linear algebra
matrix
LaTeX
linear algebra unsolved

Let $A$ be a real nXn matrix such that  $AA^{t}=A^{t}A$. Then $TrAA^{t}$ is the sum of the squares of the absolut values of the eigenvalues of $A$.(sry for not using latex, i'm not very accustomed to it...)

The key is that $A$ is unitarily diagonalizable; there is some unitary $U$ such that $U^{*}AU=D$ is (complex-valued) diagonal.
If such a $U$ exists, $AA^{T}=AA^{*}=UDU^{*}UD^{*}U^{*}=UDD^{*}U^{*}$; its eigenvalues are the entries of $DD^{*}$, which are the squares of the absolute values of the eigenvalues of $A$.

Now we need to prove that such a $U$ exists: if $A$ is a complex $n\times n$ matrix with $AA^{*}=A^{*}A$, $A$ is unitarily diagonalizable.
We use Schur's lemma; there exists a unitary $U$ such that $U^{*}AU=T$ is upper triangular. Now $TT^{*}=U^{*}AA^{*}U$ and $T^{*}T=U^{*}A^{*}AU$, so $TT^{*}=T^{*}T$. For $i,j\le n$, let $t_{ij}$ be the $ij$-entry of $T$: $t_{ij}=0$ if $i>j$. The $ij$-entry of $TT^{*}$ is $\sum_{k=1}^{n}t_{ik}\overline{t_{jk}}= \sum_{k=\max(i,j)}^{n}t_{ik}\overline{t_{jk}}$ and the $ij$-entry of $T^{*}T$ is $\sum_{k=1}^{n}\overline{t_{ki}}t_{kj}= \sum_{k=1}^{\min(i,j)}\overline{t_{ki}}t_{kj}$.
Equating these for the diagonal elements $i=j$, $\sum_{k=1}^{i-1}|t_{ki}|^{2}=\sum_{k=i+1}^{n}|t_{ik}|^{2}$. The rest is simple; we get that $t_{1k}=0$ for $k>1$ by the first equation, then $t_{2k}=0$ for $k>2$ by the second, and so on. The only solutions to the system have $t_{ij}=0$ for $i<j$, and thus $T$ is diagonal.

The converse of this theorem is also true, and it's easy.

Maybe, solution to the problem can be simplified.

1) Suppose the spectrum $\sigma(A)$ of $A$ is real, then since $A$ is a normal operator, it is diagonalizable in some orthonormal basis $(e_{k})_{k=1}^{n}$ of $\mathbf R^{n}$ with corresponding eigenvalues $\lambda_{k}$.

Let $(\cdot,\cdot)$ be the canonical inner product of $\mathbf R^{n}$, then for every $B \in M_{n}(\mathbf R)$ the magnitude $\sum_{k=1}^{n}(Be'_{k},e'_{k})$ does not depend on selection of an othonormal basis $(e'_{k})_{k=1}^{n}$ of $\mathbf R^{n}$ and equals $\mathrm{Tr}\,(B)$.

Thus $\mathrm{Tr}\,(A^{t}A) = \sum_{k=1}^{n}(A^{t}Ae_{k},e_{k}) = \sum_{k=1}^{n}(Ae_{k},Ae_{k}) = \sum_{k=1}^{n}\lambda_{k}^{2}$.

2) Suppose that $\sigma(A) \setminus \mathbf R \neq \varnothing$. I think one can modify the method above but I don't know how.


You forgot so simplify the radical in the denominator, that would give us [hide]$1+\frac{1(x-\sqrt{x}}{(x+\sqrt{x})(x-\sqrt{x}}\rightarrow\frac{x-\sqrt{x}}{x^2-x}\rightarrow\frac{x-\sqrt{x}}{x(x-1)}$
So the answer is $1+\frac{x-\sqrt{x}}{x(x-1)}$[/hide]


[quote="minsoens"]I think you should assume $ a_1 \le a_2 \le... \le a_n$, not $ a_1 \ge a_2 \ge... \ge a_n$. Then your proof is alright.[/quote]
yes, I'm sorry. I am corrected  :wink:


Find out all the solutions for the equation:

8x(2x^2-1)(8x^4-8x^2+1)=1, for x in (0,1).

bye bye

we can see that if we denote x=cost we have cos2t=(2x^2-1) and cos4t=(8x^4-8x^2+1) and the eqs is equivalent to: cost*cos2t*cos4t=1/8 and the sol are cos \pi/9 and cos 2\pi/7.

cheers! :D :D :D

hmmm seems to me that someone forgot about cos \pi/3 ?! :D

hello, rewriting your equation in the form
$ (2x\minus{}1)(\minus{}1\minus{}6x\plus{}8x^3)(\minus{}1\minus{}4x\plus{}4x^2\plus{}8x^3)\equal{}0$
we easily get the solutions
$ x_1\equal{}\frac{1}{2}$
$ x_2\equal{}\frac{1}{2}\left(\frac{1}{2}(1\plus{}i\sqrt{3})\right)^{\frac{1}{3}}\plus{}\frac{1}{2^{\frac{2}{3}}(1\plus{}i\sqrt{3})^{\frac{1}{3}}}$
$ x_3\equal{}\minus{}\frac{1}{6}\plus{}\frac{7^{\frac{2}{3}}}{3\cdot2^{\frac{2}{3}}(1\plus{}3\sqrt{3}i)^{\frac{1}{3}}}\plus{}\frac{1}{6}\left(\frac{7}{2}(1\plus{}3\sqrt{3}i)\right)^{\frac{1}{3}}$
Sonnhard.

Dr. Sonnhard Graubner, please tell me. Do you have some software mfor these algebraic computations?
Thnkas.


quadratics
algebra
polynomial
system of equations
algebra proposed

Let $A\neq 0$ be an integer. Solve the following system of equations over ZxZxZ.

\begin{eqnarray*} x+y^2+z^3&=& A \\ \frac1{x} + \frac1{y^2} + \frac1{z^3} &=& \frac{1}{A}\\ xy^2z^3 &=& A^2 \end{eqnarray*}

I think I've solved the problem in all $\mathbb{R}$, and not only in $\mathbb{Z}$.

The system has solutions (in $\mathbb{R}$) only if $A=-1$.
In that case, the only four solutions (in $\mathbb{R}$) are:
$x=-1$, $y=\pm 1$, $z=1$.
$x=1$, $y=\pm 1$, $z=-1$.

[hide="Proof"]
To simplify, I'll call $v_1=x$, $v_2=y^2\ge 0$, $v_3=z^3$, and the three equations are now:
E1) $v_1+v_2+v_3=A$
E2) $(1/v_1)+(1/v_2)+(1/v_3)=1/A$
E3) $v_1 v_2 v_3 = A^2>0$
Thanks to E3, we know that $v_2>0$ and $v_1 v_3>0$.

If we multiply E2 and E3, we get $v_2 v_3+v_1 v_3+v_1 v_2=A$.
If $i\ne j \ne k \ne i$, $v_i (v_j + v_k) + v_j v_k=A$, and using E1 and E3, $v_i (A-v_i)+(A^2/v_i)=A$.
So, for every $i$, we have a quadratic equation, $A^2-(v_i-v_i^2)A-v_i^3=0$, with two possible solutions: $A=v_i$ or $A=-v_i^2$.

Case 1) There's no $i$ with $A=v_i$, so $A=-v_1^2=-v_2^2=-v_3^2$.
That means $|v_1|=|v_2|=|v_3|=t>0$ and $A=-t^2$.
We know that $v_2>0$ and $v_1 v_3>0$, so there are two possibilities:
Case 1.1) $v_1=v_2=v_3=t>0$ and $A=-t^2$. Impossible (E1: $3t=-t^2<0$).
Case 1.2) $-v_1=v_2=-v_3=t>0$ and $A=-t^2$. Only possible if $t=1$ (E1: $-t=-t^2$; E2: $-1/t=-1/t^2$; E3: $t^3=t^4$), but then, $A=v_1=v_3$, which should be in case 2.

Case 2) There's at least one $i$ with $A=v_i$. In this case, the equations are: E1, E2: $v_j+v_k=0$; E3: $v_j v_k=A$.
We have $v_j=-v_k$ and $A=v_i=-v_j^2=-v_k^2<0$. So, $i\ne 2$, because we know that $v_2>0$ and $v_1 v_3>0$. There are two possibilities left:
Case 2.1) $v_2=-v_3=t>0$ and $A=v_1=-v_2^2=-v_3^2=-t^2<0$.
Case 2.2) $v_2=-v_1=t>0$ and $A=v_3=-v_2^2=-v_1^2=-t^2<0$.
Both are only possible if $t=1$ (E1 and E2 are not restrictive, but E3: $t^3=t^4$). That gives us the only two solutions:
1) $v_1=-1$, $v_2=1$, $v_3=1$, $A=-1$.
2) $v_1=1$, $v_2=1$, $v_3=-1$, $A=-1$.
[/hide]

We set $x=a, \,\, y^2 =b, \,\, z^3=c$.
The given equations are written now in the form :

\begin{eqnarray*} a+b+c&=& A \\ \frac1{a} + \frac1{b} + \frac1{c} &=& \frac{1}{A}\\ abc &=& A^2 \end{eqnarray*}

and the second gives : $\frac1{a} + \frac1{b} + \frac1{c} =\frac{1}{A} \Rightarrow ab+bc+ca = A$

Consider the polynomial $P(x)=(x-a)(x-b)(x-c)$ that has $3$ roots in $\mathbb{R}$.

After expanding

$\\ P(x) = x^3 -(a+b+c)x^2+(ab+bx+ca)x-abc \Rightarrow \\ \\ P(x) =x^3 -Ax^2+Ax-A^2 \Rightarrow P(x) =(x-A)(x^2+A)$

But since $P(x)$ has $3$ roots, $A$ must be negative and let $A=-B^2 , \, B>0$

Then $P(x) =(x+B^2)(x-B)(x+B)$ with roots $-B^2, \,\, B, \,\, -B$.

Now set $B=C^6$ .Since $b=y^2 >0$ hence we take two cases.

[b]1.[/b] $x=-B, \,\, y^2=B, \,\, z^3 = -B^2$ that has the solution

$\\ \boxed{x=-C^6, \,\, y= \pm C^3, \,\, z = -C^4 }$

[b]2.[/b] $x=-B^2, \,\, y^2=B, \,\, z^3 = -B$ that has the solution

$\\ \boxed{x=-C^12, \,\, y= \pm C^3, \,\, z = -C^2 }$

where $\boxed{C= \sqrt[12]{-A}}$.

 :)

As usual, I messed up at the last line :(

[quote="lordWings"]
There are two possibilities left:
Case 2.1) $v_2=-v_3=t>0$ and $A=v_1=-v_2^2=-v_3^2=-t^2<0$.
Case 2.2) $v_2=-v_1=t>0$ and $A=v_3=-v_2^2=-v_1^2=-t^2<0$.
Both are only possible if $t=1$ (E1 and E2 are not restrictive, but E3: $t^3=t^4$).
[/quote]
Actually, E3 isn't restrictive either: $t^4=t^4$. Every $t$ gives a solution.

So, the system has solutions (in $\mathbb{R}$) only if $A=-t^2<0$. ($t=(-A)^{1/2}$).
In that case, there are four solutions (in $\mathbb{R}$):
$x=-t^2$, $y=\pm t^{1/2}$, $z=-t^{1/3}$
$x=-t$, $y=\pm t^{1/2}$, $z=-t^{2/3}$


superior algebra
superior algebra unsolved

Prove that $ SL_n$ is semisimple. Calculate its Weyl group. and

2) (Bruhat decomposition?) prove that any element in $ SL_n$ can be expressed as $ b_1wb_2$ with $ b_1,b_2$ in the Borel group and $ w$ (a representative of) a uniquely determined element of the Weyl group.  Of the subsets $ BwB$, which are open? closed?

First of all the Weyl Group is $ S_n$ right?


Prove that a triangle ABC is right-angled if and only if
$sinA + sinB + sinC = cosA + cosB + cosC +1$

well, it's really easy to prove it one way. If it is right-angled wolog $A=90^o$ have $sinA=cosA+1$ and $sinA=cosB$ and $sinB=cosA$. Now the other part is difficult... :?

Easy problem ...  :)
If we have $(sinA+sinB)-(cosA+cosB)=1+cosC-sinC$,then:
$2cos\frac{A-B}{2}(sin\frac{A+B}{2}-cos\frac{A+B}{2})=2cos\frac{C}{2}(cos\frac{C}{2}-sin\frac{C}{2})$ and we have $sin\frac{A+B}{2}=cos\frac{C}{2}$and $cos\frac{A+B}{2}=sin\frac{C}{2}$ so we have:
$2cos\frac{A-B}{2}(cos\frac{C}{2}-sin\frac{C}{2})=2cos\frac{C}{2}(cos\frac{C}{2}-sin\frac{C}{2})$ so we have :
1.$cos\frac{C}{2}-sin\frac{C}{2}=0$ so $C=90$
2. $cos\frac{A-B}{2}=cos\frac{C}{2}$ so $\frac{A-B}{2}=\frac{C}{2}$
so $A=90$.
3.$cos\frac{A-B}{2}=cos\frac{C}{2}$ so $\frac{A-B}{2}=-\frac{C}{2}$
so $B=90$.
So we are done.


Let set $ T \equal{} \{1,2,3,4,5,6,7,8\}$. Find the number of all nonempty subsets $ A$ of  $ T$ such that $ 3|S(A)$ and $ 5\nmid S(A)$, where $ S(A)$ is the sum of all the elements in $ A$.

[quote="pbornsztein"]There is no way to select a subset of $ T$ with at least $ 15$ elements, so you probably asked for $ S(A)$ to be the sum of the elements in $ A$ :wink:

Pierre.[/quote]
Edited

[hide="hint I"]$ 1 \le S(A) \le 1\plus{}2\plus{}\cdots\plus{}8\equal{}36 \Rightarrow S(A)\equal{}3,6,9,12,18,21,24,27,33,36$[/hide]
[hide="hint II"]Let $ \#An\equal{}$ the number of subset of $ A$ with $ S(A)\equal{}n$ then $ \#A36\equal{}1, \#A(36\minus{}i)\equal{}\#Ai$[/hide]

I can't understand one thing,why there is always someone who post a hint?
Tell me the reason.If you've solved it then post your complete proof,if not as you've already seen(i hope),i posted it in Proposed and Own Problems,hence i know solution,and if nobody will solve it during the week,i'll post a hint.OK?

Sorry for the inconvenience.

I ll post the complete solution:
$ 1 \le S(A) \le 1 \plus{} 2 \plus{} \cdots \plus{} 8 \equal{} 36 \Rightarrow S(A) \equal{} 3,6,9,12,18,21,24,27,33,36$
Let $ \#An \equal{}$ the number of subsets of $ A$ with $ S(A) \equal{} n$ then $ \#A36 \equal{} 1, \#A(36 \minus{} i) \equal{} \#Ai$
$ \#A3 \equal{} \#A33 \equal{} 2$
$ \#A6 \equal{} 4$
$ \#A9 \equal{} \#A27 \equal{} 7$
$ \#A12 \equal{} \#A24 \equal{} 10$
$ \#A18 \equal{} 14$
$ \#A21 \equal{} 13$
$ \#A36 \equal{} 1$
$ TOTAL \equal{} 70$

Yes,your proof is correct,but i think that you must show all subsets,that you have counted.
There is one easier proof,that use much less counting,and i'll post it if nobody do it,until next week,OK?


Can anyone explain to me the interpretation of the ordinal property and the cardinal property.

I know that ordinal represents number, it doesn't have meaning in and of itself.

I know that cardinal represents an interpretation, it has meaning in and of itself.

Can anyone explain this a little further, if at all possible in intuitive terms and not a mathematical representation?

Thank you.

Ordinals are [url=http://en.wikipedia.org/wiki/Well-order]well-ordered[/url] and cardinals just have a [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url].  The finite ordinals and the finite cardinals look the same, but the infinite ordinals and the infinite cardinals look very different.  How "mathematical" do you want this explanation to be?


I have posted it http://www.mathlinks.ro/viewtopic.php?t=280934 but receiving no reply.
This is an old problem from China TST(I forget the year)
It is indeed very hard but nice :D


function
trigonometry
logarithms
complex analysis
complex analysis unsolved

Given two entire functions $f$ and $g$ that satisfy $f^{2}(z)+g^{2}(z) = 1$ for all $z$, there exists an entire function $h$ such that $f(z) = \cos h(z)$ and $g(z) = \sin h(z)$.

I'm not very good at complex analysis, but let's see what happens.

Define the two entire functions
\begin{eqnarray*}a(z) & = & f(z)+i g(z) \\ b(z) & = & f(z)-i g(z) \, . \end{eqnarray*}
Note that $a(z) b(z) = f(z)^{2}+g(z)^{2}= 1$.  In words, $a$ and $b$ are reciprocals of each other.  In particular, $a$ and $b$ are never zero.

Because $a$ is never zero, by somebody's theorem (Picard?), there is an entire function $h$ such that $e^{ih(z)}= a(z)$.  Then we have $e^{-ih(z)}= \frac{1}{a(z)}= b(z)$.  Thus
\[\cos h(z) = \frac{e^{ih(z)}+e^{-ih(z)}}{2}= \frac{a(z)+b(z)}{2}= f(z) \, . \]
Similarly for $\sin h(z)$.

Neat. The existence of $h$ is just the [url=http://mathworld.wolfram.com/MonodromyTheorem.html]monodromy theorem[/url]. Since $a(z)\ne 0$, the (potentially multi-valued) analytic function $\log a(z)$ can be extended along any path in $\mathbb C$, hence it is single-valued in $\mathbb C$.


Find every $n$, such that there exist a permutation $a_1,...,a_n$ of ${1,2...,n}$ with property that all differences $|a_k-k|$ are different.

In http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1157858#1157858 pbornsztein gave a proof that we must have $ n \equiv 0, 1 \pmod 4$, and a link was given to http://mathcentral.uregina.ca/mp/previous2001/aug02sol.html which has a construction for all such $ n$.  Sequence [url=http://www.research.att.com/~njas/sequences/A075866]A075866[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] enumerates permutations with this property.  This seems rather harder than Preolympiad level ;)


In each line and column of a table $ (2n \plus{} 1)\times (2n \plus{} 1)$ are written arbitrarly the numbers $ 1,2,...,2n \plus{} 1$. It was constated that the repartition of the numbers is symmetric to the main diagonal of this table. Prove that all the numbers on the main diagonal are distinct.

Let $ a_{i,j}$ be the number in the $ i$th row and the $ j$th column. You're saying that if $ x\neq y$ then $ a_{x,y}\equal{}a_{y,x}$?

[hide="Yes,"]
Excluding the numbers on the main diagonal, each number must appear an even number of times since they are on symmetric halves; but each number appears an odd number of times on the board, so each number must appear on the main diagonal.
[/hide]


[quote="tranthanhnam"]:D  EM MUON GUI BAI NHUNG KHONG BIET KI HIEU TOAN HOC O DAU CA
THANH VIEN MOI MU , => RAT LUNG TUNG[/quote]
Ban phai hoc su dung tex cai da

[quote="manutd"]Cho hỏi nếu không có ký hiệu toán học thì dùng các tex như thế nào. Tôi muốn biết tất cả các code của ký hiệu toán học thì vào đâu. Cám ơn![/quote]
ben nay thi dung cap dolar,ban chua thanh thao thi co the tham gia mot thoi gian ben diendantoanhoc.net de cho biet cong thuc.

[quote="nguyentrongtin"]Co the chi minh danh vai ki hieu don gian de post bai ko vay[/quote]

Đã bảo là qua diễn đàn toán học :diendantoanhoc.net học gõ thử đã

ban hay vao lick nay va test thu nhe http://diendantoanhoc.net/forum/index.php?showtopic=1235


calculus
derivative
function
logarithms
real analysis
real analysis unsolved

What is the derivative $ dx$ of the function:

$ t(x)\equal{}g(x)^{f(x)}$ ?

An easy way to think about such derivatives is using "logarithmic differentiation," as follows:

$ \ln t \equal{} f \ln g \implies$
$ \frac{t'}{t} \equal{} \frac{fg'}{g} \plus{} f' \ln g \implies$
$ t' \equal{} g^f \left( \frac{fg'}{g} \plus{} f' \ln g \right)$.

This can also be found by applying the chain rule and product rule to $ e^{f(x) \ln g(x)}$.


Hmm, so are you asking how to simplify those exact expressions with those exact angles, or just those expressions but with any angles? That is, simplifying something like cos^2 30 - sin^2 30 can be said to be easier than the same expression with another angle (say 23) in place of 30 (simply since the numberlcal values of sin 30 and cos 30 are pretty well-known).

Fierytycoon

[quote="Rep123max"]Can somebody explain to me how to simplify something like sin 3(pi) and cos^2 30 (degrees)-sin^2 (30)degrees.[/quote]

In order to solve these problems, you'll need to memorize some of the important trig relationships.  This'll help you recognize a pattern that will simplify into something else.

#1: for the first one, you need to know about periodicity.

Ever seen a graph of the sin or cosine function?  It looks like a wave.  Every 2pi units, it repeats.  Similarly, if you look on the unit circle, if you go around the circle 2pi units, you'll get back to where you started (because in the unit circle, the radius is 1 so the circumference is 2pi).

Thus, sin(x) = sin(x+2pi) = sin(x+4pi) = sin(x+6pi) = ....
and sin(x) = sin(x-2pi) = sin(x-4pi) = ...

In your case, you've got sin(3pi) which is equivalent to sin(3pi - 2pi) = sin(pi).

From there, you can say that sin(pi) = 0.

#2: Okay, for this one, you need some formulas.  One, to be precise.

The double-angle formulas are a way of telling you what the sin/cos/tan of an angle twice as big as angle a will be in terms of a.

cos 2a = 2cos:^2: - 1 = 1 - 2sin:^2: a = cos:^2: a - sin:^2: a

The last one is the one we want.  cos 2a = cos:^2: a - sin:^2: a

In our case, we've got cos:^2: 30 - sin:^2: 30 = cos  2(30) = cos 60

Cos 60 is well-known as [b]1/2[b/].

As you can see, a knowledge of the relationships between sines, cosines, and tangents is important.  If you learn some of the formulas and write the rest down somewhere where you can consult it later, you'll do fine.  Just keep working on problems like this and you'll start to recognize these relationships.  If you've got any more problems, post 'em here.  Hope this helps. - mathfanatic

Oh, and Fierytycoon - as it turns out, you could just as easily solve cos:^2: 23 - sin:^2: 23 - it becomes cos 2(23) = cos 46.  Granted, you can't find a pretty numerical value, but it's simplified as much as it can be.

[quote]Granted, you can't find a pretty numerical value, but it's simplified as much as it can be.[/quote]

Yeah, that is what I meant, since the numbers used by the OP were 'nice', so no formula (for cos(2x) in this case) was really needed.

Fierytycoon

Fierytycoon: where in MA are you from?  west or east?

So where are you from: eastern or western MA?

Fierytycoon


Tom's age is $ T$ years, which is also the sum of the ages of his three children.  His age $ N$ years ago was twice the sum of their ages then.  What is $ \frac {T}{N}$?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

[quote="t0rajir0u"]Tom's age is $T$ years, which is also the sum of the ages of his three children.  His age $N$ years ago was twice the sum of their ages then.  What is $\frac{T}{N}$?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$[/quote]
[hide]Say the kids ages are x,y, and z. So Tony's age is x+y+z. N years ago means that now it will be T-N and x+y+z-3N. And he was twice their ages so now we have a equation. T-N=2(x+y+z-3N). Simplyfing this we get now that T-N=2(x+y+z)-6N. Since x+y+z=T substitute this value in to get T-N=2T-6N. So T=5N. Since it asks for T/N it is 5 or D.[/hide]


I need some help on an issue i cant realy understand:

Example:

I have to objects (A & B). I have the speed of both of them

A: 1m / 1h = 1 m/h
B: 1m /3h = 0,33 m/h

Now if i want to find the average speed of both objects im getting different results:
1. 2m / 4h = 0.5 m/h (Total distance over total time)
2. (1+0.33)/2 = 0.665 m/h (average of both speeds)

could some one please explain me why i get the difference and which will be the correct answer

Thanks in advance

I think 2 is right...
If you were given distance and time, you'd use $ \text{average speed}\equal{}\frac{\text{total distance}}{\text{total time}}$.
But considering you're given speeds, you should be able to just average them...

1 is certainly not right because for one thing, you're not given distance or time, but rate, and because you added the fractions incorrectly  :lol:

[quote="grn_trtle"]I think 2 is right...
If you were given distance and time, you'd use $ \text{average speed} \equal{} \frac {\text{total distance}}{\text{total time}}$.
But considering you're given speeds, you should be able to just average them...

1 is certainly not right because for one thing, you're not given distance or time, but rate, and because you added the fractions incorrectly  :lol:[/quote]

I'm really confused with this issue.
I dont see where i added the fraction incorrectly??
If you look at it the other wy around all together the 2 objects move 2m in 4h so that makes an avarage speed (or rate) of 0.5 m/h...
something is not correct in one of the results but i dont know what ?!?!

Ah ha!  I found it!  :lol:

It's just because in the second one, you should be weighting your averages, because the second speed "takes more time" (3 h) than the other (1 h).

Thus, to balance this, we must multiply the one that takes more time by 3.  Since we now have four things to add, the denominator must be 4.

$ \frac{1 \plus{} 3 \left( \frac13 \right)}{4} \equal{} \frac{1\plus{}1}{4} \equal{} \boxed{\frac12}$

This gives your desired result.

If it goes the same distance, use ernie's method. If it goes the same time, a simple average works.


calculus
integration
calculus computations

Evaluate: $ \sum_{n\equal{}0}^\infty (\int_{0}^{\frac{1}{2}} x^{n} dx)$

[quote="jmerry"]A standard principle of problem solving: whenever you write down an iterated sum/integral, you usually intend to switch the order.[/quote]
You can do that?  :o

First you can calculate the part of the integral, then you would calculate the part of the infinite seriies,
You would find $ \sum_{m\equal{}1}^{\infty} \frac{1}{m\cdot 2^{m}}$

Can we do that? Sure. Any combination (sum/sum, sum/integral, or integral/integral) can be switched, provided certain conditions hold. There are two standard sufficient conditions that allow the switch:
(a) Everything's positive; in this case, we accept $ \infty\equal{}\infty$ as a valid equality.
(b) The sum/integral of the absolute values is finite.

For this problem, both of those conditions are true. Positivity is obvious, and we can get the absolute convergence by comparing to a geometric series.
If you go through with it, the inner sum of the $ x_n$ should be easy to evaluate exactly.

One other note: if we replace that upper limit $ \frac12$ with $ t$ for some $ t\in (\minus{}1,1)$, we get a nice power series expression equating that series $ \sum_{m\equal{}1}^{\infty}\frac1m t^m$ with a standard function.


geometry rectangle SFFT MATHCOUNTS AMC USA(J)MO USAMO

Can anyone explain to me what Sam's Favorite Factoring is?, or what is Clever Manipulation? In several of the forums, people had been using this for problems such as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=favorite+favourite+factoring&t=76601]HERE[/url]. as you can see what [i]mathgeniuse^ln(x)[/i] wrote.  :?: Thanks![/list]

It's [b]Simon's[/b]favorite factoring trick.  ;)

I believe it's [b]Simon's[/b] factoring trick:

$xy+ax+ay+a^2=(x+a)(y+a)$

You can do stuff like

$\frac 1x+\frac 1y=\frac 17$
$\frac{x+y}{xy}=\frac 17$
$7(x+y)=xy$
$xy-7x-7y+49=49$
$(x-7)(y-7)=49$

Clever manipulation is, well, clever.

Simon's (complexzeta) favorite factoring trick is $a+b+ab+1=(a+1)(b+1)$

Clever manipulations come in many forms. (a-b)(a+b)=a^2-b^2. Another is a^3+b^3=(a+b)(a^2+b^2-ab). Another is a^3-b^3=(a-b)(a^2+b^2+ab). They help in state a good amount. Last year on the state team round one problem required simon's favorite factoring trick. It was 1/a+1/b=2/15. How many values are there of a and b.

Yeah, that was the example I mentioned. I think it was also on some sprint round from some year at either states or nationals. (I'm very precise, huh.)

That doesn't necessarily require Simon's Favorite Factoring Trick though, it's possible to solve for a or b and then put it in on a graphing calculator

[quote="pgpatel"]Chapter sprint actually. Anyways the problem was, the area of a rectangle that isn't a square is equal to the perimeter. What is the area?
 :ninja:  :ninja:  :ninja:[/quote]

I think using SFFT would be slower than simply knowing it's 3 and 6.

When I did it I didn't know the factoring trick so I spent 6 minutes on that one problem. Not all of us can guess and check that fast...

[quote="pgpatel"][quote="samath"]Actually, I think Ignite  just meant knowing that the answer was 3 and 6 because it's such a common problem.[/quote] Yes, I realize that but my post came out wrong last time. I guess it should be more like, not all of us have such a great intuition.[/quote]

I remember that because it was our first problem in Algebra that required Algebra. (We learned it using guess and check method though >_>)

[quote="pgpatel"]Clever manipulations come in many forms. (a-b)(a+b)=a^2-b^2. Another is a^3+b^3=(a+b)(a^2+b^2-ab). Another is a^3-b^3=(a-b)(a^2+b^2+ab). [/quote]

Those are standard manipulations.  Here's a clever one, although it is quite a bit past Mathcounts level:

Prove that $n^4 + 4^n$ is never prime for positive integers $n$.

[hide="Solution"]
If $n$ is even then the entire expression is even but larger than $2$, so it cannot be prime.

If $n$ is odd, we can write $n = 2k+1$.

$n^4 + 4 \cdot 4^{2k} = n^4 + 4 \cdot 2^{4k}$.  Substitute $m = 2^k$.  Then we have $n^4 + 4m^4$.

Here comes the clever manipulation (unless you are already familiar with the Sophie Germaine Identity.)

$n^4 + 4m^4 = (n^2)^2 + (2m^2)^2 = (n^2)^2 + 2(n^2)(2m^2) + (2m^2)^2 - 2(n^2)(2m^2) = (n^2 + 2m^2)^2 - (2mn)^2 = (n^2 + 2mn + 2m^2)(n^2 - 2mn + 2m^2)$

Then you only need to prove neither of the factors is equal to $1$, which is tedious and unimportant.  Hence $n^4 + 4^n$ is always composite.  QED.
[/hide]

 ;)

Subtracting and adding the same quantity is a standard way to find clever manipulations.  Try it on a few problems, you'll find it makes life easier.  Another technique, clever substitutions:

$a$ and $b$ are positive reals such that $a^3 - b^3 = 24, a - b = 2$.  Find all possible values of $a + b$.

[hide="Solution"]
Recall the standard factoring $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.  From this we can divide the two equations to get $a^2 + ab + b^2 = 12$.  This is the standard manipulation.  Here comes the clever one:  Substitute $x = a + b, y = ab$.  The first equation becomes

$x^2 - y = 12$

And the second, when squared (as $a^2 - 2ab + b^2 = 4$) becomes

$x^2 - 4y = 4$

Recall that the problem asks for all possible values of $a + b = x$.  So we need to eliminate $y$.  Simple:

$4x^2 - 4y = 48$
$3x^2 = 44$
$\boxed{x = \pm \sqrt{ \frac{44}{3} } }$
[/hide]

Note that I solved this problem without solving for $a, b$ individually.  That is because the manipulation I used was clever.   ;)

There's a very nice thread about this in the South Carolina forum.  It was designed as a tutorial by a past USAMO participant, and is an EXCELLENT resource.  It is located [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75738]here[/url].


calculate :     $ \sum_{i\equal{}1}^n\frac{1}{(2i\minus{}1).2i}\minus{}\sum_{i\equal{}1}^n\frac{1}{n\plus{}i}$

:)

What do you mean calculate? Simplify? :S Because n isn't chosen... want it to infinity ? But I think that would just be - infinity..

$ \sum_{i\equal{}1}^n\frac{1}{(2i\minus{}1)2i}\equal{}\sum_{i\equal{}1}^n\frac{1}{2i\minus{}1}\minus{}\frac {1}{2i}
\equal{}\sum_{i\equal{}1}^n\frac{1}{2i\minus{}1}\plus{}\frac {1}{2i}\minus{}2\frac {1}{2i}\equal{}H_{2n} \minus{}H_n$

$ \sum_{i\equal{}1}^n\frac{1}{n\plus{}i}\equal{}\sum_{i\equal{}n\plus{}1}^{2n}\frac{1}{i}\equal{}H_{2n}\minus{}H_n$

$ H_n$ is meant for the nth harmonic number,

$ H_n \equal{}\sum_{i\equal{}1}^{n}\frac{1}{i}$


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[color=blue] [size=150]$12{nd}$  Iberoamerican Olympiad[/size]
Guadalajara, MEXICO. [1998] [/color]

[color=blue]Problem 4 [/color]
Let $n$ be a positive integer. Consider the sum $x_1y_1 + x_2y_2 +\cdots + x_ny_n$,
where that values of the variables $x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n$ are either 0 or 1.

Let $I(n)$ be the number of 2$n$-tuples $(x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n)$
such that the sum of the number is odd, and let $P(n)$ be
the number of 2$n$-tuples $(x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n)$
such that the sum is an even number. Shot that:
$\frac{P(n)}{I(n)}=\frac{2^n+1}{2^n-1}$

$\text{\LaTeX}{}ed$ by Carlos Bravo - [carlosbr]

I think that this solution works:

Proof:

First we establish the following [u]Lemma:[/u]
For all finite sets with size $n$, the number of subsets with an even size is equal to the number of sets with an odd size, making the number of subsets for each type $\frac{2^n}{2}=2^{n-1}$.
The number of even-size subsets is $\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose 2i}$ and the number of odd sets is $\sum_{i=0}^{\lfloor\frac{n-1}{2}\rfloor}{n\choose 2i+1}$.  These can be shown to be equal since
$\displaystyle 0=0^n=(1+(-1))^n=\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose 2i}-\sum_{i=0}^{\lfloor\frac{n-1}{2}\rfloor}{n\choose 2i+1}$
which makes the two equal.  Since there are $2^n$ sets total, we have the Lemma.

Now to proof the suggestion, we will prove the following analogous problem:
The ratio of subsets $A$ and $B$ from an $n$-element set that satisfy $\mid A\cap B\mid$ being even and subsets $A$ and $B$ from an $n$-element set that satisfy $\mid A\cap B\mid$ being odd is $\displaystyle\frac{2^n+1}{2^n-1}$.
We will sum over the size of $\mid A\mid$.  If the size of $\mid A\mid$ is $k$, let $E_k(n)$ be the number of sets $B$ such that $\mid A\cap B\mid$ is even.
We have $E_0(n)=2^n$, and for some given $k>0$, we must have an even number of those $k$ elements in $B$.  The elements not used can either be in or out of the set.  Thus, using our Lemma above we have
$E_k(n)=2^{k-1}\cdot 2^{n-k}=2^{n-1}$
Now, since $\displaystyle P(n)=\sum_{i=0}^{n}{n\choose i}E_i(n) =2^{n-1}+\sum_{i=0}^{n}2^{n-1}{n\choose i}=2^{n-1}+2^{2n-1}$.
So we have $I(n)=2^{2n}-P(n)=2^{2n-1}-2^{n-1}$.

Finally, $\displaystyle\frac{P(n)}{I(n)}=\frac{2^{2n-1}+2^{n-1}}{2^{2n-1}-2^{n-1}}=\frac{2^n+1}{2^n-1}$
as desired.

[quote="carlosbr"]Let $n$ be a positive integer. Consider the sum $x_1y_1 + x_2y_2 +\cdots + x_ny_n$,
where that values of the variables $x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n$ are either 0 or 1.

Let $I(n)$ be the number of 2$n$-tuples $(x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n)$
such that the sum of the number is odd, and let $P(n)$ be
the number of 2$n$-tuples $(x_1, x_2,\ldots, x_n, y_1, y_2,\ldots, y_n)$
such that the sum is an even number. Shot that:
$\frac{P(n)}{I(n)}=\frac{2^n+1}{2^n-1}$
[/quote]

Put $a_i=x_i\cdot y_i$. Note that $a_i=1$ iff $x_i=y_i=1$ (in other case $a_i=0$). Put $F(x)=(x+3)^n$. If we write $F(x)=\sum_{k=0}^n s_kx^k$ for $(s_1,s_2,...,s_n)\in \mathbb {Z}^n$, we note that $s_k$ is the number of sequences $a_1,a_2,...,a_n$ with sum $k$, and hence $P(n)=\sum _{2|r} s_r$ while $I(n)=\sum_{2\not {|} r} s_r$. Note that $P(n)+I(n)=F(1)=4^n$ and $P(n)-I(n)=F(-1)=2^n$. Then $P(n)=2^{2n+1}+2^{n-1}$ and $I(n)=2^{2n-1}-2^{n-1}$, and we conclude the desired.

[hide="Alternate Solution"]
Since each $x$ and $y$ can take two values we have that there are a total of $2^{2n}$ ordered $2n$-tuples, moreover we have that $P(n)+I(n)=2^{2n}$($*$). Next, we consider the $2n+2$ tuple,
\[(x_1,x_2,\cdots,x_{n+1},y_1,y_2,\cdots,y_{n+1})\]
If $x_{n+1}=0$, then regardless of what $y_{n+1}$ is we will have a total of $P(n)$ even $2n+2$ tuples, the same goes for when $y_{n+1}=0$. If both $x_{n+1}$ and $y_{n+1}$ are equal to $1$ then there'll be a total of $I(n)$ even $2n+2$ tuples. Hence $P(n+1)=3P(n)+I(n)=2P(n)+2^{2n}$(by $*$)  :ninja: .

Next, we claim that $P(n)=2^{2n-1}+2^{n-1}$,

[i]Proof by Induction:[/i]

[i]Step 1:[/i] Clearly, $P(1)=3$ so this step is done.
[i]Step 2:[/i] Assume that $P(n)=2^{2n-1}+2^{n-1}$.
[i]Step 3:[/i] We need to prove that $P(n+1)=2^{2n+1}+2^n$, note that,
\[2^{2n+1}+2^n=2^{2n}+2^n+2^{2n}=2P(n)+2^{2n}=P(n+1)\]
The last step is given by our previous result that $P(n+1)=3P(n)+I(n)=2P(n)+2^{2n}$. Lastly we have,
\begin{eqnarray*}
 \dfrac{P(n)}{I(n)}&=&\dfrac{P(n)}{2^{2n}-P(n)}\\
 \dfrac{P(n)}{I(n)}&=&\dfrac{2^{2n-1}+2^{n-1}}{2^{2n}-(2^{2n-1}+2^{n-1})}\\
 \dfrac{P(n)}{I(n)}&=&\dfrac{2^{n-1}(2^n+1)}{2^{n-1}(2^n-1)}\\
 \dfrac{P(n)}{I(n)}&=&\boxed{\dfrac{2^n+1}{2^n-1}}
\end{eqnarray*}
[/hide]

We are looking for the number of pairs $x,y \in \mathbb F_2^n$ such that $\langle x,y \rangle=0$ (here $\mathbb F_2$ is the field with two elements and $\langle - , - \rangle$ is the standard scalar product).
Fix $y$ and consider the linear map $\langle - , y \rangle: x \mapsto \langle x,y\rangle$ from $\mathbb F_2^n \to \mathbb F_2$. If $y=0$, then it is the zero map and thus has kernel $\mathbb F_2^n$; otherwise, it is surjective and we find its kernel to have dimension $n-1$, thus $2^{n-1}$ elements. In total, varying $y$, we get $2^n+2^{n-1}(2^n-1) = 2^{n-1}(2^n+1)$ pairs $x,y$ such that $\langle x,y \rangle=0$, solving the problem.


Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \le \frac {27}{(p \plus{} 2)(a\plus{}b\plus{}c)^2}$.

Dear Vasc,Is the same hard 2?
[quote="Vasc"]Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \ge \frac {27}{(p \plus{} 2)(a \plus{} b \plus{} c)^2}$.[/quote]

Dear Vo, this is Hard 2:
[quote="Vasc"]Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \le \frac {27}{(p \plus{} 2)(a \plus{} b \plus{} c)^2}$.[/quote]

and this is Hard 5:
[quote="Vasc"]Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac 1{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{pc^2 \plus{} a^2 \plus{} b^2} \ge \frac {27}{(p \plus{} 2)(a \plus{} b \plus{} c)^2}$.[/quote]


For any three real numbers $ a$, $ b$, and $ c$, with $ b\neq c$, the operation $ \otimes$ is defined by:
\[ \otimes(a,b,c) \equal{} \frac {a}{b \minus{} c}
\]What is $ \otimes(\otimes(1,2,3), \otimes(2,3,1),\otimes(3,1,2))$?

$ \textbf{(A)}\ \minus{}\!\frac {1}{2}\qquad
\textbf{(B)}\ \minus{}\!\frac {1}{4}\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ \frac {1}{4}\qquad
\textbf{(E)}\ \frac {1}{2}$

[quote="#H34N1"]For any three real numbers $a$, $b$, and $c$, with $b\neq c$, the operation $\otimes$ is defined by:
\[\otimes(a,b,c)=\frac{a}{b-c}\]
What is $\otimes(\otimes(1,2,3), \otimes(2,3,1),\otimes(3,1,2))$?

A. $-\frac{1}{2}$
B. $-\frac{1}{4}$
C. $0$
D. $\frac{1}{4}$
E. $\frac{1}{2}$[/quote]

[hide="solution"]$\otimes(1,2,3)=-1$
$\otimes(2,3,1)=1$
$\otimes(3,1,2)=-3$

$\otimes(-1,1,-3)=\frac{1}{4}$[/hide]


linear algebra
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A problem with result similiar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=86676]this one[/url].
Suppose $C$ is a skew-symmetric $n\times n$ matrix with real coefficients. Prove that there exist symmetric matrices $S_{1},S_{2}\in\mathcal M_{n}(\mathbb R)$ that $C=S_{1}S_{2}-S_{2}S_{1}$

[quote="harazi"]Nice! Use the fact that any matrix is a product of two symetric matrices![/quote]
How can we prove that ?  :roll:

It's already proved on the forum, search for some of my topics and alekk's too.

I can't prove the problem with the Lemma you told  :( , Although I have another solution. :)

My solution:
We can proceed by induction.
The case $n=2$ is very easy.
In general, assume $n=k$ the proposition is true.
We note a skew-symmetric matrix is of the form $a_{ii}= 0$ and $a_{ij}=-a_{ji}$ for $i \neq j$. Let us write the matrix in the form
$\left( \begin{array}{cc}0 & a \\-a^{T}& A \\ \end{array}\right)$
Now, by induction hypotehsis we know that $A = S_{1}S_{2}-S_{2}S_{1}$ for some symmetric matrix of size k*k.
Then suppose $B_{1}= \left( \begin{array}{cc}x & b \\ b^{T}& S_{1}\\ \end{array}\right)$, $B_{2}= \left( \begin{array}{cc}x & c \\ c^{T}& S_{2}\\ \end{array}\right)$
where $x,b,c$ are to be determined.
Now computing $B_{1}B_{2}-B_{2}B_{1}$, we get it is equal to
$\left( \begin{array}{cc}0 & a' \\-a'^{T}& S_{1}S_{2}-S_{2}S_{1}\\ \end{array}\right)$
where $a' = xc+bS_{2}-xb-cS_{1}$
We hope it is equal to $a$, so that induction can be completed.
Set $c = 0$(zero vector), and take $x$ any value other than eigenvalues of $S_{2}$, then $S_{2}-xI$ is invertible, so we can take $b = a(S_{2}-xI)^{-1}$. Induction is now completed.


Prove that any finite set of points in $n$ dimentional space such that the $n-k$-hyperplane defined by $n-k+1$ points has another point of the set in it has the property of being confined to a $n-k$-hyperplane.  $k$ is below $n$ and is arbitrary.  This is a generalization of a collerary of Sylvester's theorem, so i'm confident that a proof exists. Note that $k$ and $n$ are integers if it wasn't immediatly obvious.

Well, it's not true, unfortunately :).

Take $n=3$ and $k=1$ for instance. Then the problem reads: "If a finite set of points in space is such that every plane determined by three of them contains at least four of them, then all points are contained in a plane." A counterexample to this would be two skew lines in space, each containing at least three points, say.


[hide="Answer to #1"]1. The first set is 1089-1024=65
The second is 99-1=98
For the third, 1000000 is not a prime and 1000100 is not a prime. At the most if the two following numbers are prime (I'm not sure)1000099-1000001=98 so the largest is 98.[/hide]

5. This one is easy since r% interest is compounded twice to get from 100 to 110, so just multiply and simple algebra.

EDIT:
[hide="2"]an intuitive method. we can split them into a group of 2 and 8. thats $ \binom{10} {2}$ then we can split the group of 8 into a group of 6 and 2. thats $ \binom{8} {2}$. then the group of 6 is $ \binom{6} {2}$. and so on. so its $ \binom{10} {2} \binom{8} {2} \binom{6} {2} \binom{4} {2}=45*28*15*6=\text{some \#}$[/hide]

[quote="illusiat"]3. The subtraction problem MATH-COU=NTS can be made true if each of the letters is replaced by a different digit and the letter groups are read as 4-digit and 3-digit integers. If C=3, how many possible values are there for U?[/quote]

Ok so put it like this

MATH
-COU
_____
  NTS

So, since NTS has no thousands digit, M must be one. Since C  is 3, A must be two to be less than 3. So,

12TH
-3OU
____
  NTS

O  must be 0, to make T-O=0, so that means that N is 9 because 12-3=9.

So U can be 4,5,6,7, or 8. so 5.

[quote="illusiat"]6. There are 6 distinct ordered pairs of nonzero digits (b,x) such that the sum of the digits of the number b^x is equal to the sum b+x. What it the sum of all of the b^x possibilities?

answers: 98, 945, 5, 0, 4.88, 777[/quote]

i have no idea, could someone help!!!

Did you use some logic while listing them?

Brute force doesn't seem too efficient.

I used a little bit of logic, and I also knew from mem. that 2^2=2+2. In addition, I knew that all 5*5*5...etc ended in 25 so I could easily find 5^2 and 5^3.
The rest was brute force. It isn't that bad, and without a calc. it only took 1 1/2 minutes.

[quote="abacadaea"]EDIT:
[hide="2"]an intuitive method. we can split them into a group of 2 and 8. thats $ \binom{10} {2}$ then we can split the group of 8 into a group of 6 and 2. thats $ \binom{8} {2}$. then the group of 6 is $ \binom{6} {2}$. and so on. so its $ \binom{10} {2} \binom{8} {2} \binom{6} {2} \binom{4} {2} = 45*28*15*6 = \text{some \#}$[/hide][/quote]

Ummm...
It's actually 9 people can be chosen to be partnered with the first person, then 7 for the next unpaired person, and so on...
So the answer is $ 9!! = 945$.


Without using Schroeder-Bernstein Theorem prove that $ [a,b] \sim (a,b) \sim [a,b)$

Based on the question title, it presumably is asking for a proof that an open, half-open and closed interval in $ \bf R$ have the same cardinality.  It's appeared before (maybe in the context of other things) -- the first idea is to go from closed to half-open.  You have to start by thinking about where to send the extra endpoint.


Here is a fun little problem I ran across. Maybe it is from a competition. I was just wondering the best way to go about it.

Please forgive me if it's not in the correct category.

I figured it has something to do with the gcd(360,270)=90, gcd(360,216)=72, gcd(216,270)=54.

I got 90 as the solution, then I though maybe 108, but I am not sure. Any hints?.

"The country of Big Wally has a railway which runs in a a loop 1080 km long. Three companies, A, B and C run trains on the track and

plan to build stations. Company A will build three stations, equally spaced at 360 km intervals. Company B will build four stations at 270

km intervals and Company C will build five stations at 216 km intervals.

The government tells them to space their stations so that the longest distance between consecutive stations is as small as possible. What

is this distance in kilometers?"

[hide]The answer is $ 54$.

Firstly, we note that the $ 54$ is valid; this is obvious from the problem description, since $ 270 \minus{} 216 \equal{} 54$.

Now, assume a smaller answer is valid. If this is the case, then the greatest common divisor would have to be less than 54 in order to yield a smaller answer, but no greatest common divisor is less than 54, so the answer is $ 54$.[/hide]

Thank you. I originally thought perhaps it was 54 because that is the smallest gcd of the three. But I then thought that was too small when I tried to draw it out. Thank you again. I just thought this was a fun little problem.

How can it be $ 54$? You can't get the longext distance to be less than $ 1080/12\equal{}90$.

For what it's worth. Someone ran this through MatLab, brute force, and arrived at 174.

A lot of these competition problems involve some sort of observation. This problem is supposed to be done without the aid of a calculator.

I am sure there is some clever trick or observation.

According to the Matlab output, place a B station 6 km from A and a C station 102 km from the same A.

I have attached a diagram of the loop.  I am not 100% confident with this, though.  When I try plotting the respective distances between

the stations, I end up with negatives in some instances and the distances between the stations do not add up correctly. I could be

mistaken.


superior algebra
superior algebra solved

Suppose $\phi: G\to G'$ is a homomorphism, where $G,G'$ are two groups. Can we say that \[\phi^{-1}(\phi(g)) = g\cdot\ker\phi\] for all $g\in G$?

I think it should be true. In fact, "$\supseteq$" is obvious, but can we prove "$\subseteq$"? I need some help. Thx.

Pick some $x \in \phi^{-1}(\phi(g))$. Then $\phi(x)=\phi(g)$, $\phi(x-g)=0$, $x-g \in Ker(\phi)$, and $x \in g+Ker(\phi)$.

Follows immediately from the definitions ...

$\phi^{-1}(\phi(\{g\})) = \{x \in G : \phi(x) \in \phi(\{g\})\}= \{x \in G : \phi(x) = \phi(g)\}$
$= \{x \in G : g^{-1}x \in \ker(\phi)\}= \{x \in G : x \in g \ker(\phi)\}= g \ker(\phi)$


Let $ X,Y,Z$ be points outside triangle $ ABC$ such that $ \angle BAZ\equal{}\angle CAY, \angle CBX\equal{}\angle ABZ, \angle ACY\equal{}\angle ABX$.

Prove that $ AX,BY,CZ$ are concurrent.

let angle BAZ = angle CAY = A
    angle ABZ = angle CBX = B
    angle BCX = angle ACY = C

by sine formula , we can easily get

(AB sinBAX) / (ACsinCAX) = BX / XC = sin C / sin B

therefore , sinBAX / sinCAX = (sinC / sin B) * ( AC/AB )

similarly ,   sinCBY / sinABY = (sinA / sin C) * ( AB/BC )
                 sinACZ / sinBCZ = (sinB / sin A) * ( BC/AC )

multiplying this three quation ,

(sinBAX / sinCAX)*(sinCBY / sinABY )*(sinACZ / sinBCZ ) = 1

By trigo. form of ceva's theorem , AX . BY . CZ are concurrent.

E. g. http://www.mathlinks.ro/Forum/viewtopic.php?t=88448

  darij


Q. What is the probability that a randomly chosen chord on the unit circle has length exceeding $\frac{\sqrt{3}}{2}$?

These three solutions are given:

Solution 1. Choose a point A on the circle. Inscribe an equilateral triangle ABC in the circle. Randomly select a point D on the circle, AD is the random chord. AD exceeds $\frac{\sqrt{3}}{2}$ if D lies on arc BC, so the probability is $\frac{1}{3}$.

Solution 2. Given a random point in the circle, there is exactly one chord that has that point as its midpoint. If the midpoint lies on the circle with the same center as the unit circle with radius $\frac{1}{2}$, the length of the corresponding chord is $\frac{\sqrt{3}}{2}$. Thus, if the midpoint lies inside this small circle the chord will exceed $\frac{\sqrt{3}}{2}$. The small circle has $\frac{1}{4}$ the area of the large circle, so the probability is $\frac{1}{4}$.

Solution 3. Since the circle is symmetric all around, let us examine a radius of the circle. Choose a random point on this radius, and call this point the midpoint of the random chord. Each point in the circle has exactly one chord that has that point as a midpoint. Note that the chord's length exceeds $\frac{\sqrt{3}}{2}$ if the point chosen is between the center of the circle and the midpoint of the radius. Thus, the probability is $\frac{1}{2}$.

Pretend you are the teacher and you are grading these three papers. Which paper would you give full credit to? What would you comment about the other two papers?

The solutions are all correct, [b]given differing methods of randomly choosing a chord.[/b]

A visual plot of the expected distribution of chord midpoints reveals a different distribution for each solution.  The problem is unclear about precisely how a 'random chord' should be selected.

:)

Solution 1: This one I like the best because it seems to choose all chords with equal probabilities because it selects the two endpoints.

Solution 2: This one uses midpoints. However, in this solution, you aren't randomly choosing a chord, but it's midpoint. If we look, we find that as we go outward, the lengths become more likely to be chosen due to the fact that there are more midpoints that generate chords with those lengths.

Solution 3: This one i like the least. Trying to generate random points by choosing a radius and then a point on that radius immediately gives us problems. It chooses points in the center more often than the others! (Try it out. The points get more crowded as you move in.) Therefore, it is biased towards the longer lengths and so I'd grade this one the lowest.

It is Betrand's paradox. ;)

http://en.wikipedia.org/wiki/Bertrand's_paradox_(probability)


geometry
vector
geometric transformation
rotation
parallelogram
geometry unsolved

Equilateral triangles $ ABM,CDP$ are drawn outside a convex quadrilateral $ ABCD$, and equilateral triangles $ BCN,DAQ$ are drawn inside $ ABCD$. If no three of points $ M,N,P,Q$ lie on a line, prove that $ MNPQ$ is a parallelogram.

Do you know how to use complex number to solve geomertry problem ?
I will be very easy if you know that .
Assign A(z1) , B(z2)  , C(z3) , D(z4) .

Let w = cos(pi/3) + i*(pi/3) .

Then M = (z1 - z2)*w + z2
        P  = (z3 - z4)*w + z4
        N  = (z3 - z2)*w + z2
        Q  = (z1 - z4)*w + z4

MN = M - N = (z1 - z3)*w = Q - P = QP
This is similar to the case that vector MN = vector QP
Thats complete the prove.

Yes, easy indeed  :D

What about a synthetic proof?

The idea is Rotation (here$ R_{(O)}^{\alpha},\alpha=60^{\circ}$,and $ O$ as the center),one of the keys of solving these kinds of questions.Be aware that we consider the angles in counterclockwise.$ R_{(A)}^{60^{\circ}}(ADB)=(AQM) (1)$.thus  $ R_{(C)}^{60^{\circ}}(BCD)=(NCP) (2)$.thus $ (1),(2) \Longrightarrow BD=MQ=PN (3)$
$ (1),(2) \Longrightarrow$ angle between$ DB,MQ$=angle between$ DB,PN=60^{\circ},\Longrightarrow MQ \parallel PN (4)$.thus $ (3),(4) \Longrightarrow MQPN$ is parallelogram,and it's done!


Inside square $ ABCD$ consider a point $ M$. Prove that the points of intersection of the medians of triangles $ ABM,BCM,CDM$ and $ DAM$ form a square.

true by an homotethy with cnter M and ratio 2/3


Given $ n\equal{}\{1, 2, 3, 4, 5, 6\}$, find the product of the six possible fractions of the form $ \frac{2}{n}$. Express your answer as a common fraction.

We need $ \frac{2}{1}\cdot\frac{2}{2}\cdot\frac{2}{3}\cdot\frac{2}{4}\cdot\frac{2}{5}\cdot\frac{2}{6}$.  Second term is 1, the first and fourth terms negate each other, so we're left with $ \frac{2}{3}\cdot\frac{2}{5}\cdot\frac{2}{6}\equal{}\frac{4}{45}$


function
real analysis
real analysis unsolved

Let $U,V$ be (open or closed) intervals in $R$, and let $f: U \implies V$ be a function which is strictly increasing (ie, if $x,y \in U$ and $x<y$ then $f(x) < f(y)$) and onto. Prove that $f$ and $f^{-1}$ are continuous.

I first fix $\epsilon > 0$ (don't you do that in every problem? ...)
I must find some $\delta$ such that if $d(x, x_{0})<\delta$, $x,x_{0}\in U$, then $d(f(x),f(x_{0})) < \epsilon$. But this method seems too tedious.

There is a nice theorem in my text that says:

Let $E$, $E'$ be metric spaces and $f: E \implies E'$ a function. THen $f$ is continuous if and only if, for every open subset $U$ of $E'$, the inverse image $f^{-1}(U) = \{p \in E: f(p) \in U\}$ is an open subset of $E$.

Perhaps I can use this theorem for this problem?

Help is appreciated.

This will not be very hard (maybe tedious, right), but you will have to do with the local definition of continuity, since globally, there's really not much you can know.

julien_santini, can you be a little more detailed? Does my methods work, or am I completely on the wrong track?


Could anybody give me the definition of surjective function bijective function ,please?

Lol that's easy. check in a book :rotfl:

Now I suggest you, read "techniques in combinatoric" some chineses' book first chapter fifth section. But if you want I to say you, if m=n then    f(m)=f(n) then it is surjective verse is bijective.
Thanx.


geometry
3D geometry
combinatorics unsolved
combinatorics

Let $ S$ be a set of $ 43$ positive integers not exceeding $ 100$. For each subset $ X$ of $ S$, denote by $ t_X$ the product of elements of $ X$. Prove that there exist two disjoint subsets $ A, B$ of $ S$ such that $ t_At^2_B$ is a cube of a natural number.


geometry
geometric transformation
rotation

[quote="ch1n353ch3s54a1l"][quote="DaCraze"]What angle is formed by a clock at 4:15?[/quote]

[hide]Hour hand: each hour that passes by is 360/12=30 degrees. For each minute that passes by, it's 30/60=.5 degrees. So the degree formed by the hour hand is $30\times 4 + 15\times .5=127.5$.
Minute hand: Each minute that passes by is 360/60=6 degrees. So the degree formed by the minute hand is $6\times 15=90$.
The degree formed by the hour and minute hand is $127.5-90=\boxed{37.5}$ degrees. [/hide][/quote]

Im pretty sure this is wrong it dosent make sence at all

...ok...fine.
There are 360 degrees in an angle. There are twelve hours in a full rotation. Therefore each hour is 360/12=30 degrees for the hour hand. Since there are 60 minutes in an hour, the thirty degrees in one hour is divided by 60 so that each minute is 30/60=1/2 degree for the hour hand.
There are 60 minutes in a full rotation, so each minute is 360/60=6 degrees.

It might be helpful to be more specific about your question in the topic instead of putting just 'question'.

i agree with ch1n353ch3s54a1l, since a clock's hour hand doesnt stand still while the minute hand moves 360 degrees, then just jumps forward an hour. In short, at 4:15, the hour hand isnt exactly on the 4, but between 4 and 5.


I see most AoPS users have GMAIL accounts, and I would like to get one.  Unfortunately, an invitation is needed.  Several people have told me their friends sent them an invite.  Is that possible?  If so, could someone please send me an invite?  My email is mysmartmouth@hotmail.com as of right now.

Thanks,

Just in case someone cares, there are lots of sites in the web that give directly invitations (and you don't have to do a post here!). If you speak spanish, well, googlemania.com is a good place to start, but if not, look in google itself, probably you will find something.


Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0.$  Let $m/n$ be the probability that $\sqrt{2+\sqrt{3}}\le |v+w|,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide]The solutions of the equation $z^{1997}=1$ are the 1997th roots of unity and are equal to $\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)$ for $k=0,1,\ldots,1996.$  They are also located at the vertices of a regular 1997-gon that is centered at the origin in the complex plane.

WLOG, let $v=1.$  Then

\begin{eqnarray*} |v+w|^2 &=& |\cos\left(\frac{2\pi k}{1997}\right)+i\sin\left(\frac{2\pi k}{1997}\right)+1|^2\\ &=& \left|\left[\cos\left(\frac{2\pi k}{1997}\right)+1\right]+i\sin\left(\frac{2\pi k}{1997}\right)\right|^2\\ &=& \cos^2\left(\frac{2\pi k}{1997}\right)+2\cos\left(\frac{2\pi k}{1997}\right)+1+\sin^2\left(\frac{2\pi k}{1997}\right)\\ &=& 2+2\cos\left(\frac{2\pi k}{1997}\right) \end{eqnarray*}

We want $|v+w|^2\ge 2+\sqrt{3}.$  From what we just obtained, this is equivalent to $\cos\left(\frac{2\pi k}{1997}\right)\ge \frac{\sqrt{3}}2.$  This occurs when $\frac{\pi}6\ge \frac{2\pi k}{1997}\ge -\frac{\pi}6$ which is satisfied by $k=166,165,\ldots,-165,-166$ (we don't include 0 because that corresponds to $v$).  So out of the 1996 possible $k$, 332 work.  Thus, $m/n=332/1996=83/499.$  So our answer is $83+499=582.$[/hide]

Choosing two roots just means that you're choosing a configuration.  So it doesn't matter what one of the roots is, it only matters what the position of the other root is in relation to it.

Well, you could consider every single case (i.e. consider $v$ if it is the root of unity corresponding to $k=1,2,\ldots 1996).$  Doing so would give you an equal probability (by symmetry).  So then you would have that the probability of picking $k=0$ was $\frac 1{1997}$, the probability of $k=1$ was $\frac 1{1997}$ and so forth.  So we would have $1997\frac 1{1997}P+1997\frac 1{1997}P+\cdots+1997\frac 1{1997}P$ (1997 of these total) where $P$ is the probability that $|v+w|\ge\sqrt{2+\sqrt{3}}$ for each of the values of $v$ picked.  It would just break down to $P$, as you can see.

That's why I said WLOG (without loss of generality...we weren't losing generality because each vertex is, for the sake of this problem, indistinguishable from each other and we could therefore just pick one of them to be $v$).

That might have been a bit overkill on the explanation...

Yes, you could also say that if you pick another point, you can just rotate it to make it 1 (this is basically another formulation of saying it is symmetric).

Well, the idea is that if we replace $v$ and $w$ with $e^{it}v$ and $e^{it}w$, then the modulus of their sum is $|e^{it}v+e^{it}w|=|e^{it}(v+w)|=|e^{it}||v+w|=|v+w|$ (e.g, the modulus is an invariant, in engel-speak).  So we can pick $t$ such that $e^{it}v=1$.

[quote="Elemennop"]Why can you assume $v=1$ ? Can't both $v$ and $w$ be different ones that aren't equal to $1$?

edit> Can any other points you can just rotate one of them around until one of them is $1$ ? Is that why you can assume $v=1$?[/quote]

Looking at the complex plane graphically, $|v - w|$ depends only on the placement of $v$ in relation to $w$ - the $||$ gives you a length, so it does not depend on the particular root you use for $v$.  Yes, you can just rotate so that $v = 1$ for any pair that you choose, which doesn't change the problem.

How come when you square both sides, you don't have to consider the negative square root in your final answer?

i.e. Can't $ \sqrt{2\plus{}\sqrt{3}} \le |x|$ also imply the negative value?

No.  The square root function is defined to pick only the positive root.

Hmm here's what I got...

Let the two roots be $ cis\left(\frac{2n\pi}{1997}\right)\equal{}cis(x),cis\left(\frac{2m\pi}{1997}\right)\equal{}cis(y)$ with $ 0\le n,m\le 1996$ and $ n>m$.

Then computing the given expression we get (after using the pythagorean identity and cosine subtraction formula)

$ \cos(x\minus{}y)\ge\cos(\pi/6)\implies\minus{}\pi/6\le 0\le 2(n\minus{}m)\pi/1997\le \pi/6\implies 1\le n\minus{}m\le166\le1997/12$

For $ m\equal{}0,n\in\{1,2,...,166\}$ (166 elements)

$ m\equal{}1,n\in\{2,3,...,167\}$ (166 elements)

$ ...$

$ m\equal{}1830,n\in\{1831,1832,...,1996\}$ (166 elements)

$ m\equal{}1831,n\in\{1832,1833,...,1996\}$ (165 elements)

$ ...$

$ m\equal{}1995,n\in\{1996\}$ (1 element)

So the total number of valid pairs is $ 1831\cdot 166\plus{}1\plus{}2\plus{}3\plus{}...\plus{}165\equal{}1831\cdot 166\plus{}165\cdot166/2$

This is out of $ \binom{1997}{2}\equal{}1997\cdot1996/2$

So the answer is $ \frac{2\cdot1831\cdot166\plus{}165\cdot166}{1997\cdot1996}$.

i don't feel like computing this monstrous sum by hand so i used a calculator but the answer it gives does not agree with the official one... :( (it is pretty close, though, only by like 0.0xxxx or something)... what's wrong with my solution???

Urgh when I did this I included zero, same as the first solution, but i still dont understand why you dont include zero.

[quote="joml88"]Let $v$ and $w$ be [u][b]distinct[/b][/u], randomly chosen roots of the equation $z^{1997}-1=0.$  Let $m/n$ be the probability that $\sqrt{2+\sqrt{3}}\le |v+w|,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$[/quote]

That's why.


inequalities
parameterization
algebra proposed
algebra

For a positive constant $ a$, solve the following inequality in $ x$.

$ x^3 \plus{} \left(a \plus{} \frac {1}{a} \plus{} 1\right)x < \left(a \plus{} \frac {1}{a} \plus{} 1\right)x^2 \plus{} 1.$

Let $ b\equal{}a\plus{}\frac 1a$, then $ (x\minus{}1)(x^2\minus{}bx\plus{}1)<0$.


algebra
polynomial
Rational Root Theorem

Hi everyone,
  I've spent a fair amount of time trying to figure out the solutions to a few math proof questions on my textbook, but still can't quite solve it. Could anyone please help me?
  1.) Prove that there is no largest negative real number.
  2.) Prove that Sq.root(2)+Sq.root(3) is irrational
  3.) Find a simplifying expression for the product  [1-1/(2^2)]*[1-1/(3^2)].....[1-1/(n^2)] and verify its validity for all integers n greater or equal to 2.
  Thank you so much.
  Tracy.

   My thoughts on them

  For Q1, I reasoned that by the equation -1*10^(-n), the greater the value of n, the larger the -ve value. Since n increases indefinately, there is no single largest -ve real number.
For Q2, I tried to prove by contradiction, assuming first that the sum is a rational number in the form of P/Q where P,Q are integers without a common factor.
For Q3, I realized that the simplified expression is (n+1)/n when n is greater or equal to 4, but for 2 or 3, the expression just won't work.

1) Take any negative real number $k$.  Then $k/2>k$, and $k/2$ is also negative.

2) Use the rational root theorem.  $\sqrt{2}+\sqrt{3}$ is a root of $(x^2-5)^2-24=x^4+\ldots+1$.  By the rational root theorem, the only possible rational roots are $\pm 1$.  But neither of these work (check it)

3) You are incorrect.  You're looking for $(n+1)/2n$, $n\geq 2$.

[quote="blahblahblah"]  By the rational root theorem, the only possible rational roots are $\pm 1$.  [/quote]

isnt it more that the only integer solutions can be $\pm1$ ...

but anyways it is obvious that $\sqrt 6$ is irrational (or u can show it the same way as for $\sqrt 2$). then $(\sqrt 2 +\sqrt 3)^2 = 5+2\sqrt 6$ which is irrational as well.  since rationals are closed on squaring $\sqrt 2 +\sqrt 3$ is irrational ...


geometry
3D geometry
sphere
calculus
integration

Let B be a sphere with radius 1, centered at the origin, in R3.  Let S be a square in the x=1 plane with corners at (1,0,0), (1,1,0), (1,1,1), and (1,0,1).  Let M be a mapping from S to B that takes the point (1, y, z) to the point of intersection between B and the line that passes through (1, y, z) and the origin.  What is the area of  the image of S (under M)?

--Dan

Well, I know what the region looks like ... if only I remembered the integral for surface area

Actually, I think I can do it without calculus ... I think.

[hide]Consider the cube with vertices at ( :pm:1, :pm:1, :pm:1).  Each face of the cube is composed of 4 squares in positions equivalent to that of the square we are considering.  If we map the entire cube onto the sphere with the mapping we're using (project towards the center) , we get the entire sphere.  Additionally, there is no overlap.  Thus, each of these 24 squares makes up exactly 1/24th of the surface area of the sphere, or :pi:/6.[/hide]


vector
linear algebra
linear algebra unsolved

Could $(\mathbb R,+)$ be a vector space over the field $(\mathbb C,+\times)$ with a scalar product.

Just check the rules for vector spaces. It would be hard to tell what was going on because the vectors would look the same as the scalars, but there's no problem.

[quote="The Captain"]No. Cos that would not be closed under multiplication.
(Am I missing something ?)[/quote]

To The Captain : Your proof is wrong. The scalar product is not the ordinary product you have.

$(\mathbb{C},+)$ and $(\mathbb{R},+)$ are two isomorphic additive groups (consider them as vector spaces over $\mathbb{Q}$). Now the vector space structure of $(\mathbb{C},+)$ over $(\mathbb{C},+,\times)$ can be transported to $(\mathbb{R},+)$.


I've passed all of the topics except for Mastery of Counting Fundamentals Level 5. I can't seem to get a problem in that subject, so does Alcumus end for me here? Also, there seems to be a subject called 'Principle of Inclusion-Exclusion', which can't generate problems also. But that subject doesn't appear on the report list, so are these glitches?

I have the exact same problem as you.
Except for the fact that I have passed "Mastery of Counting Fundamentals Level 5"
And I am not sure whether the developers just haven't put in any new questions for "Principle of Inclusion-Exclusion" or there is a glitch as you mentioned.

Glitch has been fixed now. Thanks. All old gliches from Alcumus have also been fixed. Lock, please.


Let [tex]ABC[/tex] be a triangle with incenter [tex]I[/tex].  Let the incenter of the triangle touch sides [tex]BC[/tex], [tex]CA[/tex], and [tex]AB[/tex] at points [tex]A_1, B_1, C_1[/tex], respectively. Segments [tex]AA_1, BB_1[/tex] meet at point [tex]G[/tex]. Let circle [tex]\omega_A[/tex] be the circle with center [tex]A[/tex] and radius [tex]AB_1[/tex]. Define circles [tex]\omega_B, \omega_C[/tex] similarly. Now, circles [tex]\omega_A, \omega_B, \omega_C[/tex] are externally tangent to circle [tex]\omega_1[/tex], while they are internally tangent to circle [tex]\omega_2[/tex]. Let [tex]O_1[/tex] and [tex]O_2[/tex] be the centers of [tex]\omega_1[/tex] and [tex]\omega_2[/tex], respectively. Also, let lines [tex]A_1 B_1[/tex] and [tex]AB[/tex] meet at point [tex]C_2[/tex], and let lines [tex]A_1 C_1[/tex] and [tex]AC[/tex] meet at point [tex]B_2[/tex]. Prove that points [tex]I[/tex], [tex]G[/tex], [tex]O_1[/tex], and [tex]O_2[/tex] lie on a common line which is perpendicular to line [tex]B_2 C_2[/tex].

This is just the well-known theorom that the Soddy Line (the line passing through the incentre, Gergonne point, inner Soddy centre, outer Soddy centre and de Long Champs point) is perpendicular to the Gergonne line (perspectrix of ABC with its contact triangle). I think you can prove this using the Sondat Theorom. Look here for more information: http://mathworld.wolfram.com/SoddyLine.html


calculus
integration
analytic geometry
vector
calculus computations

I just gave an exam today in a Calculus III class. Here's one question.

Compute $ \int_0^{\infty}\int_x^{\infty}\frac1{(x^2\plus{}y^2\plus{}1)^2}\,dy\,dx$

(If you know your typical "mainstream" American calculus books, you should be able to spot which chapter this came from. The exam covered the whole chapter - double and triple integrals, coordinate systems, and changes of variable.)

I was asked exactly this question the first time I was tutoring Calc III and it definitely threw me off since I hadn't looked at which topics they had covered that week. Once I realized that it was switching to polar it wasn't bad at all.

[quote="HilbertThm90"]Once I realized that it was switching to polar it wasn't bad at all.[/quote]
Yes, that's the whole point. 1605272 gets full credit (as do quite a few of the people in my class, but then I have some grading to do.)

[quote="Kent Merryfield"](If you know your typical "mainstream" American calculus books, you should be able to spot which chapter this came from. The exam covered the whole chapter - double and triple integrals, coordinate systems, and changes of variable.)[/quote]
Which book are you using?  I'll just venture a guess and go with Marsden and Tromba?  My favorite change of variable problem is $ \int_{\minus{}\infty}^{\infty}e^{\minus{}x^{2}}\,dx$.  You should give them this problem, with the hint of turning it into a double integral.

Mardsen and Tromba is not what I meant when I talked about "mainstream" calculus books. I was talking about a book that you use for a three semester sequence, starting with Calculus I, and for which the last chapters, the ones done in the third semester, cover mutivariable and vector topics. It could be Stewart, Edwards & Penny, Larson, ... What we're actually using is the "shrunken" Stewart: [i]Essential Calculus.[/i]

I assigned the integral of the Gaussian as a homework problem; it's in the exercises.

Slightly more than 20% of my students got full credit on the test question, with several others just making silly little mistakes.

Wouldn't three semesters make them trimesters? :P  At my school, Calc I and II are usually done separately from multivariable calculus.  We use completely different books: Calc I and II for humanities/pre-med uses one book (not sure which), Calc I and II for engineers uses one of the Stewart books (not the big one), Honors Calculus uses Spivak, Calc III uses Marsden and Tromba, and Honors Multivariable Calculus uses Colley.  So my school doesn't really do the same sequence as yours.


we all know formulae of  $ \sum_{k\equal{}1}^n k,\sum_{k\equal{}1}^n k^2$ .
How to computer $ \sum_{k\equal{}1}^n k^i , i\equal{}3,4,5,6...$ ?
Thanks

Posted many times before, for instance here:

http://www.mathlinks.ro/viewtopic.php?t=204509

Paper mentioned there is now available here:

http://web.archive.org/web/20060901095523/http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf


I am not sure if I should post this here, because I am already in college and am not applying to one. I am in my freshmen year, doing a bachelor's in Business Administration with concentration in Accounting but I am thinking about getting either a Bachlor's Degree and Econonmics or getting an MBA in the same 5 year program which is oferred in my University. I was wondering what is better, these 2 bachelor's degrees or a bachelor's and an MBA?

The MBA would open up more doors for you than a second undergraduate degree. Note that many of the best MBA programs require real-world work experience after college as an entrance requirement, so an MBA program that lets go straight in from studying your undergraduate degree may not be considered as strong an MBA as those other programs.

Really? In my college I cen get my bachelor's and MBA at the same time when graduating after 5 years. So then do you think that the MBA is better in this case, or the second bachelor's? Also, I spoke to the department head of Economics and he told me that I can go to a ph.d program in Economics after getting a Bachelor's without needing a master's degree. Is it like that everywhere?

I did not really understand what you meant blahblah.

What I mean is that increasing numbers of people seem to be jumping straight from their Bachelor's to their PhDs.  Of course, this is mostly in the pure sciences, as master's seem to be far more prevalent than doctorates in business admin.

It is correct that in many fields (and economics is one of them), the usual degree program to enter after an undergraduate degree is a Ph.D. program, with no stopping to get a master's degree along the way. That is especially true of mathematics.

that hapepned to my parents... my dad once told me that he did two bachelors and he regret not taking a chance to do MBA, but that was partly because 1. he was in korea, and they have a weak scholarship program 2. he was partl incapable.

So therefore, which do you think is better 2 bachelor's or a bachelor's and MBA?

It is well know that a bachlor and a MBA is better than two bachelor, anyways there is a site that is talking about it and the degree offered today and their priority.
james

What does it have to do with an online degree. Also, what if the 2 degrees are earned at the same time? Like a bachelor's and MBA, tolkanadult, you are saying the degree is worse because of this?

I was wondering if you can go to a phd program in economics if you have an MBA?

I would say bachelors and MBA, because it shows that you know advanced matierial, and can deal ith harder stuff.  But I don't know that much, so I may be wrong.

[quote="hello"]Also, what if the 2 degrees are earned at the same time? Like a bachelor's and MBA, tolkanadult, you are saying the degree is worse because of this?[/quote]

It may be that an MBA will be less valuable if it is easy to get into that MBA program. Some MBA degrees get part of their value, it appears, by requiring work experience in the business world as an entrance requirement, and also by requiring high entrance test scores and high grades in an undergraduate degree program beforehand. But you would have to ask people locally and possibly in other places (by email) what they think about the reputation of the programs you are considering.

To answer your other question, yes, it is possible to enter a Ph.D. program in economics after gaining an MBA degree. If a Ph.D. in economics is really your goal, you may want to go directly into that kind of program and skip the MBA.

One of my regrets was to enter the MBA program right after college. Without prior work experience I could hardly relate to, or learn from, the many lectures on real life business scenarios. Apparently the business schools recognized that common problem. Soon after my time most, if not all, credible MBA programs started to require (or very highly recommend) 2 or more years of work experience. A quick check on the Internet yields the following information:

Harvard: Average work experience of MBA students is 4 years.

U of Penn: Required essay "Please describe your career progress to date".

UCLA: "Candidate should bring a set of skills and show clear potential for managerial leadership", and employment history is requested.

USC Self Evaluation question: "Are you ready to be a student again?"

The above programs should know how to keep up with the current business trend.

Well I want to get an accounting degree and economics. But if I just get a BA in business administration, will I then be able to enter a phd program an econ because I dont know if I will try to get an econ degree?

The obsolescence of master's degrees is a relatively recent phenomenon (one generation or so).  It used to be very common to go through the master's as prerequisite en route to a PhD.  Now, the distinction is that Master's are unfunded, and unfundable, graduate school degrees, whereas PhD programs offer at least the possibility of financial support.  Master's programs are generally money-making enterprises or sources of cheap teaching labor, at the universities that offer them.  Most PhD programs also offer a token master's degree upon fulfillment of doctoral candidacy requirements, for students who may opt not to finish the thesis.

[quote="tokenadult"]To answer your other question, yes, it is possible to enter a Ph.D. program in economics after gaining an MBA degree. If a Ph.D. in economics is really your goal, you may want to go directly into that kind of program and skip the MBA.[/quote]

Some schools have combined Ph.D + MBA programs, where the doctorate is usually in economics, organizational behavior or other business-related subject.

[quote="fleeting_guest"] Now, the distinction is that Master's are unfunded, and unfundable, graduate school degrees, whereas PhD programs offer at least the possibility of financial support.  Master's programs are generally money-making enterprises or sources of cheap teaching labor, at the universities that offer them. [/quote]

This is true with the exception of the new class of Professional Science Master's degrees.  More information about them is here: http://www.sciencemasters.com/ .

Thanks to fleeting_guest and to smbelcas for describing some of the various possibilities open to the person who opened this thread.

Yes, thanks to everyone who contributed. So I am wondering, even though I am getting off topic by a lot, is it likely to become a CFO after being an accountant (later getting an MBA or something like that) or do you have to be a finance major?

Also, if you are an accountant, is it possible to have any jobs more on the financial side? Wall Street related jobs?

I cant major in every subject so thats why I am asking if I study one thing, I can also work in a somehow related field.

Hello! It has been my experience that once you have your degree(s) and your first job, outside academia, that it is the quality of your work that will cause you to advance much more than the school and degrees that you have.

That said, the MBA seems to be the new financial services union card. Practically everyone we know who is an employee has recently acquired or is in the process of receiving the MBA. Look at all the Exec MBA programs out there. They may build on experience, but the students are there to punch their cards.

If you do not have a leaning towards teaching, I'd go for the combined BBA/MBA. Since they would be from the same school, you would not need to worry that there would be a lot of boring overlap. You are interested in CFO -just take a good number of finance & econ classes within the BBA program.

And read the IBD!

I got only a few oissues of the IBD, I read some of the Wall Street Journal as much as I can. Is IBD the best magazine for investing or what? I want to read some investing books (not complicated, since I have very little background) can you recommend any? Also, should I read investing blogs, if I can even understand them?

I know TONS about this but a MBA can sometimes be useless (from a career advancement perspective)... it all depends on what you want to do after you graduate

what do you want to do after graduation, career wise?

if it's investment banking I would apply for the jobs first. If you get accepted (very hard to do) for one of them TAKE IT (if this is ur thing). A lot of these places will sponsor you (pay you) to get your MBA. However, if you are not sure what you want to do I'd definitely rec. the MBA (at a good school) as it will make you look a lot better than other candidates

If you want to advance a lot in say the accounting career? Become a controller or above  I think an MBA is important.


Define a sequence $ \{a_n\}_{n=1}^{\infty}$ as following: $ a_1= \frac{1}{2}$ and
$ a_{n+1}=\frac{a_n ^2}{a_n^2- a_n +1}$ for all positive integer $ n$ . Prove that
$ a_1 + a_2+ a_3+ ...+ a_n <1$ for every positive integer $ n$


:blush:  :blush:  :blush:  use THE  IAG  withe
(b+c)(c+d)(d+a)(a+b) sup ou -- 16abcd

I think a,b,c,d positive real numbers (else it is not correct), and you needed to prove
$(a+b)(b+c)(c+d)(d+a)\ge 16abcd.$
It is obviosly $a+b\ge 2a^{1/2}b^{1/2},b+c\ge 2b^{1/2}c^{1/2},...$
Multiplyed it give needed result.


The sum of three different positive unit fractions is 6/7.  What is the least number that could be the sum of the denominators of these fractions?

It has to be this one that all of us miss... but no one answers it...

i was... it was about the problem... although it didn't have a solution.

anyways... since you want something...

i might be going in the wrong direction ... but what the heck...

6xyz = 7(xy + yz + xz) ???

will this get me anywhere?

Probably not. Algebra might not be the easiest way here.. a little bit of casework should get you the answer. Don't forget each fraction must be 1/a positive integer.. hint: [hide]can all three of those be 1/4 or smaller?[/hide]

OH... Ok... now i get it.. you all three can't be 1/4 or smaller was a big hint. so... then after i read the hint... i tried 1/2... got 5/12...

then one of the 2 numbers left had to be greater than 1/6... so i tried 1/3... and 5/14 - 1/3 = 1/42 yay! 2+3+42 = 47.

Thanks!

Careful.  The 5/14 is supposed to read 5/12.  So, 5/12-1/3=1/12 :arrow: 2+3+12=17.  We all make math miscalculations getting total different answers sometimes.

Actually, frodo, it's the other way around -- the 5/12 should have read 5/14.

[hide]I am using numbers which work easily for these fractions(even thought all the fractions are equal) from 6/7 we can't do anything except for 1/1, which is too big and 1/7 which is not enough. from 12/14, we can only do 1/2 or 1/7 and 1/14. so it doesn't work; neither does 18/21. for 24/28, we can use 1/2, 1/4 and 1/7 1/28, but they add up to 25/28 which is too much. for 30/35, we can use 1/7, 1/5 and ,1/35 not enough numbers. for 36/42 we cam use: 1/2,1/3,1/6,1/7,1/14, 1/21 and 1/42. we have to use three numbers, one must be 1/2, leaving 15/42, which must be two number and only combitntion that works is 1/2,1/42 and 1/3. adding up to 47.[/hide]


trigonometry
vector
inequalities
function
inequalities solved

Let $ A$, $ B$, $ C$ be the angles of a triangle $ ABC$. Prove that for any real numbers $ x$, $ y$ and $ z$, we have

$ x^{2}\plus{}y^{2}\plus{}z^{2}\geq 2yz\cos A\plus{}2zx\cos B\plus{}2xy\cos C$.

When does equality occur?

This one can be prove by vector:(vector(y)+vector(x)-vector(z)) \geq 0 where vector is put on sides of triangle ABC.
The equality is x,y,z are sides of a triangle have angles:A,B,C :)

Nice solution, but I think you meant

(vector(x)+vector(y)+vector(z))^2 \geq 0

instead of (vector(y)+vector(x)-vector(z)) \geq 0.

[b]EDIT:[/b] This problem has also been discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 .

  Darij

I prove inequality $(*)\ \ \sum x^2\ -\ \sum 2yz\cos A\ \ge \ 0$, where $x,y,z>0$.

For the function $f(x)=x^2-2(y\cos C+z\cos B)x+y^2+z^2-2yz\cos A\ge 0$

we have $\triangle '=(y\cos C+z\cos B)^2\ -y^2\ -\ z^2\ +\ 2yz\cos A\ =$

$-y^2\sin^2 C-z^2\sin^2 B+2yz(\cos B\cos C+\cos A)=$

$-y^2\sin^2 C-z^2\sin^2 B+2yz\sin B\sin C=-(y\sin C-z\sin B)^2\le 0$.

Thus inequality (*) is true and we have equality $\Longleftrightarrow \frac{\sin A}{x}=\frac{\sin B}{y}=\frac{sin C}{z}$.

By the substitutions $x=au,\ y=bv,\ z=cw$ where $u,\ v,\ w$ are positive,

the inequality (*) becomes $(u-v)(u-w)a^2+(v-w)(v-u)b^2+(w-u)(w-v)c^2\ge 0$.

The last inequality is equivalently with $\sum (y\cos C+z\cos B-x)^2\ge 0.$

You use the relation det $\left(\begin{array}{ccc} -1&\cos C&\cos B\\\\cos C&-1&\cos A\\\\cos B&\cos A&-1\end{array}\right)=0$.

for to find the condition of equality $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}.$


inequalities
Gauss
algebra
polynomial
inequalities proposed

[quote="silver1989"]Let a,b,c,k$\geq 0$.Prove that:
  $\frac{1}{(ka+b)^{2}}+\frac{1}{(kb+c)^{2}}+\frac{1}{(kc+a)^{2}}\geq \frac{9}{4(k+1)^{2}(ab+bc+ca)}$[/quote]

 No!  Sorry ! I thought I had proved it , but no! Last moment I saw that the given inequality is different , than what I proved. I do not cancel the post however , because somebody else would probably see a better idea on this way.

 We have ( BCS- Englel -Andreescu - ...  form) :

$\frac{1^{2}}{(ka+b)^{2}}+\frac{1^{2}}{(kb+c)^{2}}+\frac{1^{2}}{(kc+a)^{2}}\geq \frac{(1+1+1)^{2}}{(kb+c)^{2}+(kb+c)^{2}+(kc+a)^{2}}\geq \frac{9}{(k+1)^{2}(a^{2}+b^{2}+c^{2})}$,

 because after the simplifications in the denominator and grouping , we can use the fact that:

  $ab+bc+ca \leq a^{2}+b^{2}+c^{2}$.
 But I'm afraid this method does not help. I,ll try again later :( .

[u] Babis[/u]

[quote="stergiu"]

$\frac{(1+1+1)^{2}}{(kb+c)^{2}+(kb+c)^{2}+(kc+a)^{2}}\geq \frac{9}{(k+1)^{2}(ab+bc+ca)}$

[/quote]

You may inspect the inequality above. It's wrong.

Mmm... I still hope that there is
$\frac{1}{(ka+b)^{2}}+\frac{1}{(kb+c)^{2}}+\frac{1}{(kc+a)^{2}}\geqslant\frac{9}{(1+k)^{2}(ab+bc+ca)}$
silver1989 might typed it wrongly or the one above is still a conjecture. Take $k=0$ it's easy and take $k=1$ it's a famous problem of Iran 1996.

right. Take $k=2$ it's the inequality in
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=141336[/url]

[quote="silver1989"]Yes!This problem is right.It was deduced  from IRAN 96[/quote]
You say
$\frac{1}{(ka+b)^{2}}+\frac{1}{(kb+c)^{2}}+\frac{1}{(kc+a)^{2}}\geqslant\frac{9}{(k+1)^{2}(ab+bc+ca)}$
is right? So how to prove it?

for k = 1, the 'iran96' inequality is very famous.
for k = 2, i solved here http://www.mathlinks.ro/Forum/viewtopic.php?t=141336

i think that this result isn't true for big k, but i'm not certainly about it..

i will try to find one counter-example or one solution.. later i come back!

This is my solution:
 First,from IRAN 96:
   $\sum\frac{1}{(a+b)^{2}}\geq \frac{9}{4\sum ab}$(1)
 Now.let a+b=x
             b+c=y
             c+a=z
 (1)$\Leftrightarrow \sum(x-y)^{2}(\frac{2}{xy}-\frac{1}{z^{2}}) \geq 0$(2)
  We need solve:
    $\sum\frac{1}{(ka+b)^{2}}\geq \frac{9}{4(k+1)^{2}\sum ab}$(3)
To solve this problem,let ka+b=x
                                       kb+c=y
                                       kc+a=z
After the transformation,You can obtain:
(3)$\Leftrightarrow f(k)\sum(x-y)^{2}(\frac{k^{2}+1}{xy}-\frac{k}{z^{2}})+g(k)(\sum\frac{xy}{z^{2}}-3) \geq 0$($f(k),g(k) \geq 0$,you can check it)(4)
 From $(2) and k^{2}+1\geq 2k$,We can prove (4) easily

But you know
for Iran's problem we set
$x=a+b, y=b+c, z=c+a$
so $x,y,z$ can be the lengths of three sides of a triangle (when we prove it, it's neccessary to use this condition).

but when you set
$x=ka+b, y=kb+c, z=kc+a$
they might not be anymore!

[quote="shalex"]
$\frac{1}{(ka+b)^{2}}+\frac{1}{(kb+c)^{2}}+\frac{1}{(kc+a)^{2}}\geqslant\frac{9}{(1+k)^{2}(ab+bc+ca)}$
[/quote]
Yes that's correct for all $k\ge0$ if $a,b,c$ form a triangle. Wlog assume $k\ge1$ (otherwise replace $k$ with $\frac1k$). Because the substitution $x+y=ka+b$ etc. as suggested by Little Gauss [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=799699#799699]in the other thread ($k=2$)[/url] yields $x=\frac{(k+1)a+(k-1)(b-c)}2$ etc, which are positive for $\frac{k+1}{k-1}a+b\ge c$ etc. So if $a+b\ge c$ etc., we are assured that $x,y,z\ge0$ and thus we can apply Iran 96 to obtain
$\frac{1}{(ka+b)^{2}}+\frac{1}{(kb+c)^{2}}+\frac{1}{(kc+a)^{2}}\geqslant\frac{9}{(k+1)^{2}(ab+bc+ca)+(k-1)^{2}(\sum ab-\sum a^{2})}$ $\geqslant\frac{9}{(k+1)^{2}(ab+bc+ca)}$.

In the remaining case it seems that, unless many other ineqs, this one can still be [i]very [/i]tight. It is still easy to solve for certain sub-cases but I don't see how to prove it for all of them. It seems though that, in spite of the 'first glance' impression, it is true for [b]all [/b]$a,b,c$ even for big $k$. I wouldn't be too surprised if it could even be 'bunched out' as by Arqady and e.lopes for the $k=2$ case. But I won't do it. :P

I have a new idea.
First,I'll prove the case :$1\leq k\leq 2$
 I have:$y+k^{2}z\geq kx$
            $\Rightarrow y+2z+(k^{2}-2)z\geq kx$(1)
Hence,I only prove in case x>y+z.
          (1)$\Rightarrow y+2z>(k-\frac{k^{2}-2}{2})x>x$
               $\Rightarrow 3(y+z)>2x$
Then,I prove $\frac{(x-y)^{2}}{z}\leq \frac{(x-z)^{2}}{y}$
                    $\leftrightarrow x^{2}-2(y+z)x+y^{2}+yz+z^{2}\leq 0$(2)
The equation (2) has a large root:$y+z+\sqrt{yz}\geq y+2z \geq x$
Therefore,I have (2)
Plus $s_{y}+s_{z}\geq 0$,I have the inequality which I need  prove
Now, I'll prove in case $k\geq 4$
Transform the inequality to the form:
$\sum(x-y)^{2}(\frac{k^{2}+1}{xy}-\frac{k}{z^{2}}+(k-1)^{2}(\frac{z}{x^{2}y}+\frac{z}{xy^{2}}+\frac{1}{xy}))\geq 0$
Let $T_{z}=\frac{k^{2}+1}{xy}-\frac{k}{z^{2}}+(k-1)^{2}(\frac{z}{x^{2}y}+\frac{z}{xy^{2}}+\frac{1}{xy})$
Similar for $T_{x},T_{y}$
Then,I'll prove $T_{z}+T{y}\geq 0$
I have $(k-1)^{2}(\frac{y}{xz^{2}}+\frac{1}{xz})$
           $\geq \frac{ ((k^{2}-2k)y-2kz)+y+k^{2}z)}{xz^{2}}\geq \frac{kx}{xz^{2}}=\frac{k}{z^{2}}$
Therefore,I have $T_{z}+T_{y}\geq 0$

But in case 2<k<4,I can't prove it.
Who can help me?

Use directly expanding or EMV. See here for more http://www.mathlinks.ro/Forum/viewtopic.php?t=141336

Thank you gackiem.I have just proved in case$k\geq 2$
I used:
 $\frac{(k^{2}+1)}{xy}\geq\frac{k}{y^{2}}$
And $(k-1)^{2}(\frac{y}{xz^{2}}+\frac{1}{xz})\geq \frac{k}{z^{2}}-\frac{2k}{xz}$
And $\frac{k^{2}+1}{xz}\geq\frac{2k}{xz}$


inequalities
triangle inequality
inequalities theorems

How to prove the triangle inequality, which states that $ |x|-|y|\le|x+y|\le|x|+|y|$?

Thank you for helping. :P

Since $ xy\leq |xy|$, we have
$ 2xy\leq 2|xy|$
$ \Longleftrightarrow x^{2}+y^{2}+2xy\leq |x|^{2}+|y|^{2}+2|xy|$
$ \Longleftrightarrow (x+y)^{2}\leq (|x|+|y|)^{2}$
$ \Longleftrightarrow |x+y|\leq |x|+|y|$.
Thus use this, $ |x|=|(x+y)-y|\leq |x+y|+|-y|\Longleftrightarrow |x|-|y|\leq |x+y|$.

Triangle inequality holds also for complex number and vector.


trigonometry
trig identities
tangent double angle

Hello there,

I am having trouble simplifying the following trigonometric identities and would appreciate some help. My work has been shown below.

Thanks!

---

1. $ \frac {\sin^3 \theta \plus{} \cos^3 \theta}{sin \theta \plus{} cos \theta} \equal{} 1 \minus{} \sin \theta \cos \theta$

I am unsure of what I should do. Should I multiply the LS by its conjugate?

2. $ \frac {\sin \theta \cos \theta}{\cos^2 \theta \minus{} \sin^2 \theta} \equal{} \frac {\tan \theta}{1 \minus{} \tan^2 \theta}$

LS = $ \frac{\sin \theta}{\cos \theta}{\cos 2\theta}$

This is as far as I got. What else can I do?

[hide="1)"]
$ \frac {\sin^3 \theta \plus{} \cos^3 \theta}{\sin \theta \plus{} \cos \theta}$

$ \equal{} \sin^2\theta \minus{} \sin\theta\cos\theta \plus{} \cos^2\theta$

$ \equal{} 1 \minus{} \sin\theta\cos\theta$[/hide]

[hide="2)"]
$ \frac {\tan \theta}{1 \minus{} \tan^2 \theta}$

$ \frac {\frac {\sin\theta}{\cos\theta}}{\frac {\cos^2\theta \minus{} \sin^2\theta}{\cos^2\theta}}$

$ \equal{} \frac {\sin\theta\cos\theta}{\cos^2\theta \minus{} \sin^2\theta}$[/hide]

You can factor $ sin^3 \theta \plus{} cos^3 \theta \equal{} (sin \theta \plus{} cos \theta)(sin^2 \theta \minus{} sin \theta cos \theta \plus{} cos^2 \theta)$
This should give it up...

For second one note that $ sin \theta cos \theta \equal{} \frac {sin 2\theta}{2}$ and $ cos^2 \theta \minus{} sin^2 \theta \equal{} cos 2\theta$
Then use tangent double angle formula...


limit
integration
real analysis
real analysis unsolved

Determine whether $ \lim_{x\to 1^{\minus{}}} \sum_{n\equal{}0}^{\infty} (\minus{}1)^n x^{2^n}$ is convergent or not, and evaluate it if converges.

I believe that it diverge. for $ x_{n} \equal{} 1 \minus{} \frac {1}{2^{n}}$
but it's not enough to say that serie diverge. I'm sure that there is some explanation .. :)

This reminds me of [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4767[/url] ;)

in the problem that Fedja mentioned i.e. find the behaviour of $ f(x) \equal{} \sum_n x^{2^{n}}$ near $ 1^\minus{}$, it was useful to compare this sum to its continuous counterpart $ g(x) \equal{} \int_0^{\infty} x^{2^y} dy \equal{} \int_0^{\infty} \exp(\minus{}|\ln(x)| 2^y) dy$. In this case it was easy to find the behaviour of $ g$ with a simple change of variable like $ \theta \equal{} |\ln(x)| 2^y$, and I think that it should be the same in your problem. Write it as $ f(x) \equal{} \sum_n x^{2^{2n}}(1\minus{}x^{2^{2n}})$ and compare it to $ g(x) \equal{} \int_0^{\infty} x^{2^{2y}}(1\minus{}x^{2^{2y}}) dy$. ( because $ \phi(a) \equal{} x^a(1\minus{}x^a)$ is monotonous, there should not be a lot of problem, I think.. )

Hmm, alekk... I think your post is a bit confusing; which one do you suggest to prove this way: convergence or divergence? :?

I don't know  :lol:  Because $ \phi(a) \equal{} x^a(1 \minus{} x^a)$ is increasing then decreasing and bounded ($ x$ fixed), one can see that
$ f(x) \equal{} \sum_{p \equal{} 1}^{\infty} x^{2^{2p}}(1 \minus{} x^{2^{2p}})$ satisfies $ C_1 \plus{} \int_0^{\infty} \phi(x,2^{2y}) dy < f(x) < C_2 \plus{} \int_0^{\infty} \phi(x,2^{2y}) dy$. So it remains to study $ \int_0^{\infty} \phi(x,2^{2y}) dy$ when $ x \to 1^ \minus{}$? A change of variable like $ \theta \equal{} |\ln(x)| 2^{2y}$ shows that $ 0 < \int_0^{\infty} \phi(x,2^{2y}) dy < C \int_0^{\infty} \frac {e^{ \minus{} \theta}(1 \minus{} e^{ \minus{} \theta})}{\theta} d\theta < \infty$. So it is uniformly bounded. Now remains to see if it converges to $ \frac{1}{2}$


algebra
polynomial
induction
algebra proposed

Find all polynomials $P\in \mathbb R[x]$such that:
\[P(x^2)+x \cdot (3P(x)+P(-x))=(P(x))^2+2x^2 \quad  \forall x\in  \mathbb R\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76271]View all problem from Vietnam NMO 2006[/url]

To the given equation
$P(x^2)+x \,(3P(x)+P(-x))=(P(x))^2+2x^2,\, \forall x\in \mathbb{R}$
we substitute $x$ by $-x$ and we get
$P(x^2)- x \,(3P(-x)+P(x))=(P(-x))^2+2x^2$.

Now we take the difference of these two equations and after factoring we have

$x \,(4P(x)+4P(-x))=(P(x))^2-(P(-x))^2$ or

$(P(x)-P(-x)-4x)\cdot (P(x)+P(-x))=0 ,\, \forall x\in \mathbb{R}$.

Then either $P(-x)=P(x)-4x$ or $P(-x)=-P(x)$.

$1)$ If $P(-x)=-P(x)$ then $P(x^2)-x^2=(P(x)-x)^2$ and
$P(x)= a_{2n-1}x^{2n-1}+ a_{2n-3}x^{2n-3}+...+ a_{3}x^{3}+ a_{1}x$

$2)$ If $P(-x)=P(x)-4x$ then $P(x^2)-2 x^2=(P(x)-2 x)^2$ and
$P(x)= a_{2n}x^{2n}+ a_{2n-2}x^{2n-2}+...+ a_{2}x^{2}+ 2 x$

We will find the polynomials $R(x)$ that satisfy $R(x^2)= R^2(x)$.

[u]Proposition[/u]: The polynomials that satisfy the relation $R(x^2)= R^2(x)$ are
of the form $R(x)=a x^{n}$ for  some integer $n$ and $a \in \{0,1\}$ or of the
 form $R(x)=c$ , $c \in \{0,1\}$
We will proceed by induction on $degR(x)$.

$\cdot$ For $n=1 , \, R(x)=ax+b$ and the relation $R(x^2)= R^2(x)$gives
$R(x)=x$ or $R=1$ or $R=0$.

$\cdot$ We assume that the proposition is true for $degR(x)\leq k$.

$\cdot$ Let $R(x)$ be a polynomial with $degR(x)=k+1$
Then $R(x)=x^{k+1}+Q(x)$ where $Q(x)$ is a polynomial with $degQ(x)=m \leq k$

$R(x^2)=R^2(x) \Longrightarrow Q(x^2) = Q(x)( Q(x)+ 2x^{k+1})$.

If  $Q(x)$ is not the zero polynomial then it divides $Q(x^2)$ and let
$\frac{Q(x^2)}{Q(x)}=A(x)$. Then $degA(x)=m < k+1 =deg( Q(x)+ 2x^{k+1})$
But this is false so $Q(x)$ is the zero polynomial , $R(x)=x^{k+1}$ and the induction is complete.

Now for the first case $P(-x)=-P(x)$ and $P(x^2)-x^2=(P(x)-x)^2$ we have
that

$R(x)= P(x)-x \Longrightarrow P(x)= a x^{2k-1}+x$ or $P(x)= 1+x ,a\in \{0,1\},k \in \mathbb {N^*}$

and for the second case  $P(-x)=P(x)-4x$ and $P(x^2)-2 x^2=(P(x)-2 x)^2$
we have that

$R(x)= P(x)-2x \Longrightarrow P(x)= a x^{2k}+2x ,a\in \{0,1\} ,k \in \mathbb {N}$.

I am not sure about the induction and typos.
I'll check it later  :)


I hope you will like this inequality. I think it is very hard.

If x,y,z are real numbers such that x+y+z=1 and -1 \leq x,y,z \leq 1
and a,b,c are the sides of a triangle prove that

(xa+yb+zc)(ya+zb+xc)(za+xb+yc) \geq (a+b-c)(b+c-a)(c+a-b)

For the special case when x-1=y=z=0 it reduces to the well known inequality

 abc \geq (a+b-c)(b+c-a)(c+a-b)

which corresponds to R \geq 2r.

Thank you very much, Manlio. This really seems very hard.

Very nice inequality.
    Put a=v+w, b=u+w, c=u+v with u,v,w>=0. Also, note 1-x=A, 1-y=B, 1-z=C. Then A,B,C>=0 add up to 2 and the inequality becomes:
  (Au+Bv+Cw)(Av+BW+Cu)(Aw+Bu+Cv)>=8uvw.
  Now remember that ((a_1+b_1)(a_2+b_2)(a_3+b_3))^1/3>=(a_1a_2a_3)^1/3+(b_1b_2b_3)^1/3. Thus,
  ((prod (Au+Bv+Cw))^1/3>=(prod(Au+Bv))^1/3+C(uvw)^1/3>=A(uvw)^1/3+B(uvw)^1/3+C*(uvw)^1/3=2(uvw)^1/3. The conclusion follows.

Your ability to solve inequalities (to find the key of the solution) is really very very very extraordinary.

Thank you very much Harazi.


AMC
USA(J)MO
USAMO
combinatorics unsolved
combinatorics

Given a set of "n"positive numbers.For each of its non empty subset consider the sum of all the subsets' numbers.
Prove that you can divide those sums onto "n" groups in such a way, that the least sum in every group is not less than half of the greatest sum in the same group.
please help!!! :maybe:  :blush:

USAMO 1996

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=353051#353051


Let$ a;b;c\geq 0$.Prove that:
    $ \sum\frac{1}{a^2\plus{}ab\plus{}b^2}\geq\frac{9}{(\sum a)^2}$

[quote="karis"]Let$ a;b;c\geq 0$.Prove that:
    $ \sum\frac {1}{a^2 \plus{} ab \plus{} b^2}\geq\frac {9}{(\sum a)^2}(*)$[/quote]
[b][u]Solution :[/u][/b]
We have :
$ a^2\plus{}ab\plus{}b^2\equal{}(a\plus{}b\plus{}c)^2\minus{}(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)c$
$ b^2\plus{}bc\plus{}c^2\equal{}(a\plus{}b\plus{}c)^2\minus{}(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)b$
$ c^2\plus{}ca\plus{}a^2\equal{}(a\plus{}b\plus{}c)^2\minus{}(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)a$
Suppose $ a\plus{}b\plus{}c\equal{}1$. We have :
$ (*) \leftrightarrow \sum\frac{1}{1\minus{}(ab\plus{}bc\plus{}ca)\minus{}a} \geq 9$
$ \leftrightarrow 1\minus{}4(ab\plus{}bc\plus{}ca)\plus{}9abc \geq 0$
$ \leftrightarrow (a\plus{}b\plus{}c)^2\minus{}4(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\plus{}9abc \geq 0$
$ \leftrightarrow a^3\plus{}b^3\plus{}c^3 \plus{} 3abc \geq a(b^2\plus{}c^2)\plus{}b(c^2\plus{}a^2)\plus{}c(a^2\plus{}b^2)$
It is true because it is inequality 's Schur.
We are done.

[quote="conan_naruto236"]
Suppose $ a \plus{} b \plus{} c \equal{} 1$. We have :
$ \sum\frac {1}{1 \minus{} (ab \plus{} bc \plus{} ca) \minus{} a} \geq 9$
$ \leftrightarrow 1 \minus{} 4(ab \plus{} bc \plus{} ca) \plus{} 9abc \geq 0$
[/quote]
Why? :huh:
[quote="karis"]Let$ a;b;c\geq 0$.Prove that:
    $ \sum\frac {1}{a^2 \plus{} ab \plus{} b^2}\geq\frac {9}{(\sum a)^2}$[/quote]
Try $ \sum_{cyc}\frac {1}{a^2 \plus{} ab \plus{} b^2} \equal{} \sum_{cyc}\frac {(3c \plus{} a \plus{} b)^2}{(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)}\geq$
$ \geq\frac {25(a \plus{} b \plus{} c)^2}{\sum(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)}.$ :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6364

[quote="dgreenb801"]Try $ \sum_{cyc}\frac {1}{a^2 \plus{} ab \plus{} b^2} \equal{} \sum_{cyc}\frac {(3c \plus{} a \plus{} b)^2}{(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)}\geq$
$ \geq\frac {25(a \plus{} b \plus{} c)^2}{\sum(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)}.$ :wink:
How does this prove the question?[/quote]
It's Cauchy-Schwartz. :wink:

[quote="dgreenb801"]I know it's Cauchy, but how do we show
$ \frac {25(a \plus{} b \plus{} c)^2}{\sum(3c \plus{} a \plus{} b)^2(a^2 \plus{} ab \plus{} b^2)} \ge 9$?[/quote]
It's obvious after expanding. :wink:

oh,pqr thechnique works here
We put $ a \plus{} b \plus{} c \equal{} 1$,$ \sum ab \equal{} q,abc \equal{} r$
The ineq becomes
$ q(3q \minus{} 1)^2 \plus{} 9r \plus{} 1 \minus{} 4q$ after expanding :lol:
And it is obivious since $ 9r \plus{} 1 \minus{} 4q\ge 0$(schur ineq)
The equality holds when $ a \equal{} 0,b \equal{} c \equal{} \frac {1}{2},a\equal{}b\equal{}c$


Two different numbers are chosen from the sequence of consecutive odd integers 51, 53, 55, …,95, 97, 99. Which of the following could not be their sum?

a. 161 b. 168 c. 186 d. 196 e. None''.

[hide="Solution"]Take the whole sequence $ \mod 2$ and you'll see that adding any two numbers will always give you $ 0$, so the answer is clearly $ \boxed{\textbf{(A)}\ 161}$, which is odd.[/hide]

[hide="SOLUTION"]Sice we are asked to add two odd numbers, the sum must be even.
The answer is thus $ A$, the only odd number.
 :) [/hide]

[hide]Clearly, an odd number plus an odd number equals an even number.  Therefore, the answer is A.  :)[/hide]

Do we really need 3 identical solutions? :lol: