vector
calculus
derivative
geometry
linear algebra
linear algebra unsolved

Hiiii people!

I have a question. With vectors. It's like ${A.B= k}$, where A,B are vectors. So. how can i solve this of A? the vector A, not just the modulus, i want the vector A.

similary  ${A\times B=C}$, how can i solve for A? the vector ones!?

someone knows!? it this a easy question?! if yes, sorry for the stupid question

thanks you all guys

Which of the infinitely many solutions do you want?

Hi Sir JMerry

In physics we have $P = F/A$

Where P is pressure,  F is force, A is area.

we have a derivative form to this.

but, its also true

$PA = F$

Since F is a vector, A is a vector and P is a vector.

we can rewrite this using vectors, it is true? yes

so. using this i solve for P.
but before that i need to solve this equations i made with dot and cross product

i can make pressure using Del operator, etc...

but i want to make the reverse. it this possible?

thanks jerry

The point is that there isn't -one- solution to solve for; there's a line or a plane. In the case of the dot product, the set of solutions is a plane. For the cross product, it's either empty or a line.

As for your pressure equation, pressure is scalar. It just doesn't make sense to assign a direction to it.

dear jmerry

Force is a Vector. Area is a vector, normal to the plane. and Pressure is a vector too!

Pressure can be a vector make all sense

Pressure has the same direction of the Area

this really make all sense. dont you think?! I think we can make pressure vector. in this case the solution is simple, we can get a unit vector in the same direction of A.

I know that $A.B = A.C$ dont need B=C . the same for cross product.

I just Ask these thinks because i never see a vector equation, solved for one vector.

i thougth that a Vector Solve exist. but... is just a impossible thing, hm... thanks

EDIT:  see http://www.grc.nasa.gov/WWW/K-12/airplane/pressure.html

if a pressure is not a vector, why put all this arrows?
dont you think we can make a pressure be a vector in the sabe direction of the vector $-A$?


probability
calculus
integration
analytic geometry

Two distinct points are selected at random form the six-point set $ \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)\}$. what is the probability that the distance between them on a Cartesian grid is a rational number? Express your answer as a decimal.

First, find the number of possibilities of choosing $ 2$ points from the $ 6$ to select from. There are $ \binom{6}{2}\equal{}15$ ways. Now, let's look for the distances that are rational numbers. Since all of the distances are less than any integral Pythagorean triples, only distances along the grid would be rational. In order for a distance to be along a grid line, the two points chosen must either share an $ x$-coordinate or a $ y$-coordinate. Since there are actually more such pairs of points then there aren't, let's count the pairs that don't share either an $ x$-coordinate or a $ y$-coordinate. These are:

$ (1,1)$ and $ (2,2)$
$ (1,1)$ and $ (2,3)$
$ (1,2)$ and $ (2,1)$
$ (1,2)$ and $ (2,3)$
$ (1,3)$ and $ (2,1)$
$ (1,3)$ and $ (2,2)$

Notice that we don't need to consider pairs for which the first point has an $ x$-coordinate of $ 2$, since the order doesn't matter because we're picking pairs.

Thus, $ \frac{6}{15}\equal{}\frac{2}{5}$ of the points do [i]not[/i] have an rational distance in between. Therefore, $ 1\minus{}\frac{2}{5}\equal{}\fbox{3/5}$ of the points [i]do[/i] have a rational distance in between.


algebra
polynomial
topology
Galois Theory
superior algebra
superior algebra unsolved

Let $K$ by any field and $L$ an algebraic extension of $K$. Then we define $\mathrm{Poly}_{K}(L) : = \{ f(x) \in K[x] \ | \ \exists s \in L: \ f(s)=0 \}$.

If $A,B$ are algebraic extensions of $K$, show that:
\[A \cong_{K}B \iff \mathrm{Poly}_{K}(A)=\mathrm{Poly}_{K}(B) \]

Assume we're working inside a fixed algebraic closure $\overline K$ of $K$. Also, I'm assuming that $K$ is infinite. The case when it is finite (or more generally, perfect, so that all extensions involved are separable) is easier to settle.

We first reduce the problem to the case when $A,B$ are finite extensions of $K$.

Indeed, assume we've proven the assertion in this case. Let $F\subset A$ be a finite set, and let $K_{F}$ be the normal closure of $K(F)$ (in $\overline K$). Now, $\mathrm{Poly}_{K}(K_{F}\cap A)=\mathrm{Poly}_{K}(K_{F}\cap B)$. By the finite statement (assumed true), we get a $K$-isomorphism $\sigma_{F}: K_{F}\cap A\to K_{F}\cap B$. In order to get a $K$-isomorphism $\sigma: A\to B$, we can take a limit point of the $\sigma_{F}$ as $F$ ranges through the finite subsets of $A$, in the following precise sense: extend $\sigma_{F}$ to the whole of $A$ by sending all elements of $A\setminus K_{F}$ to $0$, for example. The $\sigma_{F}$ so extended become maps from $A$ into $B$ whose restrictions to $K_{F}\cap A$ are $K$-homomorphisms. Since each element of $A$ has finitely many conjugates over $K$, these maps are elements of a product of finite sets, and if we endow these finite sets with the discrete topology, the product becomes compact. The [url=http://planetmath.org/?op=getobj&from=objects&id=3250]net[/url] $(\sigma_{F})_{F}$ has a convergent subnet (every net in a compact space has a convergent subnet), and the limit of such a convergent subnet is easily seen to be a $K$-isomorphism of $A$ onto $B$.

We are now left with having to prove the statement for finite extensions $A,B$ of $K$.

For every element of $A$, one of its conjugates over $K$ lies in $B$. This means that $A$ is the union of $\sigma(B)\cap A\ (*)$ as $\sigma$ ranges through the (finitely many) $K$-homomorphisms of $B$ into $\overline K$. Since the $K$-linear space $A$ is a union of finitely many subspaces $(*)$, one of these must be $A$ itself (this holds because $K$ was assumed infinite). In other words, we can find a $K$-homomorphism $\sigma: A\to B$. Similarly, we can find one from $B$ into $A$, so $A$ and $B$ must have the same degree over $K$ and any such $\sigma$ is in fact an isomorphism.

When $K$ is perfect (so, in particular, this covers the case when $K$ is finite), reduce the problem to the case of finite extensons as above, and then use the fact that a finite separable extension is finite to see that $A=K(\alpha),B=K(\beta)$ for roots $\alpha,\beta$ of the same irreducible polynomial in $K[X]$.

P.S.

Charlydif used the same linear subspace argument [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100479]here[/url].


number theory
prime numbers
number theory proposed

Find all $m,n\in N ^*$,m <> n such that:
            $\varphi (m)|n ,\varphi (n)|m$ :P

We will use well known formula $\varphi(n)=p_1^{k_1-1}(p_1-1)...p_s^{k_s-1}(p_s-1)$.

We have two trivial solutions $(m,n)=(2,3)$ or $(m,n)=(1,2)$.
Suppose $m,n\neq 2$. Let $m=2^am_0$ and $n=2^bn_0$, $m_0$ and $n_0$ are odd numbers.
[b]1. [/b] $m_0=1$ and $n_0=1$ - impossible.

[b]2. [/b] $m_0=1$ and $n_0>1$. We have $\varphi(m)=2^{a-1}$ $\Rightarrow$ $b\geq a-1$. Moreover, $2^{b-1}\varphi(n_0)\,|\,2^{a}$ $\Rightarrow$ $\varphi(n_0)=2^{s}$ and $s=0,1,2$. So we obtain following pairs $(m,n)$:
$(2^a,3\cdot 2^{a-1})$, $(2^a,3\cdot 2^{a})$ and $(2^a,5\cdot 2^{a-1})$.

[b]3. [/b] $m_0>1$ and $n_0>1$. Then $\varphi(n_0)$ and $\varphi(m_0)$ are even numbers. Assume WLOG $a\leq b$. We have $2^{b-1}\varphi(n_0)\,|\,2^am_0$ $\Rightarrow$ it is necessary $a=b$ and $\varphi(n_0)$ isn't divisible by 4. Since $a=b$ we obtain by the same way that $\varphi(m_0)$ isn't divisible by 4. It implies that $n_0$ and $m_0$ are powers of prime numbers, i.e. $n_0=p^k$ and $m_0=q^l$. Assume WLOG $p\leq q$. We know that $p-1$ isn't divisible by 4. If $p-1\neq 2$ then $p-1$ has odd prime divisor $r$, $r<p$. Since $\varphi(n)\,|\,m$ we conclude $r\,|,m$ -- impossible. So $p=3$. If $k>1$ then $3\,|\,\varphi(n)$ $\Rightarrow$ $q=3$. And we easily derive $k=l$, $l=k+1$ or $k=l+1$, hence we obtain following pairs $(m,n)$:
$(2^a3^k,2^a3^{k+1})$.
Suppose $k=1$. If $l>1$ then $q\,|\,n$ $\Rightarrow$ $q=3$. And again we obtain the same pairs. If $l=1$ then we derive from $\varphi(m)\,|\, n$ that $q-1\,|\,6$, i.e. $q=7$. So additional pairs $(m,n)$:
$(3\cdot 2^a,7\cdot 2^a)$.

Answer: $(2^a,3\cdot 2^{a-1})$, $(2^a,3\cdot 2^{a})$, $(2^a,5\cdot 2^{a-1})$, $(2^a3^k,2^a3^{k+1})$, $(3\cdot 2^a,7\cdot 2^a)$.

I hope these are all cases ;)


Let's take up simple proving skills by induction, however simple, these can be tough for students not comfortable with induction.

1) Show that $n^2<3^n$ for all integers $n\geq 0$.

2) Prove that $2^n+3^n<4^n+2$ for all integers $n\geq 0$.

3) (Challenge) Prove that for integers $n>1$, $\frac{9^{3n-4}+3^{3n-4}+1}{13}$ is always an integer.

1.  [hide]The inequality holds for the base cases n = 0, 1, and 2, since 0^2 < 3^0, 1^2 < 3^1, and 2^2 < 3^2.  Thus we need only prove the inductive step for .

Assume the inequality holds for n = k, i.e. k^2 < 3^k.  We will prove that it holds for n = k+1.  Multiplying the inequality k^2 < 3^k by 3 yields 3k^2 < 3^(k+1).  It then suffices to prove that (k+1)^2 < 3k^2.

Rearranging (k+1)^2 < 3k^2 yields

2k^2 - 2k - 1 > 0

k^2 - k - 1/2 > 0

(k-1/2)^2 > 3/4

k - 1/2 > /2

k >

So for k  2, (k+1)^2 < 3k^2 < 3^(k+1), and the inductive step is complete. [/hide]

Okay, here's an attempt at this problem. (nr. 3)

Hint: [hide] 3^3 = 27 = 1 (mod 13) [/hide]

Solution: [hide]

 is , so we have a base case.

Assume that  is an integer, or equivalently that

. We show that

But our expression is

We know that  divides , so it just remains

to show that  divides

We have that this is

Now we note that

Also,

Thus, 13 divides the expression and our inductive hypothesis is true, proving the claim.

[/hide]

My attempt at 3:   I'll just prove the induction step. [hide]

.

Now

is equivalent to:

It's easy to check that

 and

Thus

Q.E.D.

[/hide]


1) ine. equivolent to:
$ 2(a^3\plus{}b^3\plus{}c^3)\plus{}3abc \geq \sum_{sym}{a^2b}\plus{}a^3\plus{}b^3\plus{}c^3$
or
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \geq \sum_{sym}{a^2b}$
which is right by Schur inequality.
2) ine. equivolent to:
$ a^3\plus{}b^3\plus{}c^3\plus{}\sum_{sym}{a^2b} \geq a^3\plus{}b^3\plus{}c^3\plus{}6abc$
or
$ \sum_{sym}{a^2b} \geq 6abc$
which is right by AM-GM inequality.
3) ine. equivolent to:
$ a^3\plus{}b^3\plus{}c^3\plus{}6abc \geq \sum_{sym}{a^2b}\plus{}3abc$
or
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \geq \sum_{sym}{a^2b}$
which is right by Schur inequality.


guys,
can you give me some links to lecture notes on elementary cosmology and accretion processes?

I think your best bet is to search university websites and hope to find something.


linear algebra
matrix
induction
graph theory
combinatorics proposed
combinatorics

In the class, there are $ 15$ boys and $ 15$ girls. On March $ 8$, some boys made phone calls to some girls to congratulate them on the holiday ( each boy made no more than one call to each girl). It appears that there is a unique way to split the class in $ 15$ pairs (each consisting of a boy and a girl) such that in every pair the boy has phoned the girl. Find the maximal possible number of calls.

For a class with $ N$ boys and $ N$ girls, the maximal number of calls is $ \frac{N\left(N\plus{}1\right)}{2}$.  If more calls are made, the partition is not unique.

Consider the boys numbered in any order as $ b_{1},b_{2},...,b_{N}$, and the girls numbered as $ g_{1},g_{2},...,g_{N}$, so that  boy $ b_{i}$ is paired up with girl $ g_{i}$ in the unique partition, for all $ i\equal{}1,2,3,...,N$.  Consider now an $ N\times N$ matrix such that, in position $ \left(i,j\right)$, a 1 is placed iff boy $ b_{i}$ called girl $ g_{j}$.  Obviously, the diagonal has all 1's since boy $ b_{i}$ called girl $ g_{i}$ for all $ i\equal{}1,2,...,N$ (since they are paired).  Now, positions $ \left(i,j\right)$ and $ \left(j,i\right)$ cannot both have 1 for any $ i\neq j$, since otherwise boy $ b_{i}$ could also be paired up with girl $ g_{j}$, and boy $ b_{j}$ with girl $ g_{i}$, making the partition not unique.  So, out of the $ N^{2}\minus{}N$ elements not in the main diagonal of the matrix, at most half have 1's.  The total number of ones (and hence of phone calls) is then no larger than $ N\plus{}\frac{N^{2}\minus{}N}{2}\equal{}\frac{N\left(N\plus{}1\right)}{2}$.  Assume now that, for any ordering of boys as $ b_{1},b_{2},...,b_{N}$, and girls as $ g_{1},g_{2},...,b_{N}$, boy $ b_{i}$ called exactly the $ i$ girls numbered as $ g_{1},g_{2},...,g_{i}$.  It is a trivial exercise in induction to prove that boy $ b_{i}$ is then paired up with girl $ g_{i}$ in the unique possible partition.  So for exactly $ \frac{N\left(N\plus{}1\right)}{2}$ calls the partition is unique, and this is the maximal number that we were looking for.

This is wrong.[hide]I disagree. The problem is equivalent with a balanced bipartite graph ($ V_{1}\equal{} V_{2}\equal{} 15$) which has to be acyclic (this is equivalent with the uniqueness). So it can have at most $ 30\minus{}1 \equal{} 29$ edges (generally $ 2N\minus{}1$), and an example can be easily constructed.[/hide]

I give in detail the next example of $ \frac{15\cdot16}{2}\equal{} 120$ phone calls:

$ b_{i}$ makes $ i$ calls: he calls $ g_{1},g_{2},...,g_{i}$.  The sum of these calls is then $ 1\plus{}2\plus{}...\plus{}15 \equal{} 120$.

Now, $ b_{1}$ can only be paired with $ g_{1}$, since he didn't call any other girl.

$ b_{2}$ called $ g_{1}$ (already taken) and $ g_{2}$, and no other girl.  So he can only be paired with $ g_{2}$.

$ b_{3}$ called $ g_{1}$ (taken), $ g_{2}$ (taken) and $ g_{3}$, and no other girl.  So he can only be paired with $ g_{3}$.

And so on, until $ b_{15}$, who called all the girls, but can be paired only with $ g_{15}$, since all other girls are already taken.  So the partition is unique... or so it seems to me.

Could you be so kind as to point where my mistake lies?  Thanks in advance...

You are obviously right sorry, I made a stupid mistake :oops: , the cyclic thing is completely wrong.

No need to be sorry, it is good to have somebody look at your solutions critically (otherwise if a real mistake was made, I could have never known about it).  So actually, thanks...    :)


trigonometry
ratio
Pythagorean Theorem
geometry

A circle of radius r is inscribed in a sector which is one-sixth of a circle of radius 1.  Find r.

[quote="chickendude"][hide]

Draw the larger sector, and the circle within it
From the center of the smaller circle, draw a radius to the point of tangency to the radius of the larger circle
Then, connect the centers of the two circles

This will form a right triangle (with the right angle at the point of tangency)

one base is $r$
let the other base be $x$
The hypotenuse is $R-r$ or $1-r$

using the pythagorean theorem

$(1-r)^{2}= r^{2}+x^{2}$
$1-2r = x^{2}$

also, using triangle trig
$tan(30) = \frac{r}{x}= \frac{r}{\sqrt{1-2r}}= \frac{1}{\sqrt3}$

$r\sqrt3 = \sqrt{1-2r}$

$3r^{2}= 1-2r$

$(3r-1)(r+1) = 0$

$r = \frac{1}{3}$ since $r=-1$ is extraneous
[/hide][/quote]
ALright thanks man, all i did was read your first fact and how the hypostenuse was 2r, well add the r to it and 2r+r=1 (the radius of the circle) so r=1/3, but your extra steps chekc the problem

[quote="chickendude"]how did you know that the hypotenuse was $2r$?[/quote]

30-60-90 triangle, hypotenuse is 2r if the opposite of 30 degress is r.

Yeah - or you could have seen that one of the angles of the right triangle was exactly one half the angle of the sector (60 degrees) because that line that we designated as the hypotenuse bisects that angle, so the triangle had to be 30-60-90. That means the ratios had to to be $1-\sqrt{3}-2$ If you look at the diagram (here it is if you didn't make draw it) You can just add $2r+r$ to get the radius of the larger circle, which is one. So $3r=1 \Rightarrow r=\frac{1}{3}$


[quote="GameBot"]Counting with Venn Diagrams and Subtraction Part 1

We tackle a couple problems with 2-circle Venn diagrams.  We also do the problems by correcting for overcounting with subtracting, which is a beginning step on a more advanced counting technique called the Principle of Inclusion-Exclusion.[/quote][hide]Venn Diagrams are a little easy[/hide]

Yeah, but when you have to solve one fast, it's a different story. :)
But this lesson was pretty easy.

this as well as a humourus video was actually very helpful :starwars:  :yoda:

[quote="Saphira 7"]I can't see the far right side of my screen where he's doing the work; I don't know if this is the case for anyone else, but it is pretty infuriating. :mad:[/quote]
I agree with that, I cannot see the screen.
[size=200] Part 1 [/size]
19 dog
14 cat
6 both
5 none

$19-6=\boxed{13\text{ Dogs only}}$
$14-6=\boxed{8\text{ Cats only}}$

[asy]
size(6cm,6cm);
draw((0,0)--(14,0)--(14,10)--(0,10)--(0,0));
draw(circle((5,6),3.5));
draw(circle((9,6),3.5));
label("Cat",(4,7));
label("Only",(4,6));
label("8",(4,5));
label("Both",(7,7));
label("6",(7,6));
label("Dog",(10,7));
label("Only",(10,6));
label("13",(10,5));
label("Neither: 5",(14,0),NW);[/asy]

There is going to be $8+6+5+13=\boxed{32\text{ total houses}}$ and
$8+6+13=\boxed{27\text{ pet houses}}$

[size=200] Part 2 [/size]
42 friends
23 know math
21 good looking
18 don't know math, and are not good looking

We use the formula $n+r-both+Neither=Total$ to find:

$23+21-both+18=42$
$44-both+18=42$
$62-both=42$
$both=\boxed{20}$

To find the number of people that know math only, we would have $23-20=3$
To find the number of people that are only good looking, we would have $21-20=1$
$3+1+20+18=4+20+18=22+20=42$ so it works :)

[asy]
size(6cm,6cm);
draw((0,0)--(14,0)--(14,10)--(0,10)--(0,0));
draw(circle((5,6),3.5));
draw(circle((9,6),3.5));
label("Know",(4,7));
label("Math", (4,6));
label("Only",(4,5));
label("3",(4,4));
label("Both",(7,7));
label("20",(7,6));
label("Good",(10,7));
label("Looking", (10,6));
label("Only",(10,5));
label("1",(10,4));
label("Neither: 18",(14,0),NW);[/asy]

Edit:  Fail I put 9 in the Cat only Venn diagram section.  Thank rrusycyzk for the answer at the end there, so I could check my work and added part 2.
Edit 2:  My Venn Diagram for part 2 failed XD


I will put up some practice problems

1.  26 - 4a = 9a

2.  8x = -11(6 - x)

3.  29 + q = -2(q - 13)

4.  -2(c + 7) - 30 = 9c

These are some easy problems..
Try them :cool:

[hide]1. 26 - 4a = 9a
add 4a to both sides and the equation becomes 26 = 13a and so a = 2

2. 8x = -11(6 - x)
you distriube the -11 and the equation becomes 8x = -66 + 11x and so if you subtract 11x from both sides, you get -3x = -66 and so x is 22

3. 29 + q = -2(q - 13)
same ways as above 29 + q = -2q+ 26, subtract 26 from both sides and you get q + 3 = -2q, subtract q from each side and you get 3 = -3q and so q = -1

4. -2(c + 7) - 30 = 9c
-2c - 14 - 30 = 9c
add 2c to both sides and add -14 to -30 and you get:
-44 = 11c
and c equals to -4

if I did the multiplication of addition wrong, sorry. I'm not very good at getting the numbers right. [/hide]

[quote="shinwoo"]Very Well!

You got them all right!

but try to hide your solutions, okay?[/quote]

huh what do you mean, I did, didn't I. it's showed up in "hide" mode on my computer. is it not showing up right?

[quote="Alfian Rahman"]My solution

Number 1
[hide]

$ 9a \equal{} 26 \minus{} 4a$
$ 26 \equal{} 9a \plus{} 4a$
$ 26 \equal{} 13a$
$ a \equal{} 2$

So, $ a$ is $ 2$
[/hide]
Number 2
[hide]

$ 8x \equal{} \minus{} 11(6 \minus{} x)$
$ 8x \equal{} \minus{} 66 \plus{} 11x$
$ \minus{} 11x \minus{} 8x \equal{} \minus{} 66$
$ \minus{} 3x \equal{} \minus{} 66$
$ x \equal{} 22$

So, $ x$ is $ 22$
[/hide]
Number 3
[hide]

$ 29 \plus{} q \equal{} \minus{} 2(q \minus{} 13)$
$ 29 \plus{} q \equal{} \minus{} 2q \plus{} 26$
$ 29 \minus{} 26 \equal{} \minus{} 2q \minus{} q$
$ 3 \equal{} \minus{} 3q$
$ q \equal{} \minus{} 1$

So, $ q$ is $ \minus{} 1$
[/hide]
Number 4
[hide]

$ \minus{} 2(c \plus{} 7) \minus{} 30 \equal{} 9c$
$ \minus{} 2c \plus{} 14 \equal{} 9c \plus{} 30$
$ 14 \plus{} 30 \equal{} \minus{} 2c \minus{} 9c$
$ 44 \equal{} \minus{} 11c$
$ c \equal{} \minus{} 4$

So, $ c$ is $ \minus{} 4$
[/hide]

I am just a student of elementary school in Indonesia[/quote]

Hi, I noticed you are new. I think it would be better if you would leave these threads alone, since they are answered already. Also, they are about 4 years old. It would be called "reviving old topics". It is okay because you didn't know that.  :wink:

I am sorry, I don't know about that :oops_sign:

Hi "arbala", OK it's a nice try since it's your first post but please see the previous posts and the dates. I mean it's already answred and if there is something you couldn't understand from the past solutions you may ask about it and its been very old to be back again.

Thanking your understanding (or PowerOfPi will be angry  :P )

Sorry I did not see the time.

When I started to see the problem I started doing it.  :blush:


Let $ x_1;x_2;...;x_n$ be $ n$ arbitrary real numbers chosen in a given closed interval $ [a;b]$.Prove that:
$ \sum^{n}_{i\equal{}}(x_i\minus{}\frac{a\plus{}b}{2})^2 \geq \sum^{n}_{i\equal{}1}(x_i\minus{}\frac{1}{n}\sum^{n}_{i\equal{}1}x_i)^2$

$ \sum_{i \equal{} 1}^{n}(x_{i} \minus{} \frac {a \plus{} b}{2})^2\ge \sum_{i \equal{} 1}^{n}(x_{i} \minus{} \frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i})^2$
$ \Leftrightarrow \sum_{i \equal{} 1}^{n}(x_{i} \minus{} \frac {a \plus{} b}{2})^{2} \minus{} (x_{i} \minus{} \frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i})^2\ge 0$
$ \Leftrightarrow \sum_{i \equal{} 1}^{n}(\frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i} \minus{} \frac {a \plus{} b}{2})(2x_{i} \minus{} \frac {a \plus{} b}{2} \minus{} \frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i})\ge 0$
$ \Leftrightarrow (\frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i} \minus{} \frac {a \plus{} b}{2})(2\sum_{i \equal{} 1}^{n}x_{i} \minus{} n\frac {a \plus{} b}{2} \minus{} n*\frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i})\ge 0$
$ \Leftrightarrow (\frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i} \minus{} \frac {a \plus{} b}{2})(\sum_{i \equal{} 1}^{n}x_{i} \minus{} \frac {n(a \plus{} b)}{2})\ge 0$
$ \Leftrightarrow n(\frac {1}{n}\sum_{i \equal{} 1}^{n}x_{i} \minus{} \frac {a \plus{} b}{2})^2\ge 0$
which is always true, and the result follows.


I solved this but I'm not sure if this is right or not.

Problem:

$x - \frac{1}{x} = 2 i \sin \theta$

Then what does $x^2 - \frac{1}{x^2} = $?

[hide="My answer"]

$(x - \frac{1}{x})(x - \frac{1}{x}) = x^2 - 2 + \frac{1}{x^2} = -4 \sin^2 \theta$

Adding four to both sides, we get:

$x^2 + 2 + \frac{1}{x^2} = -4 \sin^2 \theta + 4\\
\sqrt {(x + \frac{1}{x})^2} = \sqrt {4 (- \sin^2 \theta + 1)}\\
x + \frac{1}{x} = \sqrt {4 (1 - \sin^2 \theta)}\\
= \sqrt{4(\cos^2 \theta)}\\
= 2 \cos \theta$

Since $x^2 - \frac{1}{x^2} = (x + \frac{1}{x})(x - \frac{1}{x})$, multiplying our original given and what we just found, we get:

$2i \sin \theta \cdot 2 \cos \theta = 4i \sin \theta \cos \theta = 2i \sin (2 \theta)$
[/hide]

Is this right? :huh:


function
calculus
derivative
analytic geometry
calculus computations

Given that $ x = e^t$, show that

            $ \frac {d^2y}{dx^2}=e{}^-{}^2{}^t(\frac {d^2y}{dt^2}-\frac {dy}{dt})$

Any help greatly appreciated.

Is $ y$ a function of $ x$?  What is $ y$?

JRav

The problem doesn't state. I assume we are to take y as a function of x.

I can't find $ \frac{dy}{dx}$ because I have no idea what $ y$ is.  I assume it's a function of $ x$ but I don't know what that function is.  This means  $ y\equal{}y(x)\equal{}y(x(t))$.

How is this question not clear? It's asking how some derivatives are related under a change of variables (under the usual convention that $ y$ is seen as a function of two different coordinate systems, although this is wrong  :wink:)

jimi, why don't you show us what you think you're supposed to do.  (You obviously have a first clue, judging by the title of the original post.)  Here is a not very helpful hint: "In order to compute a second derivative, we first need to compute [what?].  In order to compute [what?], we first use [something].  Then ...."

JBL

$ \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=e{}^-{}^t.\frac{dy}{dt}$

now I suppose I'm to differentiate both sides with respect to x but how do I do that with the RHS?

By the product rule plus the chain rule.

$ \frac {d}{dx}\left( \frac{dy}{dt} \right) \equal{} \frac {d^2y}{dt^2}\cdot\frac {dt}{dx}$

And differentiating $ e^{ \minus{} t}$ is the same.

Kalle

Could you please explain how you got that? I am a little unclear of certain uses of the chain rule.
Thanks.

Again, it's nothing more than a straight application of the chain rule.  It it helps, think of it as
$ \frac{d}{dx}\left(\frac{dy}{dt}\right)\equal{}(y''(t))(x(t))$

What JRav wrote is wrong and you should ignore it (unless it gets fixed).

The easiest way to see why what Kalle wrote is the chain rule is to define $ z(t) \equal{} y'(t)$ (or $ z(t) \equal{} \frac {dy}{dt}$, if you prefer).  So, after taking the first derivative you have $ \frac {dy}{dx} \equal{} e^{ \minus{} t} \cdot z(t)$ (although actually I wouldn't have switched the $ x$ to an $ e^{ \minus{} t}$ just yet, since you're about to take a derivative with respect to $ x$).  Then what Kalle wrote is just $ \frac {dz}{dx} \equal{} \frac {dz}{dt} \cdot \frac {dt}{dx}$, which should look familiar to you.

Don't get too closely wedded to the particular variable and function names that get used in things like the chain rule :)

I pointed this out in another thread (maybe one of jimi's?).  If it helps, think of it like this:
$ \frac{dz}{dx}\equal{}\frac{dz}{\not dt}\cdot\frac{\not dt}{dx}\equal{}\frac{dz}{dx}$
Note that this is nonsense and does not actually mean anything.  But if it helps you see why it works, then use it.

Thank you Kalle, JBL and JRav for your patience. I get this now, thanks to you guys; much appreciated.


I did it two ways, and my answers don't agree...

[hide="Using the sum of the first n squares formula"]

Sum of the first $2n$ squares $=\frac{1}{6}2n(2n+1)(4n+1)$.

Now we need to subtract twice the sum $(2^{2}+4^{2}+...+(2n)^{2})$
$=\frac{4}{3}n(n+1)(2n+1)$

and then add $(2n)^{2}$ because we only needed to subtract this bit once.

$S=\frac{1}{3}(n(2n+1)(4n+1)-4n(n+1)(2n+1)+12n^{2})$
$S=\frac{n}{3}(8n^{2}+6n+1-8n^{2}-12n-4+12n)$
$S=\frac{n}{3}(6n-3)$
$S=n(2n-1)$.[/hide]

[hide="More directly"]
The required sum is equal to
$1+\sum^{n}_{1}(2k+1)^{2}-(2k)^{2}$
$1+\sum^{n}_{1}4k+1$
$n+1 \sum^{n}_{1}4k$
$n+1+2n(n+1)$
$2n^{2}+3n+1$[/hide]

[quote="tim1234133"]Now we need to subtract twice the sum $(2^{2}+4^{2}+...+(2n)^{2})$
$=\frac{4}{3}n(n+1)(2n+1)$ [/quote]

Is this even right?

[quote="Franz Joseph"]Evaluate:

$1^{2}-2^{2}+3^{2}-4^{2}+..........+(2n+1)^{2}$

2 methods (or more) to do this, see if you can find more than one way.[/quote]

I don't get your sum :| I can hardly understand how a sqared number can be negative. If $(2n+1)^{2}$ is somewhat the general therm of your sum then it should look something like $(-1)^{k}(2n+1)^{2}$. I may be wrong, but please explain this to me if you don't mind :blush:

To icx: $(2n+1)^{2}$ is not the general term, it is the final term. The sum is $1-4+9-16+...-(2n)^{2}+(2n+1)^{2}$, where the series terminates here.

[quote="tim1234133"]I hope so  :maybe:
- my reasoning was as follows:
$(2^{2}+4^{2}+...+(2n)^{2})=4(1^{2}+2^{2}+...+n^{2})=4\times \frac{1}{6}n(n+1)(2n+1)$, but we want twice that, which is why the 6 is replaced by a 3.[/quote]

Then it should look something like

$\frac{2}{3}n(n+1)(2n+1)$

I wanted to subtract [b]twice[/b] that value $=2 \times 4 \times \frac{1}{6}n(n+1)(2n+1)$

Having said that, do you know the answer to this? Are either of my solutions correct?

tim I believe your second answer is correct.  Let's just factor a lot.  The sum becomes $\ (1-2)(1+2)+(3-4)(3+4)+\ldots (2n-1-2n)(2n-1+2n)+(2n+1)^{2}$.  The first part of each of these factors is going to be $\-1$ therefore the sum is equal to $\ (2n+1)^{2}-(1+2+3+4+\ldots 2n-1+2n = (2n+1)^{2}-n(2n+1) = (2n+1)(n+1)= \boxed{2n^{2}+3n+1 }$

[quote="lordWings"]How about a graphic solution? Find the red area.[/quote]
The red area in the first picture is
$(2n+1)^{2}-(2n)^{2}+\dots+3^{2}-2^{2}+1^{2}=\sum_{k=1}^{n}(4k+1)=(2n+1)(n+1)$
The red area in the second picture is
$(2n+1)^{2}-(2n)^{2}+\dots+3^{2}-2^{2}+1^{2}=\sum_{k=1}^{2n+1}k=(2n+1)(n+1)$


inequalities
function
algebra proposed
algebra

Find all functions f defined on the positive integers with values in the same set and such that  for all $n$
\[f(n+1)>\frac{f(n)+f(f(n))}{2}.\]

[i]Edited by Myth[/i]

Prepare for a battle!

First of all, we will prove that $f$ is an increasing function.
Let $m_1=\min f(n)$, $f(n_1)=m_1$. If we suppose that $n_1>1$ then $f((n_1-1)+1)>\frac{f(n_1-1)+f(f(n_1-1))}{2}$ is a wrong inequality! So $f(1)=m_1$ and $f(n)>m_1$ for all $n>1$.
Suppose we proved that $f(1)<f(2)<...<f(n)<f(k)$ for all $k>n$. Let $m_{n+1}=\min_{k>n}f(n)$, $m_{n+1}=f(n_{n+1})$. We have $m_{n+1}=f(n_{n+1})>\frac{f(n_{n+1}-1)+f(f(n_{n+1}-1))}{2}$. From here we see
1) $f(n_{n+1}-1)<m_{n+1}$
or
2) $f(f(n_{n+1}-1))<m_{n+1}$
From 1) we conclude $n_{n+1}-1\in\{1,2,...,n\}$. It follows that $n_{n+1}-1=n$, so $n_{n+1}=n+1$.
From 2) we conclude $f(n_{n+1}-1)\in\{1,2,...,n\}$. And again it follows $n_{n+1}-1=n$.

As $f$ is increasing, it follows that $f(n)-n$ is nondecreasing. Suppose $f(n)-n\geq 2$ for some $n$ (so for all $km\geq n$ too :) ), then $f(n+1)-f(n)>f((f(n))-f(n+1)\geq f(n+2)-f(n+1)$. It means that $f(k+1)-f(k)$ for $k\geq n$ is an infinite decreasing sequence, but $f(k+1)>f(k)$ !!!

Therefore, answer is $f(k)=k$ for $k=1,...,n-1$, $f(k)=k+1$ for $k\geq n$.

I hope, it is correct :) Because it is 4:08 here, and I am so stupid this year :(


counting
distinguishability
geometry
geometric transformation
rotation
function
ceiling function

[quote="iamagenius"]in how many ways can four identical red chips and two identical white chips be arranged in a circle?

why is the answer not:

$\frac{\frac{6!}{4!2!}}{6}=2.5$?[/quote]

because there can not be 2 and a half ways to rearrange something

[quote="iamagenius"]obviously. but the method is correct is it not? the numerator represents how many different arrangements in a line. and in a circle you just divide by the number of objects to account for rotations.[/quote]
They are indistinguishable so the only ways are the two white chips together, them seperated by one, and them seperated by two. So 3?

[quote="iamagenius"]in how many ways can four identical red chips and two identical white chips be arranged in a circle?

why is the answer not:

$\frac{\frac{6!}{4!2!}}{6}=2.5$?[/quote]

Notice that in this problem flips and rotations cannot be accounted four.

Here is a hint.  Try to consider symmetry.

bpms is right.
it is three.
this picture may help.
sorry that it sucks so much though :oops:

[quote="hikaru123"]This result is right. But can you give the answer for the problem with m red chips and n white chips??[/quote]

That is going to require a lot of case work.

i'm probably wrong on this,
but so far, for m red chips and n white chips,
i get the equation:
[hide]
(m+n-1)!/(m!)(n!) then take the ceiling function
[/hide]
please tell me i'm wrong on this
because as i am typing this, i already found
something wrong with my equation.............

[quote="davidlizeng"]i'm probably wrong on this,
but so far, for m red chips and n white chips,
i get the equation:
[hide]
(m+n-1)!/(m)(n) then take the ceiling function
[/hide]
please tell me i'm wrong on this
because as i am typing this, i already found
something wrong with my equation.............[/quote]
How did you get it?
Wouldn't that give 15 for the original problem?  :maybe:

ooops..
didn't put the !s in the denominator.
edited..

I'm so dumb..........  :wallbash:  :fool:


calculus
integration
limit
trigonometry
logarithms
real analysis
function

Evaluate

\[\lim_{n\to\infty}\int^\frac{\pi}{2}_0 \frac{\sin ^2nx}{1+x} dx\]

The limit is $\frac12\ln(1+\frac{\pi}2)$.

We can write $\int_0^{\frac{\pi}2} \frac{\sin^2(nx)}{1+x}dx = \frac12\int_0^{\frac{\pi}2}\frac1{1+x}dx - \frac12\int_0^{\frac{\pi}2}\frac{\cos(2nx)}{1+x}dx$.

The first integral can be readily computed to the value given above. The second integral tends to zero as $n\to\infty$ by the Riemann-Lebesgue Lemma.

You don't really need the heavy machinery of the Riemann-Lebesgue Lemma on that second integral; one integration by parts and a simple estimate will do the trick.

$\int_0^{\frac{\pi}2}\frac{\cos(2nx)}{1+x}dx = {\frac1{2n}\frac{\sin(2nx)}{1+x}|_0^{\frac{\pi}2}+ \frac1{2n}\int_0^{\frac{\pi}2}\frac{\sin(2nx)}{(1+x)^2}dx}$

$|\int_0^{\frac{\pi}2}\frac{\cos(2nx)}{1+x}dx| \le \frac1{2n}\int_0^{\frac{\pi}2} 1\cdot dx$

and that goes to zero as $n\to\infty.$

In the language of Schwartz distributions (generalized functions), the sequence of functions $\sin^2(nx)$ tends weakly to $\frac12$ as $n\to\infty.$

Kent Merryfield,

Indeed, my solution is exactly same as yours.

Kunny

you can have the same result using some non-trivial fact:

$ \lim_{n\to\infty}\int^a_b f(x)g(nx) dx = \frac{1}{T}\int^T_0 g(x) dx
\int^a_b f(x) dx $, where function $g(x)$ is T-periodic and $f,g$ are continious on $[a,b]$  (See IMC1994,A5)

eugene,

It is interesting! Thank you  :)

kunny


modular arithmetic
number theory proposed
number theory

We examine the following two sequences: The Fibonacci sequence: $F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 }$ for $n \geq 2$; The Lucas sequence: $L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2}$ for $n \geq 2$. It is known that for all $n \geq 0$
\[F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, \]
where $\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}$. These formulae can be used without proof.

Prove that $1+L_{2^{j}}\equiv 0 \pmod{2^{j+1}}$ for $j \geq 0$.

First note that
$$L_n^2 = \left(\alpha^n + \beta^n\right)^2=\alpha^{2n} + \beta^{2n} + 2(\alpha\beta)^n = L_{2n} + 2(-1)^n\, .$$
So, for all $k\geq 1$
$$L_{2^{k+1}} +1 = L_{2^k}^2 - 1\, .$$
Use induction on $j$ to finish the problem. It's easy to check for $j = 0,  1$. Assuming that the claim is true for some $j\geq 1$ we get
$$L_{2^{j+1}} +1 = L_{2^j}^2 - 1 = (2^{j+1}m -1)^2 - 1 = 2^{2j+2}m^2 - 2^{j+2}m \equiv 0 \pmod{2^{j+2}}\, .$$


calculus
integration
conics
parabola
geometry
limit
analytic geometry

For $a>0$, let $l$ be the line created by rotating the tangent line to parabola $y=x^{2}$, which is tangent at point $A(a,a^{2})$, around $A$ by $-\frac{\pi}{6}$.
Let $B$ be the other intersection of $l$ and $y=x^{2}$. Also, let $C$ be $(a,0)$ and let $O$ be the origin.

(1) Find the equation of $l$.

(2) Let $S(a)$ be the area of the region bounded by $OC$, $CA$ and $y=x^{2}$. Let $T(a)$ be the area of the region bounded by $AB$ and $y=x^{2}$. Find $\lim_{a \to \infty}\frac{T(a)}{S(a)}$.

1) the slope of $ y \equal{} x^2$ at $ \left(a,a^2\right)$ is obviously $ 2a \equal{} \tan\alpha$.  The slope of $ l$ is obviously
\[ \tan\left(\alpha \minus{} \frac {\pi}{6}\right) \equal{} \frac {2a \minus{} \frac {1}{\sqrt {3}}}{1 \plus{} \frac {2a}{\sqrt {3}}} \equal{} \frac {2\sqrt {3}a \minus{} 1}{2a \plus{} \sqrt {3}}.
\]
Since it passes through point $ \left(a,a^2\right)$, its equation is
\[ y \equal{} a^2 \plus{} \frac {\left(2\sqrt {3}a \minus{} 1\right)\left(x \minus{} a\right)}{2a \plus{} \sqrt {3}}.
\]
2) Call  $ \left(b,b^2\right)$ the other point of intersection of $ l$ and $ y \equal{} x^2$.  Then, obviously
\[ b \plus{} a \equal{} \frac {b^2 \minus{} a^2}{b \minus{} a} \equal{} \frac {2\sqrt {3}a \minus{} 1}{2a \plus{} \sqrt {3}},
\]

\[ a \minus{} b \equal{} 2a \minus{} \frac {2\sqrt {3}a \minus{} 1}{2a \plus{} \sqrt {3}} \equal{} \frac {4a^2 \plus{} 1}{2a \plus{} \sqrt {3}}.
\]
It is also obvious, using the area of a trapezoid, that
\[ T(a) \equal{} \frac {\left(b^2 \plus{} a^2\right)\left(a \minus{} b\right)}{2} \minus{} \int_b^ax^2dx \equal{} \frac {a^3 \minus{} 3a^2b \plus{} 3ab^2 \minus{} b^3}{6} \equal{} \frac {\left(a \minus{} b\right)^3}{6},
\]
and trivially
\[ S(a) \equal{} \int_0^ax^2dx \equal{} \frac {a^3}{3},
\]
or
\[ \frac {T(a)}{S(a)} \equal{} \frac {\left(a \minus{} b\right)^3}{2a^3},
\]
and since trivially
\[ \lim_{a\rightarrow\infty}\frac {a \minus{} b}{a} \equal{} 2,
\]
the value of the limit we are looking for is $ \frac {2^3}{2} \equal{} 4$.

That's right. :)

$ l: \ y \equal{} (a \plus{} b)x \minus{} ab\Longrightarrow T(a) \equal{} \int_b^a \{(a \plus{} b)x \minus{} ab \minus{} x^2\}\ dx$

$ \equal{}\minus{} \int_b^ a (x \minus{} a)(x \minus{} b)\ dx \equal{} \frac {1}{6}(a \minus{} b)^3.$


Sincerely, I don't have the slightest idea for this problem:
   Suppose that $a,b$ are distinct positive integers and that the numbers $[a^n x]$ with $n>0$ natural and $x$ a fixed real number give at most $log_a {\sqrt{b}}$ residues mod $b$. Prove that $x$ is rational.

Any comment on this tremendously hard problem ?  :wink:

Define $\delta_n = \lfloor a^{n+1}x \rfloor - a \cdot \lfloor a^{n}x \rfloor \in \{0,1,\dots, a-1\}$. It represents the $n+1$-th digit after decimal point of $x$ in base-$a$. To show $x$ is rational, it is equivalent to show it is rational in base-$a$. It then suffices to show the sequence $\{\delta_n\}$ is eventually periodic. Consider a directed graph $G$ on the different residue classes $mod~b$. Make an edge from $r_1~(mod~b)$ to $r_2~(mod~b)$ iff there exists $n > 0$ such that $\lfloor a^nx \rfloor \equiv r_1~(mod~b)$ while $\lfloor a^{n+1}x \rfloor \equiv r_2~(mod~b)$. Notice that $r_2 \equiv ar_1 + \delta_{n+1}~(mod~b)$, so $\delta_{n+1}$ is uniquely determined by the edge $r_1r_2$. As $n$ ranges from $1, 2, \dots$, we can imagine traversing the graph $G$ along these edges. We need to show that our "route" is eventually periodic.

We will show that $G$ does not contain two different cycles (including self-loops and cycles of length two) that share a vertex. Assuming this claim, then $G$ decomposes into vertex-disjoint cycles and other vertices which are connected in a directed tree. It is not difficult to see that for such a graph, any infinite path must be eventually periodic -- once we leave a cycle, we can never come back and so we move farther away from the root of the tree. In fact, this claim about disjoint cycles is equivalent to the original problem. If two cycles $C_1$ and $C_2$ share a vertex $v$, then we can start the path at $v$ and go along the sequence $C_1, C_2, C_1, C_2, C_2, C_1, C_2, C_2, C_2, C_1, \dots$ without ever repeating. We can then "reverse-engineer" an irrational number $x$ that corresponds to such a path.

Now let's prove the claim. Suppose $C_1$ and $C_2$ are simple cycles that share a vertex, corresponding to residue $r$. If $m_1$ and $m_2$ are the length of $C_1$ and $C_2$, then there are integers $q_1, q_2$ such that $0 \leq q_i < a^{m_i}$ and $a^{m_1} r + q_1 \equiv r \equiv a^{m_2} r + q_2 ~(mod~b)$. Write $d = (r, b)$, then $d | q_1, q_2$ and moreover $a^{m_1} r' + q_1' \equiv r' \equiv a^{m_2}r' + q_2' ~(mod~b')$, with $q_i' = \frac{q_i}{d}~$, $r' = \frac{r}{d}~$, $b' = \frac{b}{d}$. Rewriting, we have $(a^{m_1}-1)r' + q_1' \equiv 0 \equiv (a^{m_2}-1)r' + q_2' ~(mod~b')$. Since $(r',b') = 1$, $b'$ must divide the determinant $(a^{m_1}-1)q_2' - (a^{m_2}-1)q_1'$. In other words, $b$ divides $\Delta = (a^{m_1}-1)q_2 - (a^{m_2}-1)q_1$. Here is where we use the condition that at most $\log_{a} \sqrt{b}$ residues are present. We must have $m_1, m_2 \leq \log_{a} \sqrt{b}$, thus $|\Delta| < a^{m_1 + m_2} \leq b$. It follows that $\Delta = 0$. Let $k = (m_1, m_2)$, then there must exist $0 \leq t < a^{k}$ such that $q_i = t \cdot \frac{a^{m_i}-1}{a^k-1}$. This means the function $x \to a^{m_i}x + q_i$ is just the function $x \to a^{k}x + t$ iterated $\frac{m_i}{k}$ times. Thus the cycle of length $m_i$ from $r$ to itself consists of $\frac{m_i}{k}$ repetitions of a smaller cycle, contradicting the assumption that $C_1, C_2$ are simple unless $m_1 = m_2 = k$. The claim is proven.

What is the motivation for your phenomenal solution? Was it hard to think of? Thank you very much and beautiful solution!

[quote = MathPanda1]What is the motivation for your phenomenal solution? Was it hard to think of?Thank you very much and beautiful solution! [/quote]
[quote = FlyingCucumber]My goodness how did you come up with that? [/quote]

Thanks for your compliments :)

Since the condition is given in terms of $a^n x$, it is quite natural to work in base-$a$. But in defining $\delta_n$ I am certainly motivated by some classical problems regarding fractional parts. For example, there is a problem (much discussed on this forum) that asks one to deduce $b$ to be perfect power of $a$ from $a^n-1 | b^n-1$. In the solution one expands the ratio into power sums and apply finite differencing to obtain a linear recurrence. This idea of looking at how consecutive terms of the sequence behave also plays an important role here, leading up to the graph-theory point of view, which really helps conceptualizing the problem. After that it takes a bit of faith to formulate the crucial claim, but once you believe in its truth the rest should be routine. Of course, the condition $\log_{a} \sqrt{b}$ fits nicely at the end, lending credibility to the strategy.

Looking back, I would say this is a great problem, combining multiple small steps from different areas of olympiad math. Plus, the result is simple, beautiful and somewhat intimidating. I would personally look for these attributes in rating a problem.

Here's a possible extension:
Find all real numbers $x$ such that there exist distinct positive integers $a,b$ such that the numbers $[a^n x]$ with $n>0$ natural give at most $log_a {\sqrt{b}}$ residues mod $b$.

If $|x| > 1$ is a rational number, write $x = \frac{p}{q}$ with $|p| > q$. I claim that $a = q + 1$ and $b = (q+1)\cdot |p|$ works. In fact, we have $b \geq a^2$, so it suffices to show $\lfloor a^nx \rfloor \equiv \lfloor ax \rfloor ~(mod~b)$ for every $n$. This is because $(a^2 - a)x = (q+1)q \cdot \frac{p}{q} = \pm b$, and so $(a^n-a)x$ is always an integer multiple of $b$.

On the contrary, suppose $x \in (0,1]$. Let $k \in \mathbb{N}$ denote $\lfloor \log_{a} \sqrt{b} \rfloor$. Moreover, define $n_0$ as the smallest $n$ for which $a^n x \geq 1$. Then we have $a^{n_0} x \leq a$. Consider the numbers $\{\lfloor a^nx \rfloor: n_0 \leq n \leq n_0+2k-1 \}$. They are distinct positive integers bounded above by $b$, so they give at least $2k$ distinct residues $(mod~b)$, contradicting assumption!

Suppose $x \in [-1, 0)$. Define $n_1$ as the smallest $n$ for which $a^n x \leq -1$. Then we have $a^{n_1} x \geq -a$. Similarly consider the numbers $\{\lfloor a^nx \rfloor: n_1 \leq n \leq n_1+2k-1 \}$. They are distinct integers between $-b$ and $-1$, so again we have a contradiction.

Finally $x = 0$ obviously works, so we have obtained a complete characterization.

Wow, congratulations on such a great proof of the extension. How did you come up with this part of the proof? Thank you very much for all your help!

[quote]How did you come up with this part of the proof?[/quote]
Mostly by trial and error. At first I took $a = q, b = p$, a choice that makes $a^nx$ integers. This works if $p > q^2$. Thus all integers $>1$ works, and so does $\frac{5}{2}, \frac{7}{2}, etc.$ It is also easy to see $x = \frac{1}{2}$ wouldn't work. So the tricky case was $x = \frac{3}{2}$. Once I figured out that particular case, the rest was easy to follow.


[quote="The soul of rock"]Let a,b,c,d be the positive numbers. Prove that \[2(a^{4}+b^{4}+c^{4}+d^{4}) \ge \sum_{cyclic}abc(a+b+c-d)\] [/quote]
It is equivalent to $2(a^{4}+b^{4}+c^{4}+d^{4})+8abcd \ge (\sum_{cyclic}abc)(\sum_{cyclic}a)$.

This is weaker than the $n=4$ case of Vasc's "general Schur" http://www.mathlinks.ro/Forum/viewtopic.php?t=14906 which states

$(a^{4}+b^{4}+c^{4}+d^{4})+12abcd \ge (\sum_{cyclic}abc)(\sum_{cyclic}a)$ :)

Yes, I know, I also create "the general Schur for n=4" too. But, there is any solution without using SurranYi?
I have one nice solution for the inequality
$(a^{4}+b^{4}+c^{4}+d^{4})+12abcd \ge (\sum_{cyclic}abc)(\sum_{cyclic}a)$ . Can you find another solution?

let me guess .
your inequality is equivalent with $\sum_{sym}(a^{2}-b^{2})^{2}+\sum_{sym}(a-b)^{2}(c-d)^{2}\geq 0$ :wink:

Opps! I didn't tell about that. My result is like Vacs! Sorry, but as I promisse , I'll not post my solution here. It's my new method.

so you're saying that you have a different proof for Vasc's inequality?

Yes, that was extracly what I said (but only for 4 variables :(  :( )!


floor function
induction
Recursive Sequences

Let $k$ be a given positive integer. The sequence $x_n$ is defined as follows: $x_1 =1$ and $x_{n+1}$ is the least positive integer which is not in $\{x_{1}, x_{2},..., x_{n}, x_{1}+k, x_{2}+2k,..., x_{n}+nk \}$. Show that there exist real number $a$ such that $x_n = \lfloor an\rfloor$ for all positive integer $n$.

Let $ a$ satisfy the equation
\[ \frac {1}{a} \plus{} \frac {1}{a \plus{} k} \equal{} 1,
\]
i.e.,
\[ a \equal{} \frac {\sqrt {k^2 \plus{} 4} \minus{} k \plus{} 2}{2}.
\]
Since $ k$ is a positive integer, it is clear that both $ a$ and $ a \plus{} k$ are irrational and thus produce a pair of complementary [url=http://en.wikipedia.org/wiki/Beatty_sequence]Beatty sequences[/url] $ y_n \equal{} \lfloor an\rfloor$ and $ \lfloor (a \plus{} k)n\rfloor \equal{} y_n \plus{} kn$.
These sequences form a partition of the positive integers, implying that each $ y_n$ is different from any of $ y_m \plus{} km$. But this matches the definition of the sequence $ x_n$, implying that $ x_n \equal{} y_n \equal{} \lfloor an\rfloor$ (a rigorous proof can be done by induction on $ n$).


A point P is marked inside the regular Hexagon ABCDEF such that the angle BAP is equal to angle DCP = 50 degrees. Find angle ABP.

[hide="Hint"] consider the regular 18-agon [/hide]

[hide="Solution"]Angle ABP=80 degrees[/hide]

Let's draw a circle such that $ B$ is center and $ BA\equal{}BC$ are radii.We see that $ P$ lie the circle.Hence, $ BA\equal{}BP\equal{}BC$ and $ m(ABP)\equal{}80^{o}$


Let $ P(x) \equal{} x^3 \plus{} 14x^2 \minus{} 2x \plus{} 2008$
Prove that:
$ \exists c \in Z$ :   $ P(c) \vdots 101$


function
calculus
integration
inequalities
number theory unsolved
number theory

$\pi(n)$ is the number of primes that are not bigger than $n$. For $n=2,3,4,6,8,33,\dots$ we have  $\pi(n)|n$. Does exist infinitely many integers $n$ that $\pi(n)|n$?

Here's an argument which makes me think the answer is "yes":

Assume there is an N s.t. for all n>=N \pi(n) doesn't divide n. We reach such an n and then we start increasing n with step 1. We can see that if \pi(n+\pi(n))=\pi(n) then when we increase n to n+\pi(n) we will eventually reach a number which is 0(mod \pi(n)), so the answer to the initial quastion would be "yes" in this case. We have assumed that the answer is "no", so there must be an N s.t. \pi(n+\pi(n))>\pi(n) for all n>=N. This can be translated to \pi(pn+n)>n or, in other "words", pn+1<=pn+n for all sufficiently large n (n>=N, for example). This doesn't yield an obvious contradiction, but it looks too strong. If it were true, we would have known about it.

This isn't a proof, but it seems a good enough argument to show us what we should prove. I'm not convinced, however. What are your opinions?

The following is not my solution, but it surely is memorable:
   We prove that for any m>=2 we can find n such that mpi(n)=n. Take m>1. Since pi(mk)/mk has limit 0 when k tends to infinity, there is a maximal k such that pi(mk)/mk>=1/m ( k=1 verifies this inequality). If we have equality, we are done. Otherwise, pi(mk)>k. Since k is maximal, we have pi(mk+m)/(mk+m)<1/m and thus k>=pi(km+m)>=pi(km)>k, false.

This is just beautiful! I did struggle on this problem for a day or so, and the proof was 3-lines long! Where did you find this?

I found it in one of my materials and I just love it.

Not solution , just a small remark :

We can replace the function $ \pi\left(n\right)$ by any function $ f: \mathbb{N}\to\mathbb{Z}^{+}$ which is increasing and satisfies $ f\left(n\right)=O\left(n\right)$ when $ n$ tends to infinity.

For example, $ f\left(n\right)=c$ for a constant positive integer $ c$ satisfies the requirement of the problem.

The following is almost the same idea as the above solution (post #3). But I feel it is more revealing (easier to visualize).

[hide]
Lemma: For a non-decreasing function $ f: \mathbb N\to \mathbb N$, if for some positive integers $ a < b$, $ f(a)\ge a$ and $ f(b)\le b$, then there exists at least one integer $ c\in [a,b]$ for which $ f(c) \equal{} c$

To visualize this: consider the graph of $ f(x)$ and $ g(x) \equal{} x$ ($ g: \mathbb N\to \mathbb N$). At some point $ f$ is above $ g$ and after some interval it is below $ g$. together with $ f$ non-decreasing, $ f$ and $ g$ must intersect in that interval.

for some integer $ k > 1$, the function $ f(x) \equal{} k\pi (x)$ satisfies the condition of the above lemma (take $ a \equal{} 2, b\to \infty$). Hence there is some $ c$ for which $ f(c) \equal{} c$ or $ k\pi(c) \equal{} c$ or $ \pi (c)|c$. But there are infinitely many $ k > 1$. So there are infinitely many such $ c$.
Done.[/hide]

[quote="Akashnil"]
To visualize this: consider the graph of $ f(x)$ and $ g(x) \equal{} x$ ($ g: \mathbb N\to \mathbb N$). At some point $ f$ is above $ g$ and after some interval it is below $ g$. together with $ f$ non-decreasing, $ f$ and $ g$ must intersect in that interval.

[/quote]

Excuse me,
 $ f(x)$ and $ g(x) \equal{} x$ ($ g: \mathbb N\to \mathbb N$) are not continuous functions. so they may not intersect. or even if continuous,may not intersect at an integral point.
am i right ? :maybe:  :?:  :maybe:

Notice that if you consider the function $ h(x) \equal{} f(x) \minus{} x$ because $ f$ is non-decreasing we have the inequality
\[ h(x \plus{} 1) \minus{} h(x)\geq \minus{} 1
\]
So when $ g(a) > 0$ , $ g(b < 0)$ then there will be a $ c\in[a,b]$ s.t. $ g(c) \equal{} 0$ which proves the lemma.

[quote="ith_power"]Excuse me,
 $ f(x)$ and $ g(x) \equal{} x$ ($ g: \mathbb N\to \mathbb N$) are not continuous functions. so they may not intersect. or even if continuous,may not intersect at an integral point.
am i right ? :maybe:  :?:  :maybe:[/quote]
Note that what I said was "To visualize this" . It wasn't a rigorous proof. The functions aren't continuous, they aren't even defined on the reals.

If you think of their graphs as dots on integral points their will be a slanted wall of dots of $ g(x)$ and $ f(x)$ begins above it and it can't jump below the wall because it is increasing.

To prove the lemma rigorously see post #10

Today I was trying this problem and observed some stuffs writting a "Haskell programme". Now, let $f(x)=x-k\pi(x)$ where $k>2$ some natural number. So $f(2)<0$. Now $f(x+1)=f(x)+1$ whenever $x+1$ is prime otherwise $f(x+1)=f(x)+1-k$ and so $f(x+1)-f(x)+1$ but also $f(x)\sim x(1-\frac{k}{log x}) >0$ so after some stage we must have that $f(x)>0$ so  if there doesn't exist a $x$ with $f(x)=0$ then exist some $x$ so that $f(x)<0$ and $f(x+1)>0$ and so $f(x+1)-f(x)>2$. So that's the contradiction. So for any $k$ we get some $x$ so that $f(x)=0$ and so there must be infinitely many $x$ so that  $\pi(x)|x$.


ARML AMC AIME USA(J)MO USAMO geometry USAMTS

What do you reccommend are the best books to prepare for ARML, AIME and USAMO? For for particular topics, such as geometry, algebra, etc.?

I already ordered all of AoPS books as well as the solution manuals, but I think that will not be enough. At the next ARML practice session I plan on buying the newer ARML book as well as the older one, but what else do I need for other competitions. What do you guys use?

Well, a good place to start would be the books published by MAA that include the actual problems from the exams.  If you want to practice for USAMO, kalva.demon.co.uk is an invaluable resource.

also, usamtS previous solutions give you an idea of how to write a good proof.

it doesn't make sense to simultaneously prepare for ARML, AIME
and USAMO as these are at totally different levels of difficulty.

my advice:

- don't spend too much time preparing for competitions with
relatively simple problems (ARML), spending time on olympiad-style
(theory, proofs, and written solutions) material will lead to more understanding.

- work your way up from whatever level is currently challenging,
upward (but keeping an eye on more difficult material and trying
your hand at that every once in a while).

- don't be overly masochistic or puritanical about not looking at solutions.
Sometimes you learn more reading other people's solutions
than developing your own (better yet is comparing the two).

- the AIME is hard and unusual in its way -- well worth learning
to solve the problems (and this will train you for all the lower-level
contests at the same time) but it doesn't lead to "transferable
skills" to the extent that solving proof problems (olympiads) does.

- develop a sense of comfort,
competence, confidence and control at each level and move
on (and at each level you're at, keep looking at and trying
some problems from the next level).  Master things but don't
keep repeating things you know other than for any intrinsic
pleasure that may bring (like musical practice for those who
enjoy it).  Be on the lookout for new things and don't be afraid
to devote time to really understanding them.

- aim high.  Keep an eye on the USAMO and IMO if those are
at all plausible for you.

[quote="splintercell990"]what else do I need for other competitions. What do you guys use?[/quote]

After you're through with AoPS 1 and 2, go ahead and work on The Art and Craft of Problem Solving and Problem-Solving Strategies. I haven't had time to work through either of them, but the few problems I got a glimpse of were good. Then there's the USAMO and International Olympiads series - one for every year. Then there's a series called Mathematical Contests that give the questions and solutions for national and international olympiads around the world.

If you'd like something good for geometry, Geometry Revisited is a classic. Work your way through that, take the AoPS Olympiad Geometry course...and you're set on geometry up through the IMO. There's a special series on geometric transformations by the MAA, so you can also get those 3 books if you're interested.

On combinatorics, consider 102 Combinatorics Problems and A Path to Combinatorics for Undergraduates by Andrescu and Feng.

And finally on a more general basis, Mathematical Olympiad Treasures and Mathematical Olympiad Treasures and good.

These are a few titles from my personal library though. Let me know if you find other good ones out there. Although, the ones I just named should keep you busy from now til the end of high school. :-P

Just curious, but what competition would Geometry Revisited be most appropriate for? From what I have heard, the people who use this book are preparing for either USAMO, MOP or the IMO, because the material in the book is too advanced for the AMC 10/12 or the AIME. If anyone has this book, feel free to share your ideas as well...

I have never understood why Geometry Revisited is so useful. The problems don't usually have solutions, and it is so incredibly boring that I can't read more than a page a day. On top of that, I learn absolutely nothing useful for competitions. I think that it would be better as a reference book than as something that you read.

I also didn't like (or purchase or use) Geometry Revisited
all that much in high school.  It's just that there weren't
too many sources in print other than the MAA books.
Today there is about 10-20 times more available
competition literature.

Note, though, that competition problems *are* largely
the traditional type of problems in Euclidean geometry
(given some configuration, prove something about it)
and sources that are not specifically competition-focused
are fine even if you're only interested in contests.
I hope the inner beauty of the subject isn't lost among
the contest-ness of it all!

Actually let me strengthen the last comment.

The non-competition sources for geometry are usually a better place to learn it.  There are various topics the contest materials introduce as ad hoc
problem solving strategies without explaining the inner logic or the motivation behind it. I'm thinking of, among other things, the treatment of inversion (e.g. lack of emphasis on the linear-fractional aspect of things or the complex cross-ratio), the short shrift given to projective geometry considerations, and the lack of systematic treatment of affine geometry (ie. the fact that some problems pertain only to ratios of lengths and areas, and that this is an important thing to look for in finding a solution).

Well, Geometry Revisisted is definitely suited for Olympiad preparation. If you're just looking at AMC/AIME/ARML problems right now, I don't recommend bothering with it yet.

It's pretty good in my opinion *shrugs*, but I haven't looked at it extensively yet. Just skimmed the topics and looked a few select problems.

[quote="themonster"]Probably the first couple chapters of Geometry Revisited are good for AMC or AIME level problems.  Like Ceva's Theorem.  I was never taught this in my Geometry class in school and I suspect that most people weren't either.  It often comes up on AIME and AMC problems, and is given a nice coverage in Revisited.[/quote]

I believe a thorough read of that book (reading every section with understanding and doing most of the exercises before reading solutions) cleared up a weakness I'd long had in olympiad geometry. This MOP, I think I missed just two geometry problems out of all of 11 tests - namely brutal problems like G8. Short answer geometry is a completely different creature in that essentially every solution is highly calculational; i.e. it is not essential to have a wide base of theorems and be able to recognize cyclic quads, concurrence, etc. The mention of Ceva's theorem as often on AIME and AMC, in my opinion, should include the fact that it is one of the most common means of proving concurrence in olympiads. (Some others being trig-Ceva, homothety, radical axis, and construction+calculation.) In fact, for the USAMO, which is usually populated by "low-tech" problems, Ceva + angle chasing + constructions are frequently sufficient for complete solutions.

That said, I agree that some topics don't quite receive the attention they deserve from an olympiad standpoint - projective transformations being one.


geometry
geometric transformation
rotation

could somebody explain rotations to me? ( definition) as well as giving examples. More specifically how could you map a line around a point? thanks alot  :D

if you want to find the loci of points equidistant from a point you would draw a circle witht the point being the centre and would alter the radius depending on haw far from the point the loci were supposed to be.
I am not sure if that is what you meant......(i think mapping a line around a point is probably drawing a circle around it)

Locus (or loci for plural) is a set of points. (e.g. a circle is the locus of points on the plane equidistant from a fixed point called the center.)


modular arithmetic
quadratics
Pythagorean Theorem
geometry

[quote="miyomiyo"]Right triangle $ ABC$ has integer side lengths and hypotenuse $ AB$.  Point $ D$ is drawn on $ AB$ such that $ AD \equal{}\frac{1}{c}$.  A perpendicular to $ AB$ is drawn from $ D$ to a point $ P$ such that $ AP \equal{} ab$.  If $ PB$ is of integer side length prove that $ PB\plus{}1$ and $ PB\minus{}1$ cannot both be prime.[/quote]
[hide] Using the Pythagorean Theorem, we find that $ PD^{2}\equal{}a^{2}b^{2}\minus{}\frac{1}{c^{2}}$ and $ PB^{2}\equal{}a^{2}b^{2}\plus{}c^{2}\minus{}2$.

$ c^{2}\equal{}a^{2}\plus{}b^{2}\Longrightarrow PB^{2}\equal{}a^{2}b^{2}\plus{}a^{2}\plus{}b^{2}\minus{}2$

Case 1: Both $ a$ and $ b$ are even.
$ PB^{2}\equal{}a^{2}b^{2}\plus{}a^{2}\plus{}b^{2}\minus{}2\equiv2\pmod4$. Quadratic residues in $ \mod4$ are $ 0,1$. Thus $ PB$ will not be an integer.

Case 2: At least one of $ a$ and $ b$ is odd.
$ PB^{2}\equal{}a^{2}b^{2}\plus{}a^{2}\plus{}b^{2}\minus{}2\equiv1\pmod2$. If $ PB^{2}$ is odd, then $ PB$ is either odd or irrational. If $ PB$ is odd, then $ PB\minus{}1$ and $ PB\plus{}1$ are both even. There are no 2 consecutive even prime numbers.[/hide]


quadratics
number theory proposed
number theory

Find all integers which are the sum of the squares of their four smallest positive divisors
(P.S. do you know from where this problem was taken?)
good luck!

I think it is from BMO 1987?

Let $ 1 \equal{} d_1 < d_2 < ... < d_k \equal{} n$ be the divisors of $ n$ and $ n \equal{} {d_1}^2 \plus{} {d_2}^2 \plus{} {d_3}^2 \plus{} {d_4}^2$

If $ n$ is odd then $ d_i$ are all odd, hence $ n$ is the sum of 4 odd numbers and hence $ n$ is even.. contradiction!

Hence $ n$ is even and $ d_1 \equal{} 1, d_2 \equal{} 2$.
If $ n \equiv 2 (mod4)$, then $ n$ is not divisible by $ 4$ and hence $ n \equal{} 5 \plus{} {d_3}^2 \plus{} {d_4}^2 \equiv 2 (mod4)$, from that we get: $ \{{d_3}^2, {d_4}^2\} \equiv \{1, 0\} (mod4)$.
If we have: $ d_3 \equal{} 2m$ then $ m$ is a divisor $ < d_3$ and hence $ d_3 \equal{} 4$ or $ d_3 \equal{} 2 \equal{} d_2$, contradiction.
If $ d_4 \equal{} 2m$ then $ d_3 \equal{} m$ and hence $ n \equal{} 5 \plus{} 5m^2$ is divisible by $ m$, hence $ m \equal{} 1$ (which leads to $ d_3 \equal{} d_1$, contradiction) or $ m \equal{} 5$ which leads to $ n \equal{} 130$

If $ n$ is divisible by 4 then $ 4 \in \{d_3, d_4\}$, WLOG assume that $ d_3 \equal{} 4$ and $ d_4 \equal{} d$, then $ n \equal{} 1 \plus{} 4 \plus{} 16 \plus{} d^2 \equiv 0 \equiv d^2 \plus{} 1 (mod4)$ and hence $ d^2 \equiv \minus{} 1 (mod4)$ which is impossible because $ \minus{} 1$ is not a quadratic residue $ (mod4)$.

Hence the only positive integer is 130.

Thanx for the correction, I was too hurry and I lost a case...

um... $ n\equal{}130\Rightarrow 1^2\plus{}2^2\plus{}5^2\plus{}10^2\equal{}130$

This is the only one.

hi !
it`s from Iran 1998 . we have two cases for $ d_3$ :
Case 1 : $ d_3$ = 4 (NickNafplio`s soulotion is true for this case)
Case 2 : $ d_3$ = p (prime number) ==> $ n \equal{} 5 \plus{} p^2 \plus{} d_4 ^2$ . so $ d_4$ is even ==> $ d_4 \equal{} 4$ or $ d_4 \equal{} 2p$.
if $ d_4 \equal{} 4$ ==> $ d_3 \equal{} 3$ ==> n=30 Con.
if $ d_4 \equal{} 2p$ ==> $ n\equal{}5\plus{}5p^2$ ==> p=5 ==> n=130


limit
real analysis
real analysis unsolved

Prove that
\[\lim_{n\to\infty}e^{-n}\sum^{n}_{j=0}\frac{n^{j}}{j!}= \frac{1}{2}\]

(of course, there are proofs not involving the Central Limit Theorem)

Let $X_{n}$ be a random variable with a Poisson distribution and mean $n.$ The sum in question is $P(X_{n}\le n)=P(X_{n}<n+1).$ (The second equality follows since $X_{n}$ is integer valued.)

What to we know about the Poisson distribution? We know that $X_{n}=\sum_{k=0}^{n}Y_{k}$ where the $Y_{k}$ are independent identicially distributed random variables with mean $1$ and variance $1.$ From here on out, that's all we'll use - that the $Y_{k}$ have mean $1$ and variance $1$ and are integer valued. Dump the rest of the information.

$X_{n}$ has mean $n$ and variance $n.$ Define $Z_{n}=\frac{X_{n}-n}{\sqrt{n}}.$ The quantity we are looking for can be written either as ${P(X_{n}\le n})=P(Z_{n}\le 0)$ or $P(X_{n}\le n+1)=P\left(Z_{n}<\frac1{\sqrt{n}}\right).$

At this point, we know by the central limit theorem that $Z_{n}$ converges in distribution to a standard normal, which means that $P(Z_{n}\le 0)$ tends to $\frac12.$

Followup question: the central limit theorem seems like overkill here. Would the weak law of large numbers suffice?


I have recently STOLEN some old Gazeta Matematica issues from my high school's attic (don't worry; nobody was using them anyway  :D  :D). Here's a problem I found there (not that tough, but interesting):

Solve in integers xy(x^2+y^2)=2z^4

It's proposed by Titu Andreescu.

[quote="grobber"]I have recently STOLEN.[/quote]

Does your Mum know that  :D

Back to this cute problem.

A try : only (n,n,n) is solution with n  Z

Obviously it is enough to consider x,y > 0.

If (p*x,p*y,z) is a solution then p|z and (x,y,z/p) is also a solution.

So it's enough to consider x and y such as (x,y)=1.

Let p > 2 prime, s.a. p^a | x. Then p | z but since (p,y)=(p,x+y)=1, p^(4a) | x so a=0 and contradiction.

The case p=2 is similar.

The only solution is x=y=1 in this case. Hence the conclusion.

I posted as guest (don't know how this could happen ??) so I can't edit my answer  :?

The solutions are (n,n, \pm n) with n  Z.

Yup, but you forgot the trivial ones: (0,n,0) and (n,0,0). It's not important anyway.

Sure my mom knows about the stealing. She's the one who taught me how to steal in the first place! :D  :D  :D Just kidding. I learned how to steal all by myself. :D


The positive reals $x_1, x_2, ... , x_n$ have harmonic mean $1$. Find the smallest possible value of $x_1 + \frac{x_2 ^2}{2} + \frac{x_3 ^3}{3} + ... + \frac{x_n ^n}{n}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=16636
http://www.mathlinks.ro/Forum/viewtopic.php?t=51416

  darij

First, taking $x_1=\cdots=x_n=1$, we obtain the value $S\triangleq \textstyle \sum_{1\le k\le n}\frac1k$. We now show this is indeed the lower bound. Note that,
\[
\frac{x_i^i}{i} +\underbrace{ \frac{1}{i}+\cdots+\frac{1}{i} }_{i-1\text{ times}} \ge x_i \implies \frac{x_i^i}{i} \ge x_i - \frac{i-1}{i}.
\]
So,
\[
\sum_{1\le i\le n}\frac{x_i^i}{i}\ge \sum_{1\le i\le n}x_i - \sum_{1\le i\le n}\frac{i-1}{i} \ge \frac{n^2}{\sum_i x_i^{-1}} - \sum_{1\le i\le n}\frac{i-1}{i}  =S,
\]
using the fact the HM of $x_i$ is 1.


AMC
AIME
USA(J)MO
USAMO
USAMTS
Putnam
AoPS Books

I like math, but I don't have any experience in it. Except for what has been taught in school, my education is elementary. I'm in my first year of Pre-Calc on the accelerated track. I wish I could do the problems on the beginner's section of the high school board, but I can't. So, for those experienced math students, what steps did you take to get to the level you're at now and what is your background? What advice about preperation (etc.) would you give a newbie like me? How would I be able to improve from the little-skill math student I am to one who can do several of the problems on these forums?

I suggest getting the Art of Problem Solving books (vol. 1 & 2) from this site. These will get you to the AIME level and USAMTS. Then order the Art and Craft of Problem Solving from Amazon or something. That covers the AIME, USAMO, IMO, USAMTS, and Putnam levels, which is about as hard as you can get. ;)

I second your opinion, Shobber. You should have complete control over your school maths first: school maths should really be a piece of cake, and afterwards you can start working on harder things.

a lot of the competition problems, even in getting starting, require tricks or clever manipulations to solve them and would be more advanced than what you would cover in a typical class. if you are serious into problem solving, get AoPS vol. 1 and go from there.

[quote="shobber"]Do your school math first. After you master it, you can surely master the beginning forums.[/quote]
Shobber, that's not very true.
Or more precisely, it is true in China. Try to fetch a math textbook from other countries and compare with yours...you will see then.

I agree with the others. Master your school math first. Then, study the Art of Problem Solving Books.

Thanks everyone for the replies.

I have to say that I already perform pretty well in my school's math classes. As I have said, I'm on the most accelerated track in my school's math department. My problem is not from having poor math abilities but stems from lack of experience with certain types of math-ones I don't see in school.

I'll definitely look into those book series. They're expensive, but if they are worth as much as you guys have said they are, then maybe I'll save up to get them.

I realize I'm a little late in the game. But, hopefully, I'll get at a level in my math problem-solving abilities that I'm happy with. For some re-assurance, can anyone tell me about their backrounds in math and how ya'll got to the level you're at now?

Joker, you beat me to making this thread.  :roll:

I find myself in a similar situation, except I'm further along in my high-school carreer.

Last year ( grade 11 ), I found myself doing mediocre in school, but finally got a real outside interest: comp-sci. I spent a little over 9 months learning everything and anything I could on the topic. I also wrote any contests I could on the matter.  I wrote the very fun CCC, but it was my first contest of any kind, and did poorly because of first-contest kind of errors.  I practiced more, and ( along with singular ( Alex Wice ) ) we got to final round ECOO ( an Ontario team based computer competition ).

Near the end of the year, Alex mentioned that after seeing me master C and a few other programing languages, that I should be able to pick up math if I tried.
So at the begining of the summer, I started messing around with math, and found that comp-sci feels like child's play, and that THIS is where the fun is.

The problem is, I am now entering grade 12. While university acceptance and such WILL NOT be an issue, I feel disappointed that even if I finish off the AoPS books that I now own , that I will likely miss a chance to actually write the Canadian Invitational, just because this is my first and only time through a year of contests, and the Canadian open is less than two months away.
( for reference to my GENERAL problem solving ability, I got a 97 of the Fermat last year, the only contest I wrote. At the time I was failing grade 11 math, but since then I have caught up, no worries my school performance is now nothing to worry about =D )

I am quite happy with exploring math for my own  sake, I wouldn't have gotten into it if I didn't enjoy it. I just fear that I will have a hard time EVER being able to solve the beautiful IMO problems that I get to read on this site.

I would also like any advice for a late-starter who is very enthusiastic.
 (Actually just looking over last year's Fermat, I think I would do a bit better, but I don't know how the Fermat compares to many other contests, it remineds me of the Canadian open and the Euclid)

[quote="rzsolt"]
I would also like any advice for a late-starter who is very enthusiastic.
[/quote]

Try contests, then study solutions.  Start with AMC12/Fermat, then progress to COMC/Euclid, then AIME/CMO, (and if you are there, USAMO/APMO/IMO)

I started being interest in math last year exactly at this time. I was in the EXACT situation as Joker.

I began studying the AOPS books and let me tell you where I am now. I am a sophmore and I will probably have no trouble qualifying for AIME. I have been inconsistent in my studies and have gotten this far. If you are consistent in your studying, I assure you, you will be USAMO worthy in a year or two.

It is still hard for me now but no pain no gain right?

To be honest, I've lost hope. Math isn't gonna help you get into a better college and you do extremely well. So I have decided to kinda let it drop for a while. It's your decision though.


superior algebra
superior algebra unsolved

Prove that for each $n>2$, the number of the elements of order $n$ in a finite group is even.
PS. I like the idea used in this problem and I've made a more advanced problem out of it, which will appear on the forum later. :D

There is an involution $x \to x^{-1}$ on the set of elements of order $n$, which has no fixed points because of $n > 2$.


Given the true statements: (1) If $ a$ is greater than $ b$, then $ c$ is greater than $ d$ (2) If $ c$ is less than $ d$, then $ e$ is greater than $ f$. A valid conclusion is:

$ \textbf{(A)}\ \text{If }{a}\text{ is less than }{b}\text{, then }{e}\text{ is greater than }{f}\qquad \\
\textbf{(B)}\ \text{If }{e}\text{ is greater than }{f}\text{, then }{a}\text{ is less than }{b}\qquad \\
\textbf{(C)}\ \text{If }{e}\text{ is less than }{f}\text{, then }{a}\text{ is greater than }{b}\qquad \\
\textbf{(D)}\ \text{If }{a}\text{ is greater than }{b}\text{, then }{e}\text{ is less than }{f}\qquad \\
\textbf{(E)}\ \text{none of these}$

[hide="My Solution"]
We know for the two statements that their contrapositives are also true. I will denote the contrapositives with '.

(1)' If $ c \leq d$, then $ a \leq b$.
(2)' If $ e \leq f$, then $ c \geq d$.

Clearly, none of the options work, so the answer is $ \fbox{\textbf{(E)}}$.
[/hide]


Just curious.

ANyone wanna share comments about their fav ladies on AoPS?

haha this might interest u 7h3.d3m0n.117

This thread is mad sketchy.

gauss1181 = Creepo.

[quote="gauss1181"]haha this might interest u 7h3.d3m0n.117[/quote]

How does [i]this[/i] work?

*raises one eyebrow quizzically*

....

[i]What[/i] and [i]Why[/i] is this even posted, or, even thought of?

Just [i]HOW[/i] is Henry a girl name?

My favorite girl is gauss1181. :)

I don't think he was aiming at me for being a girl.

Anyway, I'm going to lock this for being too..err. I think Ignite said it best.

[quote="Ignite168"]
This thread is mad sketchy.
[/quote]


Give $a,b,c \in R^{+}$. Prove that:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\geq 2$  :)

this one is quite easy!
note that :$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}+2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\geq 2$
$\longleftrightarrow \sum_{cyc}a(a-b)(a-c)+2(\sqrt{abc(a+b)(b+c)(c+a)}-2abc)\geq 0$ which is obvious. equality occurs at $(a,b,c)\sim (0,1,1)$ and permutations

Woa, well done, super fast, i just posted just hour ago. Oke, so maybe we will come to a more complex one:
Find the maximum of k such that the inequality below is true for $\forall a,b,c \in R^{+}$
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\geq 2$  :)

I guess you mean minimum $k$ right?
it's $\sqrt{2}$ with equality at $(a,b,c)\sim (1,1,1),(0,1,1),(1,0,1),(1,1,0)$ :wink:

Yes, indeed, use the same technique. A good example for Schur inequality. However, that's still not the point i want to reach. Maybe you want too try the another one. For every $n \in R^{+}$, find the minimum of $k$ such that this inequality holds for $a,b,c \in R^{+}$:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{ka^{n}b^{n}c^{n}}{(a+b)^{n}(b+c)^{n}(c+a)^{n}}\geq 2$  :)

directly we see that for $n\in [0,1],k_{min}=2^{3n-1}$
but for n>1 I haven't thought yet.

for n>1 the inequality can not hold $\forall a,b,c \in \mathbb{R}^{+}$ no matter what k we choose :)
just put $a=b$ then multiply both sides with $2a(a+c)^{2}$ then divide by $c$ then you will be left with $(c-a)^{2}+2^{2n}\frac{a^{n+1}c^{n-1}}{(a+c)^{2n-2}}-4a^{2}\geq 0$ clearly false for $c\rightarrow 0$

Oke, well done  :D . I won't go more, maybe keep this secret for another problem. :)

ok another thing :
I've noticed that the (n-1)EV theorem gives that the expression achieves minimum when two variables are equal. just keep $a+b+c,a^{3}+b^{3}+c^{3}$ constant :wink:

Oke, i deduced that too with ABC , so the power here is not so important. :)

[quote="Mimoide"]Could you tell me please if there is any topic where I could find about the last you mentioned, either here or in the "theorems" section?  :blush:[/quote]
I found only [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=440463#440463[/url].


For $ a\in\mathbb R$ and $ n\in\mathbb N$ ,$ n\geq 3$ compute:
\[ \lim_{x\to\infty}x\left(2x\minus{}\sqrt[n]{x^{n}\plus{}ax^{n\minus{}1}\plus{}1}\minus{}\sqrt[n]{x^{n}\minus{}ax^{n\minus{}1}\plus{}1}\right)\]

the answer is $ \minus{}\frac{1}{n}(\frac{1}{n}\minus{}1)a^{2}$
you only must use the next fact
$ (1\plus{}x)^{a}$when $ x\to 0$ is
equal to $ 1\plus{}ax\plus{}\frac{a(a\minus{}1)}{2}x^{2}\plus{}o(x^{2})$

[color=darkblue]Using the substitution $ x: \equal{}\frac{1}{x}$ the proposed exercise becomes a particular case of the [b]limit $ 11$[/b] from[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=166710 :

$ 11\blacktriangleright\sum_{k \equal{} 1}^{n}a_{k}\equal{} 0$ and $ p\in Z$, $ p\ge 2$ $ \implies$ $ \lim_{x\to 0}\frac{1}{x^{2}}\cdot\left(\sum_{k \equal{} 1}^{n}\sqrt [p]{1\plus{}a_{k}x\plus{}b_{k}x^{2}\plus{}\ldots}\minus{}n\right) \equal{}\frac{1}{p}\cdot\sum_{k \equal{} 1}^{n}b_{k}\minus{}\frac{p\minus{}1}{2p^{2}}\cdot\sum_{k \equal{} 1}^{n}a^{2}_{k}$.


[quote="ragnarok23"]Assume a regular hexagon $ABCDEF$ with side length $8$. The midpoints of sides $AB$, $CD$, and $EF$ are $X$, $Y$, and $Z$, respectively. Find the area of $\triangle XYZ$. Express your answer in simplest radical form.[/quote]
[hide]Draw this and you see that it is an equilateral triangle and 4 trapezoids. Draw the rightmost and leftmost altitudes of any trapezoid and you have a rectangle and 2 30-60-90s. From here you can get that the length of a side of the triangle is 12. So the area is $36\sqrt{3}$ with the formula $\frac{s^{2}\sqrt{3}}{4}$.[/hide]

[quote="ragnarok23"]Assume a regular hexagon $ABCDEF$ with side length $8$. The midpoints of sides $AB$, $CD$, and $EF$ are $X$, $Y$, and $Z$, respectively. Find the area of $\triangle XYZ$. Express your answer in simplest radical form.[/quote]
[hide][img]http://img291.imageshack.us/img291/1364/hexagonsub3.png[/img]

If we make a diagram of the problem, and add lines everywhere to make $24$ equilateral triangles, we see that the triangle is $\frac{3}{8}$ of the hexagon.

The area of a regular hexagon is $\frac{3s^{2}\sqrt{3}}{2}.$ The area of this hexagon is $96\sqrt{3},$ so the area of $\triangle XYZ$ is $36\sqrt{3}.$[/hide]

[hide] you can see if you draw the midpoints and the connections in the problem that there are 3 trapezoids in the hexagon. you can find, that the midpoint divides the hexagon's side in half, becoming 4. You also can notice that the altitude from a vertice of the trapezoid forms a 30-60-90 triangle. you know this because the altitude forms 2 90-degree angles on both sides. since the hexagon has an interior angle of 120 degrees, you only have to worry about the side that actually has a straight line (not the cut off side). subtract 90 from 120, and get 30. you know that the OTHER side has TWO 90 degree angles. so you have a 30-60 triangle. 180-90=90, and then you have a 30-60-90 triangle. From here, you can figure out that one side of the equilateral triangle is 12, since 4*2+2*2=4*3=12. then, you use the formula for equilateral area, s^2rt.3 /4, and thus you get $144\sqrt3/4$, and then it ends up becoming 36root3. :D[/hide]

[hide]$36\sqrt3$

divide the hexagon into little equilateral triangles, and the large triangle is 3/8 of the hexagon, so

$\frac38 * \frac{3*8^{2}*\sqrt3}{2}= 36\sqrt3$[/hide]


number theory
prime factorization
number theory proposed

Let $ a$ be a positive integer. A permutation $ \pi$ of the set $ \left\{1,2,...,n\right\}$ will be called a [i]v-permutation[/i] if for every divisor $ d$ of $ n$ and for arbitrary $ i$ and $ j$ from the set $ \left\{1,2,...,n\right\}$ satisfying $ d\mid i \minus{} j$, we have $ d\mid\pi\left(i\right) \minus{} \pi\left(j\right)$.

For every $ n$, we denote by $ v\left(n\right)$ the number of all v-permutations of the set $ \left\{1,2,...,n\right\}$.

Find the prime factorization of $ v\left(n\right)$, given that of $ n$.

A nice problem! Do you have a complete solution?

Here's a partial solution in case only one prime $ p$ divides $ n$.

Let $ S\equal{}\{1,...,n\}$.

In case $ n\equal{}p$, we have obviously $ p!$ v-permutations, for no two numbers in $ S$ are congruent $ \mod p$, and all are congruent $ \mod 1$.

In case $ n\equal{}p^\alpha$, $ \alpha>1$, we use the following reasoning : it is obviously necessary that $ \pi$ take equivalence classes $ \mod p^{\alpha\minus{}1}$ to equivalence classes $ \mod p^{\alpha\minus{}1}$. But it is also sufficient, because if two numbers are congruent $ \mod p^{\alpha\minus{}1}$, they will also be congruent $ \mod p^k$ where $ 1 \leq k < p^{\alpha\minus{}1}$. There are $ p^{\alpha\minus{}1}$ such equivalence classes in $ S$, each containing $ p$ elements. Any permutation of the equivalence classes is possible : there are $ p^{\alpha\minus{}1}!$ such permutations. Then any permutation of the numbers within those classes is possible, and for each of the classes there are $ p!$ such permutations.

Hence the total number of v-permutations is then $ p^{\alpha\minus{}1}!({p^{\alpha\minus{}1}})(p!)$.

The more general case, where more primes divide $ n$, seems very hard. Maybe I just haven't thought about it enough.
Please let me know if you find a mistake in my reasoning.

No, your solution for the $ \alpha > 1$ case is not right. You also need to consider the case of $ d \equal{} p^k$ with $ k < \alpha \minus{} 1$ (since in this case, we have more pairs $ \left(i,j\right)$ satisfying $ d\mid i \minus{} j$). Hence, no, not every permutation of the $ p^{\alpha\minus{}1}$ equivalence classes modulo $ p^{\alpha\minus{}1}$ is possible.

The case of $ n$ being not a prime power is not the hardest part of the problem.
[hide="Hint"]Prove that $ v\left(ab\right)\equal{}v\left(a\right)v\left(b\right)$ for any coprime $ a$ and $ b$.[/hide]

  darij

You are right.
Is $ (p!)*(p!)^2*...*(p!)^{\alpha\minus{}1} \equal{} (p!)^{\alpha(\alpha\minus{}1)/2}$ correct?

Uhm, when doing this problem I could convince myself that $ v(p^k) \equal{} (p!)^{\frac {p^k \minus{} 1}{p \minus{} 1}}$

Are you sure of your result?

Anyway,  [hide="here is my proof"]
You divide your $ p^k$ numbers into $ p$ boxes: the first box goes from $ 1$ to $ p^{k \minus{} 1}$, the second box from $ p^{k \minus{} 1} \plus{} 1$ to $ 2\cdot p^{k \minus{} 1}$ and so on.

Now every part must contain a complete residue system modulo $ p^{k \minus{} 1}$. So you can put your numbers into your boxes in $ (p!)^{p^{k \minus{} 1}}$ ways. Then you order the number in the first box. Obviously you have $ v(p^{k \minus{} 1})$ ways of doing so. Then the order in the other boxes must necessarily be the same because if $ p^{k \minus{} 1}|i \minus{} j$ then $ p^{k \minus{} 1}|\pi (i) \minus{} \pi (j)$

So $ v(p^k) \equal{} v(p^{k \minus{} 1})\cdot (p!)^{p^{k \minus{} 1}}$[/hide]


Prove:
$(p-1)! \equiv-1 (mod\ \ p)$ Where p is prime
When I started working this one out I used mod 5 as my example
so
$(5-1)! \equiv 4*3*2*1 \equiv 24 \equiv-1 (mod\ \5)$
So it works in mod 5, now suppose we use a generator: 2, and raise it to powers and multiply these powers of two together:
$2^{1}*2^{2}*2^{3}*2^{4}\equiv 2*4*3*1$
Which is the same as the original $(5-1!)$ so:
this should also be  $\equiv$ to $-1$
But $2^{1}*2^{2}*2^{3}*2^{4}\equiv 2^{10}$
Since $10 \equiv 0\ \ (mod\ \ 5)$
$2^{1}*2^{2}*2^{3}*2^{4}\equiv 2^{10}\equiv 2^{0}\equiv 1$

Why can't you substitue congruences into the exponents?
for example: $2^{0}\not\equiv 2^{5}\ \ (mod\ \ 5)$ Even though $0 \equiv 5$
In fact, maybe $n^{a}\equiv n^{b}\ \ (mod\ \ p)$ as long as $a \equiv b\ \ (mod\ \ (p-1))$ I can see why this conjecture doesn't work when $a=b$ but would this work if $a \neq b$??
help on either problem is much appreciated  :lol:

I don't get what you're asking.  Why should you be able to 'substitute congruences into the exponents'?  Why isn't subtraction commutative?  Why isn't the derivative of the product equal to the product of the derivatives?

The simple answer is that for mathematics to be consistent, some things have to work out in a way that you don't necessarily want them to.

[quote="wigianto"]In fact, maybe $n^{a}\equiv n^{b}\ \ (mod\ \ p)$ as long as $a \equiv b\ \ (mod\ \ (p-1))$ I can see why this conjecture doesn't work when $a=b$ but would this work if $a \neq b$??
help on either problem is much appreciated  :lol:[/quote]

There is actually a statement like this.

[b]Fermat's Little Theorem:[/b]  Let $p$ be a prime and $a$ an integer such that $p \not | a$.  Then

$a^{p-1}\equiv 1 \bmod p$

In any case, the original theorem can be proven without recourse to generators.

[hide="Big hint"] Consider the product $(a \cdot 1) \cdot (a \cdot 2) \cdot ... \cdot (a \cdot (p-1))$.  ;) [/hide]

[quote="Karth"]First problem:
[hide]True by Wilson's Theorem  :D [/hide][/quote]

A silly solution.  Obviously the problem asks to prove Wilson's Theorem, which can be done the way I hinted to, or by considering multiplicative inverses $\bmod p$.

Just to hint to the multiplicative inverses method Qiaochu mentioned...

[hide]
Can you show $[p-1]!^{2}\equiv 1$?[/hide]


if $p$ is a prime. prove
a)if $p\equiv 1 (mod 4)$ then $p=a^2+b^2 (a,b\in\mathbb{Z})$.

b)if $p\equiv 3 (mod 4)$ then $p\not =a^2+b^2 (a,b\in\mathbb{Z})$.

a is a non-trivial result.  One way to do it is [hide]to show that if p=4k+1, then there is an n for which p divides n^2+1=(n+i)(n-i).  If p did not factor in Z(i), it would have to be a complex number.[/hide]

b is a near-trivial result.  It follows from [hide]any square is congruent to 1 or 0 mod 4[/hide].[/hide]


linear algebra
matrix
algebra
polynomial
induction
invariant
vector

Let A, B be positive definite matrix over F  (=R, C)
such that A^2=B^2.
Show that A=B.

Thanks a lot :)

We may assume $ A\equal{}diag(\lambda_1,\cdots,\lambda_1,\cdots,\lambda_k,\cdots,\lambda_k)$ with $ 0<\lambda_1<\cdots<\lambda_k$. $ \lambda_i^2\equal{}\lambda_j^2$ implies $ \lambda_i\equal{}\lambda_j$ then $ A$ is a polynomial in $ A^2$. $ B,A^2$ commute then $ B,A$ commute and $ B\equal{}diag(B_1,\cdots,B_k)$ where $ B_i$ is a PDS matrix. We can diagonalize the $ B_i$ leaving $ A$ invariant.Thus $ B_i\equal{}diag(\mu_1,\cdots,\mu_p)$ with $ \mu_j>0,\mu_j^2\equal{}\lambda_i^2$ ; then $ \mu_j\equal{}\lambda_i$.

Excuse me, does this work ?

For V is n-dim inner product space.
We may consider self-adjoint,
positive semidefinite operators T, U on V such that TT=UU

Claim T=U

By induction on n
n=1, V=<x>, if T(x)=ax, U(x)=bx
TT=UU implies that a^2=b^2 and then a=b since T, U are positive semidefinite.
Hence T=U

Suppose that the assertion holds for n>1

Since all the eigenvalues of T and U are non-negative (precisely, the same).
If all eigenvalues is 0, T=U trivially.
Wlog: W=N(TT)=N(UU) is a proper subspace of V

Say V=W⊕Z where Z is the orthonormal complement of W

Note that W is T, U-invariant, so is Z
Let T'=the restriction of T on W
    T~=the restriction of T on Z

It is clear that T', U' are self-adjoint, positive semidefinite operators on W,
and T'T'=U'U', so are T~, U~.

By ind. hypothesis. T'=U', T~= U~

This yields that T=U on V
                          #

Hi wuxr,
1) My proof is still valid  when $ A,B$ are  only positive semidefinite.
Moreover we have the same result if we replace $ A^2\equal{}B^2$ with $ A^k\equal{}B^k$ where $ k$ is a positive integer.
2) In your proof if $ W\equal{}\{0\}$ then what do you do ?
3) A  result valid for any matrices $ A,B$:
If $ AA^*\equal{}BB^*$ then there exists a (non unique) unitary matrix $ U$ s.t. $ A\equal{}BU$.

[quote="loup blanc"]We may assume $ A \equal{} diag(\lambda_1,\cdots,\lambda_1,\cdots,\lambda_k,\cdots,\lambda_k)$ with $ 0 < \lambda_1 < \cdots < \lambda_k$. $ \lambda_i^2 \equal{} \lambda_j^2$ implies $ \lambda_i \equal{} \lambda_j$ then $ A$ is a polynomial in $ A^2$. $ B,A^2$ commute then $ B,A$ commute and $ B \equal{} diag(B_1,\cdots,B_k)$ where $ B_i$ is a PDS matrix. We can diagonalize the $ B_i$ leaving $ A$ invariant.Thus $ B_i \equal{} diag(\mu_1,\cdots,\mu_p)$ with $ \mu_j > 0,\mu_j^2 \equal{} \lambda_i^2$ ; then $ \mu_j \equal{} \lambda_i$.[/quote]

Hi, loup blanc
Thanks for your help.
After your warning, I understand the problem of 2) you point.
Would you mind explain in detail:

then $ A$ is a polynomial in $ A^2$. $ B,A^2$ commute then $ B,A$ commute and $ B \equal{} diag(B_1,\cdots,B_k)$ where $ B_i$ is a PDS matrix. We can diagonalize the $ B_i$ leaving $ A$ invariant.

Thanks a lot!  :)

1) $ A\equal{}P(A^2)$ where $ P$ is a polynomial s.t. $ P(\lambda_i^{2})\equal{}\lambda_i$.
2) $ BA^2\equal{}B^3\equal{}A^2B$ ; $ AB\equal{}P(A^2)B\equal{}BP(A^2)\equal{}BA$.
3) Here we have a decomposition of the vector space  $ E\equal{}E_1\oplus\cdots\oplus{E_k}$ s.t. $ A_i\equal{}A_{|{E_i}}\equal{}\lambda_i{I}$  ; then $ BA\equal{}AB$ implies $ B\equal{}diag(B_1,\cdots,B_k)$ is a general result. $ B_i$ symmetric is obvious.
$ B_i\geq{0}$ because if $ X\equal{}[0,\cdots,X_i,0,\cdots,0]^T$ then $ X^TBX\equal{}X_i^TB_iX_i\geq{0}$.
4) With a new orthonormal basis of $ E_i$ we diagonalize $ B_i$ and $ A_i$ is invariant.

[quote="loup blanc"]1) $ A \equal{} P(A^2)$ where $ P$ is a polynomial s.t. $ P(\lambda_i^{2}) \equal{} \lambda_i$.
2) $ BA^2 \equal{} B^3 \equal{} A^2B$ ; $ AB \equal{} P(A^2)B \equal{} BP(A^2) \equal{} BA$.
3) Here we have a decomposition of the vector space  $ E \equal{} E_1\oplus\cdots\oplus{E_k}$ s.t. $ A_i \equal{} A_{|{E_i}} \equal{} \lambda_i{I}$  ; then $ BA \equal{} AB$ implies $ B \equal{} diag(B_1,\cdots,B_k)$ is a general result. $ B_i$ symmetric is obvious.
$ B_i\geq{0}$ because if $ X \equal{} [0,\cdots,X_i,0,\cdots,0]^T$ then $ X^TBX \equal{} X_i^TB_iX_i\geq{0}$.
4) With a new orthonormal basis of $ E_i$ we diagonalize $ B_i$ and $ A_i$ is invariant.[/quote]

Hi, loup blanc
I still have a doubt.
If A=
[1 0 0]
[0 2 0]
[0 0 2]

then, what is the polynomial $ P$ such that $ P(A^2)\equal{}A$

Thanks a lot.

$ P(1)\equal{}1,P(4)\equal{}2$. Using Lagrange interpolation you obtain $ P(x)\equal{}1/3x\plus{}2/3$. This method works because $ A$ is diagonalizable. Else it is more complicated ; use Hermite interpolation.

[quote="loup blanc"]$ P(1) \equal{} 1,P(4) \equal{} 2$. Using Lagrange interpolation you obtain $ P(x) \equal{} 1/3x \plus{} 2/3$. This method works because $ A$ is diagonalizable. Else it is more complicated ; use Hermite interpolation.[/quote]

Thanks, I completely understand.
The way showing the commutation of A, B is so nice.
Is this skill usual?

[quote="loup blanc"]$ P(1) \equal{} 1,P(4) \equal{} 2$. Using Lagrange interpolation you obtain $ P(x) \equal{} 1/3x \plus{} 2/3$. This method works because $ A$ is diagonalizable. Else it is more complicated ; use Hermite interpolation.[/quote]

Sorry, one more thing

what is Hermite interpolation?

[quote="loup blanc"]Hi wuxr,
 My proof is still valid  when $ A,B$ are  only positive semidefinite.
Moreover we have the same result if we replace $ A^2 \equal{} B^2$ with $ A^k \equal{} B^k$ where $ k$ is a positive integer.
[/quote]
 You can see that If $ A,B$ are Hermitian matrices such that $ A^{2k \plus{} 1} \equal{} B^{2k \plus{} 1}$ where $ k$ is a positive integer then $ A \equal{} B$ Are you agree ?

Hi wuxr,
"what is Hermite interpolation?"
See Wikipedia !
HiQuybac,
"You can see that If A,B are Hermitian matrices such that $ A^{2k \plus{} 1} \equal{} B^{2k \plus{} 1}$ where $ k$ is a positive integer then $ A \equal{} B$. Are you agree ?"
Yes I do


analytic geometry
vector
linear algebra
matrix

I have 2 questions:

1. By giving the basis, describe a 2-dimensional subspace of $ R^3$ that contains none of the coordinate vectors $ (1,0,0), (0,1,0), (0,0,1)$.

How about the plane $ x \equal{} 5$, over all $ z$ and $ y$? That's a 2D plane, it's a subspace, and it doesn't contain the coordinate vectors. But how can I describe this using a basis?

2. How can I find a basis for the subspace of $ R^4$ described as the set of vectors for which $ x_1 \plus{} x_2 \plus{} x_3 \equal{} 0$ and $ x_3 \plus{} x_4 \equal{} 0$?

3. How can I find a basis for the subspace of $ R^4$ described as the subspace spanned by $ (1,1,1,1), (1,2,3,4)$, and $ (2,3,4,5)$? How many points are required to determine a line in 4D space? In $ n$-dimensional space?

Thanks

$ x\equal{}5$ is not a subspace. A subspace must be closed under addition and matrix multiplication and this set satisfies neither.

Geometrically, a two-dimensional subspace of $ \mathbb{R}^3$ is a plane [b]that passes through the origin.[/b]

It's not that hard to avoid the standard basis elements. Here's one example: $ \{(x,y,z): x\plus{}y\plus{}z\equal{}0\}.$ But I didn't give a basis for that - I'll let you do that. (Hint: think the way you do in solving systems of linear equations. This is a homogeneous system - it's just that there's only one equation, already in row echelon form. There are two free variables.)

For 2: again, think row echelon form and free variables.

For 3: If those three vectors were independent, they would be a basis of their span. But are they independent?

And two points suffice to determine a line in any dimension $ \ge 2.$

Kent, can you elaborate on what you said about using row echleon form and free variables? Can I write the 2 constraints in question 2 as $ x_1 \plus{} x_2 \plus{} x_3 \equal{} 0$ and $ x_3 \plus{} x_4 \equal{} 0$ as a matrix with first row $ [1 1 1 0]$ and second row $ [0 0 1 1]$?

Yes you can. And $ \begin{bmatrix}1&1&1&0\\0&0&1&1\end{bmatrix}$ is already in row echelon form (not quite reduced), so almost no computations are needed. There are pivots in the first and third columns, hence the second and fourth variables are free.

and no, the three vectors in question 3 are not linearly independent. In fact, the first plus the second gives the third.

I guess my confusion here is what is a systematic method of computing a basis?

So what I have for question 2,

is that $ x_2, x_4$ are the free variables. Let $ x_2 \equal{} 1$, and $ x_4 \equal{} 0$. We get $ (\minus{}1, 1, 0, 0)$ for $ (x_1, x_2, x_3, x_4)$. If we let $ x_2 \equal{} 0$ and $ x_4 \equal{} 1$, we get $ (1, 0 , \minus{}1, 1)$ for the vector x. Now b/c we have two equations, each reduces the dimension of $ R^4$ by one, so we have a dimension of 2. So these are the two vectors for the basis?

Also for the first question,

I wrote the matrix $ [111]$, and let $ x \equal{} 1, y \equal{} 1$. So I got $ z \equal{} \minus{} 2$. Then I let $ x \equal{} 1, y \equal{} \minus{} 1$, and got $ z \equal{} 0$. So I think the vectors $ (1, 1, \minus{} 2)$ and $ (1, \minus{} 1,0)$ span the subspace?

But I'm still confused with question 3, the one with the subspace spanned by $ (1,1,1,1), (1,2,3,4)$, and $ (2,3,4,5)$. The three points determine a plane, right? How should I proceed from there?

If a set is not independent, then at least one vector can be written as a linear combination of the others in the set. You should be able to see how to do that here.  Omit that vector from the set, since doing so does not change the span. Is the remaining set independent? If it is, you have your basis. If not, then repeat the process and throw out another vector.


Show that for all positive integers $n$, \[\lfloor \sqrt{n}+\sqrt{n+1}\rfloor =\lfloor \sqrt{4n+1}\rfloor =\lfloor \sqrt{4n+2}\rfloor =\lfloor \sqrt{4n+3}\rfloor.\]

First, i will show, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$, what is equivalent to that:
$ \sqrt{4n+3} -  \sqrt{4n+1} < 1  \\
4n+3+4n+1-2\sqrt{(4n+1)(4n+3)} < 1  \\
4n+2-\sqrt{(4n+1)(4n+3)}< 1/2$
(Of course, this difference is greater than $0$.)
It is obvious, that $4n+5/3 < \sqrt{(4n+1)(4n+3)}$ for all $n>0$, so you can say:
$4n+2-(4n+5/3)<1/2$, what is true.
So it is shown, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$.
Now, because of $\sqrt{4n+1} \leq \sqrt{4n+2} \leq \sqrt{4n+3}$ and $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$, it's also true, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+2} \right\rfloor =\left\lfloor \sqrt{4n+3} \right\rfloor$
For $ \sqrt{n}+\sqrt{n+1} $ holds the same condition as for $\sqrt{4n+2}$, i.e. $ \sqrt{4n+1} \leq  \sqrt{n}+\sqrt{n+1}  \leq \sqrt{4n+3} $, so that you can also say, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{n}+\sqrt{n+1} \right\rfloor =\left\lfloor \sqrt{4n+3} \right\rfloor$, so the proposition is proven.

To prove that $ \left\lfloor \sqrt{4n\plus{}1} \right\rfloor \equal{} \left\lfloor \sqrt{4n\plus{}3} \right\rfloor$, actually we only need to show that there is no integer in $ (\sqrt{4n\plus{}1},\sqrt{4n\plus{}3}]$, i.e., there is no perfect square in $ (4n\plus{}1,4n\plus{}3]$, which is quite obvious.

Hello,

This is kinda old but I liked this problem and I will give a very beautiful solution for it:

$\begin{array}{l}
 n = \sqrt {n^2 }  < \sqrt {n\left( {n + 1} \right)}  < \sqrt {\left( {n + 1} \right)^2 }  = n + 1 \\
  \\
  \Rightarrow 2n < 2\sqrt {n\left( {n + 1} \right)}  < 2n + 2 \to (1) \\
  \\
 2\sqrt {n\left( {n + 1} \right)}  = \left( {\sqrt n  + \sqrt {n + 1} } \right)^2  - \left( {2n + 1} \right) \to (2) \\
  \\
 (2)in(1) \Rightarrow 2n < \left( {\sqrt n  + \sqrt {n + 1} } \right)^2  - \left( {2n + 1} \right) < 2n + 2 \\
  \\
  \Rightarrow 4n + 1 < \left( {\sqrt n  + \sqrt {n + 1} } \right)^2  < 4n + 3 \\
  \\
  \Rightarrow \sqrt {4n + 1}  < \sqrt n  + \sqrt {n + 1}  < \sqrt {4n + 3}  \to (3) \\
  \\
 \exists k \in Z^ +  :k^2  \le 4n + 1 < \left( {k + 1} \right)^2  \\
  \\
 But \\
  \\
 4n + 2 \wedge 4n + 3\left( { \sim  \equiv } \right)2,3\left( {\bmod 4} \right) \Rightarrow 4n + 2 \wedge 4n + 3 \ne \left( {k + 1} \right)^2  \\
  \\
  \Rightarrow k^2  \le 4n + 1 < 4n + 2 < 4n + 3 < \left( {k + 1} \right)^2  \\
  \\
  \Rightarrow k \le \sqrt {4n + 1}  < \sqrt {4n + 2}  < \sqrt {4n + 3}  < k + 1 \\
  \\
 (3) \Rightarrow \sqrt {4n + 1} ,\sqrt {4n + 2} ,\sqrt {4n + 3} ,\sqrt n  + \sqrt {n + 1}  \in \left[ {k,k + 1} \right) \\
  \\
  \Rightarrow \left\lfloor {\sqrt n  + \sqrt {n + 1} } \right\rfloor  = \left\lfloor {\sqrt {4n + 1} } \right\rfloor  = \left\lfloor {\sqrt {4n + 2} } \right\rfloor  = \left\lfloor {\sqrt {4n + 3} } \right\rfloor  = k \\
 \end{array}$


:blush: ....

Let $q,r,p$ be three prime numbers.If $\frac{p}{q}+\frac{q}{r}+\frac{r}{p}$ is a natural number prove that $p=q=r$

Make the solution as simple as possible  :lol: .

[hide]

$\frac{p^{2}r+pq^{2}+qr^{2}}{pqr}$ is an integer. $p^{2}r \equiv qr^{2}\equiv 0 \mod r \implies r|pq^{2}\implies r=p \vee (q^{2}=r^{2}\implies q=r)$

case 1) $r=p \rightarrow \frac{r^{2}+q^{2}+qr}{pq}$ is integer.
$qr \equiv q^{2}\equiv 0 \mod q \implies q|r^{2}\implies q=r \implies q=p=r$

case 2) $q=r \rightarrow \frac{p^{2}+pr+qr}{pq}$ is an integer. $p^{2}\equiv pr \equiv 0 \mod p \implies p|qr \implies (p=q \vee p=r) \implies p=q=r$[/hide]

[hide]Assume, without loss of generality, that $p\neq q$ and $p\neq r$.  Since p, q, and r are prime, $(p,q)=1$ and $(p,r)=1$
 Let $\frac{p}{q}+\frac{q}{r}+\frac{r}{p}=x$
Multiplying by a common denominator, we get  $p^{2}r+r^{2}q+q^{2}p=xpqr$
 Taking this mod p, we have that $r^{2}q\equiv 0 \bmod{p}$  Since r and q share no common factors with p, this is impossible because the left side cannot be divisible by p.  Thus either q or r is equal to p, a contradiction.  WLOG, assume $q=p$  By using this same argument on r, we find that either q or p is equal to r, which means that all three must be equal.[/hide]


limit
integration
calculus
calculus computations

$f:[0,1]\rightarrow R$ continous

$u_n=\prod_{k=1}^{n}(1+1/nf(k/n))$

It is known that $\lim u_n=exp(\int_{0}^{1}f(t)dt)$

Prove that

$\lim (\frac{u_n}{exp(\int_{0}^{1}f)})^n=exp(-1/2\int_{0}^{1}f^2)$

We are just to apply $\ln(1+x)=x-x^2/2+O(x^3)$.