Solve the equation $ f(x \plus{} t) \equal{} f(x) \plus{} f(t) \plus{} 2\sqrt(f(x))\sqrt(f(t))$

I've read the solution and did consider the approach of letting t = 0 and then deducing that f(x) = c/4 where c is some constant but didn't reach the conclusion because for some reason I didn't want to square both sides involving a negative quantity on one side.

What I did try was letting t = x and then letting t = 2x and so on and then finding the pattern that f(nx) = (n^2)f(x) and then proving it by induction. Then substituting 1 for x, I came to the conclusion that f(x) = cx^2 was a solution to the functional equation. But looking back now, if I plugged in -t in place of t, then there would be no solutions. Would it be correct if I restricted the domain or is f(x) = 0 the only solution?

I don't follow.  $ t \equal{} x$ gives $ f(2x) \equal{} 3 f(x)$.  What do you mean by "no solutions" if you plug in $ \minus{}t$?

[quote="Snipez"]What I did try was letting t = x and then letting t = 2x and so on and then finding the pattern that f(nx) = (n^2)f(x) and then proving it by induction. Then substituting 1 for x, I came to the conclusion that f(x) = cx^2 was a solution to the functional equation.[/quote]
Besides the computation error noted in the above post, there is another flaw: you can't come to the conclusion that $ f(x) \equal{} cx^2$ for all $ x$. Substituting $ x \equal{} 1$ into the formula you found gives you that $ f(n) \equal{} f(1)n^2$ for all [b]positive integers[/b] $ n$. Your conclusion was that $ f(x) \equal{} cx^2$ for all real numbers $ x$. Real numbers and positive integers are very different.

[quote="Snipez"]...for some reason I didn't want to square both sides involving a negative quantity on one side.[/quote]
It's completely valid to square both sides of a true equation. However, as you (perhaps inadvertently) noted, there is a loss of information. If you know $ a \equal{} b$, then trying to conclude X from $ a^2 \equal{} b^2$ alone means that X needs to be true when $ a \equal{} \minus{}b$ also (and it may not be). However, that it is a "loss of information" is not nearly as bad as it may sound; most steps in correct proofs will do this in some way. It's not always a bad idea to try to keep track of these losses in the process of solving a problem though, because if you hit a wall you have something to scrutinize.

[quote="t0rajir0u"]I don't follow.  $ t \equal{} x$ gives $ f(2x) \equal{} 3 f(x)$.  What do you mean by "no solutions" if you plug in $ \minus{} t$?[/quote]

Oh sorry, the term on the right should read $ 2\sqrt(f(x))\sqrt(f(t))$. What I meant was from the induction it seemed like $ f(x) \equal{} x^2$ was a solution. By inspection, $ (x \plus{} t)^2 \equal{} x^2 \plus{} t^2 \plus{} 2xt$ so it seemed that $ f(x) \equal{} x^2$ worked until I substituted a -t in place of t.

@MellowMelon: Ah okay, so now that the question is fixed (I'll try to edit the original post), then the induction only works for positive n (sorry if that is by definition)? And you are right, I didn't really consider negative values seeing as how I replaced t with x, 2x, and so on.

Right I think it's the loss of information issue that prevented me from squaring both sides. But since we're starting with $ \minus{} a \equal{} b$, is it more "safe" to simply square to get $ a^2 \equal{} b^2$? I mean as in your example, going the other way around we have to be careful to remember that either $ a \equal{} b$ or $ a \equal{} \minus{} b$. Thanks for the replies.

My solution. Any comments are more than welcome, and suggestions.

$ f(x + y) = f(x) + f(y) + 2\sqrt {[f(x)f(y)]}$
Setting $ x = y = 0, f(0) = f(0) + f(0) + 2\sqrt {[f(x)f(y)]}$
$ 0 = f(0) + 2|f(0)|$.
Now Case 1:  $ f(0) \geq 0$ implies $ 0 = f(0) + 2f(0)$. (not absolute value). Easily conclude $ f(0) = 0$.
Now Case 2:  $ f(0) < 0$  implies $ 0 = f(0) - 2f(0)$. Still, $ f(0) = 0$.

Let $ x = - y$.
$ f(0) = f(x) + f( - x) + 2\sqrt {[f(x)f( - x)]} = 0$ This is equivalent to $ [\sqrt(f(x)) + \sqrt(f( - x))] = 0$. Thus is the domain permits it,
$ f(x) = f( - x)$. Therefore the function is $ \text{even}$.
So it is $ \text{possible}$ using $ - t$ works for under the square root.

The rest is trivial, let $ y = x$, $ y = 2x$ etc... and indeed $ f(x) = \lambda x^2$ is a solution.

$ \text{\sigma}$.
$ \text{\pi}$.
$ \text{\zeta}$.

I had to add these three lines. I made a typo with the $ \lambda$ originally and that came up. Now I understand why and it is awfully neat. :D.

$ f(x) \equal{} cx^2$ is not, in fact, a solution for precisely the reason Snipez identified.  It is not true that $ (x \plus{} t)^2 \equal{} x^2 \plus{} t^2 \plus{} 2 |xt|$.

Your mistake is when you conclude that $ \sqrt {f(x)} \plus{} \sqrt {f( \minus{} x)} \equal{} 0$.  Because square roots are non-negative (the presence of the square root in the original equation forces $ f(x) \ge 0$ for all $ x$), this actually means $ f(x) \equal{} 0 \forall x$.

Grr (whats the code for the embarassed smiley?), what a waste of time. :D I left the post running while I went to eat and didn't get to see his post.

Is it possible the solution can be put together quickly just by where I erred and your pointing out how $ f(x) \geq 0 \forall x$ gives $ f(x) = 0$ from the $ \sqrt$?

Is it possible the domain could have been specified to allow for the $ \sqrt$ to be valid? Or unless quite strictly clarified in the problem statement, don't assume it?

No.  Writing $ \sqrt {f(x)} \plus{} \sqrt {f( \minus{} x)}$ can be done with no assumptions about the domain of $ f(x)$, and the problem as written implies that $ x, t$ range over all reals, which implies (rather implicitly) that $ \sqrt{ f(x) } \sqrt{ f(t) }$ is well-defined for all $ x, t$.


geometry
rectangle
number theory
relatively prime
combinatorics unsolved
combinatorics

let p,q is relatively prime number. a subset S of{0,1,2...} is called [i]dien [/i] if 0 belong to S and for any n belong to S then n+p,n+q belong to S.find the number subset [i]dien[/i]

this was posted before and its on the harder end of a shortlist. (http://www.kalva.demon.co.uk/short/soln/sh00c6.html)
what you do is:
find a bijection between passes between paths through lattice points on an m*n right triangle staying completely within the triangle from one end to the other. once youve got that using easy number theory stuff. consider any such path in a m*n rectangle and consider a series of (m+n) shifts of that path. after you prove precisely one of them satisfied the conditions, then youre done since there are then (a+b, b) / (a+b) where (k,l) = k!/l!(k-l)!.


superior algebra
superior algebra solved

Let $(G,\cdot)$ a group. Show that $f: G\to G$, $f(x)=x^{-1}$ is an isomorphism iff $(G,\cdot)$ is abelian.

I've got the first part but the second not yet.  :(

What is $f$¿ ;) ($x \mapsto x^{-1}$ I think)

Yes, ZetaX. I'm really sorry beacuse I forgot that.  :blush: Edited now.

$f$ is bijective (since $f \circ f = id_{G}$). So, $f$ is an isomorphism iff $f$ is a homomorphism iff $(xy)^{-1}= x^{-1}y^{-1}$ for all $x,y \in G$. But this equation is equivalent to $y^{-1}x^{-1}= x^{-1}y^{-1}$, i.e. $yx = xy$.

I don't get why $(xy)^{-1}=y^{-1}x^{-1}$  :maybe: sorry maybe I tiered cause it's almost midnight...


calculus
derivative
analytic geometry
function
vector
calculus computations

Given a function $ f(\mathbf{r})$ in two dimensions, use the Chain Rule to e xpress its partial derivatives with respect to Cartesian coordinates $ (x,y)$ in terms of its partial derivatives with respect to polar coordinates $ (r, \theta)$. From the relationship between the basis vectors in these coordinate systems, deduce that
\[ \nabla f \equal{} \frac{\partial f}{\partial r} \mathbf{e}_r \plus{} \frac{1}{r} \frac{\partial f}{\partial \theta} \mathbf{e}_{\theta}.\]

I understand that $ \mathbf{e}_x, \mathbf{e}_y$ are the basis vectors in the directions of the $ x$ and $ y$ axes, respectively. But what does $ \mathbf{e}_r$ and $ \mathbf{e}_{\theta}$ mean?

Also, the chain rule should be $ \frac{\partial f}{\partial x} \equal{} \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} \plus{} \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x}$ and likewise for $ \frac{\partial f}{\partial y}$? I make substitutions after that in terms of $ r$ and $ \theta$ but they don't seem to work out to something nice

please read this
[url]http://www.myoops.org/ans7870/18/18.013a/textbook/chapter09/section04.html[/url]


calculus
integration
trigonometry
calculus computations

find:
$ \int_{0}^{\pi}(\int_{x}^{\pi}\frac{\sin y}{y}dy )dx$


My school textbook teaches that to find, for instance, the maximum profit you can make within a set of constraints in a business or whatever, you first graph the constraints and shade the common region. You then take the coordinates of the vertices of the region and find the profit each of them would yield. The largest would be the maximum profit.

But it didn't explain why this works, and that's what I'm wondering. Thanks in advance!

Nice... That is the graphic method BTW

Now, what are we doing? Suppose we are doing with 2 variables, put them on a graph (normal axes, each one represents one of the variables). Now, we should (with the given conditions) enclose an area in which those values can be.
For example, find the maximum of $ 2a\minus{}b$ if
$ a,b\ge0$
$ \minus{}3a\plus{}2b\le2$
$ 2a\minus{}3b\le3$
$ a\plus{}b\le6$

We draw lines which separate for each of these conditions such that they separate the plane into 2 parts, one which satisfies the condition, one that isn't (for every point inside it) - in other words, the line is the equality case (example $ a\plus{}b\equal{}6$, so $ b\equal{}6\minus{}a$ and so on)

Now for the given example, we are left with a polygon, with vertexes $ (0,0),(1.5,0),(4.5,1.5),(2,4),(1,0)$

We know that $ a$ and $ b$ are inside or on the sides of that polygon. Now use the fact that minimum and maximum of a linear function are on its ''ends'', so we get that one of the vertexes is the solution, or in our example: $ (4.5,1.5)$

Note: this is not a good method for many variables and/or higher degree conditions

@Ihatepie: Well, I know "how" to do it, I just didn't know "why".

@Bugi: Thanks! That was really helpful :)

[quote="isabella2296"]But it didn't explain why this works, and that's what I'm wondering. Thanks in advance![/quote]

[quote="Bugi"]That is the graphic method BTW[/quote]
Hi, there is a proof which shows that optimality happens at one of the vertices of the polygon (or an edge if there are redundancies) and never at an interior point. It has to do with convex sets and convex hulls (extremal theory), maybe beyond the scope of this forum. Sorry, I can't think of a source for this info...  :ewpu:

EDIT: For a good linear programming book, an access to Chvatal's Linear Programming book is great. :)

@Izzy: no problem :)

@10000th: as I said, linear function has its maximum on one of its ends, so if we draw a segment through a point which we look at such that one of the endpoints is one of the vertexes, and the other lies on one of the sides, we see that the value at that interior point is less or equal to at least one of the segment endpoints, which are vertex and point on a side (or vertex again)

[quote="Bugi"]Note: this is not a good method for many variables and/or higher degree conditions[/quote]
There's a reason why linear programming is called "linear". The function and the constraints must be linear.

[quote="Bugi"]@10000th: as I said, linear function has its maximum on one of its ends, so if we draw a segment through a point which we look at such that one of the endpoints is one of the vertexes, and the other lies on one of the sides, we see that the value at that interior point is less or equal to at least one of the segment endpoints, which are vertex and point on a side (or vertex again)[/quote]FYI, my post was for isabella2296 and... I'm sorry that you are repeating this stuff again but I think I know about this subject more than you think I do ;)

@brainomega: Exactly. That makes Integer Programming a much harder set of problems, in general. Gomory cuts :D

@10000th: OK, no need for any panic  :wink:  I just wanted to write it in the thread, so people could know, and it was addressed to you, because you said you can't find a source

@brainomega: hahahaha yeah I forgot it was linear programming...


I've been trying to get TeXnic Center and miktex on my computer, but now that I've finally downloaded them, I can't get them to work together. When I open TeXnic Center, after the tips, it runs the configuration wizard, which according to the guide on AopsWiki, I should give it the path of where I downloaded miktex. However, since I got the newer version of miktex (2.6), I cannot just copy and paste the given default location, and the newer version has a lot of subfolders, and I don't know which one to put.

Hi.

I upgraded to 2.6 last week, and if I recall correctly, the only difference in the path is that 2.5 is changed to 2.6

So the default folder should be
C:\Program Files\MiKTeX 2.6\miktex\bin

Thanks!  Because I was looking in the AoPSWiki and it said to go to some other path which didn't work.  I just found the bin folder manually and crossed my fingers.  It worked together in the end.

I have the same problem because I downloaded miktex2.7 and it doesn't work with Texnic Center.But if i change 2.5 in 2.7 it doesn't work.What can I do?

It does work with TexnicCenter. You have to tell the wizard (Build -> Define Output Profiles -> Wizard (bottom left)) where to find latex. This is something like C:\Program Files\MiKTeX 2.7\miktex\bin. Check that in this directory (or similar) it contains latex.exe - if it does then that is the one you want.


Let a building be $ h$ tall and be $ S$ width.
Find the least value of $ v_o$ (initial velocity) to throw a small stone up over the building.
(ignore the air resistance)

You can resolve $ v_{0}$ into $ v^{x}_{0}$ and $ v_{0}^{y}$.  We know $ v^{x}_{0}$ will be constant throughout the flight but $ v_{0}^{y}$ will reduce to $ 0$ halfway into its trajectory.  Assume that this occurs halfway across the building (horizontal distance $ S/2$).  Now you just need to create your equations and solve them.

Let $ a$ be the height above the top of the building that the stone will reach at the apex of its flight over the center of the building. That's a height of $ h \plus{} a$ above ground level, and to throw a stone to that height requires an initial vertical velocity of $ v_y$ such that $ v_y^2 \equal{} 2g(h \plus{} a).$

Now, assume the stone has constant horizontal velocity $ v_x.$ It would then take a time of $ t \equal{} \frac {S}{2v_x}$ for the stone to move from the center of the building to the edge. For it to barely clear the edge, it must fall a distance of $ a \equal{} \frac12gt^2 \equal{} \frac12g\left(\frac {S}{2v_x}\right)^2$ in that time. Hence, $ a \equal{} \frac {gS^2}{8v_x^2}.$

The square of the initial velocity at ground level is $ v_x^2 \plus{} v_y^2,$ and we wish to minimize that.

$ v_x^2 \plus{} v_y^2 \equal{} v_x^2 \plus{} 2g(h \plus{} a) \equal{} v_x^2 \plus{} 2g\left(h \plus{} \frac {gS^2}{8v_x^2}\right)$

$ \equal{} 2gh \plus{} v_x^2 \plus{} \frac {g^2S^2}{4v_x^2}$

By AM-GM, we have that $ v_x^2 \plus{} \frac {g^2S^2}{4v_x^2}\ge gS,$ with the minimum occurring when $ v_x^2 \equal{} \frac {gS}2.$ Putting that into the expression we have, we see that the minimum value of $ v_x^2 \plus{} v_y^2$ is $ g(2h \plus{} S)$,or:

The minimum initial velocity is $ \boxed{\sqrt {g(2h \plus{} S)}}.$

This is the same as the minimum velocity needed to throw the stone straight up to a height of $ h \plus{} \frac {S}2.$ It's as if we added half the width of the building to its height. The same formula works even if $ h \equal{} 0:$ to jump over a canyon of width $ S,$ you must reach an initial velocity of $ \sqrt {gS}.$


calculus
derivative
algebra solved
algebra

Consider f(x) = 3 \sum x^2 y^2 - 2 \sum x^2 y z with y and z fixed.

The second derivative of f(x) is 3y^2 + 3z^2 - 2yz is nonnegative because 3y^2 + 3z^2 \geq 6yz \geq 2yz.

Thus, f(x) is concave up, so its maximum is at one of the endpoints, x = 0 or x = 1. We extend the argument to y and z. Thus, the given value's maximum is obtained when x, y, z = 0 or 1.

Case: At least one of x, y, z = 0. WLOG, let x = 0.

Then we must prove 3y^2z^2 \leq 3, which is trivial.

Case: None of x, y, z = 0.

Then x = y = z = 1.

In that case, the given value is 9 - 6 = 3 \leq 3.


Find all functions  f:R -> R  such that

f(x+y) = f(x) + f(y) , for all x,y:R  and f(x) > or = 0  and  x > or = 0.

I got the solutions from Mathematics Excalibur (Vol 8, no:1), but I was not quite understand about it. Can U solve it in a simpler understanding?

Thank You.

[quote="ken_himura1984"]  and f(x) > or = 0  and  x > or = 0.
[/quote]

Do you perhaps mean: f(x)>=0 if x>=0 ?

Misha

Induction easily shows that f(k)=k*f(1), for every integer k.  I would imagine that k*c is the only solution, but because it says real numbers, I'm not sure.

Edit: In fact, it's just as easily shown by induction that f(m/n)=m*f(1/n) for every integers m and n.  Because of the density of the rational numbers, this should imply that f(x)=x*c is the only solution....but I may be wrong about this.


trigonometry
function
inequalities
rearrangement inequality

Find the minimum of expresion $ \frac{\sin^3 x}{\cos x}\plus{}\frac{\cos^3 x}{\sin x}$ for $ 0<x<\frac{\pi}{2}$

I think the expression doesn't have minimum.

Denote $ f(x) = \left(\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\right)$ and use Cauchy-Schwarz:
\begin{align*}
\sin(2x)f(x) &= (\sin x\cos x+\sin x\cos x)\left(\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\right) \\
&\ge (\sin^2 x+\cos^2 x)^2 \\
&= 1
\end{align*}

so $ f(x) \ge \frac{1}{\sin 2x} \ge \frac{1}{1} = 1 = f\left(\frac{\pi}{4}\right)$.

Oh, I made a mistake!

Here is my solution:

$ \frac {\sin ^ 3 x}{\cos x} \plus{} \frac {\cos ^ 3 x}{\sin x} \equal{} \frac {(\sin ^ 2 x \plus{} \cos ^ 2 x)^2 \minus{} 2\sin ^ 2 x\cos ^ 2 x}{\sin x \cos x}$

$ \equal{} \frac {1}{t} \minus{} 2t\ \left(0 < t\leq \frac {1}{2}\right)$, where $ t\equal{}\sin x\cos x\equal{}\frac{1}{2}\sin 2x$, since $ y \equal{} \frac {1}{t},\ y \equal{} \minus{} 2t$ are decreasing functions for $ 0 < t\leq \frac {1}{2}$.

Thus the minimum value is $ 1$ when $ t \equal{} \frac {1}{2}\Longleftrightarrow x \equal{} \frac {\pi}{4}$.

by Am-GM $ \frac{sin^3x}{cosx} \plus{} sinx.cosx \geq 2sinx^2$ and similarly $ \frac{cos^3x}{sinx} \plus{} sinx.cosx \geq 2cos^2x$
add them up we get $ \frac{sin^3x}{cosx}\plus{} \frac{cos^3x}{sinx}\plus{}2sinx.cosx \geq 2sinx^2 \plus{}2cos^2x$
note that $ sinx^2\plus{}cos^2x \geq 2sinx.cosx$
$ \frac{sin^3x}{cosx}\plus{} \frac{cos^3x}{sinx} \geq sinx^2\plus{}cos^2x\equal{}1$

[hide="Yet another Solution"]The sequences $ (\sin^2 x, \cos^2 x)$ and $ \left(\frac {\sin x}{\cos x} \plus{} \frac {\cos x}{\sin x}\right)$ are sorted similarly, so by Chebychev's $ \frac {\sin^3 x}{\cos x} \plus{} \frac {\cos^3 x}{\sin x} \ge \frac {1}{2}(\sin^2 x \plus{} \cos^2 x)\left(\frac {\sin x}{\cos x} \plus{} \frac {\cos x}{\sin x}\right) \equal{} \frac {1}{2}\left(\frac {\sin x}{\cos x} \plus{} \frac {\cos x}{\sin x}\right)\ge 1$.

Similarly, one may use Rearrangement for $ (\sin^3 x,\cos^3 x)$ and $ \left(\frac {1}{\cos x} , \frac {1}{\sin x}\right)$.[/hide]

I think that Kunny is referring example 18 in Arthur Engel's book, Problem Solving Strategies... (though I can't remember this problem...)

I seem to remember this was in AoPS 2 as well.

Let $ s \equal{} \sin x$ and $ c \equal{} \cos x$.

Note that $ s > c \implies s^3 > c^3, \; s^{ \minus{} 1} < c^{ \minus{} 1}$, and that $ s < c \implies s^3 < c^3, \; s^{ \minus{} 1} > c^{ \minus{} 1}$.

By the [[Rearrangement Inequality]] over the two lists $ (s^3, c^3)$ and $ (c^{ \minus{} 1}, s^{ \minus{} 1})$, we find that:

$ s^3 c^{ \minus{} 1} \plus{} c^3 s^{ \minus{} 1} \geq s^3 s^{ \minus{} 1} c^3 c^{ \minus{} 1} \equal{} s^2 \plus{} c^2 \equal{} 1$

[b]Edit:[/b] [i]Looks like this was in one of Bertram's posts, too[/i]

$ (c\minus{}s)^2(c^2\plus{}cs\plus{}s^2)\ge 0\implies s^4\plus{}c^4\ge cs (c^2\plus{}s^2)\equal{}cs\implies \frac{s^3}{c}\plus{}\frac{c^3}{s}\ge 1$

For a basc solution, use AM-GM with $ \sin 2x\equal{}2\sin x\cos x$.

[hide="Solution"]
Notice that\[ \frac {\sin^3 \theta}{\cos \theta} \plus{} \frac {\cos^3 \theta}{\sin \theta} \equal{} \frac {\sin^4 \theta \plus{} \cos^4 \theta}{\cos \theta\sin \theta}\]Yet, $ \cos \theta\sin \theta \equal{} \frac {\sin 2\theta}{2}$ and\[ \cos^4 \theta \plus{} \sin^4 \theta \equal{} (\sin^2 \theta \plus{} \cos^2 \theta)^2 \minus{} 2\cos^2 \theta\sin^2 \theta \equal{} 1 \minus{} \cos \theta\sin \theta\sin 2\theta\]Then, we have that the term is $ \frac {2}{\sin 2\theta} \minus{} \sin 2\theta$.  Clearly, by maximizing $ \sin 2\theta$, we minimize $ \frac {2}{\sin 2\theta}$ and minimize $ \minus{} 2\sin 2\theta$, and hence, minimize the term.  Therefore, since $ \sin 2\theta\le 1$, we get that the minimum occurs when\[ \sin 2\theta \equal{} 1\implies \frac {\sin^3 \theta}{\cos \theta} \plus{} \frac {\cos^3 \theta}{\sin \theta} \equal{} \boxed{1}\][/hide]

$ 1\blacktriangleright\ \ E \equal{} \frac {\sin^3 x}{\cos x} \plus{} \frac {\cos^3 x}{\sin x} \equal{}$ $ \frac {\sin^4 x \plus{} \cos^4 x}{\sin x\cos x\left(\sin^2x \plus{} \cos^2x\right)} \equal{}$ $ \frac {t^4 \plus{} 1}{t(t^2 \plus{} 1)} \equal{}$ $ \frac {(t^3 \minus{} 1)(t \minus{} 1)}{t(t^2 \plus{} 1)} \plus{} 1 \equal{}$

$ \frac {(t \minus{} 1)^2(t^2 \plus{} t \plus{} 1)}{t(t^2 \plus{} 1)} \plus{} 1\ge 1$ , where $ t \equal{} \tan x$ . We have equality iff $ t \equal{} 1$ , i.e. $ x \equal{} \frac {\pi}{4}$ .

$ 2\blacktriangleright\ \ E \equal{} \frac {\sin^3 x}{\cos x} \plus{} \frac {\cos^3 x}{\sin x} \equal{}$ $ \sin ^2x\cdot\tan x \plus{} \cos^2x\cdot\cot x$ . Prove easily that $ \left(\sin^2x\minus{}\cos^2x\right)(\tan x\minus{}\cot x)\ge 0$ .

In conclusion, $ E\ge \frac 12\left(\sin ^2x\plus{}\cos^2x\right)(\tan x\plus{}\cot x)\equal{}\frac 12\cdot (\tan x\plus{}\cot x)\ge 1$ , i.e. $ E\ge 1$ with equality iff $ x\equal{}\frac {\pi}{4}$ .


lets a b c and d integers  such that a²+b²+c²+d²=1989
  and lets     n²=max{a,b,c,d}  and   m=a+b+c+d
find the values of m and n
(x² =x*x)

Do you mean find $ \max{n}$ and $ \max{m}$ because we can have e.g. $ 1^2 \plus{} 4^2 \plus{} 6^2 \plus{} 44^2 \equal{} 2^2 \plus{} 6^2 \plus{} 10^2 \plus{} 43^2 \equal{} 1989$? (Note that, in fact, $ \max{n} \equal{} 44$ because $ \lfloor\sqrt {1989}\rfloor \equal{} 44$.)

Edit: actually, we can find $ \max{m}$ quite easily too (because $ a \plus{} b \plus{} c \plus{} d < \equal{} 2\sqrt {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}$ by QM-AM where $ \lfloor 2\sqrt {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}\rfloor \equal{} \lfloor 2\sqrt {1989}\rfloor \equal{} 89$, but $ 20^2 \plus{} 22^2 \plus{} 23^2 \plus{} 24^2 \equal{} 1989$ so we can have $ a\plus{}b\plus{}c\plus{}d\equal{}89$).

sorry i don't like your proof ,  i'm not convinced!   :roll:

Which part of the proof do you not like?
I completely understand both proofs, but the second one may be confusing.

QM-AM is the Quadratic Mean - Arithmetic Mean inequality, which is a semi-well-known inequality. It's part of the QM-AM-GM-HM (Quadratic, Arithmetic, Geometric, Harmonic means) inequality. QM-AM states that the square root of the arithmetic mean of the squares of a group of positive real numbers is greater than or equal to the arithmetic mean of the numbers. In other words,

$ \sqrt {\frac {a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2}{n}}\geq \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}{n}$

where $ a_i > 0$ for all integers $ 1\leq i\leq n$. Letting $ n \equal{} 4$, $ a_1 \equal{} a$, $ a_2 \equal{} b$, $ a_3 \equal{} c$, and $ a_4 \equal{} d$ gives us

$ \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}\geq \frac {a \plus{} b \plus{} c \plus{} d}{4}$

$ \frac {\sqrt {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}}{2}\geq \frac {a \plus{} b \plus{} c \plus{} d}{4}$

$ 2\sqrt {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}\geq a \plus{} b \plus{} c \plus{} d$

We know that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1989$, so we have $ 90 > 2\sqrt {1989} > 89\geq a \plus{} b \plus{} c \plus{} d$, since $ a$, $ b$, $ c$, and $ d$ are positive integers (non-zero). Therefore the theoretical maximum of $ m \equal{} a \plus{} b \plus{} c \plus{} d$ is 89, but we must prove it.

Note that $ 20^2 \plus{} 22^2 \plus{} 23^2 \plus{} 24^2 \equal{} 1989$ and $ 20 \plus{} 22 \plus{} 23 \plus{} 24 \equal{} 89$. Since we can't have $ a \plus{} b \plus{} c \plus{} d\geq 90$, we have that the maximum $ a \plus{} b \plus{} c \equal{} d \equal{} m$ can be is 89.

i understand the  proof  but it seems that you just discuss one  possibility !!! that what i mean

We only need one possibility. We know that $ max(m)<90$, and since $ m$ is an integer, the theoretical greatest $ m$ could be is 89, and it can be that, so $ max(m)\equal{}89$...

We don't always need to find all possibilities to prove something...

no i didn't mean that and i know that we don't have to find all possibilities (because that's why we need  maths!!!!). what we talking about is that he gives a case (exemple) of the problem, and if it is the only possibility , prouve it  taht is the purpose.

Note that $ a^2,b^2,c^2,d^2 \geq 0$ so then this implies that $ a^2,b^2,c^2,d^2\leq \lfloor\sqrt {1989}\rfloor \equal{} 44$. Then show that $ a \equal{} 44$ is possible and we're done.

yes it is possible
  for exemple  {a,b,c,d}={1,4,6,44}
 it is easy to find the valious of n :
 we have  4(n²*n²)then n>=5
and n²=<44 then n=<6
so we have n=5 or n=6

Aw, we all missed $ n^2\equal{}max(a,b,c,d)$. We all thought it was $ n\equal{}max(a,b,c,d)$. :D

The maximum that $ n^2$ can be is 44, so the maximum $ n$ is $ \sqrt{44}\equal{}2\sqrt{11}$ (note that $ n$ doesn't have to be an integer).

Since 44 is an integer, $ \sqrt{44}$ is too.

/me rolls eyes.

Just because an integer is an integer doesn't mean the square root of the integer is an integer.

You said that $ n$ was an integer, but it doesn't have to be.

ho  says that?!!!!
i didn't do you know you make me sik really sik
how old are you my boy??

[quote="ouboujlal"]no n is an integer because a b c and d are !!!!  open you eyes next time !!! :P[/quote]

And as of this post I am approximately $ 14.09885$ years old.

Think about it, if $ n\equal{}2$, what integer squared is equal to two?


trigonometry
search
geometry unsolved
geometry

Given a triangle ABC. Let M and N are the midpoints of the sides BC and AC respectively,
$ \angle {MAC}\equal{}\alpha$, $ \angle {NBC}\equal{}\beta$. Find the maximum of
                             $ P\equal{}\sin \alpha\plus{}\sin \beta$
( 3 angle $ A,B,C$ are variables)

Sorry for my bad English. I hope u can understand this problem and help me to solve it.

The maximum value of P is $ \frac{2}{\sqrt{3}}$
This is a problem in Bulgaria 1997 (around 4). You can see its proof in many documents (search on Internet). A well-known document is "Mathematical Olympiads 1997-1998: Olympiad Problems from Around the World".


We have positive numbers a,b,c,d, and: $ a\plus{}b\plus{}c\plus{}d\equal{}1$. Prove, that:

$ \frac{a}{b}\plus{}\frac{b}{a}\plus{}\frac{c}{d}\plus{}\frac{d}{c} \leqslant \frac{1}{64abcd}$

Good luck :D

[quote="Sylwek"]We have positive numbers a,b,c,d, and: $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove, that:

$ \frac {a}{b} \plus{} \frac {b}{a} \plus{} \frac {c}{d} \plus{} \frac {d}{c} \leqslant \frac {1}{64abcd}$
[/quote]
$ \left(\frac {a}{b} \plus{} \frac {b}{a} \plus{} \frac {c}{d} \plus{} \frac {d}{c}\right)abcd\equal{}(ac\plus{}bd)(bc\plus{}ad)\leq
\left(\frac{ac\plus{}bd\plus{}bc\plus{}ad}{2}\right)^2\equal{}$
$ \equal{}\left(\frac{(a\plus{}b)(c\plus{}d)}{2}\right)^2\leq\left(\frac{\left(\frac{a\plus{}b\plus{}c\plus{}d}{2}\right)^2}{2}\right)^2\equal{}\frac{1}{64}.$

I solved it the same way, congratulations :)


I don't think it's [i]only[/i] for 7th graders.  I never did it though .  [url=http://www.tip.duke.edu/SandW/SummerAndWeekend.htm]Here's[/url] a link to their site though...google is a wonderous resource...  From what the site says, it appears to be open to more grades than just 7th.

[quote="juicybooty911"]is the Duke TIP program only for 7th graders?[/quote]
im taking it this summer and im in 9th grade so no... i think its 7-10 but im not sure

Duke tip is for 4-5th and 7th graders in terms of recognition. For last a few years, if you were in 7th grade and your standard test score was in the 95% in your grade level, you would be qualified to register with Duke tip system (if you are in one of the southeast 16 states) and then you could take SAT I or ACT. For SAT I, if your score was greater or equal to 650 on either verbal or math, you would qualify for grand recognition.

Duke tip summer classes are for students in 7th to 10th grade. The classes are in different levels and have different requirement for certain levels of SAT scores.

I just got back from TIP.  I think it is 7-12.

The standard grade for testing on the SAT or ACT in the Duke region is seventh grade, but the distance learning classes and summer classes are open to people in other grades, and open to people from other regions.

[url=http://www.tip.duke.edu/]Duke University Talent Identification Program[/url]

Duke Summer Studies, the main TIP program (in my opinion), is for 7-10 grade.  There is also TIP Leadership, which is for 11th graders.

[quote="iceui"]Where are the similar programs in Illinois? Thanks![/quote]

The similar programs in Illinois are mostly in Evanston or nearby. They are run by the [url=http://www.ctd.northwestern.edu/index.html]Center for Talent Development[/url], the regional talent search and distance learning program for the Midwest.

I attended the Duke TIP summer program in 7th grade.  I don't think I learned very much and the tuition of couple thousand dollars is a bit too much.  But overall it was a fun camp.

[quote="stupidkid"]I attended the Duke TIP summer program in 7th grade.[/quote]

Which course did you take? More generally, what are the better courses at TIP, and which are the ones to avoid?

At risk of a little subject drift, I'd like to ask the same question about the CTY and CTD summer programs? Which are the most challenging? Which are not so challenging? Which are the most interesting?

I took a course called Restless Earth.  It was mostly over ecology and some geology.  I really don't know which are the good courses, but I heard from my roommate that psycology is pretty interesting.

I went to the TIP summer program the summer after my 8th grade year (that would be, um... back in 2000).  I took a class called "Mathematical Problem Solving" -- oddly enough (considering my interests now) not my first choice when I applied.  I don't know if they still run that course, but I certainly got a lot out of it.  I credit it with actually getting me interested in math problem solving.  Anyone who is going and considers themselves to be at the introductory level of problem solving should see if TIP is offering a course like this.

From my observation while I was there (not direct personal experience, so take this with a grain of salt), the strictly school-curricular courses are [i]not[/i] the best choices.  My impression was that the only people who actually learned most of the material presented in those classes were those who already knew it or could have easily mastered the material without the classes at all.  For the first few days of TIP I was envious of the kids who were taking precalc and advancing beyond me in the school curriculum, but I soon realized that I was getting far more out of the problem-solving class than the others were out of their curricular classes.  If you want to learn high school geometry, precalc, or what have you -- why bother with an expensive summer program?  If you are going to a program like this, pick a class for which you can't just learn the material anywhere.  Choose a subject for which mentoring and interaction with other interested and intelligent pupils significantly helps.

They still have Mathematical Problem Solving, but it was only the second term this year, which was unfortunate, because I wanted to it but couldn't fit it into my schedule.  But it's definitely good for MathCounts!


function
trigonometry
integration
real analysis
real analysis solved

Integrate: csc xcot xsec x dx

I came up with -csc x + (1/sinx) + C

I wasn't sure if it was ok to convert sec x to (1/sinx)

Whoa, hold the phone.

$\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}}$

$\sec{\theta} = \frac1{\cos{\theta}}$

Therefore $\cot{x} \cdot \sec{x} = \frac1{\sin{x}} = \csc{x}$

So now you do $\int{\left(\sin{x}\right)^{-2}}$ which is the same as $\int{\csc^{2}{x}}$ which obviously equals:

${-\cot{x}}$

Ahh ok, so sec =  (1/cos x)

I'm having trouble following you though. What happened to (csc x) in the original problem?

its still there. thats why i used $\csc^{2}x$


search
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probability and stats

In a large genetic study, scientists found that 550 out of 1000 people carry a mutation that may make them more succeptible to a rare disease.
a/ Assuming that this sample was taken at random, and assuming that people either do (called mutant) or do not (called non-mutant) have the mutation, write down the appropriate probability distribution to describe this experiment.
b/ From this data, what is the probability that fewer than 50% of the people in the greater population have the mutation?

I solve part a/ as follow,
From the text, the probability that an individual have the mutation is $ p\equal{}0.55$. Consider a group of $ N$ people.
Let $ f(k)$ be the probability that $ k$ out of $ N$ people have the mutation. Then $ f(k)\equal{}C_n^kp^k(1\minus{}p)^{N\minus{}k}$.
How to solve part b/? It seems that I have no idea of what to do.

I surmise you are expected to assume that $ p\equal{}0.5$ and find the probability that out of 1000 random people at least 550 have the mutation. This is certainly [b]not[/b] what is formally asked but I do not see how to make sense out of the question as written either. You'll, probably, have to talk to your teacher about what was meant.

Note that in a) you, most likely, put the things "upside down". I would say that the probability $ p$ is not known to you (it is just the true ratio of the mutant people to all people in the population) and $ N\equal{}1000$ (the size of the sample). The idea to use the binomial distribution is correct as long as the total population greatly exceeds $ 1000$.


Hi! I'm planning to join a math circle but I'm not quite sure about the opinions of others.  Do you think that joining a math circle is worth it?  Is its price and its opportunities are worth it as well?  How much commitment is needed?  Do you think it's possible to factor this in with 7 APs, sports, and other extracurricular activities?

Please reply back! Thanks :)

Also, can someone give me a brief overview of iTest?  When is the next round start, and how to become involved in it?

That sort of depends on where you live (hence what math circle you're talking about).  Although it's new and rather different from the SFBA math circle, Seattle's math circle, for example, is free and requires no commitment.  So... where do you live?

I go to two math circles, and I can say that it is surely worth it. First fo all, you decide your level of commitment. You attend 1-2 hour lectures and get problem sets in the end. Its up to you to decide whether you want to do the problems later. I've found that a math circle can introduce you to many topics which you don't know the existence of, and which tend to be very interesting to discover. So if you like math and want to learn more about it, I strongly suggest to join a local math circle if you have one in your area.

I'm thinking about attending the Albany Math Circle.  I live around there


Solve for $ x$:

$ 5\minus{}[x\plus{}(3\minus{}2x)]\equal{}2[x\minus{}3(2\minus{}x)]$.

$ 5\minus{}x\minus{}3\plus{}2x\equal{}8x\minus{}12$
$ x\equal{}2$


How would you define "image"? Non-complex numbers don't have an "image" - they are all abstract qualities of mathematics.

Alone, I've never heard of an "image" of a complex number.

However, if you have a function that takes a complex number, for example, $ f(a \plus{} bi) \equal{} 2a \minus{} bi$, then the image of a complex number, $ z$, is the value the function maps it to, $ f(z)$.

So, in this case, given the function above, the image of $ 3 \plus{} 2i$ would be $ f(3\plus{}2i) \equal{} 6 \minus{} 2i$.

In general, you can define the image of anything as the result of a function.  So the image of 3 using the function $ f(x) \equal{} x^2$ is 9.

This can be generalized to define the images of sets and functions.

I'm guessing that original poster is simply misreading an abbreviation.

The expression $ \text{Im}(z),$ or $ \Im(z),$ is read as the [b]imaginary part[/b] of $ z,$ not as the "image" of $ z.$

$ \text{Im}(3\minus{}4i)\equal{}\minus{}4.$

$ \text{Im}(z)\equal{}\frac{z\minus{}\overline{z}}{2i}.$

$ z\equal{}\text{Re}(z)\plus{}i\,\text{Im}(z).$

$ \text{Im}(zw)\equal{}\text{Re}(z)\,\text{Im}(w)\plus{}\text{Im}(z)\,\text{Re}(w).$

And other similar identities.

he might mean image upon inversion of plane, for which Wikipedia has a fairly good explanation


function
calculus
LaTeX
real analysis
real analysis unsolved

Hi guys,

I'm not sure if this is calculus or not but i have the following problem.

I've been told that the following problem can be solved on excel fairly easily... but i'm unsure what i should do with it. I briefly remember my "Maths" teacher mentioning a table of results on excel, and then a graph.

Could anybody guide me through the solution so i can practice some examples?

Here is a screenshot of the problem:

http://www.putfile.com/pic.php?img=3201934

Thanks

I don't know how to use Excel, so I can't guide you on how to set up what they want. But I get the impression that they want you to generate a table of values of the function at various points in the intervals (time $x$ and angle $\theta$). Then you use the table to create a graph and a best fit line. Then use that to solve for $x$ when $v=2, \theta = 45$.

If you don't know how to use Excel either, I'm sure you can find guides online; try Google.

A tip for this forum: it's much better to type the problem rather than link to a picture of it. People will be more likely to help you and more appreciative because you made it easier for us. Also, typing up the problem forces you to notice all the details in it -- might be helpful for you from time to time as well.

There's a link on the sidebar to a LaTeX guide, so that you can learn the syntax for putting math stuff in the forum. (You won't need to download anything.)


Chandra has four bowls.  Each one is a different color (red, blue, yellow, green).  She also has exactly one glass the same color as each bowl.  If she chooses a bolw and a glass from the cupboard, how many pairings are possible? One such pairing is a blue bowl and a yellow glass.

[hide]each bowl has 4 possibilities for pairing. So its 4 x 4 = $16$ possibilities.[/hide]

[hide] There are 4 bowls that can go with the first glass, 4 that can go with the second glass, etc. 4x4=16[/hide]

[hide="answer"]bowl     cup
bowl     cup
bowl     cup
bowl     cup
each bowl maps to 4 cups so 4 x 4 = [size=150][b]16[/b][/size]  [/hide]


algebra
polynomial
number theory
least common multiple
function
Euler
vector

that  $ \mathbb{Q}(\xi_m) \bigcap \mathbb{Q}(\xi_n) \equal{} \mathbb{Q}(\xi_{gcd(m,n)})$

I said that since $ \xi_m \equal{} e^{(2\pi*i/m)}$ and $ \xi_n \equal{} e^{(2\pi*i/n)}$

Recall that $ \gcd(x^m \minus{} 1, x^n \minus{} 1) \equal{} x^{\gcd(m, n)} \minus{} 1$.

I don't think this'll help.
We need all [url=http://en.wikipedia.org/wiki/Cyclotomic_polynomial#Cyclotomic_polynomials]cyclotomic polynomials[/url] irreducible.
Then $ [\mathbb{Q}(\zeta_n): \mathbb{Q}] = \deg \Phi_n = \varphi (n)$ for all $ n\in\mathbb{N}$.

We also need $ \varphi (m)\varphi(n) = \varphi(\gcd(m,n))\varphi(\text{lcm}(m,n))$ for all $ m,n\in\mathbb{N}$.
But the [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]Euler function[/url] $ \varphi$ is multiplicative,
so by factoring $ m,n$ into coprime prime powers we just need to show
$ \varphi (p^\alpha)\varphi(p^\beta) = \varphi(p^{\min\{\alpha,\beta\}})\varphi(p^{\max\{\alpha,\beta\}})$ for any prime $ p$ and $ \alpha,\beta\in\mathbb{N}_0$.
This is trivial.

And now
$ [\mathbb{Q}(\zeta_m): \mathbb{Q}(\zeta_{\gcd(m,n)})] = {\varphi(m)\over \varphi(\gcd(m,n))}$
gives $ \mathbb{Q}(\zeta_m)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ generated by $ \zeta_m^k$ for $ k = 0,\ldots,{\varphi(m)\over \varphi(\gcd(m,n))} - 1$.
Analog $ \mathbb{Q}(\zeta_n)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ is generated by $ \zeta_n^k$ for $ k = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1$.
So the extension $ \mathbb{Q}(\zeta_m,\zeta_n)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ is generated by the elements $ \zeta_m^k\zeta_n^l$
for $ k = 0,\ldots,{\varphi(m)\over \varphi(\gcd(m,n))} - 1,l = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1$.

We find $ u,v\in\mathbb{Z}$ with $ um + vn = \gcd(m,n)\Rightarrow {1\over \text{lcm}(m,n)} = {\gcd(m,n)\over mn} = {v\over m} + {u\over n}$
${ \Rightarrow \zeta_{\text{lcm}(m,n)}} = \zeta_m^v\zeta_n^u\in\mathbb{Q}(\zeta_m,\zeta_n)$

${ \Rightarrow [\mathbb{Q}(\zeta_m,\zeta_n): \mathbb{Q}(\zeta_{\gcd(m,n)})]\ge [\mathbb{Q}(\zeta_{\text{lcm}(m,n)}}): \mathbb{Q}(\zeta_{\gcd(m,n)})] \\
\hspace*{1.67in} = {\varphi(\text{lcm}(m,n))\over \varphi(\gcd(m,n))} \\
\hspace*{1.67in} = {\varphi(m)\over \varphi(\gcd(m,n))}\cdot {\varphi(n)\over \varphi(\gcd(m,n))}$.

Therefore the elements $ \zeta_m^k\zeta_n^l$ for $ k = 0,\ldots,{\varphi(m)\over \varphi(\gcd(m,n))} - 1,l = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1$
form in fact a basis of $ \mathbb{Q}(\zeta_m,\zeta_n)$ over $ \mathbb{Q}(\zeta_{\gcd(m,n)})$ and are therefore linear independent over $ \mathbb{Q}(\zeta_{\gcd(m,n)})$.

But now for $ x\in \mathbb{Q}(\zeta_m)\cap\mathbb{Q}(\zeta_n)$ we can write

$ x = \sum_{k = 0}^{{\varphi(m)\over \varphi(\gcd(m,n))} - 1}a_k\zeta_m^k = \sum_{l = 0}^{{\varphi(n)\over \varphi(\gcd(m,n))} - 1}b_l\zeta_n^l$ with some $ a_k,b_l\in \mathbb{Q}(\zeta_{\gcd(m,n)})$.
and the above gives $ x = a_0 = b_0\in\mathbb{Q}(\zeta_{\gcd(m,n)})$

[quote="olorin"]

And now
$ [\mathbb{Q}(\zeta_m): \mathbb{Q}(\zeta_{\gcd(m,n)})] \equal{} {\varphi(m)\over \varphi(\gcd(m,n))}$
gives $ \mathbb{Q}(\zeta_m)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ generated by $ \zeta_m^k$ for $ k \equal{} 0,\ldots,{\varphi(m)\over \varphi(\gcd(m,n))} \minus{} 1$.

[/quote]

ok, this is the only part of your proof that I don't get.  How did you know what generated the extension by simplying observe the index of the extension?

Whenever you have a field extension $ K(a)/K$ generated by one element $ a$,
then $ K(a)\cong K[X]/(f)$, where $ f$ is the minimalpolynomial of $ a$ over $ K$.
Also $ K(a)$ has degree $ n: \equal{}\deg f$ over $ K$ and $ 1,a,a^2,\ldots, a^{n\minus{}1}$ forms a basis of the $ K$-vector space $ K(a)$.

[quote="olorin"]I don't think this'll help.
We need all [url=http://en.wikipedia.org/wiki/Cyclotomic_polynomial#Cyclotomic_polynomials]cyclotomic polynomials[/url] irreducible.
Then $ [\mathbb{Q}(\zeta_n): \mathbb{Q}] = \deg \Phi_n = \varphi (n)$ for all $ n\in\mathbb{N}$.

We also need $ \varphi (m)\varphi(n) = \varphi(\gcd(m,n))\varphi(\text{lcm}(m,n))$ for all $ m,n\in\mathbb{N}$.
But the [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]Euler function[/url] $ \varphi$ is multiplicative,
so by factoring $ m,n$ into coprime prime powers we just need to show
$ \varphi (p^\alpha)\varphi(p^\beta) = \varphi(p^{\min\{\alpha,\beta\}})\varphi(p^{\max\{\alpha,\beta\}})$ for any prime $ p$ and $ \alpha,\beta\in\mathbb{N}_0$.
This is trivial.

And now
$ [\mathbb{Q}(\zeta_m): \mathbb{Q}(\zeta_{\gcd(m,n)})] = {\varphi(m)\over \varphi(\gcd(m,n))}$
gives $ \mathbb{Q}(\zeta_m)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ generated by $ \zeta_m^k$ for $ k = 0,\ldots,{\varphi(m)\over \varphi(\gcd(m,n))} - 1$.
Analog $ \mathbb{Q}(\zeta_n)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ is generated by $ \zeta_n^k$ for $ k = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1$.
So the extension $ \mathbb{Q}(\zeta_m,\zeta_n)/\mathbb{Q}(\zeta_{\gcd(m,n)})$ is generated by the elements $ \zeta_m^k\zeta_n^l$
for $ k = 0,\ldots,{\varphi(m)\over \varphi(\gcd(m,n))} - 1,l = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1$.

We find $ u,v\in\mathbb{Z}$ with $ um + vn = \gcd(m,n)\Rightarrow {1\over \text{lcm}(m,n)} = {\gcd(m,n)\over mn} = {v\over m} + {u\over n}$
${ \Rightarrow \zeta_{\text{lcm}(m,n)}} = \zeta_m^v\zeta_n^u\in\mathbb{Q}(\zeta_m,\zeta_n)$

${ \Rightarrow [\mathbb{Q}(\zeta_m,\zeta_n): \mathbb{Q}(\zeta_{\gcd(m,n)})]\ge [\mathbb{Q}(\zeta_{\text{lcm}(m,n)}}): \mathbb{Q}(\zeta_{\gcd(m,n)})] \\
\hspace*{1.67in} = {\varphi(\text{lcm}(m,n))\over \varphi(\gcd(m,n))} \\
\hspace*{1.67in} = {\varphi(m)\over \varphi(\gcd(m,n))}\cdot {\varphi(n)\over \varphi(\gcd(m,n))}$.

Therefore the elements $ \zeta_m^k\zeta_n^l$ for $ k = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1,l = 0,\ldots,{\varphi(n)\over \varphi(\gcd(m,n))} - 1$
form in fact a basis of $ \mathbb{Q}(\zeta_m,\zeta_n)$ over $ \mathbb{Q}(\zeta_{\gcd(m,n)})$ and are therefore linear independent over $ \mathbb{Q}(\zeta_{\gcd(m,n)})$.

But now for $ x\in \mathbb{Q}(\zeta_m)\cap\mathbb{Q}(\zeta_n)$ we can write

$ x = \sum_{k = 0}^{{\varphi(m)\over \varphi(\gcd(m,n))} - 1}a_k\zeta_m^k = \sum_{l = 0}^{{\varphi(n)\over \varphi(\gcd(m,n))} - 1}b_l\zeta_n^l$ with some $ a_k,b_l\in \mathbb{Q}(\zeta_{\gcd(m,n)})$.
and the above gives $ x = a_0 = b_0\in\mathbb{Q}(\zeta_{\gcd(m,n)})$[/quote]
It's nice! Thank you! :lol:

Friendly advice: never quote an entire post  :wink: .


show that If $\sqrt{2}$ is written in decimal notation ,there is at least one nonzero digit at the interval of $1,000,000-th$ and $3,000,000-th$ the digits. :)


Let $ a,b,c$ be positive numbers satisfying $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$.Prove that
\[ \frac {a^2b \plus{} 1}{(2a \plus{} b)^2} \plus{} \frac {b^2c \plus{} 1}{(2b \plus{} c)^2} \plus{} \frac {c^2a \plus{} 1}{(2c \plus{} a)^2}\ge \frac {2}{3}
\]
P/s:I edited


Καλημέρα

Θα ήθελα να ρωτήσω αν υπάρχει καμιά μαζεμένη λίστα με διάφορες αντικαταστάσεις στα ολοκληρώματα έτσι ώστε να μπορώ να τα υπολογίζω αναλυτικά χωρίς χρήση mathematica. Κάποιες βασικές αντικαταστάσεις τις ξέρω αλλά δε με καλύπτουν στην περίπτωση του ολοκληρώματος:

$ \int{\frac{1}{x^2\sqrt{a\minus{}\frac{b}{x^2}}}}dx$

Για το συγκεκριμένο ολοκλήρωμα ξέρω τη λύση (όποιος θέλει ας το δει σαν άσκηση) αλλά με ενδιαφέρει να ενταχθεί στα πλαίσια μιας συστηματικής μεθοδολογίας υπολογισμού ολοκληρωμάτων και όχι σα μια ευφυή σκέψη που είδα σε κάποιο βιβλίο χωρίς συνέχεια.

[quote="gpapargi"]Θα ήθελα να ρωτήσω αν υπάρχει καμιά μαζεμένη λίστα με διάφορες αντικαταστάσεις στα ολοκληρώματα έτσι ώστε να μπορώ να τα υπολογίζω αναλυτικά χωρίς χρήση mathematica. Κάποιες βασικές αντικαταστάσεις τις ξέρω αλλά δε με καλύπτουν στην περίπτωση του ολοκληρώματος:[/quote]

Off the top of my head, μου έρχονται τα "Στοιχεία Ανώτερης Μαθηματικής Ανάλυσης" του Πάλλα... old-fashioned μεν, αλλά έχει τις [i]άπειρες[/i] αντικαταστάσεις και καλύπτει εκτενέστατα το θέμα.

Αύριο που θα ανοίξει η σχολή, θα κοιτάξω να δω και αν οι Abramowitz-Stegun έχουν καλή κάλυψη σε στοιχειώδες επίπεδο ή αν είναι πιο εξειδικευμένοι.

Cheerio,

Durandal 1707

Έχουμε τίποτα νεότερο;
Το συγκεκριμένο ολοκλήρωμα βγαίνει αμέσως με αντικατάσταση $ x \equal{} \frac {1}{y}$

Προέρχεται από τη διαφορική εξίσωση που προκύπτει στην προσπάθεια να βρεθεί η εξίσωση τροχιάς που κάνει ένα σώμα στο βαρυτικό πεδίο. Ο Τραχανάς κάνει το μετασχηματισμό πάνω στη διαφορική λέγοντας ότι από τη μορφή της φαίνεται ότι η «φυσική μεταβλητή» της   εξίσωσης είναι η $ \frac {1}{x}$.
Όμως η αιτιολόγηση αυτή δε με καλύπτει γιατί σε άλλες εξισώσεις που η «φυσική μεταβλητή» είναι η ίδια… το κόλπο δεν πιάνει.
Στο βιβλίο του Landau βγάζει το «στοιχειώδες ολοκλήρωμα» όπως το αποκαλεί κάνοντας όπως ήταν αναμενόμενο την ίδια αντικατάσταση.

Δεκτό και βοήθημα της Γ λυκείου ή και σελίδα με αντικαταστάσεις στο δίκτυο (θα ήταν το προτιμότερο), δυστυχώς εγώ δε βρήκα κάτι.

Sorry George, ξεχάστηκα... τον Abramowitz άστον καλύτερα, τώρα που τον ξαναξεφύλλισα είναι πολύ περισσότερο geared στις ειδικές συναρτήσεις παρά σε κλειστά ολοκληρώματα. Εξακολουθώ να σου προτείνω τον Πάλλα που έχει πολλές τεχνικές, ορισμένες εκ των οποίων αρκετά εξειδικευμένες (κουράγιο μόνο με την καθαρεύουσα αν σ'ενοχλεί). Έχει και τις άπειρες ασκήσεις για εξάσκηση...

Πιο στοιχείωδες (άλλα ok) είναι το Calculus του Stewart, εκεί θα βρεις αρκετές τεχνικές αντικατάστασης (αλλά τίποτα ιδιαίτερο, καμία σχέση με τον Πάλλα).

Ωστόσο, δεν είμαι σίγουρος τί ακριβώς ψάχνεις... σίγουρα δεν υπάρχει μία μαγική λίστα με αντικαταστάσεις που να έχει όλες τις μορφές ολ/των και να μπορείς να παίρνεις πάντα τη σωστή. Π.χ. η συγκεκριμένη αντικατάσταση φαίνεται αρκετά εύκολα στη μορφή που δίνεις το ολοκλήρωμα μόλις παρατηρήσεις το διαφορικό $ d(1/x) \equal{} \minus{}dx/x^2$ που σου κλείνει το μάτι... γενικά όμως... πρόβλημα ;)

Does this help?

Durandal 1707

Παναγιώτη ευχαριστώ πολύ. Τελικά έβγαλα άκρη. Γράφω την ιστορία γιατί είναι χρήσιμη.

Πήγα στο Σαββάλα και βρήκα το δεύτερο τόμο του Πάλλα ενώ το κομμάτι που μας ενδιαφέρει είναι στον πρώτο τόμο. Ξεφύλλισα το δεύτερο τόμο. Ωραίο βιβλίο κλασσικό με μεγάλη ύλη μέσα. Έκανε μόνο 10.45 ευρώ (ο Β τόμος).
Τον πρώτο τόμο τον βρήκα στην Αίθρα. Εκεί ο πρώτος τόμος έκανε 50 Ευρώ και ο δεύτερος 46 αν και είναι δίπλα στο Σαββάλα!!!

Έχει κανείς υπόψη του που μπορεί κανείς να βρει το βιβλίο σε λογική τιμή; Γενικά ξέρει κανείς βιβλιοπωλείο εκτός της Αίθρας που να έχουν παλιά κλασσικά βιβλία σε κάπως λογική τιμή; Η Γεωμετρία του Κανέλου  στην Αίθρα κάνει 85 ευρώ γιατί είναι παλιό!!!

Τελικά βρήκα βιβλίο με αντικαταστάσεις στα ολοκληρώματα. Του Σπανδάγου (ιδιοκτήτη της Αίθρας) με 13 ευρώ. Ίσως δεν είναι τόσο για να καταλάβεις έννοιες αλλά είναι γραμμένο καθαρά στο στυλ: μορφή τάδε -> αντικατάσταση τάδε. Για κάποιον που έχει καταλάβει από αλλού τις έννοιες και θέλει απλά να μπορεί να αντιμετωπίζει ολοκληρώματα που εμφανίζονται μπροστά του στη φυσική κάνει.
Εκεί στη σελίδα 105 γράφει ότι τα ολοκληρώματα της μορφής
$ \int{\frac{dx}{(x-d)^n \sqrt{ax^2+bx+c}}}$
Υπολογίζονται με την αντικατάσταση $ w=\frac{1}{x-d}$
Αν κάνεις πράξεις το δικό μου ολοκλήρωμα τελικά φτάνεις στη μορφή $ \int{\frac{dx}{x\sqrt{x^2-1}}}$ όπου για d=0, n=1, a=1, b=0 και c=-1 προκύπτει από τη γενική μορφή.

Γενικά πιστεύω ότι μπορεί να υπάρξει λίστα με αντικαταστάσεις ίσως όχι τόσο μαγική αλλά σίγουρα πολύ χρήσιμη. Πχ πιστεύω πως το mathematica πρέπει να έχει τέτοια λίστα μέσα του και να κάνει απλά κάποιο pattern matching για να καταλάβει τη μορφή. Πως αλλιώς θα μπορούσε να βγάζει παράγουσα;


logarithms
function
real analysis
real analysis unsolved

Suppose that $f(x)=\sum_{n\geq 0}{a_{n}e^{inx}}$ belongs to $L^{1}(0,2\pi)$. Prove the existence of a sequence $(n_{k})_{k\geq 1}$ such that $\frac{n_{k+1}}{n_{k}}$ converges to $1$ and the series of general term $a_{n_{k}}$ converges absolutely.

I don't have a solution, but would like to point out that the title is misleading: Fourier coefficients [b]can[/b] be big in $L^{1}$ (something like $a_{n}\approx 1/\log(e+|n|)$, which is not absolutely summable along any sequence $n_{k}$ as in the statement.) The key assumption here is that $a_{n}=0$ for $n<0$, that is, $f$ is in the Hardy space $H^{1}$.

How about this for a counterexample: let $g(x)=\sum_{n\ge 0}b_{n}e^{inx}$, where $b_{n}=\frac{1}{\sqrt{n+1}\log^{p}(n+e)}$ for some $p>1/2$. This function is (just barely) in $L^{2}$. Let $f(x)=g(x)^{2}$, then the Fourier coefficients of $f$ (call them $a_{n}$) are obtained by convolving the sequence $(b_{n})$ with itself: $a_{n}=\sum_{k=0}^{n}b_{k}b_{n-k}$. All terms in this sum are positive, and the middle third of them, $\frac{n}{3}\le k\le \frac{2n}{3}$, are of size about $\frac{1}{n\log^{2p}n}$. Hence $a_{n}\ge\frac{c}{\log^{2p}n}$ for some $c>0$. For any sequence $(n_{k})$ as in the statement we have $a_{n_{k}}\ge \frac{c'}{\log^{2p}k}$, which is a divergent series.

 :huh:

I attach here the solution of the proposer.

My mistake had nothing to do with Fourier series: I thought that the condition $\frac{n_{k+1}}{n_{k}}\to 1$ implies $n_{k}\le Ck^{1+\epsilon}$. In fact, it implies $n_{k}\le C(1+\epsilon)^{k}$.
 :blush: $\times 10^{3}$


algebra
polynomial
superior algebra
superior algebra unsolved

I haven't been able to solve this problem, can anybody help?

Prove: If K is a finite extension of the field F of degree 2^n (for some positive integer n), a prime polynomial of odd prime degree over F is also prime over K.

If it were not prime, could it have a prime factor of odd degree? (consider the fields obtained by adjoining a root of such to either K or F)

If $ \alpha$ is the root of the prime polynomial $ f(x)$ in $ F[x]$,where should show that $ f(x)$ is irreducible over $ K[x]$
i.e, we should show that the degree of  $ \alpha$ is still $ deg f$.
You may consider $ E\equal{}F[\alpha],KE\equal{}K[\alpha]$,you can show that $ [KE: F]\equal{}2^n deg f$,because $ (deg f,2)\equal{}1$

Thanks for your responses; they led me to the answer.


I have attached the problem that I need help with.  Any help would be greatly appreciated!  Thanks!

i think this is the wrong section. try the calculus or high school section.

[hide="sketch of a solution"]
Let $g(t)=1-\frac 1t$. Then $g(g(t))=1-\frac{1}{1-\frac 1t}=1-\frac{t}{t-1}=\frac{1}{1-t}$ and $g(g(g(t)))=t$.

Substitute $x=t$, $x=1-\frac 1t$, and $x=\frac{1}{1-t}$ into the differential equation. Let $a=f^\prime (x)$, $b=f^\prime \left( 1-\frac 1x \right)$, $c=f^\prime \left( \frac{1}{1-x} \right)$. You will have a system of three linear equations in $a$, $b$, and $c$.

Solve the system of equations for $a$ (in terms of $x$ and $\lambda$) to get an explicit formula for $f^\prime (x)$. Then just set it equal to $\frac{\lambda}{2}$ and solve for $x$.
[/hide]


inequalities
function
inequalities proposed

Let $ 0 \leq a,b,c \leq 1$. Find maximum of $ M\equal{}a(b\minus{}c)^2\plus{}b(c\minus{}a)^2\plus{}c(a\minus{}b)^2$

Let $ 0 \leq a,b,c \leq 1$. Find maximum of $ M \equal{} a(b \minus{} c)^2 \plus{} b(c \minus{} a)^2 \plus{} c(a \minus{} b)^2\plus{}a^2b^2c^2$

[quote="dragon1"]Let $ 0 \leq a,b,c \leq 1$. Find maximum of $ M \equal{} a(b \minus{} c)^2 \plus{} b(c \minus{} a)^2 \plus{} c(a \minus{} b)^2 \plus{} a^2b^2c^2$[/quote]
It's the same.  :wink:

[quote="arqady"][quote="dragon1"]Let $ 0 \leq a,b,c \leq 1$. Find maximum of $ M \equal{} a(b \minus{} c)^2 \plus{} b(c \minus{} a)^2 \plus{} c(a \minus{} b)^2$[/quote]
$ \frac {\partial^2 M}{\partial a^2}\geq0,$ $ \frac {\partial^2 M}{\partial b^2}\geq0,$ $ \frac {\partial^2 M}{\partial c^2}\geq0.$
Hence,
$ \max M \equal{} \max\{M(0,0,0),M(0,0,1),M(0,1,1),M(1,1,1)\} \equal{} M(0,1,1) \equal{} 2.$[/quote]
I don't understand your idea.please tell me more about it  :blush:
thanks

tuandokim, $ M$ is a convex function of $ a,$ of $ b$ and of $ c.$ Hence, the maximum value of $ M$ occurs on an endpoints of the segment $ [0,1]$ id est, when $ \{a,b,c\}\subset\{0,1\}.$ :wink:


In a triangle ABC
if A=3B

Prove that $ bc^{2}$ = $ (a\minus{}b)^{2}(a\plus{}b)$

$ cosA \equal{} cos3B$

$ cosA \equal{} cosB(4cos^2B \minus{}  3)$

Now cosine rule should do it.

Conjurer,
please try to give an algorithm/solution which can be worked out easily.
solving the equation you got would drive anyone nuts :wacko:

P.S: [hide="hint"]

first try to prove this.
if in a triangle PQR, P=2Q
then
      $ p^{2}\equal{}q(q\plus{}r)$

the above-mentioned fact simplifies the given problem.[/hide]

Hello Keerthan.
Anyway that relation is easy to prove by constructing an angle bisector and using the angle bisector theorem.

thus far the problem has been relatively simple.
NOW I wish to find a condition that holds good when A=n.B (n is a natural)
It is possible to show that knowing a condition for A=n.B, we can find a condition for A=(n+1)B.
[hide="for further reference"]
the proof can be found in a book which I will name if anyone requires. It is a rather simple proof, but assumes that the condition(equation) is homogeneous.[/hide]
BUT is it possible to find the condition itself as an equation involving n,a,b,c.
If not, WHY?


Two students, $P$ and $Q$ are playing a number game in which numbers from $1$ to $50$ inclusive are written down on a piece of paper. $P$ starts the game by crossing out a number. Then $Q$ crosses out a number, followed by $P$ crossing out another number, and so on. This is done until only two numbers are left on the sheet of paper. If the sum of the two numbers is not divisible by $3$, $P$ wins. If the sum of the two numbers is divisible by $3$, $Q$ wins. What is $Q$'s strategy to win?

The sum of the numbers from 1 to 50 is divisible by 3. Q's strategy is to remove any number such that the sum of the remaining numbers is a multiple of 3. Q's choices should be as follows.

P's choice      Q's choice
(mod 3)           (mod 3)
0                               0
1                               2
2                               1

Between 1 and 50 there are 16 numbers which are equivalent to 0 mod 3, and 17 each equivalent to 1 and 2. Thus, whatever P's choice, Q is guaranteed at least one number which he can choose to leave a multiple of 3. Q therefore wins.

yep. the 4th president got it right heh :P


let $b > a > 0$ and $\delta > 0$ be real numbers. Then for any given positive $r \in \mathbb{R}$, proof that
$(\frac{b+\delta-a}{b-a} \cdot \frac{b^{r+1}-a^{r+1}}{(b+\delta)^{r+1}-a^{r+1}})^{\frac{1}{r}} > \frac{b}{b+\delta}.$The lower bound in inequality is best possible.

anyone try? I'm thinking howto solve it.......


group theory
abstract algebra
superior algebra
superior algebra solved

Give an example of group G and a nonempty subset H of G such that HH is a subset of H and H is not a subgroup of G

[hide="what i did: :oops: "]H ought to be infinite[/hide]


How many topologies are there on a $n-element$ set?

Discussed - inconclusively - [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63469]here.[/url]

Appears to be an unsolved problem for general $n.$

But what is the cardinal of the set of all topologies on the set of natural numbers?
is this an open question ,too?


topology
advanced fields
advanced fields unsolved

in r2 are the open balls a base for every topology ?

Could you give a numerical example of a base on a topology ?

No. Take the topology $\{\emptyset, \mathbb{R}^{2}\}$


[quote="MCrawford"]How do you know for sure you can't produce a greater range?[/quote]

because it restricts you to nonnegative integers and the smallest integer you can use is 0

dinoman wrote:[hide]33

the 5 numbers you can have is 0, 18, 19, 20, 33[/hide]

 :gamma:  :varepsilon: YOU V :varepsilon:  :gamma: Y SU :gamma:  :varepsilon:  OI= IT?

Could I not lower the 18 and raise the 33?

[hide]A median of 19 tells us the middle number (in the case of 5 numbers, the third largest) is 19.  A mean of 18 tells us the average of the 5 numbers is 18.

To get the largest range, we need to minimize the least and maximize the greatest of the 5 numbers.  Because the numbers must be nonnegative integers, the least is 0.

The numbers are also distinct, so no repetition is allowed.  To maximize the largest number, the smallest numbers will need to be at a minimum, so we can assume a second smallest number of 1.  This leaves us with the set of numbers {0,1,19,x,y} where 19 < x < y and (0+1+19+x+y)/5 = 18

Solving for x and y, we get x+y = 70.  Picking the smallest possible x (20), we can solve y=50.

The set of numbers {0,1,19,20,50} satisfies the requirements

1) a median of 19: the middle number is 19

2) a mean of 18:  (0+1+19+20+50)/5 = 90/5 = 18

Therefore the maximum range is 50.[/hide]

Given a list of 5 distinct nonnegative integers with mean of 18 and median of 19, what is the greatest possible range?

[hide]

They add up to 90.

_ _ 19 _ _ = 19

0 1 19 20 _ = 40

0 1 19 20 50 = 90

50

[/hide]


What state are you from and is the countdown round official? :?

-In Nevada, no. :(

However, I did would've came in first no matter what. :D

>Also, check out my other topic:
 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33291]State Competition Scores[/url]

In Washington, countdown round is official.

In SC it is not, which is a pity!  If it was official, I'd be going to the Nats!
Can a moderator please combine these multiple threads, along with the polls?

In kentucky, it is not official. But if it were, there would be 4 people in the same school going to nationals each year!

GA
NO

Either way i would go to nats.
as 1st Place!

CA: No... my rank wouldn't change even if it were official though.

U can't really say that u would be going to Nats if it was unofficial, b/c the random bracket can allow the 14th place to enter the top 4 by pairing him with like 15 and 16 and pairing the top guys like 1 and 2 together. Ladder format or weighted bracket is much more fair. I'm assuming that all nonofficial states have a random bracket format.

UTAH--yes.  I'm glad, otherwise I wouldn't be going to nats--5th to 2nd.

the way it worked in texas last year was the tenth place person on the written would face off against ninth. winner of that would face eight place. etc...
and the countdown was offficial, but the top 4 on written ended up being 4 to go to nationals anyways. i think one guy went from 8th to 5th and barely lost the countdown to the 4th place guy which would have put him on the national team..

I don't mean to be offensive to any of you, but this is my honest opinion:

The countdown round should never be official at the state level. Suppose there is a huge dropoff in score from 1st to 10th place (not unbelievable) and the tenth place manages to get, say, 4th place in the countdown (hard, but also not unbelievable). Suppose that this person says that their countdown is excellent and they always do the best of anyone at nationals on the videos. Now send them to nats (after all, they won the countdown round). They take the written test and get 150th. I don't see these people getting to the countdown round.

In other words, the written test is much more important that the countdown, and we can't let those who have no chance of making the national countdown but would win if they did to go to nationals.

I think that countdown should NOT be official, however, it is important to be good at Countdown if you want to do well at Nationals. Countdown skills (speed) can help you do problems faster under pressure and therefore give you more time to check your answers at the end, but then Written Round skills (accuarcy) are a lot more important. I remember in 7th grade when I was really really really good at countodwn (I owned everyone in my school). However, I wasn't put on the team so I didn't get to do Countdown at chapter and came in a close 11th at state, so I didn't make Countdown there either. So, I decided to sacrifice that speed and work on my accuarcy, which turned out to be a good choice. This year, I got owned at chapter countodwn (because it was the first time I've actually done it in front of a huge crowd), and placed worse individually than I did last year at chapter, but at least I improved my score (43 to 44, 4th place to 5th place). However, at state, it all payed off; I got the same improvement of 1 point, but it moved me up 9 ranks (probably because this year's State test was so much harder). This year, I made the CA team with a 40 (26/7) and got 2nd place.

Tennessee

no

Thank god, cuz i can do good in the written but when it comes to people watching me i miss the questions.

In CA, the 10th place person (or was it 8th or 9th place?) beat 5 or 6 people, then got stopped at either 3rd or 4th... Then I beat the 3rd place person and then the 1st place person beat me... the 3rd place person beat the 1st place person (the third and first from this year were both in the countdown last year) last year at countdown... very interesting...

In SoCal, the 10th (or maybe 9th) place person made it to the final round and then I beat him. Still- 10th written, 2nd countdown - doesn't make sense. Anyway, countdown is unofficial in both NoCal and SoCal.

In Washington, the 9th place individual beat 5 people in a row, but lost to the 4th place individual. The Countdown round is official. From then on, it was pretty boring.

[quote="yif man12"]The top people not winning countdown doesn't make it shameful. It was generally boring because nobody got on a roll and it was just plain slow from the get-go. Too many unanswered questions.

And yes, I was there. And I totally got killed in a countdown competition by my friend 40-11. We just did the questions before the kids answered. Very embarrassing.[/quote]

Well the nervous factor makes it kinda hard...and what made it boring was that no one was really any good (hence, 9th place won)

[quote="eryaman"][quote="yif man12"]Last year was cool when that really hot and sexy kid won countdown.[/quote]Hmmm...I have a hunch that you're the "hot and sexy kid".[/quote]

Hey, I can't deny that statement.  :D

[quote="yif man12"]Last year was cool when that really hot and sexy kid won countdown.[/quote]

Do you really think so? I didn't think much of her...

Owned.

NJ's Countdown was disappointing. My school's five competitors got 1st (me), 5th, 6th, 7th (an individual), and 20th. Anger. It really sucks for the 5th place guy, because he's in eighth grade, he started in 4th place, and he came in 5th last year, too. Same with the 6th place guy, who moved from 9th to 6th, and is also in eight grade.

Ouch. Just ouch. I don't think that was called for.

As a result of my immaturity...

[img]http://w3bdevil.com/forums/Owned-Connect4.jpg[/img]

It was pretty amusing (for awhile) to watch my teammates practicing the countdown. We didn't have an answer key, so my teacher would ask me the answer after about five seconds but more often than not, she had to call time after 45 seconds after nobody answered... Now that's just sad.

But three of us still made the top ten at regional, and I got second at state (but lost countdown to the freakishly fast 6th place person).

Treething- If you look carefully, you'll see red has already won.  :D

And who got...
[img]http://w3bdevil.com/forums/Owned-Connect4.jpg[/img]
...this time???  :D


In a triangle ADC, DC = 65 and Altitudes DB and CE are 33 and 63 respectively. Prove that AB and AE can not be integers.

Provide separate proof for:
In a Triangle A'D'C', D'C' = 65k, altitudes D'B' and C'E' are 33k and 63k respectively. Is there any k so that A'B' and A'E' are integers? If not, explain why. If so, find all such k.

Please help, this is really urgent.

The problem isn't so urgent anymore, but can anyone please provide a solution?

You can calculate $ AB$ and $ ED$ with Pythagore . The triangles $ ABD$ and $ AEC$ are similar and you can calculate $ m$ and $ n$ wich are not integer ( $ 8m$ and $ 8n$ are integer ) .


A square of side length 2 has semicircles drawn on each of its sides. An elastic band is then stretched tightly around the figure (square and 4 semicircles). What is the length of the elastic band (in this stretched position)?

[quote="Totally Zealous"][hide]Doesn't the elastic band form a square? With the vertices the points of tangency? Since the radius is 1, the elastic band has diagonals of 4. Solve to get $\boxed{8\sqrt{3}}$[/hide]
[img]http://content.answers.com/main/content/wp/en/thumb/5/53/222px-C-homer.png[/img][/quote]

What? draw a diagram and you will see

It doesn't form a square, SplashD's answer's right.

[quote="SplashD"][hide]Draw tangents to get 4 line segments and 4 quarter circles and get an answer of $\boxed{4\sqrt{2}+2\pi}$[/hide][/quote]

I'm not getting how you got 4 quarter circles? how do you know that's where the tangents hit the semicircles? and how do you know how long the tangents themselves are?

Well tangents are parallel to the sides of the square.

Can someone post a picture?  Here is my picture, but it's a work in progress.  I drew the tangents, but where do I draw the curves to satisfy the problem?

[quote="Inspired By Nature"][quote="SplashD"]Oh, sorry. My previous statement was false. Just think of a square with it's corners removed and replaced with quarter circles.
Sadly I don't have any good drawing tools.[/quote]
Be creative, i used paint and powerpoint. :P
See if you can solve the problem using this diagram.[/quote]

From the diagram:
[hide]
We have 4 quarter-circles that form one full circle with diameter 2.  Therefore circumference is $2\pi$.  Then we add up the line segments, which are equal to the line segments of the inscribed square, which has side lengths $\sqrt{2}$.  Four of these side lengths is $4\sqrt{2}$.  Adding in the $2\pi$ from earlier, we get $\boxed{4\sqrt{2}+2\pi}$
[/hide]

By the way, how did you use Powerpoint for that diagram?? :maybe:

[quote="mysmartmouth"]
By the way, how did you use Powerpoint for that diagram?? :maybe:[/quote]
I started with powerpoint. I made the square and the semi-cirlces on ppt, and then i copied it to paint, and finished with the rest of the lines. :D


Hi folks,

If you are interested in being part of the Georgia ARML Team, then this is the post for you. Georgia has a tradition of doing very well at the ARML competition and picking the team is one of the most difficult things that we (the coaches) have to do every year. If you don't know what ARML is, check here [url]http://www.arml.com/index.php[/url].

The 2 biggest tournament/tests that we look at are the AMC/AIME and the state varsity competition although we look at results from every tournament throughout the year. We will choose about 33-34 students for the Georgia A and B teams. Each ARML team is made up of 15 students, so this represents 2 full teams and a few alternates. In addition, we have the Georgia C team that is made up of students that are interested, but didn't get selected to the A/B team. (Yes, I know the naming rules are changing this year. I'm just using the old names.)

In the past USAMO qualifiers are near automatic selections, as are the top 15 or so at the state tournament. [b]This is not automatic![/b] Once those folks are taken into consideration, the ARML coaches start looking at the next group from the state tournament. We look for good results from other tournaments throughout the year that confirm your ability. We are also looking for multiple tournament results to let us know that you are interested enough to participate.

As you can tell, the state tournament plays a key role in the selection of the team. We make our final selections at the tournament and announce the team at the awards ceremony. If you aren't at the tournament, it's hard to qualify! I believe that all of the team invitations have been sent for the state tournament. If you aren't sure if you school has been invited, EMAIL me and I'll let you know. I don't maintain the actual list, but I've got it and know about how far down the list we went. Each school is limited to 4 competitors, but we invite additional individuals [b]for ARML consideration only[/b]. These students are typically not eligible for awards at the state tournament. We will be sending these individual invitations in the next couple of weeks. If you believe that you will not be coming as part of a school team and believe that we should consider you for the team, please send me an email. Please include what tournaments you have gone to, what you placing/score was, our AMC/AIME scores (if you took them) and any other information that I should know. (School, grade, etc.)

Thanks,
Tom
TFulton42@gmail.com <---- new email, but the old one still works.

Does anyone know where previous years' Varsity State Contests may be found? Thanks

Sorry for the late notice, but I will not be able to attend the practice tomorrow.

I will give John Calhoun my forms for submission.

Thanks,

Oliver

P.S. I think I will be able to attend all other practices.

Oliver, you should probably email one of the coaches; they're probably more likely to check their emails than to check this thread before tomorrow.

Hey folks,

Eric, if you get this, please contact me. If anyone else sees or talks to Eric, please have him contact me.

Thanks,
Tom


geometry
inequalities
trigonometry
area of a triangle
Heron\u0027s formula
triangle inequality
algebra

[quote="MCrawford"]A triangle with sides of integer lengths has two sides equal to 5 and 8 units respectively.  Find the largest and smallest possible areas of the triangle.[/quote]
[hide]
By the Triangle Side Inequality, the smallest the third side can be is 4 and the largest is 12.

Applying in Heron's Formula:

$\sqrt{(8.5)(8.5-5)(8.5-8)(8.5-4)}=\sqrt{(8.5)(3.5)(0.5)(4.5)}=\sqrt{66.9375}\approx8.18$

That's the smallest area.

$\sqrt{(12.5)(12.5-12)(12.5-8)(12.5-5)}=\sqrt{(12.5)(0.5)(4.5)(7.5)}=\sqrt{210.9375}\approx14.5237$

That's the largest area.[/hide]

I hope that's right.  :?

but just because the side is the longest, or the shortest does not make the area the smallest or largest,

[hide="Solution"]let the 3rd side be x, then using heron's formula, we get $s=\dfrac{13+x}{2}$

then substituting those in, and using the difference of squares, the area simplifies to

$A=\dfrac{1}{4}\sqrt{(169-x^2)(x^2-9)}$ and by the triangle inequality, 3>x>13

using a calculator to test values, we find that s=4 gives us the least area, and s=9 gives the greatest area[/hide]

[quote="riddler"][hide="Sol"]
Why doesnt this work,
Area of triangle=$\dfrac{1}{2}*5*8\sin{\theta}$
When can sin be maximum???? $sin90^{\circ}$
So, Largest area=$20$
When $sin0=0$,
Area is smallest so equal to $0$
[/hide][/quote]

What you did wrong is that you can't have an angle in a triangle with measure 0.  ;)

[quote="riddler"][hide="Sol"]
Why doesnt this work,
Area of triangle=$\dfrac{1}{2}*5*8\sin{\theta}$
When can sin be maximum???? $sin90^{\circ}$
So, Largest area=$20$
When $sin0=0$,
Area is smallest so equal to $0$
[/hide][/quote]

and for your maximum: alright but when the angle is 90° then the thirth side is of length $\sqrt{5^2+8^2}=\sqrt{89}$, this is not an integer...

[quote="Altheman"][hide="Solution"]let the 3rd side be x, then using heron's formula, we get $s=\dfrac{13+x}{2}$

then substituting those in, and using the difference of squares, the area simplifies to

$A=\dfrac{1}{4}\sqrt{(169-x^2)(x^2-9)}$ and by the triangle inequality, 3>x>13

using a calculator to test values, we find that s=4 gives us the least area, and s=9 gives the greatest area[/hide][/quote]
From $A=\dfrac{1}{4}\sqrt{(169-x^2)(x^2-9)}$, is it possible to find the value of $x$ that maximizes/minimizes the area without calculator? (Of course, we can do it by hand, but do we have to guess and check anyways?)


prove that :

$\mathbb{C\otimes}_{\mathbb{R}}\mathbb{C}\cong_{\mathbb{R}}\mathbb{R}^{4}$

tensor products commute with direct sums and $\mathbb{C}= \mathbb{R}\oplus \mathbb{R}$, so this is trivial.

please write your solution completely.
thanks.


inequalities
geometry
geometric transformation
trigonometry
inequalities unsolved

Prove or disprove:
If $x,y,z$ are the sides of a triangle then $x^3+y^3+z^3+2xyz\ge x^2(y+z)+y^2(x+z)+z^2(y+x)$

[quote="leepakhin"]Substituting $x=a+b$ and so on we will get $LHS-RHS=-8abc$. Therefore, your inequality is wrong. :([/quote]
the inequality is reversed. :)

I`m just translating
http://www.zaba.ru/cgi-bin/tasks.cgi?tour=russia.kvant.2001.var2

The translation is wrong. The problem should be:

[color=blue][b]Problem.[/b] Can the sides a, b, c of a triangle satisfy

$a^3+b^3+c^3+2abc\geq a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)$ ?[/color]

The answer is [b]no[/b]. In fact, for every triangle, we have

$a^3+b^3+c^3+2abc<a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)$,

because

$a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)-\left(a^3+b^3+c^3+2abc\right)$
$=\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)>0$

(since b + c - a > 0, c + a - b > 0, a + b - c > 0, because of b + c > a, c + a > b and a + b > c by the triangle inequality).

  Darij

I have another solution.
If $a,b,c$ can be a triangle.
then we have
\[ \cos A+\cos B+\cos C=1+4\sin \frac{A}2\sin \frac{B}2\sin \frac{C}2>1 \]
Which is eqaul to
\[ x^3+y^3+z^3+2xyz< x^2(y+z)+y^2(x+z)+z^2(y+x) \]


ratio
projective geometry
geometry proposed
geometry

Let hexagon ABCDEF circumscribed (O).
$ AF\cap DE \equal{} H, AB\cap CD \equal{} I, BC\cap ED \equal{} K, CD\cap EF \equal{} G, BC\cap EF \equal{} Q, ED\cap AB \equal{} P, KG\cap FP \equal{} N, QN\cap AB \equal{} M$.
Prove that: BF,IH,KG,EM concurrent.
[url=http://img12.imageshack.us/i/71330511.png/][img]http://img12.imageshack.us/img12/3623/71330511.png[/img][/url]
I don't know if the problem had.
If the problem had I sorry.

Let $ X,Y,Z,T,U,V$ be the points of tangency of $ (O)$ with $ AB,BC,CD,DE,EF,FA$ respectively. Denote by $ W_1 \equiv XY\cap UV, W_2 \equiv XZ \cap VT$ and $ W_3 \equiv UZ \cap VT$. By Pascal's theorem, we have that $ W_1,W_2,W_3$ are collinear. Thus, the polars of $ W_i$ with respect to $ (O)$, i.e. $ BF,IH,KG$ concurs at a point $ S$.

Now we prove that $ S,M,E$ are also collinear. Cross ratio works here. I will post this part later due to some business... :blush:


quadratics
summer program
PROMYS
number theory
modular arithmetic

The standard book for college classes on number theory seems to be [url=http://www.amazon.com/exec/obidos/tg/detail/-/0471625469/qid=1081695924/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/102-5438630-0929740?v=glance&s=books&n=507846]this one[/url]. Unfortunately, it's very expensive. If you want something cheaper but not as in-depth, you can try [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883856204/qid=1081695992/sr=8-3/ref=sr_8_xs_ap_i3_xgl14/102-5438630-0929740?v=glance&s=books&n=507846]this one[/url].

[quote="JS1527"]Three Questions:

(1) Does anyone know a website that explains this really well?

(2) If this were to be taught in a college class, what would this class be called?  What textbook would it use?

(3) Am I the only one who has never been taught this in high school?

Thanks![/quote]

(1) I hope other people jump in with Web sites. As I recall, Cut the Knot does a pretty good job explaining modular arithmetic, which I looked up online when I was living overseas three years ago and didn't have access to good math libraries.

(2) Modular arithmetic (congruences mod m) is usually taught in a number theory course. Some aspects of it might also be taught in an abstract algebra course. One popular book that I have checked out from my friendly local public library right now with a good explanation of modular arithmetic is The Book of Numbers by Guy and Conway.

(3) Most United States high school students learn little or nothing of modular arithmetic with the formal notation developed by Gauss. Kids in China in "key" schools like the Hua Loo-keng School in Beijing learn modular arithmetic in sixth grade.

Hope this helps! I hope other people will jump in with suggestions for studying the topic of modular arithmetic.

[quote="JS1527"](3) Am I the only one who has never been taught this in high school?[/quote]

I don't remember if modular arithmetic is part of the NYC curriculum, but it's a standard part of any competition training program/math team curriculum.

Thanks for all the advice!

I looked at my library's online catalog and they have the book that ComplexZeta suggested, but its unable to be requested (arghh).  So I ordered the book that tokenadult suggested and another one, The Theory of Numbers by Henry Neal McCoy. <--- dont know anything about this one, but it seemed like the only one that was a
"textbook" explanation of number theory.  If anyone has website suggestions that would be very helpful, but thanks for all the info!

We host a introductory number theory course here at AoPS.  Nearly half the course is devoted to modular arithmetic.  We will eventually host more advanced number theory courses that will cover theorems related to modular arithmetic such as Fermat's little theorem, Euler's theorem, and Wilson's theorem as well as quadratic reciprocity (which is just plain cool) and number theoretic functions.

[quote="ComplexZeta"]Actually, Idiot_Without_A_Village was (is?) putting together a [url=http://www.xanga.com/home.aspx?user=learn_number_theory]website[/url] on number theory, so you could try checking that out.[/quote]

He gave up after about two entries.

I recommend Hardy and Wright's <u>Introduction to the Theory of Numbers</u>.

Its very comprehensive, but so packed w/ detail that it can be hard to read at times. (It's what they recommend you buy at PROMYS)

The problem with Hardy and Wright is that it doesn't have any exercises. The only problems are unsolved.


trigonometry
inequalities unsolved
inequalities

$ \frac{\sin A}{1\plus{}\sin\frac{A}{2}}\equal{}\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{1\plus{}\sin\frac{A}{2}}\equal{}\frac{2\sin\frac{A}{2}\sqrt{1\minus{}\sin^{2}\frac{A}{2}}}{1\plus{}\sin\frac{A}{2}}$.
Let $ f: (0,1)\to\mathbf{R},f(x)\equal{}\frac{2x\sqrt{1\minus{}x^{2}}}{1\plus{}x}$
We have $ f(x)\geq\frac{1}{\sqrt{3}},\forall x\in(0,1)$
Let $ x\equal{}\sin\frac{A}{2},y\equal{}\sin\frac{B}{2},z\equal{}\sin\frac{C}{2}$.
Then $ f(x)\plus{}f(y)\plus{}f(z)\geq 3\cdot\frac{1}{\sqrt{3}}\equal{}\sqrt{3}$

[quote="nmtvn"]Consider the triangle ABC, show that: $ \frac{sinA}{1\plus{}sin\frac{A}{2}}\plus{}\frac{sinB}{1\plus{}sin\frac{B}{2}}\plus{}\frac{sinC}{1\plus{}sin\frac{C}{2}}\geq\sqrt{3}$[/quote]
I think, it's wrong: try $ A\equal{}2^{\circ}$ and $ B\equal{}C\equal{}89^{\circ}.$ :wink:


heLp pLease.....

pm me...

0r repLy here pLease..

i need heLp with the trig0n0metric tabLe..

are y0u famiLiar with that??..

[quote="miyomiyo"]Yes, but what in specific about the trigonometric table?[/quote]

trig table? you mean memorizing the common and simple trig functions of like sin and cos?

[quote="Glen Mann"]theres a trick for remembering the easy degree values of sin and cos, 0, 30, 45, 60, 90.

in that order, for cos we have

$\frac{\sqrt{4}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{0}}{2}$

you can see a pattern emerge, 4 3 2 1 0. For sin, the pattern is simply the otehr way around, 0 1 2 3 4.[/quote]

but that of course is assuming the trig is in degrees...in radians then id have to convert  :weightlift:

but still thats a good trick  :D


whenever I try to see problems, it says that I need to "input valid student id at position 1". does that mean anything, or is it just a glitch?

Try making a private message with the admin., see what they can do.
Or look in previous blogs, there might be some info.  :rotfl:


I did not want to use the SOS method in this case I do not choose any other
from the SOS method appears, I always want to find a method of thinking to solve  problems can only use the SOS
I look forward to discussing all of this thinking
We have asumme: $ xy \plus{} yz \plus{} zx \equal{} 1$ and $ x\geq y\geq z$.
We have:$ \frac {x \plus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{x \plus{} y} \plus{} \frac {y \plus{} z}{z \plus{} x} \plus{} \frac {z \plus{} x}{y \plus{} z} \plus{} \frac {z \plus{} x}{x \plus{} y} \plus{} \frac {x \plus{} y}{z \plus{} x}\geq\ 5 \plus{} \frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{xy \plus{} yz \plus{} zx}$
$ \Leftrightarrow \sum_{cyc}\frac {(x \minus{} z)^2}{y^2 \plus{} 1}\geq \frac {\sum (x \minus{} y)^2}{2}$
$ \Leftrightarrow \sum_{cyc}\frac {1 \minus{} x^2}{1 \plus{} x^2}(y \minus{} z)^2\geq 0$ $ (1)$
Let :$ S_a \equal{} \frac {1 \minus{} x^2}{1 \plus{} x^2}$,similary $ S_b,S_c$
Becau:$ 1\geq xy\geq y^2$ So $ y,z\leq 1$.So $ S_b,S_c\geq 0$.
Because: $ a^2S_b \plus{} b^2S_a\geq 0$(easy prove)
So $ (1)$ is true.

[quote="trejos"]Let $ x$, $ y$ and $ z$ be non negative reals, such that there are not two simultaneously equal to $ 0$. Show that

$ \frac {x \plus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{x \plus{} y} \plus{} \frac {y \plus{} z}{z \plus{} x} \plus{} \frac {z \plus{} x}{y \plus{} z} \plus{} \frac {z \plus{} x}{x \plus{} y} \plus{} \frac {x \plus{} y}{z \plus{} x}\geq\ 5 \plus{} \frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{xy \plus{} yz \plus{} zx}$
 and determine the equality cases.[/quote]
It's only: $ \sum\ \frac{x}{y\plus{}z} \geq\ \frac{(x\plus{}y\plus{}z)^2}{2(x\plus{}yz\plus{}zx)}$ (obviously trues)  :maybe:

[quote="trejos"]Let $ x$, $ y$ and $ z$ be non negative reals, such that there are not two simultaneously equal to $ 0$. Show that

$ \frac {x \plus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{x \plus{} y} \plus{} \frac {y \plus{} z}{z \plus{} x} \plus{} \frac {z \plus{} x}{y \plus{} z} \plus{} \frac {z \plus{} x}{x \plus{} y} \plus{} \frac {x \plus{} y}{z \plus{} x}\geq\ 5 \plus{} \frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{xy \plus{} yz \plus{} zx}$
 and determine the equality cases.[/quote]
Yes, SOS is probably overkill for it.
\[ \sum \frac {2x \plus{} y \plus{} z}{y \plus{} z} \geq 5 \plus{} \frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{xy \plus{} yz \plus{} zx} \iff 2 \sum \frac {x}{y \plus{} z} \geq \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx}
\]
But from Cauchy,
\[ \sum \frac {x}{y \plus{} z} \equal{} \sum \frac {x^2}{xy \plus{} zx} \geq \frac {(x \plus{} y \plus{} z)^2}{2(xy \plus{} yz \plus{} zx)}
\]
EDIT: I see that [b]nguoivn[/b] has noticed it before. :)