Solve the following simultaneous equations to get value of (x+y):

$ y^{2}x\equal{}15x^{2}\plus{}17xy\plus{}15y^{2}$
$ x^{2}y\equal{}20x^{2}\plus{}3y^{2}$

Not much help but x is a positive integer between [hide]1 and 999, inclusive[/hide]

Real solution only $ x\equal{}19,y\equal{}95$.

EXAMPLE
$ y^{2}x(20x^{2}\plus{}3y^{2})\equal{}x^{2}y(15x^{2}\plus{}17xy\plus{}15x^{2})$
Use equalities equation .
@ to Rust : $ (x,y)\equal{}(0,0)$ is a solution so you are wrong.

[quote="Gib Z"]Hey sorry to revive old threads, but could someone give a detailed solution instead of just the answer please? I need working :([/quote]
Sure.
if $ y\equal{}0$ $ \rightarrow$ $ x\equal{}0$, so $ (0, 0)$ is also a solution... (but not if you solve it in the interval $ {1, ... , 999}$, tough I consider that fact useless because this system can be solved in $ R$).
Let $ a\equal{}\frac{x}{y}$. If we divide both equations by $ xy$ we obtain the system:
$ x\equal{}20a\plus{}\frac{3}{a}$ $ (1)$
$ y\equal{}15a\plus{}\frac{15}{a}\plus{}17$ $ (2)$
If we divide $ (1)$ by $ (2)$ we obtain : $ a\equal{}\frac{20a\plus{}\frac{3}{a}}{15a\plus{}\frac{15}{a}\plus{}17}$ ...
In the end, when we expand we have that $ 15a^{3}\minus{}3a^{2}\plus{}15a\minus{}3\equal{}0$ $ \leftrightarrow$ $ (5a\minus{}1)(3a^{2}\plus{}3)\equal{}0$. So, $ a\equal{}\frac{1}{5}$.
Then we just replace $ y$ by $ 5x$ and we obtain the exact solution which Rust said...


Suppose that $ f: \mathbb{R} \rightarrow \mathbb{R}$ be two times differentiable and $ \forall x \in \mathbb{R}  ; 2f(x\plus{}1)\equal{}f(x)\plus{}f(2x)$. Prove that $ f$ is a constant.

If the condition were "twice [b]continuously[/b] differentiable", the problem would be easy: since $ 4f''(2x)\equal{}2f''(x\plus{}1)\minus{}f''(x)$, we would immediately conclude that  $ \max_{[\minus{}a,a]}|f''|$ does not exceed $ 3/4$ times itself for all $ a\ge 2$, i.e., that $ f''\equiv 0$, after which there would be almost nothing to do. As it is, the problem is less obvious because it may happen that $ f''$ is unbounded on every interval :?


Thee tiles are marked $ X$ and two other tiles are marked $ O$. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads $ XOXOX$?

$ \textbf{(A)}\ \frac{1}{12}\qquad
\textbf{(B)}\ \frac{1}{10}\qquad
\textbf{(C)}\ \frac{1}{6}\qquad
\textbf{(D)}\ \frac{1}{4}\qquad
\textbf{(E)}\ \frac{1}{3}$

[hide]

There are a total of 5!=120 arrangements of 5 X's and O's.

for $ XOXOX$, There are 3 X's for the first spot, 2 O's for the second spot, 2 X's for the third spot, 1 O for the 4th spot, and the last X for the last spot. Therefore there are
$ 3!\cdot 2!$ arrangements of XOXOX.

$ 3!\cdot 2!\equal{}6\cdot 2\equal{}12$

$ 12/120\equal{}1/10$

[/hide]

[hide="Solution"]The desired pattern can happen in $ 3!\cdot2!\equal{}12$ ways.
The total number of patterns is $ 5!$
We have $ \frac{12}{120}\equal{}\boxed{\textbf{(B)}\ \frac{1}{10}}$[/hide]

grn_trtl and 27r assumed that the $ X$'s and $ O$'s were distinguishable, which is fine.  You can also count them as not being distinguishable.
[hide]Therefore, there are $ \frac{5!}{3!2!}\equal{}10$ ways to arrange the $ X$'s and $ O$'s.  There is only way to make the permutation $ XOXOX$, so the answer is $ \boxed {\frac{1}{10}}$.[/hide]

[hide] There are $ \binom{5}{3}\equal{}10$ ways in which to place the Xs. Only one is the desired outcome. Thus, it's $ \textbf{(B)}\ \frac{1}{10}$.[/hide]

You can also simply use basic combinatorics,

there is a $\frac{3}{5}$ chance you take X, assuming you, did there's $\frac{1}{2}$ chance of taking "O", then $\frac{2}{3}$ chance of X again, and $\frac{1}{2}$ chance of O

which multiplies and cancels out nicely leaving $\frac{1}{5} \cdot \frac{1}{2}$ for total of $\frac{1}{10}$


sorry,edit
Let $a,b,c$ are real numbers  :(

Thank you Son and Can for your problem and solution.

oh... I am amazing because your solution is very nice  :P
Have any proof for it?  :roll:

[quote="zweig"]toanhocmuonmau > One question : I saw in other topic "LHS - RHS", what does it mean ?  :)[/quote]
LHS=left hand side
RHS=right hand side
:)


algebra
polynomial
linear algebra
linear algebra unsolved

Does anybody know a solution? I didn´t find anywhere in the site, so I posted here.

Let A e X  be real matrices with positive entries, of dimensions nxn and nx1, in that order. Show that if (A^2)X=X implies AX=X.

Is there anybody thinking about this problem?

Suppose $ A^2X\equal{}X$, then $ A^2X\minus{}X\equal{}0\Rightarrow (A^2\minus{}1)X\equal{}0$, so $ x^2\minus{}1$ is a factor in the characteristic polynomial. This then means that $ (x\minus{}1)(x\plus{}1)$ is a factor in the characteristic polynomial, so $ (A\plus{}1)X\equal{}0$ and $ (A\minus{}1)X\equal{}0$ $ \Rightarrow AX\equal{}\minus{}X$ and $ AX\equal{}X$.

There may be some mistake with jumping to that being a factor of the characteristic polynomial, though...

Yes, I think there is some mistake. You can´t ensure x^2-1  is a factor in the
characteristic polynomial. You can assure that characteristic polynomial has a factor ((x-1)^k)((x+1)^m), K+m<n+1. An idea is to prove that -1 can´t be an eigenvalue of A in these conditions.

You can use [b]Perron–Frobenius Theorem[/b]: [quote]Any square matrix with positive entries has a unique
eigenvector with positive entries (up to a multiplication by a positive scalar), and the
corresponding eigenvalue has multiplicity one and is strictly greater than the absolute
value of any other eigenvalue.[/quote]

We obviously have that $ Y\equal{}(A\plus{}I)X\neq 0$ and that $ (A\minus{}I)Y\equal{}0\Rightarrow AY\equal{}Y\Rightarrow A^2Y\equal{}Y$. Because $ Y$ has positive entries it follows from the theorem above that there exists $ \lambda>0$ so that $ \lambda X\equal{}Y\Rightarrow \lambda X\equal{}AX\plus{}X\Rightarrow$ $ AX\equal{}(\lambda\minus{}1)X\Rightarrow A^2X\equal{}(\lambda\minus{}1)^2X\equal{}X\Rightarrow$ $ \lambda\equal{}2\Rightarrow 2X\equal{}AX\plus{}X\Rightarrow AX\equal{}X$. q.e.d.


Question 1:
Can this be factor? if yes, can someone show me how?
14m^2 + 4mn - 15n

Question 2:
can someone show me how to do step 3 and explain how you did it?
step 1: 4x^2 + 12x + 9
step 2: (2x)(2x) + 2 (2x)(3)   + (3) (3)
step 3: ??

Thank you all

For your second part there, it turns out to be:

$(2x+3)^3$

how you get that? I knew the answer is (2x+3)^2
I want to know how you get the answer. explain it please

Thank you

First off, you can tell that it will be in the form (x + y)(w + z) because the middle and last factor are both positive.
Well, for this one, you can see that the first and last coefficients are perfect squares, so you could first try and factor it that way.
If that doesn't work, you can do as you did and see what numbers the first and last term factor into and just use guess and check to see which ones work.

Question one:
4x^2 + 24x + 9
(2x)(2x) + 4 (2x)(3) + (3) (3)

Question two:
4x^2 + 12x + 9
(2x)(2x) + 2 (2x)(3) + (3) (3)

To me, both Question one and Question two return (2x+3)^2
How does that work? Some one kindly show me how to do it in step by step?

I don't understand if you miswrote Question 1 or what.. but something went wrong.

And for question 1, I do not believe that it is factorable, I can't see a scenario where you could have a $m^2$ , $mn$ , and $n$ but no '$m$' or $n^2$ variable. I tried to find ways where the n would cancel out, but didn't get anywhere.

Number 1 is [b]not[/b] factorable.  Do you know about the discriminant test?  A quadratic is not factorable is $b^{2}-4ac$ is not a perfect square.  Have you learned about the quadratic formula?
$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=x$, as you can see if $b^{2}-4ac$ is not a perfect square, the solutions to $x$ are a radical.

Yeah, I did the discriminant test, it was just weird because of the variables.

Yes towersfreak2006, you understood exactly want I was asking for, thank you so much.  :lol:


[quote="Albanian Eagle"]prove that if $a,b,c$ are positive reals such that $a+b+c=1$
$\sum{{a}/{(2-a)}}+ab+bc+ca<1+abc$[/quote]
$\sum \frac{2}{2-a} +\sum ab < 4+abc$ (1).
Let $1-a=x$,$1-b=y$,$1-c=z$,$p=xyz=ab+bc+ca-abc$,then $x,y,z \in (0,1)$.$x+y+z=2$ and $m=xy+yz+zx=ab+bc+ca+1$.
Rewrite (1) as
$\sum \frac{2}{1+x} +xyz <4$
$\leftrightarrow$ $\frac{2(7+m)}{3+m+p} +p <4$
$\leftrightarrow$ $m-1 > \frac{p^2}{2-p}$
$\leftrightarrow$ $ab+bc+ca > \frac{p^2}{2-p}$
$\leftrightarrow$ $p+abc > \frac{p^2}{2-p}$
$\leftrightarrow$ $abc > \frac{2p(p-1)}{2-p}$,which is true because $abc >0$ and $p \leq  \frac{8}{27} < 1$.

this is true,
the inequality holds for all  $a+b+c \le 1$

This inequality is strong, for example:a~0,7;b~0,3;c=0,5

well maybe its general view may sound familiar
$\sum_{i=1}^n \frac{x_i}{1+S-x_i} + \prod_{i=1}^n (1-x_i) \le 1$
for all $x_i \in [0,1]$ if $S$ denotes $x_1+x_2+...+x_n$


Results:
Iurie Boreico aka Iura 34=gold
Alexandr Zamorzaev aka Sasha 32=gold
Simion Filip aka DusT 23=bronze
Vanea Borsenco 18=bronze
Andrei Galusca 17=bronze
Liubomir Chiriac aka sailor 16=bronze

Freemind I don't know who you are.
Would you be so kind to  tell your name?

Congratulations to the Moldavian team!
Especially, to Boreico and Zamorzaev who got gold!
Prefere to remain unknown for some time.

yes, congratulations for the first two gold medals for the Moldovian team and for the first time place in the first 20 countries :)

If i'm not wrong this topic and my reply are posted in the Result's of the 45th IMO 2004 so to which else do you think i'm refering to? :D  :D
BTW i know about his other two gold medals and they are of course great achievements too.  :)

[quote="Valentin Vornicu"][quote="Iris Aliaj"]Great job Moldova! Go on like that! ;)
Congratulations on your gold medal Sasha. Bravo!  :)[/quote] ahem, to which exactly of the three (3!!) gold medals (IMO, BMO and JBMO) that Sasha won this year are you referring to? :D:D[/quote]

Actually, Sasha got this year four gold medals. The three one that you specified, and one at Romanian MO.

if i recall correctly, sasha is 15... and he has three more gold medals to win :)
so, congratulations again (he was sitting in front of me, during the awarding, so i already made them to him, face to face)...

Ma_go, how did you solve at IMO?
Actually, Iura (Moldova) is relatively at the same age as Sasha (a little elder) and also got gold.
Iura also participated the previous year at IMO. He was 14 and got SILVER!!
He needed one more point to get gold.

Congrats again, medalist!
Iura, could you tell me how many points did you get at each problem? But Sasha or the others?

[quote="freemind"]Actually, Sasha got this year four gold medals. The three one that you specified, and one at Romanian MO.[/quote] oh forgotten about that ... indeed 4 gold medals in 1 year is the record! :) (previously held by reid barton with 3 gold medals in one year at one imo, one ioi and one in informatics balkan olympiad :))

If I remember well:
Iura 7 4 2 7 7 7
Sasha 7 7 2 7 3 6
DusT 6 7 2 7 0 1
Liubomir Chiriac 7 1 0 3 5 0
About other I don't remember

Congrats for the 2nd day!
 :) Iura, What do you think about this IMO? How do you appreciate the level of the problems?

Thanks to everyone for congratulations! :)

A good job,guys!
I suppose all winners are from moldovan-turkish lyceum?

Yes, all 6 participants are from Moldo-Turkish liceum.

[quote="Alex V. Bibin"]Congratulations and good luck!
Where will study the winners after graduating?[/quote]
Well for graduating there is still some years to pass.
The two gold-medalist finished the 9th grade so they habe 3 years still.
the 4-bronze medalists finished theree of them 10th grade and one 11th grade.
it's really cool!

Once again congratulations to the moldavian team, you rule fellas!!!!!!!!

Waw ... i'm really impressed about your results

I was a member of Moldova's team some years ago, and at that time we were very happy about our bronze medals :)
And now Moldova has gold medals, super

Marcel T shoud be verrrrry happy

Who was the leader and deputy leader of Moldova's team?
Was Marcel in?

Leader - Valeriu Baltag
Deputy leader - Liubomir Chiriac

Liubomir Chiriac? from ministry of education?  :?


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 10
USAMTS

AMC will be coming fairly soon. Are you guys looking forward to it?

Yes. This time I will not be dumb/lazy and take the AMC 10.

I am. I can see just how silly of an idiot I am.  :D

[quote="Elemennop"]Nope, absolutely terrified of it.[/quote]

haha you must be kidding  :lol:

the AMC is terrified by you.

I'm going to take the 10. I have 2 get 19 correct...not too hard, but im still nervous. Maybe i should have selected the 12 to take...noooo!!! Whatever... my goal is to make aime this year...

*crosses fingers and prepares to die at AMC*

I'm taking the 10 also.

I need 20+, since I'm definitely NOT leaving any blank or risking having just 19 right.

Yeah, but it's better to leave something blank than get it wrong... btw it used to be 2.5 for each blank answer...now it is 1.5   :o  :o  :o

oh really. wasn't aware of that. is that change for the AMC 12 as well?

Yep.  Hopefully the contest is fun.  Yeah, you could take the 10 A and then the 12 B (or switch the order).  Or, assuming that your school has enough copies of each test, you could just wait until when they're handing the tests out to decide whether to take the 10 or the 12—if enough tests have been ordered, you really don't have to decide until seconds before you open the test packet.

Happy New Year in, um, Ireland, everyone, by the way.

Will you get to know your score for the AMC A before you take the AMC B?

Happy New Year, everyone  :)

My goals for the AMC and AIME:

AMC: 120+
AIME: 10+

If I satisfy both, I'll hopefully make USAMO if there are no significant deviations in the index (considering that the index has crossed 230 only once in the past 5 years or so).

[quote="now a ranger"]Well I don't know if they will.

However, I will write down my answers on a seperate sheet of paper and check online on unl.amc.edu hopefully.

I'm only taking AMC 10A though.

 :lol:[/quote]

Or just circle them in your test booklet.

My goals:
AMC12: 130+
AIME: 9+
USAMO: 1+

[quote="Xantos C. Guin"][quote="Arvind_sn"]So has anyone made MOP? I know Miles has and Arnav, I hope so...did u make it Santosh?[/quote]

No, I didn't even make the USAMO. But I will make the USAMO this time. MOP is out of the question unless I improve proof writing skills and give up GHP.[/quote]

sorry to be off topic, but what is GHP?

and umm yeah USAMO is out of the question for me in 8th grade  :huh: only in geometry 1st year haha

Govenor's Honors Program. It is a 6 week program where students get to work in their areas of expertise.

Meh. I'm pretty confident I'll make USAMO, but about MOP,  :rotfl:

[quote="Boy Soprano II"]Good luck to everybody on the AMC A tomorrow!

[hide]Even to the Alabamans  :P [/hide]
Get some sleep tonight![/quote]

That was good advice. I think that I did well.

I got a good night's sleep

I feel asleep at 9:55 and woke up at 7 AM for school.

better than my usual sleep repertoire  :P

heh

I got pwned.

only a few points of the cutoff  :(  :(  :(

My score was exactly 114. I checked answers already. Well maybe I left one blank, not sure. doubt it.

114.0

[quote="drunner2007"][quote="Karth"]Dude I got pwned.

 :noo:

oh no!s... there go my chances for usamo :'(

If my crappy aime score increases, then maybe...[/quote]

what was your score around?

100-110, 110-120, ...

I have a feeling that this test was a little harder than last year. maybe it has something to do with the usamo expansion. since they're accepting more people, it wouldn't have been a bad idea to make amc a little harder.[/quote]

it was a 100+ for sure, but not sure if it's good enough for usamo...

i sorta thought the test was easier than past amcs i took...

but i sound contradictory

oh well

You might think the test was easier because you've gotten better.

Anyhow AIME counts lots more than AMC for USAMO purposes.  Remember that you can differ from another person by at most 50 index points from the AMC, but by up to 150 index points via AIME.

[quote="Benjamin Hu"]I was confident I made 100+ until my teacher told me I got 99 and qualified by percentage. :(  :mad:[/quote]
There went my chances of USAMO.


[quote="Utkirstudios"]Let a,b,c be positive real numbers.Prove that:
\[ \frac{{a^{2}-bc}}{{4a^{2}+4b^{2}+c^{2}}}+\frac{{b^{2}-ac}}{{4b^{2}+4c^{2}+a^{2}}}+\frac{{c^{2}-ab}}{{4a^{2}+4c^{2}+b^{2}}}\ge 0 \]
and find all equality cases.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123742


integration
limit
function
inequalities
real analysis
calculus
rudin

Find $ \int_{0}^{n}(1-\frac{x}{n})^{n}e^{x/2}\ dx$ for $ n \to \infty$

Certainly on $ [0,n]$ we have $ 0 \leq (1-\frac{x}{n})^{n}\leq 1$. Because this is true for every $ n$, $ (1-\frac{x}{n})^{n}e^{x/2}\leq e^{x/2}$ on $ [0,\infty)$. Also $ \lim_{n}\ (1-\frac{x}{n})^{n}e^{x/2}= e^{-x/2}$

This suggests strongly to apply the Dominated Convergence Theorem. But can we? The set on which we integrate is not fixed...  :blush:

Conceptual feedback is very welcome; the calculations are straightforward. Thats why I did not put it in the computations section.

We can make it a fixed set; just multiply by the characteristic function of $ [0,n]$ and integrate over the positive real line.

But then, $ e^{x/2}$ is no good as a dominating function, since it is not integrable.

You need this inequality: $ \left(1-\frac{x}{n}\right)^{n}\le e^{-x}$ for $ 0\le x\le n.$

With that in place, we can say that $ \left(1-\frac{x}{n}\right)^{n}e^{x/2}\chi_{[0,n]}(x)\le e^{-x/2}$ on $ [0,\infty).$

Now, we have the conditions of the DCT met.

[Where did you get this - an exercise in "green Rudin," perhaps?]

[quote]Thats why I did not put it in the computations section.[/quote]
I'd like to see the computations section kept to the level of a standard calculus sequence or differential equations course. Any mention of the Dominated Convergence Theorem has no place there. You posted this in the correct forum.

[quote="Kent Merryfield"]
[Where did you get this - an exercise in "green Rudin," perhaps?]
[/quote]

My paperback is red. But it came out of his 'Real and Complex Analysis'


I would be very thankful if any of you post any useful theorems you know.

There are a lot of theorems.  My personal favorite is Jensen's Inequality.

http://mathworld.wolfram.com/JensensInequality.html

for geometry, stewart, ceva, menelaus, ptolemy

Binomial Theorem, Chicken Mcnugget Theorem

learn every theorem in both aops books...

Thanks everyone.  Even though I knew about bionomial theorem, I searched for some information.  My only problem is that I can't find information about the Chicken Mcnugget Theorem and its applications.  Can any of you refer me to a book, link, etc.?  Thanks in advance.  By the way, any more theorems?

[url=http://mathworld.wolfram.com/HallsTheorem.html]Hall's Theorem[/url] doesn't seem very useful, but that may because I have never seen its applications.

Hmm....well, thank you.  By the way, everything is useful for something, especially things in mathematics.  Well, I suppose more theorems would be good...:D

Hmm...this seems to have interesting applications.  Do you know any problems that this theorem could be applied to?

[quote="RC-7th"]About the chicken mcnugget theorem:

If McDonalds sells McNuggets in boxes of m and n, then the largest number of chicken nuggets that you can't buy is mn-m-n.

For example, if you can buy McNuggets in boxes of 3 and 5, then the largeest number that you can't get is 15-3-5=7.[/quote]

I would like to correct you. m and n have to be relatively prime. A minor error, but still.  ;)

[quote="RC-7th"]About the chicken mcnugget theorem:

If McDonalds sells McNuggets in boxes of m and n, then the largest number of chicken nuggets that you can't buy is mn-m-n.

For example, if you can buy McNuggets in boxes of 3 and 5, then the largeest number that you can't get is 15-3-5=7.[/quote]

What do you mean by the maximum number of nuggets I can buy?  How does this work?  How is this limit created?  How is this theorem proven?  How would I use this theorem in a problem not revolving around nuggets?


Let a, b, c be positive integers. Prove that the inequality
\[ (x\minus{}y)^a(x\minus{}z)^b(y\minus{}z)^c \ge 0
\]
holds for all reals x, y, z if and only if a, b, c are even.

[quote="onlylove_math"][quote="STARS"]Let a, b, c be positive integers. Prove that the inequality
\[ (x \minus{} y)^a(x \minus{} z)^b(y \minus{} z)^c \ge 0
\]
holds for all reals x, y, z if and only if a, b, c are even.[/quote]
Are U sure your problem true?I don't think it allways true for all reals x, y, z and a, b, c be positive integers[/quote]
Oh my friend,i think you didint understand the question :P "
"holds for all reals x, y, z if and only if a, b, c are even."
If a, b, c are even."the proof is very easy
But the first case is hard
I only have some ideals :Let$ x\equal{}0,y>0,z<0$ and check $ a,b,c$
But i cant finish it .sorry :blush:

[hide="Solution"]
1) a,b, and c are odd. We take y>x>z and then inequality doesn't hold
2) Two of them are odd. WLOG b and c are odd. We take y>z>x and then inequality doesn't hold
3) One of them is odd. WLOG a is odd. We take y>x and then inequality doesn't hold
4) a,b and c are even. The inequality holds for every x,y and z.

I can take different values for x,y and z because it says that inequality is true for EVERY x,y and z.
[/hide]


Hi,
Could anyone help me the following question? Thx!!

By the ball centered at $ x$ and having radius $ r$ in which the set $ S_{x,\delta} \equal{} \{y: \rho (x,y)<r\}$

Prove that if $ 0< \epsilon < \delta\minus{}\rho (x,z)$, then $ S_{z,\epsilon} \subset S_{x,r}$

The containment, in terms of points: if $ y\in S_{z,\epsilon}$, then $ y\in S_{x,r}$. What do you have that might be useful? The triangle inequality.


I was solving a problem on one of my class message board, and I went back to add something, and as I typed, the letters started erasing.

So if I had
Example

It became

Thisple

HELP! This has never happened before!

It must be a local problem. Quick try to fix: restart your browser and then your computer if it still doesn't work.

Did you use Word?
If so, check at the bottom, see if the 3 letters OVR are in black instead of white. Then you've accidentally hit the Insert key (next to Backspace.)

That means when you type in blah and then put your mouse in the middle of the word and start typing, you get

b[u]hi[/u]b, or something similar.

@louiethechewy123: No, it was on AoPS, but thanks anyway.

Huh, I didn't restart the brower but it's fine now

Wait, I just discovered that the INSERT key works on AOPS too....  :o

It's not specific to AoPS, it's your Web browser.

Just so you know, this issue is solved for me. :wink:


Let $ a_i, b_i > 0$, $ i \equal{} 1,2,\cdots,n$. if $ ( \sum\limits_{i \equal{} 1}^na_i^p)^{\frac {1}{p}}\ge ( \sum\limits_{i \equal{} 1}^nb_i^p)^{\frac {1}{p}}$ for $ p \equal{} 1,2 and \plus{}\infty$.
prove or disprove that it holds for all positive integer $ p$.

Sorry, this is not true. Take $ a_1\equal{}4, a_2\equal{}3, a_3\equal{}3$ and  $ b_1\equal{}4,b_2\equal{}4, b_3\equal{}1$.


parameterization
number theory
Diophantine equation
relatively prime
number theory open

prove that there are infinetely many solutions in positive integers:
x^3+y^5=z^7

This is just an extension of the observation that $ x^{n\minus{}1} \plus{} y^{n\minus{}1} \equal{} z^n$ has the rather simple solution $ x \equal{} y \equal{} z \equal{} 2$.  The technique is as follows:  given the Diophantine equation

$ x^p \plus{} y^q \equal{} z^r$

where $ r$ is relatively prime to $ pq$, find positive integers $ u, v$ such that $ pqv \plus{} 1 \equal{} ru$ (which is possible by Bezout's lemma).  Then

$ x \equal{} a(a^p \plus{} b^q)^{qv}, y \equal{} b(a^p \plus{} b^q)^{pv}, c \equal{} (a^p \plus{} b^q)^u$

is a two-parameter family of solutions.


Let $ p$ be a prime. In case there exists an integer $ n$ such that $ n^3 \plus{} n^2p$ is a perfect cube, show that $ n < p$.

Is the above statement true (haven't been able to prove it). If not, any ideas of another upper bound on $ n$ (as a function of $ p$)?

$ 64^3 \plus{} 64^2\cdot 61 \equal{} 80^3$ is the smallest counterexample.  $ 216^3 \plus{} 216^2\cdot 127 \equal{} 252^3$ is next and $ 729^3\plus{}729^2\cdot 271 \equal{} 810^3$ is third.

You're right, thanks. Any idea on any bound on $ n$?

I think you can show $ n < p^2$.  Otherwise, $ 27(n^3 \plus{} pn^2)$ would lie strictly between $ (3n \plus{} p \minus{} 1)^3$ and $ (3n \plus{} p)^3$, which contradicts $ n^3 \plus{} pn^2$ being a cube.

Thanks!

Anything tighter? Or any other restriction to the choise of $ p$ or $ N$? I've shown that $ n$ must be a cube, at least if $ n$ is not divisible by $ p$. Perhaps this is impossible (yes it is, I found out). Futhermore, $ n < (p/3)^{3/2}$ is a tighter bound.


Let $ a,b,c\geq0$.Prove that:
$ \sqrt{a^2\plus{}bc}\plus{}\sqrt{b^2\plus{}ca}\plus{}\sqrt{c^2\plus{}ab}\leq \frac{3}{2}(a\plus{}b\plus{}c)$

[quote="Sunkern_sunflora"]Using Cauchy-schwarz,
[</sum>(a^2+bc)^1/2]^2 <= [3*(1/4)][</sum>2a^2+2bc] <= [3*(3/4)][a+b+c]^2
or, [</sum>(a^2+bc)^1/2] <= 3/2(a+b+c)..........Done!
Is my proof OK?? :?:  :!:[/quote]
I think you wrong  :wink:
[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ \sqrt {a^2 + bc} + \sqrt {b^2 + ca} + \sqrt {c^2 + ab}\leq \frac {3}{2}(a + b + c)$[/quote]
this is my proof

assume $ a + b + c = 2$

we need prove that :

$ \sqrt {a^2 + bc} + \sqrt {b^2 + ca} + \sqrt {c^2 + ab}\leq 3$

assume $ a\le b\le c$

by cauchy-schwarz
$ \sqrt {a^2 + bc} + \sqrt {b^2 + ca} + \sqrt {c^2 + ab}\le\sqrt {2(a^2 + bc + b^2 + ca)} + \sqrt {c^2 + ab}$

thus, we need prove that $ \sqrt {2(a^2 + bc + b^2 + ca)} + \sqrt {c^2 + ab}\le 3$

setting $ F(a,b,c) = \sqrt {2(a^2 + bc + b^2 + ca)} + \sqrt {c^2 + ab}$

$ F(0,a + b,c) - F(a,b,c) = \frac {4ab}{\sqrt {2[(a + b)^2 + c(a + b)]} + \sqrt {2(a^2 + b^2 + ac + bc)}} - \frac {ab}{c + \sqrt {c^2 + ab}}$

thus, we need prove that
$ F(0,a + b,c)\ge F(a,b,c)$ $ \longleftrightarrow 4c + 4\sqrt {c^2 + ab}\ge \sqrt {2[(a + b)^2 + c(a + b)]} + \sqrt {2(a^2 + b^2 + ac + bc)}$

by cauchy-schwarz
$ \sqrt {2[(a + b)^2 + c(a + b)]} + \sqrt {2(a^2 + b^2 + ac + bc)}\le$
$ \le2\sqrt {2(a^2 + b^2 + ac + bc+ab)}$

then, we is will proven $ 2\sqrt {2(a^2 + b^2 + ac + bc + ab)}\le 4c + 4\sqrt {c^2 + ab}$

$ \longleftrightarrow \sqrt {2(a^2 + b^2 + ac + bc + ab)}\le 2c + 2\sqrt {c^2 + ab}$

$ \longleftrightarrow a^2 + b^2 + ac + bc\le 4c^2 + 4c\sqrt {c^2 + ab} + ab$

because $ a\le b\le c \rightarrow a^2 + b^2 + ac + bc\le 4c^2$

$ \rightarrow a^2 + b^2 + ac + bc\le 4c^2 + 4c\sqrt {c^2 + ab} + ab$ is right

$ \rightarrow F(0,a + b,c)\ge F(a,b,c)$

thus, we is will proven  $ F(0,a + b,c)\le 3$

by Mixing Variable theorem we need  $ F(0,x,y)\le 3$ with $ x + y = 2$

$ \longleftrightarrow \sqrt {2(x^2 + xy)} + y\le 3$

$ \longleftrightarrow \sqrt {2(x^2 + x(2 - x))} + 2 - x\le 3$

$ \longleftrightarrow (x - 1)^2\ge 0$ (it right)

have done~  :wink:


function
geometry
geometric transformation
reflection
floor function

Clever.

[hide]Note that $ [a] \plus{} [a \plus{} 1/n] \plus{} \ldots \plus{} [a \plus{} (n \minus{} 1)/n] \equal{} [na]$ for any $ a$ and positive integer $ n$.

Add $ [\frac {3x \plus{} 3}{6}]$ to both sides. By the identity above,
\[ [\frac {3x \plus{} 3}{6}] \plus{} [\frac {3x \plus{} 5}{6}] \plus{} [\frac {3x \plus{} 7}{6}] \equal{} [\frac {3x \plus{} 3}{2}],
\]
and
\[ [\frac {3x \plus{} 3}{6}] \plus{} [\frac {x \plus{} 2}{2}] \equal{} [\frac {3x \plus{} 3}{6}] \plus{} [\frac {3x \plus{} 6}{6}] \equal{} [\frac {3x \plus{} 3}{3}] \equal{} [x \plus{} 1].
\]
Every term dies.[/hide]

(topic title edited to reflect problem better)

I'm sorry for the topic title , but my english isn't very good . :(

[hide]Rearrange,

$ [(3x\plus{}5)/6] \plus{} [(3x\plus{}7)/6] \minus{} [(3x\plus{}6)/6] \equal{} [(3x\plus{}3)/2] \minus{} [x\plus{}1]$

Let $ y \equal{} x/2\plus{}1$, thus,

$ [y\minus{}1/6] \plus{} [y\plus{}1/6] \minus{} [y] \equal{} [3y\minus{}3/2] \minus{} [2y\minus{}1]$.

Let [y] = a, then we have three cases,

C1: LHS = a-1 if $ a\le y<a\plus{}1/6$.
C2: LHS = a if $ a\plus{}1/6\le y<a\plus{}5/6$.
C3: LHS = a+1 if $ a\plus{}5/6\le y<a\plus{}1$.

What is RHS for each of those cases? Over the case's range,

C1:
$ 3a\minus{}3/2 \le 3y\minus{}3/2 < 3a\minus{}1$, thus, $ [3y\minus{}3/2] \equal{} 3a\minus{}2$
and
$ 2a\minus{}1 \le 2y\minus{}1 < 2a\minus{}2/3$, thus, $ [2y\minus{}1] \equal{} 2a\minus{}1$
difference is $ a\minus{}1 \equal{} LHS$.

C2:
Break into two cases, $ a\plus{}1/6\le y<a\plus{}1/2$
$ 3a\minus{}1 \le 3y\minus{}3 < 3a$, thus, $ 3a\minus{}1$,
and
$ 2a\minus{}2/3 \le 2y\minus{}1 < 2a$, thus, $ 2a\minus{}1$,
difference is $ a \equal{} LHS$.

AND

$ a\plus{}1/2/le y< a\plus{}5/6$,

$ 3a\le 3y\minus{}3 < 3a\plus{}1$, thus, $ 3a$,
and
$ 2a\le 2y\minus{}1<2a\plus{}2/3$, thus, $ 2a$,
difference is $ a \equal{} LHS$.

C3:
$ 3a\plus{}1 \le 3y\minus{}3 < 3a\plus{}3/2$, thus, $ 3a\plus{}1$,
and
$ 2a\plus{}2/3 \le 2y\minus{}1 < 2a\plus{}1$, thus, $ 2a$,
difference is $ a\plus{}1 \equal{}LHS$.

done.[/hide]

Question:  does Hermite's identity account for every nontrivial linear dependence among functions of the form $ \lfloor \frac{an \plus{} b}{c} \rfloor$?

$ [a]\plus{}[a\plus{}\frac{1}{3}]\plus{}[a\plus{}\frac{2}{3}]\equal{}[3a]$

$ [a]\plus{}[a\plus{}\frac{1}{2}]\equal{}[2a]\ ,\ (\forall)a\in \mathbb{R}$

$ \Rightarrow [a\plus{}\frac{1}{3}]\plus{}[a\plus{}\frac{2}{3}]\minus{}[a\plus{}\frac{1}{2}]\equal{}[3a]\minus{}[2a]\Leftrightarrow [a\plus{}\frac{1}{3}]\plus{}[a\plus{}\frac{2}{3}]\plus{}[2a]\equal{}[3a]\plus{}[a\plus{}\frac{1}{2}]$

Put $ a\rightarrow \frac{x\plus{}1}{2} \Rightarrow$ the identity .


parameterization
vector
topology
advanced fields
advanced fields theorems

Consider a differentiable manifold $M$ and a differentiable map: $\phi: M \times \mathbb{R}\rightarrow M$, such that $\forall x \in M, \phi(x,0)=id_{M}$ and $\forall s,t \in \mathbb{R}, \phi(\phi(x,s),t)=\phi(x,s+t)$.

Let $X(x)$ denote the tangent vector at $x$ given by the curve $t \rightarrow \phi_{x}(t)$.

Is $X$ a $C^{\infty}$-vector field, or is $X$ only $C^{1}$ ?

Related to your other question:  You can take X to have the minimum regularity required to generate a flow (using ODE existence thm).

On the other hand, it's possible that if you start with a differentiable flow, then its vector field might have even worse regularity than that, but I'm not sure.


group theory
abstract algebra
graph theory
combinatorics unsolved
combinatorics

hi chaps,

my name is karolina and this my first post.

i was wondering if there's any relation between the group of automorphisms of a graph and those of its subgraphs. i was thinking in particular of edge-erased subgraphs :maybe:

any insight would be mostly appreciated :)

1. Welcome :)

2. This is probably the wrong forum for this problem. Consider the College Playground section, maybe under "Superior Algebra - Groups, Fields, Rings Ideals".

3. I guess "edge-erased subgraphs" just means subgraphs with the the same vertex set?

4. The question is very open-ended and I'm not sure what you're looking for. If $ X$ is a graph, $ G \equal{} \mbox{Aut}(X)$ its automorphism group, $ Y$ a subgraph of $ X$ with the same vertex set and $ H \equal{} \mbox{Aut}(Y)$, then we may have $ H \leq G$, we may have $ G \leq H$, and we may have that neither inclusion holds. Both $ G$ and $ H$ are obviously subgroups of $ S_{V(X)}$ (the full permutation group on the vertex set) but beyond that I don't think much can be said in such generality.

hi :)
[quote="randomgraph"]
3. I guess "edge-erased subgraphs" just means subgraphs with the the same vertex set? .[/quote]
yes
[quote="randomgraph"]
4. The question is very open-ended and I'm not sure what you're looking for. If $ X$ is a graph, $ G \equal{} \mbox{Aut}(X)$ its automorphism group, $ Y$ a subgraph of $ X$ with the same vertex set and $ H \equal{} \mbox{Aut}(Y)$, then we may have $ H \leq G$, we may have $ G \leq H$, and we may have that neither inclusion holds. Both $ G$ and $ H$ are obviously subgroups of $ S_{V(X)}$ (the full permutation group on the vertex set) but beyond that I don't think much can be said in such generality.[/quote]

thank you for replying. it was that, i was wondering if something more specific was known. in case  either $ H \leq G$ or $ G \leq H$  can we apply lagrange's theorem?

[quote="karolinaH"]
thank you for replying. it was that, i was wondering if something more specific was known.
[/quote]
I don't think it's a problem of what is known, but rather that this is such a general scenario that nothing much is necessarily true in all cases.

[quote]
in case  either $ H \leq G$ or $ G \leq H$  can we apply lagrange's theorem?[/quote]
Of course - why not?  :) whenever one (finite) group contains another, the order of the smaller divides that of the larger.

Is it true (easy?) that every finite group is the automorphism group of some graph?

Suppose we take any two groups $ G$ and $ H$ (or, if the answer to the preceding question is "no," then any two groups $ G, H$ that are the automorphism groups of some graphs).  Is it true that I can find a graph $ \mathcal{G}$ with a subgraph $ \mathcal{H}$ on the same vertex set such that the automorphism group of $ \mathcal{G}$ is $ G$ and the automorphism group of $ \mathcal{H}$ is $ H$?

[quote="JBL"]Is it true (easy?) that every finite group is the automorphism group of some graph?[/quote]
This is a result of Babai, but I don't think it's trivial.  A few friends of mine were convinced it was open until they asked a professor, and they're fairly good at graph theory.

As for your second question, I think it should be possible to construct the subgraph $ \mathcal{H}$ to be "highly asymmetric" with respect to the rest of $ \mathcal{G}$ - that is, to construct them more or less separately, but I haven't thought about it.

thanks for the replies,  even if the graph X and subgraphs Y only differ by one edge we can't conclude that either $ G\leq H$ or $ H\leq G$, can we?

jbl  what's the answer to your first question?

@t0rajir0u: no, that's a much older result (Frucht, 1939). IIRC you can even restrict to cubic graphs. I need to think about this but I'm not sure it's all that hard: starting with graphs with colored edges and maybe directed as well, making some kind of Cayley graph should work. Now you need to decorate the edges with simple subgraphs so as to represent the coloring and direction. This is very vague and I'm not at all sure it's true, but I guess this is the general approach.

If I were Laci Babai, I might have taken offense at "This is a result of Babai, [b]but[/b] I don't think it's trivial"  :lol: I think he has similar results but I can't remember what they are.


Let $n,a,b\in Z^+$, $n$ divides $a+b$, $n^2$ divides $a^2+b^2$. Prove that $n^m$ divides $a^m+b^m$ for all $m\in Z^+$

[quote="dondigo"]Yes, but using the tranformation we can use two-step induction.
$a^m+b^m=(a^{m-1}+b^{m-1})(a+b)-ab(a^{m-2}+b^{m-2})$.
If the thesis it true for $m-1$ and $m-2$, then it's true for $m$.
The first step is fulfilled (it's true for $m=1$ and for $m=2$)[/quote]
I don't see how this addresses Warut's concern...

Okay, new proof  :D

[hide="Proof"]
We have $n | a+b$ and $n^2 | a^2 + b^2$.  Squaring the first expression gives $n^2 | a^2 + 2ab + b^2 \implies n^2 | 2ab$.  If $n$ is odd, then $n^2 | ab$ and $a^m + b^m$ can be expressed as $(a+b)^m - ab(...)$, where the expression in ellipsis is given by the binominal theorem (I'm too lazy to write it out), so we are done.

If $n$ is even, write $n = 2k$ and $4k^2 | 2ab \implies 2k^2 | ab$.  Since we know $2k | a + b$, we know they must have equal parity, and since we know their product is even, $2 | a, b$, so write $a = 2x, b = 2y$.  If $k, x, y \equiv 1 \bmod 2$, we use the proof for odd $n$ and we are done; if not, repeat this reasoning until they are, so we have the desired result for all $n$. [/hide]


limit
integration
logarithms
calculus
real analysis
real analysis unsolved

Evaluate $ \lim_{x\rightarrow \plus{}\infty} \frac{{\int_0^x}ln(1\plus{}t^6)dt}{x\minus{}xlnx}$

you need to find asymptotical behavior of $ \int_{0}^{x} \ln (1\plus{}t^{6})dt$ just integrate it by parts .
another way is : use L'Hospital rule .

hello, let $ f(x)=\int_{0}^x\ln(1+t^6)\,dt$, then integration by parts gives us
$ {-6x-\arctan(\sqrt{3}-2x)+2\arctan(x)+\arctan(\sqrt{3}+2x)-\frac{1}{2}\ln(1-\sqrt{3}x+x^2)+\frac{1}{2}\sqrt{3}\ln(1+\sqrt{3}x+x^2)+x\ln(1+x^6)}$ and $ h(x)=x-x\ln(x)$,applying L'Hospital's rule and calculating $ \lim_{x \to \infty}\frac{f^'(x)}{g^'(x)}$ we get that the searched limit is -6.
Sonnhard.

try to integrate by parts  , to find main asymptotycal member :
thus $ \int_{0}^{x}\ln (1\plus{}t^{6})dt \equal{} t\ln(t^{6}\plus{}1)|_{0}^{x} \plus{}o\left( \int_{0}^{x}\ln (1\plus{}t^{6})dt\right)$
thus limit will be $ \minus{}6$ since $ t\ln(t^{6}\plus{}1)|_{0}^{x} \approx 6x\ln x$


floor function
induction
number theory proposed
number theory

Let $n$ be a positive integer. If in the binary representation $n$ has exactly $s$ digits $1$ then prove that in the set $\{C_n^k|0\le k\le n\}$ are exactly $2^s$ odd numbers.

Let $s(n)$ be the amount of digits 1 in the binary representation of $n$. First we will prove that $2^a \| n!$ where $a=n-s(n)$.

Note that clearly $a=\lfloor \frac{n}{2} \rfloor+\lfloor \frac{n}{2^2} \rfloor+...$. Also note that $\lfloor \frac{n}{2^k} \rfloor$ is $n$ with its $k$ rightmost digits erased, so some ith digit of $n$ that is a 1, represents $2^i$ and contributes to the sum as $2^{i-1}+...+2+1=2^i-1$. So in total the sum gives $n$ minus 1 for each 1-digit present in $n$, in other words, $n-s(n)$. That proves our assertion.

Now the binomial coefficient is odd iff the maximal coefficients of powers of $2$ dividing $k!$ and $(n-k)!$ add up to the maximal coefficient of a power of $2$ dividing $n!$, that is $n-s(n)=k-s(k)+(n-k)-s(n-k)$, so $s(n)=s(k)+s(n-k)$. Clearly for any way of dividing the 1-digits of $n$ into two subsets we get numbers $k$ and $n-k$ satisfying this and there are $2^{s(n)}$ ways of doing it, so what's left to prove is that if $k, n-k$ satisfy $s(n)=s(k)+s(n-k)$ then their 1-digits must be a partition of the 1-digits of $n$.

Now we do it by induction on the digits of $n$. Suppose it is true for $n$ of up to $x$ digits (the base case is obviously true), so now take any $n$ of $x+1$ digits. You can write it as $2^x+n_2$, where $n_2$ has at most $x$ digits. Now the first $x$ 1-digits of $k, n-k$ must be a partition of the 1-digits of $n_2$ by the inductive hypothesis, and it's now easy to see that any representation of the $2^x$ not being an x+1th 1-digit gives more than one new 1-digit. So the proof is complete.

Can this be extended for any prime $p$ ?

Sure, it can be extanded but it depends on what you are asking as a extansion because $1$ is many different things $\mod [2]$... ;)

Anyway, all that follows from Lucas' theorem which states that for any prime $p$, if $n = n_0+n_1p + \cdots n_tp^t$ is th expansion of $n$ in base $p$ and if $k \leq n$ is such that $k=k_0 + k_1p + \cdots + k_tp^t$ (the extansion of $k$ may have been prolonged so as to  reach $t$) then :
$C_n^k = C_{n_0}^{k_0} \cdot C_{n_1}^{k_1} \cdots C_{n_t}^{k_t} \mod [p].$

Therefore $C_n^k \neq 0 \mod [p]$ if and only if $n_i \geq k_i$ for all $i$.

In particular, if $p=2$ then $C_n^k$ is odd iff $k_i =0$ for all $i$ such that $n_i=0$ (that is for $t+1-s$ subscripts, according to the original statement). The $s$ others $k_i$ (those for which $n_i = 1$) can be $0$ or $1$. It leads to $2^s$ such numbers as desired.

Now, you may choose your generalisation...

Pierre.


[quote="namish"]I am using Andrew Wiles' proof of FLT as a basis for my science project.  I plan to take some sort of detour in his proof, trying to get from one place in his proof to another, using my own procedure.  I'm not sure what the best way to do this is.  Right now, I've been trying to follow his proof, and I've been reading books on Galois theory and modular forms.  For trying to change something that was worked on obsessively for eight years by Wiles, I'm afraid that I won't be able to do anything to this proof.  There is a great lack of organization to this approach, too.  Any suggestions that will help me make some progress?[/quote]

Many people have made improvements on Wiles' original arguments but there is no known detour around the heavy machinery (which would take years to learn). Galois theory and modular forms are the easy part of the proof. There is also algebraic geometry, modular curves, algebraic number theory, p-adic analysis, homological algebra, and commutative algebra.


[quote="Mr.Ocax"]I don't know what's wrong with me, but I'm have problems factoring/simplifying the following equation:

1) $a(2b+2d-b^2+d^2)=2ad$

$2b+2d-b^2+d^2=2d$

$2b-b^2+d^2=0$

Of course, all thi is done assuming a doesn't equal 0

2) $a(2b+2d-b^2+d^2)=2ad$

$2ab+2ad-ab^2+ad^2=2ad$

$2ab-ab^2+ad^2=0$

Then, does a=1?  Why are there two ways of factoring this thing?  And then turn out to be different.  HELP.. someone explain, please!

Mr.Ocax[/quote]
Nothing wrong. In case one you omit the case $a=0$ whereas in case two you include it.


calculus
integration
limit
calculus computations

I'm trying to calculate the following integral : $ \int_0^\infty \frac {1}{\sqrt {2\pi a^3}} \exp( - \frac {a^2}{2x})}{\exp( - \lambda x) dx$ , we can see that $ \frac {\partial^2 I}{\partial a \lambda} - a \lambda I = 0$ so maybe $ I$ have the form $ \exp( - \frac {a^2 + \lambda^2}{2})$ with some factor so that $ \lim I_{a \rightarrow 0^ + } = + \infty$   :(  ? any idea will be welcome  :idea:

hello, what do we know about $ a$ and $ \lambda$?
Sonnhard.


Given a finite sequence $ S \equal{} (a_1,a_2,\ldots,a_n)$ of $ n$ real numbers, let $ A(S)$ be the sequence
\[ \left(\frac {a_1 \plus{} a_2}2,\frac {a_2 \plus{} a_3}2,\ldots,\frac {a_{n \minus{} 1} \plus{} a_n}2\right)
\]of $ n \minus{} 1$ real numbers.  Define $ A^1(S) \equal{} A(S)$ and, for each integer $ m$, $ 2\le m\le n \minus{} 1$, define $ A^m(S) \equal{} A(A^{m \minus{} 1}(S)).$  Suppose $ x > 0$, and let $ S \equal{} (1,x,x^2,\ldots,x^{100})$.  If $ A^{100}(S) \equal{} (1/2^{50})$, then what is $ x$?
$ \textbf{(A) } 1 \minus{} \frac {\sqrt {2}}2\qquad \textbf{(B) } \sqrt {2} \minus{} 1\qquad \textbf{(C) } \frac 12\qquad \textbf{(D) } 2 \minus{} \sqrt {2}\qquad \textbf{(E) } \frac {\sqrt {2}}2$

Cool problem. :)

[hide="Answer"]Working out the first few terms for successive $A$'s, we see that the one term in $A^{100}(S)$ is $\left(\frac{(x+1)}{2}\right)^{100}=\frac{1}{2^{50}}$. Simplify to find $x=\sqrt{2}-1\Rightarrow \boxed{B}$. Finding terms is not difficult, either. Notice that the terms of $A(S)=\left\{\frac{1+x}{2},\frac{x(1+x)}{2},\frac{x^2(1+x)}{2},...\right\}$, and the terms of $A^2(S)=\left\{\frac{(1+x)^2}{2^2},\frac{x(1+x)^2}{2^2},\frac{x^2(1+x)^2}{2^2},...\right\}$. We simply keep raising the $\frac{1+x}{2}$ to successive powers.[/hide]

[hide]Note that, neglecting the denominator of 2^n, terms add together as in pascal's triangle (or alternately in the form of n choose k). If you're not convinced, try averaging a few terms yourself. Now this means that, without the denominator, A100(S) = (100 0) 1 + (100 1) x + (100 2) x^2 + (100 3) x^3 + .... + (100 99) x^99 + (100 100)x^100. Notice that this is the binomial expansion of (1+x)^100. Now add in the denominator and you get (1+x)^100 / 2^100 = 2^-50, therefore (1+x)^100 = 2^50, 1+x = sqrt.(2), x = sqrt.(2) - 1 (negative squareroot doesn't work because it asks for a positive solution).[/hide]

[quote="SamE"][hide]We have $A(S)=\left(\frac{1+x}2+\frac{x+x^2}2+\cdots+\frac{x^{99}+x^{100}}2\right)$. We can factor this as $\left(\frac{1+x}2\right)(1,x,x^2,\ldots,x^{99})$. In fact, since everything is proportional, the $\frac{1+x}2$ just stays there. We can prove, by induction (if we wanted to), that $A^k(S)=\left(\frac{1+x}2\right)^k(1,x,x^2,\cdots,x^{100-k})$. So $A^{100}(S)=\left(\frac{1+x}2\right)^{100}=\frac1{2^{50}}$. Simplifying, this gives us $1+x=\sqrt{2}$, so $x=\sqrt{2}-1$. $(B)$.[/hide][/quote]
This is basically when I came up with too when I looked at it.  That seems strangely easy for a problem 23, once you get past the slew of notation (it took me a while to figure out that each time you apply the sequence transformation, you take off one term  :P ).


topology
advanced fields
advanced fields unsolved

Give an example of two compact subspaces of a non-Hausdorff space, whose intersection is not compact.

Consider an infinite set S with discrete topology. This is obviously not compact.
Add two more pointsA,B to S and declare the two subsets {A} U S, {B} U S are open.
Now, {A} U S, {B} U S are both compact. Their intersection is S with discrete topology which is not compact.


geometry
geometric transformation
reflection
analytic geometry
vector
graphing lines
slope

Oh my god... The math teachers at my school mess up a ton. And i mean that.
On a test, she thought that (4,4), (-2,4), and (-2,0) were collinear, and didn't listen to my friend's arguments and told him to sit down.he was like :mad:   :censored:    :cursing:  :furious:

And she also teaches stuff really badly so the class is like:

Teach: gives notes, "oh, copy this down" and everyone is like zzz because geo is so boring
Students: talk talk
T: If yo are going to talk like this, i might as well not teach
S: (celebrate) Yeah!!!
T: That was very disrespectful...(other stuff i forget)

But it was funny.

Anyone else have bad teachers? I'd like to hear

[color=green][i]Moderator says[/i]: the title of this topic has been edited at the request of the original poster.  Also, both I and the original poster would appreciate it if discussion focused along the lines suggested by TZF and ZetaX, and I may delete or edit posts to encourage this.[/color]

Instead of telling an uninteresting story that reflects your immaturity, please [i]at least try[/i] to think of something that might produce a meaningful discussion. Some improvements:

- What is the best way to deal with teachers such as the one you described, or others with similar mentality?
- What kind of skills can be gained/developed through situations like these?
- What are some productive things that one can do while sitting through a class like this (without being blatantly disrespectful)?
- What should be the teacher's role in the classroom? What should be the relationship between teachers and students?
- What constitutes disrespect (on either students' or teacher's part)? Under what circumstances, if any, is disrespectful classroom behavior justifiable?

The bext thing to do is nicely tell your teacher that so think he made a mistake.

Epic fail, infinite 101 on the topic starting post...
I am in his school and I suffer the same abuse at the hands of our math teacher. Our math teacher is retarded. Seriously. Telling her nicely is totally useless and pulls a participation grade down. She only likes people who have been in her class before. She hates smart people who try to bring up any point other than what she or the book are saying. And she reads the book wrong, too. As for these, TZF,

- What is the best way to deal with teachers such as the one you described, or others with similar mentality?
No one is as retarded she is. I understand trying not losing a battle with students, but the evidence is in her holy grail, the book.
- What kind of skills can be gained/developed through situations like these?
Keeping your mouth shut to [color=white]keep your participation score above a D[/color] not argue with *superiors*
- What are some productive things that one can do while sitting through a class like this (without being blatantly disrespectful)?
infinite?
Sitting through a class for me is just spazzing for 30 minutes.
- What should be the teacher's role in the classroom? What should be the relationship between teachers and students?
Teacher should teach proper material without restricting student's questions like a dictator. Student should listen and not blatantly defy the teacher.
- What constitutes disrespect (on either students' or teacher's part)? Under what circumstances, if any, is disrespectful classroom behavior justifiable?
Teacher = Not respecting student statements; restricting creativity
Students = Again, blatantly defying the teacher without any proof

I was thinking about making a thread similar to this. But not with the angle that infinite has taken.

To end my post, I would also like to say that she suffers from grammatical problems. She took off 21 points off of many people's tests on this problem:

Here are the triangles $ \delta ABC$ (insert coordinates here)and $ \delta DEF$(insert coordinates here). Prove that they are either congruent or not congruent. The corresponding angles are congruent.

Lots of people used ASA. Why? Well, it says that you have all of the angles congruent, according to normal interpretation. However, our math teacher states that you have to prove that the angles are congruent, although it is stated in the problem.

I think the worst sort of teacher is one who refuses to be corrected by his/her students.

Thankfully, almost all of my teachers will acknowledge their  mistakes and correct them immediately, and are usually pretty gracious about it.

My math teacher gives "mistake points" for anyone who finds a mistake on his quizzes, tests, boardwork, etc.  I think it's a really great policy (though it's one he only uses with his advanced classes)

My linear-algebra teaching assistant did not know what a mixed product was. I had to tell her what it was, i.e. the scalar product of A with (B cross product with C). And also tell her that three vectors of $ R^3$ were coplanar iff their mixed product = 0.

Nevertheless, I still found H = -7/2 as she did.

Oh my god, I didn't know the term "mixed product", either. Am I stupid¿

@EggyLv.999, infinite101: You may be right (there are a _lot_ of bad teachers out there), but still "retarded" is not a proper choice of words ;)
I don't know the american system very well, but here your parents can meet the teachers (either on a regular or occasional basis) to talk about problems arising. Maybe having your parents talk to her could help. Or talking to some superiors of her ;)

But i truly feel like she has all the power in the world. She literally takes off ten participation points for pointing out her mistakes.Nor does she acknowledge them.

Hi infinite101 and others;

I sympathize with your problem. For what it is worth consider that it is a good learning experience. Later on, you will find yourself in situations on the job where you will meet up with an irrational person who has authority over you. Do not confront her, it is a waste of time and perhaps her goal. Answer her questions the way she wants them answered, get through the class with a good grade. Consider it only a minor rough spot in your travels and it will pass.

Ugh I have the same problem with so many teachers.  Maybe it's just my personality but I cannot stand writing the teacher's "right answer" when it's wrong.   :roll:  I kept telling my science teacher that the slope at a point (x,y) was not y/x (it was linear but didn't pass through origin).  But she would not listen to me.  Usually after around 20 minutes of explaining she would somehow kind of get it, but this time she told me she wasn't so good with higher level math and to ask the physics teacher.

My favorite recent class in science:

T: All right (0,6), the slope at this point is 6/0 which is 0 because anything divided by zero is zero!
S: Uhhh... anything divided by zero is undefined!
T: Really?  No...
S: Yeah really!
T: Ohh ok undefined, so anyway let's just call that zero.

I suggest you go talk to a teacher who will easily understand the mistakes she made and then ask them to talk to her.  Some teachers will just not listen to anything their students have to say.

Well, I was lucky that my linear algebra TA could actually listen to us students. But... on second thought, we were both right to an extent.

Remember, Quebecers have to go through 2 years of college before going to university. As far as I know it, mixed products are learned in college-level linear algebra first. This is why I said that it was a linear algebra thing. And I was correct at the same time.

But we review scalar, vectorial and mixed products in multi-variable calculus and this is where my TA was right. This is where her claim stems from.

I have sort of the same problem, but spread into different classes (one teacher doesn't listen to students, one teacher doesn't have good grammar skills, another makes mistakes). I agree that if they are all in the same class, it can get out of hand. I guess if what the teacher talks about is something you already know, you can watch what he/she writes or listen a little, but think a level deeper into what he/she says. For example, when your teacher is talking about radicals and says you can't have a negative radicand with square roots, "imagine" imaginary numbers and see if the properties of radicals apply to imaginary numbers as well.

@ Truffles:

Glad you take our side.

T:0 times undefined is -1!
S (me- i've gotten lots of points deducted for this stuff): No it's not
T: Yes it is. We're moving on.
S: But...

@PoP:

There is no way to think deeper. This isn't algI, where the teacher was okay and actually told us that "You will learn this in Alg 2." This is geo, which we swill get no more of for the rest of our pre-college education. Therefore, she lets us know everything that is in the book. Things out of the book do not "comply with religious beliefs and are immediately thrown out as slander against the good name of The Holy Textbook of Geometry."-me... And the problems are of the "apply a formula" sort, nothing to actually think about.

Note: this is an unorganized slew of random yet truthful arguments against our math teacher's despotic ways.

Oh and a list of average correction times
Arithmetic error- 3 minutes
ACTUAL ERROR on test- 0 minutes (T: I'm right, you're wrong)
ACTUAL ERROR on notes- 10-20 minutes (refusal to accept until all of her favorites agree)

She plays favorites by "Who has been in my class the most". Some people have had her twice already. Me and infinite? 0 times. One of my friends gets to do AIME and AMC12 during notes :mad:

She gave me a B+ for participation despite being the person who was most active in all sorts of debates and other stuff.

Try pretending that she is right, and when you have to use geo on a test thats not hers do the right way. That's what I did when my math teacher insisted that my method for solving equations was wrong.  :)  and knod and when you leave her class then you can do it the correct way.

You do have a choice here:
The choice is called the principal.  Go tell the person who's in charge of her and maybe she'll back down or get fired.
Or does the teacher union meddle in these things too?

Hi infinite101;

Good luck and may the force be with you.

Hi.
Today something happened to me that was similar to infinite's situation.

So for social studies, we were learning about Ancient China.  A student asked the teacher a question and she said the incorrect answer (I know because I'm chinese and she is american).  I told her (nicely, I think) the correct answer and she said :"Do you think I don't know when I'm the social studies teacher?"
I wanted to reply :"Do you think I don't know when I'm the chinese dude here?"
I later corrected her again and she sent me out to the hall :mad: .

[quote="Altheman"] Another thing that occurs to me is that you might as well bring some AMC problems and do them during class. I gained very little from my math classes in high school so this is what I did. This solves two problems. First, you will not be wasting your time. Second, it will keep you from bothering your teacher.[/quote]

Sorry if this kind of drifts off the main focus of this topic, but how would you do something like that? My "overly-curious" classmates always manage to find a way to peek at what I'm doing (rather what anyone's doing) and ask me weird stuff about them. That really makes me feel uncomfortable.

Also, wouldn't the teacher notice you doing problems other than those in class? How would you go about that?

Thank you, and sorry again for going a bit off topic.

@limac:  Just point out to them that you really, truly feel that you (and your classmates) will be better off if you do those problems.  In addition, think about it from the teacher's perspective.  If he/she does not have to worry about you making him/her look bad in front of all the students, that's a plus and the teacher should have enough sense to see it as such.

@all:  I know most people here think (and probably are) at the top of their class, but that is no reason to pick a debate with your teacher over working some that may be beyond what other students in your class may be able to do.

The school system in America is thus; make sure the students who are middling and who may not understand up to standards, and deliberately ignore the students who already know the information, for that reason alone.

Remember, they have a whole class to teach, and they would rather not be publicly embarrassed.  If you were teaching, and you were corrected by a student you perceive as at the top of the class, how would you feel?

Turn your experiences here into a constructive one.  Instead of trying to find fault with the system, make your own system.

EDIT:  1400th post!

I had a math teacher last year (6th grade), and no one really like her at all. This was a seventh grade class, and I was a sixth grader with two other sixth graders (because we are awesome :lol:). She would not listen to us at all. I guess some people are biased, and you can't help it. She pulled my grade down for no apparent reason (I saw none) but even so, we all managed to get As in that course. Even though we were in a seventh grade class in sixth grader, we had also already learned that material, and so it got kind of boring at times. I mean, we were smarter than all of the class (us three sixth graders). So I guess you can't go without some bias with some smart-Alec people in your class that are one grader lower that the class you are teaching.

But that is my experience.

Well, one teacher that I had didn't mind. I just told him about the problem that I was doing in an enthusiastic way. However, I should say that I had a teacher who did not like me doing this (I brought in a math book that was harder than what we were doing in class and she sorta glared at me). So I settled into just doing homework in class (this was enough to keep me occupied anyways).

[quote="zapi2007"]
Remember, they have a whole class to teach, and they would rather not be publicly embarrassed.  If you were teaching, and you were corrected by a student you perceive as at the top of the class, how would you feel?

Turn your experiences here into a constructive one.  Instead of trying to find fault with the system, make your own system.
[/quote]

Well the OP did ask how to deal with it, they weren't just blatantly complaining.  I've never understood why the teacher would feel embarrassed if the student corrected them.  I can see that they have a sense of pride.  But the teacher being embarrassed doesn't justify ignoring student's responses.

[quote="Truffles"]But the teacher being embarrassed doesn't justify ignoring student's responses.[/quote]

Clearly, some of them feel it's justified.  Unless you can somehow change their perspective, good luck. (I've had this problem before, and it's not fun  :| )

I know some teachers are arrogant, and dont care of what other people say.

One teacher got sacked because they were doing that way too much. :lol:

But you can't change them.

I would just talk less, and listen more, while day dreaming, but keeping one ear tuned in to class incase they call on you.

Thats what i did, and it works perfectly!

I'm so glad this became a proper discussion :) I'm going to address a particular type of bad teacher in this post.

This type of teacher has a sort of inferiority complex when it comes to dealing with good students. An apparently arrogant or authoritarian attitude might be their defense mechanism for handling this.

In the middle of a class, the teacher is under pressure to try to explain some material successfully in a certain amount of time. If a good student interupts their thoughtflow to challenge one of their results, they might be thrown off. If they are unable to think on their feet and react gracefully, it will set off their inferiority complex, and they will perceive the student as being confrontational or disrespectful.

Similarly, if a student that they know to be a good student is not paying attention in class, then this teacher might perceive the student as looking down on the class in a passive-aggressive way.

[u]Potential ways of dealing with:[/u]

[b]1.[/b] Ask your questions after class early on in the year. This way, they won't feel as uncomfortable because you're not challenging them in front of the class, but hopefully it will implicitly communicate to them that they should be more careful.

[b]2.[/b] Rather than telling them that they're wrong, ask them to explain a particular step. If they're obviously not getting it, or you sense them getting annoyed, tell them you'll ask them after class.

[b]3.[/b] If you're going to be doing other math in their class and you ask them what it is, try to communicate to them that you're doing it because you like math, not because their class is bad.

[b]4.[/b] Tell them about how you like math, or about math team, or stuff like that after class. Make sure you come across as genuinely interested and not like you're bragging. This communicates to them that you view them as a math teacher, and not as some screw-up chump, which diffuses their inferiority complex to some extent.

[b]5.[/b] If they get frustrated because you use a different method, entertain them. You know what you're doing anyway, and there's nothing that you can possibly gain by trying to argue to them that your method is better. For example, if the class is learning factoring and you're told to solve $ x^2 \equal{} 256$, then just subtract 256 and factor. Obviously you can take the squareroot, but it's a waste of time to try to argue why that's better when the problem is intended to be an exercise in factoring.

Hi;

I couldn't disagree more. Now we are asking a student to baby an adult through an entire year.

Why not just come to a simple conclusion. Those students have a bad life presently, there is little or nothing to be done about it. She might fail them on a whim. A teacher if he/she expects respect than they have to act more maturely than their students. They have to be better prepared and much less thin skinned than a child.

She is over exercising her authority and there isn't anything they can do about it. They can act more intelligently than their teacher by realizing that this too shall pass.

[b]bobbym[/b], I don't understand what your objection is. Obviously a teacher [i]shouldn't[/i] have a bad attitude. The problem is that many [i]do[/i], and it's therefore up to the students (or parents) to deal with it. There's no way around this.

It happens that, with many bad teachers, it's less work to try to be sensitive to their thought process than to keep putting up with things the way they are. It's also nicer and more mature (there's nothing wrong with a student doing something mature).

Remember that teacher's with bad attitudes are usually not bad [i]people[/i] -- there's often something to gain, and certainly nothing to lose, by being on good terms with them. Note that I'm not suggesting that anyone suck up to their teacher or anything like that.

[b]Everyone:[/b] Back in post #2 of this thread, I suggested the following as a possible topic of discussion:
[color=darkblue][i]-What kind of skills can be gained/developed through situations like these?[/i][/color]

Some of the skills that I've developed through this kind of situation are:
- The ability to judge people's personalities and predict their emotions
- The interpersonal and communication skills to deal with people in an optimal way
- The judgment to determine what the optimal way is

For me, these weren't skills I could've picked up by reading a book (unlike the subject matter of the classes). In that sense, it was a great opportunity for me -- I don't regret it at all.

Hi TZF;

Not objecting at all. Just pointing out that your points 1 - 5 are best done by her therapist. We and her students are not qualified to try and get her through rough times. One would presume that a teacher can steer her own ship. Asking students to relate to that individual is not going to have the desired effect.

Everyone has given them sound advice. This was mine, dummy up and look passive. Do what she says. Get through the class. Get a good mark. She is not the first irrational person they are going to meet. Their bosses will be worse. They don't have to like her and she doesn't have to like them.

Now what the becheesus is wrong with that? Why have I drawn more fire than the only living turkey on Thanksgiving?

Perhaps you're being misunderstood because you write things like [quote="bobbym"]I couldn't disagree more.[/quote] and then later elaborate that [quote="bobbym"]Not objecting at all.[/quote] :roll:

By the way, this is not the first time in this thread that you've asserted X and then later claimed that you've never asserted any such thing.  This behavior is ... mystifying, and if you want other people to be able to make sense of what you write, you should stop.  It's also not the case that you "drew fire" at all -- TZF's post was calm and of moderate tone, and didn't criticize you at all.

Hi JBL;

[quote]It's also not the case that you "drew fire" at all -- TZF's post was calm and of moderate tone, and didn't criticize you at all.
[/quote]

I know that. I thought my comment on the end was humorous. It was not in any way designed to insult or accuse TZF.

I didn't change my tone at all. I was pointing something out, I am not sure that is the same as objecting. Reading a little further you would see that I don't agree with his points 1 - 5 and why. Also there was a whole post between those two quoted comments. If you pull out single statements out of different posts some of the continuity is lost.

I haven't been asserting X and then saying I haven't. I have been asserting x and others have been assuming X. I have stated that several of the posters who have replied directly to me did not read my post,

[quote]By the way, this is not the first time in this thread that you've asserted X and then later claimed that you've never asserted any such thing. This behavior is ... mystifying, and if you want other people to be able to make sense of what you write, you should stop.[/quote]

When making an accusation like that you might have been kind enough to have pointed out where exactly I did that. Am I being pulled up because I don't agree with the mainstream?

Another poster was out line with me and you said nothing. I was left to deal with it myself. We settled it because he is a gentlemen. I have not been out of line with anyone here and you say I am nonsensical. Isn't that an out of line comment?


Given are four coplanar points A, B, C and D such that

AB/CD = BC/AD = CA/BD = \sqrt3.

Prove that triangle ABC is equilateral and that D is its center.

cute problem
using vectors with D as origin,we get that D must be the centroid of the triangle ABC.now tis easy to proceed

I'll write it down somewhat more rigorously:

For any points A, B, C, D, where G is the centroid of ABC, we have

(#) AD2 + BD2 + CD2 = AG2 + BG2 + CG2 + 3DG2

In triangle ABC, AG2 = (2b2 + 2c2 - a2)/9 and the similar equations.

Plugging the conditions and this relations in, equation (#) gives DG2 = 0.
So D must be the centroid of triangle ABC.

Hence we have AB2 = 3CD2 and the similar equations.
This gives c2  = (2a2 + 2b2 - c2)/3,
so a2 + b2 = 2c2 (and the similar equations). Now it is obvious that a = b = c.


inequalities
limit
calculus
calculus computations

[quote="ryan1988"]For any two real sequences $ a_{n}, b_{n}$ prove that

$ \lim_{{n}\to{\infty}}{\sup(a_{n}+b_{n})\leq \lim_{n\to \infty}{\sup{a_{n}}}+\lim_{n\to \infty}{\sup{b_{n}}}}$[/quote]
Lets say $ (a_{n}),(b_{n})$ are bounded so I dont have to deal with the extended reals.

[hide="Hint"]
Show that,
$ \sup \{a_{i}+b_{i}|i\geq n\}\leq \sup \{a_{i}|i\geq n\}+\sup \{b_{i}|i\geq n\}$
And finish it off by taking a limit.
[/hide]

To prove that $ L$ is the supremum of some set, you have to prove that $ L$ is an upper bound of that set and that no smaller number is an upper bound. But if you want to prove that the supremum of some set is less than or equal to $ L$ you only need to do half as much work - you only have to show that $ L$ is an upper bound of that set.

In sylow_theory's hint, you must show that the right hand side of his inequality is an upper bound of the set on the left.

[quote="ryan1988"]Can u give more details?[/quote]
Once you established that fact that Kent said. Use the following theorem:

[b]Theorem:[/b] Let $ (s_{n})$ and $ (t_{n})$ be convergent sequence such that $ s_{n}\leq t_{n}$ then $ s\leq t$ where $ s=\lim s_{n}$ and $ t=\lim t_{n}$.


modular arithmetic
number theory solved
number theory

We work in $\mathbb Z_{43}$. I'll denote $7$ with $\alpha$. Clearly, we have $-6=\alpha^{-1}$. We want to prove that for all primes $p>1$ we have $\alpha^p+\alpha^{-p}=1$. Notice that $\alpha$ satisfies the equation $\alpha^3+1=0$. Since $\alpha^6=1,\ \alpha^p+\alpha^{-p}$ only depends on the residue of $p$ modulo $6$, and given that $p>3$ is prime, we have either $p\equiv 1\pmod 6$ or $p\equiv 5\pmod 6$. Finally, it's clear that being the roots of the equation $x^2-x+1=0$, the residues $\alpha,\alpha^{-1}$ satisfy $\alpha+\alpha^{-1}=1,\alpha^5+\alpha^{-5}=1$, and we're done.


[quote="pianoforte"]A lone farmer owns a farm with only cows and chickens. If there's a total of 31 heads and 88 legs on the farm, how many cows and chickens are there?[/quote]

I can almost swear someone posted a problem along the lines of this one before.

[hide]$x$ is the number of cows and $y$ is the number of chickens.

$x+y=31$
$4x+2y=88$

$x+y=31$
$2x+y=44$

$x=13$

$13+y=31$
$y=18$[/hide]

[quote="pianoforte"]I meant for the farmer to count too( I never said that the head and legs count were from only the chickens and the cows). I didn't notice that a chicken has the same amount of body parts as the farmer. My fault :([/quote]

Allright then...  :)

[hide]$x+y=30$
$4x+2y=86$

$x+y=30$
$2x+y=43$

$x=7$

$7+y=30$
$y=23$[/hide]


inequalities
calculus
integration
Euler
LaTeX
arithmetic sequence
number theory

Sharing some really interesting questions I got from an amazing book. Post your answers/thoughts here! Let's share the joy of solving the qns together...

[b]Qn1[/b]
If $a,b,c$ form an arithmetic sequence, where a,b,c are real numbers uch that $2b^2 + ac = 1$, prove that

 $\frac{1}{a^2 + 1}, \frac{1}{b^2 + 1}, \frac{1}{c^2 + 1}$

form an arithmetic sequence

[b]Qn2[/b]
If $n=2^a.(2^{a+1} - 1)$, where $a$ is a positive integer and $2^{a+1} - 1$ is a prime number, express in terms of $n$ the sum of all positive integers that divide $n$.

will be back to add 2 more, do check back! :D

[b]Qn3[/b]
Prove that the inequality

$ab + bc + ca + \frac{b^2c^2}{a^2} + \frac{c^2a^2}{b^2} + \frac{a^2b^2}{c^2} \le \frac{b^4c^4}{a^2} + \frac{c^4a^4}{b^2} +  \frac{a^4b^4}{c^2}$

holds for all non-zero real numbers $a,b,c$

[b]Qn4[/b]
If $x, y$ and $z$ are positive real numbers such that $x + y + z = 3$, prove that $\frac{1}{\sqrt x} + \frac{1}{\sqrt y} + \frac{1}{\sqrt z} \ge 3$

so anyone can help with the solution to these problems? the answers cannot be found in the book..wish someone could show me the light... :oops:

number 4.
if x+y+z=3, then sqrtx+sqrty+sqrtz is less than or equal to 3. Therefore, assuming equality for worst case scenario, label sqrt x sqrt y and sqrt z a b and c respectively. Therefore, it remains to prove that 1/a+1/b+1/c>=3
using AM-GM, we get that the GM is 3/(cbrt abc). Since it is well known that equal a, b, and c make the largest possible abc given an addition limit (which can also be deduced through a quick labeling of variables,) a=b=c=1 so 3/cbrt(abc)=, 3. so 1/a+1/b+1/c>=3

1st one.

you know $\frac{a+c}{2}=b$(*), square it to get $b^2$, then plug that junk into $2b^2+ac=1$ to get $a^2+4ac+c^2=2$ (i).

we wanna show that $(\frac{1}{a^2 + 1}+ \frac{1}{c^2 + 1})/2= \frac{1}{b^2 + 1}$

adding the fractions on LHS you get $\frac{a^2+2+c^2}{2(a^2c^2+a^2+c^2+1)}$.  Then using eqn (i) you have $a^2+c^2=2-4ac$, so it becomes $\frac{4-4ac}{2(2-4ac+a^2c^2+1)}$=$\frac{2-2ac}{3-4ac+a^2c^2}=\frac{-2}{ac-3}$ (ii).

to simplify $\frac{1}{b^2 + 1}$, substitute (*) for $b$ to get $\frac{4}{a^2+2ac+c^2+4}$, then using (i) again we have it =$\frac{4}{6-2ac}$, which is equal to (ii).

[hide="i got number four"]
By AM>= HM, we have:
$\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \geq \frac{3}{\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}}$

Hence: $\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq \frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$

By cauchy, we have:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2 \leq (x+y+z) \cdot 3 = 9$
$\sqrt{x}+\sqrt{y}+\sqrt{y} \leq 3$

Hence $\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \geq 3$

$\text{QED}$[/hide]

2nd one, let $p:=2^{a+1}-1$, then the sum of the divisors of $n$ are

$1+p+p(2^1+ \ldots +2^a) + (2^1+ \ldots + 2^a)$, since $p$ and $2^a$ are relatively prime (no integral solutions to 2^{a+1}-1=2).

Summing and factoring, we get the sum to be $(p+1)(2^{a+1}-1)=(2^{a+1})(2^{a+1}-1)$.  Since $n=(2^a)(2^{a+1}-1)$, the sum is $2n$

I don't know if this author expected you to do this, but Euler proved that if $2^k - 1$ equals a prime number $p$, then the number $2^{k-1} p$ is a perfect number. [url=http://www.cut-the-knot.com/do_you_know/numbers.shtml#perfect]Says right here[/url]. So, since our number is $n = 2^{k-1}(2^k - 1) = 2^{k-1} p$, we know $n$ is a perfect number and the sum of the divisors of any perfect number $n$ is always $2n$, thats what makes it perfect.
Edit: I guess it's pretty easy(cats...) beat me to the posting, and made me look stupid simultaneously :D.

[quote="shobber"]

By cauchy, we have:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2 \leq (x+y+z) \cdot 3 = 9$
$\sqrt{x}+\sqrt{y}+\sqrt{y} \leq 3$

[/hide][/quote]

Sorry, but whats cauchy?  :blush:

omigosh, i feel so stupid in here... :(

Cauchy-Schwarz inequality:

$(a_1^2+a_2^2+a_3^2+...+a_n^2)(b_1^2+b_2^2+b_3^2+...+b_n^2)\geq (a_1b_1+a_2b_2+a_3b_3+...+a_nb_n)^2$

let's say $n=3$ and we replace the $b$'s with the number $3$, we get:

$(a_1^2+a_2^2+a_3^2)(9+9+9)\geq (3a_1+3a_2+3a_3)^2$ or more easily $3(a_1^2+a_2^2+a_3^2)\geq (a_1+a_2+a_3)^2$

AM-GM inequality:

$\frac{x_1+x_2+x_3+...+x_n}{n}\geq \sqrt[n]{x_1x_2x_3...x_n}$

it just means that the arithmetic mean of $n$ terms $\geq$ the geometric mean of $n$ terms...

in the easiest form we have for example: $\frac{a+b}{2}\geq \sqrt{ab}$

a=b=c=1 gives counterexample to #3, is there some other restriction

Number 2 is just a description of a perfect number. Guess many people have tried to get $2n$ by using algebraic expression, but no need to do all that work, if it is a perfect number.

[quote="Scrambled"]
AM-GM for three variables states:
(a+b+c)/3>=cbrtabc
could someone tex that up please?[/quote]

$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$,  for $a,b,c \geq 0$

[quote="Scrambled"][quote="beaucoa"][quote="Scrambled"]
using AM-GM, we get that the GM is 3/(cbrt abc). Since it is well known that equal a, b, and c make the largest possible abc given an addition limit (which can also be deduced through a quick labeling of variables,) a=b=c=1 so 3/cbrt(abc)=, 3. so 1/a+1/b+1/c>=3[/quote]

Thanks for your help! Could you kindly explain further what is AM-GM and the rest quoted, if possible?  :blush:[/quote]
AM-GM for three variables states:
(a+b+c)/3>=cbrtabc
could someone tex that up please?[/quote]

Scrambled, you still haven't learned $\LaTeX$?

[quote="mathical88"][quote="cats..."]1st one.
$\frac{2-2ac}{3-4ac+a^2c^2}=\frac{-2}{ac-3}$ (ii).
.[/quote]

how does this happen too? could you show the missing steps?  :oops:[/quote]
missing step:

$=\frac{-2(ac-1)}{(ac-3)(ac-1)}=$

[quote="mathical88"][quote="shobber"]
By AM>= HM, we have:
$\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \geq \frac{3}{\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}}$
[/quote]

how does this happen? i mean, the AM-GM inequality does not show this...i think there are some steps missing, and i'm afraid i dun know them... :([/quote]

If we use:

$(a+b+.....+n)(\frac{1}{a}+ \frac{1}{b}+.....+\frac{1}{n}) \geq n^2 $

Can you work it out from here?

[quote="seamusoboyle"]

If we use:

$(a+b+.....+n)(\frac{1}{a}+ \frac{1}{b}+.....+\frac{1}{n}) \geq n^2 $

Can you work it out from here?[/quote]

yep, i can, thx! but is there such a formula? where did it come from? does it have a name?

cauchy-schwartz inequality.

what is the correct version of question #2 supposed to be?

edit: i copied a proof of cauchy from a previous post.

heres one proof of c-s:

$f(x):=(a_1x-b_1)^2+(a_2x-b_2)^2+...+(a_nx-b_n)^2$ >= 0 .  Expanding gives

$(a_1^2+a_2^2+...+a_n^2)x^2-2(a_1b_1+a_2b_2+\ldots+a_nb_n)x+(b_1^2+b_2^2+b_3^2+...+b_n^2)$.

Since $f(x)$ >= 0, the discriminant must be  <= 0.  Setting the discriminant <=0 gives the desired inequality.  Then the equality case comes from the definition of $f(x)$.

[quote="beaucoa"][b]Qn3[/b]
Prove that the inequality

$ab + bc + ca + \frac{b^2c^2}{a^2} + \frac{c^2a^2}{b^2} + \frac{a^2b^2}{c^2} \le \frac{b^4c^4}{a^2} + \frac{c^4a^4}{b^2} +  \frac{a^4b^4}{c^2}$

holds for all non-zero real numbers $a,b,c$
[/quote]

very sorry for the mistake in here! i juz checked, i typed the qn wrongly...  :blush:

it should be

Prove that the inequality

$ab + bc + ca + \frac{b^2c^2}{a^2} + \frac{c^2a^2}{b^2} + \frac{a^2b^2}{c^2} \le \frac{b^4+c^4}{a^2} + \frac{c^4+a^4}{b^2} +  \frac{a^4+b^4}{c^2}$

holds for all non-zero real numbers $a,b,c$

[quote="beaucoa"][quote="beaucoa"][b]Qn3[/b]
Prove that the inequality

$ab + bc + ca + \frac{b^2c^2}{a^2} + \frac{c^2a^2}{b^2} + \frac{a^2b^2}{c^2} \le \frac{b^4c^4}{a^2} + \frac{c^4a^4}{b^2} +  \frac{a^4b^4}{c^2}$

holds for all non-zero real numbers $a,b,c$
[/quote]

very sorry for the mistake in here! i've juz checked; i've typed the qn wrongly...  :blush:

it should be

Prove that the inequality

$ab + bc + ca + \frac{b^2c^2}{a^2} + \frac{c^2a^2}{b^2} + \frac{a^2b^2}{c^2} \le \frac{b^4+c^4}{a^2} + \frac{c^4+a^4}{b^2} +  \frac{a^4+b^4}{c^2}$

holds for all non-zero real numbers $a,b,c$[/quote]

[quote="cats..."]$ ab + bc + ca + \frac{b^2c^2}{a^2} + \frac{c^2a^2}{b^2} + \frac{a^2b^2}{c^2} \le \frac{b^4+c^4}{a^2} + \frac{c^4+a^4}{b^2} + \frac{a^4+b^4}{c^2} $

do a.m-gm on rhs, like $\frac{b^4}{a^2}+\frac{c^4}{a^2} \ge 2\cdot \frac{b^2c^2}{a^2}$, then simplify, it comes out to $\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+\frac{a^2b^2}{c^2} \ge ab+bc+ca$.

multiply everything by 2, so you got $2(\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+\frac{a^2b^2}{c^2}) \ge 2(ab+bc+ca)$

use am-gm on lhs for pairs of terms, like $\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\ge2c^2$, then the inequality becomes $a^2+b^2+c^2 \ge ab + bc +ca$, which is true.[/quote]

Hmm...sorry for this really demanding request of mine, but could you kindly please elaborate your working further? I'm afraid I still do not understand it until now...   :blush:  :(

[quote="mathical88"][quote="shobber"]
By AM>= HM, we have:
$\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \geq \frac{3}{\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}}$
[/quote]

how does this happen? i mean, the AM-GM inequality does not show this...i think there are some steps missing, and i'm afraid i dun know them... :([/quote]

well , here it goes

By AM-GM ,

$\frac 1a +\frac 1b + \frac 1c \geq \frac{3}{\sqrt[3]{abc}}$  and also $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$ combine both you will see the result


calculus
derivative
vector
real analysis
real analysis unsolved

Show that
$ \left(\frac{\mathbf{x}}{\left\|\mathbf{x}\right\|}\right)'\equal{}\frac{(\mathbf{x}\times\mathbf{x'})\times\mathbf{x}}{\left\|\mathbf{x}\right\|^{3}}$.

You can use the fact that $ (\mathbf{a}\times\mathbf{b})\times\mathbf{c}\equal{}(\mathbf{a}\cdot\mathbf{c})\mathbf{b}\minus{}(\mathbf{b}\cdot\mathbf{c})\mathbf{a}$.

[quote="JRav"]Show that
$ \left(\frac {\mathbf{x}}{\left\|\mathbf{x}\right\|}\right)' \equal{} \frac {(\mathbf{x}\times\mathbf{x'})\times\mathbf{x}}{\left\|\mathbf{x}\right\|^{3}}$.

You can use the fact that $ (\mathbf{a}\times\mathbf{b})\times\mathbf{c} \equal{} (\mathbf{a}\cdot\mathbf{c})\mathbf{b} \minus{} (\mathbf{b}\cdot\mathbf{c})\mathbf{a}$.[/quote]

\[ \left(\frac {\mathbf{x}}{\left\|\mathbf{x}\right\|}\right)' \equal{}\frac{d}{dx}\left[ \mathbf{x} \left( \mathbf{x}\cdot\mathbf{x}\right)^{ \minus{}{\scriptstyle \frac{1}{2}} } \right]\]
\[ \equal{} \mathbf{x'} \left( \mathbf{x}\cdot\mathbf{x}\right)^{ \minus{}{\scriptstyle \frac{1}{2}} } \plus{}\mathbf{x}\frac{d}{dx}\left[\left( \mathbf{x}\cdot\mathbf{x}\right)^{ \minus{}{\scriptstyle \frac{1}{2}} }\right]\]
\[ \equal{} \mathbf{x'} \left( \mathbf{x}\cdot\mathbf{x}\right)^{ \minus{}{\scriptstyle \frac{1}{2}} } \minus{}{\scriptstyle \frac{1}{2}} \mathbf{x}\left( \mathbf{x}\cdot\mathbf{x}\right)^{ \minus{}{\scriptstyle \frac{3}{2}} }\left( \mathbf{x'}\cdot\mathbf{x}\plus{}\mathbf{x}\cdot\mathbf{x'}\right)\]

\[ \equal{}\frac{\mathbf{x'} \left( \mathbf{x}\cdot\mathbf{x}\right) \minus{}\mathbf{x}\left( \mathbf{x'}\cdot\mathbf{x}\right)}{\left\|\mathbf{x}\right\|^{3}}\]

\[ \equal{} \frac {(\mathbf{x}\times\mathbf{x'})\times\mathbf{x}}{\left\|\mathbf{x}\right\|^{3}}\]

Thanks; my problem was in precisely how to handle the chain rule applied to the dot product (whether or not it was allowed).


Would it be ok to write $ p^q\plus{}q^p\equal{}p\plus{}q (\mod{pq})$ instead of $ p^q\plus{}q^p \equiv p\plus{}q (\mod{pq})$?
Also, can we deduce from the above that $ p^q\plus{}q^p\equal{}p\plus{}q$ has a solution for some prime $ p,q$?

[quote="Los"]Would it be ok to write $ p^q \plus{} q^p \equal{} p \plus{} q (\mod{pq})$ instead of $ p^q \plus{} q^p \equiv p \plus{} q (\mod{pq})$?
Also, can we deduce from the above that $ p^q \plus{} q^p \equal{} p \plus{} q$ has a solution for some prime $ p,q$?[/quote] If it is congruent modulo $ pq$, then you should write $ \equiv$.(Or so I think!)

1)  The second would be preferred.

2)  No.  For example, $ x^2 \equiv 2 \bmod 7$ has an integer solution, but $ x^2 \equal{} 2$ does not.  And in your example, note that $ p^q > p, q^p > q$ hence the LHS is strictly greater than the RHS.  In fact, your example is a system in disguise:  it really means

$ p^q \equiv p \bmod q$
$ q^p \equiv q \bmod p$


topology
real analysis
real analysis unsolved

Characterize those sets $ A\subset\mathbb R$ having the following property: for every nowhere dense set $ B\subset\mathbb R$, the set $ A\plus{}B\equal{}\{a\plus{}b\ |\ a\in A,\ b\in B\}$ is also nowhere dense.

Let's say a set is "small" if it satisfies the condition of the problem. Say it is "big" otherwise.

1. $ A$ is small if and only if its closure $ \overline{A}$ is small. ( proof: $ \overline{A}\plus{}B \subset \overline{A\plus{}B}$ )

2. Every unbounded set is big. ( proof : Assume $ A$ is bounded above. Let $ (a_n)$ be sequence in $ A$ such that $ a_n \minus{} a_{n\minus{}1} > 1$ and let $ (q_n)$ be an enumeration of rational numbers in $ [0,1]$. The set $ B \equal{} \{\minus{}a_n \minus{} q_n : n \in \mathbb{N}\}$ is nowhere dense. But $ A\plus{}B$ contains all rational numbers in $ [\minus{}1,0]$. )

lemma. $ \overline{A}\plus{}\overline{B} \equal{} \overline{A\plus{}B}$ if $ A$ is bounded. ( proof : if $ x$ in $ \overline{A\plus{}B}$, there exist sequences $ a_n$ in $ A$ and $ b_n$ in $ B$ with $ a_n \plus{} b_n$ converging to $ x$. Passing to convergent subsequence of $ a_n$, we may assume $ a_n$ converges, then $ b_n$ converges too)

3. If $ A$ is countable and compact then $ A$ is small. ( proof: Suppose $ B$ is nowhere dense. Bair Category Theorem says $ A\plus{}\overline{B}$ has empty interior. By the previous lemma, $ A\plus{}\overline{B} \equal{} \overline{A\plus{}B}$. Therefore $ A\plus{}B$ is nowhere dense.)

4. If $ A$ is uncountable and closed then $ A$ is big.

proof: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167044]every uncountable closed set contains Cantor set[/url] shows $ A$ contains a Cantor set $ C \equal{} \cap_n C_n$ where $ C_n$ is the union of intervals in $ \mathcal{C}_n$ and we may assume all intervals in $ \mathcal{C}_n$ have length $ <1/3^n$ and that $ C_0 \equal{} [0,1]$. (see the construction of $ \mathcal{C}_n$ from the link).

We will construct another Cantor set $ D$ such that $ D$ is nowhere dense while $ C\plus{}D$ is an interval. [url=http://demonstrations.wolfram.com/TheSumOfTwoCantorSets/]The illustration of sum of two standard Cantor set being an interval[/url] suggests how to construct $ D$.

Inductively define for each $ n$ a family $ \mathcal{D}_n$ of $ 2^n$ disjoint intervals : Define $ \mathcal{D}_0 \equal{} \{ [0,1]\}$ For each interval $ I\equal{}[x\minus{}y,x\plus{}y]$ in $ \mathcal{D}_n$, delete the center middle part of very small length $ a_n >0$, leaving two intervals $ I_1 \equal{} [x\minus{}y, x\minus{} a_n/2]$ and $ I_2 \equal{} [x \plus{} a_n/2, x\plus{} y]$, the remaining $ 2^{n\plus{}1}$ intervals form the next collection $ \mathcal{D}_{n\plus{}1}$. Let $ D \equal{} \cap_n D_n$ where $ D_n$ is the union of all intervals in $ \mathcal{D}_n$. Then $ D$ is compact and nowhere dense.

It is possible to give small enough values to $ a_n$ for all $ n$ such that the following two conditions hold:
(1) the intervals of $ \mathcal{D}_n$ have length $ \ge 1/3^n$ (hence each interval in $ \mathcal{D}_n$ is longer than each in $ \mathcal{C}_n$)
(2) $ a_n$ is smaller than the length of each interval in $ \mathcal{C}_{n\plus{}1}$.

Consider $ I \in \mathcal{C}_n$ and $ J \in \mathcal{D}_n$. Suppose $ I_1, I_2 \in \mathcal{C}_{n\plus{}1}$ belongs to $ I$ and $ J_1, J_2 \in \mathcal{D}_{n\plus{}1}$ belongs to $ J$. Condition (1) says $ J$ is longer than $ I$ and condition (2) says the gap between $ J_1$ and $ J_2$ is shorter than the intervals $ I_1$ and $ I_2$. Therefore (draw a picture) $ I\plus{}J \equal{} (I_1 \cup I_2) \plus{} (J_1 \cup J_2)$. This implies $ C_n \plus{} D_n \equal{} C_{n\plus{}1} \plus{} D_{n\plus{}1}$ and hence we have $ C_n \plus{} D_n \equal{} C_0 \plus{} D_0 \equal{} [0,2]$ for all $ n$.

We also have $ C\plus{}D \equal{} [0,2]$ : if $ 0 \le x \le 2$, $ x$ is in $ C_n\plus{}D_n$ for all $ n$, so $ (x \minus{} C_n) \cap D_n$ forms a decreasing sequence of nonempty compact sets whose intersection is $ (x \minus{} C) \cap D$, so $ (x \minus{} C) \cap D$ is nonempty and $ x$ is in $ C\plus{}D$. This completes proof.

5. $ A$ is small if and only if its closure is countable and compact. (proof: it follows from 1, 2, 3 and 4)


Prove that $ \boxed {\ \{x,y\}\subset [0,1]\ \implies\ (1 \plus{} xy)^2\ \ge\ \left(1 \minus{} x \plus{} x^2\right)\left(1 \minus{} y \plus{} y^2\right)\ }$ .

Give a geometrical interpretation of this inequality.

I don't know the geometric interpretation but it is obvious by expanding:

$ LHS\minus{}RHS\equal{}xy(x\plus{}y\plus{}1)\plus{}x(1\minus{}x)\plus{}y(1\minus{}y)$

[quote="Virgil Nicula"]Prove that $ \boxed {\ \{x,y\}\subset [0,1]\ \implies\ (1 \plus{} xy)^2\ \ge\ \left(1 \minus{} x \plus{} x^2\right)\left(1 \minus{} y \plus{} y^2\right)\ }$ .[/quote]
[b]O.K.[/b] [u]Otherwise[/u], $ LHS \minus{} RHS \equal{} 3xy \plus{} (x \plus{} y)(1 \minus{} x)(1 \minus{} y)\ge 0$ with equality iff $ \{x,y\} \equal{} \{0,1\}$ or $ x \equal{} y \equal{} 0$ .

[b][u]Consequence[/u].[/b] $ \{x,y,z\}\subset [0,1]\implies (1 \plus{} xy)(1 \plus{} yz)(1 \plus{} zx)\ge (1 \minus{} x \plus{} x^2)(1 \minus{} y \plus{} y^2)(1 \minus{} z \plus{} z^2)$ .


function
analytic geometry
graphing lines
slope
conics
parabola
calculus

Your bus is leaving your bus stop at a constant acceleration of A ft/sec2. You are running to catch the bus. Your velocity is constant, and at the moment the bus begins to move you are at a distance B ft. from the bus. What is your minimum running velocity that will allow you to catch the bus. (Hint: think backwards to find a velocity and then a position function for the bus.)

At the moment you catch the bus (call it time $t_{0}$), your speed matches the speed of the bus. If you were any faster than that, you could have slowed down a little and still caught up.

Call your speed $v.$ By what I said, $v=At_{0}.$ But we also have that $v$ is the constant speed needed to cover the distance $B+\frac12At_{0}^{2}$ in time $t_{0}.$ Hence

$At_{0}=\frac{B+\frac12At_{0}^{2}}{t_{0}}$

$2At_{0}^{2}=2B+At_{0}^{2}$ or $t_{0}=\sqrt{\frac{2B}{A}}$

The question asked for the speed. That would be

$v=A\sqrt{\frac{2B}{A}}=\sqrt{2BA}.$

Let's chech the units of measurement. $B$ is in feet and $A$ is in $\frac{\text{ft}}{\sec^{2}}$ Thus $\sqrt{2BA}$ is in ft/sec, which is the correct units for a velocity.

Of course, all this problem asked for was the slope of the tangent line to a parabola that passes through an exterior point (that, conveniently enough, lies on the axis of the parabola.)

Your other problem - the deer in the headlights - also involves a tangent line to a parabola through an exterior point. In that other problem, the point is not on the axis of the parabola.


function
real analysis
real analysis unsolved

Prove or disprove:there is a real function $f$ which is zero on a set whose complementary is of Lebesgue measure 0 and such that $f$ takes all real values on any nontrivial interval.

What do you mean by nontrivial interval?
do you mean if the interval,there exist a $x$ belong to the interval such that $f(x)\neq 0$?

A nontrivial interval is an interval not reduced to a point.

Such functions do exist, but I'm going to use the Axiom of Choice.

First, we want a set $S$ of measure zero whose intersection with every interval has continuum cardinality. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=category&t=104561]Here[/url] the real line is decomposed into a set of category $1$ and a set of measure $0$. The set of measure $0$ used there will do.

Now that we have $S$, we construct $f$ by transfinite induction. First take $f=0$ on $\mathbb R\setminus S$. Now well-order the set $\mathcal I\times\mathbb R$ (where $\mathcal I$ is the collection of all non-degenerate intervals on the real line) so as to make it order isomorphic to the first ordinal of continuum cardinality.

If $I\in\mathcal I$ and $r\in\mathbb R$, then in the step $(I,r)$ of the transfinite inductive process we pick a point $x$ in $S\cap I$ which has not already been used (this is always possible because up to this point we've only used fewer than continuum points, and $S\cap I$ has continuum cardinality) and put $f(x)=r$. When we're done we let $f=0$ on the points we haven't used in the induction.

P.S.

The category analogue can be constructed similarly. In other words, there is a real function which takes non-zero values on a set of category $1$ and which takes all real values in every non-degenerate interval.

clearly it suffices to find a function that takes all the value in $[0,1]$ for example in any non trivial interval. Call a real $x=\overline{a.a_{1}a_{2}a_{3}...}$ good if, for all $n \geq n_{0}$, $a_{2n+1}=0$.

Now, if $x$ is 'good', take $n_{0}$ the least integer that works in the previous definition and define $f(x)= \overline{0.a_{2 n_{0}}a_{2 n_{0}+2}a_{2 n_{0}+4}...}$. If $x$ is not good, $f(x)=0$.

Then $f$ works because the set of 'good' real is of null measure.


blogs
AoPSwiki
modular arithmetic
MATHCOUNTS
number theory

[hide="hint"]
Try out the first few terms. like $7, 7^{2}, 7^{3}, ...$ See a pattern in the units?[/hide]

[quote="Art of Owna"]mod 1000 that, use fermats little theorm.. but i get 7^(-1)  mod 1000... I dont know what to do from here..[/quote]
Well ...

If you happen to notice that $7*11*13=1001\equiv 1 \bmod 1000$ you will get the answer right away.
[hide]$11*13=143$ that is what $7^{-1}\bmod 1000$ would be :)

[hide="because"]$\phi (1000)=400$ and $7^{2000}\equiv 1 \bmod 1000$
[hide]$7^{2000}\equiv 1 \equiv 10001 \equiv 7*11*13 \bmod 1001$
[/hide]
[/hide]

[/hide]

[hide="Solution"]
By Euler's Totient Theorem:
\[7^{ \varphi(1000) }= 7^{400}\equiv 1 \bmod 1000 \implies 7^{1999}= 7^{2000-1}\equiv 7^{-1}\bmod 1000 \]
But wait, a negative exponent in mods? Indeed, we want the modular inverse of $7$ in $\bmod 1000$. Look at (I think it was) [b]t0rajir0u[/b]'s blog for a fuller discussion of all the fun things we can do in modular arithmetic, or just look on Google.

Now, since $\gcd(7,1000)=1$ we know that $7$ has a unique inverse in $\bmod 1000$ such that $7 \cdot 7^{-1}\equiv 1 \bmod 1000$. The quickest way to calculate this, I think, is to understand that if we call the $\bmod 1000$ inverse of $7$ as $a$, then:
\[7 \cdot a = 1000k+1 \implies a = \frac{1000k+1}{7}\]
for some $k \in \mathbb{Z}_{7}$. We immediately find that $1001=7 \cdot 143$, so our final answer is thus:
\[7^{1999}\equiv \boxed{143 \bmod 1000}\]
[/hide]

[quote="Glen Mann"]what exactly is this whole mod thing? :blush:[/quote]

modular arithmetic you can AoPSWiki it but here are the basics

$x$ is said to be congruent to $y \pmod z$ if and only if $x-y$ is divisible by $z$...

of course mods can be used to find the last digit of random exponential expressions like $9^{2007}$ which may appear in MATHCOUNTS (they usualy have problems like those)


[quote="Mitzah"]What is the number of subsets with 20 elements of a set with 52 elements?[/quote]

[hide]I got
$\frac{32\cdot33}{2}=528$ if the numbers are all distinct[/hide]

is set {1,2} different from set {2,1}

[quote="robinhe"][hide]Yes, a subset is just a set so the order does not matter.

I got $\binom{52}{20}$ because isn't it just 52 choose 20 for a subset of 20?[/hide][/quote]

That's what I thought as well...

[quote="robinhe"][hide]Yes, a subset is just a set so the order does not matter.

I got $\binom{52}{20}$ because isn't it just 52 choose 20 for a subset of 20?[/hide][/quote]

That's probably right. I was thinking of the wrong situation..  :?

[quote="Mitzah"]What is the number of subsets with 20 elements of a set with 52 elements?[/quote]

[hide="is this really that easy?"]${52\choose20}=\text{huge number}$ (it doesn't fit on my calculator). [/hide]


inequalities
trigonometry
inequalities unsolved

Given $x,y,z>0$ and $xy+yz+zx=1$ prove that
$\frac{x}{xy+1}+\frac{y}{yz+1}+\frac{z}{zx+1}\ge \frac{3\sqrt{3}}{4}$
Please help me! :)  :)

K81o7 Can please you show me clearly how to use it ?Thanks

[quote="K81o7"]Let $f(a)=\frac{1}{a}$ and use Weighted Jensen's...then use Cauchy, or just AM-QM...[/quote]
"K81o7", it was the joke? :D

[quote="prithwijit"]W.L.O.G assume that $x \le y\le z$.
I can show that $\frac{x}{1+xy}+\frac{y}{1+yz}+\frac{z}{1+zx}= \frac{y}{1+xy}+\frac{z}{1+yz}+\frac{x}{1+zx}+\frac{(x-y)(y-z)(z-x)}{(1+xy)(1+yz)(1+zx)}$. I chose $\alpha$, $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma = \pi$ and $x = \tan(\alpha/2)$, $y = \tan(\beta/2)$, $z = \tan(\gamma/2)$. Now $\frac{(x-y)(y-z)(z-x)}{(1+xy)(1+yz)(1+zx)}\ge 0$. If $x = y = z = 1/\sqrt{3}$ then $\frac{x}{1+xy}+\frac{y}{1+yz}+\frac{z}{1+zx}=\frac{3\sqrt{3}}{4}$.[/quote]
For your assumption $x\leq y\leq z$ the expression $(x-y)(y-z)(z-x)$ is negative (or $0$) so you rather can't say that $\frac{(x-y)(y-z)(z-x)}{(1+xy)(1+yz)(1+zx)}\ge 0$. BTW, the inequality is cyclic, not symmetric, so you can't WLOG assume that $x\leq y\leq z$.
Besides of it, I don't understand your solution. What gives you the substitution $x=\tan(\alpha/2)$ etc? And why from $x = y = z = 1/\sqrt{3}\implies \frac{x}{1+xy}+\frac{y}{1+yz}+\frac{z}{1+zx}=\frac{3\sqrt{3}}{4}$ you deduce $\frac{x}{1+xy}+\frac{y}{1+yz}+\frac{z}{1+zx}\geq\frac{3\sqrt{3}}{4}$ for any $x,y,z$ :?: Is it a next joke :?:

[quote="prithwijit"]$x = 1$, $y = 2$, $z = 3$. $x-y =-1$; $y-z =-1$; $z-x = 2$. $(x-y)(y-z)(z-x) = (-1)(-1)(2) = 2 > 0$.  :D[/quote]
I'm very sorry for this horrible error :blush:  :blush:  :blush:. I mistoke it with $x\geq y\geq z$. But still the inequality is not symmetric.

Thank you for your interest, but is it true ??, I hope it is true but I can't do it, Thanks! :)

I wasn't joking at all...but maybe I screwed up my use of weighted jensen's...

By Weighted Jensen's, if $f(a)=\frac{1}{a}$, \[xf(xy+1)+yf(yz+1)+zf(zx+1) \ge (x+y+z)f(\frac{x(xy+1)+y(yz+1)+z(zx+1)}{x+y+z})\]  \[=\frac{(x+y+z)^{2}}{x^{2}y+y^{2}z+z^{2}x+x+y+z}\] Yeah that definitely doesn't look like the expression I came up with...fair enough it was a joke then  :D
I probably did something like $\frac{x}{xy+1}+\frac{y}{yz+1}+\frac{z}{zx+1}\ge 3(\frac{x+y+z}{xy+yz+zx+3})$ which is clearly invalid...though if we could show that, it would prove the inequality...

I'm working on this thing...

[quote="spanferkel"]Anyway, I think for weighted Jensen the weights must be constants.  :wink:[/quote]

Nope, variables can involve with the weights.

Indeed, that's not needed. But even if you want them constant... just take fixed x,y,z constants and prove it for those. As x,y,z were arbitrary, you proved it for all...

Hope everybody help me to solve it! Thanks  :maybe:  :maybe:

By substituting $x\rightarrow1/x,y\rightarrow1/y,z\rightarrow1/z$ the constraint becomes $x+y+z=xyz.$ Now substituting $x\rightarrow\tan\alpha,y\rightarrow\tan\beta,z\rightarrow\tan\gamma$ we get \[\sum\frac{y}{xy+1}=\sum\frac{\cos\alpha\sin\beta}{\sin\alpha\sin\beta+\cos\alpha\cos\beta}=\sum\frac{\cos\alpha\sin\beta}{\cos(\alpha-\beta)}\] and the inequality will be a corollary from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105247]here[/url].

bodan your inequality is corollary of mine certainly not mine is corollary of yours,please check it!  :D  :D

[quote="gemath"]Given $x,y,z>0$ and $xy+yz+zx=1$ prove that
$\frac{x}{xy+1}+\frac{y}{yz+1}+\frac{z}{zx+1}\ge \frac{3\sqrt{3}}{4}$
[/quote]
It's true of course!


The set $ \{5, 8, 10, 18, 19, 28, 30, x\}$ has eight members. The
arithmetic mean of the set's members is 4.5 less than $ x$. What is
the value of $ x$ ?