$\forall n\in\mathbb{N}, \quad 7\left(\sum_{i=1}^n h_i^2\right)^2 - 4\sum_{i=1}^n h_i^3 \geq 0$ where
$\sum_{i=1}^n h_i = 1$ and $h_i \geq 0$.

Some cheap ideas that don't solve the problem (yet):

1. Lagrange multipliers leads to cubic equations.

2. This can be rewritten in homogeneous form as

$ 4\left(\sum_{i=1}^nh_i\right)\left(\sum_{i=1}^nh_i^3\right)\le7\left(\sum_{i=1}^ nh_i^2\right)^2. $

3. The 4 and the 7 (or $ \frac47, $ or $ \frac74 $ ) look unusual. Do you have any insight as to where they're coming from?

The homogeneous form is exactly the original form of the problem. I just simplified it when I posted it.
The number 4 and 7 may be a little strange.
They come from an on-going study of point process but excuse me that I can't go to details here.
However I don't think them crucial. With some numerical simulations,
it seems to me that the inequality is still correct when 7 is replaced by 6.

[quote="bzhang"]The inequality is:
$\forall n\in\mathbb{N}, \quad 7\left(\sum_{i=1}^n h_i^2\right)^2 \ge 4\left(\sum_{i=1}^n h_i^3\right) \left(\sum_{i=1}^n h_i\right)$ if $h_i \geq 0$.[/quote]
This inequality cannot be correct for all $n$.

Pick $h_i = \frac{1}{i}$.

Then $\sum_{i=1}^n h_i^2$ and  $\sum_{i=1}^n h_i^3$ have finite limits as $n \to \infty$, but $\sum_{i=1}^n h_i \to \infty$.

This problem is actually well posed, since we have the constraint
$\sum_{i=1}^n h_i = 1$ and $h_i\geq 0$.

However, no formal proof has been found up to now for $n\geq 3$.

No, bzhang, hpe's counterexample completely demolishes the problem for all sufficently large $n.$ The way you posed it in the first place is equivalent to the homogeneous version; you can't escape the force of the counterexample by trying to claim a difference.

Let's do this: let $S_n=\sum_{i=1}^n\frac1i.$ Note that $S_n\approx \ln n.$

Let $h_i=\frac1{S_ni}.$

Then $\sum_{i=1}^nh_i^2=\frac1{(S_n)^2}\sum_{i=1}^n\frac1{i^2}\approx \frac{c_1}{(\ln n)^2}$ for large $n$ and for some $c_1$

and $\sum_{i=1}^nh_i^3=\frac1{(S_n)^3}\sum_{i=1}^n\frac1{i^3}\approx \frac{c_2}{(\ln n)^3}$ for large $n$ and for some $c_2$

Then we're left trying to assert that

$\frac{7c_1^2}{(\ln n)^4}-\frac{4c_2}{(\ln n)^3}\ge 0.$

That's clearly false for sufficiently large $n.$


The atomic number of Indium is 49. What is the most likely formula for the oxide of Indium?

This question is weird. According to the book's solution, Indium is in group 3, but it obviously is not.

$ \ce{In}$ belongs to group IIIA so it would probably form (via periodicity):

[hide=" Indium Oxide"]

$ \ce{In2O3}$[/hide]

Do this without periodic table. Start by writing the electronic configuration of Indium:

$ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 4d^{10} \, 5s^2 \, 5p^1$

from which we can see at a glance that

1) Indium is in block p
2) Indium is in period 5
3) Indium is in group 13

Therefore, Indium will very likely have the oxidation state $ \plus{}3$. So, indeed $ \ce{In2O3}$ is a formula for indium oxide.


integration
function
calculus
calculus computations

Calculate the arc length of the equation $ y\equal{}\frac{4}{5}x^\frac{5}{4}$ on the interval $ [0,4]$.

I know you set it up by doing
$ L\equal{}\int^{b}_{a}\sqrt{1\plus{}(\frac{dy}{dx})^2}dx$

$ \frac{dy}{dx}\equal{}x^\frac{1}{4}$

$ L\equal{}\int^{4}_{0}\sqrt{1\plus{}(x^\frac{1}{4})^2}dx\equal{}\int^{4}_{0}\sqrt{1\plus{}x^\frac{1}{2}}dx$

...but I can't get anywhere from there...

Most arclengths cannot be computed in closed form. There are rare exceptions.

This appears to be one of the rare exceptions (and one that I wasn't all that familiar with, at least, not compared to the more commonly seen 2/3 and 3/2 powers).

Try letting $ u\equal{}x^{\frac12},$ rewritten as $ x\equal{}u^2.$ And then after that, let $ v\equal{}u\plus{}1,u\equal{}v\minus{}1.$

Or skip the middleman and make that one substitution: $ v\equal{}1\plus{}x^{\frac12},x\equal{}(v\minus{}1)^2,dx\equal{}2(v\minus{}1)\,dv.$

I get, using Kent's solution,
$ L= \frac{4}{15}(1+\sqrt{x})^{3/2}(3\sqrt{x}-2)   |_{x=0}^{x=4}=\frac{16\sqrt{3}}{5}+\frac{8}{15}$ units...

Guys, if you have problems integrating functions, perhaps MAPLE is of great help..you can either buy it or "illegally"download it in the internet..your choice...Because in Maple, not only it gives the answer but the worked-out solution as well..of course, there are a lot more features of MAPLE other than integration..try it! :lol:


geometry
circumcircle
projective geometry
modular arithmetic
Euler
geometry proposed

assume that ABC is acute traingle and AA' is median we extend it until it meets circumcircle at A". let $AP_a$ be a diameter of the circumcircle. the pependicular from A' to $AP_a$ meets the tangent to circumcircle  at A"  in the point $X_a$; we define $X_b,X_c$ similary . prove that $X_a,X_b,X_c$ are one a line.

From Desargue's theorem, it follows that we need to show that the triangles formed by the tangents to the circumcircle in $A\",B\",C\"$ on the one hand, and by the perpendiculars from $A'$ to $AO$ and the like on the other hand are perspective.

Let the perpendiculars through $B',C'$ to $BO,CO$ respectively intersect at $T$, and let the tangents to the circumcircle in $B\",C\"$ intersect at $S$. I'll show that $ST$ passes through $G$ (the centroid).

Let $P$ be $B\"C\"\cap BC$. $SG$ is the polar of $P$ wrt $(O)$, so we need to show that $T$ lies on this polar. A quick angle chase shows that $T$ is, in fact, the intersection of the tangents through $B',C'$ to the nine-point circle. We thus need to show that if $T'$ is the intersection pt of the tangents in $B,C$ to $(O)$, then $T'$ lies on the polar of $P$ wrt $(O)$, but this is just a degenerate case of Pascal's theorem.

The solution of the problem is a strightforward consequence of the following lemma RL110904 and an opportune iterate application of Desargues theoreme.

Lemma RL110904

Given a ciclic quadrialteral ABCD, let K be the intersection of AC and BD and L be the intersection between the tangent to c trhough C and D. If R and S are two point on BD and AC respectively s.t. RS//AB, then the intersection of the perpendiculars from R to the diameter through B and from S to the diameter through A stay on line KL.

Proof.

this lemma derive from the well-known fact that (which is in turn consequence of Pascal theorem) the diagonals of two dual quadrilaterals are concurrents.

Let me restate the problem so that it can fit my solution:
[color=darkblue][b]"Let $ ABC$ be triangle inscribed in circumcircle $ (O)$. Denote $ A_1,B_1,C_1$ the intersections by the medians from $ A,B,C$ and $ (O)$, respectively. Let $ A_3,B_3,C_3$ be the midpoints of $ BC,CA,AB$, respectively. The perpendicular from $ A_3$ to $ AO$ intersects the tangent from $ A_1$ wrt $ (O)$ at $ X_a$. Define the same for $ X_b,X_c$. Prove that $ X_a,X_b,X_c$ are collinear." [/b][/color]
[i]Proof[/i]:

[b]Lemma $ 1$:[/b] Let $ AB$ be the chord of $ (O)$. Denote $ C$ is an arbitry point on the greater arc $ AB$. Let $ M$ be the midpoint of $ AB$. Denote $ E$ the intersection by $ CM$ and $ (O)$, $ D$ by the intersection of $ CO$ and $ (O)$. The perpendicular from $ M$ to $ CD$ intersects the tangent from $ E$ wrt $ (O)$ at $ K$. Let $ KN$ be another tangent form $ K$ to $ (O)$ ($ N\in (O)$). Then $ N,M,D$ are collinear.

Indeed, let $ S$ be the projection of $ M$ onto $ CD$. Denote $ X$ by the pole of $ MS$ wrt $ (O)$. Since $ K\in$ the polar of $ X$ wrt $ (O)$ hence the polar of $ K$ passes $ X$, which implies that $ X\in NE$, which also implies $ X\equiv CD\cap NE$. Let $ Y\equiv NE\cap MS\Longrightarrow (XYEN) \equal{} (XSDC) \equal{} (XSCD) \equal{} \minus{} 1\Longrightarrow YS,EC,DN$ are concurrent. But $ M$ already is the intersection of $ CE,YS$, thus $ M\in ND$.

[b]Lemma $ 2$:[/b] Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$, $ H$ is its orthocenter. Denote $ A_3,B_3,C_3$ the midpoints of $ BC,CA,AB$, respectively. $ A_1,B_1,C_1$ be the intersections of $ AA_3,BB_3,CC_3$ with $ (O)$, respectively. Denote $ A_0,B_0,C_0$ be the second intersections of circumcircles of diameters $ AH,BH,CH$ with $ (O)$, respectively. Prove that: $ A_1A_0,B_1B_0,C_1C_0$ are concurrent.

Indeed, let $ D,E,F$ be the projections of $ H$ onto $ BC,CA,AB$, respectively. Consider the inversion through pole $ H$, power $ \overline {HD}.\overline {HA} \equal{} \overline {HE}.\overline {HB} \equal{} \overline {HF}.\overline {HC} \equal{} k^2$. It is very well- known that $ \mathcal {I}(H,k^2): A_3\mapsto A_0$, $ B_3\mapsto B_0$, $ C_3\mapsto C_0$. Therefore, it is so obvious that $ \{B_3,C_3,B_0,C_0\}$ is the set of concyclic points. Also, $ (C_3C_1,C_3B_3)\equiv (C_3C,C_3B_3)\equiv (CC_1,CB)\equiv (B_1C_1,B_1B_3) \pmod \pi$, which implies $ \{C_1,C_3,B_3,B_1\}$ is a set of concyclic points. Now, we have $ B_1C_1,B_3C_3,B_0C_0$, respectively are the radical axes of $ \{(B_1C_1B_3C_3), (O)\}$, $ \{(B_1C_1B_3C_3),(B_3C_3B_0C_0)\}$, $ \{(B_0C_0B_3C_3),(O)\}$. Therefore $ B_1C_1,B_3C_3,B_0C_0$ are concurrent, let $ Y_1$ be the intersection of these lines. We have $ \overline {Y_1B_3}.\overline {Y_1C_3} \equal{} \overline {Y_1B_0}.\overline {Y_1C_0}\Longrightarrow \mathcal {P}_{(Y_1/(A_3B_3C_3))} \equal{} \mathcal {P}_{(Y_1/(O))}$. With the same define and argument for $ Y_2,Y_3$. We conclude that $ Y_1,Y_2,Y_3$ are collinear and further the line contains them is the radical axes of the Euler circle and $ (O)$. We have $ Y_1\equiv B_0C_0\cap B_1C_1$, $ Y_2\equiv A_0B_3\cap A_1B_1$, $ Y_3\equiv A_0C_0\cap A_1C_1$, $ Y_1,Y_2,Y_3$ are collinear therefore $ \triangle A_0B_0C_0$ and $ \triangle A_1B_1C_1$ are $ 2$ perspective triangles. Then by [b]Desgaurte theorem[/b], $ A_1A_0,B_1B_0,C_1C_0$ are concurrent.

Back to our problem.
Let $ X_aA_0,X_bB_0,X_cC_0$ let other tangents form $ X_a,X_b,X_c$, respectively to $ (O)$. Let $ A',B',C'$ be the intersections of $ AO,BO,CO$, respectively with $ (O)$. Then by [b]lemma $ 1$[/b], we conclude that $ A'A_0,B'B_0,C'C_0$ passes through $ A_3,B_3,C_3$, respectively, moreover, they also concurrent at $ H$- the orthocenter of $ \triangle ABC$. Thus, $ A_0,B_0,C_0$ now also play roles that they are respectively the second intersections of circumcircles of diameters $ AH,BH,CH$ with $ (O)$. Apply [b]lemma $ 2$[/b], we conclude that $ C_0C_1,A_0A_1,B_0B_1$ concurrent. But, $ C_0C_1,A_0A_1,B_0B_1$, respectively are the pole of $ X_c,X_a,X_b$ wrt $ (O)$. We conclude $ X_a,X_b,X_c$ are collinear.
Our proof is completed then.

Yep, very nice problem. Take a look at [url=http://www.mathlinks.ro/viewtopic.php?p=1256346#1256346]a very nice solution by Nguyen Minh Ha here[/url].


floor function
inequalities
triangle inequality
combinatorics proposed
combinatorics

Let $ n$ be a positive integer. $ n$ line segments of length $ 1,2,\dots,n$ are given. Show that the number of ways to choose three of the segments so that they form a triangle is equal to
\[ \frac 16\left\lfloor\frac n2\right\rfloor\left(\left\lfloor\frac n2\right\rfloor\minus{}1\right)\left(2n\minus{}2\cdot (\minus{}1)^n\minus{}3\right)\]

I have proven it using quite a lot of algebra and summation:
[hide="Solution"]Let the shorter leg of the triangle be $ k$. For each $ 2\geq k\geq\lfloor\frac{n}{2}\rfloor$, we can let the difference between the other two sides be $ 1$, $ 2$, ..., $ k\minus{}1$ to obtain a non-degenerate triangle (by the triangle inequality); so pairs of the two other side lengths are $ (k\plus{}1,k\plus{}2)$, ... $ (n\minus{}1,n)$, $ (k\plus{}1,k\plus{}3)$, ..., $ (n\minus{}2,n)$, ..., $ (k\plus{}1,2k)$, ... Thus, for each such $ k$, there are $ (n\minus{}k\minus{}1)\plus{}(n\minus{}k\minus{}2)\plus{}\cdots \plus{}(n\minus{}2k\plus{}1)\equal{}\frac{(k\minus{}1)(2n\minus{}3k)}{2}$ ways to choose segments.

For $ \lfloor\frac{n}{2}\rfloor \plus{}1\leq k\leq n\minus{}2$, there are similarly $ (n\minus{}k\minus{}1)\plus{}(n\minus{}k\minus{}2)\plus{}\cdots \plus{}1\equal{}\frac{(n\minus{}1\minus{}k)(n\minus{}k)}{2}\equal{}\binom{n\minus{}k}{2}$ ways to obtain a non-degenerate triangle. (When $ n$ is even, these two cases are the same for $ k\equal{}n/2$, but we have avoided this in choosing $ k$.)

We can get the desired expression by straightforwardly evaluating the following sums, followed by some algebraic manipulation:

\[ \sum_{k\equal{}1}^{\lfloor\frac{n}{2}\rfloor} \frac{(k\minus{}1)(2n\minus{}3k)}{2} \plus{} \sum_{k\equal{}\lfloor\frac{n}{2}\rfloor \plus{}1}^n \frac{(n\minus{}1\minus{}k)(n\minus{}k)}{2}\][/hide]
Can anyone else (perhaps nayel) share a "slicker" solution (if I think of one, I will post it)?


3 times tossing is used then there are possible 8 outcomes.

                                         HHH HHT HTH HTT THH THT TTH TTT
As the coin is fair we can consider that the probability of each outcome is also equal. For simplicity we can consider that if the same thing is repeated 8 times we can expect to get each possible sequence once.

Suppose a coin is tossed n times. And the same process is repeated 2^n times. How many sequence i will get which contains a consequnce of H(head) of length at least k.

Some Result is here:
n=4   k=1 the answer is 15.
n=4   k=2 the answer is 8
n=4   k=3 the answer is 3

n=6   k=2 the answer is 43

My problem is how can i find it by formula or mathmetically?? anyone help me.....

will this help?

[hide]go to this website:[url]http://mathworld.wolfram.com/CoinTossing.html[/url][/hide]

 :P backup every thing you say with evidence :lol:


Let a,b,c are real numbers satisfies that a,b,c>0 and a^2+b^2+c^2+abc=4.
Find the max value of (a+b+c)

By Cauchy-Schwarz, $ (1 \plus{} 1 \plus{} 1)(a^2 \plus{} b^2 \plus{} c^2) \equal{} 12 \minus{} 3abc\geq(a \plus{} b \plus{} c)^2\implies\sqrt {12 \minus{} 3abc}\geq a \plus{} b \plus{} c$.

Also, by AM-GM, we have $ 4 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} abc\geq4\sqrt [4]{a^3b^3c^3}\implies1\geq abc$, so plugging this into our first inequality, we have

$ \sqrt9 \equal{} \boxed3\geq a \plus{} b \plus{} c$, which can be achieved when $ a \equal{} b \equal{} c \equal{} 1$.

Edit: Sorry, I mixed up a sign.

[quote="math154"]By Cauchy-Schwarz, $ (1 \plus{} 1 \plus{} 1)(a^2 \plus{} b^2 \plus{} c^2) \equal{} 12 \minus{} 3abc\geq(a \plus{} b \plus{} c)^2\implies\sqrt {12 \minus{} 3abc}\geq a \plus{} b \plus{} c$.

Also, by AM-GM, we have $ 4 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} abc\geq4\sqrt [4]{a^3b^3c^3}\implies1\geq abc$, so plugging this into our first inequality, we have

$ \sqrt9 \equal{} \boxed3\geq a \plus{} b \plus{} c$, which can be achieved when $ a \equal{} b \equal{} c \equal{} 1$.[/quote]
From $ \sqrt {12 \minus{} 3abc}\geq a \plus{} b \plus{} c$ and $ \sqrt {12 \minus{} 3abc}\geq 3$, we can't infer that $ 3\geq a \plus{} b \plus{} c$


anyone can post math book of USA(high school curriculum)?
did i have any mistake in my expression? my english is too bad but i'll try to improve it.
thank u very much!

There is no single math textbook for the high school curriculum. Generally, there is one textbook for each class. Furthermore, not all high schools use the same math textbooks; they have freedom to choose which one best suits them.

how about secondary curriculum, can u post it?(math book)
thank u very much

The US doesn't require schools at any level to use one standard textbook.

thank a lot, did u have any book about function equation?

Yes. While there is no standard textbook, all textbooks of the same level will contain the same information.

For functions, pick a standard Algebra I or Algebra II textbook, depending on your level. Some of the common brands I have used are Holt (Holt, Rinehart, and Winston) and McGraw-Hill; you can ask around for others or look them up online.


$x,y,z \geq 0$ and $x+y+z=1$
Find max and min of $x^3 + y^3 + z^3$

[quote="Altheman"][hide]let $f(x)=x^3$

$f(x)$ is increasing wth positive numbers, so it is concave up

then by jensen's

$\frac{\sum f(x_i)}{n} \ge f\left(\frac{\sum{x_1}}{n}\right)$

so...

$\frac{x^3+y^3+z^3}{3} \ge \left(\frac{1}{3}\right)^3$

$x^3+y^3+z^3 \ge \frac{1}{9}$

now we have to show the minimum $\boxed{\frac{1}{9}}$ will occur, and using the equality condition for jensen's, we have $x=y=z=\frac{1}{3}$ for this min to occur

for the max

$x^3+y^3+z^3=(x+y+x)^3-3(x+y)(y+z)(x+z)$

so $x^3+y^3+z^3=1-3(1-x)(1-y)(1-z)$

and clearly the last term has to be negative or zero because none of the variables can be greater than 1, or the others will be negative, so when one of them =1, that gives the max at $\boxed{1}$[/hide][/quote]

No offence about your max-method, but [hide="I would not work with Jensen."]Just use Power Mean on {x,y,z}, getting \[ \sqrt[3]{\frac{x^3+y^3+z^3}{3}} \geq \frac{x+y+z}{3} \] So, \[ x^3+y^3+z^3 \geq \frac{1}{9} \] with equality when $x=y=z=\frac{1}{3}$...?[/hide]

We could also use Holder's Inequality:

[hide]
$x + y + z \leq (1 + 1 + 1)^{2/3}(x^3 + y^3 + z^3)^{1/3} \Rightarrow x^3 + y^3 + z^3 \geq \frac{1}{9}$

[/hide]

[quote="1234567890"]What's Holder's? :?[/quote]
It's a generalization of Cauchy-Schwartz: for $1/p + 1/q = 1$ with $p,q > 1$ and two positive sequences $a_1, a_2, ..., a_n$ and $b_1, b_2,..., b_n$, we have

$\sum^n_{k=1} a_n b_n \leq \left(\sum^n_{k=1}a_n^p\right)^{1/p}\left(\sum^n_{k=1}b_n^q\right)^{1/q}$

Note that this reduces to Cauchy when $p = q = 2$.

[quote="Altheman"]i suppose that it is not the most simple method, but it works, and i suppose i essentially derived the power-mean for powers 3 and 1, i had not looked at that AoPS section yet...[/quote]
In fact, I think you can use Jensen's Inequality to derive both the PM-AM inequality and the AM-GM inequality.


[quote="1234567890"]I was wondering what would be good to do in a math club with students at beginning levels. Ideas for the first day?[/quote]

Are you thinking of starting one?

We will have our first meeting soon, actually.

The first day is rather simple. Introduce yourself. Introduce the other officers/captains/volunteers/coaches or whatever. If you need to vote for these, or somehow decide them.

Talk about the various competitions you'll be participating in. Format, difficulty, qualification, etc... Try to sell the kids on participating. Bribe them with promises of getting money in the MMPC or other prizes. Give them a syllabus with the competitions if you want.

If this hasn't taken up most of your time, work on some simple problems in preparation for contests, like the MMPC part I. Make these the easiest problems - put it on the board and see if someone can solve it and explain it. Then add extra comments.

This is what I like to do on the first day.

you might want to show them some easy problems to see if they enjoy doing them or if they get frustrated by them.

What my old Mathcounts coach had on the first day was an easy question, in which we solved individually. Then we had a crazy hard question, in which we solved as a group, with the help of the coach. This demonstrated teamwork.

We also set goals individually.

We basically review over contest procedures and what our club is interested in doing this year.


probability
real analysis
real analysis theorems

Given the distinction of countably infinite and uncountably infinite sets, is there any sort of similar "countable" and "uncountable" distinction when dealing with infinitesimals?

I've always thought of epsilon as 1/infinity, but maybe that's just a bad definition, considering the ambiguity of "infinity."  Otherwise, it seems like it could have applications (albeit not very practical ones) when dealing with probabilities.

This idea does have applications in probability theory. You might be interested in Edward Nelson's book Radically Elementary Probability Theory. http://www.math.princeton.edu/%7Enelson/books/rept.pdf

Epsilon, as conventionally used (in, say, the definition of a limit) is not a fixed value, but rather a variable quantity that can be chosen "as small as you like."  If thinking of it as the multiplicative inverse of infinity is a useful heuristic for you, that's fine, but you should be aware that it is nothing more formal than a heuristic.

There are numerous number systems which include "more than one infinity."  The ordinals, to which the cardinals (and so the distinction between countable and uncountable sets) are related, are a good example of such a system.  Unfortunately, they aren't a field (they don't even have commutative addition) so that things like additive and multiplicative inverses don't exist.

A good example of a system in which those things do exist are the surreal numbers, an invention largely due to John Conway.  They include all of the ordinals (although with a better-behaving arithmetic) but are also a field, so that infinitesimals of different sizes do exist.

More generally, there exist non-standard models for the axioms of the real numbers under which various different sorts of infinitesimals exist.  Unfortunately, I don't know anything more about them.

does nonstandard analysis relate to this. i have a book on it if you want to see how the author deals with infintesimals

Indeed, the whole idea of nonstandard analysis is to rigorously construct a field of numbers that contains infinitesimals. I've actually read a fair amount about this so if you have any particular questions I'll try to answer. :)


When three numbers are added two at a time, the three distinct sums are 32, 48, and 46. What is the least of the three numbers?

we have \[ a\plus{}b\equal{}32\] \[ a\plus{}c\equal{}48\] \[ b\plus{}c\equal{}46\] If we add the three equations together, we get \[ 2a\plus{}2b\plus{}2c\equal{}126\] \[ a\plus{}b\plus{}c\equal{}63\]  Subtracting any of the original equations from this gives us one of the numbers, and we find the smallest one is $ \boxed{15}$

The last part of the solution- we know the largest and second largest numbers sum to 48. So 63-48=15


A line $ l$ is tangent to a circle $ S$ at $ A$. For any points $ B,C$ on $ l$ on opposite sides of $ A$, let the other tangents from $ B$ and $ C$ to $ S$ intersect at a point $ P$. If $ B,C$ vary on $ l$ so that the product $ AB \cdot AC$ is constant, find the locus of $ P$.

Let us rewrite the problem in order to use the common triangle notation

[b]Problem: [/b]Let $ (I,r)$ be a fixed circle tangent to a fixed line $ \mathcal{L}$ at $ X.$ $ B,C$ are points on $ \mathcal{L}$ such that $ X \in \overline{BC}$ and $ BX \cdot CX \equal{} k^2 \equal{} \text{const}.$ Tangents from $ B,C$ to $ (I)$ different from $ \mathcal{L}$ meet at $ A.$ Find the locus of $ A.$

Since $ BX \equal{} p \minus{} b \ , \ CX \equal{} p \minus{} c  \Longrightarrow  (p \minus{} b)(p \minus{} c) \equal{} k^2 \equal{} \text{const}$

From the identity $ r \equal{} \sqrt {\frac {(p \minus{} a)(p \minus{} b)(p \minus{} c)}{p}}$ we get

$ r^2 \equal{} \frac {(p \minus{} a)k^2}{p} \equal{} \left(1 \minus{} \frac {a}{p}\right )k^2 \ \Longrightarrow \ \frac {p}{a} \equal{} \frac {k^2}{k^2 \minus{} r^2}$

$ a \cdot h_a \equal{} 2rp  \Longrightarrow  h_a \equal{} \frac {2rp}{a} \equal{} \frac {2r \cdot k^2}{k^2 \minus{} r^2} \equal{} \text{const}$

$\Longrightarrow$ A-altitude has constant length, hence the locus of $ A$ is a parallel line to $BC.$


modular arithmetic
quadratics
number theory unsolved
number theory

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

1) Numbers $1^2$, $2^2$, ..., $\left(\frac{p-1}{2}\right)^2$ give distinct residues modulo $p$. Indeed, if $a^2=b^2$ modulo $p$ then $p\,|\,a-b$ or  $p\,|\,a+b$ - impossible for $1\leq a,b\leq \frac{p-1}{2}$, $a\neq b$.
2) Since $p=1$ modulo 4, it follows that for each square residue $r$ number $-r$ is square residue too (well known fact).
3) From 1 and 2 we conclude all square residues $1^2$, $2^2$, ..., $\left(\frac{p-1}{2}\right)^2$ can be partitioned into $\frac{p-1}{4}$ pairs $\left(a^2,b^2\right)$ s.t. $a^2+b^2=0$ modulo $p$, it means $\left\{\frac{a^2}{p}\right\}+\left\{\frac{b^2}{p}\right\} = 1$. Therefore, answer is $\frac{p-1}{4}$.

[u][b]The author of this posting is : mecrazywong[/b][/u]
____________________________________________________________________

Myth, can you explain what are square residues to me? sorry for my poor knowledge in number theory........  :blush:

[u][b]The author of this posting is : Valiowk[/b][/u]
____________________________________________________________________

It's easy to show that the values of $k^2 \pmod {p}$, where $k = 1, 2, \ldots, \frac{p-1}{2}$ are distinct. From Euler's Criterion, $a$ is a quadratic residue iff $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$. Since $\frac{p-1}{2}$ is even, $-a$ is a quadratic residue iff $a$ is a quadratic residue. Hence we may pair up our quadratic residues into $\frac{p-1}{4}$ disjoint pairs such that the sum of each pair $\equiv 0 \pmod{p}$. Thus, $\sum_{k=1}^{\frac{p-1}{2}} \left\{ \frac{k^2}{p} \right\} = \frac{1}{2} \frac{p-1}{2} = \frac{p-1}{4}$.

EDIT: Now I know the feeling when somebody posts more quickly than you ;) . A number $r$ is a quadratic residue mod $n$ if there exists an integer $k$ such that $k^2 \equiv r \pmod{n}$.

[u][b]The author of this posting is : siuhochung[/b][/u]
____________________________________________________________________

[quote="mecrazywong"]Thanks, both Myth and Valiowk![/quote]

sam, you should have paid attention in cj's lesson... he has talked about this question in the lesson on 27/11


Wow.  I messed up #1, so I only got 3 right.  I also should have gotten #4, 420 is just too obvious.

Please don't make new threads about this; we already have at least two.


$x,y,z$ are roots of the following cubic equation.

\[t^3-t^2-2t+1=0\]

Add 'em up and it's obvious, only I don't know of a rigorous way to show that those are the only solutions.

[quote]$x$,$y$,$z$ are roots of the following cubic equation.

$t^3-t^2-2t+1=0$[/quote]

Just to clarify, since I was confused at first: The above means $x=y=z=t$, where $t$ is a root of the given equation (at first, I thought you meant that $x$, $y$, $z$ were three distinct values, given by the roots of the equation.)

If you allow $x$, $y$, $z$ to be complex numbers, though, there are more solutions than that.

Yakko

Because of the "symmetry" of the equations: each is exactly the same, except that x, y, and z are substituted for eachother.  So, you can't really tell which is which.

But that doesn't mean they have to be equal, it just means that our solutions will come in groups of three or six when the three variables aren't all equal.

I think I do have a proof of it, though, if we only seek real solutions:

[hide]Rearrange the equations so one variable is always on only one side:
$y^3+y^2+2y= 2z^3+1$
$x^3+x^2+2x= 2y^3+1$
$z^3+z^2+2z= 2x^3+1$

Now, consider two different solutions, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$.  We can assume without loss of generality that $x_1 > x_2$.  Now, $f(t) = t^3+t^2+2t$ is an increasing function of t, so this implies $x_1^3+x_1^2+2x_1 > x_2^3+x_2^2+2x_2$ and so $y_1 > y_2$ and then similarly $z_1 > z_2$.  Now, assume $(x, y, z)$ is a solution.  Then we know $(y, z, x)$ is as well, by symmetry.  If $x \ge y$, we have, by applying the result arrived at above, that $x \ge y \ge z \ge x$ and so they are all equal, and a similar result if $x \le y$, so the only real solutions satisfy $x = y = z$ and they are as kunny claimed.[/hide]


vector calculus function geometry conics ellipse parameterization

Let $ C$ be a simple closed curve in the $ (x,y)$ plane. Show by considering the vector function of position $ \bold{\underline{F}}$, defined by $ \bold{\underline{F}} \equal{} (y \minus{} z, z \minus{} x, x \minus{} y)$, that the area $ A$ enclosed  by $ C$ is given by

$ A \equal{} \frac {1}{2} \ointop_C (xdy \minus{} ydx)$.

Use this result to find the area enclosed by the ellipse $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1$, where $ a$ and $ b$ are positive constants.

The elliptic paraboloid is described by the equation

$ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} \frac {z}{c}$,
where $ c$ is a positive constant. With $ \bold{\underline{F}}$ as defined above, determine the flux of $ \bold{\underline{F}}$ through that part of the surface which lies between the planes $ z \equal{} 0$ and $ z \equal{} d$ where $ d$ is a positive constant.

----------------------------------------------------------------------------

This is what I have so far:

Applying Stokes theorem:

$ \oint_C \bold{\underline{F}}.d\bold{\underline{S}} \equal{} \iint_S (\nabla \times \bold{\underline{F}}).d\bold{\underline{S}} \equal{} \iint_S \minus{}2.d\bold{\underline{S}} \equal{} \minus{}2A$

$ A \equal{} \minus{}\frac{1}{2} \oint_C \bold{\underline{F}}.d\bold{\underline{S}}$
$ A \equal{} \minus{}\frac{1}{2} \oint_C (ydx \minus{} xdy)$
$ A \equal{} \frac{1}{2} \oint_C (xdy \minus{} ydx)$

I'm not sure if this is correct, my mathematics could be completely wrong. Any help would be great, thanks.

Your computations are correct. This is usually done with Green's theorem and with 2-dimensional vector field $ \vec F(x,y)\equal{}\frac{1}{2}\langle \minus{}y,x\rangle$. But whatever.
For the second part, you can parametrize the surface using $ x$ and $ y$ as parameters and compute the surface integral directly. (One can also use the fact that $ \mathrm{div}\vec F\equal{}0$ to simplify things, but it's not necessary.)

For the second part, and unless I am mistaken, using Gauss's theorem, and since the divergence of $ \vec F$ is zero, then the flux on the given paraboloid is equal but with opposite sign to the flux on the ellipse $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} \frac {d}{c}$.  So you only need the integral over the surface of the ellipse of $ \vec F \cdot d\vec S$.  The parametrization for the surface integral would then be $ x \equal{} a\rho\cos t$, $ y \equal{} b\rho\sin t$, where $ 0\leq\rho\leq\sqrt {\frac {d}{c}}$, and $ 0\leq t < 2\pi$, and where $ dxdy \equal{} ab\rho d\rho dt$ for the purposes of integration.  The function to integrate is then $ a^2b\rho^2\cos t \minus{} ab^2\rho^2\sin t$, which yields 0 when integrated over $ t$, or the flux is 0.

[quote="tom_m"]how exactly would you parametrize the surface?[/quote]
$ \vec r(x,y) \equal{} \langle x,y, cx^2/a^2 \plus{} cy^2/b^2\rangle$, where $ (x,y)$ lies in a certain ellipse ($ xy$-projection of the surface).


group theory
abstract algebra
superior algebra
superior algebra theorems

IF $ N\lhd G$ such that $ N$ and $ G/N$ are solvable then $ G$ is solvable.

Let the solutions (correct word¿) be
\[ \{e\}\equal{} N_{0}\lhd N_{1}\lhd ...\lhd N_{a}\equal{} N\]
and
\[ \{e\}\equal{} M_{0}\lhd M_{1}\lhd ...\lhd M_{n}\equal{} G/N\]
with cyclic factors.
By one of the isomorphism theorems, the subgroups $ V$ of $ G/N$ correspond bijectively to those subgroups $ U$ of $ G$ containing $ N$. More exactly, if $ \pi: G\to G/N$ is the projection, this bijection is given by $ U\to\pi(U)$ and inverse map $ V\to\pi^{\minus{}1}(V)$. This map preserves normal subgroups and $ (A/N)/(B/N)\cong A/B$ for $ A,B$ subgroups of $ G$ and $ B$ normal.
Thus we can take
$ N'_{0}\equal{}\pi^{\minus{}}1 (M_{0}) \equal{} N$,
$ N'_{1}\equal{}\pi^{\minus{}}1 (M_{1})$,
$ N'_{2}\equal{}\pi^{\minus{}}1 (M_{2})$,
...
$ N'_{n}\equal{}\pi^{\minus{}}1 (M_{n}) \equal{} G$
and get that the chain \[ N\equal{} N'_{0}\lhd N'_{1}\lhd ...\lhd N'_{n}\equal{}G\] has only cyclic factors.
Putting this chain together with the first chain mentioned above, we get the following chain which solves $ G$:
\[ \{e\}\equal{} N_{0}\lhd N_{1}\lhd ...\lhd N_{a}\equal{} N \equal{} N'_{0}\lhd N'_{1}\lhd ...\lhd N'_{n}\equal{}G\]

[quote="ZetaX"]Let the solutions (correct word¿) be
[/quote]
I think in English its called "sub-normal series of groups" its called "sub-normal" because each following group has a preceding group as its normal subgroup. There is also something called "normal series of groups" in which all the groups in the sequence are normal subgroups of the full group you're working in.


Does anyone know when the list of perfect scorers will be posted on the AMC website? :lol:

I think it will be posted in December, but I'm not sure.

:huh:
waaaaaaaaaa?
my math teacher said Nov 22... r u sure?
BTW: POWER OF PI, COME ON FTW!
(im lonely... :D )

It's already past November 22. Furthermore, some people wouldn't have take the test by then.

EDIT: I found something. [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC8/2009-amc8TM.pdf]Here[/url] is the teachers' manual. On page 5, right side, it says "Please Note". Underneath, it says that all results should be mailed out by January 16 (I'm thinking earlier than that too), so the Perfect Scores should be up by then.

ah, thank you Power of Pi. I was also wondering when they would put up perfect scores.. I hope I am one :)


inequalities
function
inequalities proposed

Let $ I$ be an interval in the real numbers, and let $ f: I\to\mathbb{R}$ be a function such that for all $ a,b\in I$ and $ 0<t<1$, $ tf(a)\plus{}(1\minus{}t)f(b)\geq f(ta\plus{}(1\minus{}t)b)$.
Let $ x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n,a_1,a_2,\dots,a_n$ be $ 3n$ real numbers such that $ x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n\in I$, $ x_1\geq x_2\geq \dots\geq x_n$, $ y_1\geq y_2\geq \dots\geq y_n$ and $ a_1,a_2,\dots,a_n\geq0$.
Given that \[ a_1x_1\geq a_1y_1\] \[ a_1x_1\plus{}a_2x_2\geq a_1y_1\plus{}a_2y_2\] \[ \cdot\] \[ \cdot\] \[ \cdot\] \[ a_1x_1\plus{}a_2x_2\plus{}\dots\plus{}a_{n\minus{}1}x_{n\minus{}1}\geq a_1y_1\plus{}a_2y_2\plus{}\dots\plus{}a_{n\minus{}1}y_{n\minus{}1}\] \[ a_1x_1\plus{}a_2x_2\plus{}\dots\plus{}a_nx_n\equal{}a_1y_1\plus{}a_2y_2\plus{}\dots\plus{}a_ny_n\]

Prove that $ a_1f(x_1)\plus{}a_2f(x_2)\plus{}\dots\plus{}a_nf(x_n)\geq a_1f(y_1)\plus{}a_2f(y_2)\plus{}\dots\plus{}a_nf(y_n)$.

Let's start with $ a_i\in\mathbb N$ for all $ i$. Then it's directly Karamata, applied to $ x \equal{} (\underbrace{x_1,...,x_1}_{a_1},...,\underbrace{x_n,...,x_n}_{a_n})$ and $ y \equal{} (\underbrace{y_1,...,y_1}_{a_1},...,\underbrace{y_n,...,y_n}_{a_n})$. It is easy to see that $ x\succ y$.

Next, suppose $ a_i\in\mathbb Q^ \plus{}$ for all $ i$. Then we can multiply everywhere with a common denominator to reduce it to the integer case.

The general case ($ a_i\in\mathbb R^ \plus{}$ for all $ i$) follows by continuity.  :)

[quote="bakerbakura"]That is actually my proof, but I wanted to know if there is a direct proof which does not use continuity[/quote]

Using Abel's Identity and by convexity it is done. It is classical proof.

Let $f: \mathbb{I}\to \mathbb{R}$ be a convex function, $a, b, c, d\in \mathbb{R}^+$, $x, y, z, w\in \mathbb{I}$ such that $$x\ge z$$ $$w\ge y$$ $$a+b=c+d$$ $$ax+by=cz+dw$$ then $$af(x)+bf(y)\ge cf(z)+df(w)$$

Can this inequality be generalized to $n$ variables.


If the function $ f(x)$ defined in real numbers satisfies the condition $ f(s^2\plus{}f(t))\equal{}(s\minus{}2t)^2f(s\plus{}2t)$, find the value of $ f(10)$.

[hide]Putting $ \minus{}s$ instead of $ s$, we get

$ (2t\minus{}s)^2f(2t\plus{}s)\equal{}(2t\plus{}s)^2f(2t\minus{}s)\iff{f(2t\plus{}s)\over(2t\plus{}s)^2}\equal{}{f(2t\minus{}s)\over(2t\minus{}s)^2}$

Since this must hold for arbitrarily chosen $ s,t$, we get $ f(x)\equal{}cx^2$, where $ c$ is a constant. Plugging that into the original equation, we get

$ c(s^2\plus{}ct^2)^2\equal{}c(s^2\minus{}4t^2)^2$

So obviously $ c\equal{}0$ or $ c\equal{}\minus{}4$. Both functions $ f(x)\equal{}0$ and $ f(x)\equal{}\minus{}4x^2$ satisfy the original equation, so $ f(10)\in\{0,\minus{}400\}$[/hide]

Whoops, forgot that $ f(x)$ is not constant...

So the answer would be $ \minus{}400$.

Thanks!


I have a post i want to edit in the ohio forum, which mod should I contact?

[quote="r15s11z55y89w21"]If it isn't too late, you should be able to edit it. Otherwise, in this case, you should contact the Ohio forum moderator.[/quote]

1. too late
2. can't find the moderator

already contacted a diff. mod, but he says he can't edit it

Can you just quote yourself and change it and post it as a new post with a note saying what you did?

The limit for editing is 24h, by the way.

It has to be the last post. And as far as the time limit is concerned, I think it's the same as with editing, but I'm not sure- Valentin will probably give a better answer.

[quote="PowerOfPi"]I thought this might be related:

What is the limit for deleting a post? Does it have to be the last post?[/quote]
I think the time limit is 1 hr if it's the last post. That's happened to me.

uhhh so anybody want to tell me who to contact?

[quote="albert1993"][quote="myyellowducky82"]@albert1993
[quote="r15s11z55y89w21"]If it isn't too late, you should be able to edit it. Otherwise, in this case, you should contact the Ohio forum moderator.[/quote][/quote]

and what if there is no ohio forum moderator???

actually i'll just contact an administrator... they can do everything right??[/quote]

yes, they can

[quote="PowerOfPi"]I thought this might be related:

What is the limit for deleting a post? Does it have to be the last post?[/quote]
Actually, if it's the last post, there is no limit. (I posted something about a year ago and there is still a little "delete" box on the post!)

Well, try deleting it then  :P

There's still an edit button on it too, but you can't edit it - same holds for deletion: there is a button, but if the time allowed for deleting expires, you get the "you can't delete it, ask a mod" message.

Just post a new message and quote your old one then correct it.