How many ppl qualify for USAMO and do I have a chance?

I scored a 124 on the AMC 10 and got 8 right on the Alternate AIME.

AMC score means nothing almost, except that it allows you to qualify for AIME. Your 8 on the AIME Alter. might qualify you for the USAMO depending on what the floor score is. Every year around 250 people qualify for the USAMO.

The easiest way to qualify for USAMO is to move to Wyoming. ;)

lol on the Wyoming :) DPopov, I totally disagree.  If you're in 11th or 12th grade, your AMC score means a lot, because your USAMO index is 10 times the AIME score, plus the SMC score.  So even if you got an 11 on the AIME I this year, if you got only a 120 on the AMC, then you wouldn't qualify.  On the other hand, if you're in 10th grade or below, then I agree :)

But spt10v is in 10th grade or under, so it doesn't matter for him.


a,b,c \geq 1
Prove that:

 \sqrt (a-1)+ \sqrt (b-1)+ \sqrt (c-1) \leq  \sqrt (c(ab+1))

First of all  I perform the substitutions a-1 = x^2 , b-1 = y^2 , c-1 = z^2
After few calculation, the relationship becomes (x+y+z)^2  =< (1+z^2)*(1+(1+x^2)*(1+y^2))
It's easy to show (brute calculation) (a+b)^2 =< (1+a^2)*(1+b^2)  for all a,b real ,  so
(x+y+z)^2 =< (1+z^2)*(1+(x+y)^2) =< (1+z^2)*(1+(1+x^2)*(1+y^2)) (q.e.d)
Diogene


[quote="ifai"]Suppose a+b=1 where a and b are positive numbers.
(i) Show that $ab\leq\frac14$.
(ii) Find the minimum value of $\left(a+\frac1a\right)\left(b+\frac1b\right)$.[/quote]
[hide="(i)"]By AM-GM,
$\frac{a+b}{2} \ge \sqrt{ab}$
$\frac{1}{2} \ge \sqrt{ab}$
$\frac{1}{4} \ge ab$
$ab \le \frac{1}{4}$[/hide]
[hide="(ii)"]Expanding the second expression, we have:
$ab+ \frac{1}{ab} + \frac{a}{b} + \frac{b}{a}$
For the expression to be of minimum value, $a=b=\frac{1}{2}$
So we are left with $\frac{1}{4}+ 4 + 1 + 1 = \frac{25}{4}$
So the minimum value of $(a+\frac{1}{a})(b+\frac{1}{b}) = \frac{25}{4}$[/hide]

[quote="ifai"]Suppose a+b=1 where a and b are positive numbers.
(i) Show that $ab\leq\frac14$.
(ii) Find the minimum value of $\left(a+\frac1a\right)\left(b+\frac1b\right)$.[/quote]

(i) [hide]
By AM-GM, we have \[ \frac{a+b}{2} \ge \sqrt[2]{ab},  \] so \[ ab \le \left( \frac{1}{2} \right) ^2 = \frac{1}{4},  \] and we are done.
[/hide]

(ii)

[hide]We need to minimize \[ \left( \frac{a^2+1}{a} \right) \left( \frac{b^2+1}{b} \right) = \frac{(a^2+1)(b^2+1)}{ab}.  \] The maximum of the denomenator is $\frac{1}{4}$, so we need to minimize \[ 4(a^2+1)(b^2+1). \]

I have to go...
[/hide]

Continueing on Chess64's work...
[hide]
Minimization occurs when $a=b=\frac{1}{2}$, so we have $4\left(\frac{1}{4}+1\right)\left(\frac{1}{4}+1\right)=4\left(\frac{5}{4}\right)^2=\frac{25}{4}$. :)
[/hide]

Masoud Zargar

Sad, most of you didn't prove the fact that, in (ii) we must have $a = b = 1/2$ to get the minimum value. Simple proofs do exist though. (Scorpius119 is the only one who tried to prove this.)

I have just derived something involving the minimization.
We know that
$\\\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\\=\left(\frac{a^2+1}{a}\right)\left(\frac{b^2+1}{b}\right)\\=4(a^2+1)(b^2+1)$
Let $a_1=a$, $a_2=b$, and $b_1=b_2=1$. If we use Cauchy-Schwarz, we get
$\left(\sum_{i=1}^na_ib_i\right)^2\le\left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)\\(a+b)^2=a^2+b^2+2ab=a^2+b^2+\frac{1}{2}\le 2(a^2+b^2)\\\implies\frac{a^2+b^2+\frac{1}{2}}{a^2+b^2}=1+\frac{1}{2(a^2+b^2)}\le 2\\\implies 2(a^2+b^2)=1\implies a^2+b^2=\frac{1}{2}\\\implies (a+b)^2\le 2(0.5)=1\\\implies a^2+b^2+2ab=1\le 1$.
Minimization occurs when $a=b=\frac{1}{2}$.  This may not be a direct proof of the minimization, but it seems to me that it is the right way.  Anyone have an idea. :(

EDIT: Nope, this is messed up. :(
Masoud Zargar

[hide="prompting"]Try to think of the following equation: $a^2+b^2=1-2ab$. [/hide]

${(a+\frac 1a})(b+\frac 1b)=\frac{(a^2+1)(b^2+1)}{ab}=\frac{(a^2+\frac 14+\frac 34)(b^2+\frac 14+\frac 34)}{ab}$

By AM-GM, $a^2+\frac 14\geq \frac a2$, and similarly for $b$

So we get ${(a+\frac 1a})(b+\frac 1b)\geq \frac{(\frac a2+\frac 34)(\frac b2+\frac 34)}{ab}$

It shouldn't be too hard to finish from here...

[quote="ifai"]Suppose a+b=1 where a and b are positive numbers.
(i) Show that $ab\leq\frac14$.
(ii) Find the minimum value of $\left(a+\frac1a\right)\left(b+\frac1b\right)$.[/quote]

(ii)

[hide="not completely rigorous but oh well"]Expanding, we find \[ ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}.  \] Since the maximum of $ab$ is $\frac{1}{4}$, the minimum of $\frac{1}{ab}$ is $4$. Now we have \[ \frac{17}{4} + \frac{a}{b} + \frac{b}{a} . \] AM-GM on $\frac{a}{b} + \frac{b}{a}$ yields \[ \frac{a}{b} + \frac{b}{a} \ge 2\sqrt{1} \implies \frac{a}{b} + \frac{b}{a} \ge 2,  \] so the minimum is \[ \frac{25}{4}. \][/hide]

What you did is no better than any of the other incorrect posts on this topic, because you fail to justify

$ab+\frac 1{ab}\geq \frac{17}{4}$

[quote="ifai"]Suppose a+b=1 where a and b are positive numbers.
(i) Show that $ab\leq\frac14$.
(ii) Find the minimum value of $\left(a+\frac1a\right)\left(b+\frac1b\right)$.[/quote]

(i) $ab\leq \left(\frac{a+b}{2}\right)^2=\frac{1}{4}.$

(ii) $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)=ab+\frac{1}{ab}+\frac{b}{a}+\frac{a}{b}$

$\geq ab+\frac{1}{ab}+2\sqrt{\frac{b}{a}\cdot \frac{a}{b}}=ab+\frac{1}{ab}+2\geq \frac{25}{4}.$

The equality holds when $a=b$ and $ab=\frac{1}{4}$ and $a>0,b>0\Longleftrightarrow a=b=\frac{1}{2}.$

Alternative solution for (ii)

$\left(a+\frac{1}{a}\right)\left(b+\frac1b\right)=\left\{(\sqrt{a})^2+\left(\frac{1}{\sqrt{a}}\right)^2\right\}\left\{(\sqrt{b})^2+\left(\frac{1}{\sqrt{b}}\right)^2\right\}$

$\geq \left(\sqrt{ab}+\frac{1}{\sqrt{ab}}\right)^2=\frac{25}{4}.$

Most people just can get 25/4 but can't give the proof.

Why? Can't you understand the behaviour of the graph of $y=x+\frac{1}{x}$ at $0<x<1?$  ;)


calculus integration LaTeX rotation derivative trigonometry

hey everyone...  i'm new to using latex, and I'm really getting used to it, however, i cant seem to find the command that will let me type a "]" during definite integration, that will let me put a superscript and subscript to notate my upper and lower bounds...  if anyone who knows what im talking about can help me out, I'd appreciate it, thanks...

You mean $\int^{b}_{a}f(x) \ dx$? This is done with the following code
[code]$\int^b_a f(x) \ dx$[/code]

no...  i am well aware how to do that...  ok after you find the anti-derivitive of f(x), the next step is to express F(x) enclosed with "[ ]" where the right side "]" has b and a as superscripts and subscripts.  it's essentially a definite integration symbol using $]$ instead of $\int$

You use delimiters.  TeX/LaTeX has the commands \left and \right which when applied before a symbol, scales the symbol so that its vertical size matches the contents in between.  So for example,

\left( \sum_{k=1}^n \cot^{-1}(k) \right)^2

looks like this:

$\left( \sum_{k=1}^{n}\cot^{-1}(k) \right)^{2}$

But since TeX/LaTeX uses the pairing of \left and \right to determine what goes "in between," these commands must always be used together.  So if you have notation where you need only one delimiter, you have to bracket it with an "empty" or "phantom" delimiter.  For example

\left. \frac{x^2}{2} - 5x \cos x \right]_{x=0}^5

looks like this:

$\left. \frac{x^{2}}{2}-5x \cos x \right]_{x=0}^{5}$

the special command \left. with the period tells TeX where the left-hand phantom delimiter should be placed, and therefore gives the correct height of the right delimiter (so that it is at least as tall as the fraction.  If you had instead typeset

\frac{x^2}{2} - \left. 5x \cos x \right]_{x=0}^5

you would see

$\frac{x^{2}}{2}-\left. 5x \cos x \right]_{x=0}^{5}$

because none of the expressions in between the \left. and \right] are taller than the number 5.

Also, look at this: $\left.\cos x\vphantom{\frac{x}{x}}\right|_{x=1}^{x=2}$ ;) Another way to do the same would be to declare the size of your delimeter explicitly: $\Bigl.\cos x\Bigr|_{x=1}^{x=2}$

\vphantom? $This\ is\ \vphantom{hidden}.$

Nice command! :D

The command \vphantom creates a box whose width is zero and height is equal to the contents following the command.  So for example,

\sqrt{\vphantom{x^2}1+x}

$\sqrt{\vphantom{x^{2}}1+x}$

produces output that has slightly more room inside the square root than the expression

\sqrt{1+x}

$\sqrt{1+x}$

The reason why you might do such a thing is if you had an expression like

\sqrt{x^2} + \sqrt{x+1}

$\sqrt{x^{2}}+\sqrt{x+1}$

which, when typeset, doesn't look so nice because the tops of the square roots are not horizontally aligned.  You could use either \mathstrut or \vphantom to correct this (in this case, \vphantom works better).

Similarly, there is a command called \hphantom which creates a box whose width is equal to the contents following the command, but has height and depth zero.  Finally, there is the command \phantom which creates a box whose dimensions are exactly that of its contents, but the contents don't get printed.  An example where you could use this is if you wanted to write

The square root operator \sqrt{\phantom{x}}

The square root operator $\sqrt{\phantom{x}}$.

Otherwise, you get $\sqrt{}$ which looks funny.  There's your TeX lesson for today.


inequalities
search
function
inequalities unsolved

[quote="tunganh"]$a,b,c \in R$. Prove that:
$a^{2}+b^{2}+c^{2}\geq \sqrt{3(a^{3}b+b^{3}c+c^{3}a)}$[/quote]
See here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6026[/url].

As for the second inequality ,because a,b,c are the sides of a triangle we have: $a=x+y$,$b=y+z$ and $c=x+z$ ,where $x,y,z>0$.So using those notations the given inequality is equivalent to:
\[\sum_{cyc}(y+z)^{2}(z+x)(y-x)\geq0\leftrightarrow\sum_{cyc}(y^{2}+2yz+z^{2})(yz+xy-x^{2}-xz)\geq 0\] $\leftrightarrow$ \[\sum_{cyc}(y^{3}z+y^{2}xy-y^{2}x^{2}-y^{2}xz+2y^{2}z^{2}+2y^{2}zx-2yzx^{2}-2yz^{2}x+z^{3}y+z^{2}xy-z^{2}x^{2}-z^{3}x)\geq 0\] $\leftrightarrow$ \[\sum_{cyc}xy^{3}\geq\sum_{cyc}x^{2}yz\] $\leftrightarrow$ $\sum_{cyc}\frac{xy^{3}}{xyz}\geq \sum_{cyc}x$ $\leftrightarrow$ $\sum_{cyc}\frac{y^{2}}{z}\geq \sum_{cyc}x$ which is true by Cauchy, and we are done.

[quote="Svejk"]As for the second inequality ,because a,b,c are the sides of a triangle we have: $a=x+y$,$b=y+z$ and $c=x+z$ ,where $x,y,z>0$.So using those notations the given inequality is equivalent to:
\[\sum_{cyc}(y+z)^{2}(z+x)(y-x)\geq0\leftrightarrow\sum_{cyc}(y^{2}+2yz+z^{2})(yz+xy-x^{2}-xz)\geq 0 \]
$\leftrightarrow$
\[\sum_{cyc}(y^{3}z+y^{2}xy-y^{2}x^{2}-y^{2}xz+2y^{2}z^{2}+2y^{2}zx-2yzx^{2}-2yz^{2}x+z^{3}y+z^{2}xy-z^{2}x^{2}-z^{3}x)\geq 0 \]
$\leftrightarrow$
\[\sum_{cyc}xy^{3}\geq\sum_{cyc}x^{2}yz \]
$\leftrightarrow$ $\sum_{cyc}\frac{xy^{3}}{xyz}\geq \sum_{cyc}x$ $\leftrightarrow$ $\sum_{cyc}\frac{y^{2}}{z}\geq \sum_{cyc}x$ which is true by Cauchy, and we are done.[/quote]
Do you have other solutions for this inequality?

[quote="N.T.TUAN"][quote="tunganh"]Another inequality:
$a,b,c$ are three sides of a triangle. Prove that:
$T= \sum a^{2}b(a-b) \geq 0$[/quote]
This is [url=http://www.kalva.demon.co.uk/imo/isoln/isoln836.html]IMO 1983/B3[/url]
search, please! :D[/quote]
But the solution here is the same as Svejk's solution. Do you have other solutions?

I used search function and have http://www.mathlinks.ro/Forum/viewtopic.php?p=366624&sid=edff0df08e893dfe79c45897ce514418#p366624


algebra function domain conics parabola geometry rectangle

Let $f(x) = {\sqrt{ax^2+bx}}$. For how many real values of a, is there atleast one positive value of b for which the domain of f and the range of f are the same set.

a) 1   b) 2    c) 3   d) 0    e) infinitely many

This is interesting...

[hide]Clearly, $a = 0$ is a value that works.

Since the range is always $[0, \infty)$, so is the domain in this case.

We know that $ax^2 + bx$ is a parabola with x-intercepts at $0$ and $-\frac{b}{a}$.

But we want $a$ to be negative because if the parabola is facing up, then we have a clash of domain and range because the range will include negatives...

And so we are looking at the part of the parabola bound by the intercepts such that the maximum to $y = 0$ is $-\frac{b}{a}$.

We know the value of $x$ which will allow this is $-\frac{b}{2a}$, so we plug that into $f(x)$ and set $f(-\frac{b}{2a}) = -\frac{b}{a}$, setting domain equal to range, and we get $a = -4$ as the only other value for which the domain and range are the same.

And $B$.
[/hide]

[quote="vidyamanohar"][quote="white_horse_king88"]This is interesting...

Clearly, $a = 0$ is a value that works.

Since the range is always $[0, \infty)$, so is the domain in this case.

[/quote]

Is it necessary that the range should be always $[0, \infty)$.

Can't we have a situation like, a part of the parabola whose square root function curve is joining the two opposite vertices of a rectangle y=m, y=n and x=m, x=n making the domain and range as [m, n]?

Will it be possible to prove or disprove this idea wrong?
[/quote]

Sorry for the confusion, I meant for $a=0$, the range is such. Clearly, the range is not always $[0, \infty)$, as shown by the second solution.

[hide]
Hmm, Jarey...choice C was the correct answer on the AMC where this appeared.  But in this post the answer should be B since the choices were switched around for some reason...[/hide]


How many positive two-digit integers are there in which each of the two digits is prime?

There are $ 4$ prime digits: $ 2$, $ 3$, $ 5$, and $ 7$.  Thus, there are $ 4$ choices for the first prime digit, and $ 4$ choices for the second prime digit.  Multiplying yields $ 4 \times 4 \equal{} \boxed{16}$.


My friend asked me about a hard problem in inequalities:
[quote]a,b,c >0.
$X=\sum{\frac{ba^2}{c}}$
$Y=\sum{\frac{ca^2}{a}}$
Find  u  min and  0 <= u  <= 1 to make this inequality true:
(1- u )X +  u. Y  <=  $(\sum{a^2})$[/quote]

Maybe $Y=\sum{\frac{ca^2}{b}}$ and$(1-u)X+uY\geq \sum{a^2}$?

No! Because It's not nice and interesting ...


function
calculus
integration
real analysis
real analysis unsolved

Give an example of bounded continuous function f on (0, +inf), such that
f(x)->0 as x->+inf. But f not in L^p(0, +inf) for any p>0.

(1/x almost works, but 1/x is not bounded).

$\frac{1}{x}$ not being bounded is not a big issue, since you can shift its singular point to the left, as in $\frac{1}{x+1}$. However, this function is in $L^{p}(0,\infty)$ for $p>1$. You want $f$ to go to $0$ VERY slowly.

Try $f(x)=\frac1{\log(x+2)}$. That 2 could be replaced by any real number greater than 1, of course.

The integral of $f^{p}$ diverges by limit comparison to $\frac1x$ for all $p$.


Title says it all. Decide whether or not you want to consider yourself Asian.

Asians: +
Non-Asians: -

(please don't take the +/- as a racist statement  :wink: )

rules same as before, no double posting, blah blah blah.

+1

Non Asian ftw. 0. starting now b/c of survivor thing?

Aren't you not allowed to post twice in an hour? So yeah go back to the 0 before.

Aw...let's see someone not asian besides me post here...

Why do you guys argue about this??

Does the race really matter??
Well, I am Asian, but why would you want to have a tug of war with race?
I mean we are same and created equal..

I am going to go with Neutral.

This isn't a competition to see what race is better, it is a friendly competition in which teams are easy to make. Also we can get sort of an estimate as to how many Asians vs. non-Asians there are on the board. If this is too racially insensitive of a game, a mod can delete it (I personally don't think it is, but that's not for me to decide).

0

[hide="uhh...what comes after 129 again?"] [hide="hold on...im starting from one again. this might take a while"]1,2,3,4... [hide="just kidding"]130[/hide] [/hide][/hide]

this isn't a game anymore so much as you asians being irritating...

ah..
whatev...
i'm not a twinkie..
i'm a fob...

(if you are korean, you may understand this, if you don't, pm me about it)

anyway..
131..
gaspness
a palindrome!!!

kind of like pop, race car, and go hang a salami, I'm a lasagna hog :|

if we already posted, are we alowed to post again?

[size=167][b]一百三十二[/b][/size]

[hide="BEAT THAT SUCKERS!!!!"]JK :D [/hide]

Oh my god somebody lock/delete this thread now.

132

Is it just me or is 1=2 the only Non-Asian?

the non asians have gone to do non asian things. come on 1=2, what are you still doing here?


inequalities algebra function domain calculus derivative

Let $ a,b,c>0, a\plus{}b\plus{}c\equal{}1$. Prove that
$ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1) \ge 8$

[quote="NguyenDungTN"]Let $ a,b,c > 0, a \plus{} b \plus{} c \equal{} 1$. Prove that
$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \ge 8$[/quote]

First, I think the condition you mean is $ a \plus{} b \plus{} c \equal{} 3$.  Second, even in this case, the problem is not true.  For example, if $ a \equal{} 2$ and $ b \equal{} c \equal{}1/2$, then
\[ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \equal{} 5 \cdot \frac{5}{4} \cdot \frac{5}{4} \equal{} \frac{125}{16} < 8.\]

[quote="nsato"][quote="NguyenDungTN"]Let $ a,b,c > 0, a \plus{} b \plus{} c \equal{} 3$. Prove that
$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \ge 8$[/quote]

First, I think the condition you mean is $ a \plus{} b \plus{} c \equal{} 3$.  Second, even in this case, the problem is not true.  For example, if $ a \equal{} 2$ and $ b \equal{} c \equal{} 1/2$, then
\[ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \equal{} 5 \cdot \frac {5}{4} \cdot \frac {5}{4} \equal{} \frac {125}{16} < 8.
\]
[/quote]

Thank you, nsato! But this inequality is true if $ ab\plus{}bc\plus{}ca\equal{}3$, problem from Agady.

$ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\ge \frac{125}{16}$ is true for $ a\plus{}b\plus{}c\equal{}3$, $ a,b,c\ge 0$.

The domain is a closed and bounded subset of $ R^3$, and both the functions and their partial derivatives are continuous. Grad g is not zero... So we can use lagrange multipliers...

It ends up that we need to solve $ \frac{a}{a^2\plus{}1}\equal{}\frac{b}{b^2\plus{}1}\equal{}\frac{c}{c^2\plus{}1}$ and $ a\plus{}b\plus{}c\equal{}3$.

Subtract one from another, and clear the denominator which is nonzero, $ (a\minus{}b)(ab\minus{}1)\equal{}0$.

i) if $ a\equal{}b\equal{}c$, then $ a\equal{}1$, giving $ 8$.

ii) if $ a\equal{}b$, $ a\ne c$, then $ a\equal{}b$, $ ac\equal{}bc\equal{}1$. Then $ c(a\plus{}b\plus{}c\minus{}3)\equal{}0\equal{}1\plus{}1\plus{}c^2\minus{}3c\equal{}(c\minus{}2)(c\minus{}1)$. If $ c\equal{}1$, $ a\equal{}b\equal{}c$, so $ c\equal{}2$. Then $ a\equal{}b\equal{}\frac{1}{2}$, which gives $ \frac{125}{16}$.

iii) if $ a\ne b$, $ b\ne c$, $ c\ne a$. $ ab\equal{}bc\equal{}ca\equal{}1$, so $ a\equal{}b\equal{}c\equal{}1$, contradiction.

Then we check the endpoints.

iv) If $ c\equal{}0$, Then $ (a\minus{}b)(ab\minus{}1)\equal{}0$, $ a\plus{}b\equal{}3$. This gives $ (a,b,c)\equal{}(1.5,1.5,0),(\frac{3\minus{}\sqrt{5}}{2},\frac{3\plus{}\sqrt{5}}{2},0)$. which give $ \frac{169}{16}$ and $ 9$, respectively.

v) If $ c\equal{}b\equal{}0$, then $ a\equal{}3$, giving $ 10$.

So the max and min are $ \frac{169}{16}$, and $ \frac{125}{16}$, respectively.


limit
function
algebra
domain
calculus
real analysis
real analysis theorems

Hi all, I'm new.. I hope someone can help me, I don't understand this... my prof touched on it briefly, and I read the book but still don't quite understand it.

Okay, why are the following both continous at a point?

lim          f(x) = x^2 - 9  =  0
x --> 3

and

lim          f(x) = | t - 2 |  =  1
x --> 3

So in both cases, if f(x) > 0 but <= 3, then it's continous? I know it's a simple way of trying to understand it, but I don't understand the defintion of continuity at a point. Can someone explain it to me, in the simplest/easiest way possible?

Thanks in advance...

Definition of continuity: Given $\epsilon>0$, there exists a $\delta>0$ such that whenever
\[ 0<|x-a|<\delta\Rightarrow |f(x)-f(a)|<\epsilon \]
then $f(x)$ is continuous at $x=a$.
Other versions I have seen are like:
$f(x)$ is continuous at $x=a$ if they meet the following conditions:
1) $\lim_{x\to a}f(x)$ exists
2) $\lim_{x\to a}f(x)=f(a)$

First, to see the intuitive meaning, think of continuity like a line that has no 'broken gaps', no holes, that the graph is a 1-piece curve. For example, the graph of $y=x$ is a straight slanted line that has no holes, no matter how much you zoom in the curve, you cannot find a hole. That's what continuity is about. How is continuity linked to limits? Limits talks about approaching a certain value but continuity talks about approaching a certain value in a nice smooth curve with no holes. This is somewhat what I think continuity means.

Second, let's see the not-so beautiful side of continuity, a function that is not continuous. For example, let $f(x)=\frac{x^2-1}{x-1}$. This function is not continuous at $x=1$ because it has a hole in it when you graph it. Why? Because you cannot divide by zero. If you try to graph it, it shows also a straight line similar to $y=x+1$ but with one exception: it has a hole at $x=1$. There are other kind's of discontinuities like the graph of $y=\frac1x$

Third, if you want to see how to work with $\delta-\epsilon$, I'll show you in a follow-up post. Hope you understand the intuitive meaning of continuity that way you can get more comfortable with it.

Hi, thanks for the reply. I think I get what you mean by continuity, graphically-speaking. But I'm not familiar with the delta/epsilon signs.. I guess I need to review on those ><
But I guess I'm trying to understand it numerically... so on the example you gave, since f(x) = 2 as it approaches 1, and f(1) = 2 and not 1, that's why it's not continuous? And on my examples from my first post, is the domain between 0 to 3..? So if it falls within that domain, then it's continuous?

The basic idea of the $\epsilon - \delta$ definition is this: a continuous function can't jump around too much in short spaces. In the definition of continuity, you can pick as small an $\epsilon$ as you want, so that the function is only allowed to vary by a tiny bit, and it is still possible to find a little interval around $x$ where the function doesn't change by more than $\epsilon$.

As 10000th user said, f a function jumps, then it obviously can't be continuous at that point. Can you think of any other ways that a function might not be continuous?

In his first example of a discontinuous function, the problem was that the function wasn't defined at $x = 1$.

In your example, the function is defined everywhere; it is a parabola. It is continuous because in a sense, the more you "zoom in" near $x=3$, the flatter the function will look. For example, in the interval $[3 - 1/1000, 3 + 1/1000]$, the function varies by about 3/250. Not a lot, eh? If you want to make this rigorous, you'd have to use the $\epsilon-\delta$ approach -- find $\delta$ in terms of $\epsilon$ so that inside the interval $[3 - \delta, 3 + \delta]$, the function varies by less than $\epsilon$.

Well, I think you have got the meaning of continuity in the previous posts.

    a) In post 2(a typo): delete from the defintion this : $0<|x-a|.....$ and write $|x-a|, \delta$

    b) You must imagine continuity like this
 << you have 10 balls which touch each other and through them on the space. If the keep touching  with the initial order ,  then the function of their position is continuous. Otherwise, not. >>

    c) The delta - epsilon definition of the limit is really difficult. I had finished the Uni and only after it I could -easily- solve problems with this way. The best way is  to see many solved problems in some good calculus books(spivak ect).

[u]    babis[/u]

Thanks guys...  I do understand what it is now, but it's just when solving numerical problems I don't see it unless I graph it...  the explanations were helpful, but I guess you all are thinking as mathematicians were I think as an average student trying to pass Calculus :D

[quote="stergiu"]
    c) The delta - epsilon definition of the limit is really difficult. I had finished the Uni and only after it I could -easily- solve problems with this way. The best way is  to see many solved problems in some good calculus books(spivak ect).
[/quote]

Yes, I haven't even learned this yet I think.. my prof didn't mention the delta-epsilon defn., all he said was that if "f" is defined on an open interval containing "a," then "f" is continous at "a"
if lim x --> a    f(x) = f(a)

So I guess I'm trying to understand what he meant by this numerically, if I solve  for f(x) and it's not equal to f(a) or if it doesn't fall within that domain, then it's continous... I don't even know if that's the correct way of thinking.. but that's how I understood it...

Anyway, thanks again guys.. this forum is very helpful!

The definition with the limit is actually the same -- guess why? It's because the limit is formally defined using epsilons and deltas in a similar way.  :P

Those proofs are difficult to learn. Don't feel discouraged because math is difficult. If your work is hard, it means you're learning from it (usually).


geometry
inequalities
triangle inequality

176. Show that if $h_{a}$, $h_{b}$ and $h_{c}$ are the altitudes of a triangle, then

$\frac{1}{h_{a}}< \frac{1}{h_{b}}+\frac{1}{h_{c}}$

194. A point is selected inside an equilateral triangle. From this point perpendiculars are dropped to each side. Show that the sum of the lengths of these perpendiculars is equal to the altitude length.

[hide="176"]
Let the triangle area be K
$a * h_{a}= 2K$
$b * h_{b}= 2K$
$c * h_{c}= 2K$

$\frac{1}{h_{a}}+\frac{1}{h_{b}}= \frac{a+b}{2K}> \frac{1}{h_{c}}= \frac{c}{2K}$

Since, $a+b > c$ by the triangle inequality
[/hide]

[hide="194"]Connecting the point to the vertices, we form 3 triangles (the point and two vertices for each). Let the altitude of the equilateral triangle be $h$, let the side length be $a$ and the altitudes of the three smaller triangles be $h_{1}$, $h_{2}$, and $h_{3}$. We can write the area of the equilateral triangle in two ways; in terms of $h$ and in terms of $h_{1}$, $h_{2}$, and $h_{3}$:

$A=\frac{a\cdot h}{2}=\frac{a(h_{1}+h_{2}+h_{3})}{2}$. Dividing out $\frac{a}{2}$ from both sides, we have the equality

$h_{1}+h_{2}+h_{3}=h$. This is what we desired and our proof is complete.

$\mathbb{QED}$[/hide]

[hide="176"]From Triangle Inequality, $a<b+c$. Area=A=$\frac{ah_{a}}{2}\implies \frac{2A}{h_{a}}<\frac{2A}{h_{b}}+\frac{2A}{h_{c}}\implies \frac{1}{h_{a}}<\frac{1}{h_{b}}+\frac{1}{h_{c}}$.[/hide]

[hide="194"]Let s be the sidelength, and $h_{a},h_{b},h_{c}$ be the perpendiculars. The Area of the equilateral=s*alititude/2=the areas of the sum of the individual triangles$=\frac{s(h_{a}+h_{b}+h_{c})}{2}\implies$ altitude$=h_{a}+h_{b}+h_{c}$.[/hide]


calculus
integration
calculus computations

The parts of this have appeared on this forum several times previously.


How do you do this?

Circle A and circle B have radii of 12 and 16 respectively. let points C and D be the points of intersection of the circles. If the length of arc CD of circle A is 4 pi, find the area of the region where the 2 circles overlap[/i]

Dude, u just posted that awhile ago.

How deep do the circles intersect each other? Like is circle B intersecting circle A a BIG chunk or a little chunk?  :D

Yea sorry about that I thought I had only clicked once but apparently I double-clicked. And I have no idea, I just typed the question word for word with all the information

[hide] 33pi - 36 root 3 - 36[/hide]

[quote="toadoncart"]How deep do the circles intersect each other? Like is circle B intersecting circle A a BIG chunk or a little chunk?[/quote]
You are given the length of arc $ CD$ on circle $ A$, which is enough to figure out the overlap.

o, i c. my geometry sux. :P   i think it might have something to do with the sectors.......


im stuck on a problem i got on a test recently, something that would probably have some beautiful simplicity to solve, but i can't see it. the problem is (# is the in integrand, b-a is the lower and upper limits, respectively, so b-a# is a definite integral)

0-6pi#cos(pi+x/12)^2

and i had to evaluate the integral using u substitution. it doesn't work though, not the ways i tried it...

im thinking u should equal some trig identity that isn't cosine...

or i don't know

oh man, im a n00b... is there a way to draw the integrands?? cause that would be cool..

these types of problems belong here:

[url]http://www.artofproblemsolving.com/Forum/index.php?f=296[/url]

and do you mean $\int_{0}^{6\pi}\cos\left(\pi+\frac{x}{12}\right)^2 \, dx$

yes, so how do you do that. writing the problem i mean. but also the problem.

First using $\cos (\pi+\theta)=-\cos \theta,$ rewrite the integrand, then use half-double angle formula; $\cos ^ 2 {\theta}=\frac{1+\cos 2 \theta}{2}.$


find all polynomial $P(x)$ satisfies $P(P(x))+x^2=P(x^2+x)$

Let $deg(P)=n$ The highest power of LHS is $n^2$, the highest power of RHS is $2n$. Thus, $n \leq 2$.

So $P(x) = ax^2 + bx + c$.

Consider the coefficient of the $x^4$ term: $a^3 = a \rightarrow x = -1, 0, 1$
Consider the coefficient of the constant term: $ac^2 + bc + c = c \rightarrow ac + b = 0$ or $c = 0$

Case 1: $a=0 \rightarrow b=0 or c=0$ The first case is contradictory. The second case will yield $f(x) = x$

Consider the coefficient of the $x^3$ term: $2a^2b = 2a \rightarrow ab = 1$ (given that a is not 0)
Consider the coeffcient of the $x^2$ term: $ab^2 + 2a^2c + ab + a + 1 = a + b$

So it is easy to check the following cases:
Case 2: $a=1 \rightarrow b=-c=1 or c=0, b=1$
Case 3: $a=-1 \rightarrow b=c=-1 or c=0, b=-1$

I was hoping on smart way in this problem :)


If $ A$ is a dense subset of a metric space $ (M,d)$, show that $ (A,d)$ and $ (M,d)$ have the same completion (isometrically).

Does it not follow from the definition of completion of a metric space S as a complete space which has a dense subset isometric to S and the observation that completions are always identifiable under isometry?

And a dense subset of a dense subset is still a dense subset.  :maybe:

[quote="§outh§tar"]Does it not follow from the definition of completion of a metric space S as a complete space which has a dense subset isometric to S and the observation that completions are always identifiable under isometry? [/quote]

Why does it necessarily follow?

Is (A,d) isometric to a dense subset of the completion of M?

Are dense sets necessarily isometric?  The problem my friend and I are having is that we can't prove that $ \hat{A}\equal{}\hat{\bar{A}}$.  Since $ A$ is dense in $ M$, $ \bar{A}\equal{}M$ so $ \hat{\bar{A}}\equal{}\hat{M}$, so $ M$ and $ \bar{A}$ have the same completion, but not necessarily $ A$.

I admit I haven't worked out the problem and I'm going off intuition but I don't think it should be this hard.

I'm not saying dense sets are isometric.

You have an isometry from M to a dense subset of its completion C. Is the image of A under this isometry dense in C? We know that isometries preserve Cauchy sequences..

Now refer to what I said earlier.

If we have an isometry from $ M$ to a dense subset of its completion, then I believe the image should be dense.  So $ f(A)$ is dense in $ \hat{M}$, with completion $ \hat{M}$.  Does this imply that $ A$ has $ \hat{M}$ as a completion as well?

By the definition of completion (I gave above), yes.


Let $R$ be a convex region symmetrical about the origin with area greater than $4$. Show that $R$ must contain a lattice point different from the origin.

This is the 2-dimensional case of Minkowski's theorem, and except of the lack of imagination for dimensions $>3$, I see no reason why this case is easier.

The $n$-dimensional version is:
Given a convex region $R$ symmetrical to the origin in the $n$-dimensional space. Show that if $R$ has volume greater than $2^n$, then $R$ contains a lattice point different from the origin.


function
integration
calculus
real analysis
real analysis solved

P8

a<b are reals. f:[a, b]->R is a continuous function s.t. |f(x)| <= | \int{a,x} f(t)dt| for all x in [a, b]. Prove that f(x)=0 for all x in [a, b].

Use only : |f| is bounded and its integral exists
|f(x)| =< |Integral(a =< t =< x){f(t).dt}| =< Integral(a =< t =< x){|f(t)|.dt}|
Let n be a natural such that n > b-a , and note u = (b-a)/n < 1. Note also M = sup(a =< t =< a+u){|f(t)|}
Manifestly, |f(x)| =< Integral(a =< t =< a+u){|f(t)|.dt}|  =< u*M for all x in [a,a+u] , so :
0 =< M =< u*M  ===>  M=0 ===> f(x)=0 on [a,a+u] ,  and f(x) =< Integral(a+u =< t =< x){|f(t)|.dt}| for all x in [a+u,b]
By using the same way it is easy to see f(x)=0 on [a+u,a+2*u], ... ,f(x)=0 on [a+(n-1)*u, a+n*u=b]
Finally f(x)=0 everywhere  on [a,b] ..


Can anyone give a list of big high school math competitons in Indiana besides AMC?

I think thats a little too hard.  I dont know a single person that passed it.

I mean like Indiana only competitons, those people from other states can do too.

in-state competitions aren't gonna get you anywhere, like ICTM.

But the point of this was INDIANA math competitons.

Like I said, USAMO is a national one.  Besides, its too hard and Ill probalby never pass it.

Does anyone have any advice or can post old amc probelmsA

[quote="vikas"]

Does anyone have any advice or can post old amc probelmsA[/quote]

Yea whats 1+1.

[quote="veryfriendly"]i know there is a rose-something competition that exists in Indiana don't know the exact name[/quote]

rose hulman?  I will be going to that.

Behold the magical [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php#in]url[/url] that tells all!

hey do you guys know how/where you can take the

USNCO, USAPHO, or the USABO...

it seems like my school doesnt offer it...

Indiana Mathleague and whoa!!!!! WHO ARE ALL THESE NEW PPL??

My sister took the usa chemistry olympiad
next year, im taking physics olymp.

erm so how can we take it? I want to take it too..

well, the semifinals are this saturday, so its probably too late to take it this year.


algebra
polynomial
binomial theorem
algebra unsolved

Let $P(x)$ be a polynomial of degree $n$, $n \in \mathbb{N}\/$, such that $P(k)=2^{k}\/$ for each $k =1,2, \ldots, n+1\/$. Determine $P(n+2)\/$.

This does not really belong to the olympiad section (too easy: a complex polynomial of degree $n$ is entirely determined by $n+1$ of its values ...)

Here's one approach.  Define the polynomial $Q$ by the equation \[Q(x) = 2 \sum_{j=0}^{n}\binom{x-1}{j}\, .\] By the binomial theorem, we have $Q(k) = 2^{k}= P(k)$ for all integers $k$ from $1$ to $n+1$.  Furthermore, $Q$ is a polynomial of degree $n$.  Thus, by julien_santini's observation, the polynomials $Q$ and $P$ must be identical.

So we have \begin{eqnarray*}P(n+2) & = & Q(n+2) \\ & = & 2 \sum_{j=0}^{n}\binom{n+1}{j}\\ & = & 2 \left[\sum_{j=0}^{n+1}\binom{n+1}{j}-1\right] \\ & = & 2 \left[ 2^{n+1}-1 \right] \\ & = & 2^{n+2}-2 \, . \end{eqnarray*}

Nice solution. I give these solution half year ago in Russian forum.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that there exists exactly one subgroup order  $n$ in the group $(\mathbb{Q}/ \mathbb{Z},+)$.

Let $H$ be a subgroup of order $n$.
Take for every element the least positive representant and take the smallest element $x$ by this.
Then all other elements are multiples of $x$ (otherwise it contradicts the choose of $x$), so $x$ generates $H$. Now the order of $x$ is exactly its denominator in reduced form, thus $x$ has denominator $n$ and then also $x=\frac{1}{n}$ (since otherwise theres a smaller one).

Another way to put it: since any two elements are commensurable, every finitely generated subgroup is generated by one element. Since every element has finite order and every finite group is finitely generated, each group of order $n$ is generated by a single element of order $n$. All elements of order $n$ are of the form $\frac kn$, and all such elements are contained in the group $\left(\frac1n\mathbb{Z}\right)/\mathbb{Z}$, which has $n$ elements.


geometry incenter circumcircle trigonometry trapezoid inradius ratio

Let $ABC$ be a scalene, acute-angled triangle with orthocenter $H$, incenter $I$. The lines $CH$ and $CI$ intersect the circumcircle of $\triangle ABC$ at $D$ and $L$, respectively. Prove that $\angle CIH$ is a right angle $\longleftrightarrow$ $\angle IDL$ is a right angle.

Let (O) be the circumcircle, R the circumradius, and r the inradius. For any acute triangle $\triangle ABC$, $CH = 2R \cos C.$ (If the angle $\angle C$ was obtuse, we would have $CH = -2R \cos C.$) WLOG, assume $\angle A < \angle B$. Then

$\angle ICH = \angle ICB - \angle HCB = \frac{\angle C}{2} - (90^\circ - \angle B) = \frac{\angle B - \angle A}{2}$

The angle $\angle CIH$ is right iff $\frac{CI}{CH} = \cos \frac{B - A}{2}.$ Thus we have to show that $\angle IDL = 90^\circ$ is right if this condition is true. L is the midpoint of the circumcircle arc AB opposite to the vertex C. Let K be the other end of the circumcircle diameter KL. The line $KL \perp AB$ is the perpendicular bisector of AB. The diameter KL forms an isosceles trapezoid CDLK with the altitude CD. Since the diagonal CL is the bisector of the angle $\angle C$, the incenter $I \in CL$. Since the angle $\angle KDL = 90^\circ$ on the diameter KL is right, the incenter $I \in KD$ lies also on the other diagonal KD. Therefore, the incenter is identical with the diagonal intersection $I \equiv CL \cap KD$ of this isosceles trapezoid. Therefore, the diacentral line IO is the perpendicular bisector of both CD and KL, hence, it is parallel to the side AB (both perpendicilar to KL). Let $M \equiv AB \cap KL$ be the midpoint of AB. Then $r = OM = R \cos C$ and $CH = 2R \cos C = 2r.$ Using Euler's formula $\frac{1}{r} = \frac{1}{R + d} + \frac{1}{R - d}$ or $R^2 - d^2 = 2rR,$ where

$d = OI = OL \tan \widehat{OLI} = OL \tan \widehat{KLC} = R \tan \frac{B - A}{2}$

$R^2 \left(1 - \tan^2 \frac{B - A}{2}\right) = 2rR$

$\frac r R = \frac{\cos^2 \frac{B - A}{2} - \sin^2 \frac{B - A}{2}}{2\cos^2 \frac{B - A}{2}} = \frac{\cos (B - A)}{1 + \cos (B - A)}$

Combining the 2 expressions for the inradius-circumradius ratio,

$\frac{\cos (B - A)}{1 + \cos (B - A)} = \cos C = -\cos(B + A)$

$\cos (B - A) + \cos (B + A) = -\cos(B - A) \cos(B + A)$

$2 \cos B \cos A = -\frac{\cos (2B) + \cos (2A)}{2} = 1 - \cos^2 A - \cos^2 B$

$(\cos B + \cos A)^2 = 1,\ \ \cos B + \cos A = 1$

Let F be the foot of the C-altitude. The parallel bases of the isosceles trapezoid CDLK are in the ratio

$\frac{CD}{KL} = \frac{CH + 2 HF}{KL} = \frac{2R \cos C + 4R \cos A \cos B}{2R} =$

$= \cos C + 2 \cos A \cos B = \cos (B - A)$

The diagonals CL = KD cut each other in the same ratio and their length is

$\frac{CL}{KL} = \frac{KD}{KL} = \cos \widehat{DKL} = \cos \widehat{DCL} = \cos \frac{B - A}{2}$

In addition, the incenter I lies on a circle (L) with radius

$LI = LB = LC = \frac{c}{2 \cos \frac C 2} = \frac{2R \sin C}{2 \cos \frac C 2} = 2R \sin \frac C 2$

Consequently,

$\cos (B - A) = \frac{CI}{LI} = \frac{CI}{2R \sin \frac C 2},\ \ CI = 2R \sin \frac C 2 \cos (B - A)$

$\frac{CI}{CH} = \frac{2R \sin \frac C 2 \cos (B - A)}{2r} = [1 + \cos (B - A)] \sin \frac C 2 =$

$= 2 \cos^2 \frac{B - A}{2} \cos \frac{B + A}{2} = (\cos B + \cos A) \cos \frac{B - A}{2} =$

$= \cos \frac{B - A}{2} = \frac{KD}{KL}$

As a result, the triangles $\triangle KDL \sim \triangle CIH$ with the equal angles $\angle DKL = \angle DCL \equiv HCI$ are similar by SAS. Hence the corresponding angles $\angle CIH = \angle KDL \equiv IDL = 90^\circ$ are equal and right.

[color=darkred][b]Enunciation.[/b] Let $ABC$ be a scalene and acute triangle. Denote the  intersections $D$, $L$ of the open rays $(CH$, $(CI$ respectively with the circumcircle of the given triangle. Prove that[/color]
$\boxed {\ DI\perp DL\Longleftrightarrow IH\perp IC\Longleftrightarrow r=R\cos C\ }\Longleftrightarrow\cos A+\cos B=1\ .$

[b]Proof.[/b] Suppose $a>b$. Remember some well-known relations in a triangle $ABC$ :
$m(\widehat {HCI})=m(\widehat {CLO})=\frac{A-B}{2}\ ;$ $IC\cdot IL=2Rr\ ;$ $r=IC\sin \frac C2\ ;$ $m(\widehat {ILD})=90^{\circ}-(A-B)\ ;$ $CL=2R\cos \frac{A-B}{2}\ ;$ $DL=2R\sin \frac{A-B}{2}\ ;$ $IC^2=\frac{ab(p-c)}{p}\ ;$ $(p-a)(p-b)(p-c)=pr^2\ ;$ $1-\cos C=2\sin^2\frac C2\ ;$ $\sin \frac C2=\sqrt{\frac{(p-a)(p-b)}{ab}}\ ;$ $\cos A+\cos B+\cos C=1+\frac rR\ .$

$1.\blacktriangleright$ $\boxed {\ DI\perp DL\ }$ $\Longleftrightarrow$ $DL=IL\cos \widehat {ILD}$ $\Longleftrightarrow$ $2R\sin \frac{A-B}{2}=IL\sin (A-B)$ $\Longleftrightarrow$ $R=IL\cos \frac{A-B}{2}$ $\Longleftrightarrow$ $R\cdot IC=2Rr\cos \frac{A-B}{2}$ $\Longleftrightarrow$ $IC=2r\cos \frac{A-B}{2}$ $\Longleftrightarrow$ $1=2\sin \frac C2\cos\frac{A-B}{2}$ $\Longleftrightarrow$ $1=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}$ $\Longleftrightarrow$ $1=\cos A+\cos B$ $\Longleftrightarrow$ $1+\cos C=1+\frac rR$ $\Longleftrightarrow$ $\boxed {\ r=R\cos C\ }\ .$

$2.\blacktriangleright$ $\boxed {\ IH\perp IC\ }$ $\Longleftrightarrow$ $IC=HC\cos \frac{A-B}{2}$ $\Longleftrightarrow$ $IC=2R\cos C\cos\frac{A-B}{2}$ $\Longleftrightarrow$ $IC=CL\cos C$ $\Longleftrightarrow$ $IC=(CI+IL)\cos C$ $\Longleftrightarrow$ $IC(1-\cos C)=IL\cos C$ $\Longleftrightarrow$ $IC^2(1-\cos C)=2Rr\cos C$ $\Longleftrightarrow$ $\frac{ab(p-c)}{p}\cdot 2\sin ^2\frac C2=2Rr\cos C$ $\Longleftrightarrow$ $\frac{2ab(p-c)}{p}\cdot \frac{(p-a)(p-b)}{ab}=2Rr\cos C\Longleftrightarrow$
$\prod (p-a)=Rrp\cos C\Longleftrightarrow pr^2=Rrp\cos C$ $\Longleftrightarrow$ $\boxed {\ r=R\cos C\ }\ .$


Oznaczmy przez $( a_n )_{n = 1}^{\infty}$ rosnacy (strictly $>$) ciag liczb rzeczywistych.
Udowodnij ze
\[ \sum_{n = 1}^{\infty} \frac{1}{a_n} < + \infty \Rightarrow \sum_{n = 1}^{\infty} | \frac{a_n}{a_{n + 1}} - \frac{n + 1}{a_{n + 1}} | < + \infty  \]

P.S: Nie mam zielonego pojecia jak "to"  zaatakowac.

Takie cos powinno dzialac:

1) Korzystasz z tw. (cwiczenie 8, Rudin "Podstawy analizy matematycznej" - dzial ciagi i szeregi):

Jesli $\{b_n\}$ jest monotoniczny i ograniczony, a $\sum a_n$ jest zbieżny to $\sum a_n b_n$ jest zbiezny

2) Na mocy tego wystarczy wykazac, ze $\sum |a_n - n-1|$ jest zbiezne (tym samym dostaniemy mocniejszy rezultat)

3)  $\sum \frac{1}{a_n} < + \infty$ implikuje $a_n = \Omega (n^{1+\epsilon})$. I to na moj gust zalatwia sprawe.

W razie czego gdyby powyzsze bylo totalnym bezsensem tlumacze sie pozna pora i ogolna sennoscia  :P  :D

W 3) miales na mysli $\frac{1}{a_n}$ zamiast $a_n$ ? (wynikaloby to z uzycia $a_n$ w 2))
2) nie zadziala : podstaw $a_n = n^2$...
Natomiast z 1, i 3 chyba da sie jakies rozwiazanie sklecic ... :roll:
Czekam jutro ... ;)

Jak tak teraz sie przyjrzalem to czy to zadanie jest w ogole prawdziwe?

Wezmy ten przyklad, ktory podales u gory $a_n=n^2$, a wtedy dostajesz $\lim_{n \to \infty} \frac{a_n -n-1}{a_{n+1}}=\lim_{n \to \infty} \frac{n^2-n-1}{n^2+2n+1}= \lim_{n \to \infty} \frac{1-\frac{1}{n}-\frac{1}{n^2}}{1+\frac{2}{n}+\frac{1}{n^2}}=1$

Najwyrazniej nie jest prawdziwe ;) (shame on me  :oops:). Natomiast dalbym czesc mozgu odpowiedzialna za matematyczne myslenie uciac ze dokladnie to pytanie pojawilo sie w CMB (Canadian Mathematical Bulletin)...

To tak przy okazji sie zapytam skad ty wziales CMB ? Jest to jakos dostepne latwo w internecie, czy po prostu ruszyles do odpowiedniej biblioteki  :)

Jest dostepne online ;) N.p take a look here :
http://endeavor.macusa.net:591/macusa/mathpropress/FMPro?-db=problems.fp5&-format=mainresult.html&-lay=all&ProblemName=CMB&-find=
Na owej stronie (problemcorner.org) znajdziesz problemy z innych czasopism, chocby z AMM ;)

Wielkie dzieki - problemy wygladaja ciekawie  :D  :D  :D odnalazlem tez poprawna wersje powyzszego: powinno byc:

$\sum |\frac{n}{a_n}-\frac{n+1}{a_{n+1}}| < \infty$

Oznaczmy $\sum \frac{1}{a_n} = M$

Szacujemy:

$\sum |\frac{n}{a_n}-\frac{n+1}{a_{n+1}}| = \sum |\frac{n}{a_n}-\frac{n+1}{a_n}+\frac{n+1}{a_n}-\frac{n+1}{a_{n+1}}| < \sum |\frac{n}{a_n}-\frac{n+1}{a_n}|+|\frac{n+1}{a_n}-\frac{n+1}{a_{n+1}}| = \sum |\frac{1}{a_n}|+|\frac{n+1}{a_n}-\frac{n+1}{a_{n+1}}| = M + \sum |\frac{n+1}{a_n}-\frac{n+1}{a_{n+1}}|$

Liczba w module jest dodatnia, czyli

$\sum |\frac{n+1}{a_n}-\frac{n+1}{a_{n+1}}| = \sum \frac{n+1}{a_n}-\frac{n+1}{a_{n+1}} = \frac{1}{a_1} + \sum \frac{1}{a_n}$

Więc cała suma ograniczona jest przez $2M+\frac{1}{a_1}$.

Jest tylko jeden malusienki bledzik na samym koncu:
\[ \sum_{n = 1}^{\infty} \left( \frac{n + 1}{a_n} - \frac{n + 1}{a_{n + 1}} \right) = \left( \frac{2}{a_1} - \frac{2}{a_2} \right) + \left( \frac{3}{a_2} - \frac{3}{a_3} \right) + \ldots = \frac{2}{a_1} + M  \]

Nie. Jedno z tych $\frac{2}{a_1}$ wchodzi do $M$.


hello i'm new here and don't know where this might go...but anywayshere is a good problem i just solved recently:
if in a game there are only 2 possible scores $a$ and $b$(relativly primes), (u can choose to get either one each time) then:
i)what is the largest score that can not be made by these numbers?(proof needed)
ii)what is the number of impossible scores?(proof needed)

please don't say for part (i), based on the Chicken McNugget Theorem it has to be $ab-a-b$--it would be unacceptable :)

good luck. :)

The coin exchange problem of Froebenius (what is the chicken MacNugget theorem???)  is solved here :

http://www.animath.fr/cours/arithm.pdf

Have a look at pages 30-32.

Pierre.

P.S. If you have solved this problem then post it in the proposed and own problem section, not in the unsolved problem one.

sorry about that,
but since i don't know any french, can u translate that? (i'm interested)
i'm more interested in the solution for the second part

thanx

i don't have much time right now but
the answer for the second part is:
$\frac{(a-1)(b-1)}{2}$


Support
probability
geometry
calculus
derivative
linear algebra

This is an serious and sensitive question. Do you agree with induced abortion?

[quote="JBL"]So, first, as moderator, I'd like to remind anyone who posts on this topic to be civil or I will be forced to use some combination of editing, deleting or locking, which I'd rather not do.

Second, I have no idea what you mean by "induced abortion."  Perhaps you could give a more detailed description of the circumstances of abortion you are asking about?[/quote]

  You don't understand? Maybe I've made mistake. I don't know English very much, so I looked it up in my Chinese-English dictionary, my dictionary told me that the word I wanted to say is the phrase "induced abortion", if that's wrong, that should be the problem of my dictionary, I should say sorry.
  For the phrase "induced abortion", the word "abortion" in it I think everyone may understand. For the word "induced", I've looked it up in the dictionary again, the meaning in Chinese is difficult to understand. So, I think if the reader have no idea at the word "induced", you can just ignore it and only look at the word "abortion".
  I'm sorry for my poor English. :blush:

I think that abortions are bad, but at the same time if somebody gets a child they don't want, they might take bad care of them so it's a tough issue. I don't believe though, that this is a very important issue in politics because I don't that it is a big enough problem in the current world.

How come youe sn is Alexander Lenin? just wondering.

[quote="hello"]How come youe sn is Alexander Lenin? just wondering.[/quote]
  Could you say in complete sentence? I can't understand that because of my poor English :blush: .

[quote="hello"]I mean your name on the left side on the screen on this aops forum is Alexander Lenin, but you are from China. I was wondering how you came up with such a screen name?[/quote]

Hmm......1, Just for fun; 2, for Alexander, just because I adore him; 3, for Lenin, I also respect him, and just give a last name

[quote="hello"]What do you think about abortions?[/quote]

I voted strongly disagree. Although in my country that's lawful. I can tell you, in my country, abortion is not only lawful, even that's advocated in certain extent, that's really terrible, I'm sad about the fact :(

Because China has a large population. Wasn't there a law before that the ideal family was with one male child in the family. Oh well, I think I will go sleep now, I am in Philly.

[quote="hello"]Wasn't there a law before that the ideal family was with one male child in the family. [/quote]

You are wrong about with one male child, it should be changed into "the more male children are, the better"

slightly support...
I think if the parents all agree on that, then it is ok. After all, it is their right to do it.

STRONGLY SUPPORT.  Now that doesn't mean that I am pro abortion and that everyone should get an abortion--NOT AT ALL!  We need to make sure to take every precaution to prevent an abortion, BUT in the end, if one is necessary, I support the WOMAN'S RIGHT TO CHOOSE!

I think that conservative right-to-life advocates are just being very silly--don't the 1500 soldiers who've died in Iraq have the right to life?

Although I do agree on a womens right to choose...this gets a little touchy.

Having sex has its consequences.  Pregnancy is one of them.  If they can't afford to have the baby... Are too young...etc.  they should go with a much better alternative which is to put the child up for adoption.

The other thing is that the baby is considered "unborn" at that stage...So people think it really doesn't matter.  But imagine what would life be like if you were aborted.  :?

I dunno...I have mixed feelings on this... I don't consider it murder, but I do find it morally incorrect.

I don't see where all this stuff about a woman's right to choose is coming from though.  If you really don't want kids go get a gonangiectomy.  You have a right to choose there.

To have unprotected sex or not.... You have a right to choose there.

NOW...

If the mother is in the condition in which the baby is in someway life threatening, causes psychological damage (rape, etc) I can understand why an abortion should be appropriate.   This is the only case in which I completely support abortion.

About those soldiers:

When you sign up for the military, you should know that you are putting your life on the line.  Regardless of whether it is wartime or not.  It is a decision you make.  It is a very honorable decision aswell.  Becoming a soldier is a decision YOU make. It is NOT the decision your parents make.

what about third world countries where people get married when they are 12 or 13? (typically for the money the person they married with gives their family money)

The only thing I have found offensive on this forum was the person who said "what's the difference between a rape-baby and a normal baby?" (paraphrased, sorry). Honestly, I find this EXTREMELY offensive. This was obviously a male who had never had a female friend who was raped. What difference is it? What difference is being reminded of this pain, every day for the rest of your life? What difference is carrying the child of a rapist? What difference is resenting that baby, and hating it, because it is half... HIM?
And if you don't think of it from the woman's point of view, from the child's then:
What difference is it to bear the weight of having a rapist father? What difference is it not knowing your heritage? What difference is it, having a mother that, deep down, wants to cry when they look at you?

Think about that.

[quote="bleumoose"]The only thing I have found offensive on this forum was the person who said "what's the difference between a rape-baby and a normal baby?" (paraphrased, sorry). Honestly, I find this EXTREMELY offensive. [/quote]
BleuMoose, I think that you are misunderstanding the intentions of Dark_Shadow, that specifically said,

[quote="Dark_Shadow"]You state that rape is rare, and abortion is okay in those cases. Otherwise, you're killing a child.
What makes a "rape baby" any less valuable than a "non-rape" baby?
Either it's about the morality of abortion, as aborting a fetus, or it's about killing a child, in which case the mother shouldn't matter.[/quote]

I think that despite the emotional implications, from a moral point of view, he is right. If you have two kids (both have already born) and one is product of a rape, while the other is product of conscensual sex, would you kill the first one because of that reason? I presume your answer would be no, but then, what is the difference from before the birth and after the birth. Despite what the kid and the mother can feel afterwards, morally and phisically, what makes different one of the morulae from the other?

What Dark_Shadow pretends (or what I understand that he pretends) is to argue that you should not make a difference in the law from an abortion product of a rape and an abortion for any other reason.

[quote="bleumoose"] This was obviously a male who had never had a female friend who was raped. What difference is it? What difference is being reminded of this pain, every day for the rest of your life? What difference is carrying the child of a rapist? What difference is resenting that baby, and hating it, because it is half... HIM?
And if you don't think of it from the woman's point of view, from the child's then: What difference is it to bear the weight of having a rapist father? What difference is it not knowing your heritage? What difference is it, having a mother that, deep down, wants to cry when they look at you? Think about that.[/quote]

I cannot answer for Dark_Shadow, but in my case, I know women as well as males that have been raped (yes, boys get raped once in a while too!), and I have personally seen the pain that a rape creates. I honestly believe that rape should be a capital crime, a crime punishable with death, because it produces a lot of damage, more than what a non-raped person can ever imagine. Nevertheless, there you are punishing the direct guilty of a crime, and rationally, the society has ever punish a criminal in a "proportional" measure to the damage that they produce to another person, institution or population.

On the other hand, when you argue about the abortion issue, your arguments are completely emotional, and you laws shouldn't be ruled by emotions but by rational arguments.

Another issue why there shouldn't be any difference between rape and no rape during abortion is because of the lying risk. I totally agree that rapist should be punish as far as law can punish them, THEY ARE CRIMINALS, nevertheless, let's think a moment about this: Suppose that the law rules that just deep health issues or rape can be motives to have an abortion and  that Norman and Tracey (ficticius names, they came to my mind because of linear algebra: [b]Norm[/b]an and [b]Trace[/b]y) have been dating for a few months, they decide to have sex and Tracey got pregnate. She knows she is not allowed to have an abortion, even when she may want to, then she goes and tell Norman. Norman is a complete pusillanimous and he just try avoid the situation, probably puts in doubt the fidelity of Tracey and at the end, just tell her that it is not his child and that he is not going to talk to her again. Tracy in that moment feels a natural but unmesurable rage, she wants to get rid of "that thing" on her uterus, but legally she can't, unless... she had been raped... and the rapist would be Norman.... the perfect revange, don't you think. Now, I consider than Norman is a coward and a couple of other things that I don't mention because this is not the most appropiated place, but he is not a criminal. The lack of braveness and responsability is not a crime. But he may be treated as a criminal because if non-criminal actions. Probably if abortion had been legal on any situation, Tracey would just go and practice one, and afterwards just hate Norman for the rest of her life (what is totally acceptable) and probably tell the story to every new girlfriend that the guy has afterwards (what I also consider totally acceptable), but not try (and potentially succed) to send him to prision because of something he didn't do.

[quote="bleumoose"]The fictitious story you just gave is EXACTLY the reason I believe all abortion should remain legal. That, and the fact that many women would be getting unsafe abortions anyways. My argument was meant to be nothing more than emotional. Many of the things I say on this forum are just something else for people to consider, and I want the people on this board that are discussing abortion to consider the emotional pain of rape and carrying a baby who is the product of a rape.[/quote]

OK, then I also misunderstood you a little. As I told, I consider that the big people that write the laws should not be emotional when they write them, as well as the judges should not be emotional when they rule. Nevertheless, it is always acceptable to touch the sencitive fibers of the people to make a point.

Well...  Maybe you shouldn't be emotional, but I wouldn't feel especially safe if the people writing the laws had all the emotional capacity of bricks.  It's not just about making a point.  It can be said the emotions and empathy is what prevents the ends from justifying the means.

Okay, sorry for misinterpreting you djimenez. It's just that it sounded like you were 'dumbing down' your arguments because of ash's age...ah well. I hope we have that cleared up then. Now on to more debate...

[quote="ash101189"]i am pro choice, but i do think it is killing an actual human. it doesnt matter if the baby is conscious or not of its surroundings! by killing it you are taking away the opportunity for it to be conscious.[/quote]
The thing is--this is going to sound weird--[i]I don't believe it's wrong[/i] to deprive something that, at the current moment, isn't and has never been conscious. I don't believe the future 'exists', I believe our choices create the future and that no one can know for sure what will happen in the future. So since the only state the fetus has as of now ever existed in is that of unconsciousness, it's not wrong to make sure it will never be conscious. And I know this will probably make no sense to you because we are approaching it from totally different viewpoints. You believe these fetuses have whole lives ahead of them. I don't believe there's anything ahead of us, there's only the past and the present. Arg, this is getting too philosophical. I hate this whole issue, it's so hard to phrase an argument for any side in a way that would justify a federal law.

[quote="LynnelleYe"]The thing is--this is going to sound weird--[i]I don't believe it's wrong[/i] to deprive something that, at the current moment, isn't and has never been conscious.[/quote]Exactly my point. Now, the only problem that I see in that position, that is half practical and half philosophical, is that, even when I don't think that there is any wrong to do an abortion on an early stage (your only killing a few cells, like when you accidentally cut your finger while chopping onions... painful but not inmoral), the argument can be used to justify a libertine sexual behavior, and a [i]I don't have to worry about the consecuences[/i] argument to defend it.[quote="LynnelleYe"]I believe our choices create the future and that no one can know for sure what will happen in the future. So since the only state the fetus has as of now ever existed in is that of unconsciousness, it's not wrong to make sure it will never be conscious.[/quote]You know, I had never thought it in that way, but it make a lot of sence.

[quote="MathFiend"][/quote][quote="JBL"]
However, Mathfiend, you never really answered the question that I find most critical: you said that you believe foetuses are children.  Fine.  What about blastocysts?  I want to know exactly where you draw the line between life and not life, before which my (hypothetical) wife can acceptably get rid of without crossing any moral lines.[\quote]
I can't answer your first question, what is a blastocyst? That would help me answer your question.[\quote]
A blastocyst is a group of 40 to 150 undifferentiated (well, mostly -- there are "outer" and "inner," but they don't have distinct functions yet) cells.  It is approximately the size of the dot you can make with a pencil on a piece of paper.  It is the fourth stage that cells pass through after ovulation: ova, zygote (that being the sperm and egg joined, still just one cell), the morula (which goes up to about 32 or 40 cells) and then blastocyst.

[quote][quote]As for your ring, you'd probably be better off carrying around a condom, as far as unwanted pregnancy and STDs are concerned.  (That sentence is of course really about statistical realities, not about your own personal willpower, so please don't take it personally.)[/quote]
Ok, for the second part, I totally disagree, and it makes me a little angry. After all I've said, can't you see that my point is to discourage sex before marriage? How does carrying around a condom do that? It is condoning exactly what I am against! Why on earth would I carry a condom to make me resist the temptation, that is simply like tying your hands behind your back and holding a freshly baked still gooey chocolate chip cookie in front of your face, but just out of reach and saying "want it?" It simply puts the temptation right there. Let's (hypothetically) say I would carry a condom in my purse rather than wearing the ring on my finger. Someone says sex, I think [i]"hey, there's a condom in my purse, I could go have sex and be safe about it."[/i] Now, if I have my ring, I am reminded that I will not have sex. See a difference? One completely disagrees with what I have said, and the other makes complete sense.[/quote][/quote][quote="MathFiend"][/quote]
So, in an information vacuum, I would agree with you -- knowing nothing else about the world, what you just said sounds reasonable.  And, in fact, if you look at studies made of teens who have taken virginity pledges and/or had abstinence only sex education versus those who have had a thorough sex education (info about contraception, etc.) the first group does tend to have sex for the first time slightly later than the second group.  However, they are just as likely to get pregnant before marriage and more likely to acquire an STD.  So, your ring (again, not literally yours), if examined as a tool to improve public health (and lower abortions), accomplishes exactly nothing.  In fact, this is the single thing that most upsets me about the pro-life movement.  They claim to want to prevent abortions, but they constantly set barriers in front of exactly those things (contraceptive education, morning-after pill, etc.) that would [i]reduce[/i] abortions.  If they were really serious, they would be working with Planned Parenthood and NARAL, not against them.

Also, and this is really a question about you personally with a message about teens in general, do you really think that having a condom around would increase your libido so much that you would forget all about not having sex?  I mean, that doesn't sound serious to me -- either you have the willpower and carrying the condom won't hurt it, or at some point you'll reach a situation where you want to have sex before you're married.  If you find yourself in the second situation, how much better to have a condom handy?  Besides, the process of getting the darn thing open and on might provide you just enough time to reconsider.

[quote="djimenez"][quote="ZennyK"][quote="Psycho"]for some reason, i strongly support it. i mean, if the parents don't want the child, how will they be able to raise it in a good way then?  :?
and what if that person later finds out his parents didn't want him at all. how should he feel then?[/quote]
So it's better to murder a person than to face hardship?[/quote]
ZennyK, Define person. because as many of us had said before, we do not consider a few non-differenciated cells a human. I do think that it is better to smash a few cells like the ones you smash when you have an ich in your skin, or when you accidentally cut your finger when you are cutting the onion, than to sentence the human been that those cells can become to, to a misserable life or a hardship he/she didn't ask for!

Well, if he realize that his/her parrents didn't want him when he is a child, well, that can be very traumatic, but I would say that in the adult age, it would not be nice, but it would not be that harsh either, I mean, it is just a matter of maturness![/quote]
When the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it. To terminate it would be to kill a human, regardless of whether or not it could "survive." Many abortions take place after differentiation occurs.

[quote="LynnelleYe"][quote="ZennyK"]
When the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it.[/quote]
Why?[/quote]
Because then the embryo must develop into a human being.

[quote="ZennyK"][quote="LynnelleYe"][quote="ZennyK"]When the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it.[/quote]Why?[/quote]Because then the embryo [b]must[/b] develop into a human being.[/quote]Why [b]must[/b] it? I mean, is it the differenciation of the cells in the embrion the thin line between a human been and a bunch human of cells? I am not so sure!


number theory
greatest common divisor
least common multiple
inequalities
relatively prime
triangle inequality

[quote="Scrambled"][quote="Misha123"]Find all pairs of positive integers a,b so that gcd(a,b),lcm(a,b) and ab are sides of a triangle!

To avoid misunderstanding:
gcd: greatest common divisor
lcm: least common multiple[/quote]
If the numbers are relatively prime, then lcm(a,b) is ab and therefore it can work because the triangle is isosceles. Therefore there are an infinite amout?[/quote]

That makes sense, look at the rest though.

we have gcd(a,b)lcm(a,b) = ab

So therefore

gcd(a,b) <= lcm(a,b) <= ab

Hence we must have gcd(a,b) + lcm(a,b) > ab

We then have: gcd(a,b) + lcm(a,b) = gcd(a,b) + ab/(gcd(a,b))

It is obvious that gcd(a,b) <= ab/2, hence if gcd(a,b) >= 2, ab/gcd(a,b) <= ab/2 and gcd(a,b) <=ab/2.  Summing, we get LHS <= ab, which is a contradiction.  It gets a tiny bit more complicated if we can have degenerate triangles (we have to include the triangle (2,2,2)) but Scrambled's answer encompasses all solutions.

[quote="Magnara"][quote="blahblahblah"]gcd(a,b) <= lcm(a,b) <= ab

Hence we must have gcd(a,b) + lcm(a,b) > ab[/quote]

Why is that?[/quote]

Apologies if it was not sufficiently clear.  The first statement is obvious, the second is the triangle inequality, and is what we want to prove.

Thanks for your comments. And JBL: your comment about the infinity of solutions is exactly what I would say about this topic. I just post the solution I had: it's rather easy(I hope you will like it):

gcd(a,b):=g.
As already said: ab=>lcm(a,b)=>max(a,b)=>min(a,b)=>g
So we must prove: ab/g+g=>ab. I make the substitution: a=a_1g,b=b_1g:
a_1b_1g+g=>a_1b_1g^2. You can rewrite it: g<=1+1/(a_1b_1). Now we have two cases: g=1 and g=2,a_1=b_1=1. It is easy to prove that both cases work.

Thanks,once more time, for your interest!

[quote="JBL"]Okay, this is going to be obnoxious, but it's a brief English-usage lesson.  The primary meaning of the word "hence" is "because of this" or "as a result."  (It also has a secondary meaning of "in the future" and an archaic usage of "away from here," as in "Get you hence!")  Your usage of the word "hence" implied that the triangle inequality was a direct consequence of the inequality you posted above it.

Scrambled -- often you will come across a math problem that asks you to find "all of something" when there are an infinite number of things.  You were, of course, correct that there were an infinite number of solutions, but what you needed to do was give a description of all solutions (such as, "All pairs of relatively prime integers") and then also show that there are no other solutions.  Knowing there are infinitely many of something is not necessarily very useful.[/quote]

JBL, I'm well aware that my comment was unclear, but I don't believe that it was incorrect.  I stated that gcd <= lcm <= ab.

It follows (or, hence) that[i] to solve the problem[/i], we need to show that these three qualities satisfy the triangle inequality.  I was in a hurry, and didn't include that little bit, but I believe that it was otherwise correct.  To better organize it, I should have said, hence a,b,c are sides of a triangle iff a+b > c by the triangle inequality, or whatever.

I'm not saying that what you said was incomprehensible or completely off the mark; I understood what you meant.  However, lacking a phrase such as "we need to show," the clear meaning of your sentence is that the triangle inequality follows from the ordering of those three numbers.  I don't think I need to defend my comment beyond that -- I was attempting to alleviate a possible cause of the confusion.


[quote="emilgouliev"]What am I supposed to put there to show the number of comments in the blogs?

I thought it was %, but that only deletes the next character.[/quote]Small bug. It will be fixed very soon :)


Hello All,

I was trying to answer [url=http://www.mathlinks.ro/viewtopic.php?t=208229]this post[/url]. My example is:

\[ f(x)=e^{-e^\cot(x)}\]

As you see, $ \cot(x)$ appears very small, so people with not very good eyes have problems reading it. Please, suggest me, how to write this formula so that it will be larger.

Thank you!
Sergey Markelov
markelov@mccme.ru

[quote="stevem"]There are 4 sizes of characters in maths mode, in descending order of size: displaystyle, textstyle, scriptstyle, scriptscriptstyle. The cot x term is in scriptscriptstyle so you can make it one size larger by using scriptstyle: $ f(x) \equal{} e^{ \minus{} e^{\scriptstyle\cot x}}$[/quote]

Stevem,

thank you very much!

Sergey Markelov


If a daddy bull weighs 1,200 pounds and eats twelve bales of hay each day, and a baby bull, who weighs 300 pounds eats three bales of hay each day, how much hay then should a mommy bull eat if she weighs 800 pounds?

[hide]Since bulls are only male, the only way for there to be a mommy bull is if the bulls are homosexual, in which case the mommy bull would eat 8 bales, (assuming the equation relating weight and hay is linear) :rotfl:[/hide]

Really??? It was a gay bull??? I was only joking :rotfl:

wait a minute does that mean bulls are gay or do they reproduce asexually or can they not reproduce except for theeir cells or only in this problem

No, they reproduce with cow. Cows are only femle and bulls are only male. :)

oh thanks i wasn't sure if i should put them on the list with those what are they...mules?


[b]hi everybody,
i am quite new to inequalities and got great help from the many marathons running here. I was  more interested in marathon on triangle and trigonometry inequalities.so i was just wondering if we could sstart a marathon on the same.[/b]

Take a look at some of the problems being posted by jeez123 in [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]HSB[/url]. No inequalities, but some trigonometry.


How to create new symbols that are required in $\LaTeX$.  I have to create new symbols which represent meet and join as used in the text book of "Introduction to Lattice Theory" by "Gaber Szasz" which resembles $\smallfrown$, $\smallsmile$ but has some more height than them. Or they are the symbols $\cap$ and $\cup$ but with extra space in between the open ends

I suggest you check out the [url=http://www.tug.org/tex-archive/info/symbols/comprehensive/]Comprehensive LaTeX Symbol List[/url].  It's possible your symbol already exists.  If not, it has instructions how to create new symbols, but I should warn you that it is quite difficult to do so.

I have verified $\LaTeX$ Symbol list.  But, I am unable to find these symbols there.  Also, there i am unable to find any procedure for creating new symbols.

Well, I hope this example will help:

% put this in preambule
\def\grpedeset{\hbox{$\Gamma$\kern-5truept\raise3.3truept\hbox{\tiny $\Delta$}}}

% somewhere in document you can use
\grpedeset

Try to find more about \kern, \raise.....
It's very usefull.


[hide="Hint"]  I have solved a general version of this problem a couple of times on Mathlinks, so I will just provide the following hint: Complex Numbers and Geometric Sums.[/hide]


[quote="GameBot"]How many non-empty subsets does a three-element set have?[/quote]

Let the elements be a, b, and c. We have:

a
b
c
ab
ac
bc
abc

This makes a total of 7.

For each element we have 2 choices when making a set, in or out. Thus there are 2^3=8 sets. However, the null set is not an option so it is 8-1=7.

We use the formula which is (2^n)-1 where n is the number of numbers in the set.

The answer is 2^3-1=7


how many terms are in the expansion of $ (2a\minus{}5b\plus{}3c)^{35}$

[hide="Hint"]I think the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]Multinomial Theorem[/url] would help.[/hide]
But I'm too busy/lazy to try it right now  :lol:

All of the terms in the expansion are of the form $ a^{k_a}b^{k_b}c^{k_c}$, where $ 0\leq k_a,k_b,k_c\in\mathbb{Z}$. We know that $ k_a \plus{} k_b \plus{} k_c \equal{} 35$, and the number of solutions to this is $ \dbinom{35 \plus{} 3 \minus{} 1}{3 \minus{} 1} \equal{} \boxed{666}$. :diablo:  :diablo:  :diablo: says I'm correct.

Wait math154 how do you know to do that? Balls and urns?

Yes...search it if you don't know what it is. It is like the most asked topic on AoPS.


A farmer was cleaning out his big old dusty farm shed.

He dragged out an old water container which was full of old dirty water and it weighed 12 pounds.

The farmer put something in the container and then it weighed less. What did he put in the container?

A hole?

(The message is too small. Please make the message longer before submitting.)

another good 1

[hide]alkali metals it exploded and therefore weighed less[/hide]


Có hai môi trường cần phân biệt, môi trường chữ thường và môi trường toán. Nếu là môi trường toán thì cần đặt giữa hai dấu đô la. Chi tiết xin xem tại LaTeX help tại diễn đàn này.

[quote="Nbach"]Cac ban Viet Nam oi,chi giup minh cach go LaTex tren mathlinks voi .Kem that day.vao tu hoi toi h ma chua biet
cach go the nao ?[/quote]
Xem o day nay http://www.mathlinks.ro/LaTeX/AoPS_L_About.php. Khong lam duoc thi vao day ma hoi http://www.mathlinks.ro/Forum/index.php?f=123.


[hide]
Let $p=2006$, $S=\sum_{n=1}^{89} \frac{1}{1+\tan^p n^\circ}$. Reverse the summation to get

\[ S=\sum_{n=1}^{89} \frac{1}{1+\tan^p (90-n)^\circ}=\sum_{n=1}^{89} \frac{1}{1+\cot^p n^\circ}=\sum_{n=1}^{89} \frac{\tan^p n^\circ}{1+\tan^p n^\circ}  \]

Adding the two term-by-term, $2S=\sum_{n=1}^{89} 1=89\Rightarrow S=\boxed{\frac{89}{2}}$ (and independent of $p$).
[/hide]

What do you mean by "reverse the summation"?
-Swapnil

"Reversing the summation" means writing the terms in reverse order. For example, if $A=1+2+3+\cdots +n$, then $A$ also equals $n+(n-1)+(n-2)+\cdots +1$.

that is an ingenious solution, can you point in the direction of where to get help with tricky sums to infinity?

This is a great problem and I certainly wouldn't be able to get in short amount of time. But here is quick tip on approaching this type of problem.

1. It's pretty hard to find $1 + \tan^{2006} n$ for any angle without calculator. And even with it, it's not very fun. So, we need to consider general case of $\tan^{p}$ or whatever you want to call the exponent.

2. Note that summation goes from 1 to 89. Moreover, we have $\tan (n) = \cot (90-n)$ which fits the given interval. So, it might be a good idea to rewrite the sum BACKWARD.

3. After you done #2, compare two sum (which are same) in terms of $\tan$. Do you see anything that's nice? Find this by adding/subtracting if you need to. The desired sum should follow.

Note that this is exactly what he did except I explained how you would approach this (without seeing his solution).


function
calculus
derivative
combinatorics solved
combinatorics

suppose for a>=b>=c>0
find number of sols to a+b+c=n, for some integer n
can we generalize?

we claim that the number of ordered n+1-tuples (x1,...,xn+1) with x1>=x2>==...>=xn+1>0 and sum m equals: m-1 C n.
proof 1. let yi =xi-1 then sum yi = m-(n+1) and yi>=0
now consider an m*n chessboard and consider a player to be at bottom left corner of this table. he can move either horizontally toward right or upward. the number of ways that he can reach the top right hand corner equals: m+n C m
but each path is a partition of the number m as the sum of n+1 nonnegative integers.
hence the number of ordered partitions into nonnegative integers will be
n+m C n.
proof 2.
we use generating functions. the number of ordered partitions of m into n+1 non-negative integers is coefivient of x^m in the expansion of
(1/1-x)^n+1 = 1-x)^-n-1  which equals m+n C n.

I do not think that order matters here..., because y_(n+1)>=y_n>=...>=y_1
I wrote a small program to verify and the answers do not match your formula, Ali.

I think that the number of solutions is:

Nr=1/n! f^(n) (0) where
f^(n) (0) is the n-th derivate of the function f(t)=1/(1-t)^3 in point t=0.
cheers! :D :D

And of course the problem can be generalized!

ax+by+cz=n with x,y,z fixed, x,y,z are natural numbers!
And you apply the same thing to this one, to find the number of it's solutions!
cheers! :D :D

the derivitive thing is pretty kool!!  :D
if i computed things correctly, if a+b+c=n, for a>=b>=c>0,

f^(-1)(t)=(-3)(1-t)^(-4)
...
f^(-n)(t)=(-3)(-4)...(n+2)(1-t)^(-n+3)
so Nr=(n+2)!/((2!)(n!))=Nr=(n+2)C2.  quiet similar to what Ali had, but i dont it is correct either.

for a+b+c=n, we have for n=3,4,...
Nr=1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,48,52

anyways, Ali had a point, Nr should be the number of partitions with at most 3 parts
so a(n) satisfy the recursion a(n)=1+a(n-2)+a(n-3)-a(n-5) (why??)

bugzpodder, my solution is corect iff we don't have the condition a>=b>=c(now I have seen it :( )
I computed all the solutions! for example n=4
your answer is 1, when mine is 3:
2+1+1=4 (your only solution)
and my solutions: 1+1+2,1+2+1,2+1+1. I think that all  we need now to compute is which number of solutions we sould substract from my number!

cheers! :D :D

[quote="bugzpodder"]suppose for a>=b>=c>0
find number of sols to a+b+c=n, for some integer n[/quote]
First of all, we will compute number of solutions $x+y+z=n$, $x,y,z\in\mathbb{N}$.
It is a well known problem and answer is $A=C_{n-1}^2=\frac{(n-1)(n-2)}{2}$. Now substract from $A$ threefold number of all solutions, for which $x,y,z$ are not pairwise distinct numbers. It is $[(n-1)/2]$ (take into consideration that if $n$ is divisible by 3 then we have solution with equal $x,y,z$.) Remaining triples can be uniquely ordered, and since each such triple has 6 different permutations, we obtain final answer:
\[\frac{\frac{(n-1)(n-2)}{2}-3\left[\frac{n-1}{2}\right]+2\xi_n}{6}+\left[\frac{n-1}{2}\right],\]
where $\xi_n=1$  if $n$ is divisible by 3, and $\xi_n=0$ otherwise.


Find the volume in cubic inches of the solid generated by revolving a region bounded by an equilateral triangle of side 4 inches about one side.

I would use calculus but I can't in this forum.  When you revolve some region around an axis, you are essentially creating a cylindrical-type shape, which has volume $ V\equal{}\pi r^{2}h$ of course.  Therefore, you want to do something similar here.

The problem is especially designed so that calculus is not necessary to solve this problem.

[hide="Hint"]Cones[/hide]

[hide="Solution."]
You are essentially creating two cones; the height of each is half the side length of the equilateral triangle and the radius of each is the same as the triangle's height.  Therefore, the total volume is $ \frac{4\pi}{3}( 2\sqrt{3} )^2$.  That comes out to ${ \boxed{16\pi}}$.[/hide]

[hide]If we unravel the 3D shape, we obtain a triangular prism with the equilateral triangle as its base, and the height equal to the path that the centroid of the triangle travels through while revolving. The centroid of the triangle is $ \frac{2\sqrt{3}}{3}$ units away from the axis of revolution. The volume of the object is then just the area of the base times that length, or:
\[ \frac{\sqrt{3}}{4} \cdot 4^2 \cdot 2 \pi \cdot \frac{2\sqrt{3}}{3} \\
\equal{}\boxed{16\pi}\][/hide]


calculus
integration
limit
logarithms
calculus computations

Find the value of $a$ such that $\lim_{n\to\infty}\frac{3}{2}\int_{-\sqrt[3]{a}}^{\sqrt[3]{a}}\left(1-\frac{t^{3}}{n}\right)^{n}t^{2}\ dt=\sqrt{2}\ \ (n=1,2,\cdots).$

I got $a = \ln(\sqrt2+\sqrt3)$.

[b]Benorin[/b], you have to discard the negative sign in your answer for an obvious reason.

[quote="Ramanujan"][quote="maokid7"]well...this might seem a little unnecessary but could you please show how you arrived at your answers.[/quote]

Here $\lim_{n \to \infty }\left( \int f_{n}(x) \, dx \right)= \int \left( \lim_{n \to \infty }f_{n}(x) \right) \, dx$.[/quote]

That is one way... also notice that

$\frac{3}{2}\int_{-\sqrt[3]{a}}^{\sqrt[3]{a}}\left(1-\frac{t^{3}}{n}\right)^{n}t^{2}\ dt=\left[ \frac{\left(n-t^{3}\right)^{n+1}}{2n^{n}(n+1)}\right]_{t=-\sqrt[3]{a}}^{\sqrt[3]{a}}$


What and how do you find the answer of:

Find the coefficient of $ x^{25}$ in $ (x^2\plus{}x\plus{}3)^{10}$.

Thanks!

[hide]there is no $ x^{25}$ term in $ (x^2\plus{}x\plus{}3)^{10}$; the highest term is $ x^{20}$.

but if that was indeed the question, then it's $ \boxed{0}$.[/hide]

[quote="HollyDrakes256"]What and how do you find the answer of:

Find the coefficient of $ x^{25}$ in $ (x^2 \plus{} x \plus{} 3)^{10}$.

Thanks![/quote]
[hide]
Well, just by looking at that, we see that the highest term of x is $ x^20$.  Therefore, the coefficient of $ x^25$ is 0.  >_>[/hide]

[quote="AdithyaGanesh"][quote="HollyDrakes256"]What and how do you find the answer of:

Find the coefficient of $ x^{25}$ in $ (x^2 \plus{} x \plus{} 3)^{10}$.

Thanks![/quote]
[hide]
Well, just by looking at that, we see that the highest term of x is $ x^{20}$.  Therefore, the coefficient of $ x^{25}$ is 0.  >_>[/hide][/quote]

Can't be lazy.  You have to put the number in {}'s.

As for how you'd find the answer in a problem that isn't this stupid, let's try something like the coefficient of $ x^{12}$ in $ (x^2\plus{}x\plus{}3)^{10}$.

In this case, you have to take terms and bunch them together to create $ x^{12}$. With consideration to the powers of $ x$, we have something like $ 2a\plus{}b\equal{}12$, where $ a$ is the number of $ x^2$ terms and $ b$ is the number of $ x$ terms.

$ a$ goes from $ 0$ to $ 6$, and for each $ a$ we get some $ b$ value that could leave constant terms to be multiplied in. At each value of $ a$, we have $ \binom{10}{a}$ ways to pick the $ x^2$s, $ \binom{10\minus{}a}{b}$ ways to pick the $ x$s, and an extra $ 3^{10\minus{}a\minus{}b}$ constant term. That is, the coefficient is \[ \sum_{a\equal{}0}^{6}\binom{10}{a}\binom{10\minus{}a}{12\minus{}2a}3^{a\minus{}2}\equal{}\boxed{115965}\]


A number is randomly selected from the integers 1 through 9.
What is the probability that the number is less than 7 and
divisible by 3? Express your answer as a common fraction.

The only ones that work are $ 3$ and $ 6$.  There are $ 9$ total numbers, so $ \boxed{\frac{2}{9}}$


integration
calculus
function
real analysis
real analysis unsolved

iwhere $ F(x)\equal{}\int_a^x f$.
Also f:[a,b] -> R.
I think my biggest problem here is that I can't say f if bounded (i.e. that it has a finite maximum).
any ideas on how I can show this?

thanks

Do you know how to show this:
If $ \mu$ is some positive measure and $ f$ is integrable and $ \lambda(E) \equal{} \int_E f \,d\mu$, then for any $ \epsilon > 0$ there is a $ \delta > 0$ such that $ \mu(E) < \delta$ implies $ |\lambda(E)| < \epsilon$?

There is a more general statement, namely that if $ \nu$ and $ \mu$ are measures such that $ |\mu|(E) \equal{} 0$ implies $ \nu(E) \equal{} 0$ then the above condition holds.

huh, not really...and this looks more complicated.  :maybe:
Basically, i'm supposed to get this from the definitions of uniform continuity.

Well, then, do you know that $ f$ can be approximated in the integral norm by functions that ARE bounded? For example, simple functions.

my take on this was to show that
$ |F(y)\minus{}F(x)| \leq \int_x^y |f|$
which is easy to show using triangular inequality. Now, the right hand side is equal to zero when y=x, and it's an increasing function. But to get uniform continuity I need to show
$ \forall \epsilon>0, \exists \delta>0, \forall x,y, |x\minus{}y|<\delta \Rightarrow |F(x)\minus{}F(y)|<\epsilon$
i'm having trouble connecting these last two steps...

You indicated that you understand how to do with with $ f$ bounded... so do you see how to connect that with my previous post?

true.
I'm sorry, I missed your argument.
Ok, so I know how to do with bounded ones. But what I don't know is how to extend that - i.e. I don't know how I could approximate
the function in the integral norm by functions that are bounded.

What do you mean by "integrable"? Do you mean "Riemann integrable" or "Lebesgue integrable"?
Hint: every Riemann integrable function is bounded.

This may help clear up a few things:

http://forums.somethingawful.com/showthread.php?noseen=0&threadid=2843892&pagenumber=1


vector
geometry
parallelogram
function
limit
real analysis
real analysis unsolved

Here is a quite hard problem:
  Let $ (E,||)$ be a real normed vector space with the property that for any $x$ in $E$ the set of those $y$ such that $||x+y||^2=||x||^2+||y||^2$ is stable my multiplication with real scalars. Then prove that the norm comes from a scalar product.

Oh, it's not that hard.. :)

The aim will be to show that for all $x,y\in E$ the parallelogram identity holds, i.e. $\|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2)$. Whenever this is true, we'll say that $P(x,y)$.

First of all, let's fix some $x,y\in E$ for which $\|x+y\|^2=\|x\|^2+\|y\|^2$ holds (in this case, we'll say that $Q(x,y)$). It's very easy to prove, by the given hypothesis, that for any $\alpha,\beta,\gamma,\delta\in\mathbb R$, if we put $u=\alpha x+\beta y,\ v=\gamma x+\delta y$, then $P(u,v)$.

Now assume we're given $x,y\in E$. According to the paragraph above, in order to show that $P(x,y)$, it suffices to show that $x,y$ lie in a subspace of $E$ generated by two vectors $z,t$ s.t. $Q(z,t)$. To this end, we look for vectors of the form $z=x,t=\alpha x+y$. It would all be over if we could prove the following: there is a real $\alpha$ s.t. $Q(x,\alpha x+y)$.

Consider the function $f:\mathbb R\to\mathbb R$, defined by $f(\alpha)=\|x+\alpha x+y\|^2-\|\alpha x+y\|^2$. We want to show that $\|x\|^2$ lies in the range of $f$. However, it's easy to prove (exercise :)) that given $a\in\mathbb R$ and $x,y$ vectors in some normed vector space, we have $\lim_{n\to\infty}(\|(n+a)x+y\|-\|nx+y\|)=a\|x\|\ (*)$. it's clears now how to use $(*)$ to show that as $\alpha\to\pm\infty,\ f(\alpha)$ goes to $\pm\infty$ respectively, so, since $f$ is continuous, $\|x\|^2$ must lie in the range of $f$.

I hope it's ok.
I saw $(*)$ as an exercise in a book a long time ago. It's a very good book, actually, "Teoreme si Probleme de Analiza Matematica", by Sorin Radulescu and Marius Radulescu.

Ok if you say so it's not that hard. Yet, I had no chance in proving it... :(


topology
Functional Analysis
advanced fields
advanced fields unsolved

I have a question on some notation that appears in Hilbert Spaces.

I came a cross a piece of a proof that says

[quote]
To prove uniqueness it suffices to establish that if $ M_1$ and $ M_2$ satisfy (1) and (2), then so also $ M\equal{}(M_1 \plus{} M_2)^\minus{}$ satisfies (1) and (2). ......
[/quote]

Here $ M_1$ and $ M_2$ are subspaces of the Hilbert space $ H.$ I do not understand what the symbol "$ ^\minus{}$" means in $ (M_1 \plus{} M_2)^\minus{}.$

You can find the original paper in [url=http://www.springerlink.com/content/k367q0w08j702123/fulltext.pdf]here[/url] and this is the part on page 390, proof of Prop 1.7.

[quote="lenny"]I do not understand what the symbol "$ ^ \minus{}$" means in $ (M_1 \plus{} M_2)^ \minus{} .$[/quote] I did not look at the article but I would bet my house ;) that it means the topological closure of the subspace $ M_1\plus{}M_2$ in your Hilbert space.

Ok, but what is the difference between the topological closure and just the "closure"? In other words, what role does the term topological play in here? What is the topology in the hilbert space?

Closure and topological closure are synonymous in this context. A Hilbert space is a metric space induced by its scalar product and as such a topological space.


Given system equation:
$ \left\{ \begin{array}{l}x \plus{} xy \plus{} y \equal{} m \plus{} 2 \\ x^2 y \plus{} xy^2  \equal{} m \plus{} 1 \\\end{array} \right.$
Find m such that the system equation above has one root.

hmm so
from the equations, i know that the top one is 1 more than the bottom one
(x^2y) + (xy^2) + 1 = x+ xy +y
move everything to the left
(x^2y) + (xy^2) - x - y - xy + 1 = 0
factor
xy(x+y-1) - (x+y-1) = 0
(xy-1)(x+y-1) = 0

if it only has 1 set of roots then

xy-1=0,  xy=1
x+y-1=0,  x+y=1

sub these 2 back to the equations

i got m=0

[quote="lgd0612"]i got m=0[/quote]
I don't think so because if m=0 then we have
$ \begin{array}{l}\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}c}{xy(x \plus{} y) \equal{} 1}  \\{x \plus{} y \plus{} xy \equal{} 2}  \\\end{array}} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}x \plus{} y \equal{} 1 \\ xy \equal{} 1 \\ \end{array} \right.(*) \\ \end{array}$
But system of equation (*)  has no solution :?:
May be you forget $ (x \plus{} y)^2  \minus{} 4xy \ge 0$

[quote="Clear_sky"]
$ (x \plus{} y) \plus{} xy \equal{} m \plus{} 2$

$ xy( x \plus{} y ) \equal{} m \plus{} 1$

$ a^2 \minus{} (m \plus{} 2)a \plus{} (m \plus{} 1) \equal{} 0$

$ D \equal{} m^2 \equal{} 0$

$ m \equal{} 0$[/quote]
Are you sure ?Have you read my reply above ?


Congratulations to Arne for becoming a moderator in the Olympaid section.
Anybody knows where is our dear [b]Myth[/b]?
It seems that he is not any more too interested in the forum.
We missed him a lot. :?

Actually I was a moderator in the past, I just had a break  :D

I guess Myth is very busy at the moment. Haven't heard from him for a long time.

Myth is still visiting this site. I saw his name on the "who is online" list.

yes indeed , i have just seen his name in the Who is online  :D  :roll:
(at this very moment)

A new congratulation
hey guys we have a new admit .
Lets say our loyalty to him  :)
[color=orange]nsato[/color]
:clap2:  :omighty:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46953[/url]

Hey I think Valentin idea has changed ,cause  he had said it  is impossible to became an admin :D

[quote="math92"]Can i be a administrator? [/quote]
[quote="valentin vornicu"]It is impossible.[/quote]
[quote="colts18"]What do you mean???(Why Not?)[/quote]
[quote="valentin vornicu"] :huh: Because it just is impossible.[/quote]
[quote="bubala"]But what if they make IMO and get a college degree and whatnot and AoPS hires them. Then they'd be admins[/quote]
[quote="valentin vornicu"]That's a totally different approach   But for now math92 must be careful with his spam warnings

Topic is officially closed[/quote]

you need to worry about not quadruple posting  ;)  :D

He was replying to math92.  Thus, it is implied that he meant it's impossible for math92 to become an admin (which, btw is most likely correct :D ) .

nsato is Naoki Sato, the new instructor at AoPS.  So of course he becomes an administrator ;)


What is the least positive integer with exactly five distinct
positive factors?

so the number must be in form of $ x^4$.

x is the smallest prime since the problem is asking for least positive integer.

$ 2^4\equal{}16$

answer : 16

Or you could realize that only square numbers have an odd number of distinct factors and easily see that the smallest perfect square with 5 factors is $ \boxed{16}$.


[quote="tetrahedr0n"]First, you need to prove that

$\lim_{x \rightarrow a}(f(x) + g(x)) = \lim_{x \rightarrow a}(f(x)) + \lim_{x \rightarrow a}(g(x)) $
This isn't too hard of a property to prove. Think about the definition of a limit.

Now you also need these fairly simple limits:

$\lim_{x \rightarrow a}(c) = c$ and
$\lim_{x \rightarrow a}(2x) = 2a$

Again, try to find an deltas for any epsilon.[/quote]

Hint: Break it up into two limits (each with their own delta and epsilon) and use the triangle inequality.

If you think about the definition of a limit, what you are trying to prove is that when $x \approx -3$, $5-2x \approx 11$.   Another property of limits that helps here is $\lim_{x \rightarrow a} f(x) = f(a)$ if $f(x)$ is defined at $a$.

[quote="Rep123max"]And now I actually got one more question in proving that something is not a limit. It wants me to prove that $\lim_{x\rightarrow1}(3x+4)\not=6.5$.

All I can do is show that 7 fits the definition, not that 6.5 does not.[/quote]

Why not pick  delta = 0.1?

Then |x-1| < 0.1 and 6.7 < |3(x-1) + 7| < 7.3.

Hence this fails to meet the definition of a limit.  Furthermore, no matter what we say the limit is (as long as it's not 7), we can choose delta to be arbitrarily small so that |x-1| < delta implies |3(x-1) + 7| is contained within an interval not containing that number.


Find the sum of all four-digit numbers whose all digits are odd.

This is indeed easy. :)

[hide="Solution"]
There are $ 5$ odd digits, so there are $ 5^4$ such numbers in the set (call it $ S$). Clearly, each number other than $ 5555$ corresponds to a single other number in $ S$ such that the average of the two numbers is $ 5555$ (a bijection from the $ S$ to $ S$, with a fixed point at $ 5555$). Adding all $ 5^4\minus{}1$ such numbers, we see that there are $ \frac{5^4\minus{}1}{2}$ such correspondences in total, each contributing $ 5555\cdot 2$ to the total sum. Thus, the total of the numbers (now including $ 5555$ itself) will be $ \left(\frac{5^4\minus{}1}{2}\right)\cdot 5555\cdot 2\plus{}5555\equal{}5^4\cdot 5555\equal{}3471875$.
[/hide]


[hide="maybe"]Your problem equivalent to $27\prod [a(a+b+c)-bc]\leq 8\prod a$. Now , you can use Muihard's theorem or Schur's theorem. I think so.[/hide]

Another idea.
Put $x=abc,y=ab+bc+ca$.
Then $(a-bc)(b-ca)(c-ab)=4x-x^{2}-y^{2}-2xy$. So, we need prove $100x\leq 27(x+y)^{2}$, or $f(a,b,c)=10\sqrt{abc}-3\sqrt{3}(abc+ab+bc+ca)\leq 0$. We have $f(a,b,c)-f(a,t,t)=(\sqrt{b}-\sqrt{c})^{2}(\frac{3\sqrt{3}}{4}(\sqrt{b}+\sqrt{c})^{2}(a+1)-5\sqrt{a})\leq 0$, where $a=\max{(a,b,c)}$ and $t=\frac{b+c}{2}$. Done!.

[quote="arqady"][quote="Twins"]$a,b,c>0$, $a+b+c=1$, Prove that $27(a-bc)(b-ca)(c-ab)\leq 8abc$[/quote]
Maybe the following proof better? :maybe:
If $a-bc<0$ and $b-ac\geq0,$ $c-ab\geq0$ or $a-bc,$ $b-ac$ and $c-ab$ are negative numbers  its true.
$a-bc<0$ and $b-ac<0$ it's can't be.
Let $a-bc,$ $b-ac$ and $c-ab$ are non-negative numbers.
Then $\sqrt{(a-bc)(b-ac)}\leq(1-c)\sqrt{ab}.$
Hence remain to prove that $27(1-a)(1-b)(1-c)\leq8,$ which triviality.[/quote]
My solution is similar to yours, but I want to find a nicer solution

We are to prove $\left(1-\frac{bc}{a}\right)\left(1-\frac{ca}{b}\right)\left(1-\frac{ab}{c}\right)\leq \frac{8}{27}.$ Honestly I have finished,  but my server has fade out.

I am shocked. :(

We can use Mixing vaiaber  Or Cauchy_svac ineq
Sorry it is verry long . I will post it on next sunday .   Now i verry busy :maybe:


function
real analysis
real analysis unsolved

Suppose $\mu()$ is a Radon measure in $\mathbb{R}^{n}$ on some $\sigma$-algebra $\Sigma$ of subsets of $\mathbb{R}^{n}$. Let $f: \mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ be continuous and proper (i.e. inverse images of open and compact sets are open and compact, resp.) with $f^{-1}(b)$ Borel for $b$ Borel.

Do we know $f^{-1}(f(A)) \in \Sigma$ if $A \in \Sigma$?

Sorry.  I don't know how to delete the other posts.


algebra
polynomial
geometry
incenter
analytic geometry

1.  Find a polynomial $ P(x)$ such that $ P(x)$ is divisible by $ x^2\plus{}1$ and $ P(x)\plus{}1$ is divisible by $ x^3\plus{}x^2\plus{}1$.

(I got to the part where $ P(x)$ must have a degree of at least 3, and then I got stuck  :oops: )

2.  Let $ P$ be an interior point of $ \triangle{ABC}$, and let $ x,y,z$ denote the distances from $ P$ to $ BC$, $ AC$, and $ AB$ respectively.  Where should $ P$ be located to maximize the product $ xyz$?

(I have a hunch that $ P$ is at the incenter, but I'm not sure how to prove it using analytic geometry without diving into a pit of vicious variables.)

[hide="1"] See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=125861]this old post of mine[/url].  This is an application of the Chinese Remainder Theorem.

Write $ P(x) \equal{} (x^3 \plus{} x^2 \plus{} 1) Q(x) \minus{} 1$.  Then $ P(x) \equiv Q(x) \minus{} 1 \bmod (x^2 \plus{} 1)$, so our general solution is $ Q(x) \minus{} 1 \equal{} R(x) (x^2 \plus{} 1)$.  Letting $ R(x) \equal{} 1$, we have

$ P(x) \equal{} \boxed{ (x^3 \plus{} x^2 \plus{} 1)(x^2 \plus{} 2) \minus{} 1 }$. [/hide]


I don't know how challenging this is, but here's my "homemade" challenge question.

Evaluate the limit as n approaches infinity of 10^n*100sin(9/(5*10^n)).

By subbing [tex]x=10^{-n}[/tex], the limit reduces to

[tex]\displaystyle \lim_{x\to0}\frac{100\sin\left(9x/5\right)}{x}[/tex].

This limit is indeterminant (since it approaches 0/0), so L'Hopital's rule says that this limit equals

[tex]\displaystyle \lim_{x\to0}\frac{100\cdot9/5\cdot\cos\left(9x/5\right)}{1}=180\cos0=180[/tex].

[quote="ZennyK"]Oops! I forgot part of the problem! The sine function in this problem is supposed to be in trigonometric form. So, converting the answer from degrees to radians, the limit equals :pi:.[/quote]

What do you mean "trigonometric form"?

Instead of circular, i.e. its argument is in degrees rather than radians.