Sometimes I prefer to type my posts in some text editor first and then post them using "copy and paste". Unfortunately, the text editor I use automatically adds a line break whenever I move to a new line and I have to remove those line breaks manually in the posting window. Is there any way to do it automatically? Besides, sometimes, even when using the posting window, it is a bit more convenient to have the typed text structured (especially it concernes long formulas that tend to move from line to line in the least expected places, which makes them hard to type without mistakes).

Personally, I find the solution used in $\LaTeX$ the most reasonable and satisfactory: solitary line breaks are ignored and if you want to start a new paragraph, just skip a line. Can we implement the same style?

Unfortunately the answer is no and no.

No automatic script for removing line breaks will be done, because I wouldn't be able to put line breaks where I want to then. Just use another editor (like Notepad?) which doesn't add a line break instead of just visually wrapping the text.

And no for the $\LaTeX$ syntax. Again that would really really be confusing for most people writing plain text stuff, and for the parser itself.

[quote="Valentin Vornicu"]
No automatic script for removing line breaks will be done, because I wouldn't be able to put line breaks where I want to then. [/quote]
Hmm..., isn't it possible to do it as a formatting option like it is done in some text editors. I mean just add a button called "format paragraph" or something like this that will remove line breaks from a selected portion of the text (the same way as the {hide} button automatically puts the hide tags around the selected text). This will allow to keep line breaks where you want them and remove them where you don't. Anyway, I realize that you have much more important things to attend to, so I'll see whether I can find another text editor as you suggested. Thanks for your time, Valentin!

What editor are you using now? Maybe it is just not set up correctly ... ?  :?


$a_{i}>0$
$b_{i}>0$

$\frac{a^2_1}{b_1}+\frac{a^2_2}{b_2}+\dots+\frac{a^2_n}{b_n}\geq \frac{(a_1+a_2+\dots+a_n)^2}{b_1+b_2+\dots+b_n}$

I've proved this inequality, but I don't know when it's equal.

You meant Cauchy-Schwarz ineq and the equality holds for $\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}$

Yes, but I would like to prove it using quite simple theorems, is it possible to do it without generalization?

The only prove is :

Use Cauchy-Schwarz ineq for $(\frac{a_1}{\sqrt b_1},...,\frac{a_n}{\sqrt b_n}$ and

$(\sqrt b_1,.....,\sqrt b_n)$ that's all

You can just work out and complete with applying $\frac{n(n+1)}{2}$ times $\frac{a^2_i}{b_i}b_j+\frac{a^2_j}{b_j}b_i \ge 2a_ia_j$...

[quote="Arne"]Of course you can prove it without using Cauchy, Silouan![/quote]Actually this [b]is[/b] Cauchy :D

No, Cauchy is now dead, Valentin. This is his inequality. Poor Cauchy, I don't want to imagine his face when seeing the millions of inequalities created after this idea.  ;)

Yes, like Valentin said this is Cauchy's inequality. I remember that some two teachers in article published in a magazine, presented this inequality under their name on it.

 Edit: You are right Harazi. ;)  A good mathematician won't agree with his name under this kind of inequality. :)

As if this was a great mathematical result or something... to be angry for such nothing is not a good sign...

Oh well guys, I don't think it matters.

Please stop discussing about something which is not worth a serious discussion :)

And, Harazi, I'm VERY glad to see that you are posting here again.

Ooopps, sorry, I did not see it was the "forbidden" forum. Sorry, things will come to normal from now on.

Do you mean that you will start posting here again :) or does it mean that you still won't be posting in the inequalities forum? :(


geometry
rectangle
cyclic quadrilateral
linear algebra

How to find Cyclic quadrilateral arc lengths given the angles?
Suppose given a cyclic quadrilateral ABCD (in order around the circle) inscribed in a circle, and each of its four angles.
The angles will be 1/2 of the opposite two arcs .
for example, Angle A is = to Arc BCD, Angle B is = to Arc CDA
Is it possible to find each of the individual arcs AB BC CD and AD?
Note, this cannot be done using linear algebra or matrices, since one of the four equation is repeated.

the arc-lengths are obviously not determined by the angles: e.g., consider a rectangle and a square inscribed in the circle.

[quote="pluristiq"]the arc-lengths are obviously not determined by the angles[/quote]

[url=http://en.wikipedia.org/wiki/Inscribed_angle]Oh yes they are...[/url]

Right angles in a rectangle or square inscribed in a circle both subtend a $ 180^\circ$ arc.

no, he's asking about the arcs the four vertices partition the circle into...


integration
real analysis
real analysis solved

f:R--R classe C^, 2 \pi periodic such that  \int _[0;2 \pi ]f(t)dt=0
and for any t \in [0;2 \pi ], |f(t)| \geq |f"(t)|
Find f

Let C(n) be the Fourier coefficients of f, with  C(0)= 0. Then the coefficients of f'' are -C(n)*n2 , and we can write :

 \int(0=<t=<2 \pi){f2(t) .dt}=Sum(n=-\infty,...,+\infty){|C(n)|2} >= \int(0=<t=<2 \pi){f"2(t) .dt}=Sum(n=-\infty,...,+\infty){|C(n)|2*n4}}

So we have to have C(n)=0 for |n|>1 , therefore f(t) = A*cos(t) + B*sin(t)
(.. can I have my place at " l'X ".. :D )

Good job  :) Diogene, I 'll move to solved problems.
 About l'X admission
http://www.polytechnique.edu/page.php?MID=21

That's an elite school, I got no chance on getting a place in there, kind of ashamed of myself  :blush:


Triangle ABC is inscribed in a circle of diameter 12 meters.  The angles of ABC are A=45, B=60, C=75.  What are the exact lengths of the arcs BC, AC, and AB?

I know what the answers are but I do not know how to get them.

BC=3,  AC=4, and AB=5

Please Help :rotfl: [/code]

When angle at center is $360^\circ$ length of arc is $2\pi r$
When angle at centet is $x^\circ$ leghth of arc is $\frac{\pi rx^\circ}{180^\circ}$

$m(\widehat{AOB})=\phi\Longrightarrow \boxed{\ L(\phi )\equiv\mathrm{arc}AB=R\cdot \phi\ }$ and $\boxed{\ S(\phi )=\frac{R\cdot L(\phi )}{2}=\frac{R^{2}\cdot\phi}{2}\ }$ (in radians).

[b][u]Exercises :[/u][/b] $\{\begin{array}{c}\boxed{\ x\in (0,\frac{\pi}{2})\Longrightarrow\sin x<x<\tan x\ }\ .\\\\ x\in (0,\frac{\pi}{2})\Longrightarrow \cos x<\frac{\pi}{2}-x<\cot x\ \ (\ x: =\frac{\pi}{2}-x\ )\ .\\\\ x\in (0,1)\Longrightarrow x<\arcsin x<\frac{x}{\sqrt{1-x^{2}}}\ \ (\ x: =\arcsin x\ )\ .\\\\ x\in (0,1)\Longrightarrow \sqrt{1-x^{2}}<\arccos x<\frac{\sqrt{1-x^{2}}}{x}\ \ (\ x: =\arccos x\ )\ .\\\\ x\in (0,\infty )\Longrightarrow \frac{x}{\sqrt{1+x^{2}}}<\arctan x<x\ \ (\ x: =\arctan x\ )\ .\\\\ \mathrm{a.s.o.}\end{array}$


Alright, these types of problems are really getting on my nerves. An example of one -

How many combinations of quarters, nickels, dimes, and/or pennies will make 35 cents?

Is there a trick to these? Or do you [i]have[/i] to list them out and count the number of results?  :o

Lol that was on Warm-Up 4.

Anyways, for cents $ 10\sim 49$, the formula is:

$ 3 \plus{} \frac{\lfloor \frac{c}{5} \rfloor \times \lfloor (\frac{c}{5} \minus{} 1) \rfloor}{2}$

For the other cents, you just have to list them out.  :P

UM..If you would have looked in this forum, I posted how to do those types of problems just a few days ago.  Look for the title "How many ways" by 007math

or go here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=234517


1.Find the greatest value of the expression
$ P\equal{}3(xy\plus{}yz\plus{}zx)\minus{}xyz$
where $ x,y,z$ are positive real numbers such that $ x^3\plus{}y^3\plus{}z^3\equal{}3$
2.Find the maximum value of the expression
$ \frac{x}{1\plus{}x^2}\plus{}\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}$
where $ x,y,z$ are real numbers satisfying the condition $ x\plus{}y\plus{}z\equal{}1$

[quote="apollo"]

2.Find the maximum value of the expression
$ \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}$
where $ x,y,z$ are real numbers satisfying the condition $ x \plus{} y \plus{} z \equal{} 1$[/quote]

 in fact we have $ \frac{x}{1\plus{}x^2} \leq\frac{18x}{25}\plus{}\frac{3}{50}$

[quote="apollo"]
2.Find the maximum value of the expression
$ P\equal{}\frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}$
where $ x,y,z$ are real numbers satisfying the condition $ x \plus{} y \plus{} z \equal{} 1$[/quote]

Let $ a\equal{}3x,b\equal{}3y,c\equal{}3z$ => $ a\plus{}b\plus{}c\equal{}3$
We need find the maximum value of  $ \sum\frac a{a^2\plus{}9}$

By AM-GM: $ a^2\plus{}1\ge2a$. We prove $ \sum\frac a{a\plus{}4} \le \frac3{5}$

<=> $ 3abc\plus{}7(ab\plus{}bc\plus{}ca)\le 24$ which true by AM-GM.

So $ max P\equal{} \frac9{10}$  :)

[quote="M.A"][quote="apollo"]

2.Find the maximum value of the expression
$ \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}$
where $ x,y,z$ are real numbers satisfying the condition $ x \plus{} y \plus{} z \equal{} 1$[/quote]

 in fact we have $ \frac {x}{1 \plus{} x^2} \leq\frac {18x}{25} \plus{} \frac {3}{50}$[/quote]

Maybe my solution is same as yours, M.A :lol:

Keyword is "tangent line".

[quote="dduclam"][quote="apollo"]
2.Find the maximum value of the expression
$ P \equal{} \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}$
where $ x,y,z$ are real numbers satisfying the condition $ x \plus{} y \plus{} z \equal{} 1$[/quote]

Let $ a \equal{} 3x,b \equal{} 3y,c \equal{} 3z$ => $ a \plus{} b \plus{} c \equal{} 3$
We need find the maximum value of  $ \sum\frac a{a^2 \plus{} 9}$

By AM-GM: $ a^2 \plus{} 1\ge2a$. We prove $ \sum\frac a{a \plus{} 4} \le \frac3{5}$

<=> $ 3abc \plus{} 7(ab \plus{} bc \plus{} ca)\le 24$ which true by AM-GM.

So $ max P \equal{} \frac9{10}$  :)[/quote]
Your solution have some mistakes, Lam. Notice that the inequality is for [b]real[/b] numbers. :)

You are right,Can. I was mistaken (i take it for $ x,y,z$ are positive real numbers  :( )

[quote="apollo"]2.Find the maximum value of the expression
\[ \frac {a}{1 \plus{} a^2} \plus{} \frac {b}{1 \plus{} b^2} \plus{} \frac {c}{1 \plus{}c^2}\]
where $ a,b,c$ are real numbers satisfying the condition $ a\plus{}b\plus{}c \equal{} 1$[/quote]
Now, let me post my solution :)
Case 1. If $ a,b,c \ge \minus{}\frac{3}{4}$, then using the fact
\[ \frac{x}{x^2\plus{}1} \le \frac{3}{10}\plus{}\frac{6}{25}(3x\minus{}1) \quad \forall x \ge \minus{}\frac{3}{4}\]
we can get the result. :)
Case 2. If $ a \le \minus{}\frac{3}{4}$, then by the AM-GM Inequality, we get
\[ \frac{c}{c^2\plus{}1} \le \frac{1}{2}\]
Hence, the inequality is valid if
\[ \frac{b}{b^2\plus{}1} \plus{}\frac{1}{2} \le \frac{9}{10}\]
\[ \Leftrightarrow 2b^2\minus{}5b\plus{}2 \ge 0\]
\[ \Leftrightarrow b \le \frac{1}{2} \vee b \ge 2\]
It suffices to consider the case $ \frac{1}{2} \le b \le 2$. Similarly, it suffices for us to consider the case $ \frac{1}{2} \le b,c \le 2$, from this, we get
\[ a\equal{}1\minus{}b\minus{}c \ge \minus{}3\]
Thus
\[ \frac{a}{a^2\plus{}1} \le \minus{}\frac{3}{10}\]
Hence
\[ \frac {a}{1 \plus{} a^2} \plus{} \frac {b}{1 \plus{} b^2} \plus{} \frac {c}{1 \plus{}c^2} \le \minus{}\frac{3}{10} \plus{}\frac{b}{b^2\plus{}1}\plus{}\frac{c}{c^2\plus{}1} \le \minus{}\frac{3}{10}\plus{}\frac{1}{2}\plus{}\frac{1}{2}\equal{}\frac{7}{10}<\frac{9}{10}\]
Moreover, let $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$, we get
\[ \frac {a}{1 \plus{} a^2} \plus{} \frac {b}{1 \plus{} b^2} \plus{} \frac {c}{1 \plus{}c^2}\equal{}\frac{9}{10}\]
thus
\[ \max P\equal{}\frac{9}{10}\]
we have done. :)

My best regard.

[quote="apollo"]1.Find the greatest value of the expression
$ P \equal{} 3(xy \plus{} yz \plus{} zx) \minus{} xyz$
where $ x,y,z$ are positive real numbers such that $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$
[/quote]
Notice that there exists $ x,y$ such that
\[ z(x\minus{}1)(y\minus{}1) \ge 0\]
\[ \Rightarrow \minus{}xyz \le z\minus{}z(x\plus{}y)\]
\[ \Rightarrow P \le 3(xy\plus{}yz\plus{}zx)\plus{}z\minus{}z(x\plus{}y)\equal{} z\plus{}xy\plus{}2(xy\plus{}yz\plus{}zx)\]
By the AM-GM Inequality, we get
\[ z \le \frac{z^3\plus{}2}{3}\]
\[ xy \le \frac{x^3\plus{}y^3\plus{}1}{3}\]
\[ \Rightarrow z\plus{}xy \le \frac{x^3\plus{}y^3\plus{}z^3\plus{}3}{3}\equal{}2\]
Also
\[ xy\plus{}yz\plus{}zx \le \frac{x^3\plus{}y^3\plus{}1}{3}\plus{}\frac{y^3\plus{}z^3\plus{}1}{3}\plus{}\frac{z^3\plus{}x^3\plus{}1}{3} \equal{}3\]
Thus
\[ P \le z\plus{}xy\plus{}2(xy\plus{}yz\plus{}zx) \le 2\plus{}2\cdot 3 \equal{}8\]
Equality holds if and only if $ x\equal{}y\equal{}z\equal{}1.$
We have done. :)

[quote="apollo"]1.Find the greatest value of the expression
$ P \equal{} 3(xy \plus{} yz \plus{} zx) \minus{} xyz$
where $ x,y,z$ are positive real numbers such that $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$[/quote]
By hypothesis $ x^3\plus{}y^3\plus{}z^3\equal{}3$, so $ x\plus{}y\plus{}z\leq 3$ (by AM-GM)
$ P\minus{}8\equal{}3x(y\plus{}z)\plus{}yz(3\minus{}x)\minus{}8\leq 3x(3\minus{}x)\plus{}\frac{1}{4}(3\minus{}z)^3\minus{}8\equal{}\minus{}(x\minus{}1)^2(x\plus{}5)\leq 0$ :)


limit
function
calculus
calculus computations

1) How do you go about proving the properties of limits such as:

$ \lim_{x\rightarrow a}(f(x)\plus{}g(x)) \equal{}\lim_{x\rightarrow a}f(x)\plus{}\lim_{x\rightarrow a}g(x)$

or

$ \lim_{x\rightarrow a}c\cdot (f(x) ) \equal{} c\lim_{x\rightarrow a}f(x)$

Using only the definition of a limit?

2) Is it legitimate to prove that a form is indeterminate by showing a sort of contradiction? Ex/

$ 0a \equal{} 0\Longrightarrow\frac{0}{0}\equal{} a$

But an exact value cannot equal an infinite number of values at the same time. Thus $ \frac{0}{0}$ is indeterminate.

#1 is the classic very first exercise in a course that does epsilon-delta proofs. I'll let someone else write it up. It may be your first exposure to the idea of using $ \frac{\epsilon}{2}$ (or something like that) in a proof.

As for #2: your argument is irrelevant. The notion of a $ \frac{0}{0}$ indeterminate form is [b]not[/b] about literally zero divided by literally zero; it's about what happens when you divide a very small number by a very small number. That anything can happen is best shown by a selection of simple examples: say (as $ x\to0$) $ \frac{x^{2}}{x},\frac{x}{x^{2}},$ and $ \frac{x}{x}.$

Actually my real problem is understanding the epsilon-delta definition. From wolfram, we have:

"A function $ f(x)$ is said to have a limit $ \lim_{x\rightarrow a}f(x) \equal{} b$ if, for all $ \epsilon > 0$ , there exists a $ \delta > 0$  such that $ |f(x)\minus{}b| <\epsilon$ whenever $ 0 < |x\minus{}a| <\delta$ ."

So if you make x closer and closer to a, |x-a| will get closer and closer to $ \delta$ and thus |f(x) -b| will become closer and closer to 0??? I don't think that's right, I'm so confused  :huh:.

Edit: Perhaps it is necessary to show that $ \delta \equal{} n\cdot\epsilon$ since $ x\rightarrow a$ at a different rate than $ f(x)\rightarrow b$. Where did that idea come from? Since $ x$ is related to $ f(x)$ (one affects the other), we have the same sort of relationship with epsilon and delta. Hopefully I'm on the right track.

Another version of the same definition, using word instead of symbols:
We can make $ f(x)$ arbitrarily close to $ b$ by choosing $ x$ sufficiently close to $ a$.

The $ \epsilon$ is how we quantify "arbitrarily close", and the $ \delta$ is how we quantify "sufficiently close". No matter how small $ \epsilon$ is, we need to be able to find a $ \delta$ for it- for specific functions, you will often be able to use a formula. Here, you'll have to use a $ \delta$ that comes from the limits you assumed (the RHS).
Also, don't forget the "whenever"; it has to work for every close enough $ x$, not just some of them.


calculus
integration
trigonometry
calculus computations

what is a good substitution to calculate $ \displaystyle\int{\frac{1}{\sqrt{x+\sqrt{x}}}dx}$?

I will do it this way. Pull out $ \sqrt{x}$ and use substitution we change it to antiderivative of $ \int \frac{1}{\sqrt{x\plus{}1}} dx$ (with some constant factor), whose antiderivative is obvious.

But $ \sqrt {x}$ becomes $ x^{1/4}$ when pulled out. The substitution $ u \equal{} x^{1/4}$ turns the integral into something like $ \int\frac {x^2}{\sqrt {x^2 \plus{} 1}}dx$ (ignoring the constants). Trig substitution ensues.

A good start would be to let $ u\equal{}x^{1/4},$ written "backwards" as $ x\equal{}u^4.$ The integral would then become:

$ \int \frac{4u^3}{\sqrt{u^4\plus{}u^2}}\,du\equal{}\int\frac{4u^2}{\sqrt{u^2\plus{}1}}\,du.$

Perhaps that form will give you the idea for what do do next, such as a trigonometric or hyperbolic substitution.

[Edit: redundant with mlok's post.]

$ 1 \plus{} \sqrt{x} \equal{} u^2 \equal{}> x \equal{} (u^2\minus{}1)^2 \equal{}> dx \equal{} 4u(u^2\minus{}1)du$, and so

$ P \equal{} \int x^{\minus{}1/4}(1 \plus{} \sqrt{x})^{\minus{}1/2}\,dx \equal{} \int (u^2\minus{}1)^{\minus{}1/2} \cdot u^{\minus{}1} \cdot 4u(u^2\minus{}1)du$

$ \equal{}> P \equal{} 4 \int \sqrt{u^2\minus{}1}\,du$.


[quote="GameBot"]What is the sum of the two counting numbers whose squares differ by 25?[/quote]

$ a^2\minus{}b^2\equal{}25$
$ (a\minus{}b)(a\plus{}b)\equal{}25$
$ a\plus{}b\equal{}25, a\minus{}b\equal{}1$ (since b can't be 0)
a+b=25.

You can also think about Pythagorean triplet {5,12,13}


Let $ f(x)\equal{}\frac{x^3}{3}\minus{}x$, we have $ f'(x)\equal{}x^2\minus{}1\equal{}(x\plus{}1)(x\minus{}1)$.
$ \therefore f(\minus{}1)<b<f(1)\Longleftrightarrow\minus{} \frac{2}{3}<b<\frac{2}{3}$.

[quote="Meatman"]Find all parameters $ b$ such that the equality  $ \frac {x^3}3 \minus{} x \equal{} b$ has 3 roots[/quote]
Draw graph of function $ f(x)\equal{}\frac{x^3}{3}\minus{}x$
[img]http://i246.photobucket.com/albums/gg93/NguyenDungTN/graph.jpg[/img]
So equation has 3 root if and only if $ \frac{\minus{}2}{3}<b<\frac{2}{3}$


For each positive integer n, a(n) is the smallest positive integer such that (a(n))! is divisible by n. Find all n such that 3a(n) = 2n.

problem 421345 at: http://www.mathematik-olympiaden.de/Aufgaben/42/4/42134b/42134b.html

A try for this cute problem. Since I don't read german :( , i don't know if there is a solution in the above web site. Here is mine  :) (not sure at all it's correct btw).

I think n = 9 is the only solution.

n = p1^a1 * .. * pk^ak with pi prime

Lemma 1 :
   there exist pi, pi | a(n)

If not, it was useless to go until a(n) because a(n) doesn't give any new pj

Lemma 2 :
   a(n) = bk * pk with bk <= ak
or (exclusive)
   a(n) = a(n / pk^ak)

We have to get every needed powers of pk. Either we have to reach some multiple of pk to have every pk we need, either we have to go further to have the other powers of pj. In the last case, we don't have to care any more of the power of
pk and a(n) = a(n / pk^ak).
In the first case, a(n) = bk * pk and obviously bk <= ak

Lemma 3 :
   a(n) = bi * pi with bi <= ai

Apply lemma 2 recursively.

So the equation to solve is 3*bk*pk = 2 prod pi^ai

Now 3 is one of the pi.
If pk != 2 because 2^a > a.
Since 2*p^a > p*a for p >= 4, pk = 3, bk = 2 and the only solution is n = 9.

Correct ?

There are easy way.

It is clear a(n)=max{a(p1^a1),...,a(pk^ak)} and a(pj^aj)=bj*pj, bj \leq aj, j=1,...,k.

Thus lemma1, lemma2 and lemma3 are self-evident.

Dude, why did you wait for 5 days to give your self-evident and easy solution  :?

It is too simple problem which didn't attract my attention


calculus
integration
algebra open
algebra

prove that the application $ f: \mathbb{N} \to \mathbb{N}$ defined by
$ f(x)\equal{}$[$ 1\plus{}\frac{1}{2}\plus{}...\plus{}\frac{1}{n}$]
is surjective
 :arrow: [x] is the integral value of x. :arrow:

It is obviosly for any $ f(n)\equal{}[\sum_{i\equal{}1}^n a_i ]$ if $ 0<a_i\le 1$ and $ \sum_i a_i \equal{}\infty$.


[quote="perfect_radio"]Let $a,b \in \mathbb N^\ast = \left\{ 1,2,3,\ldots \right\}$, $a<b$, $a$ does not divide $b$. Solve the equation
\[ a \left\lfloor x \right\rfloor - b \left( x - \left\lfloor x \right\rfloor \right) = 0 .  \]

[i]Ioan Ucu Crişan[/i][/quote]
Since $a\lfloor x\rfloor\in\mathbb Z$, $b\{x\}\in\mathbb Z$, so $x=\lfloor x\rfloor+\frac yb$, $0\le y<b$, $y\in\mathbb Z$

So $a\lfloor x\rfloor=y$, so $a|y$, so the solutions are $x=t+\frac{at}b$ or $\boxed{x=\frac{(a+b)t}b}$ where $t\in\mathbb Z$ and $0\le t\le\left\lfloor\frac ba\right\rfloor$ ($a$ does not divide $b$)


Let $ q_{n}$ denote the greatest prime divisor of $ n$ and $ p_{n}$ denote the $ n$th prime number.  Show that $ \sum_{n=2}^\infty \frac{1}{nq_{n}}=\sum_{n=1}^\infty \frac{1}{p_{n}^{2}}\prod_{j=1}^{n}\bigg(1-\frac{1}{p_{j}}\bigg)^{-1}$.

Given the greatest prime divisor $ p_{m}$ of a positive integer, we know that the set of all these possible integers is simply $ A_{m}=\{p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{m}^{\alpha_{m}+1}|\alpha_{i}$ are all non-negative integers for all $ i\}$.

Hence we have

$ \sum_{n=2}^\infty \frac{1}{nq_{n}}=\sum_{k=1}^\infty \frac{1}{p_{k}}\sum_{a\in A_{k}}\frac{1}{a}$

$ =\sum_{k=1}^{\infty}\frac{1}{p_{k}}s_{1}s_{2}\cdots s_{k-1}(s_{k}-1)$

where $ s_{i}=1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\cdots = \frac{1}{1-\frac{1}{p_{i}}}=\left (1-\frac{1}{p_{i}}\right )^{-1}$.  Also, $ s_{k}-1=\frac{1}{p_{k}}\left (1-\frac{1}{p_{k}}\right )^{-1}$.  This means that the sum becomes

$ \sum_{k=1}^\infty\frac{1}{p_{k}}\prod_{i=1}^{k}\left (1-\frac{1}{p_{i}}\right )^{-1}\cdot\frac{1}{p_{k}}=\sum_{k=1}^\infty\frac{1}{p_{k}^{2}}\prod_{i=1}^{k}\left (1-\frac{1}{p_{i}}\right )^{-1}$

as desired.


What number comes next in this sequence:

8, 72, 46, 521, 612, ?

Yikes! Tough 1! The answer is 343, because the sequence is made up of all the cube numbers, backwards.

8-  8
27- 72
64- 46

ect.

PS: Thnx breakdancer for catching my mistake, just edited it out.

Lol, this is a pretty bad problem.  :rotfl:


function
inequalities
real analysis
real analysis unsolved

The argument in an auction theory
But, i'm not sure why it's true!

Take a risk averse von-Neumann Morgenstern utility function u(x);
such that u(0)=0, u'>0 and u"<0 (concavity is due to risk aversion).
The author argues that for all y>0 then u(y)/u'(y) >y.

It's clear that x>y ==> u(x)>u(y) from u'>0 and hence x>0 ==> u(x)>0
It's not clear to me...

For $t > 0$, $u'(t) > u'(t)+t u''(t)$.  Integrate that inequality from $t = 0$ to $t = y$ to get the desired result.

Thanks for the feedback. Suppose you have a group of n >= 4 risk-neutral bidders with no budgets constraints, each with independent randomly distributed valuations for an object up for bid with values in [0,w], each given by the same distribution function F. Assume common knowledge of these facts. What is a symmetric solution for the auction where the winner pays what the 4th highest bidder bid? Are there any special assumptions that must be made of F?


Find all functions $ f: [1;\infty)\rightarrow [1;\infty)$ which satisfy $ f(x)\leq 2x \plus{} 2$ and $ x f(x \plus{} 1) \equal{} f(x)^2 \minus{} 1$ for all $ x$.

Let $ f(x)\equal{}g(x)\plus{}x\plus{}1$, then $ \frac{g(x\plus{}1)}{g(x)}\equal{}\frac{2x\plus{}3}{x}>2$. Therefore if $ g(x)>0$ exist m, suth that $ g(x\plus{}m)>g(x)*2^m>2(x\plus{}m)\plus{}2$. If $ g(x)<0$ exist m, suth that $ g(x\plus{}m)<g(x)*2^m<\minus{}x\minus{}m\minus{}1\to f(x)<0$.
Therefore we get only case $ g(x)\equiv 0$ or $ f(x)\equiv x\plus{}1$.


geometry
perimeter
inequalities
geometry unsolved

Let $A$ be a shape in the plane and let $S(A)>0$ be its Area and $P(A)>0$ its perimeter.

Let $D(A)$ be the biggest distance between the points of $A$ .

prove that : \[\frac{P(A)}{\pi}\leq D(A) \leq \frac{P(A)}{2}\]

Actually, that's not necessarily true:  you can make a shape with a fractal pattern (The Koch Coastline comes to mind) that has an infinite perimeter but a finite diameter, which would make the inequality false.
Perhaps you meant a convex polygon/shape?  In that case, I'm pretty sure its true but can't think of how to prove it right now.  However, I believe it has something to do with [hide="hint"] the fact that if the denominator was 2, it would be a line segment, and if it was $\pi$, it would be a circle.[/hide]


Find all continuous functions $ f: \mathbb R \to \mathbb R$ such that $ f(x)\equal{}0$ if and only if $x\equal{}0$, and also
\[f(x)f(y)f(x\plus{}y)\equal{}f(xy)(f(x)\plus{}f(y)),\]
holds for all reals $x$ and $y$.

[quote="duythuc_lqd"]Find all f continuous on R.$ f: R\to R$ and $ f(x) = 0\Leftrightarrow x = 0$. And satisfy $ f(x)f(y)f(x + y) = f(xy)(f(x) + f(y))$[/quote]

Let $ P(x,y)$ be the assertion $ f(x)f(y)f(x+y)=f(xy)(f(x)+f(y))$

For $ x\neq 0$ : $ P(x,1)$ $ \implies$ $ f(1)f(x+1)=f(x)+f(1)$ and then, setting $ x\to 0$ and using continuity, $ f(1)=1$

Then $ f(x+1)=f(x)+1$ and so $ f(x+n)=f(x)+n$ and $ f(n)=n$ $ \forall x,\forall n\in\mathbb Z$

Then $ P(x,n)$ $ \implies$ $ nf(x)(f(x)+n)=f(nx)(f(x)+n)$ and so $ f(nx)=nf(x)$ $ \forall x\neq -n$ and so, with continuity, $ f(nx)=nf(x)$ $ \forall x$

So $ f(x)=x$ $ \forall x\in\mathbb Q$

So, with continuity, $ f(x)=x$ $ \forall x\in\mathbb R$


function
calculus
derivative
limit
real analysis
algebra unsolved
algebra

Let $ f(x): (a,b) \to R$ continuous such that:  $ lim_{h \to 0} \frac{f(x\plus{}h)\minus{}f(x\minus{}h)}{h} \equal{}0 \ \forall a<x<b$
Prove that $ f(x)\equal{}const$

you say that $ x_h$ is a point where $ g$ reaches it's minimum, so $ x_h$ is fixed..how can you observe a limit of a fixed number?
i mean isn't $ x_0\equal{}x_h$?

$ x_h$ is a function of $ h$. He was ambiguous when he said $ \lim x_h$, but what he meant was the limit as $ h$ goes to $ 0$ of $ x_h$. Since $ g(x)$ is not defined for $ h \equal{} 0$, he had to manually define $ x_0$ using the limit. That was the purpose of that statement.

[quote=Baronov]We shall prove that f is constant on arbitrary closed sub-interval [c,d] of (a,b). If this is proven than the claim follows.
Fix h>0. For this h let $ g(x) \equal{} \frac{f(x\plus{}h)\minus{}f(x\minus{}h)}{h}$. Since f is continuous, g is uniformly continuous on [c,d]. So there exhist m = min(g(x)). Let $ x_{h}$ is the point which achieves this minimum. Let $ x_{0}\equal{}lim x_{h}$ we have $ g(x_{0})$ tends to 0, when h tends to 0.
From here we get that for each $ \epsilon > 0$ there exhist $ h_{0}$ such that for each $ h<h_{0}$ we have $ f(x) \minus{} f(y) < h\epsilon$ for each x,y. From here f is differentiable and its derivative is 0 and so f is constant.[/quote]

I don't see any sense as it is written. Maybe, instead of $m = \min(g(x))$, it should be $m = \max|(g(x))|$, but even so, how do you get $ |f(x) - f(y)| < h\varepsilon$? You should prove some uniform convergence or something! I do not say that this approach is not possible, but the main difficulty is still ahead.

Let us fix some $x_0\in (a,b)$ and $\delta>0$ (it can be arbitrary small). Since $f(x)$ is continuous at $x_0$, there exists $\varepsilon>0$ such that $|f(x)-f(x_0)|<\delta$ whenever $|x-x_0|\le \varepsilon$. We shall prove that
$$|f(x)-f(x_0)|<\delta|x-x_0| +\delta \qquad(1)$$
for any $x\in(a,b)$. This would imply $f(x)=f(x_0),\forall x\in(a,b)$. It's enough to prove it only for $x\in(x_0,b)$, the other case is similar. Note that $(1)$ holds for $x\in[x_0,x_0+\varepsilon]$. Let
$$F:=\{y\in [x_0,b) : \text{ (1) holds for each } x\in [x_0,y]\}$$
So $[x_0,x_0+\varepsilon] \subset F.$ Let $y:=\sup F$. We will prove $y=b$. Suppose for the sake of contradiction that $y<b$. Take $\varepsilon_1$ small enough such that for each $h,|h|\le \varepsilon_1$ it holds
$$\displaystyle \frac{|f(y+h)-f(y-h)|}{2h}\le \delta.$$
Thus,
\begin{align*}
|f(y+h)-f(x_0)|&\le |f(y-h)-f(x_0)|+|f(y+h)-f(x-h)|\\
&\le \delta|y-h-x_0|+\delta +2h \delta \\
&= \delta|y+h-x_0|+\delta.
\end{align*}
This means that $\sup F\ge y+\varepsilon_1$, contradiction.

[b]Remark.[/b] We only use that $f$ is continuous at $x_0$, that is, it is enough to require continuity only at one point.


In a right triangle, parallel lines to each side are drawn through the incentre of the triangle. These lines divide each side of the triangle into three parts. Prove that the length of the middle part of the hypotenuse equals the sum of the lengths of the middle parts of the legs of the triangle.

In a any triangle $ABC$ , a inside point $P$ has the mentioned property
if and only if $\boxed {\ a\cdot [PBC]=b\cdot [PCA]+c\cdot [PAB]\ }\ \ (1)$.

In the our particular case: $a=90^{\circ}$, $P: =I\Longrightarrow$
$[PBC]=\frac 12 ar$, $[PCA]=\frac 12 br$, $[PAB]=\frac 12 cr$
and the relation $(1)$ is true: $a^2=b^2+c^2.$

Virgil Nicula, I appreciate you being a teacher, writer, general know-all et cetera, but can you make yourself hide your solutions, as a gesture of a consideration to those who want to solve the problems themselves? Thank you.

Here's a solution with angle chasing.  :)

Since $HE || AC$ then $\measuredangle ACB = \measuredangle HEB$. Since $BC || FI$ then $\measuredangle AFI = \measuredangle ACB$ $\Longrightarrow$ $\measuredangle HEB = \measuredangle ACB$.
Also $\measuredangle FGO = \measuredangle OKE = 90^{\circ}$. So $\Delta FGO \sim \Delta EKO$.
$CGOK$ is cyclic so $\measuredangle FOG + \measuredangle FOE + \measuredangle EOK = 180^{\circ} - \measuredangle ACB$.
$90^{\circ} - \measuredangle ACB + \measuredangle FOE + 90^{\circ} - \measuredangle ACB = 180^{\circ} - \measuredangle ACB$ $\Longrightarrow$ $\measuredangle FOE = \measuredangle ACB$.
Thus $FO = EO$ and for ASA $\Delta FGO \equiv \Delta EOK$ so $FG = EK$.
Do the same for $BKOH$. So we get $HI = KD$ and from this the statement $FG + HI = ED$.

Could some one tell me, please, if it's possible to convert a .fig(from Cabrì) into an allowed extension?  :)


Guys, try this:

a/b + b/c + c/a >= [(n-1)(a+b+c)+a / (n-1)(a+b+c)+b] + [(n-1)(a+b+c)+ b / (n-1)(a+b+c)+c] +
[(n-1)(a+b+c)+c / (n-1)(a+b+c) + a]

for :
(i) n = (2^(2k)+2)/3 for some integer k
(ii) for all n>=2, but n is still an integer.

note ive only solved (i).

Why do you use n-1 instead of n or x etc.
Anyway, for real n>1  as defined by you, your inequality is true by rearrangement :)

Bomb

You can generalise this inequality to any number of variables,

Bomb


solve equation: $ (2x^3\plus{}x\minus{}3)^3\equal{}3\minus{}x^3$

[hide="Partial Solution"]Subtract $ x^3$ to obtain:
$ ((2x^3 \minus{} 3) \plus{} x)^3 \minus{} x^3 \equal{} 3 \minus{} 2x^3$
$ y: \equal{} 2x^3 \minus{} 3 \implies (y \plus{} x)^3 \minus{} x^3 \equal{} \minus{} y$
$ y(y^2 \plus{} 3xy \plus{} 3x^2) \equal{} \minus{} y$

$ y \equal{} 0$ or

$ y^2 \plus{} 3xy \plus{} 3x^2 \equal{} \minus{} 1$

The first gives $ 2x^3 \minus{} 3 \equal{} 0 \implies \boxed{ x \equal{} \sqrt [3]{\frac {3}{2}}, \; \omega \sqrt [3]{\frac {3}{2}} , \; \omega ^2 \sqrt [3]{\frac {3}{2}}}$, where $ \omega \neq 1, \; \omega^3 \equal{} 1$

The second simplifies to:
$ 4x^6 \plus{} 6x^4 \minus{} 12x^3 \plus{} 3x^2 \minus{} 9x \plus{} 10 \equal{} 0$, which does not factor over the integers... I don't know how to solve this.
[/hide]


Prove that for all positive reals a, b, c and d ;

(a + b + c + d) \leq 4(a + b + c + d) + 24 \sum(abc)

where the summation is cyclic.

Hello, Arne!
Your problem is a holiday for my soul.

Let f(a,b,c,d) = 4(a^3+b^3+c^3+d^3) + 24\sum abc - (a+b+c+d)^3.

Suppose WLOG a\geq b \geq c \geq d.
Consider f(a,b,c,d) - f(a,b,c+d,0) = 24ab(c+d) - 12cd(c+d) \geq 0;
So if f(a,b,c,0) \geq 0 for all positive a,b,c then f(a,b,c,d) \geq 0 for all positive a,b,c,d.

Suppose WLOG a\geq b \geq c.
Consider f(a,b,c,0) - f(a,b+c,0,0) =24abc - 12bc(b+c) \geq 0.

So we must prove f(a,b,0,0) \geq 0, i.e. (a^3+b^3)/2 \geq ((a+b)/2)^3 !!!


How many pairs of digits $ (m,n)$ are there such that $ m\minus{}n\equiv m\plus{}n\bmod10$?

I don't know much about modular arithmetic, so I'm probably wrong, but here goes.
[hide]If $ m\minus{}n\equiv m\plus{}n\bmod10$, then after some simple algebra, $ 0\equiv 2n\bmod10$ so n can be 5 or 0. 2*10=20  [/hide]

[quote="dracs taxi"][hide="Not quite."]
  From the help I got from my Dad :blush: n mod 10 = the integer remainder of $ {n\over 10}$. For example, 34 mod 10 = 4 and 34 mod 3 = 1.[/hide]
  Now take a stab at it.[/quote]

You really have to think before you speak. Your post doesn't give us any information about solving the problem, and furthermore, vamathletes' solution is correct.

[quote="abacadaea"]using vamathletes solution, wouldnt the answer be $ \infty$:  0, 5, 10, 15, 20......[/quote]
No, because the question asks for digits, not integers. Vamatheletes is correct.

[quote="DonkeyKong"]I don't get how to the simplified version.

Like the real equation to 0=2n mod 10

P:S How do you insert an equation into you text box?[/quote]

Manipulate the m-n==m+n as if it was a normal equation (not congruence). Just as a note, this only works for integer addition and multiplication (and subtraction, which is the same as addition)

For the second question, put a LaTeX command (see the "commands" and "symbols" sections of the LaTeX tutorial) inside two dollar signs. For example,

[code]$m-n\equiv m+n\pmod{10}$[/code]

gives

$ m\minus{}n\equiv m\plus{}n\pmod{10}$.

[quote="mihail911"]if u somehow got $ \minus{}2n\equiv0\pmod{10}$ is that wrong?[/quote]
That works also, because $ \minus{}2n\equiv0\equiv2n\bmod{10}$.


Find an infinite geometric series whose sum is 6 and such that each term is four times the sum of all the terms that follow it.

Such a series does not exist (or your definition of "geometric series" is an unusual one):

Suppose $ 6\equal{}\sum_{k\equal{}0}^{\infty}q^{k}\equal{}\frac{1}{1\minus{}q}$. Then we get $ q\equal{}\frac{5}{6}$. But with this q we have
$ 4\sum_{k\equal{}n\plus{}1}^{\infty}q^{k}\equal{}4q^{n\plus{}1}\sum_{k\equal{}0}^{\infty}q^{k}\equal{}4q^{n\plus{}1}6\equal{}20\left(\frac{5}{6}\right)^{n}\neq\left(\frac{5}{6}\right)^{n}\equal{}q^{n}$

The standard definition of a geometric series allows an arbitrary first term.

Oh, and this topic doesn't belong in the college forums.

[hide]
Let our series be $ x,xr,xr^{2},...$

We know that $ x\equal{}4(\frac{xr}{1\minus{}r})$ by our infinite sum formula. (Or, more generally, $ xr^{n}\equal{}4(\frac{xr_{n\plus{}1}}{1\minus{}r})$)

Solving for $ r$, we get $ r\equal{}\frac{1}{5}$.

We know our total sum is 6, so reapplying the formula, we have $ \frac{x}{1\minus{}\frac{1}{5}}\equal{}6$

Solving for $ x$, we get $ x\equal{}\frac{24}{5}$

So our series is $ \frac{24}{5},\frac{24}{45},\frac{24}{125},...$ where the general term is $ x_{n}\equal{}\frac{24}{5^{n}}$[/hide]

[hide="solution"]$ x\plus{}xr\plus{}xr^{2}\plus{}\cdots \equal{}6$

$ x\equal{}4(xr\plus{}xr^{2}\plus{}\cdots)$

$ 5(xr\plus{}xr^{2}\plus{}\cdots)\equal{}6$

$ x\equal{}\frac{24}{5}$

$ xr\plus{}xr^{2}\plus{}\cdots\equal{}\frac{6}{5}$

$ r\equal{}\frac{1}{5}$

$ x\plus{}xr\plus{}xr^{2}\plus{}\cdots\equal{}\frac{24}{5}\plus{}\frac{24}{25}\plus{}\frac{24}{125}\cdots$
[/hide]

Bananas.

i was kinda confused by the wording. does following mean ahead or behind? i assumed behind. so i got this: assuming the series is $ a, ar, ar^{2}...$, we have $ \frac{a}{1\minus{}r}\equal{}6$ and $ ar\equal{}4a\Longrightarrow r\equal{}4$ which is impossible if this is an infinite geometric series. i know i made a stupid assumption that doesn't really make sense. :blush:

[quote="mihail911"]i was kinda confused by the wording. does following mean ahead or behind?[/quote]

Following means after, so we are talking about $ a_{n}\equal{} 4(a_{n\plus{}1}\plus{}a_{n\plus{}2}\plus{}a_{n\plus{}3}\plus{}... )$.  The usual term for "behind" is "preceding," or just "before."

oh, i see. "following" always seemed to me like $ a_{1}$ is following $ a_{2}$ stupid i know.  :lol: i kinda looked too much into it.

[quote="Gockel"]Such a series does not exist (or your definition of "geometric series" is an unusual one):

Suppose $ 6 \equal{}\sum_{k \equal{} 0}^{\infty}q^{k}\equal{}\frac{1}{1\minus{}q}$. Then we get $ q \equal{}\frac{5}{6}$. But with this q we have
$ 4\sum_{k \equal{} n\plus{}1}^{\infty}q^{k}\equal{} 4q^{n\plus{}1}\sum_{k \equal{} 0}^{\infty}q^{k}\equal{} 4q^{n\plus{}1}6 \equal{} 20\left(\frac{5}{6}\right)^{n}\neq\left(\frac{5}{6}\right)^{n}\equal{} q^{n}$[/quote]

The first term doesn't have to be 1...

1=2:
[quote="jmerry"]The standard definition of a geometric series allows an arbitrary first term.
[/quote]


factorial
algebra
polynomial
combinatorics unsolved
combinatorics

Hello,

If I'm given the value of the following sum, how do I explicitly solve for a?

x + x*(x-a) + x*(x-a)*(x-2a) + x*(x-a)*(x-2a)*(x-3a) +x*(x-a)*(x-2a)*(x-3a)*(x-4a)......etc until ....*(x-na)

a is a constant close to zero.

If I took logs I'd have n terms in x, n-1 for x-a, n-2 for x-2a...this makes me think I may need some sort of factorial tool to solve the equation?

Thanks for any help.

Jonathan.

What do you mean by "given the value of the sum"?  As a polynomial in $x$?  Or what?

I think if this sum equals (constant), find $a$ in terms of $x$ and (constant).


The first day of work, Jed stuffed $ \frac15$ of the envelopes. The second day, he stuffed $ \frac14$ of the remaining ones. The third day, he stuffed $ \frac13$ of the ones still unstuffed, and on the fourth day, he stuffed $ \frac12$ of the envelopes still left. What fraction of the envelopes are left for him to stuff on the fifth day? Express your answer as a common fraction.

Unstuffed portion:

(1/2)(2/3)(3/4)(4/5)=1/5. That is our answer.


I was doing some log problems from a pre-calculus book, and I could not solve a couple. I think they require the change of base... I'm not sure.

[size=150]Log[/size][size=75]16[/size][size=150]X[/size][size=150]+Log[/size][size=75]4[/size][size=150]X+Log[/size][size=75]2[/size][size=150]X=7[/size]
Find X

(e^x+e^-x)/2 = 1
Find X

(e^x-e^-x)/2 = 2
Find X

Second one:

[hide]
$e^x+e^{-x}=2$
$ln (e^x+\frac{1}{e^x})=ln (2)$
$ln (\frac{e^{2x}+1}{e^x})=ln (2)$
$ln (1)+ ln(e^{2x})- ln(e^x)= ln (2)$
$0+2x-x=ln (2)$

$x=ln (2)$
[/hide]

If you follow the identities used in that solution, the third problem is very similar.

[hide] ok here we go:

e^x + e^-x = 2       e^x =a
a + 1/a = 2
a = 1 so x = 0

e^x+e^-x = 4     e^x = a  same as before
a +1/a =4
a^2 -4a +1
a= 2 +/- sqrt(3)

x = ln(2+/-sqrt(3))[/hide]

This step is wrong:

$ln (\frac{e^{2x}+1}{e^x})=ln (2)$
$ln (1)+ ln(e^{2x})- ln(e^x)= ln (2)$

Note that $\log_c{a+b}\neq\log_c{a}+\log_c{b}$

[quote="Ceva Theorem"](e^x+e^-x)/2 = 1
Find X

(e^x-e^-x)/2 = 2
Find X[/quote]

Are these hyperbolic functions (sinh, cosh...)? Or am I reading stuff wrong?


Exista functii $f: \mathbb{Z}\to\mathbb{Z}$ astfel incat $f(f(x))+f(x)+x=0$? ;)

Da. E suficient sa partitionam $\mathbb{Z}^*$ in triplete $(a,b,c)$ astfel incat $a+b+c=0$.


geometry
geometric transformation
reflection
function
complex analysis
complex analysis unsolved

Without loss of generality , assume $r_{1}=R_{1}=1$ and $r_{2}=R_{2}=r >1$
Suppose $f: A(1,r)\to A(1,r)$ . We can assume that $|f(z)|\to 1\mbox{ as } |z|\to 1$
Then  we use The Schwarz reflection principle and we obtain an entire function with the following properties : $|f(z)|=1\mbox{ as } |z|=1$ and $f(0)=0\mbox{ }f(z)\neq 0 \mbox{ }\forall z\neq 0$
We show that $f(z)=\alpha z^{m}\mbox{ m is a natural number}$
Then we show that $m$ must be 1 .

[quote="eugene"]Prove that any conformal isomorphism of $\{r_1<|z|<r_2 \}$ to $\{R_1<|z|<R_2 \}$ is linear[/quote]

How about $r_1=R_1=\frac 12,\ r_2=R_2=2$, and $f(z)=\frac 1z$? It's certainly not linear. It is, however, true that $\frac{R_2}{R_1}=\frac{r_2}{r_1}$ and the conformal isomorphism has the form $f(z)=az$ or $f(z)=\frac az$, where $|a|=\frac{R_1}{r_1}$.

Ok , you're right . I think he must assume that $|f(z)|\to R_{1}\mbox{ as }|z|\to r_{1}$ :)

Thanks. Frankly speaking in the intial problem condition there wasn't said that $|f(z)|\to R_{1}\mbox{ as }|z|\to r_{1}$, but according to Grobber's remark, i think, yes, we should also add the solution $f(z)=\frac az$, where $|a|=\frac{R_1}{r_1}$.


[quote="nr1337"]It says "If you enroll in one of our online classes, information about the class, such as problem sets, solutions, and class transcripts, will be found here when the course begins" in the My Classes page.  Does this mean I'm signed up?[/quote]

No, u should probably ask Mr. rusczyk or something. When you sign up, you immediately receive an email confirmation and the class name is added as a link to the "My Classes" section where you just viewed that message.

[quote="nr1337"]Do we get an e-mail confirming that we signed up for a class?  I just want to make sure that I correctly signed up for the MathCounts Problem Series.[/quote]

You are not yet registered and you should recieve a confirmation email when you are enrolled.

Note to anyone who registers for our classes: Please exclude our emails from your spam filter.  If it says artofproblemsolving.com in the address it could be your order information.  We suspect spam filters to be a source of some miscommunications.

[quote="MCrawford"]
Note to anyone who registers for our classes: Please exclude our emails from your spam filter.  If it says artofproblemsolving.com in the address it could be your order information.  We suspect spam filters to be a source of some miscommunications.[/quote]

Which actually brings me to a question...what are the confirmation emails used for?

[quote="MithsApprentice"]
Which actually brings me to a question...what are the confirmation emails used for?[/quote]

Confirmation emails are sent to anyone who orders the books or any of our online classes.  It serves as both a receipt and lets people know that the order was processed.

Thank you.  Now that class shows up in "My Classes"


Uh, I think I have basic understanding of mods, but can someone give me some problems (basic, intermediate, hard) on mods? Preferably something useful on the AIME (like some special trick or something). Thanks.

Use the search function, there's a lot of poblem with mod.

Remembering the multiplicative property, (if a=b(modm) and c=d(modm)) then you can see that if a=b(modm) then               a^n=b^n (modm).
Then you can do things like figure out the rem. when dividing 2^100 by 17. Note that 2^4=-1(mod17), so 2^100=              (2^4)^25=(-1)^25=-1(mod17), so the positive remainder is 16.

hrm...interesting..

Uh, there was this one part to an AIME problem...and it required number theory.

It was like $19n=141+180k$....how do i do that?

$19n=141+180k$

Look at that equation in $mod$ $19$. The LHS equals zero since it is a multiple of 19. The RHS can be written as $8+9k$. You can plug in values of $k$ to quickly evaluate the RHS side in $mod$ $19$ until it also equals zero. Once you get a value for k, you compute $141+180k$ and divide that by $19$ to get the corresponding $n$.


algebra
polynomial
topology
analytic geometry
algebra unsolved

You have an infinite family $A$ of polynomials, each one having many variables from a countable set of variables. It is known that when each variable is 0 or 1, the value of each polynomial from A is also 0 or 1. Moreover, any non-empty finite set of polynomials from A has a common root in which each variable is 0 or 1. Then there exists a common root for all polynomials, with each component 0 or 1.

The only fact about polynomials we need is that each depends on only finitely many variables.
Also, the number of variables is irrelevant. Let the variables be indexed by $\mathcal{I}$.

Define $X=\{0,1\}^\mathcal{I}$, with the product topology. $X$ is compact.
For each $p\in A$, let $S_p=\{x\in X: p(x)=0\}$. We have that each $S_p$ is closed by the definition of the topology; it is restricted to a closed set in finitely many variables and free in all of the others.
Also, any finite collection of the $S_p$ has nonempty intersection. By compactness of $X$, this implies that all of the $S_p$ have nonempty intersection and we are done.

[quote="harazi"]Excuse me for the stupid question, but what is the product topology?  :([/quote] The weakest topology in which all natural projections $P_j\,: \, \prod_i X_i\to X_j$ are continuous. In other words, the base of the product topology in $\prod_{i\in I}X_i$ is the set of all "cylinders" $U_{i_1}\times\ldots\times U_{i_k}\times\prod_{\text{all other }i}X_i$ where $U_{i_j}\subset X_{i_j}$ are arbitrary open sets. Note that we take only finitely many non-trivial $U$'s.

Thank you, fedja. Could you also please explain why in this case ${0,1}^I$ is a compact?  :?

$\{0,1\}$ is compact, and there is a theorem (Tychonov's Theorem) which states that the product of compact spaces is compact. You can google that (I'm not sure about the spelling of the name; you might try with $f$ at the end instead of $v$, for example, since Russian names are usually spelled in many ways in English).

The basic open sets in the product topology: Start with the Cartesian product of $X_\alpha$. The basic open sets for this topology are the products of arbitrary open sets $G_\alpha$ in some finite collection of indices $\alpha$ and $X_\alpha$ for all other indices $\alpha$.
The product topology is the weakest topology in which projection onto one coordinate is continuous.
The key theorem used here is Tychonoff's Theorem, that the product of compact spaces is compact.

In this case, the topology used for $\{0,1\}$ is the obvious discrete topology.

I spent too long composing this- everyone posted under me.

Ok, thanks a lot. Wow, this is really strange problem!!!

The countable case would be accessible to sequential methods, but this really is a compactness result, and you'd have to come pretty close to proving Tychonoff's theorem anyway.

Where did this come from? I don't see how anyone would solve it without some knowledge of topology.

That's a non-trivial theorem called Tychonoff's theorem. The proof is not so complicated in the case of countable products of compact metric spaces, which is enough for your original problem. In this case the product topology is metrizable and the corresponding metric is $d(\bold x, \bold y)=\sum_k 2^{-k} \frac{d_k(x_k,y_k)}{1+d_k(x_k,y_k)}$ where $\bold x=(x_1,x_2,\dots)\in\prod_k X_k$ and similarly for $\bold y$ and $d_k$ is the metric in $X_k$. The convergence in the product topology is then just "coordinatewise" convergence. Cantor's diagonal process is then enough to choose a convergent subsequence from any sequence. It remains to recall that, for metric spaces, this is enough for compactness.

The general case requires using axiom of choice and completely different ideas, so look it up in some topology textbook. It is a very useful theorem.

It is my teacher of informatics who gave me this problem formulated in terms of sentences and stuff and I searched a lot for a "normal" solution, but it seems that one needs some topology to solve it. Anyway, thank you very much for your answers.  :)


logarithms
inequalities unsolved
inequalities

Let $ 0<a<b$, show that $ f(x)\equal{}\frac{b^x\minus{}a^x}{((a\plus{}b)/2)^x}$ is increasing on $ (0,\infty)$

Denote $ s \equal{} \frac {b\plus{}a} {2}$, $ d \equal{} \frac {b\minus{}a} {2}$, so $ 0< d < s$ and $ 0< a < s < b$.
Then $ f'(x)\equal{}\frac{(s\plus{}d)^x\ln \frac {s\plus{}d} {s} \minus{} (s\minus{}d)^x\ln \frac {s\minus{}d} {s}} {s^x} > 0$.

It's better to show, that $ g(x) \equal{} \ln f(x)$ is increasing.
$ g(x) \equal{} \ln {(b^x \minus{} a^x)} \minus{} x\ln {((a \plus{} b)/2)}$.
$ g'(x) \equal{} \frac {1}{b^x \minus{} a^x}(b^x\ln {b} \minus{} a^x\ln {a}) \minus{} \ln \frac {a \plus{} b}{2}\geq \frac {1}{b^x \minus{} a^x}(b^x\ln {b} \minus{} a^x\ln {b}) \minus{} \ln \frac {a \plus{} b}{2} \equal{} \ln {b} \minus{} \ln \frac {a \plus{} b}{2} \equal{} \ln \frac {2b}{a \plus{} b} > 0$
So $ g$ is increasing.

[quote="miwalin"]Let $ 0 < a < b$, show that $ f(x) \equal{} \frac {b^x \minus{} a^x}{((a \plus{} b)/2)^x}$ is increasing on $ (0,\infty)$[/quote]
$ f(x) \equal{} \left ( \frac{2b}{a\plus{}b} \right )^x \minus{} \left ( \frac{2a}{a\plus{}b} \right )^x$
$ \frac{2b}{a\plus{}b} > 1$ and hence $ \left ( \frac{2b}{a\plus{}b} \right )^x$ is increasing.
$ 0 < \frac{2a}{a\plus{}b} < 1$ and hence $ \left ( \frac{2a}{a\plus{}b} \right )^x$ is decreasing.
The conclusion follows.


I was looking at the mathcounts sprint school round, and there was one problem I couldn't do.

Number 30.

Could someone give the answer and how to get to it?

http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Sprint.pdf

[hide]
All you have to do is find AX and then subtract it from 5 (length of AM)
Since AXB is a right triangle and it shares an angle with ABC, they must be similar.
So $ AX/AB \equal{} AB/AC \equal{} 6/10.$
$ AX \equal{} 36/10 \equal{} 18/5$
$ XM \equal{} 5 \minus{} 18/5 \equal{} 7/5 \equal{} \boxed{1.4}$

[/hide]

[hide="Quick Pythagorean Way"]
Let $ BX\equal{}h$ and $ AX\equal{}q$.
Notice that $ XM\equal{}5\minus{}q$ and $ XC\equal{}10\minus{}q$.
By the Pythagorean Theorem, we get these two sets of equations:
[list]$ 64\equal{}h^2\plus{}(10\minus{}q)^2$
$ 36\equal{}h^2\plus{}q^2$
[/list]
Simplify these to get:
[list]$ 28\equal{}(10\minus{}q)^2\minus{}q^2$
$ \equal{}(100\minus{}20q\plus{}q^2)\minus{}q^2$[/list]
Therefore $ 28\equal{}100\minus{}20q \implies q\equal{}\frac{18}{5}$
Plug $ q$ into $ 5\minus{}q$ and you get $ \boxed{XM\equal{}\frac{7}{5}}$, or $ 1.4$[/hide]

[hide]
An easy way to do this problem is to first use the Pythagorean Theorem to find side BC.  BC is 8 because triangle ABC is a 6-8-10 triangle.  We can use this information to find the area of ABC which is 24.  Since 24 is the area of triangle ABC, AC x BX is 48.  We know AC is 10, so BX is 4.8.  Using the Pythagorean Theorem, we can find AX.  Therefore AX is 3.6.  So MX is equal to 5 (length of MA) - 3.6 (length of AX) which is 1.4.[/hide]


inequalities
induction
algebra
polynomial
inequalities unsolved

[b]mecrazywong[/b] seems to have a point, there's something wrong with the problem. I will solve a slightly different problem, which can hopefully relate to the corrected version of the original problem.

[b]Problem:[/b]
If $a_1,...a_n,b_1,....b_n$ are reals numbers such that $\sum a_i^2 \leq 1$ and $\sum b_i^2 \leq 1$, then
\[
(a_1^2+a_2^2+....+a_n^2-1)(b_1^2+b_2^2+.....+b_n^2-1) \leq (a_1b_1+a_2b_2+...+a_nb_n-1)^2
\]

[b]Solution:[/b]
By AM-GM
\begin{eqnarray*}
LHS &=& \left( 1-\sum_{i=1}^n a_i^2  \right) \left( 1-\sum_{i=1}^n b_i^2 \right) \\
&\leq& \left( 1- \sum_{i=1}^n \frac{a_i^2+ b_i^2}{2} \right)^2 \\
&\leq& \left( 1- \sum_{i=1}^n a_ib_i \right)^2 \\
&=& RHS
\end{eqnarray*}

Also, a similar problem appeared in the Chinese TST in 1996:

Let $n\geq 4$ be an interger, and $a_1, a_2, \ldots, a_n; b_1, b_2, \ldots, b_n$ be real numbers, satisfying $\sum_{i=1}^n a_i^2 <1$ and $\sum_{i=1}^n b_i^2 <1$. Let $A^2 = 1 - \sum_{i=1}^n a_i^2$, $B^2 = 1 - \sum_{i=1}^n b_i^2$, and $W = \frac{1}{2}\left( 1-\sum_{i=1}^n a_ib_i \right)^2$. Determine all possible real number $\lambda$, such that the polynomial $x^n + \lambda (x^{n-1} + \cdots + x^3 + Wx^2 + ABx + 1) = 0$ has only real roots.

[quote="Singular"]Constraint looks like cauchy.  Let X, Y, Z be $\sum x_{i}^2$, $\sum y_{i}^2$, $\sum x_iy_i$.

constraint is $(X-1)(Y-1) \geq (Z-1)^2$ equivalent to $XY + 2Z \geq Z^2 + X + Y$ (*)

"Lagrange Identity" says $XY - \sum_{1\leq i < j \leq n} (x_iy_j - x_jy_i)^2 = Z^2$

So constraint says $\sum_{1\leq i < j \leq n} (x_iy_j - x_jy_i)^2 \geq \sum (x_i - y_i)^2$ (**),
that is, $(*) \Leftrightarrow (**)$

Now, easy to proceed by induction.  Base case I leave as exercise.  Suppose that the constraint
implies the result for n-1.

In the crucial n-1 to n step, we are given "(*) imples result" for n-1 and wish to prove:

"$\sum_{i=1..n-1} (x_iy_n - x_ny_i)^2 + \sum_{1\leq i < j \leq n-1} (x_iy_j - x_jy_i)^2 \geq (x_n - y_n)^2 + \sum (x_i - y_i)^2$,
implies result for n".

The inequality reduces to:

$\sum_{i=1}^{n-1} (x_iy_n - x_ny_i)^2 \geq (x_n - y_n)^2 $  (using the hypothesis),

$\Leftrightarrow \sum_{i=1}^{n-1} (x_{i}^2y_{n}^2 + x_{n}^2y_{i}^2) - 2(x_{n}y_{n}x_{i}y_{i}) \geq x_{n}^2 + y_{n}^2 -2x_ny_n$

$\Leftrightarrow (\sum x_{i}^2)y_{n}^2 + (\sum y_{i}^2)x_{n}^2 -2x_{n}y_{n}\sum x_{i}y_{i} \geq x_{n}^2 + y_{n}^2 -2x_ny_n$

$\Leftrightarrow (X-1)y_{n}^2 + (Y-1)x_{n}^2 \geq 2x_ny_n(Z-1)$

But $LHS \geq \sqrt{(X-1)(Y-1)}2x_ny_n \geq RHS$ as required (2nd ineq: we were given $(X-1)(Y-1) \geq (Z-1)^2$ for n-1 by hypothesis in form (*) ), induction done.

ps. i was in a rush, so the notation is not perfect, but you should be able to understand.[/quote]

In my opinion, singular's solution has problem.

$\sum_{i=1}^{n-1} (x_iy_n - x_ny_i)^2 \geq (x_n - y_n)^2 $  (using the hypothesis),
This is wrong. When we have $A \ge B$ and $C \ge D$, we may not have $A-C \ge B-D$.

$LHS \geq \sqrt{(X-1)(Y-1)}2x_ny_n \geq RHS$  Use of AMGM assumed X-1 and Y-1 to be positive.

Dear $RaduSorici$ where did you get this  problem?If you have other problems of USA TST 2004 can you please post or send to my e-mail.

[quote="Singular"]siuhochung brings up some problems,.. i don't know if they are fatal... i think the (x-1)(y-1) business isnt fatal, because x<1 implies y<1 so the product is positive, but not sure yet and wont bother to look.

my reasoning was that this looked like a #6 on TST, so i thought it would be hard, and not just amgm, so i didnt even look

thats what i get for rushing[/quote]

i think the (x-1)(y-1) business isnt fatal, because x<1 implies y<1 so the product is positive
it's fatal.
Check it out again yourself.


Find the general solution for sinx + sin2x + sin3x=1 + cosx +cos2x.
Also, rate its toughness.

[quote="thanhnam2902"][color=darkblue]Equation $ \Longleftrightarrow (sin3x \plus{} sinx) \plus{} sin2x \equal{} (1 \plus{} cos2x) \plus{} cosx$

$ \Longleftrightarrow 2sin2x.cosx \plus{} sin2x \equal{} 2cos^2x \plus{} cosx$

$ \Longleftrightarrow sin2x(2cosx \plus{} 1) \equal{} cosx(2cosx \plus{} 1)$

$ \Longleftrightarrow (2cosx \plus{} 1)(sin2x \minus{} cosx) \equal{} 0$

$ \Longleftrightarrow (2cosx \plus{} 1)(2sinx.cosx \minus{} cosx) \equal{} 0$

$ \Longleftrightarrow cosx(2cosx \plus{} 1)(2sinx \minus{} 1) \equal{} 0$

$ \Longleftrightarrow$ $ cosx \equal{} 0$ OR $ cosx \equal{} \minus{} \frac {1}{2}$ OR $ sinx \equal{} \frac {1}{2}$

Then it's easy! :P[/color][/quote]

yeah...y din i think of it???man i stupid...

I think the identity $ \sin A \plus{} \sin B \equal{} 2\sin \left(\frac {A \plus{} B}{2}\right) \cos \left(\frac {A \minus{} B}{2}\right)$ is not well-known to American students, and without it this qustion is very hard. So I would rate this question as a 4 out of 5 in relation to AMC-12 questions.


If one-third of six apples sells for 60 cents, how many dollars will one-fourth of 40 apples cost?

One-third of six is 2.
2 apples is 60 cents.
Each apple is 30 cents.

One-fourth of forty is 10.

The cost of 10 apples is 10*.3=3 dollars

2 apples cost $ 60$ cents, so every apple is $ 30$ cents.
Therefore, $ 10$ apples cost $ 3$ dollars.


I've taken calculus and am looking to learn some linear algebra on my own. The basic books I'm looking at right now are "Linear Algebra : A Modern Introduction" by David Poole and "Linear Algebra with Applications" by Otto Bretscher. I'm also looking for texts that are more advanced and came across Linear Algebra by Hoffman & Kunze and Linear Algebra by Stephen H. Friedberg, Arnold J. Insel, and Lawrence E. Spence.

Are these books a good choice to start with? Anyone have any suggestions for some solid texts on the subject? Thanks.

I have used Poole's book, which is excellent as a first text for linear algebra.  There is also a solutions manual.

I have also used Hoffmann and Kunze, which has a pretty good selection of topics but which I dislike overall.  You might want to try Axler - it seems pretty nice as a second text for linear algebra.

EDIT: I visited the UBC library today and glanced through Axler - I still like the feel of it, but it does not cover nearly as much as Hoffmann and Kunze, and some of the omissions are reasonably significant.

Thanks for the reply. How about the book Linear Algebra by Stephen H. Friedberg, Arnold J. Insel, and Lawrence E. Spence? Anyone have any experience with it?   I heard it has the same flavor as Hoffman & Kunze but I'm not too sure how thorough it is since I don't have access to a good library.

I have no experience with it.  I should note that my (graduate) algebra prof last semester, whose opinion I respect immensely (he's incredibly bright), really likes Hoffmann and Kunze.  So maybe it's not as bad as I'm making it out to be.


Define $ a_{2}=2001$ and for $ n\ge3$ we have $ a_{n}=a_{n-1}.a_{n-2}-1$.Prove that there are infinitely many values of $ a_{1}$ so that $ a_{n}=2002$ for some $ n$ . :wink:

Is there a limitation for $ a1$, integer, real, complex?


Given $n$ a positive integer different from $0$, consider the set $E=\{1,2, \ldots,3n\}$. We partionate $E$ into three sets $A$ , $B$ and $C$. each one contains $n$ elements.
Prove that there exist $(x,y,z)$ from $A*B*C$ such that one of the numbers $x,y,z$ is the sum of the two others.

It is already on the forum. But I can't find it.
I remember that Grobber wrote something on it, but I am not sure...

No, I don't think so.. :? However, I also remember this problem. It's definitely around here somewhere :).

There was an idea with Cauchy-Davenport theorem.

he-he :
http://www.mathlinks.ro/Forum/viewtopic.php?t=26762&postorder=asc&highlight=schweitzer&start=15

Pierre.

Oh... It was so recently and I completely forgot it. It means the idea is out of head.


quite some people have seen the topic,but none give the reply... :(

[quote="little bear"]请问每天的 “problem of the day”的那道题目有记录可以查到么？我的意思是我怎么查到昨天公布的以及以前的题目呢？thanks :)[/quote]
Can you traslate what you said? I can't read Chinese. :yup:


:blush: Ok, I KNOW I am supposed to know this, buuuuuut.... I don't.

 The question that I am stuck on is~

 IDENTIFY the set of numbers that make up the range of y = x + 2 if the domain is {0, 1, 2, 4}.

  :surrender: HELP! I have no clue... Ok, first of all, what in the world is the range of Y = X + 2?!?

Plug them in

The lowest y will occur at $ x \equal{} 0$ and the highest will occur at $ x \equal{} 4$

So let's look for y

$ y \equal{} 0 \plus{} 2$

$ y \equal{} 4 \plus{} 2$

So the lowest y is 2 and the greatest y is 6

Therefore the set of numbers that make up the range are $ \{2, 3, 4, 6\}$


function
logarithms
calculus
calculus computations

Condition 1. f: R+ ---> R+

Condition 2. f(x),f'(x),f''(x)>0

Condition 3. f(x)*f''(x)=2{f'(x)}^2

Does the function satisfying all 3 conditions exist??

If the function doesn't exist, prove that.

Plz help me

From the given condition, we have $ \frac {f''(x)}{f'(x)} \equal{} 2\frac {f'(x)}{f(x)}$.

Integrate both sides, $ \ln f'(x) \equal{} 2\ln f(x) \plus{} \ln A$ $ (\because Condition 2)$.

$ \minus{} \frac {f'(x)}{f(x)^2} \equal{} B$, here $ B \equal{} \minus{} A$.

Integrate that, $ \frac {1}{f(x)} \equal{} Bx \plus{} C$, yielding $ f(x) \equal{} \frac {1}{ax \plus{} b}$, here $ a,\ b$ are arbiterary constants.

f(x)=1/(ax+b) satisfies condition no.2??

If f(x)>0 for all x, then a>0, b>0

Therefore f'(x)=-a(ax+b)^(-2)<0 ??

but the same thing...

f'(x)=-a(ax+b)^(-2)

because of a>0, (ax+b)^(-2)>0 , f'(x)<0


integration
limit
abstract algebra
calculus
real analysis
function
logarithms

I'm studying Fourier transformations in my University . I have a difficult problem . I haven't solve it . Can anyone help me :)

Let $f\in\mathfrak{L}^{p}([0,2\pi))$ with $1<p<\infty$ .
Put $c_{n}=\frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{-i n t}dt$ $\forall n\in\mathbf{Z}$ and

$s_{n}=\sum_{k=-n}^{n}c_{k}e^{ikt}$ $\forall n\in\mathbf{N}$

Prove that : $\lim_{n\to\infty}||f-s_{n}||_{p}=0$

You need to get $s_n$ into a tractable form. Have you dealt with the Dirichlet kernel yet? If not, the basic idea is to write $c_k$ as an integral, and then exchange the order of summation and integration in the definition of $s_n$. Then you can express $s_n$ as a convolution of $f$ with a useful function which is called the $n^{th}$ Dirichlet kernel.

Once you have this, put your norm in integral form and try to show why that integral should be small. Often, a good approach for this sort of problem is to split the integral into two or more parts, where you can characterize the behavior of the function precisely on each part separately.

The Dirichlet kernel isn't going to help you, not directly.

We'll need to show, to begin with, that $\|s_n\|\le C_p\|f\|_p,$ independent of $n.$ The best we can get directly is Young's ineqaulity for convolution:

$\|s_n\|_p\le\|D_n*f\|_p\le\|D_n\|_1\|f\|_p.$

Unfortunately, $\|D_n\|_1\sim \ln n$ so that approach goes nowhere.

This is a famous theorem, proved in the 1920's or 1930's by Marcel Riesz. Rather than work directly with the partial sums, he estimated the Hilbert transform of $f.$

A couple of ways to look at the Hilbert transform. Here's one. (Assume, to begin with, that $f$ is real.) Find the harmonic function $u=u(re^{i\theta})$ on the unit disk of the complex plane whose boundary values are $f(\theta).$ Then find $v$ harmonic such that $v(0)=0$ and $u+iv$ is analytic on the disk. The boundary values of $v(re^{i\theta})$ as $r\to1^-$ are the Hilbert transform, $Hf(\theta).$

On a formal level, $\widehat{Hf}(k)=i\text{ sgn}(k)\widehat{f}(k).$

(Unless it's $-i\text{ sgn}(k)$ - I never remember that detail and I'm too lazy to check right now.)

It's easy to see that $\|Hf\|_2\le\|f\|_2,$ but then in your original problem, $p=2$ was the easy case. You were always concerned with $p\ne2.$

M. Riesz proved that for $2<p<\infty,\,\|Hf\|_p\le C_p\|f\|_p,$ where $C_p=O(p)$ as $p\to\infty.$ The Hilbert transform is a self-adjoint convolution operator; from that it follows that for $1<p<2,\,\|Hf\|_p\le C_{p'}\|f\|_p$ where $\frac1p+\frac1{p'}=1.$

From there, Riesz went on to prove what you're asking for: that partial sums of the Fourier series of an $L^p$ function converge in the $L^p$ sense to the original function.

This is false for $p=1$ or $p=\infty.$ Note the behavior of the constant $C_p.$

There are many omitted steps here; there really isn't room to explain this on line. I'm going to suggest that you try looking this up in [b]the[/b] classic book: Antoni Zygmund's [i]Trigonometric Series[/i].

It's probably also in Katznelson, but go see if they have Zygmund in your library.

Thank you a lot :) .
Your message is very interesting . So my problem is really difficult  :( . I'll try to look up in that book .

Sorry, I have no idea what I was talking about. Neither of those have anything to do with how we approached convergence in the norm in my class. We actually only proved it for $L^2$, so we were able to use properties of the Hilbert space.

For $L^{2}$ , we can use Parseval's identity :) . And how can you approach the general one ???

Ah, but you can only use Parseval's identity to prove that if you're not using that to prove Parseval's identity. :P

You can't use properties of a Hilbert space on $L^p$ except when $p=2$ because that's the only one that is a Hilbert space. I don't know how to approach the other ones; we never talk about anything but $L^1$ and $L^2$ in my classes (though my real analysis prof did discuss Holder's inequality one time).

Actually, I wonder if you could use Holder here?

$L^2$ is simply not the interesting case; there the only properties of the Fourier transform we need is that the trigonometric basis is a complete orthonormal set. And in $L^2$, the Hilbert transform is, after modding out by constants, an isometry.

The only real role for Hölder's inequality is in the relatively simple fact that if the Hilbert transform is bounded on $L^p,$ then it is also bounded on $L^{p'}.$

All of the general proofs I know involve the notion of interpolation of operators. Here's the Riesz-Thorin interpolation theorem, the beginnings of the complex method of interpolation:

Suppose that $T$ is a linear operator defined on a set of functions large enough to include all of the simple functions. Suppose that $\|Tf\|_{q_0}\le M_0\|f\|_{p_0}$ and $\|Tf\|_{q_0}\le M_0\|f\|_{p_0}.$ For $0\le t\le1,$ let $\frac1{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1}$ and $\frac1{q_t}=\frac{1-t}{q_0}+\frac{t}{q_1}.$ Then $\|Tf\|_{q_t}\le M_0^{1-t}M_1^t\|f\|_{p_t}.$

This is called the complex method of interpolation because its proof uses a theorem of complex analysis related to the maximum modulus theorem called the Hadamard three lines theorem. Here's an example of the Riesz-Thorin theorem at work: Consider the operator that sends a function on the circle to it's Fourier coefficients. This operator sends $L^2$ to $\ell^2: \ \|\widehat{f}\|_2=\|f\|_2$ by Parseval. It also sends $L^1$ to $\ell^{\infty}: \ \|\widehat{f}\|_{\infty}\le\|f\|_1.$ Then it follows by interpolation that for $1\le p\le 2,\ \|\widehat{f}\|_{p'}\le \|f\|_p.$ This is known as the Hausdorff-Young inequality.

We know that that Hilbert transform is a bounded operator from $L^2$ to $L^2.$ M. Riesz's original proof involved a clever complex analysis argument involving integrals of $(u+iv)^4,\,(u+iv)^6,$ and so on around the circle in order to prove that it is a bounded operator from $L^{2k}$ to $L^{2k}.$ Interpolating between those gives the result for $2\le p<\infty.$

A more modern proof - Katznelson should have it - proves that the Hilbert transform satisfies a property called "weak type (1,1)" and then uses another interpolation method - this time, the Marcinkiewicz interpolation theorem (the real method of interpolation) - to get the result for $1<p\le 2.$


inequalities
real analysis
real analysis unsolved

I am having trouble showing why this is true. Can anybody show me how this works? Thank you.

Let A contained in R (R - all reals) be a non-empty subset which is bounded above, and set B = {x^3|x belongs to A}. Prove that sup(B) exists, and that sup(B) = (sup(A))^3.

I greatly appreciate your help.

The proof is in two parts: $\sup(B)\le (\sup(A))^{3}$ and $\sup(B)\ge (\sup(A))^{3}$. For the first part, observe that for any $a\in A$ the inequality $a^{3}\le (\sup(A))^{3}$ holds. For the second, note that $b^{1/3}\le (\sup(B))^{1/3}$ for any $b\in B$.


can you help me out these problems

Solve y

Problem 1
$ \displaystyle \sqrt{(y-2)+\sqrt{2y-5}}+\sqrt{(y-2)+3\sqrt{2y-5}}=7\sqrt{2}$

Problem 2

$ \displaystyle \sqrt{(y)+\sqrt{y}}-\sqrt{(y)-\sqrt{y}}=\frac{3}{2}\sqrt{\frac{y}{y+\sqrt{y}}}$

I hope someone solves this problem

$\sqrt{(y-2)+\sqrt{2y-5}}+\sqrt{(y-2)+3\sqrt{2y-5}}=7\sqrt{2}$

Simplify by substituting $x = 2y - 5,\ y = \frac{x + 5}{2}$.

$\sqrt{\frac{x + 1}{2} + 3 \sqrt x} + \sqrt{\frac{x + 1}{2} + \sqrt x} =7 \sqrt 2$

Imagine the left side as A + B and expand the equation by A - B.

$\frac{x + 1}{2} + 3 \sqrt x - \frac{x + 1}{2} - \sqrt x = 7 \sqrt 2 \left( \sqrt{\frac{x + 1}{2} + 3 \sqrt x} - \sqrt{\frac{x + 1}{2} + \sqrt x}\right)$

$\frac{\sqrt{2x}}{7} = \sqrt{\frac{x + 1}{2} + 3 \sqrt x} - \sqrt{\frac{x + 1}{2} + \sqrt x}$

Add this to the original equation and square the result.

$2 \sqrt{\frac{x + 1}{2} + 3 \sqrt x} = \frac{\sqrt{2x}}{7} + 7 \sqrt 2$

$2x + 2 + 12 \sqrt x = \frac{2x}{49} + 4 \sqrt x + 98$

$\frac{96x}{49} + 8 \sqrt x - 96 = 0$

$12x + 49 \sqrt x - 588 = 0$

Solve this quadratic equation for $\sqrt x$ and take the positive root.

$\sqrt x = \frac{-49 + \sqrt{2401 + 28224}}{24} = \frac{-49 + \sqrt{49 \cdot 625}}{24} = \frac{-49 + 175}{24} = \frac{126}{24} = \frac{21}{4}$

$x = \frac{441}{16},\ \ y = \frac{x + 5}{2} = \frac{521}{32}$

Check:

$y - 2 = \frac{521 - 64}{32} = \frac{457}{32}$

$\sqrt{2y - 5} = \sqrt{\frac{521 - 80}{16}} = \sqrt{\frac{441}{16}} = \frac{21}{4} = \frac{168}{32}$

$A = \sqrt{y - 2 + 3\sqrt{2y - 5}} = \sqrt{\frac{457 + 504}{32}} = \sqrt{\frac{961}{32}} = \frac{31}{4 \sqrt 2} = \frac{31 \sqrt 2}{8}$

$B = \sqrt{y - 2 + \sqrt{2y - 5}} = \sqrt{\frac{457 + 168}{32}} = \sqrt{\frac{625}{32}} = \frac{25}{4 \sqrt 2} = \frac{25 \sqrt 2}{8}$

$A + B = \frac{31 \sqrt 2}{8} + \frac{25 \sqrt 2}{8} = \frac{56 \sqrt 2}{8} = 7 \sqrt 2$

If you try a similar approach for the 2nd problem, you quickly find that the equation does not have a real root. Maybe you did not copy it correctly.

Yeah that's what Palytoxin and I got on the [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=44170]other[/url] thread that was posted.

Tip to original poster: Don't post the same question in different forums! Thanks.

put also some conditions like $2y-5 \geq 0$ ...etc...etc

After changing the sign between the 2 square roots on the left side from plus to minus, the 2nd problem is solvable. Imagine the left side as A - B and expand the equation by A + B.

$\sqrt{y + \sqrt y} - \sqrt{y - \sqrt y} = \frac 3 2 \sqrt{\frac{y}{y + \sqrt y}}$

$y + \sqrt y - (y - \sqrt y) = \frac 3 2 \sqrt{\frac{y}{y + \sqrt y}} \left( \sqrt{y + \sqrt y} + \sqrt{y - \sqrt y}\right)$

$2 \sqrt y = \frac 3 2 \sqrt y + \frac 3 2 \sqrt y \sqrt{\frac{y - \sqrt y}{y + \sqrt y}}$

$1 = 3 \sqrt{\frac{\sqrt y - 1}{\sqrt y + 1}}$

$\sqrt y + 1 = 9 (\sqrt y - 1)$

$\sqrt y = \frac{10}{8} = \frac 5 4,\ \ y = \frac{25}{16}$

Check:

$L = \sqrt{\frac{25}{16} + \frac 5 4} - \sqrt{\frac{25}{16} - \frac 5 4} = \sqrt{\frac{45}{16}} - \sqrt{\frac{5}{16}} = \frac{3 \sqrt 5}{4} - \frac{\sqrt 5}{4} = \frac{\sqrt 5}{2}$

$R = \frac 3 2 \sqrt{\frac{\frac{25}{16}}{\frac{25}{16} + \frac 5 4}} = \frac 3 2 \sqrt{\frac{25}{45}} = \frac 3 2 \sqrt{\frac 5 9} = \frac{\sqrt 5}{2}$


1. a scarlet bag contains exactly three numbers: 1,5,7. a gray bag contains exactly three numbers: 2,4,6. two distinct numbers are drawn at random from each bag. find the probability that the sum of the 2 numbers from the scarlet bag is greater than the sum of the two numbers from the gray bag.

2. there exists a right triangle in which the length of the hypotenuse is 97, and the lengths of the two legs are each positive integers. if  the perimeter of the triangle is more than 213, find the perimeter of the triangle.

[hide="2"]Thinking about the Pythagorean Triples, which are $(m^{2}-n^{2},2mn,m^{2}+n^{2})$, we let
\[97=m^{2}+n^{2}. \]
By quick trial and error, we see that it is true when $m=9$ and $n=4$.
Then the other sides can be $m^{2}-n^{2}=81-16=65$ and $2mn=2 \cdot 9 \cdot 4=72$.
The perimeter is then $97+65+72=\boxed{234}$.

For purposes of double checking, we see that $65^{2}+72^{2}=97^{2}$ with help of a calculator :) .[/hide]

[quote="iamagenius"]for #1, i thought that there is a 1/3 probability of picking a number and then 1/2 probability of picking another so that would be 1/6[/quote]
Picking 1,5 is the same as picking 5,1.

#1[hide]the two are independent.  with the scarlet bag we get 6,8,12 all with probability $\frac{1}{3}$

with the gray bag, we get $6,8,10$ with equal probability.  hence the probability of gray winning is $\frac{1}{3}$, the probability of a tie is $\frac{2}{9}$, hence hte probability that scarlet wins is $\frac{4}{9}$.
[/hide]

this reminds me of non-transitive dice - if you have time you should look this up, it's actually really cool.


Prove that any value $a+\sqrt b$ is equal to $\sqrt {c+\sqrt d}=\sqrt{ e+\sqrt{f+\sqrt g}}= \cdots $

Could you please rephrase the question? It's not clear what are you asking.


Euler
number theory proposed
number theory

Maybe well known:
$p$ a prime number, $n$ an integer. Prove that $n$ divides $\phi(p^n-1)$ where $\phi(x)$ is the Euler function.

$n$ is the order of $p \mod p^n-1$, and the order divides every $k$ with $p^k \equiv 1 \mod p^n-1$, so especially $k=\phi(p^n-1)$.

:blush:  it is in fact trivial ... it's appeared in my course as a consequence of non trivial things. Sorry not to have tought on it.

We used it as an exercise for olympiad french training the year you follow the training Alekk...(or maybe the year after?)

Pierre.


The sequences $ (a_n),(b_n)$ are defined by $ a_1 = 1,b_1 = 2$ and
$ a_{n + 1} = \frac {1 + a_{n} + a_{n}b_{n}}{b_{n}},b_{n + 1} = \frac {1 + b_{n} + a_{n}b{n}{a_{n}}$.
Show that $ a_{2008} < 5$

I don't know why latex looks like this  :(

[quote="enndb0x"]The sequences $ (a_n),(b_n)$ are defined by $ a_1 \equal{} 1,b_1 \equal{} 2$ and
$ a_{n \plus{} 1} \equal{} \frac {1 \plus{} a_{n} \plus{} a_{n}b_{n}}{b_{n}},b_{n \plus{} 1} \equal{} \frac {1 \plus{} b_{n} \plus{} a_{n}b_{n}}{a_{n}}$.
Show that $ a_{2008} < 5$[/quote] You didn't pair all of your { with } properly.

$ a_{n\plus{}1} \equal{} \frac{1 \plus{} a_n}{b_n} \plus{} a_n$

$ b_{n\plus{}1} \equal{} \frac{1 \plus{} b_n}{a_n} \plus{} b_n$

Should be easy to find the limit of $ a_n$ as n approaches infinity from there.

[quote="yanglunj"]Should be easy to find the limit of $ a_n$ as n approaches infinity from there.[/quote]  No, actually, that's not helpful at all.  (We have $ \lim_{n \to \infty} b_n \equal{} \infty$.)  I fail to see a nice way to grapple with this question, but I'll keep playing around.  In particular, it's obviously very sensitive to the initial values chosen.

Incidentally, the idea that a recent question from the Russian NMO is "Preolympiad" is pretty ridiculous.

[hide="A bit messy but with general result"]From the initial equations it's clear that both sequences are strictly increasing and positive. Rewrite the equations as

$ a_{n + 1} + 1 = {(a_n + 1)(b_n + 1)\over b_n}, b_{n + 1} + 1 = {(a_n + 1)(b_n + 1)\over a_n}$

Substitute $ A_n: = a_n + 1, B_n: = b_n + 1$ with $ A_1 = 2, B_1 = 3$

$ A_{n + 1} = {A_nB_n\over{B_n - 1}},B_{n + 1} = {A_nB_n\over{A_n - 1}}$

We need to prove $ A_n < 6$

Since $ A_n, B_n$ are obviously rationals, we can put $ A_n = {x_n\over y_n}, B_n = {u_n\over v_n}$ to get

$ {x_{n + 1}\over y_{n + 1}} = {x_nu_n\over y_n(u_n - v_n)},{u_{n + 1}\over v_{n + 1}} = {x_nu_n\over v_n(x_n - y_n)}$

From there we get the recursive system:

$ \{\begin{array}{ccc}x_{n + 1} & = & x_nu_n \\
y_{n + 1} & = & y_n(u_n - v_n) \\
u_{n + 1} & = & x_nu_n \\
v_{n + 1} & = & v_n(x_n - y_n)\end{array}$

with $ (x_1,y_1) = (2,1),(u_1,v_1) = (3,1)$

(Note to the last equation: since $ A_n$ is greater than $ 1$, we have that $ x_n > y_n$, hence all $ v_n$ are of the same sign, which is obviously positive because $ v_1 = 1$. We'll need this result later.)

Simple calculation gives $ x_2 = 6, y_2 = 2, u_2 = 6, v_2 = 1$, hence $ A_2 = 3, B_2 = 6$

From the above equations we see that for $ n\geqslant 2$, $ x_n = u_n = 6^{2^{n - 2}}$ (which is obvious from the pattern: $ x_2 = u_2 = 6, x_3 = u_3 = 6^2,x_4 = u_4 = 6^4, x_5 = u_5 = 6^8$ etc.)

On the other hand, $ y_{n + 1} - v_{n + 1} = y_n(u_n - v_n) - v_n(x_n - y_n) = y_nu_n - v_nx_n$, but for $ n\geqslant 2$, we have $ u_n = x_n$, hence

$ y_{n + 1} - v_{n + 1} = x_n(y_n - v_n)$ for $ n\geqslant 2$

From there we get

$ y_n - v_n = 6^{2^{n - 3}}\cdot 6^{2^{n - 4}}\cdot\dots\cdot 6\cdot(y_2 - v_2)$ for $ n\geqslant 3$.

$ y_n - v_n = 6^{2^{n - 2} - 1}(2 - 1) = 6^{2^{n - 2} - 1}$

As $ v_n$ is positive, we see that $ y_n > 6^{2^{n - 2} - 1}$. But then

$ A_n = {x_n\over y_n} < \frac {6^{2^{n - 2}}}{6^{2^{n - 2} - 1}} = 6$ for $ n\geqslant 3$. Values $ A_1$ and $ A_2$ also satisfy $ A_n < 6\iff a_n<5$, and the proof is thus completed.[/hide]

Can't believe I didn't see this simple beauty in the first place :)

[hide="MUCH improved solution"]Substitutions $ A_n, B_n$ are the same as above, with $ A_1 \equal{} 2, B_1 \equal{} 3$. We got

$ A_{n \plus{} 1} \equal{} {A_nB_n\over B_n \minus{} 1},B_{n \plus{} 1} \equal{} {A_nB_n\over A_n \minus{} 1}$

Take the reciprocals of both equations:

$ {1\over A_{n \plus{} 1}} \equal{} {1\over A_n} \minus{} {1\over A_nB_n}, {1\over B_{n \plus{} 1}} \equal{} {1\over B_n} \minus{} {1\over A_nB_n}$

Now put $ \alpha_n: \equal{} {1\over A_n},\beta_n: \equal{} {1\over B_n}$ with $ \alpha_1 \equal{} {1\over 2},\beta_1 \equal{} {1\over 3}$. This yields

$ \alpha_{n \plus{} 1} \equal{} \alpha_n \minus{} \alpha_n\beta_n,\beta_{n \plus{} 1} \equal{} \beta_n \minus{} \alpha_n\beta_n$

Subtracting these, we get

$ \alpha_{n \plus{} 1} \minus{} \beta_{n \plus{} 1} \equal{} \alpha_n \minus{} \beta_n$.

Therefore $ \alpha_n \minus{} \beta_n$ is constant, and is equal to $ \alpha_1 \minus{} \beta_1 \equal{} {1\over 6}$.

But $ \alpha_n \minus{} \beta_n \equal{} {1\over 6}$ yields $ \alpha_n > {1\over 6}$ (since $ \beta_n$ is positive), and that in turn means $ A_n \equal{} {1\over\alpha_n} < 6\iff a_n < 5$. QED[/hide]


Entirely aside from having the mathematical ability to solve USAMO problems, what are tips experienced Olympiad problem solvers have about actually writing out USAMO solutions?  Although it's a little early, I ask because

1)  I have terrible handwriting, and

2)  I often skip a lot of steps in my reasoning.

I'm wondering how these factors would affect the scoring of my solutions.  From writing solutions on AoPS I've already learned to be reasonably clear as far as stating important steps in my reasoning, but I often skip small steps.  I swear the important steps are there!

Thoughts?  Comments?

Sketch out a list of the things you need to prove in the order you need to prove them on scratch paper first. This is what I do for USAMTS, like solve all of the problems at once then submit solutions a couple weeks later when I have time to type them up (or it's just the last weekend lol), but it's also what I've been doing on practice USAMOs. Mess around a whole bunch, write all of your thoughts down in case one of them is useful, then when you connect the dots, go over it all in your mind. After that, sketch the order of things you want to prove as I said above, then formalize it on paper. Don't worry about scratching things out, but if half the page is scribbles, you might as well start over. If you find some major flaw in your argument while writing it up, don't throw away your proof so far, but definitely go back to your scratch paper to try to fill the gap.

Of course all of this is assuming you manage to come up with a solution/proof; otherwise I still don't know if it's worth the time to come up with a total BS proof or something close. I got away with it once on a WOOT test, but that's the only time I've ever had success when I didn't have a complete solution.

If you have a BS proof you  might get 1 or 2 points for it.
But a blank sheet of paper saves the grader some time.

This is my proof: "Because of my short proofs, you are going to finish several minutes before other graders. I think I deserve some points for my thoughtfullness."

Btw, there was this math jam a while back on writing solutions in the USAMO. Check out the transcripts.

Here is the transcript you are talking about:

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=32]How to take the USAMO[/url]


$ \sqrt {2 \dfrac{2}{3} } \equal{} 2 \sqrt { \dfrac{2}{3} }$

Is there another equation like this, where the square root of the mixed number equals the whole number part times the square root of the fractional part?

Is there a formula that can create an infinite number of equations like the one above?

hello, i think $ \sqrt{3\frac{3}{7}}=\sqrt{\frac{24}{7}}$ and why is this $ 3\sqrt{\frac{3}{7}}=\sqrt{\frac{27}{7}}$?
Sonnhard.

I think it should be $ \sqrt {3\frac {3}{8}} \equal{} 3\sqrt {\frac {3}{8}}$

In fact we have that if:

$ n\sqrt {\frac {a}{b}} \equal{} \sqrt {n\frac {a}{b}}$

$ \sqrt {\frac {n^{2}a}{b}} \equal{} \sqrt {\frac {nb \plus{} a}{b}}$

$ nb \plus{} a \equal{} n^{2}a$

$ nb \equal{} a(n^2 \minus{} 1)$

Since the two sides are equal, their prime factorisations are the same, but since $ a$ and $ b$ are coprime (otherwise the fraction $ \frac{a}{b}$ would be cancelleable), and $ n$ and $ n^2 \minus{} 1$ are coprime, this equation only has solutions when $ a \equal{} n$ and $ b \equal{} n^2 \minus{} 1 \equal{} a^2 \minus{} 1$.

Thus all radicals which have this property are given by:

$ n\sqrt {\frac {n}{n^2 \minus{} 1}}$.

I hope that this is correct, and that there aren't any other solutions which I have missed.


Compute $ \left( \frac{12}{37}\right)$, where $ \left( \frac{a}{b}\right)$ is the [url=http://en.wikipedia.org/wiki/Legendre_symbol]Legendre Symbol[/url].

[hide="Solution"]The Legendre symbol evaluates to $ 1$ if and only if $ a$ is a quadratic residue of $ b$.

$ 7^2 \equiv 12$ modulo $ 37$, so the desired answer is $ \boxed{1}$.[/hide]

For the sake of me not being bored, I'll write (a,b) instead of Legendre Symbol :P

$ (12,37)\equal{}(2,37)^2 (3,37)$

$ (2,37)\equal{}\minus{}1$, as $ 37\equiv 5\pmod 8$, but $ (2,37)^2\equal{}1$

$ (3,37)\equal{}1$, as $ 37\equiv 1\pmod {12}$

$ (12,37)\equal{}1$


[youtube]jsPBVNecOMo[/youtube]

If you want to close the window after about half the movie is done, hang in there.

This is very cool. I am probably going to write this all down. It is that cool.
EDIT: I just did!

I'm not sure if I agree, but it is extremely well made.

It's favorited 666 times.  :rotfl:

[quote="zolojetto"]hmm...thants kinda strange, um how do you have 0 posts!!![/quote]

He only posts in G&FF, where post counts don't count.

Posts don't count in GFF.
If they did, I'd have over 2000 posts.

Darn, 1=2 beat me to the punch.

miyomiyo can get this one!!! What is GFF???

This must be one of the coolest video ever! It made me happy  :lol:  and sad  :(  at the same time.

[quote="woser456"]Um, actually no.

Because of the net value in posts, I transfer my posts via connections in the Japanese banking system. From there, we sell the posts, and then buy them back for a lower price, then buy extra posts and send them back to America. So, according to the American govt. I now have x posts, but I really have x+However many extra I obtained in Japan. So, that means that I can freely take advantage of the American people with the posts I got in Japan. From there, the process is repeated, and I'm really getting more and more posts that the American govt. doesn't know about. Meanwhile, the original posts are stored with a Phillipino woman currently living in Istanbul. So, according to America, I have no posts.[/quote]

lol!

Anyway, that's a pretty cool vid.

[quote="woser456"]Um, actually no.

Because of the net value in posts, I transfer my posts via connections in the Japanese banking system. From there, we sell the posts, and then buy them back for a lower price, then buy extra posts and send them back to America. So, according to the American govt. I now have x posts, but I really have x+However many extra I obtained in Japan. So, that means that I can freely take advantage of the American people with the posts I got in Japan. From there, the process is repeated, and I'm really getting more and more posts that the American govt. doesn't know about. Meanwhile, the original posts are stored with a Phillipino woman currently living in Istanbul. So, according to America, I have no posts.[/quote]

sure what ever you say!!!(i had to read it 4 times before i understood any piece of it!) xD

Yeah, and I'm that Filipino woman with your posts.


How do you create a subheading in LaTeX? I want the main title to appear and after that, subheading 1 and subheading 2. Is there any way of setting spaces for a particular [b]area[/b] of the document? I want certain parts to be 1.5 spaced, but not others...

edit- also, is there any way of making the text go to the end of the page rather than breaking off in the middle. I want the text to appear on a 8.5 by 11 normal peice of paper, and I used the following commands in my preamble:

[code]\documentclass[a4paper]{article}
\setlength{\parindent}{0pt}
\renewcommand{\baselinestretch}{1.5}
\begin{document}[/code]

However, this still doesn't work...

I don't know the subheading.  What happens if you try something like

\title{My title \\ \small My subtitle}

Regarding line spacing: you can change the \baselinestretch within your document.

\renewcommand{\baselinestretch}{1.5}

switches to 1.5 spacing.  Here's the trick: it doesn't take effect until you change the font size.  (Don't ask me why.)  So you need to insert

\small\normalsize

in order for the new baselinestretch setting to take effect.  Then issue

\renewcommand{\baselinestretch}{1}

to change back to single spacing.  Again, you'll have to then issue \small\normalsize before the change takes effect.

If you want your output on 8.5x11 paper, why are you using the a4paper option?  Also, if you're using pdftex, you need the lines

\pdfpageheight 11in
\pdfpagewidth 8.5in

in the preamble.

Finally, to have the text run lower down the page, change the \textheight length.

\section* and \subsection* and \subsubsection* will give you subheadings and stuff like that. If you want them numbered, remove the *.


Prove that $ \frac {1}{2}(a \plus{} b)^{2} \plus{} \frac {1}{4}(a \plus{} b) >\equal{} a\sqrt {b} \plus{} b\sqrt {a}$ :)

[hide="Solution"]With simple factoring, this inequality is equivalent to:

$ \text{AM} \left( a, b \right) \left[ \text{AM} \left(2a, \; \frac {1}{2} \right) \plus{} \text{AM} \left(2b, \; \frac {1}{2} \right)\right]$ $ \geq \text{GM} \left(a, b \right)\left[ \text{GM} \left(2a, \; \frac {1}{2} \right) \plus{} \text{GM} \left(2b, \; \frac {1}{2} \right) \right]$

Which holds true by AM-GM.[/hide]


What are all the solutoins $(x,y,z)$ that satisfy $\frac1x+\frac1y+\frac1z=\frac13$ and $x\geq{y}\geq{z}$

z cant be 2 or 3.  We also have $z \leq 9$.

What follows is the outline of some nasty bruteforcing.

---

We try z=4,5,6,7,8,9.  We get that $1/x + 1/y$ has to be equal to 1/12, 2/15, 1/6, 4/21, 5/24, 2/9 respectively.

For each of the cases we have $y \le$ 24, 15, 12, 12, 10, 9.

Now we can brute force it.  The solution for y gives immediately the solution for x.

1/12 (z=4) case:

(y-12) divides 12y, so that y = 13, 14, 15, 16, 18, 20, 21, 24.

z=5 case;

(2y - 15) divides 15y, so that y = 8, 9, 10, 12, 15.

z=6 case;

(y-6) divides 6y, so that y = 7,8,9,10,12

z = 7 case;

(4y - 21) divides 21y, so that y = 6, 7, 12.

z = 8  case;

5y - 24 divides 24 y, so that y = 5, 6, 8

z= 9  case;  y = 9.

you mean integer solutions right? also can they be negative?
i dont really know how to solve this problem, but i think that i saw some earlier topic that used "Vieta's Formula's" to solve a similar problem

"Brutforcing" is not much thinking, and cleverness as writing cases and computing

If $x,y,z$ are not integers I think that this can't be easily solved (if it can be solved at all)

[quote="silouan"]yes but what we do at brutforcing?I don't know what is brutforcing. Please someone explain it to me.[/quote]
ALso call "Brute & force", its mean expand everything to get the answer. or in number theory you try some/all values which satisfy the equation.(like singular's work).

Brutforcing is only sleng, that's not some specific way of solving problems, some theorem or I don't know what....
That is only expression when you solve problem by dividing it in many cases because there is no other way witch leads to shorter and more elegant solution...

if you know how to do it with only 2 fractions, three is just one step more

$\frac{xy+xz+yz}{xyz}=\frac{1}{3} \Rightarrow xyz - 3xy-3xz-3yz=0$

$(x-3)(y-3)(z-3) = 9(x+y+z-3)$ is the best factorization I can get.

Now the only thing I can think of is some nasty casework. :ninja:

close.

You know how to solve
$\frac1x+\frac1y=\frac{a}b$

just make $\frac13-\frac1z$ the new $\frac{a}b$ and solve for $x$ and $y$ with case work for $z$


Let $a, b,c>0$ such that $ab+bc+ca=3abc.$ Prove that $a+b+c\geq 2abc+1.$ ;)

ooops, sorry! :oops: . I will check again my calculations. :blush:


Let $p$ be prime. Prove that there is infinitely much $n$ which satisfies $p \mid (2^n-n)$

I want to propose a quite complicated solution (of course, this problem also has a very simple one).
Define $n_0=2$, $n_{k+1}=2^{n_k}$, $k\geq 0$. Then for all sufficiently large $k$ we have $p\mid 2^{n_k}-n_k$ ;)

It is very simple. I wanted $n=1\pmod{p}$ and $2^n=1\pmod{p}$, hence the result :)


induction
strong induction
algebra unsolved
algebra

Two real sequence $ \{x_{n}\}$ and $ \{y_{n}\}$ satisfies following recurrence formula;

$ x_{0}\equal{} 1$, $ y_{0}\equal{} 2007$
$ x_{n\plus{}1}\equal{} x_{n}\minus{}(x_{n}y_{n}\plus{}x_{n\plus{}1}y_{n\plus{}1}\minus{}2)(y_{n}\plus{}y_{n\plus{}1})$,
$ y_{n\plus{}1}\equal{} y_{n}\minus{}(x_{n}y_{n}\plus{}x_{n\plus{}1}y_{n\plus{}1}\minus{}2)(x_{n}\plus{}x_{n\plus{}1})$

Then show that for all nonnegative integer $ n$, $ {x_{n}}^{2}\leq 2007$.

[hide="Possible start?"]We have $ \frac{y_{n}\minus{}y_{n\plus{}1}}{x_{n}\plus{}x_{n\plus{}1}}\equal{}\frac{x_{n}\minus{}x_{n\plus{}1}}{y_{n}\plus{}y_{n\plus{}1}}$
Which implies $ x_{n\plus{}1}^{2}\equal{}x_{n}^{2}\plus{}y_{n\plus{}1}^{2}\minus{}y_{n}^{2}$. Now induction may work but I really don't have the time to check. :blush: [/hide]

hmm
i think it's hard to get the implicit formula for $ x_{n}$ directly
after applying strong induction, then we can restrict the variables like this
Let $ a \equal{} 2007$ then
$ 0\leq{x_{n}}^{2}\leq a$
$ a^{2}\minus{}1\leq{y_{n}}^{2}\leq a^{2}\plus{}a\minus{}1$
(left side holds for all $ n$, whereas right side does not)

this might be a possible approach

[quote="kunny"]I got it! :)[/quote]

could you please post your solution?
i'm a high school student taken this exam
and no one of me and my friends got this problem

[quote="aviateurpilot"]$ x_{n}^{2} \equal{} y_{n}^{2} \minus{} 2007^{2} \plus{} 1$[/quote]

Well, can you show how you got this, and what to do next?  :)

I heard only very few people solved this in the test.... :|

Well, I still can't get the solution (one-line) posted. It would be great if anybody post a full one.

Well, it's quite clear the solution above is not quite understandable. Maybe we should look for a new one...


trigonometry integration function calculus derivative limit logarithms

if $ tan z$ is expanded as a laurent expansion about $ z\equal{}\frac{\pi}{2}$ what wll be the principal part?

$ \tan x\equal{}\frac{\sin z}{\cos z}.$ The sine is $ 1$ at $ \frac{\pi}2.$ The cosine has a zero there. Is it a simple zero? Because if it is, then the tangent will have a simple pole. How can you tell whether a zero is simple?

POLES of $ f(z)$ are given by $ cos(z) \equal{} 0$
so $ z \equal{} 2n\pi \pm \pi/2$   where $ n \equal{} 0,1,2,3...$
$ n \equal{} 0$ implies $ z \equal{} \pm \pi/2$
that implies $ z \equal{} \pi/2$ and $ z \equal{} \minus{} \pi/2$ are simple poles

therfore laurent expansion of $ f(z)$ is
$ f(z) \equal{} \sum_{0}^{\infty}a_n (z \minus{} \pi/2)^n \plus{} \sum_{1}^{\infty}b_n(z \minus{} \pi/2)^{ \minus{} n}$
where $ b_n \equal{} \frac {1}{2\pi i} \int_c \frac {f(z)}{(z \minus{} \pi/2)^{ \minus{} n \plus{} 1}}dz$
so $ b_1 \equal{} \frac {1}{2\pi i} \int_c\frac {tanz}{(z \minus{} \pi/2)^{ \minus{} 1 \plus{} 1}}dz$
 $ b_1 \equal{} \frac {1}{2\pi i}[ \minus{} log cosz]$
at $ \pi/2$ $ b_1\equal{}\minus{}\frac{1}{2\pi i}$
is it right?

[quote="JAS"]that implies $ z \equal{} \pi/2$ and $ z \equal{} \minus{} \pi/2$ are simple poles[/quote]
What was that that's doing the implying?

OK, tell me: are the poles of $ g(z) \equal{} \frac1{1 \minus{} \sin z}$ simple or not?

no
$ z \equal{} \pi/2$is simple pole of $ tanz$ not of $ cosz$
poles of $ \frac{1}{1\minus{}sinz}$ are given by $ 1\minus{}sinz\equal{}0$
$ sinz\equal{}1$ so $ z\equal{}n\pi\plus{}\pi/2$ where$ n\equal{}0,1,2,3...$
so$ z\equal{}\pi/2$ is simple pole
i have confusion
please clear it

I was just checking to see whether you understood the meaning of the word "simple" in the expression "simple root" and "simple pole". And you don't.

The function $ \frac1{1\minus{}\sin z}$ has no simple poles. It has poles of order 2 at $ z\equal{}2n\pi\plus{}\frac{\pi}2.$ The word "simple" means "of order 1", and this function doesn't have any of those.

And that means that what you wrote in your post 3 of this topic is just complete nonsense. All of it. Including that you were throwing around some integral formula with no understanding either of its purpose or its values.

Yes, it is true that $ \tan z$ has a simple pole at $ z\equal{}\frac{\pi}2.$ That's because $ \cos z$ has a simple root there. $ \cos\frac{\pi}2\equal{}0$ but the derivative there is $ \cos'\frac{\pi}2\equal{}\minus{}1.$ So near $ \frac{\pi}2,$ $ \cos z$ looks like $ \minus{}\left(z\minus{}\frac{\pi}2\right).$ Since $ \sin z$ looks like $ 1$ in an neighborhood of the same point, we have by simple division that the leading term of the expansion for $ \tan z$ is $ \frac{\minus{}1}{z\minus{}\frac{\pi}2}.$ There's your principal part of a Laurant expansion.

You need to slow down. The path to understanding does not lie through asking 20 or 40 questions - and I assure you that if you post 20 more questions, 19 of them will be completely ignored. And don't throw around every scrap of vocabulary you've ever heard and every formula you've ever seen. Try to understand one concept at a time, and don't post anything new until you've understood the last thing you dealt with.

I know simple pole means of order 1
for example $ \frac {1}{z \minus{} 2}$ has a simple pole at $ z \equal{} 2$
in case $ \frac {1}{1 \minus{} sinz}$
$ z \equal{} 2n\pi \plus{} \pi/2$ ok
at $ n \equal{} 0$ $ z \equal{} \pi/2$
at $ n \equal{} 1$ $ z \equal{} 5\pi/2$
at $ n \equal{} 2$ $ z \equal{} 9\pi/2$
at $ n \equal{} \minus{} 1$ $ z \equal{} \minus{} 3\pi/2$
and so on
how can we say pole of order 2 at $ z \equal{} \pi/2$
and what had you applied for finding principal part
can we use laurent expansion or not?

please give me some more exercises and clear my concept. i have discussed with all my collegues alll of tham said that $ \frac{1}{1\minus{}sinz}$ has simple pole at $ \pi/2$

Then I think you need some new colleagues. Or they said something else, like "isolated singularity" and you heard it as "simple pole." Or they just said "pole" and you added the word "simple."

Let's start by looking at the analytic function $ f(z)\equal{}1\minus{}\sin z$ about $ z\equal{}\frac{\pi}2.$ We have $ f\left(\frac{\pi}2\right)\equal{}0,f'\left(\frac{\pi}2\right)\equal{}0,$ $ f''\left(\frac{\pi}2\right)\equal{}1,$ and $ f'''\left(\frac{\pi}2\right)\equal{}0.$ From this, we have the Taylor expansion:
\[ 1\minus{}\sin z\equal{}\frac12\left(z\minus{}\frac{\pi}2\right)^2\plus{}O\left(z\minus{}\frac{\pi}2\right)^4\]
Now, do long division of power series:
\[ \frac1{1\minus{}\sin z}\equal{}\frac{2}{\left(z\minus{}\frac{\pi}2\right)^2}\plus{}O(1)\]
as $ z\to\frac{\pi}2.$

That's a pole of order 2. (With residue zero, as it turns out.)

We can also say that a point $ z \equal{} z_0$ is a pole of order $ n$ for a function $ f$ if $ f$ can be written int the form $ f(z)\equal{}\frac{\psi(z)}{(z\minus{}z_0)^n}$ where $ \psi(z_0) \neq 0$. Also, if $ \frac{1}{f(z_0)} \equal{} 0$, then the order of the pole is the smallest integer $ n$ such that $ \lim_{z \to z_0}\left| f(z) (z\minus{}z_0)^n \right|< \infty$.

thanks a lot.
its clear to me pole of $ \frac {1}{1 \minus{} sinz}$
now principal part of $ tan z$
cann't we use the laurent expansion as in post 3
and what had you applied for finding principal part.

i have studied the book goyal and gupta
in that book no example is solved as you have solved
and in schaum series there are only problems examples which have been solved are easy and there is no hint to solve tricky questions of exercise
can you suggest me any other book.

[quote="JAS"] $ b_n \equal{} \frac {1}{2\pi i} \int_c \frac {f(z)}{(z \minus{} \pi/2)^{ \minus{} n \plus{} 1}}dz$
so $ b_1 \equal{} \frac {1}{2\pi i} \int_c\frac {tanz}{(z \minus{} \pi/2)^{ \minus{} 1 \plus{} 1}}dz$
 $ b_1 \equal{} \frac {1}{2\pi i}[ \minus{} log cosz]$
at $ \pi/2$ $ b_1 \equal{} \minus{} \frac {1}{2\pi i}$
is it right?[/quote]
This part is utter nonsense. You're trying to say

$ b_1 \equal{} \frac1{2\pi i}\int_C\tan z\,dz.$

But note that that's a definite integral, a line integral. What's $ C?$ $ C$ is a closed curve that winds once counterclockwise around $ z \equal{} \frac {\pi}2,$ without catching any of the other poles inside of it.

What does $ \minus{} \log\cos z$ have to do with this? That's a possible expression for an antiderivative of $ \tan z.$ It's certainly not itself the value of the line integral in question. But wait ... $ \minus{} \log\cos z$ isn't even well defined! The logarithm has branches (it's multiple valued). And you can't choose a branch consistently if you wind all the way around a branch point. Guess what? $ z \equal{} \frac {\pi}2$ is a branch point for $ \minus{} \log\cos z.$

To be honest, the easiest way to evaluate that integral is to [b]already know[/b] the value of $ b_1$ from a separate, independent computation.

To understand Laurent series, understand power series. Power series are fundamental. Many computations concerning Laurent series can be most easily accessed via algebraic manipulations of power series.

Example of a power series computation turned in to a Laurent series computation: I would like to know the residue at $ 0$ of $ f(z) \equal{} \frac1{z^2\sin z}.$ (It's possible I need that for a computation of $ \zeta(2).$)

$ \sin z$ has a simple zero at $ 0$ but that means $ z^2\sin z$ has a zero of order 3 and hence $ \frac1{z^2\sin z}$ has a pole of order 3.

I do know that $ \sin z \equal{} z \minus{} \frac {z^3}6 \plus{} O(z^5),$ from which $ z^2\sin z \equal{} z^3 \minus{} \frac {z^5}6 \plus{} O(z^7).$

Now I put that into a straightforward long division algorithm. (Not shown here, but you should be able to reproduce it.) I get
\[ \frac1{z^2\sin z} \equal{} z^{ \minus{} 3} \plus{} \frac {z^{ \minus{} 1}}6 \plus{} O(z^1)\]
Note that that shows the whole principal part of the Laurent expansion about zero, from which we can read off that the residue is $ \frac16.$


Let $ u(x),\ v(x)$ be the condition with respect to variable $ x$ respectively. Let denote $ [u(x)\Longrightarrow v(x)]$ the proposition : $ v(x)$ holds for any $ x$ being $ u(x)$. Now suppose that  $ x$ be variable whose range is all real numbers and $ a$ be real constant number. Let $ p(x),\ q(x),\ r(x)$ be conditions such that $ p(x) : x > 1,\ q(x) : x > 1 \plus{} \frac {a}{2},\ r(x) : (x \minus{} a)(x \plus{} a) > 0$.
Find the range of $ a$ such that the proposition : $ [(p(x)\Longrightarrow r(x))\ or\ (q(x)\Longrightarrow r(x))]$ is true.

So for the proposition to be true, any value of x that holds for p(x) also holds for r(x). Since p(x) holds for x>1, and r(x) holds for x^2>a^2, then r(x) holds for any x>1, so a has to be 1 in that case. In order for the second part of the proposition to be true, any x that belongs to r(x) must also be part of q(x), so 1 + a/2 = x = sqrt(a^2). Solving this equation gives us a= -2/3, 2.

So a=, 1, 2, -2/3

is this right?


Prove tthat for any positive integers a, b : |a:sqrt:2 - b|(a + b) > 1/2.

A try for this one :

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|a:rt2: - b|*(a:rt2: + b) = |2a-b| >= 1 because :rt2: is irrational.

So is enough to prove (a+b)/(a:rt2: + b) > 1/2 which is clearly true  :mrgreen: [/hide]

Well done belenos, quick and elegant solution

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When I see a spoiler I can't resist looking in it instead of searching. But I'm sure I'm alone in this case     or am I?    [/hide]