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I pose two problems that are quite well known.The first one is so easy to solve that I label it zero!I remember giving a cute solution for the second one long time ago, but I can't remember it now!!
$ 0.$Suppose $ V$ is a subspace of $ n\times n$ matrices over a field such that all of its elements except zero are invertible. Prove that $ \text{dim}V\leq n$.
$ \\1.$Suppose $ V$ is a subspace of $ n\times n$ matrices over a field such that all of its elements are non-invertible. Prove that $ \text{dim}V\leq n^2\minus{}n$.

see also [url]http://www.mathlinks.ro/viewtopic.php?search_id=1236858374&t=44669[/url]

0. Let $ \{A_1,\dots,A_k\}$ be an independent set of matrices in $ V.$ Choose a arbitrary nonzero vector $ v$ in $ F^n.$ Then $ \{A_1v,\dots,A_kv\}$ must be an independent set in $ F^n.$ To see this:

Suppose $ \alpha_1A_1v\plus{}\alpha_2A_2v\plus{}\cdots\plus{}\alpha_kA_kv\equal{}0.$ Then $ (\alpha_1A_1\plus{}\alpha_2A_2\plus{}\cdots\plus{}\alpha_kA_k)v\equal{}0.$ But $ \alpha_1A_1\plus{}\alpha_2A_2\plus{}\cdots\plus{}\alpha_kA_k\in V,$ and every nonzero matrix in $ V$ is invertible. The only way it can send $ v$ to zero is to be the zero matrix. Then, the independence of this set of matrices shows that $ \alpha_1\equal{}\alpha_2\equal{}\cdots\equal{}\alpha_k\equal{}0.$

Hence $ \{A_1v,\dots,A_kv\}$ is an independent set in $ F^n,$ but such an independent set can have no more than $ n$ vectors. We conclude that $ k\le n$ and since we have this for all independent sets, $ \dim V\le n.$

It is less obvious to me that this bound is achieved.

As a matter of brute force inspection, I did find two two-dimensional subspaces of $ GL_2(\mathbb{Z}_2)\cup\{0\}:$

$ \left\{\begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&1\end{bmatrix}, \begin{bmatrix}1&1\\1&0\end{bmatrix}\right\}$

$ \left\{\begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&1\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}1&0\\1&1\end{bmatrix}\right\}$

These are the only possibilities in that context.

[quote="Kent Merryfield"]

It is less obvious to me that this bound is achieved.[/quote]
If you find the bound then you solved "1" :lol:

[quote="Kent Merryfield"]It is less obvious to me that this bound is achieved.[/quote]

It is more obvious to me that over every algebraically closed field, the bound can be strengthened to $ \dim V\leq 1$. ;)

The problem 1 is a particular case of the so-called Flanders theorem, which states that if $ K$ is a field with certain properties and $ r$, $ m$, $ n$ are three integers such that $ 0\leq r\leq m\leq n$, then any subspace of the vector space $ \text{Hom}\left(K^n,K^m\right)$ which consists of linear transformations of rank $ \leq r$ only must have dimension $ \leq nr$. Unfortunately, my only source for this theorem is [url=http://www.math.su.se/~mleites/Prasolov/prasLinAlg/pr-linAlg-main.dvi]Prasolov's "Problems and theorems in linear algebra"[/url] (section 8.3), where (in a very Russian tradition) almost no care is done about fields $ K$ different from $ \mathbb{R}$ and $ \mathbb{C}$. Having had only a short glance at the proof, I see it works only for infinite $ K$, but maybe it needs even more conditions, who knows. And it may also be that the Flanders theorem doesn't require the infinity of $ K$. Can anyone here shed some more light on this result?

  Darij

Infinite $ K?$ Finite $ K?$ Here are some observations. (I'm still talking about problem 0, about a subspace $ V$ in which $ 0$ is the only noninvertible element.)

If $ K\equal{}\mathbb{Z}_p,$ then the finite field of order $ p^n$ can be realized as a set of $ n\times n$ matrices, and has dimension $ n.$ So in these cases, we can have the dimension of $ V$ equal to $ n.$ (And the examples I gave above were simply realizations of the field of order $ 4.$)

If $ K\equal{}\mathbb{R}$ and $ n\equal{}2,$ then we can simply take the matrices $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}.$ In other words, the complex numbers.

And, about that algebraically closed field thing - isn't there a way to realize the quaternions as $ 2\times 2$ complex matrices? That would be a case with an algebraically closed field and $ \dim V>1.$

At this point, I realize that all of the examples I have been talking about have been division rings. If such a subspace $ V$ is closed under multiplication (a subring of $ M_n(K)$), then it would be a division ring. Does it have to be closed under multiplication?

OK, I can partly answer that from my examples above in $ M_2(\mathbb{Z}_2).$ The first example is closed under multiplication - which makes it a division ring, which makes it a field. The second example is not closed under multiplication, but it can be represented at the set of all $ AB$ where $ B$ is one specific matrix and $ A$ ranges over the set of the first example. So maybe the question could be refined.

[quote="Kent Merryfield"]Infinite $ K?$ Finite $ K?$[/quote]

Infinite $ K$. There is a "the polynomial is $ 0$ everywhere $ \Longrightarrow$ its coefficients are $ 0$" argument in the proof.

[quote="Kent Merryfield"]If $ K \equal{} \mathbb{Z}_p,$ then the finite field of order $ p^n$ can be realized as a set of $ n\times n$ matrices, and has dimension $ n.$ So in these cases, we can have the dimension of $ V$ equal to $ n.$ (And the examples I gave above were simply realizations of the field of order $ 4.$)[/quote]

Nice observation. But please write $ K \equal{} \mathbb{F}_p$ rather than $ K \equal{} \mathbb{Z}_p$, since the latter can equally mean $ p$-adic integers...

[quote="Kent Merryfield"]And, about that algebraically closed field thing - isn't there a way to realize the quaternions as $ 2\times 2$ complex matrices? That would be a case with an algebraically closed field and $ \dim V > 1.$[/quote]

Are you sure that the quaternions can be seen as $ 2\times 2$ complex matrices in the sense that there is an isomorphism between the $ \mathbb{C}$-algebra $ \mathbb{H}$ and a sub-$ \mathbb{C}$-algebra of $ \text{M}_2\left(\mathbb{C}\right)$ ? If you are talking about the representation shown in http://en.wikipedia.org/wiki/Quaternion#Matrix_representations , it only yields an isomorphism between the $ \mathbb{R}$-algebra $ \mathbb{H}$ and a sub-$ \mathbb{R}$-algebra of $ \text{M}_2\left(\mathbb{C}\right)$.

My argument why $ \dim V\leq 1$ over any algebraically closed field $ K$ was that if $ A$ and $ B$ are two matrices, then $ xA \plus{} yB$ is invertible if and only if $ \det\left(xA \plus{} yB\right)\neq 0$. But $ \det\left(xA \plus{} yB\right)$ is a homogeneous polynomial of nonzero degree in $ x$ and $ y$, and over an algebraically closed field such a polynomial has at least one zero different from $ \left(x,y\right) \equal{} \left(0,0\right)$. For this zero $ \left(x,y\right)$, the corresponding matrix $ xA \plus{} yB$ is not invertible.

  Darij

[quote="darij grinberg"][quote="Kent Merryfield"]It is less obvious to me that this bound is achieved.[/quote]

It is more obvious to me that over every algebraically closed field, the bound can be strengthened to $ \dim V\leq 1$. ;)

  Darij[/quote]
Hi,Darij,Can you tell me how to see this? :blush:

If you'll read the comment directly preceding yours, you'll note that the final paragraph contains an explanation of his claim.

Thank you,I haven't read more when I can't understand the proposition.I am sorry :blush:

[quote="Kent Merryfield"]Infinite $ K?$ Finite $ K?$ Here are some observations. (I'm still talking about problem 0, about a subspace $ V$ in which $ 0$ is the only noninvertible element.)

If $ K \equal{} \mathbb{Z}_p,$ then the finite field of order $ p^n$ can be [u]realized[/u] as a set of $ n\times n$ matrices, and has dimension $ n.$.[/quote]Would you please clarify this? I know that this may look so elementary, but I want to make sure that I've understood you the right way.

See $ \mathbb F_{p^n}$ as a vector space over $ \mathbb F_p$. It has dimension $ n$. Now take a basis $ b_1,..., b_n$ and identify $ \mathbb F_{p^n}$ with $ (\mathbb F_p)^n$ this way.
Now every $ 0 \neq a \in \mathbb F_{p^n}$ gives an $ \mathbb F_{p}$-vector space automorphism of $ \mathbb F_{p^n}$ by $ x \mapsto ax$, thus this corresponds to some invertible $ n \times n$ matrix; this is an automorphism of rings, thus of fields (if extended to $ 0$).


calculus integration conics parabola geometry videos vector

On $xy$plane the parabola $K: \ y=\frac{1}{d}x^{2}\ (d: \ positive\ constant\ number)$ intersects with the line $y=x$ at the point $P$ that is different from the origin.
Assumed that the circle $C$ is touched to $K$ at $P$ and $y$ axis at the point $Q.$
Let $S_{1}$ be the area of the region surrounded by the line passing through two points $P,\ Q$ and $K,$ or $S_{2}$ be the area of the region surrounded by the line which is passing through $P$ and parallel to $x$ axis and $K.$ Find the value of $\frac{S_{1}}{S_{2}}.$

to evaluate S1 we must do the integral:
$\int f(x)-p(x)dx$ where f(x) is the line passing through q and p and p(x) is the parabola.
We know that a circle touches the y axis and K only once, so by basic geometry we see that the distance from the point Q to the origin is equal to the distance of the point P to the origin. This distance is $d\sqrt{2}$ so $f(x)=ax+d\sqrt{2}$ just putting the point d in this equation, we finally have:
$f(x)=(1-\sqrt{2})x+d\sqrt{2})$ and we finally get that the other intersection will at the point  $-d\sqrt{2}$
$S_{1}=\int_{-d\sqrt{2}}^{d}(1-\sqrt{2})x+d\sqrt{2})-\frac{x^{2}}{d}$ i evaluated this and got $\frac{d^{2}}{6}(7+13\sqrt{2}$
But about S_2, i think you missed something. You just said that this line is parallel to the x axis. What about the height of it? There is no information about it on the problem!

Sorry for that, I have just edited.

kunny

so i got $\frac{7+13\sqrt{2}}{8}$ i am not so confident about its value but i think that is right!

Regrettably your answer is incorrect.

Your equation of the line $PQ$ seemed to be wrong.

[color=green]Heute's calculation of Integral 177[/color] - [color=red]Volume 1[/color]

Let $\mathcal K$ be the parabola of equation $y = \frac{x^{2}}{d}$ and $A \left( u, \frac{u^{2}}{d}\right), \, B \left( v, \, \frac{v^{2}}{d}\right)$ be two points on it s.t. $u < v$. The equation of $AB$ is $y = \frac{x(u+v)-uv}{d}$, so the area between $[AB]$ and $\mathcal K[AB]$ is given by \[\frac1d \int_{u}^{v}\Big( x(u+v)-uv-x^{2}\Big) \, dx = \frac{(v-u)^{3}}{6d}.\]
It's pretty obvious that $P(d,d)$, so we have that $\frac{S_{1}}{S_{2}}= \left( \frac{v_{1}-u_{1}}{v_{2}-u_{2}}\right)^{3}= \left( \frac{d-u_{1}}{2d}\right)^{3}$.

[color=green]Heute's calculation of Integral 177[/color] - [color=red]Volume 2[/color]
Let $F(x,y) = \frac{x^{2}}{d}-y$. The tangent at the point $\left( x_{0}, y_{0}\right) \in \mathcal K$ is \[\left( x-x_{0}\right) \cdot \frac{\partial F}{\partial x}\left( x_{0}, y_{0}\right)+\left( y-y_{0}\right) \cdot \frac{\partial F}{\partial y}\left( x_{0}, y_{0}\right) = 0 ,\]
so the tangent at $P(d,d)$ is $t_{P}: y = 2x-d$.
The circle $\mathcal C$ touches $t_{P}$ at $P$ and $Oy$ at $Q$, so $WP = WQ$, where $\left\{ W \right\}= Oy \cap t_{P}$. It's pretty clear that $W(0,-d)$.
Therefore, $WQ = WP = d \sqrt 5$, so $Q \left( 0, \left( \sqrt 5-1 \right) d \right)$.
Now let $R \left( r, \frac{r^{2}}{d}\right)$ be the second intersection of $PQ$ with $\mathcal K$. The equation of $PR$ is $y = \frac{x(r+d)-rd}{d}= x \cdot \frac{r+d}{d}-r$. Its intersection with $Oy$ is $\left( 0,-r \right)$, so $r = \left( 1-\sqrt 5 \right) d$.

[color=green]Heute's calculation of Integral 177[/color] - [url=http://www.youtube.com/watch?v=pVIx3KzFMx0]Epilogue[/url] *
$u_{1}= r$, so $\frac{S_{1}}{S_{2}}= \left( \frac{\sqrt 5}2 \right)^{3}= \frac{5 \sqrt 5}8$.

* Watch the video until the grand finale (The first song that the [i]almost bald[/i]** guy unsuccessfully tries to sing is [url=http://www.versuri.ro/versuri/hff_vama+veche_epilog.html]Vama Veche - Epilog[/url]).
** Nobody likes an underachiever, but he'll do, I guess...

PS: I hope I didn't mess up the computations.

Fantastic!  :) Your answer is correct, perfect_radio.

Here is my solution.

Let $A$ be the center of the circle in question and $r$ be the radius. From $\frac{d}{dx}\frac{1}{d}x^{2}=\frac{2}{d}x,$ the equation of the tangent at $P(d,\ d)$ is $y=2(x-d)+d\Longleftrightarrow y=2x-d.$ Since the normal vector $\overrightarrow{n}=(1,2),$ we have that $\overrightarrow{OA}= \overrightarrow{OP}+\overrightarrow{PA}=(d,\ d)+r\cdot \frac{1}{|\overrightarrow{n}|}\overrightarrow{n}=(d,\ d)+\frac{1}{\sqrt{5}}(-2,1),$
which is equal to the vector $(r,\ r+h),$ where $h$ is the difference of the $y$ coordinate of $Q$ and $P.$ Thus we obtain that $r=(5-2\sqrt{5})d,$ yielding $A((5-2\sqrt{5})d,\ (\sqrt{5}-1)d)$ and $Q(0,\ (\sqrt{5}-1)d).$ Then the equation of $PQ$ is $y=(2-\sqrt{5})x+(\sqrt{5}-1)d.$ Solving the system of equation for this equation and $y=\frac{1}{d}x^{2},$

we have $x^{2}+(\sqrt{5}-2)dx-(\sqrt{5}-1)d^{2}=0\Longleftrightarrow (x-d)(x+(\sqrt{5}-1)d)=0,$ yielding the second intersection point is $R((1-\sqrt{5})d,\ (1-\sqrt{5})^{2}d).$

Using the formula ${\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)dx=-\frac{1}{6}(\beta-\alpha)^{3}},$
we have

$S_{1}=-\frac{1}{d}\int_{(1-\sqrt{5})d}^{d}(x-d)(x-(1-\sqrt{5})d)=\frac{1}{6d}(\sqrt{5}d)^{3},$

$S_{2}=-\frac{1}{d}\int_{-d}^{d}(x+d)(x-d)dx=\frac{1}{6d}(2d)^{3}.$ Therefore $\frac{S_{1}}{S_{2}}=\frac{5\sqrt{5}}{8}.$


[quote="solafidefarms"]I like Rush Limbaugh and Sean Hannity almost the same, but Rush is a little more amusing.

Billy[/quote]

I LOVE to listen to Rush Limbaugh, but you forgot my favorite!  Bill O'Reilly!!!!!!!!!!

[quote="yif man12"]Hmm, I wonder what Rush is saying about McNabb, now that the Eagles are 4-0 partially because he's got TO...[/quote]

OSNAPS.  Major BURN.

I think they're all fun at which to laugh.


superior algebra
superior algebra unsolved

Show that $ {Z}_{mn}$ is isomorphic to $ {Z}_{m} \times {Z}_{n}$ if and only if $ (m,n)\equal{}1$. Thanks in advance.

Are these groups?  Rings?  Which direction are you stuck on?

Perhaps try looking at the number of elements of additive order $ (m,n)$ in each.


Can you distinguish a green bottle from another green bottle?

[quote="aluminum essence"][quote="PenguinIntegral"][quote="aluminum essence"]Here's a permutation problem from my textbook that I'm stuck with. Any help will be appreciated.  :)

Question:

How many different ways are there of arranging seven green and eight brown bottles in a row, so that exactly one pair of green bottles is side-by-side?

(Since I'm new, I'm not sure if it's the appropriate board for this. If not, please let me know)[/quote]
Count the number of ways you can permutate 6 green and eight brown bottles so that no greens are side by side, then add the remaining green the side of each green bottle.

I hope that is clear. If you need more explanation, ask.[/quote]

I'm not sure how to the first part. (i.e. finding # of ways of permutating 6 green and 8 brown so they are not side by side)

And no, none of the green bottles are distinguishable from each other (same with brown[/quote]

We have six green bottles:

GGGGGG

We need to but 8 browns in between to so they aren't together. But we want to do this with the least amount of browns possible.

GBGBGBGBGBG but we have three browns left.

We can put the next 3 browns into the lineup in 7^3 not distingushable ways.

Then the green can go be put in the lineup in 6 ways, for a total answer of 7^3*6.

Is this right??

About how to permutate 6 green and 8 brown without any green side by side, first space the green and brown so you have green,brown,green,brown,etc. and then take the extra browns and, for every spot you can place it in, that is a possible arrangement. Anyone have any arguements against doing it this way?

PenguinIntegral, I believe you would be right.

[quote]Then the green can go be put in the lineup in 6 ways, for a total answer of 7^3*6[/quote]

That would give 2058 possible arrangements, but the book says 504.

I think there are less than 7^3 ways of placing the three brown bottles, but I don't seem to be able to figure it out right now.  :(

[hide="solution"]
So you have eight brown bottles, and 9 spaces between the bottles
_B_B_B_B_B_B_B_B_
You want to choose six spaces out of the nine to put in the green bottles (one green bottle in each space) and the seventh green bottle will be forced to go with another one
Since there are 6 ways that the seventh green bottle can fit with the rest, we have
$\binom{9}{3}\cdot 6 = 504$
[/hide]

you guys have forgotten something.
the problems says EXACTLY one  pair of green...

if you have GG
and GbG....

and if you put them together you have morethan one pair of green.

G B G B G B G B G B G B G B This is the arrangement of 7 brown and 7 green bottles.
The eighth brown bottle can be placed along with any of the 7 brown bottles, in 7 ways.
Now ways to form the above arragngement( G B G B G B G B G B G B G B ) can be found out by fixing the seven green bottles and placing the 7 brown bottles in the eight available places(since 2 cannot be together).
This can be done in $\binom{8}{7} = 8$ ways.

Thus the answer to this question must be $7\cdot8=56$

[quote="aluminum essence"]
That would give 2058 possible arrangements, but the book says 504.

I think there are less than 7^3 ways of placing the three brown bottles, but I don't seem to be able to figure it out right now.

[/quote]
Are the bottles differentiable. In that case the 7 green bottles can be arranged in 7! ways and the seven brown bottles can be arranged in $\binom{8}{7}\cdot{7!}$.
Thus if the bottles can be distinguished then no. of possibble arrangements are $7!\cdot\binom{8}{7}\cdot{7!}\cdot7$

Well this not at all the answer given in your book.(Are you sure about the answer given in the book because we also in India have some publishers printng quite erroness books).  I feel more experienced people here will tel better than me.

Oops! I solved for 2 brown. Now I will now solve for 2 green.

Arranging 8 browns in a row. Then we will place 6 greens in any of the 9 places(represented by x)
  x B x B x B x B x B x B x B x B x
And placing the last green in 6 ways so that only 2 are together.

Thus the total no.of ways is $\binom{9}{6}\cdot{6}=504$

Got it. This was the answer in your book.
I must say youir book is not like mine. HiHi... :rotfl:

I get it now. Thanks to all for your help  :)

If you further feel free to ask.
By the way I was also doing permutation combination and for interested people can go for the following links for much better questions:
[url=Counting Problem??? Help...]www.artofproblemsolving.com/Forum/viewtopic.php?t=57452[/url]
OR
[url=Press Conference Problem]www.artofproblemsolving.com/Forum/viewtopic.php?t=57455[/url]


[quote="BOGTRO"]Prove or disprove: All cyclic inequalities take their maximum (or minimum) when all of the variables are equal.

I apologize if this is a stupid question to be asking in this forum.[/quote]

if $ a,b>0,a\plus{}b\equal{}6$, find $ \left(a\plus{}\frac{1}{a}\right)\left(b\plus{}\frac{1}{b}\right)$ minimum.

[quote="kuing"][quote="BOGTRO"]Prove or disprove: All cyclic inequalities take their maximum (or minimum) when all of the variables are equal.

I apologize if this is a stupid question to be asking in this forum.[/quote]

if $ a,b > 0,a \plus{} b \equal{} 6$, find $ \left(a \plus{} \frac {1}{a}\right)\left(b \plus{} \frac {1}{b}\right)$ minimum.[/quote]Would I be right in saying that the minimum is $ 2$, and it occurs at $ a \equal{} 3 \pm 2\sqrt {2}$?

Anyway, there are well-known inequalities where the maximum/minimum does not occur when all are equal. Example:

Given $ a,b,c \ge 0, ab \plus{} bc \plus{} ca \equal{} 1$, prove that $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \ge \frac {5}{2}$. The minimum occurs at $ a \equal{} b \equal{} 1,c \equal{} 0$ or any permutation.

Furthermore, there are inequalities where equality occurs more than once. An example of which is given $ x,y,z,t \ge 0, x \plus{} y \plus{} z \plus{} t \equal{} 4$, prove that $ (1 \plus{} 3x)(1 \plus{} 3y)(1 \plus{} 3z)(1 \plus{} 3t) \minus{} 131xyzt \ge 125$. Equality occurs when $ x \equal{} y \equal{} z \equal{} t \equal{} 1$ or $ x \equal{} y \equal{} z \equal{} \frac {4}{3}, t \equal{} 0$.

However, you can guess that equality occurs when one of the following happen:
1) All variables are equal
2) Some variables are equal
3) One variable is on the boundary (i.e. is equal to maximum/minimum value on interval, usually 0)

[quote="-Elixir-"][quote="kuing"][quote="BOGTRO"]Prove or disprove: All cyclic inequalities take their maximum (or minimum) when all of the variables are equal.

I apologize if this is a stupid question to be asking in this forum.[/quote]

if $ a,b > 0,a \plus{} b \equal{} 6$, find $ \left(a \plus{} \frac {1}{a}\right)\left(b \plus{} \frac {1}{b}\right)$ minimum.[/quote]Would I be right in saying that the minimum is $ 2$, and it occurs at $ a \equal{} 3 \pm 2\sqrt {2}$?

[/quote]

if $ a \equal{} 3 \pm 2\sqrt {2}$, then  $ \left(a \plus{} \frac {1}{a}\right)\left(b \plus{} \frac {1}{b}\right)\equal{}36$, not $ 2$.

[quote="-Elixir-"][quote="kuing"][quote="BOGTRO"]Prove or disprove: All cyclic inequalities take their maximum (or minimum) when all of the variables are equal.

I apologize if this is a stupid question to be asking in this forum.[/quote]

if $ a,b > 0,a \plus{} b \equal{} 6$, find $ \left(a \plus{} \frac {1}{a}\right)\left(b \plus{} \frac {1}{b}\right)$ minimum.[/quote]Would I be right in saying that the minimum is $ 2$, and it occurs at $ a \equal{} 3 \pm 2\sqrt {2}$?

[/quote]
minimum is $ 2\sqrt{37}\minus{}2$, when $ a,b\equal{}3\pm\sqrt{9\minus{}\sqrt{37}}$.

[quote="earldbest"][/quote][quote="-Elixir-"][quote="kuing"][quote="BOGTRO"]Prove or disprove: All cyclic inequalities take their maximum (or minimum) when all of the variables are equal.

I apologize if this is a stupid question to be asking in this forum.[/quote]

if $ a,b > 0,a + b = 6$, find $ \left(a + \frac {1}{a}\right)\left(b + \frac {1}{b}\right)$ minimum.[/quote]Would I be right in saying that the minimum is $ 2$, and it occurs at $ a = 3 \pm 2\sqrt {2}$?

[/quote][quote="earldbest"]

if $ a = 3 \pm 2\sqrt {2}$, then  $ \left(a + \frac {1}{a}\right)\left(b + \frac {1}{b}\right) = 36$, not $ 2$.[/quote]Yeah, rechecked it, and it's not the minimum anyhow. Must have done something wrong. My apologies.


Auston is 60 inches tall. Using the conversion 1 inch = 2.54 cm, how tall is Auston in centimeters? Express your answer as a decimal to the nearest tenth.

It is easy to see that $ 60\,\text{inches}\equiv60\cdot2.54\,\text{centimeters} \equal{} \boxed{152.4\,\text{centimeters}}$.


For right part: $f\"(a) \>0 \implies \max_{a,b,c,d \in \{1,2\}} f(a,b,c,d)$.

[quote="rogue"]For every $1\le a, b, c, d\le2$ prove the inequality
$\frac{4}{3}\le\frac{a}{b+cd}+\frac{b}{c+da}+\frac{c}{d+ab}+\frac{d}{a+bc}\le2.$[/quote]
For the left part :
$\frac{a}{b+cd}+\frac{b}{c+da}+\frac{c}{d+ab}+\frac{d}{a+bc}\geq \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}=\frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c})\geq \frac43$

that is nice Hawk Tiger !

What inspired you to use $\frac{a}{b+cd}\geq\frac{a}{b+c+d}$ :)

er,I don't know.
I just think there should be a more symmetric form. :)

nice solution, Hawk Tiger !
[quote="Hawk Tiger"]
I just think there should be a more symmetric form. :)[/quote]
Right part $\frac{a}{b+cd}+\frac{b}{c+da}+\frac{c}{d+ab}+\frac{d}{a+bc}\le2$ also admits symmetric form:

$\frac{a}{b+cd}+\frac{b}{c+da}+\ldots\le\frac{a}{b+c+d-1}+\frac{b}{c+d+a-1}+\ldots\le2!$  :lol:

[quote="Hawk Tiger"] $\frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=$ $(a+b+c+d)(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c})\geq \frac43$[/quote]

Can someone please explain me this step ??    ;)  thanks ^^

Here ,apply AM-GM inequality or apply Chauchy inequality

[quote="bodan"][quote="s1ck_psych0"]And how come:

[quote="Hawk Tiger"]
$\frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c})$[/quote]

??[/quote]

Just a typo but it's still correct: \[ \frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right)-4\geq\frac{16}{3}-4=\frac{4}{3}.  \] Pay attention, s1ck_psych0  ;)[/quote]

thx. bodanche! :pilot:

[quote="s1ck_psych0"][/quote][quote="bodan"][quote="s1ck_psych0"]And how come:

[quote="Hawk Tiger"]
$\frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c})$[/quote]

??[/quote]

Just a typo but it's still correct: \[ \frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right)-4\geq\frac{16}{3}-4=\frac{4}{3}.  \] Pay attention, s1ck_psych0  ;)[/quote][quote="s1ck_psych0"]

thx. bodanche! :pilot:[/quote]

Poveli i drugpat, visarce  ;)

[quote="bodan"][/quote][quote="s1ck_psych0"][/quote][quote="bodan"][quote="s1ck_psych0"]And how come:

[quote="Hawk Tiger"]
$\frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c})$[/quote]

??[/quote]

Just a typo but it's still correct: \[ \frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right)-4\geq\frac{16}{3}-4=\frac{4}{3}.  \] Pay attention, s1ck_psych0  ;)[/quote][quote="s1ck_psych0"]

thx. bodanche! :pilot:[/quote][quote="bodan"]

Poveli i drugpat, visarce  ;)[/quote]

mali aj ne mi se praj fraer, ne igraj igracki tuka, aj da ne se brukame   ;)

[quote="bodan"]Just a typo but it's still correct: \[ \frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right)-4\geq\frac{16}{3}-4=\frac{4}{3}.  \] Pay attention, s1ck_psych0  ;)[/quote]

I haven't yet understand from where comes the $\frac{16}{3}$ I have $\frac{16}{6}$...

But I have another solution.

We have $E=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c} = f(a,b,c,d)$
if $a=b=c=d \to E=\frac43 \ge \frac43$ so this case okey.

We can say $a \ge b \ge c \ge d$. For the moment we have $a=b=c=d$.
Let's fix $b,c,d$ and $E=f(a)$. So $f'(a) = \frac{1}{b+c+d} - \frac{b}{(c+d+a)^2} - \frac{c}{(d+a+b)^2} - \frac{d}{(a+b+c)^2}$. We see that if we grow $a$ then $E=f(a)$ grows too and we are done.

[quote="bodan"][quote="BogG"][quote="bodan"]Just a typo but it's still correct: \[ \frac{a+b+c+d}{b+c+d}+\frac{b+c+d+a}{c+d+a}+\frac{c+a+b+d}{d+a+b}+\frac{d+a+b+c}{a+b+c}-4=(a+b+c+d)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right)-4\geq\frac{16}{3}-4=\frac{4}{3}.  \] Pay attention, s1ck_psych0  ;)[/quote]

I haven't yet understand from where comes the $\frac{16}{3}$ I have $\frac{16}{6}$...

[/quote]

We have \[ \left((a+b+c)+(b+c+d)+(c+d+a)+(d+a+b)\right)\left(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}\right)\geq16  \] by Caushy  ;)[/quote]

Or simply by AM-HM  ;)


For a;b;c are positive real numbers satisfying $a \leq b \leq c$.Prove that:
$\frac{b}{c}+\frac{c}{a}\geq \frac{a}{b}+\frac{b}{a}$

it's equivalent to
$b^{2}\left(\frac1a-\frac1c\right)+a-\frac{cb}{a}\le 0$
because of convexity we need only check $(b-a)(b-c)=0$ which is trivial.


Hi,

here is the problem:

there are 1000 student lockers in a room,
student number 1 opens all the lockers then;
Student number 2 closes all the 2nd lockers then;
student number 3 changes all the 3rd lockers (i.e if it was open he/she closes it, if it was closed he/she opens it) then;
student number 4 changes all the 4th lockers (same as above) then
student number 5 changes all the 5th lockers, and so on til student number 1000

How many lockers will be open at the end, Give reasons

My method was to do the first few then work out a pattern, as it turned out the square numbers were the lockers that stayed open i.e locker 1, 4, 9, 16, etc since they will have an odd number of factors (it seemed that even number of factors resulted in the locker remaining closed).  It was using this theory that the number of open lockers is 31, since it is the last number w/ a square under 1000.  Am i right?

Problem 2

A man says in his will that he will give all his money to his children as follows;

1st born - $1000 plus 10% of what is left at this point
2nd born - $2000 plu 10% of what is left at this point
and so on, all children recieve the same.  How many children are there?

My method was trial and error, i just started at 51,000 and moved up until i hit 81,000 and the 1st and 2nd born got the same ($9000) hence there were 9 children.  The problem with this is i had no set method, do you guys know of any way to do this quickly for any set of terms?

thanks

about firs one: each locker changes status for every divisor, so lockers that will end up open are those with odd number of divisors i.e squares!


geometry
geometric transformation
dilation

i have nearly ten new words in you post and i can't find them in the dictionary.
But i still wanna have a try:
Blue implies Cu.[/hide]

yes, obviously metal A is $Cu$
$Cu+B\rightarrow Cu_xB_2$ my guess for $B$ is $S$ because of the color of the compound, but your question doesnt ask for $B$ so don'r worry about it:
$CuS+H_2SO_4\rightarrow CuSO_4+H_2S$ hopefully you know that the color of $CuSO_4$ is blue. so $D=CuSO_4$
then

$CuSO_4+NH_3\rightarrow [Cu(NH_3)_4]^{2+}+SO_4^{2-} $ so $E=[Cu(NH_3)_4]^{2+}$
there you are done
 :D
oh, by the way, i didn't balance the equation, you should be able to do that,(hopefully :) )

[quote="inteluser"]I don't think $\text{S}$ occurs too often as a gas, though.  Maybe $\text{CuO}$?

$\text{CuO} + \text{H}_2\text{SO}_4\rightarrow\text{CuSO}_4+\text{H}_2\text{O}$ ?

What's the light blue precipitate, by the way?  Anhydrous $\text{CuSO}_4$?[/quote]
no, of course $S$ can be gas,
$CuO$ is dark Red. so $B$ can not be $O_2$
and yes the blue percipitate is ${CuSO}_4$

Interesting.  Is sulfur gas $\text{S}_8$?

So is it just monatomic $\text{S}$?  $\text{S}_2$?

i believe it is in a form of $S_2$ (not 100%sure... ;) )

Copper I oxide, Cu2O, is red.  Copper II oxide, CuO, is black.

$CuS$ is black too. but that doesn't matter because the question is not asking for $B$ ;)

I'm pretty sure that A-D meant A [i]through[/i] D...

[quote="Devendra Patel"]When metal A is heated in air,it reacts with gas B to produse compound C,which is a black powder.Compound C reacts with dil sulphuric acid to give a blue soolution of compound D. Addition of aqueous ammonia dropwise to this blue solution gives a light blue precipipate that dissolves in excess ammonia to give a deep blue solution containing the complex ion E.
1) Identify the substances A to D
2) Write balanced chemical equations for the reaction of the compound C with Dilute Sulphuric acid
3) give the formula for the complex ion E[/quote]

my guesses

blue -> copper. gases in air are mostly oxygen, nitrogen, likely one is oxygen. thus:

$2\text{Cu} + \text{O}_2 \to 2\text{CuO}$, $\text{CuO} + 2\text{H}^+ \to \text{H}_2\text{O} + \text{Cu}^{2+}$, $\text{Cu}^{2+} + \text{SO}_4^{2-} \to \text{CuSO}_4$, $\text{CuSO}_4 + 2\text{NH}_3 \to \text{[Cu(NH}_3\text{)}_2\text{]}^{2+} + \text{SO}_4^{2-}$

Something makes me want to say that would be $\text{[Cu(NH}_3\text{)}_4\text{]}^{2+}$, not $\text{[Cu(NH}_3\text{)}_2\text{]}^{2+}$.

Why does anhydrous $\text{CuSO}_4$ precipitate on addition of $\text{NH}_3$?  I know $\text{CuSO}_4$ is a grayish powder anhydrous, and a blue powder as $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$, but that doesn't seem to help me.  Is $\text{CuSO}_4$ less soluble in basic solution?

EDIT: Pauling ([i]General Chemistry[/i]) states that $\text{S}_2$ is unstable and can only be formed by heating $\text{S}_8$ or $\text{S}_6$ to "high temperatures."  Ordinary sulfur ($\text{S}_8$) boils at 717.8K, so I'm not sure that sulfur gas is very common.

i guess, since Cu is a transition metal, it will form a ligand, because you have EXCESS $NH_3$, also it might be that $SO_4^{2-}$ is a conjucate base of weak acid...

[quote="frankthetank"][quote="inteluser"]Something makes me want to say that would be $\text{[Cu(NH}_3\text{)}_4\text{]}^{2+}$, not $\text{[Cu(NH}_3\text{)}_2\text{]}^{2+}$.

Why does anhydrous $\text{CuSO}_4$ precipitate on addition of $\text{NH}_3$?  I know $\text{CuSO}_4$ is a grayish powder anhydrous, and a blue powder as $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$, but that doesn't seem to help me.  Is $\text{CuSO}_4$ less soluble in basic solution?
[/quote]

I think the precipitate is $\text{Cu(OH)}_2$, which redissolves to form the tetraamine complex.  We did a lab about this in class.[/quote]
i dont think $Cu(oH)_2$ is blue, but $CuSO_4$ IS blue.

A few comments:

Since the metal is heated [i]in air[/i], the gas it reacts with should be present in air.  Oxygen is the most likely candidate since nitrogen is very unreactive in comparison and no other compounds are present in very large quantities in air.

${\rm Cu}^{2+}$ does indeed form a tetraamine complex, though its formation reaction can be split into four steps each with its own equilibrium constant.  However, you'll seldom find formulas written out for anything but the tetraamine.

Addition of ammonia could, in principle at least, lead to formation of either Cu(OH)$_2$(s) or CuSO$_4$(s) precipitates.   In the first case, ammonia acts as a base forming hydroxide ions for precipitation.  In the latter case, addition of ammonia shifts the equilibrium

$\rm HSO_4^- \rightleftharpoons H^+  + SO_4^{2-}$

to the right, increasing the concentration of free sulfate ion for precipitation with the copper.   However, copper (II) sulfate is fairly soluble, so it's not clear that the concentrations would lead to precipitation.   Copper (II) hydroxide, in constrast, has a very low solubility ($K_{sp} \approx 10^{-20}$), so it could precipitate out under even mildly basic conditions.   And it is blue too, if I'm reading my CRC handbook correctly.  (As a theoretical chemist, remembering colors of compounds is one of my weak points...)


Hi! I'm kind of new to this..so I don't know if this is actually possible.

I'm writing a book with lots of small sections - to small to justify a file each,but annoying when they're all in one file. Is there anyway to create a list in which  you can minimise the sections that you aren't working on and just see the headings?

e.g. if you go to Help, Contents of a program, there's a list of topics and you can click on the small + sign on the left and it drops down a list of items in that section.

Hope someone understands that and can help!

I assume that you want this in your editor, rather than in the final document. Which editor are you using? TeXnicCenter's navigator bar gives the sort of list you want provided you open the file as a project. WinEdt will do the same - see the snapshots [url=http://www.winedt.com/snap.html]here[/url].

Thanks for your quick answer - mot exactly what I meant, but it'll do the job!
Have a nice day.


By putting 1.80  dollars worth of gas in his Ford Ranger, the gas tank needle moves from 1/4 full to 3/8 full. What additional amount must Ben pay to finish filling the gas the tank? (Assume that the needle is always accurate.)

A. 5.40 d     B.9.00 d     C.9.90 d    D.10.80 d      E.14.40 d

[hide] He spends $ \$1.80$ for $ \frac38 \minus{} \frac14 \equal{} \frac18$ of the tank. Therefore, in order to fill the remaining $ \frac58$ of the tank, he will need $ 5(1.80) \equal{} \boxed{\$9.00}$. Thus, the answer is B.[/hide]

yep.. I got the answer as [hide="."]B[/hide]

Guys, Please do not post solutions for what has already been posted.


function
quadratics
number theory unsolved
number theory

N is set of all positive integers, and let ${p}$ a prime number..
the function ${f: N\to{{-1.0.1}}}$  which is not constant satisfies the following contitions.

(1) for all $m,n\in\mathbb{N}$ , $f(mn)=f(m)f(n)$
(2) for all  $n\in\mathbb{N}, f(n+p)=f(n)$

if we difine ${a}=\sum_{n=1}^{p-1}f(n){c^{n}}$ for complex number ${c}$ that ${c}^{p}=1,{c}\not=\1$, find all value that ${a}^{2}$ can have.

Let's assume $p$ odd.
By (2), the value depends only $\mod p$.
Let's find all possible $f$ first (neglecting that it isn't allowed to be constant).

If $f(0)=1$, then $1=f(0)=f(0 \cdot n) =f(0) f(n) = f(n)$ for all $n$, thus the function is constant $1$.
Assume there is some $m \not\equiv 0 \mod p$ with $f(m)=0$. For all $a$ there is an $n$ with $mn \equiv a \mod p$, thus $f(a)=f(mn)=f(m)f(n)=0$ for all $a$. So this gives always the constant function $0$.

So from now we can assume $f(0) \neq 1$ and $f(m) \neq 0$ for $p \nmid m$.
But if $a$ is a square $b^{2}\mod p$, we get that $f(a)=f(b^{2})=f(b)^{2}\geq 0$, thus $f(0)=0$ ($0$ is clearly a square) and $f(a)=1$ for all quadratic residues $a \mod p$.
Let $c$ be any fixed nonsquare $\mod p$. Then all nonsquares are of type $ac \mod p$, where $a$ runs through the quadratic residues $\mod p$. Thus, if $d$ is any nonsquare, we get that $f(d) = f(ac)= f(a) f(c) = f(c)$.
Depending whether we choose $f(c)=1$ or $f(c)=-1$, we get either the function \[f(n) = \begin{cases}0 \text{ iff }p|n \\ 1 \text{ otherwise}\end{cases}\] or the Legendre symbol \[f(n) = \left( \frac{n}{p}\right) = \begin{cases}1 \text{ iff }n \text{ is a quadratic residue }\mod p \\ 0 \text{ iff }p|n \\-1 \text{ iff }n \text{ is a nonsquare }\mod p \end{cases}.\] Let $\zeta_{p}= e^\frac{2 \pi i}{p}$ and set $g_{a}= \sum_{k=1}^{p-1}f(k) \zeta_{p}^{ak}$.
Now for the constant ones and the first one of the last two, it's easy to check that $g_{a}=-f(1)$.
For the remaining Legendre symbol case:
$g_{a}= f(a) g_{1}$, proof by simple checking.
Thus $g_{a}^{2}= f(a)^{2}g_{1}^{2}= g_{1}^{2}$ for all interesting $a$.
But it's well known that $g_{1}^{2}= (-1)^\frac{p-1}{2}p$.


Determine a condition (with proof) such that the initial value problem ($ y'=f(x,y)$ and $ y(x_{0})=y_{0}$) always has a unique solution $ \phi (x)$ for $ a \leq x \leq b$


Find all functions such that
$f(x) + f({1}/{(1-x)}) = {(2(1-2x))}/{(x(1-x))}$ for every real number $x$, except for $x=0$ and $x=1$.

Call $g(x) = \frac{1}{1-x}$.  We find that $g(g(g(x))) = x$.

Now, $f(x) + f(g(x)) = \frac{2(1-2x)}{x(1-x)} = \frac{2}{x} - \frac{2}{1-x}$ by the method of partial fractions.  We can get the following system by repeated substitution $x \to g(x)$.

$f(x) + f(g(x)) = \frac{2}{x} - \frac{2}{1-x}$

$f(g(x)) + f(g(g(x)) = 2(1-x) - \frac{2(1-x)}{(1-x) - 1} = 2 - 2x - \frac{2 - 2x}{-x} = \frac{2}{x} - 2x$

$f(g(g(x)) + f(x) = 2(1-x) - \frac{2}{1-x}$

Add the first to the third and subtract the second to get

$2f(x) = \frac{2}{x} - \frac{2}{1-x} - \frac{2}{x} + 2x + 2 - 2x - \frac{2}{1-x} = 2 - \frac{4}{1-x}$
$\boxed{ f(x) = 1 - \frac{2}{1-x} }$

(Edit:  $888^{th}$ post!  Lucky :D )


The lengths of six sticks are the first six prime numbers. How many ways can three of the six sticks be selected so that a triangle can be formed with the three sticks?

[hide="Length of Sticks"]
2,3,5,7,11,13
[/hide]

[b]Triangle Inequality[/b]: Given three sides of triangle ABC, $ a\plus{}b>c$

[hide="Amazing Brute Force"]
The stick with length 2 cannot be part of the triangle. (Can you see why?)

[hide="Length 3"]
3,5,7
3,11,13
[/hide]

[hide="Length 5"]
5,7,11
5,11,13
[/hide]

[hide="Length 7"]
7,11,13
[/hide]

Adding the possible triangle lengths up, we have a grand total of $ \boxed{5}$ ways.
[/hide]

I could be wrong, so feel free to correct.


topology
function
real analysis
real analysis unsolved

Prove that the closure of the set S is equal to the set of the limits of all sequences whose elements are just in S.

I'm approaching this by first proving that the set of all limits is a subset of the closure of S (done that), and then showing that the closure is a subset of set of limits. This is the part I am confused about.

This depends on the nature of the topological space you're working in. It's true in any metric space, but there are more exotic spaces in which it fails.

An example: Let $ X$ be $ \{0,1\}^{\mathbb{R}}$, with the product topology, and let $ S$ be the subset of functions $ f$ such that $ f^{\minus{}1}(1)$ is finite. No sequence in $ S$ converges to the constant function $ 1$, but the closure of $ S$ contains $ 1$, since every open set containing $ 1$ contains elements of $ S$. In fact, the closure of $ S$ is all of $ X$; every nonempty open set in $ X$ contains elements of $ S$.

An example that's similar in spirit: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=166959]two different topologies[/url] on the same space with the same convergent sequences.

Yeah, I should have mentioned that I am working with metric spaces. Anyone else have any ideas?

If $ x$ is in the closure of $ S$, every open set containing $ x$ contains at least one point of $ S$. This includes the balls of radius $ \frac1n$...


function
calculus
derivative
parameterization
integration
trigonometry
real analysis

Suppose you have a function $ f(x)$ with derivatives.
You make a sample of points of this function in a given interval and make a spline interpolation of some order.
To adjust the parameters you can also use the derivative of $ f$. But you instead use another function $ g$ for some reason.
When you make the plot of the interpolation it is very close to $ f$ as expected.
But when you derive the interpolation, the function is not close to the derivative of $ f$.

You want to know the arc-length of the curve using the interpolation, but how can this be donne?
If you make $ \delta$ small in the integral to incress precision, the derivative will get worse and worse.

Example using $ \cos(x)$ sampled with 500 points in the intervall from $ 0$ to $ \frac{\pi}{2}$ using $ x$ as derivative.
The plot of $ \|\frac{inter(x\plus{}\delta )\minus{}inter(x)}{\delta }\minus{}\cos (x)\|$ for $ \delta$'s for the value $ x\equal{}0.4$

The error is not zero on zero. And for more complex functions, the behavior is more chaotical.

And my intention were to calculate the arc-length of this spline.
But how can I get more precision, or it is not possible by definition?

I don't understand which arclength you want: of $ f$ or of the spline? If of the spline (as you write at the end), then just forget about $ f$ and $ g$, and integrate $ \sqrt{1\plus{}(h')^2}$ where $ h$ is your spline.

I understood it to mean that the OP wants the length of the spline to be close to the length of the curve $ f$, whose derivative we for some reason do not have access to.

Then I think that using $ g$ "as a derivative" is not the best idea. With a function as nice as $ \cos$, and with 500 sample points, you can get a sweet approximation without derivative. Something using more than two adjacent sample points, like in Simpson's method of integration.

Also,  $ x$ does not look much like $ (\cos x)'\equal{}\minus{}\sin x$. Did you mean $ \minus{}x$?


hello, there exists a general way to solve the equation
$ a\sin(x)\plus{}b\cos(x)\equal{}c$ dividing by $ \sqrt{a^2\plus{}b^2}$ we have
$ \frac{a}{\sqrt{a^2\plus{}b^2}}\sin(x)\plus{}\frac{b}{\sqrt{a^2\plus{}b^2}}\cos(x)\equal{}\frac{c}{\sqrt{a^2\plus{}b^2}}$ and now let $ \sin(\phi)\equal{}\frac{b}{\sqrt{a^2\plus{}b^2}}$ and
$ \cos(\phi)\equal{}\frac{a}{\sqrt{a^2\plus{}b^2}}$ and using the addition formulas
we have $ \sin(x\plus{}\phi)\equal{}\frac{c}{\sqrt{a^2\plus{}b^2}}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, there exists a general way to solve the equation
$ a\sin(x) \plus{} b\cos(x) \equal{} c$ dividing by $ \sqrt {a^2 \plus{} b^2}$ we have
$ \frac {a}{\sqrt {a^2 \plus{} b^2}}\sin(x) \plus{} \frac {b}{\sqrt {a^2 \plus{} b^2}}\cos(x) \equal{} \frac {c}{\sqrt {a^2 \plus{} b^2}}$ and now let $ \sin(\phi) \equal{} \frac {b}{\sqrt {a^2 \plus{} b^2}}$ and
$ \cos(\phi) \equal{} \frac {a}{\sqrt {a^2 \plus{} b^2}}$ and using the addition formulas
we have $ \sin(x \plus{} \phi) \equal{} \frac {c}{\sqrt {a^2 \plus{} b^2}}$.
Sonnhard.[/quote]
how would you solve for sin x afterwards? and what about $ \phi$?

can't we use the pythagorean identities? cause i don't really understand your solution :(

well i was reading some old trig books (you can find them in google book search)
AND...

since a=2, b=-1, $ \sin{\phi}\equal{}\frac{2}{\sqrt{5}}$ and $ \cos{\phi}\equal{}\frac{1}{\sqrt{5}}$

and so $ \tan{\phi}\equal{}1/2$ and solve for $ \phi$ by inverse trig.
Then $ x\plus{}\phi\equal{}\sin^{\minus{}1} {1}$ since $ c\equal{}\sqrt{5}$
you know $ \phi$ and $ \sin^{\minus{}1}{1}\equal{}90$

now Solve for x.


how is the weather in merida these days?

[quote="Mildorf"]Actual temperatures are often 100°F (37.7°C) but due to humidity can feel like 110°F+ (43.3°C+). Also, be sure to have lots of sunblock.[/quote]Wow ... it's THAT bad? :(

well, what about rain? I mean do we have to take our umbrella?

it seems very very hot weather? And in that temparature our head is going to be boiled! And in breakfast we will eat that, so how can we make our exam well :?

 :D

We wanted that they gived only mexican food for breakfast on the exam days, but someone told us that they are going to give international food (includes mexican) in all the meals, exept (maybe) for the last dinner

rain !! at 37°C !!!!!!!!! is that possible ? what do u mean exactly? is it hot , cold ?? :?  so what should we take with us, swimming suits or rain coats?  :?  :?  :?  :?

[quote="wisemaniac"]so what should we take with us, swimming suits or rain coats?  :?  :?  :?  :?[/quote]I would take both if I were you :D

im not going to the imo but i saw in the news that there are a couple of hurricanes around the mexico area!

It wouldn´t be the first time that a math olimpyad is hold near huracans, however, are you sure that there are huracans around Mexico? When there are, it starts to rain a lot in my city and that hasn´t passed yet

Here's the forecast for today in Meridia, Mexico: [url=http://www.weather.com/outlook/travel/businesstraveler/local/MXYN0117?lswe=mexico&geoCd=4&lswa=WeatherLocalUndeclared&countryName=Mexico&what=WeatherLocalUndeclared&itemCd=MX&countryCd=MX&geoCdChild=1&itemName=Mexico]forecast for Merida[/url]
Here's the  [url=http://www.weather.com/activities/travel/businesstraveler/weather/tenday.html?locid=MXYN0117&from=36hr_topnav_business]10 day forecast[/url].
Look's like it will be in the high 90's most of the time for you guys and feeling like 110 (Fahrenheit).   Looks like a fair chance for rain at least once in the next 10 days.

[quote="Palytoxin"]is the hurricane affecting there in anyway? just wondering...[/quote]No, not at all :)
Joml please next time use the [url ] tag when posting very long url's (the pages tend to gain horizontal scroll when you do that :) ).

The weather could get interesting about four days from now; Hurricane Emily is currently in the eastern Caribbean, and its projected path takes it over the Yucatan.

Merida has the protection of being on the west side of the peninsula and not being coastal, so you won't get the worst of it. On the other hand, high winds and heavy rain are likely.

Accord to USA's National Hurricane Center, Hurricane "Emily" will very near Merida on Monday morning.

http://www.nhc.noaa.gov/

 :(  :o

Is the IMO making any preparations or changes?

Hurricane Emily currently has sustained winds of 185 km/hr (115 mph). US National Hurricane Center currently projects a track directly at Merida area.  Hurricane to reach Merida area at about 1 am or 2 am Monday morning IF current forecasts are correct. Some strengthening is expected.

Use this link

http://www.nhc.noaa.gov/

and then select "Warnings and 5-Day Cone"

 :o  :o  :o  :o

Any preparations made?

- Worried IMO parent

[quote="Hip"]Any preparations made?

- Worried IMO parent[/quote]I have no idea if they are any preparations made for this. I will ask  :|

Thank you very much.

Maybe I am just a wimp (a Worried IMo Parent).     :)

[quote="Hip"]Maybe I am just a wimp (a Worried IMo Parent).     :)[/quote]I think that you should worry more about the score of your kid :D :D

Well, the competition (his part) is over and the hurricane is yet to come....

 :|

One thing is sure: They cannot leave Mexico if the hurricane is there. Stay inside the hotel and it should be safe.

I predicted that a hurricane was going to Merida and Dave (he's from Mexico) said that there was no hurricane near the area  :D
but yeah i live in a place where we get strong hurricanes all the time, so what u have to do is to stay in the hotel and everything will be fine, but dont go out because there are many dangerous things flying around.

see nhc link for latest info: http://www.nhc.noaa.gov/text/refresh/MIATCPAT5+shtml/160557.shtml?

We just had a special meeting dedicated to this (finally). It seems that everything is serious, so the program will be changed.

It also seems that there will be no closing ceremony (it depends on what state of emergency the Governor of the State Yucatan decretes tonight ...
 :? hopefully there  will be no need for ... :stretcher:

[quote="Yossarian"]Our flight is on Tues morning, so...
Looks like I can make an insurance claim!
Finally saw Valentin. Guy wearing a mathlinks shirt. Very obvious.[/quote]My flight is to Houston on Tuesday morning as well (I think many people are taking that flight :) ). Hopefully the hurricane won't affect it too much  :?

PS Yes, it is obvious ain't it?  :lol:

[quote="tywebb"]i just saw something on the official imo website http://erdos.fciencias.unam.mx/hurricane.htm which said

"Saturday, July 16th

The necessary precautions are being taken by the government of Yucatan State, in the still remote chance that the coast of Yucatan is affected by Hurricane Emily. The IMO participants are prepared, but at this time there has been no indication of any impending threat. Ceremonies and trips are still confirmed at this time."

is this out of date, or something?[/quote]

The center of the hurricane will pass some 50-100 km north of merida, but still thing are very serious. This is the worst hurricane here since 1988, and still there are quite a lot of hurricanes here. We won't be sleeping in our hotel rooms this night, but in one common room without windows (at least in FiestaAmericana).

[quote="Stijn"]We won't be sleeping in our hotel rooms this night, but in one common room without windows (at least in FiestaAmericana).[/quote]Indeed we didn't sleep (well at least some of us). It was the funniest hurricane ever :P Everybody stayed in a huge room, we have 2 pool tables, 3 ping pong tables, 2 table hockey tables, 4 table-soccer tables, mafia (and variations), poker, pillow fights, singing, one birthday and much much more. I will upload the pictures I took yesterday night in a couple of days  :P

Valentin, thanks for letting us all know  :lol: - this was what we wanted to hear and what we couldn't see on CNN or satellite images!   My prediction was close - our team would be in the big room playing cards all night and not even notice the hurricane.

A concerned but not unduly worried parent.

its good to hear news from mexico! but you guys are not the only ones being hit by massive storm
this year IChO are also having a misfortunate hit by a typhoon in Taiwan (believe me typhoons are way stronger than hurricanes) i wonder if they are OK...

[quote="Hip"]Yes,  thank you for letting us know.

I heard that an awards ceremony will take place as well.  Very good.

- Hip (a/k/a the "WIMP")  :)[/quote]It did take place :) It was quite nice actually. Everyone danced in the end, and everybody took pictures with (almost) everybody else  :lol:

I will upload some of the pictures from the "hurricane party" as we call it now :P


It takes me 15 min to run seven miles, How many miles can I run in 12 min?

We can set up the ratio $ \frac{7}{15}\equal{}\frac{x}{12}$. Solving we get:
$ 15x\equal{}84 \implies x\equal{}\frac{84}{15}\equal{}\boxed{5\frac{3}{5}miles}$

[quote="PowerOfPi"]We can set up the ratio $ \frac {7}{15} \equal{} \frac {x}{12}$. Solving we get:
$ 15x \equal{} 84 \implies x \equal{} \frac {84}{15} \equal{} \boxed{5\frac {3}{5}miles}$[/quote]
    your equation is right but id your answer right?

I'll do it another way.
You could say you run 7 miles in 15 minutes, or $ \frac{7}{15}$ of a mile in a minute. To find the miles per 12 minutes, simply multiply by twelve to get:
$ \frac{7}{15}\cdot12\equal{}\frac{28}{5}\equal{}5\frac{3}{5}\equal{}5.6             miles$
I think that is right.

[quote="smartytarty"][quote="PowerOfPi"]I'll do it another way.
You could say you run 7 miles in 15 minutes, or $ \frac {7}{15}$ of a mile in a minute. To find the miles per 12 minutes, simply multiply by twelve to get:
$ \frac {7}{15}\cdot12 \equal{} \frac {28}{5} \equal{} 5\frac {3}{5} \equal{} 5.6 miles$
I think that is right.[/quote]
Congrats  :10:  :10:  :clap2:[/quote]

Now here's where you come in Akhil. (If you don't get it, then ask akhil.)

I can tell you are pretty new, but please think about it before you press the submit button. Is your post contributing? Is your post necessary?


inequalities
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Suppose that the number of used cars sold each day at a dealershop is a random variable X, with mean 50 and variance 25.

a. Using Chebyshev's Inequality, find a lower bound for the probability that the dealer will sell more than 44 and less than 56 cars tomorrow?

b. Approximate the probability that, over the next 100 days, the dealer will sell at least 4940 cars.

a. This is $P(|X-50|\le 6)=1-P(|X-50|\ge7).$

Let $Z=\frac{X-50}5.$ This has bee normalized so that $E(Z)=0$ and $\text{Var}(Z)=1.$

$P(|X-50|\ge 7)=P\left(|Z|\ge\frac75\right)\le\left(\frac75\right)^{-2}.$

Hence, $P(|X-50|\le 6)\ge1-\left(\frac57\right)^{2}=\frac{25}{49}.$

The actual probability could easily be considerably greater than this; Chebyshev is seldom anywhere close to being a "tight" inequality.

b. The question cannot be answered unless we know how sales on different days are related. The default assumption (reasonable or not) would be independence, so in order to able to work the problem, I'll assume independence.

For the sum of 100 days of sales, the expected value is the sum of the expected values: $100\cdot50=5000.$ The variance (here's where the assumption of independence comes in) is the sum of the variances: $100\cdot25=2500.$ Hence the standard deviation is $\sqrt{2500}=50.$

Let $Y$ be the number of sales and define $Z=\frac{Y-5000}{50}.$ Now we'll quote the Central Limit Theorem and make the assumption that $Z$ is approximately standard normal.

$P(Y<4940)=P(Y<4939.5)= P\left(Z<\frac{4939.5-5000}{50}\right)$

$= P(Z<-1.21)$

From a table of normal probabilities, the probability is approximately $.113.$


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[b]Problem 1[/b]: Given $ a,b,c\ge 0$.Prove that

$ \sum_{cyc} \frac{(2a\plus{}b\plus{}c)^2}{2a^2\plus{}(b\plus{}c)^2} \le 8$

[b]Problem 2[/b]:[Phan Thành Việt] Given $ a,b,c\ge 0$.Prove that

$ \sum_{cyc}\frac{a^4}{b^2\plus{}bc\plus{}c^2} \ge \frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c}$

[b]Problem 3[/b]: Given $ a,b,c\ge 0$.Prove that
$ \frac8{27}(ab\plus{}bc\plus{}ca)^3\le(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)\le \frac1{64}(a\plus{}b\plus{}c)^6$

[hide="1"]
We have
\[ \frac{(2a\plus{}b\plus{}c)^2}{2a^2\plus{}(b\plus{}c)^2}\equal{}\frac{4a^2\plus{}4a(b\plus{}c)\plus{}(b\plus{}c)^2}{2a^2\plus{}(b\plus{}c)^2}\equal{}1\plus{}\frac{2(a^2\plus{}2ab\plus{}2ac)}{2a^2\plus{}(b\plus{}c)^2}\]
Now Cauchy gives $ (a\plus{}b\plus{}c)^2\leq \frac{3}{2}(2a^2\plus{}(b\plus{}c)^2)$, so that
\[ \sum_\text{cyc} \frac{a^2\plus{}2ab\plus{}2ac}{2a^2\plus{}(b\plus{}c)^2}\leq \frac{3}{2}\cdot \frac{a^2\plus{}b^2\plus{}c^2\plus{}4(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2}\leq \frac{5}{2}\]
which is equivalent to the desired inequality.
[/hide]

[hide="2"]
Use Cauchy to get
\[ \left( \sum_\text{cyc} a^2(b^2\plus{}bc\plus{}c^2)\right) \left( \sum_\text{cyc} \frac{a^4}{b^2\plus{}bc\plus{}c^2}\right) \geq (a^3\plus{}b^3\plus{}c^3)^2\]
so it remains to prove
\[ \left( \sum_\text{cyc} a^2(b^2\plus{}bc\plus{}c^2)\right) \leq (a^3\plus{}b^3\plus{}c^3)(a\plus{}b\plus{}c)\]
which is clear upon expanding.
[/hide]

Q1 is USAMO 03 Q5.
Here's another solution:
http://www.kalva.demon.co.uk/usa/usoln/usol035.html

[quote="dduclam"][b]Problem 1[/b]: Given $ a,b,c\ge 0$.Prove that

$ \sum_{cyc} \frac {(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \le 8$

[/quote]
http://www.mathlinks.ro/viewtopic.php?search_id=1952506903&t=53672
http://www.mathlinks.ro/Forum/viewtopic.php?t=46
http://www.mathlinks.ro/Forum/viewtopic.php?t=28678
http://www.mathlinks.ro/Forum/viewtopic.php?t=43278
http://www.mathlinks.ro/Forum/viewtopic.php?t=48989

[quote="scorpius119"][hide="1"]
We have
\[ \frac {(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \equal{} \frac {4a^2 \plus{} 4a(b \plus{} c) \plus{} (b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \equal{} 1 \plus{} \frac {2(a^2 \plus{} 2ab \plus{} 2ac)}{2a^2 \plus{} (b \plus{} c)^2}
\]
Now Cauchy gives $ (a \plus{} b \plus{} c)^2\leq \frac {3}{2}(2a^2 \plus{} (b \plus{} c)^2)$, so that
\[ \sum_\text{cyc} \frac {a^2 \plus{} 2ab \plus{} 2ac}{2a^2 \plus{} (b \plus{} c)^2}\leq \frac {3}{2}\cdot \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 4(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\leq \frac {5}{2}
\]
which is equivalent to the desired inequality.
[/hide]

[hide="2"]
Use Cauchy to get
\[ \left( \sum_\text{cyc} a^2(b^2 \plus{} bc \plus{} c^2)\right) \left( \sum_\text{cyc} \frac {a^4}{b^2 \plus{} bc \plus{} c^2}\right) \geq (a^3 \plus{} b^3 \plus{} c^3)^2
\]
so it remains to prove
\[ \left( \sum_\text{cyc} a^2(b^2 \plus{} bc \plus{} c^2)\right) \leq (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)
\]
which is clear upon expanding.
[/hide][/quote]
In truth, it seems to be true but it's not obvious! You should show the rest of the solution! I think it's wrong!

Sorry because of my mistake!  :blush: Your last inequality is true which is proved  by B.C.S
Here is my solution for the rest:
(a^3+b^3+c^3)(a+b+c)>=(a^2+b^2+c^2)^2
Then we need to prove:(a^2+b^2+c^2)^2>=2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c), which is clearly true by applying Cauchy's inequality :)


two positive integers x, y are such that  x+y = a
For which numbers a, xy can be divided by a

$y = a -x $ thus $ xy = x ( a-x) = ax -x^2 $
Thus if $ xy \equiv 0 \bmod {a} \\
\\ ax - x^2 \equiv 0 \bmod {a} \\
\\ x^2 \equiv 0 \bmod {a} $

Answer: not a sqaure-free number

obviosly a>=2

if a is a sqaure-free number then

a must divide xand y

so a=x+y>=a+a

contradiction

consider a is not sqaure-free number.

there exist a prime number p such that p^2 divides  a

set x=a/p  and y=(p-1)a/p

then x+y=a, and xy=(a/p^2)(p-1)(a)

Q.E.D.

Isn't it obvious that he means squarefree numbers¿

[quote="Rushil"]What's a sqaure free number???[/quote]

Square free number <--------the number don't have any square factor inside.

you have done nothing wrong, but you have done nothing to determine the property of a as well.


group theory
abstract algebra
superior algebra
superior algebra unsolved

1) If $ ba=a^4b^3$, prove that $ \text box{ord}(a^4b)=\text box{ord}(a^2b^3)$.

2) Find all homomorphisms from $ \mathbb{Z}_{12}$ to $ \mathbb{Z}_{30}$.

3) Let $ G$ be a finite group $ A \subset G$ and $ |A|>\frac{|G|}{2}$. Prove that each element of $ G$ can be written as product of two elements of $ A$.

4) Let $ G$ be a subgroup of $ S_n$ and $ G \not=A_n$. Prove that $ G$ has exactly $ \frac {|G|}{2}$ even elements.

5) Let $ G$ be a group, if $ |G|=2k$ and $ k$ be an odd number, prove that there exist a subgroup of $ G$ like $ H$ where $ |G: H|=2$.

2) A homomorphism $ f$ is uniquely defined by the image of $ 1$. For it to be well defined, you need that $ 12 f(1) \equal{}0$.

3) See post number 6 in http://www.mathlinks.ro/viewtopic.php?t=194818 for the general version.

4) This is wrong, it should be $ G \not\subset A_n$.

5) This was discussed before.

Thanx, but
no. [b]3[/b] : it's not an answer.
no. [b]4 [/b]: yes,  :maybe: .
no. [b]5[/b] : how can I find the answer ?

5) is wrong? It would imply that every group of order $ 2k$, $ k$ odd, is solvable.


Prove that as n gets very large, n!/n^n approaches 0.

Prove that as n gets very large, (2n)!/n^n approaches infinity.

Most likely not rigorous enought, but...

[hide]

1.  We know that [tex]n!< n^n[/tex], thus [tex]\frac{n!}{n^n} < 1.[/tex] If [tex]\frac{n!}{n^n}[/tex] approaches 0, we can write

[tex]\lim_{n \to \infty} \frac{n!}{n^n} < \frac{1}{n}[/tex] since if n approaches infinity, 1/(n) will approach 0.  Multiplying both sides by n^n, we get

[tex]\lim_{n \to \infty} n! > \frac{n^{n}}{n}[/tex]

As n gets larger, for some non-negative number x, (n-x)^(1/n) gets closer to one.  Thus, taking the nth root of both sides, we get

[tex]\sqrt[n]{n} \cdot \sqrt[n]{n-1} \cdot \sqrt[n]{n-2}... < n[/tex]

Since as n gets larger, (n-x)^(1/n) approches one, the left side becomes a product of ones.  1*1*1*1 < n.  This is true, so the fraction approaches 0 as n gets larger.

[/hide]

I would start by showing convergence or divergence:

[hide]

using the ratio test for lim n-> :inf: (n!/n^n)

q = lim n -> :inf: ( [ (n+1)n!/(n+1)^(n+1) ] * n^n/n! )

= lim n -> :inf: ( [ (n+1)n^n ] / [ (n+1)^n * (n+1) ] )

= lim n -> :inf: ( n^n/(n+1)^n ) = 0

so the first series converges by the ratio test. as to WHAT it converges to, i don't know how to show that it's one (did I already by accident?). is there maclaurin involved?

for the second series, when we apply the ratio test, we get:

q = lim n ->  :inf: ( (2n+2)(2n+1)/(n+1) ) which diverges after expansion and some l'hopitals (q =  :inf:)

[hide][/hide][/hide]

whoa, can someone explain nr1337's thing to me

Will LaTeXing it make it easier to read?  Or is there another problem?

i think LaTeX would help, thanks (tho I'm not really one to talk since I don't know LaTex myself)

It should be a bit better now.  I'll look through and see if I can find anything to explain better.  Writing has always been my worst point, so don't be surprised if you can't read my writing.  That's why I like math. :D

[hide]n^n < n! < ((n+1)/2)^n may help

n! < ((n+1)/2)^n comes from G.M. <= A.M.[/hide]

Belenos, n > (n+1)/2, so n^n > ((n+1)/2)^n.  You've got at least two or three typos/wrong things there.

Good job, Vecter, but who said anything about series?  nrleet, I have to admit that I have no idea where that 10 came from or what it is doing there.

Well, if n gets bigger, then 1/n approaches zero.  1/10^n would also approach zero.  I don't know why I thought of that, though...  I fixed it.  I still have no clue where I got that ten...  Is it right, though?

For the first question:

[hide]Rewrite [tex]\displaymath{{n!}\over{n^n}}[/tex] as [tex]\displaymath{{1}\over{n}}\cdot{{2}\over{n}}\cdots{{n}\over{n}}[/tex].

When you increase n by 1, you increment each of the denominators by 1 (hence reducing the fraction, making it approach 0); then you multiply it by [tex]\displaymath{{n+1}\over{n+1}}[/tex], which does not change the value of the expression. Hence the expression approaches 0.[/hide]

This is likely not very rigorous.

In vector's solution, he said that the second one reduced to (2n+2)(2n+1)/(n+1)... Im getting that it comes to (2n+2)(2n+1)(n^n)/(n+1)^(n+1). After someone answers that, please answer this question.
Am I allowed to say that the limit of the above expression (mine) is = to the prod of the limits of n^n/(n+1)^n and (2n+2)(2n+1)/(n+1)? I think that is what monsieur vector did, although if thats the case, the n^n/(n+1)^n he already showed to be 0.

Yes, jelyman does seem to be correct.  I guess that one needs some patching up.  Actually, it's not very hard at all -- probably the easier of the two.

yeah sorry, I did a very ad hoc cancellation before. I think it ends up as he said:

[hide]

lim n ->  :inf: [ (2n+2)(2n+1)(n^n) ] / [ (n+1)(n+1)^n ]

i assumed (wrongly perhaps?) that

lim n ->  :inf: [ (n^n)/(n+1)^n ] = 1

leaving us with this part of the limit:

lim n ->  :inf: [ (2n+2)(2n+1) ] / [ (n+1) ] which diverges

I'm not sure if that's correct though, the with the limit going to one.

[/hide]

if anything, that would solve the second problem, but what about the first?

JBL wrote:Your first was good, I believe.  For the second, [hide]you'd have to prove what you're claiming about (n/(n+1))^n.  There is a good, simple way to do this without invoking things like the ratio test.[/hide]

I don't think my first one is good. The ratio test only tells you if it converges or not, not what it converges too ... or at least that's what I remember.

anyone make any progress on these problems? i'm interested to find the answer (i actually remember doing these exact problems in BC calc, ah should've remembered...)

Hint for the first problem (in spoiler): [hide]Try to prove that n! :le: nn-1 for n :ge: 1.[/hide]

Hint for the second problem: [hide](2n)! has more than n/2 terms that are at least 3n/2.[/hide]


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Let $ a,$ $ b,$ $ c$ be positive real numbers. Setting
\[ x\equal{}\sqrt{\frac{3a^2}{b^2\plus{}bc\plus{}c^2}},\quad y\equal{}\sqrt{\frac{3b^2}{c^2\plus{}ca\plus{}a^2}},\quad z\equal{}\sqrt{\frac{3c^2}{a^2\plus{}ab\plus{}b^2}}.\]
Prove that, if $ k \ge \sqrt{\frac{10}{3}},$ then
\[ (k\plus{}x)(k\plus{}y)(k\plus{}z) \ge (k\plus{}1)^3.\]

What about the proof for the last inequality?

[quote="hedeng123"][quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be positive real numbers. Setting
\[ x \equal{} \sqrt {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}},\quad y \equal{} \sqrt {\frac {3b^2}{c^2 \plus{} ca \plus{} a^2}},\quad z \equal{} \sqrt {\frac {3c^2}{a^2 \plus{} ab \plus{} b^2}}.\]
Prove that, if $ k \ge \sqrt {\frac {10}{3}},$ then
\[ (k \plus{} x)(k \plus{} y)(k \plus{} z) \ge (k \plus{} 1)^3.\]
[/quote]

$ (k \plus{} x)(k \plus{} y)(k \plus{} z) \minus{} (k \plus{} 1)^3 \equal{} (x \minus{} 3 \plus{} z \plus{} y)k^2 \plus{} (zx \minus{} 3 \plus{} xy \plus{} yz)k \minus{} 1 \plus{} xyz$.

we can prove (by Holder Ineq)

$ x \plus{} y \plus{} z\geq 3$,

$ xy \plus{} yz \plus{} zx\geq 3$

so we only prove

$ \frac {10}{3}(x \minus{} 3 \plus{} z \plus{} y) \plus{} \sqrt {\frac {10}{3}}(zx \minus{} 3 \plus{} xy \plus{} yz) \minus{} 1 \plus{} xyz\geq 0.$[/quote]

or we only prove

$ \frac {10}{3}(x \minus{} 3 \plus{} z \plus{} y) \minus{} 1 \plus{} xyz\geq 0.$

$ \Longleftrightarrow \frac {10}{3}(\sum\sqrt {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}} \minus{} 3) \minus{} 1 \plus{} \sqrt {\prod {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}}}\geq 0$

$ \Longleftrightarrow \frac {10}{3}(\sum\sqrt {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}}) \minus{} 11 \plus{} \sqrt {\prod {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}}}\geq 0$

Note
$ \sum\sqrt {\frac {a^2}{b^2 \plus{} bc \plus{} c^2}}\geq \sqrt {\frac {(a^2 \plus{} b^2 \plus{} c^2)^3}{\sum{a^4(b^2 \plus{} bc \plus{} c^2)}}}$

we only prove

$ \frac {10\sqrt {3}}{3}\sqrt {\frac {(a^2 \plus{} b^2 \plus{} c^2)^3}{\sum{a^4(b^2 \plus{} bc \plus{} c^2)}}} \minus{} 11 \plus{} \sqrt {\prod {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}}}\geq 0$

The inequality holds,but i can not prove it.

[quote="can_hang2007"]See my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-53-v-q-b-can.html[/quote]

 :(  I can not access this link.

[quote="hedeng123"] :(  I can not access this link.[/quote]
You can download it here (I am sure that you can download it): http://cid-b2deb869e6bb1093.skydrive.live.com/self.aspx/Inequality%20Problems/Inequality%2053.rar

[quote="Soarer"]@can_hang
Actually why don't you just post your solution in your blog, directly in latex instead of putting it in a separate file? I guess it shouldn't be much trouble since you can directly copy the latex code.[/quote]
I am sorry. I like to write the solution in file more because I think the solution will have a better looking than posting directly on the blog. LaTeX of blospot (and wordpress, too) is not nice.

By the way, I like to type LaTeX very much.


In isosceles trapezoid ABCD , AB= 20, CD = 50, and the height is 28. Find the area of triangle AGD, with G the intersecting point of the diagonals AC and BD.

[hide]Note that $\triangle ABG$ and $\triangle DGC$ are similar because we can prove that  vertical angles are congruent and alt. interior angles are congruent if a transversal cuts two parallel lines. Since the the are similar if the altitude of $\triangle DCG$ is x then the altitude of $\triangle AGB$ is       $\frac{2x}{5}$ since the altitudes add up to 28

$x+\frac{2x}{5}=28$
$\frac{7x}{5}=28$
$x=20$.

 Since the altitude of $\triangle ADC$ is 28 and its base is 50, its area is 700. The altitude of $\triangle DGC$ is 20 and its base is 50 so its area is 500, therefore the area of $\triangle AGD$ is $200$.[/hide]

[quote="now a ranger"]how did you get the ratios of the similar triangles 2x/5 and x?

I understand everything except that.[/quote]

The ratios of the "bases" so 20 and 50. Basically every side and altitude of the triangle with base 20 will be 2/5s of every corresponding side and altitude.

oh ratios of similar triangles... gotcha

silly me  :lol:


limit
topology
calculus
calculus computations

Let $ (M,d)$ be a metric space, $ 0 < c < 1$ be a fixed number, and $ f: M\to M$ be such that $ d(f(x),f(y))\leq cd(x,y)$ for all $ x,y\in M$.

Select a point $ x_0\in M$ and define a sequence:

$ x_1 \equal{} f(x_0)$ if $ n \equal{} 1$ and $ x_n \equal{} f(x_{n \minus{} 1})$ if $ n > 1$.

If $ (M,d)$ is complete, prove that $ f$ has a fixed point, and prove that $ f$ cannot have more than one fixed point.

The easy part: Showing f cannot have more than one fixed points.
Suppose $ x$ and $ y$ are distinct fixed points of $ f$ and $ 0 < c < 1$. Then $ f(x) \equal{} x$ and $ f(y) \equal{} y$. And we have $ d(x,y) \equal{} d(f(x),f(y))\geq cd(x,y) \implies d(x,y) \equal{} 0 \implies x \equal{} y$.

The other part: Show the following:
1. $ x_{1}, x_{2}, ..., x_{n}$ is a cauchy sequence
2. Invoke the completeness of $ (M,d)$ to conclude the convergence of the sequence in (1), say it converges to an element $ x_{0}$ of $ M$.
3. Since the condition that $ d(f(x),f(y))\geq cd(x,y \implies f$ is continuous, conclude that
$ f(x_{0}) \equal{} f(\lim_{n \to \infty}x_{n}) \equal{} \lim_{n \to \infty}f(x_{n}) \equal{} \lim_{n\to \infty}x_{n \plus{} 1} \equal{} x_{0}$

Keywords: Banach, contraction. This is a famous theorem, and that sequence is the essence of the constructive proof.


Given that $ a, b, c, d$ are positive real numbers what is the range of

$ \frac {a\plus{}b}{a\plus{}b\plus{}c} \plus{} \frac {b\plus{}c}{b\plus{}c\plus{}d} \plus{} \frac {c\plus{}d}{c\plus{}d\plus{}a} \plus{} \frac {d\plus{}a}{d\plus{}a\plus{}b}$

[quote="user1234"]Given that $ a, b, c, d$ are positive real numbers what is the range of

$ \frac {a \plus{} b}{a \plus{} b \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} c \plus{} d} \plus{} \frac {c \plus{} d}{c \plus{} d \plus{} a} \plus{} \frac {d \plus{} a}{d \plus{} a \plus{} b}$[/quote]Note that $ \frac{a \plus{} b}{a \plus{} b \plus{} c}\equal{}1\minus{}\frac{c}{a \plus{} b \plus{} c}>1\minus{}\frac{c\plus{}d}{a\plus{}b\plus{}c\plus{}d}$ so
$ \sum_{cyc}\frac{a\plus{}b}{a\plus{}b\plus{}c}>\sum_{cyc}\left(1\minus{}\frac{c\plus{}d}{a\plus{}b\plus{}c\plus{}d}\right)\equal{}4\minus{}\frac{2(a\plus{}b\plus{}c\plus{}d)}{a\plus{}b\plus{}c\plus{}d}\equal{}2$.

Sorry I have to go for now >_<

[hide="Note that"]The inequalities $ k>\sum$ and $ k<\sum$ are homogenous for constants $ k$.[/hide]

[quote="10000th User"][quote="user1234"]Given that $ a, b, c, d$ are positive real numbers what is the range of

$ \frac {a \plus{} b}{a \plus{} b \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} c \plus{} d} \plus{} \frac {c \plus{} d}{c \plus{} d \plus{} a} \plus{} \frac {d \plus{} a}{d \plus{} a \plus{} b}$[/quote]Note that $ \frac {a \plus{} b}{a \plus{} b \plus{} c} \equal{} 1 \minus{} \frac {c}{a \plus{} b \plus{} c} > 1 \minus{} \frac {c \plus{} d}{a \plus{} b \plus{} c \plus{} d}$ so
$ \sum_{cyc}\frac {a \plus{} b}{a \plus{} b \plus{} c} > \sum_{cyc}\left(1 \minus{} \frac {c \plus{} d}{a \plus{} b \plus{} c \plus{} d}\right) \equal{} 4 \minus{} \frac {2(a \plus{} b \plus{} c \plus{} d)}{a \plus{} b \plus{} c \plus{} d} \equal{} 2$.

Sorry I have to go for now >_<[/quote]

But is this bound tight enough?
Also, there is the question of the upper bound.

[hide]
$ \frac {a \plus{} b}{a \plus{} b \plus{} c \plus{} d} < \frac {a \plus{} b}{a \plus{} b \plus{} c}$

$ \therefore 2 < \sum \frac {a \plus{} b}{a \plus{} b \plus{} c}$

Equality when $ a \equal{} b \to 0$, $ c \equal{} 1$and $ d \to \infty$

$ \sum \frac {a \plus{} b}{a \plus{} b \plus{} c} \equal{} 4 \minus{} \left(\frac {c}{a \plus{} b \plus{} c} \plus{} \frac {d}{b \plus{} c \plus{} d} \plus{} \frac {a}{c \plus{} d \plus{} a} \plus{} \frac {b}{d \plus{} a \plus{} b}\right)$

$ \frac {c}{a \plus{} b \plus{} c} \plus{} \frac {d}{b \plus{} c \plus{} d} \plus{} \frac {a}{c \plus{} d \plus{} a} \plus{} \frac {b} {d \plus{} a \plus{} b} \ge \frac {c}{a \plus{} b \plus{} c \plus{} d} \plus{} \frac {d}{a \plus{} b \plus{} c \plus{} d} \plus{} \frac {a}{a \plus{} b \plus{} c \plus{} d} \plus{} \frac {b}{a \plus{} b \plus{} c \plus{} b} \equal{} 1$

$ \therefore \sum \frac {a \plus{} b}{a \plus{} b \plus{} c} < 3$

Equality when $ a\to \infty, b\equal{}c<<\infty, d\equal{}1$

[/hide]


[hide]
Eh, it probably has something to do with the fact that $1000^{1000}$ is enormous.  I dunno, I'd have to play with it some more but that's my first thought (you also know the leftmost digits of $1000^{1000}$, of course).[/hide]

[hide]
Since $1000^{1000}$ is enormously larger than $999^{999}$, the leftmost digits are almost certainly $10$. Now for the third...

[/hide]

[hide]
Hopefully this finishes it off.

We already can tell that $1000^{1000}$ is much larger than $1^{1}+2^{2}+\cdots+999^{999}$.  But we don't yet know if it's large enough so that the rest of the sum wont affect the leftmost digits.

We can't really do much with $1^{1}+2^{2}+\cdots+999^{999}$ directly as it stands, so let's try and make a bound on it.  Perhaps we can show that it is less than $2\cdot 999^{999}$.  So we examine: \begin{eqnarray*}1^{1}+2^{2}+\cdots+998^{998}&?& 999^{999}\\ &?& (1+1+\cdots+1)\cdot 999^{998}\\ &?& 999^{998}+999^{998}+\cdots+999^{998}\end{eqnarray*} So this means that $1^{1}+2^{2}+\cdots+998^{998}< 999^{999}$ and $1^{1}+2^{2}+\cdots+999^{999}<2\cdot 999^{999}$.

So $2\cdot 999^{999}$ is larger than the sum we are concerned.  Obviously if it doesn't affect the value of the leftmost three digits then neither will the sum that we are concerned with.

Let's look at how many digits $2\cdot 999^{999}$ has compared with $1000^{1000}$ \begin{eqnarray*}D(2\cdot 999^{999}) &=& \lfloor \log_{10}\left(2\cdot999^{999}\right)\rfloor+1\\ &=& \lfloor \log\left(\left(2^{1/999}\right)^{999}999^{999}\right)\rfloor+1\\ &=& \lfloor 999\log\left(2^{1/999}999^{999}\right)\rfloor+1\\ &=& 2997\\ D(1000^{1000}) &=& \lfloor \log\left(1000^{1000}\right)+1\\ &=& 3001 \end{eqnarray*} Thus we can see that the leftmost three digits are just 100. (ok, so this required a calculator for one part...)[/hide]

Edit: Ok, I looked at the official solution and it's much better...so someone else can try to find it.

Nice.  The "official" solution uses the fact that $1000^{1}+1000^{2}+1000^{3}...+1000^{1000}$ is obviously larger than what we are looking for.  It also has an obvious value: $100100100100...$.  So, even with numbers much larger than the ones we want, we still maintain the same 3 first digits.  So the answer is $\boxed{100}$.

I think I like jb05's solution better though.

Just an interesting fact I though I'd point out:

You'd probably expect tons of 0s at the front, but according to Mathematica, the value of the expression is something like 10003...


algebra
polynomial
integration
calculus
derivative
real analysis
real analysis unsolved

Fix $n\geq 1$ . Show that there is a measure $m$ on $[0, 1]$ such that for every polynomial $P$ of degree at most $n$ we have that
$\int P dm = \sum_{k=1} ^ n P^{(k)}(\frac{k}{n})$

where $P^{(k)} (\frac{k}{n})$ is the $k$'th derivative of $P$ evaluated at $\frac{k}{n}$ for different $k$ values.

an idea:
if one knows the values of a polynomial of degree $n$ at $n+1$ points, then one knows exactly this polynomial, and thus all its derivatives.

For example, if one knows $P(0)=a_0$, $P(\frac{1}{n})=a_1$,..,$P(\frac{n}{n})a_n$. Then
$P(\frac{k}{n})=a_k=P(0)+(\frac{k}{n})P'(0)+..+\frac{(\frac{k}{n})^n}{n!}P^{(n)}(0)$.
Hence, for $1 \leq k \leq n$, this gives $n$ equations with $n$ unkown values : $P'(0), P^{(2)}(0),.., P^{(n)}(0)$. This is a Cramer system (Vandermonde system), wich shows that one can find $\alpha_0,..,\alpha_n$, independant of the polynomial $P$, such that $P'(0)=\alpha_0 a_0 + \alpha_1 a_1 + .. + \alpha_n a_n$.
One can do the same thing for $P^{(k)}(\frac{k}{n})$. Hence, this shows that the measure $m$ is a linear combination of the dirac measure $\delta_{\frac{k}{n}}$


1. The sum of Jane's age and Alice's age is 24 more than Frank's age.  The sum of Jane's age and Frank's is 10 more than Alice's age.  The sum of Frank's age and Alice's age is 20 more than Jane's age. What is the sum of Jane's, Frank's and Alice's ages?

2. If n is a natural number, let f(n) denote the number of distinct prime factors of n.  (thus f(6) = 2 since 6 = 2 x 3, f(84)=3 since 84 = 2^2 x 3 x 7, f(27)= 1 since 27 = 3^3.)  What is the maximum value of f(n) for 1 less than or equal to n lesser than or equal to 1000?

3. A container has 4 quarts of maple syrup.  One quart is removed and replaced with one quart of corn syrup and the result is thoroughly mixed.  This operation is repeated three more times.  How much maple syrup has been taken out by these four removals?

4. In a class of 10 students, 4 forgot to put their names on typed term papers.  What is the probabiliy the teacher correctly guesses which paper belongs to each student?

5. A regular hexagon is inscribed a circle of radius one.  What is the hexagon's area?

theone gets #1

mathfiend that is incorrect for #5

Mathfiend gets #4
Tare gets #5

Leeyuhang is incorrect

the Saga continues...:wink:
If you get #1 in the first one then we could have the ultimate faceoff...we might even borrow the Virtual Classroom :P

oh u ppl r so mean i had to do SAT prep and u started without me!!!!!!!!!!!!!!!!!!!!!!  :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry: :cry:

It's OK notfunny, don't cry, [b]you wouldn't've won anyways[/b] :twisted:

just yokin' with ya for making the thread so heavy, not! :lol:

me either...I was too busy w/ Algebra homework and missed the whole of round 4...

I was too busy [url=http://www.albinoblacksheep.com/flash/spankthemonkey.php]Spanking the Monkey[/url] to realize that round 5 started already for like 2 minutes :lol: but oh well that's better compared to you guys :P

The disco music is so "rad"! :) My personal best is like 673/hour but oh well that's off topic.

Anyways what now?

FOR THE TIEBREAKER: (this is between Tare and theone):

Let x be a single digit. For how many different values of x can the nine digits 2,2,3,4,4,5,6,6,x be arranged in a 3X3 array so that the sums of the three rows are equal?

3

Dam, I didn't realize this either, stupid monkey...

Congratulations theone, you win this round. Good job.

i(accidentally put 9:oops:) don't get this question does diagonals count?

Well it doesn't matter really, all you do is add them together and get 32+x then see that that needs to be divisible by 3.

You were both right... try this.

What is the largest number you can put on a calcultor by pressing each of the keys 1,2,3,y^x exactly once? For example: 3 1 y^x 2 would produce 31^2=961.

Explanation: In exponents the exponent matter more than the base so put 3 for the exponent and 21 (which is larger than 12) for the base and get the answer.

delete(i wasn't suppose to do this
but can you write it as 2^31?

if u think about it 2^3 is less than 3^2

[quote="topper"]theone breaks the tie and wins.

Ready for a new edition?[/quote]

Can you explicitly state when the contest begins for each round? Thanks.

Yes, please do. I was talking to lee on AIM at the tie, and donig hw, so I didn't know it came until he told me!

Y should I? You were laughing at me the whole time.

Ah, so you had insider information plus AIM, now that might've affected...*abruptly stops talking*

no actually it was slower that way. you ahd to press a button to switch between servers.

But I clicked post later than u by seconds.


abstract algebra
algebra
polynomial
vector
group theory
linear algebra

1. Let $ \vec{a}, \vec{b} \in \mathbb{R}^{3}$ be non-parallel and $ A: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3},$ given by
\[ A\vec{x}=\vec{a}\times(\vec{b}\times\vec{x})-\vec{b}\times(\vec{a}\times\vec{x}.)\]
(a) Show that $ A$ is linear.
(b) Find the basis of the kernel and the image of $ A$.

2. In the space of real polynomials of degree at most $ 4$ are given subspaces
\[ U=L\{ x^{4}+2x^{3}-x+1,x^{4}-2x^{3}+2x^{2}+x+1,2x^{4}+6x^{3}-x^{2}-3x+2\}\]
\[ V=\{p\in \mathbb{R}^{4}[x]; p^{\prime}(0)+p^{\prime\prime\prime}(0)=0,p(-1)=p(1)\}.\]

Find the basises of $ U, V, U \cap V$ and $ U+V$.

3. In the group $ (G,\circ)$ define
\[{ x\sim y \Leftrightarrow \mathrm{exists} \quad \mathrm{such} \quad} g\in G, \quad \mathrm{that}  \quad g\circ x \circ g^{-1}=y.\]

(a) Show, that the relation $ \sim$ is the equivalence relation.
(b) Show that all the elements of the permutation group $ S_{3}$ are cycles and write them down.
(c) Find the equivalence class of the transposition $ (1,2)\in S_{3}$.

4. Let $ A: V \Rightarrow V$ be a linear transformation and $ \{v_{1},\ldots,v_{k}\}$ the basis of its kernel.

(a) Show, that if $ \{v_{1},\ldots,v_{k}, w_{1},\ldots,w_{s}\}$ is a lineary independent set, then the vectors $ Aw_{1},Aw_{2},\ldots,Aw_{s}$ are lineary independent.

(b) Show, that $ \dim{ \ker {A^{2}}} \leq 2\dim{\ker A}$.

3.
[hide]
We will show that the relation is :
a. Symmetric, in other words if $ x\sim y$ then $ y\sim x$
$ x\sim y$ means, there exist $ g$ such that $ gx{g}^{\minus{}1}\equal{}y$ then $ x\equal{}{g}^{\minus{}1}yg\equal{}{g}^{\minus{}1}y{({g}^{\minus{}1})}^{\minus{}1}$ (Because $ g \in G$, $ {g}^{\minus{}1} \in G$, because there exist $ {g}^{\minus{}1}$ such that
$ x\equal{}{g}^{\minus{}1}yg\equal{}{g}^{\minus{}1}y{({g}^{\minus{}1})}^{\minus{}1}$ then $ y\sim x$
b. Reflexive, in other words we will prove that $ x\sim x$, note that $ G$ is a group such that there exist an identity element in $ G$, named $ e$,
$ ex{e}^{\minus{}1}\equal{}exe\equal{}x$, because there exist $ e$ such that $ ex{e}^{\minus{}1}\equal{}x$, then $ x\sim x$
c. Transitive, in other words we will prove that $ x\sim y$ and $ y\sim z$ implies $ x\sim z$. $ x\sim y$ means there exist $ g \in G$ such that $ y\equal{}gx{g}^{\minus{}1}$, $ y\sim z$ implies there exist $ h \in G$ such that $ z\equal{}hy{h}^{\minus{}1}$, we will prove that there exist $ m \in G$ such that $ z\equal{}mx{m}^{\minus{}1}$. Take $ m\equal{}hg$, then $ mx{m}^{\minus{}1}\equal{}(hg)x{(hg)}^{\minus{}1}\equal{}h(gx{g}^{\minus{}1}){h}^{\minus{}1}\equal{}hy{h}^{\minus{}1}\equal{}z$.
Because there exist $ m\equal{}hg \in G$ such that $ mx{m}^{\minus{}1}\equal{}z$, then $ x\sim z$.
Because the relation is reflexive, symmetric and transitive, then the relation is an equivalence relation.

[/hide]
Sorry if I am wrong.

And there are the [b](b)[/b] and [b](c)[/b] parts of the [b]third[/b]:
[b]
(b):[/b]
$ \left( \begin{array}{ccc} 1 & 2 & 3 \\
1 & 2 & 3 \end{array} \right) \equal{} \mathrm{identity}$

$ \left( \begin{array}{ccc} 1 & 2 & 3 \\
1 & 3 & 2 \end{array} \right) \equal{} (23)$

$ \left( \begin{array}{ccc} 1 & 2 & 3 \\
2 & 1 & 3 \end{array} \right) \equal{} (12)$

$ \left( \begin{array}{ccc} 1 & 2 & 3 \\
2 & 3 & 1 \end{array} \right) \equal{} (123)$

$ \left( \begin{array}{ccc} 1 & 2 & 3 \\
3 & 1 & 2 \end{array} \right) \equal{} (132)$

$ \left( \begin{array}{ccc} 1 & 2 & 3 \\
3 & 2 & 1 \end{array} \right) \equal{} (13)$

We see that all elements of $ S_{3}$ are indeed cycles. $ \surd$

[b](c):[/b]

1. $ g \equal{} I \Rightarrow I \circ (12) \circ I^{ \minus{} 1} \equal{} (12)$

2. $ g \equal{} (23) \Rightarrow (23) \circ (12) \circ (23)^{ \minus{} 1} \equal{} \left( \begin{array}{ccc} 1 & 2 & 3 \\
3 & 2 & 1 \end{array} \right) \equal{} (13)$

3. $ g \equal{} (12) \Rightarrow (12) \circ (12) \circ (12)^{ \minus{} 1} \equal{} \left( \begin{array}{ccc} 1 & 2 & 3 \\
2 & 1 & 3 \end{array} \right) \equal{} (12)$

4. $ g \equal{} (123) \Rightarrow (123) \circ (12) \circ (123)^{ \minus{} 1} \equal{} (123) \circ (12) \circ (321)\equal{}\left( \begin{array}{ccc} 1 & 2 & 3 \\
1 & 3 & 2 \end{array} \right) \equal{} (23)$

5. $ g \equal{} (132) \Rightarrow (132) \circ (12) \circ (132)^{ \minus{} 1} \equal{} (132) \circ (12) \circ (231)\equal{}\left( \begin{array}{ccc} 1 & 2 & 3 \\
3 & 2 & 1 \end{array} \right) \equal{} (13)$

6. $ g \equal{} (13) \Rightarrow (13) \circ (12) \circ (13)^{ \minus{} 1} \equal{} \left( \begin{array}{ccc} 1 & 2 & 3 \\
1 & 3 & 2 \end{array} \right) \equal{} (23)$

We conclude that $ (12)\sim y \Leftrightarrow y\in \left\{(12),(13),(23)\right\}.$


I tried solving for it, but I got a contradiction...

$\log_{x+1}x < 1$ for all $x > 0$; thus, a solution exists if and only if the base of the logarithm on the right is bigger than $\pi$.  There aren't going to be "nice" solutions.

hmm...i saw somewhere the notation $lg$ used for the logarithm in base 2 - which would have no solutions.

[quote="piac31"]Wikipedia says that lg can be a base 2 or base 10 log depending on location. Considering base 2 wouldn't have any solutions, I think we can assume it is base 10.

I have no clue how to prove a solution to this, but using my graphing calculator, I came up with x=1.6116419.[/quote]
that's right. it is base 10,$lg\pi=log_{10}pi$

I'm not sure what the original poster intended to challenge in this thread. The calculator works fine as shown below:

Solve for $x$ in equation $(x+1)^{\log \pi}-x = 0$.

This gives what one of previous posters stated.  Maybe the original poster can clarify bit? And by the way, everyone except JBL seems to forget that $log$ looks much nicer with slash ---> $\log$.


conics
parabola
analytic geometry
graphing lines
slope

Two parabolas have the same focus, namely the point (3, -28).  Their directrices are the $x$-axis and the $y$-axis, respectively.  Compute the slope of their common chord.


Post the cutoff for nationals from your state here.

The 4th place person on our team got about a 40, but he came down from 3rd.  The lowest scorer on the team had I think a 38

Probably about 30-25 but he came from 8th place.
4th place was probably 30.

In CA the scores were 42, 41, 41, 41. I got a 41 and was 5th :mad:

I think I got a 41, yet I only got 6th. (at least i only missed 4 problems when I went over the state.)

I hope not. That would be just embarrassing. (this year was SO much easier)

Ours was in the 35-37 range, our cutoff this year was MUCH lower compared to other states but the same as Florida o_O

Indiana ain't telling ya'll :)

Ugh, seeing all these states I would've made it in is making me sad, but there are also states I wouldn't have made it in (but would've without the last 6 minutes of my test).

7th written, 4th overall (me) got 32.
2nd written, 3rd overall got 34.

Countdown is official in AR, so  a 31 or a 32 would have got you to nats.

[quote="solafidefarms"][quote="Xantos C. Guin"]Apparently from Mr. Rusczyk's blog, CA's scores were 42,41,41,41,41,41.

Im not sure but I'm guessing that GA's scores are in the low 30's.[/quote]

As I am guessing, based on some selective scores I know from people not in the top, it's more like high 30s. Mr. Hite had 3 people in the top 10, but his team was 4th :(.[/quote]

Oh, cool. Is he coaching? It's possible since teams 1-3 might have noone in the top4.

Apparently Hightower was 16th and their best individual got 75th place. :(
Last year all the 8th graders from Hightower were atleast 71st.

[quote="easyas3.14159..."]ok, complaining won't get me  to nats but for the ten billionth time WHY OH WHY IS CA SO HARD????[/quote]

That is actually a very easy question to answer...

Honestly though, why is Indiana so good? I mean if you think of states with a high population or are on the east coast (and California) you generally think of good states. Indiana's really good -- why? And they don't just have good people, they have REALLY good people that do it not just because they have a super good coach (Mr. Fischer). A coach can only get you so far. I can understand why Indiana has great depth (Mr. Fischer), but how come it has some superstar students... and superstar coaches!

[quote="pgpatel"][quote="Fanatic"]:rotfl: once again

Actually, I thought I did worse than I actually did.

And yes, the tradition of not telling the scores was there last year. I swear. Somebody told anyways though is what I heard.[/quote]2 people did. [b]nat mc[/b] and BigHC. Mostly natmc though[/quote]
Once again, [b]I[/b] didn't tell anyone what the scores of Indiana were. I might have said what [b]I[/b] had gotten, but [b]I[/b] did not say anything about their scores. [b]Dnas[/b] did tell afterwards though. [b]Please[/b] stop falsely accusing me.  :|

[quote="pgpatel"]Blame dnas why don't you. You said your score and the cutoff so we new 4,5. Everyone knew Pardha got a 46 giving us 1. So we had 1,4,5. We could have figured out Dnas and I_Would_Say.[/quote]
I'm not blaming Dnas, he is my friend, I know him, you don't, and I didn't say 4 or 1, maybe my own score but thats it. so please, lets not quarrel. ;)

congrats if you made it to nats. you should be practicing then.  :)

Sorry for making assumption nat mc. I'll probably see you at nats.

[quote="aznness"][quote="easyas3.14159..."]ok, complaining won't get me  to nats but for the ten billionth time WHY OH WHY IS CA SO HARD????[/quote]

Also, CA tends to have a lot of asian immigrant families, and asian immigrant parents tend to push their asian immigrant childen very hard. Also, the chinese mathematics curriculum is a lot more intensive than the math taught in the US of A.

I'm 1st generation chinese :lol:[/quote]
well, so does indiana... and these problems would be fixed if they took more people from harder states...(i stand behind my policies)...

[quote="aznness"][quote="easyas3.14159..."]ok, complaining won't get me  to nats but for the ten billionth time WHY OH WHY IS CA SO HARD????[/quote]

Also, CA tends to have a lot of asian immigrant families, and asian immigrant parents tend to push their asian immigrant childen very hard. Also, the chinese mathematics curriculum is a lot more intensive than the math taught in the US of A.

I'm 1st generation chinese :lol:[/quote] FIrst Generation Indian :D

[quote="Zmastr"]America is not just caucasian people.  Indian people and Chinese people are part of America, too.  It's one big melting pot.  :D

[hide]But caucasians need to put more effort into math, like me![/hide][/quote]

Tell that hidden part to people like Adam Hesterberg.  ;)

Not all Asians are geniuses, and not all Caucasians are dumb.

Anyway:
1st generation Antarctican.

# 1997: Zhihao Liu, Wisconsin
# 1998: Ricky Liu, Massachusetts
# 1999: Po-Ru Loh, Wisconsin
# 2000: Ruozhou Jia, Illinois
# 2001: Ryan Ko, New Jersey
# 2002: Albert Ni, Illinois
# 2003: Adam Hesterberg, Washington
# 2004: Gregory Gauthier, Illinois
# 2005: Neal Wu, Louisiana

Two caucasians since 1997 :P

If we keep talking about racial stuff, I'll either move this to Round Table or lock it.

Get on topic, we're talking about cutoffs.

A word about Michigan: The cutoff was really 40, I got a 39, but had 2 targets wrong, so if you got a 39, the chances are you wouldn't have made it.

FInally, the cutoff of IN has been revealed.
[hide="drumroll please"] Sachin Schinde got a 39[/hide]

Yup, that's correct. You would've needed either 39 or 40, Sachin's split was 25/14... I think he got #29 but not #30. Mine was 26/12... both targets missed in the last set, last question changed from right to wrong at last sECOND>

After the sprint round:

Me: Sachin, what did you get?
Sachin: I'm not telling you
Me: I'll tell you mine if you tell me urs
Sachin: Ok
Me: PROMISE?
Sachin: yes
Me: ok. I got a 26
Sachin: (Silent, cuz he lost to me)
Me: WTF


geometry
3D geometry
sphere
function
topology
algebra
domain

hi, let$A=\bar{B}((0,0),1)$ thus the closed unit ball around the origin in $\mathbb{R}^2$ and let $B=A$without (0,0)

let f be function from B to A, that maps every point on $\partial{B}$ to itself

can f be continuous?

my guess not?  it has a connection with topology it seems but surely this itself can be seen much more elementary?

Sometimes, such intuitively clear questions do not have simple proofs. I don't know about this one, but think about stuff like the Jordan Curve Theorem, or Brouwer's Domain Invariance Theorem.

For a quick but definitely non-elementary proof (which works in the same conditions, in $\mathbb R^n$ for any $n\ge 2$), you can do this:

Let $\phi$ be the map sending $x\in\mathbb D^n\setminus\{0\}$ ($\mathbb D^n$ is the unit ball in $\mathbb R^n$) to $\frac x{\|x\|}$, and let $g=\phi\circ f$, where $f$ was our initial function. Now $g$ induces a non-trivial homomorphism $g_*$ from the $n-1$'th homology group of $\mathbb D^n$, which is trivial, to the $n-1$'th homology group of $\mathbb S^{n-1}$ (the unit sphere in $\mathbb R^n$), and this is impossible.

wow you probably know that i was hinting of 'simple connectivity' but thought this particular case with a nice ball must be easy

guess not, I am not familiar with that group yet

A variation, which may be more elementary (based on a solution I gave to an exam problem). This is specific to the disk and the circle.

As grobber said, this is equivalent to finding a continuous function from the disk to the circle.

Theorem: Any continuous function $f$ from a compact simply connected set $D$ (such as the unit disk) to the circle is homotopic to a constant map.

Proof: Choose a point $x_0\in D$. WLOG, $f(x_0)=1$ (we regard the circle as embedded in $\mathbb{C}$, and we can just multiply $f$ by a constant).
Let $g(x_0)=0$. For each point $x\in D$, choose any curve $\gamma$ with endpoints $x_0$ and $x$, and let $g(x)$ be a lift of $f$ on $\gamma$: $e^{ig(x)}=f(x)$, and $f$ is continuous on $\gamma$. (This is very similar to analytic continuation, if you've seen that)

$g$ is well-defined by the monodromy theorem, since any two of the defining curves are homotopic. $g$ is continuous, since it is continuous on every curve.
Since $D$ is compact, the image of $D$ in $\mathbb{R}$ is compact and therefore bounded by some constant $M$. Let $f_t(x)=\exp(i(1-t)g(x))$. $f_t$ is a homotopy between $f$ and the constant map 1.

Now, the restriction of $f$ to the circle is also homotopic to a constant, so it cannot be the identity.

I realise now I am wrong
the map should be from A to B
does that change your answers now? :o

We read past that, too; that's the question we answered anyway.


I have to note that this is very weak.

[hide="Not induction"]
Rushed and nonrigorous since I have to leave in a matter of minutes, but supplying details isn't difficult.

Rearrange to get that we need $ n^2! > n^{n^2 \plus{} 2}$ for $ n\ge 3$; we thus need that $ (n^2)!/n^{n^2} > n^2$ for $ n\ge 3$; we note that each of the pairs $ (n^2 \minus{} k)(1 \plus{} k)\ge n^2$ (where $ k$ cycles through nonnegatives up to $ \lfloor n^2/2 \rfloor$; note that the product of all the LHSes gives the numerator of the inequality and analogously for the RHSes) and that the bound becomes less sharp as $ k$ increases - multiplying out easily gives that for $ n\ge 3$ the bound becomes sufficiently nonsharp so that the quotient is $ > n^2$.
[/hide]

[quote="MNrule"]Prove, by induction, that for all integers: $ n\geq3: (n^2 \minus{} 1)! > n^{n^2}$[/quote]

no need of induction.just pair the factors of $ (n^2\minus{}1)!$ in a way to their product be greater than $ n^2$.consider $ n$ odd and $ n$ even.

This problem is specifically meant to be done by induction. Anyone?