$24)$ Suppose a square with $12 cm$ sides is revolved about a diagonal as an axis. Find the sruface area of the figure formed.

Could someone draw a diagram or picture for this also? I feel confused in a way with these types of questions.  :(

This problem has already been solved in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124636]this article[/url].

[quote="mathgeniuse^ln(x)"]If yout think about it, the figure formed will be two cones on top of one another such that the height is $6\sqrt{2}$ and the diameter of the base is $12\sqrt{2}$[/quote]

what do you mean by two cones? I don't see how two cones form.  :(

Think about the diagonal. That will be the base of the two cones. And the two vertices the diaganol it touches will be the vertex of the cones.

Thanks to Klebian and Ragnarok, I now can proclaim I understand revolving !!! I practiced about 12 problems of different kinds from past tests to make sure I could differentiate!!! Thanks to all!  :lol:


I apologize if this question has been repeated, but I couldnt find it.

I'm a bit worried because I haven't gotten a checkmark or notification that they received my answers, just because its Friday and answers can be talked about.

Can anyone help?

Thanks

just wait...it always takes a bit of time for them to check things in...

Just wondering if the checkmarks have been coming in yet...
It's been a while and I haven't gotten my checkmark.

Anyone know how long I should wait for the checkmarks to all come in?

If so many people are worrying about checkmarks, then honestly I wouldn't be worried at all.   :huh:


geometry
3D geometry
rectangle
calculus
integration
perimeter

There are primes $ p_1 \text{ and } p_2$ such that

$ p_1 \minus{} p_2 \equal{} a$ where $ a$ is the 2nd smallest multiple of 23 with $ b$ factors

where $ b$ is a cube and greater than 5 and less than 10. Find the smallest possible product  $ p_1 \cdot p_2$

oh wow 1337

btw b is 8
possible values for a are 138 184 230... 2nd smallest is 184
the primes are 7 and 191
7*191 = 1337
shuld known

Even mewto has started to copy my idea of making problems whose answers are 1337. WOW. :|

Yay! stevenmeow posted in the MO Forum!!  :D

Anothe problem!

FInd the number $ 5$ greater than $ x(x\plus{}1)$ where $ x$ is the bth triangular number, where b is of the form $ a^3$ where a is the smallest prime number!

[quote="tinytim"]Even mewto has started to copy my idea of making problems whose answers are 1337. WOW. :| [/quote]

Um... this idea is NOT original. Do not claim that you had the grand idea one day to make problems answers 1337. It just makes the answer more predictable....  :dry:

Yeah. What's the minimum area of a rectangle with integral side lengths and perimeter 2776?  :P

Alrgiht um... FInd the area under the curve $ x^2\plus{}\frac{212}{15}$ from 0 to 15

I will guess...

(drumroll)

Obviously it's 1337. (gasp) :o

Give me a harder integral... :P

edit: like "finding the area bounded by blablabla" is quite retarded just say evaluate the integral because it's basically the same thing.

edit2: I forgot I don't know calculus.  :blush:

A cyclic quad has side lengths 23, 223, 32, and 156. If one of the diagonals is 4992, what is the other diagonal?

If no one gets the right answer by Saturday night, I will post the answer.


Find a quick, easy way to prove that, for $a,b,c,d,e,f \in \mathbb{Z}$,

$(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})$

can be expressed as the sum of four squares.

I'm sure there are quicker ways to do this... here's what my solution is based off of. (Dont read until you post).
[hide=" Basis of my solution."]
$cos^{2}\theta+sin^{2}\theta = 1$.

[hide=" What in the world is he talking about?"]
3-D vectors[/hide]
[/hide]

eh... this is like the complex number proof for R^2 vectors, just you use R^4 (which is quaternions) and the modulus (another word?) is distributive...

|AB|=|A||B|

squaring gives the result (make one of the components 0 so it mimicks the R^3 vector that you give)

sam, your proof is good too, it seem that it would extend well to next dimension, the only thing is that i am not exactly sure as to how cross product is defined in R^4, intuitively, it should follow, but i have not seen a rigirous development of that

obviously there exists that integer vector that is perpendicular to the the other two (using dot product)...

[quote="t0rajir0u"][hide="Here it is"] Quaternion norm is multiplicative.  $(a^{2}+b^{2}+c^{2}+0^{2})(d^{2}+e^{2}+f^{2}+0^{2}) = (\text{Stuff}^{2}+\text{Junk}^{2}+\text{Crap}^{2}+\text{Things}^{2})$ [/hide][/quote]

Aren't there some $\text{Whatchamacallits}$ in there too? :D

[quote="Alex"] this is like the complex number proof for R^2 vectors, just you use R^4 (which is quaternions) and the modulus (another word?) is distributive...

|AB|=|A||B|

squaring gives the result (make one of the components 0 so it mimicks the R^3 vector that you give)

sam, your proof is good too, it seem that it would extend well to next dimension, the only thing is that i am not exactly sure as to how cross product is defined in R^4, intuitively, it should follow, but i have not seen a rigirous development of that

obviously there exists that integer vector that is perpendicular to the the other two (using dot product)...[/quote]

It should continue into $n$ dimensions... the cross-product squared would give $n$ squares, then the dotproduct would give the $(n+1)^\text{th}$ square. Unless the cross-product does something wierd, which i dont think it will... Then we can show that $(\sum_{k=1}^{n}a_{k}^{2})(\sum_{k=1}^{n}b_{k}^{2})=\sum_{k=1}^{n+1}m_{k}^{2}$ for integers $a,b,m$...  (if that's what you meant in continuing to $n$ dimenstions)

[hide="Alex--"]
get on AIM![/hide]

The only dimension $>3$ where one can define a cross product is 7. :o
See http://en.wikipedia.org/wiki/Cross_product

[quote="t0rajir0u"]I don't think an obvious analogue of the cross product is defined for more than $3$ dimensions.[/quote]
[quote="B23"]The only dimension  where one can define a cross product is 7.
See http://en.wikipedia.org/wiki/Cross_product[/quote]

Oh, i had only thought of the cross-product just in terms of the matrix :oops: ... So I was thinking of the "wedge product"? (Thanks [b]B23[/b] for the link  :) ).  In which case the proofs wont work, except for the sum of $8$ squares?


$ f : R \equal{}\equal{}> R$
$ f(x)\equal{}(3x\plus{}lxl)/2$
$ is$ $ f$ $ injective$ $ ?$ $ surjective$ $ ?$

$ f(x)\equal{}2x$ if $ x\geq 0$ and $ f(x)\equal{}x$ if $ x<0$
From we has
1) $ f(x)$ is surjective because for each a we has
$ f(\frac{a}{2})\equal{}a$ if $ a\geq 0$
$ f(a)\equal{}a$ if $ a<0$
2)$ f(x)$ is inject .
Other method you can see the graph of function $ f(x)$


inequalities
graph theory
combinatorics proposed
combinatorics

Consider $A=\{1,...,n\}$ .let $\mathcal E$ and $\mathcal Q$ be two permutations of these numbers.

$\mathcal E \cap \mathcal Q=\{x_1,...,x_k\}$ means that $x_1,...,x_k$ have the same place in both of the permitations.

Problem.Let $P$ be a permutation of the first $n_{th}$ natural numbers.
Find the smallest number of permutations like $x_1,...,x_k$ such that :

$\displaystyle \bigcup_{i=1}^k (P \cap x_i) \neq \emptyset$

Consider a bipartite graph on vertices $v_1, v_2, \dots, v_n$ and $w_1, w_2, \dots, w_n$ such that $v_i$ is adjacent to $w_j$ iff $j$ has NOT appeared yet in position $i$ in any of the permutations.   If the non-intersection statement is to hold, then it follows that this graph does NOT have a perfect matching (such a matching would correspond to a $P$ which failed to intersect any of the permutations).  By Hall's theorem, there must be a subset $A$ of the $v$'s which is connected to fewer than $|A|$ of the $w$'s.

This corresponds to a set $A$ of positions and a set $B$ of numbers of cardinality at least $n-|A|+1$ such that every number in $B$ has already appeared in every position in $A$ after $k$ permutations.

Since only one number can occur in a position at a time and all $|B|$ numbers have to occur in each position, we have $k \geq |B| \geq n-|A|+1$.

Since a number can only fill one position at a time and a number must fill $|A|$ positions, we have $k \geq |A|$.

Adding these two inequalities, $k \geq \frac{n+1}{2}$

Conversely, if $k$ is at least $\frac{n+1}{2}$ then we can construct such a set of permutations by letting $x_i$ take $a$ to $1+a+i (mod k)$ for each $a$ which is at most $k$.


So, I have to take the new SAT next Saturday. I was fairly pleased with my old results (790M/750V), and I was wondering if I take the new one, if I can use those old ones combined with the new writing score for most colleges. Like, say I got a 700M/700V/750W, could I still use my old SAT M/V scores coupled with the writing to have a 790M/750V/750W?

Thanks

While the old SAT score is evidence that your new SAT score wasn't a fluke in this hypothetical scenario, I believe they changed a couple things about the math and verbal sections (e.g. no analogies) for the new SAT.  For that reason I would suspect that colleges would not accept what you suggest.

That probably depends on which college you are talking about.  Each college has different requirements for different things.

Harvard explicitly says that it will give you the benefit of your best combination of subscores on the "old" or "new" SAT I. That has long been Harvard's policy for applicants who take a test more than once--that can only help you, not hurt you. I think most colleges of similar selectivity have similar policies for evaluating and combining subscores from multiple test administrations.

Note that that is good for you, but it is also good for all the applicants against whom you are competing. This means that there are a lot more applicants with 1600s (or, shall we say, 2400s) in the overall Harvard applicant pool than there are [url=http://www.collegeboard.com/prod_downloads/about/news_info/cbsenior/yr2003/pdf/2003CBSVM.pdf]single-sitting 1600 scorers[/url] each year.

Good luck on your next test.

i think most schools would look at both. it shouldn't be a problem for me tho as i expect to improve in both sections on the new one. i got a 720 math that needs to go to an 800 and i got a 590 verbal  :wallbash: . it's good the analagies were taken off; i was 25th percentile on them when i took it.  :winner_first: . with the analagies off, i hope to get a 650 english. if i don't, i'll retake until i do cuz i need a 650 english to get into clemson honors. :gleam:

Hm, I didn't know they took the best combo....I'm considering taking the new SATs again - I took it for the first time last May, and got a 1600 Math/Verbal  :)  but a low writing score (710) - stupid essay....*grumbles*  It was rather annoying since I've mostly gotten 800s on the writing section for the PSAT, since it's usually my best subject and I was looking forward to having it boost my SAT score; but my counselor advised me not to take it again,  and I'm quite certain I'll make a bunch of stupid mistakes and get lower scores for the other two sections - BUT if they take the best combo.....*talks herself in circles* Oh well, it'll prolly end up with me deciding my Saturday morning isn't worth it.  (growing teenagers do need sleep, you know....they ought to set SATs on schooldays or something)


inequalities
geometry
area of a triangle
Heron\u0027s formula
Cauchy Inequality
inequalities proposed

Show that :

$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}-1\leq \frac{a^3+b^3+c^3}{6}$

$\forall \ a, b, c >0$

 :)

[quote="cezar lupu"]Thanks Arqady. I thought so, too. Anyway I didn't see this topic until now. I saw that those two inequalities dedicated to Virgil Nicula are much more interresting than this one. :)[/quote]

Ok,Ok.!

I think these ones are nicer..!  :)

I)
Prove that in each triangle $ABC$ with medians  $m_a, m_b, m_c$  and area  $S$  the following holds:
$\frac{1}{m_am_b}+\frac{1}{m_bm_c}+\frac{1}{m_cm_a}\leq \frac{\sqrt{3}}{S}$.

II)
Find the greatest value of $M$ $\in \mathbb{R}$ such that the following ineq is true $\forall$  $x, y, z$  $\in \mathbb{R}$
$x^4+y^4+z^4+xyz(x+y+z)\geq M(xy+yz+zx)^2$.

 ;)

Hey
Where are you??

i think the above are quite good!! :D

I'm kind of busy right now. I have to study for exams. Don't worry, I'll post solutions to these nice inequalities very soon. :)  ;)

Altough i have an exam tommorow, here is a solution for the first inequality:

[color=blue][b] Problem.[/b]

   Prove that in each trinagle with medians $m_{a}, m_{b}, m_{c}$ and $a, b, c$ be the side lenghts of the triangle
   the following inequality holds true:

     \[ \frac{1}{m_{a}m_{b}}+\frac{1}{m_{b}m_{c}}+ \frac{1}{m_{c}m_{a}}\leq\frac{\sqrt{3}}{S}.  \]

  [/color]

  [i] Solution. [/i]

    Well, by the well-known relation( or by Stewart theorem  :D ) we have $m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$. So by Cauchy-Schwarz inequality we get $m_{a}\geq\sqrt{p(p-a)}$. So we have:

   \[ \frac{1}{m_{a}m_{b}}+\frac{1}{m_{b}m_{c}}+ \frac{1}{m_{c}m_{a}}\leq\frac{1}{p\sqrt{(p-a)(p-b)(p-c)}}(\sqrt{p-a}+\sqrt{p-b}+\sqrt{p-c}).  \]

    Using Heron's formula we only need to prove that \[ \sqrt{p-a}+\sqrt{p-b}+\sqrt{p-c}\leq\sqrt{3p}.  \]

    Now, by Cauchy inequality we have $\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{3(x+y+z)}$.
    By putting $x=p-a, y=p-b$ and $z=p-c$ the inequality follows.

  P.S I have found another solution using Jensen inequality. I shall post it later cause i'm studying for my exams. :)

[quote="socrates"]

II)
Find the greatest value of $M$ $\in \mathbb{R}$ such that the following ineq is true $\forall$  $x, y, z$  $\in \mathbb{R}$
$x^4+y^4+z^4+xyz(x+y+z)\geq M(xy+yz+zx)^2$.

 ;)[/quote]
[b]Socrates[/b] if you want know my solution look this
[hide]If $x=y=z$=>$M\leq \frac{2}{3}$
Let’s prove that when $M=\frac{2}{3}$ the inequality is true.
$x^4+y^4+z^4+xyz(x+y+z)\geq \frac{2}{3}(xy+xz+yz)^2$<=>$3(x^4+y^4+z^4)\geq 2(x^2y^2+x^2z^2+z^2y^2)+xyz(x+y+z)$
We have that $a^2+b^2+c^2\geq ab+bc+a$=>$(x^4+y^4+z^4)\geq x^2y^2+x^2z^2+z^2y^2\geq xyz(x+y+z)$=> $3(x^4+y^4+z^4)\geq 2(x^2y^2+x^2z^2+z^2y^2)+xyz(x+y+z)$
[color=blue]Answer:[/color]$\frac{2}{3}$
[/hide]
If no waite for [b]Cezar Lupu[/b] :D

[quote="Beat"]socrates from where did you get the first inequality?[/quote]

I don't remember exactly... :blush:
But i think it is neither too new nor well known.

The sure is, that it is a nice problem, one solution of which, is the one given by Cezar..!  :D

[quote="Cezar Lupu"]First of all, my name is not [b] Cazar [/b], it is [b] Cezar [/b]. At least this is the name from my identity Card. :)
 [/quote]
Sorry [b]Cezar[/b] :blush: ,I edited my post :) .
Also I want to see  your book too,If that is possible :)

[quote="Cezar Lupu"]Of course it is possible. But, there is a problem. I don't have it on PDF format. :([/quote]is it published? If yes, what is its name?

The book is called: Inegalitati alese in Matematica- :)

[quote="Cezar Lupu"]

        But from \[ \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{z+x}\leq\frac{3}{2}\cdot\frac{x+y+z}{xy+yz+zx},  \]
        (see also my book  :D )
  ;)[/quote]
I can't wait to see your book for such a nice inequality.
Here is my solution:
Because:
\[ \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{z+x}=\frac{x^2+y^2+z^2+3(xy+yz+zx)}{(x+y)(y+z)(z+x)} \]
\[ \le \frac{8x^2+y^2+z^2+24(xy+yz+zx)}{9(x+y+z)(xy+yz+zx)} \]
(This is because $(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz \ge \frac 8 9(x+y+z)(xy+yz+zx)$)
So we should only to prove:
\[ \frac{8x^2+y^2+z^2+24(xy+yz+zx)}{9(x+y+z)(xy+yz+zx)} \le \frac{3}{2}\cdot\frac{x+y+z}{xy+yz+zx},  \]
Which is ture because:$x^2+y^2+z^2 \ge xy+yz+zx$.

[quote="Cezar Lupu"]First of all, my name is not [b] Cazar [/b], it is [b] Cezar [/b]. At least this is the name from my identity Card. :)
 Anyway here is my solution:

 [color=blue] [b] Problem. [/b]

    Let $x, y, z$ three nonnegative real numbers. Prove that:

                             \[ x^{4}+y^{4}+z^{4}+xyz(x+y+z)\geq\frac{2}{3}(xy+yz+zx)^{2}.  \]

 [/color]

  [i] Solution. [/i]

      In my book of inequalities it is proven that for any real $t$ and for any $a, b, c$ nonnegative numbers,
      the following inequality (Schur :D ) holds:

                                            \[ a^{t}(a-b)(a-c)+b^{t}(b-c)(b-a)+c^{t}(c-a)(c-b)\geq 0.  \]

       For $t=2$ we obtain \[ x^{4}+y^{4}+z^{4}+xyz(x+y+z)\geq x^{3}(y+z)+y^{3}(z+x)+z^{3}(x+y).  \]
       Now, we only need to prove that the following inequality holds:

                                         \[ x^{3}(y+z)+y^{3}(z+x)+z^{3}(x+y)\geq\frac{2}{3}(xy+yz+zx)^{2}  \]

       It is also proved (in my book  :D ) that the following inequality holds:

      [color=green] [b] Lemma. [/b]

         Let $x, y, z, a, b, c$ six nonnegative numbers. Then, the following inequality holds:

                                   \[ \frac{x^{3}}{a}+\frac{y^{3}}{b}+\frac{z^{3}}{c}\geq\frac{(x+y+z)^{3}}{3(a+b+c)}.  \]
          [/color]

        By applying the lemma, we get
                \[ x^{3}(y+z)+y^{3}(z+x)+z^{3}(x+y)\geq\frac{(x+y+z)^{3}}{3(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})}.  \]

        But from \[ \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{z+x}\leq\frac{3}{2}\cdot\frac{x+y+z}{xy+yz+zx},  \]
        (see also my book  :D )
        we only need to prove $(x+y+z)^{2}\geq 3(xy+yz+zx)$, which is trivial. :)  ;)[/quote]
the lemma, it's well known of
if p, q > 0, p >= q+1, \[ \sum{\frac{a^p}{b^q}} \ge n^{q+1-p} \cdot \frac{(\sum{a})^p}{(\sum{b})^q}  \]

Yes, Zhaobin tht's my solution too, although here is a nicer and shorter solution from the previous one.

 [color=blue][b] Inequality.[/b]

     Let $x, y, z$ three nonnegative real numbers. Prove the inequality

                           \[ \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\leq\frac{3}{2}\cdot\frac{x+y+z}{xy+yz+zx}.  \]

  [/color]

 [i] Solution.[/i]

    the given inequality is equivalent with this one \[ (xy+yz+zx)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})\geq\frac{3}{2}(x+y+z)\Leftrightarrow  \]

          \[ \frac{xy+yz+zx}{x+y}+\frac{xy+yz+zx}{y+z}+\frac{xy+yz+zx}{z+x}\leq\frac{3}{2}(x+y+z)\Leftrightarrow  \]

          \[ \frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x}\leq\frac{x+y+z}{2}.  \]

    But this last inequality follows immediately from AM-HM $\frac{2xy}{x+y}\leq\frac{x+y}{2}$. Writing the analogues and by adding them up the conclusion follows. ;)


calculus
derivative
function
algebra
domain
limit
complex analysis

Let $f$ be an analytic function on a domain $D$ and let $w_{1}$, $w_{2}$ be two complex numberw such that $\mid w_{1}\mid=\mid w_{2}\mid =1$. For $z$ in $D$ consider the directional derivatives

$D{}_{w}{}_{1}f(z) = \lim_{t \rightarrow 0}\frac{f(z+tw_{1})-f(z)}{t}$, $D{}_{w}{}_{2}f(z) = \lim_{t \rightarrow 0}\frac{f(z+tw_{2})-f(z)}{t}$

Show that:
$\mid D{}_{w}{}_{1}f(z) \mid = \mid D{}_{w}{}_{2}f(z) \mid$

The statement seems obvious since f is analytic this means that the directional derivates are equal in every direction. But I don't know how to prove it.

You forgot to subtract $f(z)$ in the numerators; as written, these limits are usually infinite.

Rewrite $D_{w_{1}}f=w_{1}\lim_{t\to 0}\frac{f(z+tw_{1})-f(z)}{tw_{1}}=$ and similarly for $w_{2}$. Both limits are now clearly restrictions of the definition of the derivative, except that one is multiplied by $w_{1}$ and the other by $w_{2}$.


when im doing f2l, sometimes i screw up the neighboring corner that i just finished.  my friend said that there was an algorithm for that situation but i cant find it on the web.  does anyone have a website that can help?

Thanks

rubiks cube I presume?
try dan's cube station. My F2L (not intuitive) is pretty lame, but I got all the important ones

good site for cubing
http://www.cubestation.co.uk/


geometry
3D geometry
number theory unsolved
number theory

Find all the positive integers less than 1000 such that the cube of the sum of its digits  is equal to the square of such integer.

Only 1,27.
Easy to see it's impossible for 100-999
And if $1\leq n\leq9$ we can see only 1 satisfied it.
For $10\leq n\leq99$,See this equality $(a+b)^3=(10a+b)^2$
$a+b=k^2,10a+b=k^3$,only k=3 and a=2,b=7

[hide=Solution] From the equation we know that $S(n)$ must be itself a perfect square, and since $n<1000$, $1 \leq S(n) \leq 27$. Thus the only possible values for $S(n)$ are $1, 4, 9, 16$ and $25$. Moreover,
\[n \equiv S(n) (mod 9) \rightarrow n^2 \equiv n^3 (mod 9) \rightarrow n^2(n-1) \equiv 0 (mod 9)\]
but $\gcd(n^2,n-1)=1$ so $9|n^2$ or $9|n-1$, leading to $S(n)=1$ or $S(n)=9$, substituting in the equations we see that the only solutions are $1$ and $27$.


trigonometry
inequalities
inequalities unsolved

Let ABC be a triangle ,S=S(ABC).Prove that :
      a^3*b^4*c^5>=(25 \sqrt 5*(2S)^6)/27

your problem is same Namdung's problem.If you give x=tgA...

Give 1/xy+1/yz+1/xz=1.Prove that :

729(x 2 +1)^3*(y 2 +1) 2 *
(z 2 +1)     \geq  3125*x^6*y^4*z^2

can you give me a short sol for this problem?

i don't know whether namdung's problem is solved or not but i have a solution for this
the ineq $ \displaystyle \Leftrightarrow \sin^3 A \sin^2 B \sin C \leq \frac{27}{25 \sqrt{5}}$ (*)
we will prove that
$\displaystyle  \sqrt{10} \sin A + \sqrt{5} \sin B + \sqrt{2} \sin C \leq 6$
$\displaystyle  \Leftrightarrow \sqrt{10}a + \sqrt{5}b + \sqrt{2}c \leq 12R$
$\displaystyle \Leftrightarrow 10 a^2 + 5b^2 + 2c^2 + 10\sqrt{2}ab + 4\sqrt{5}ac + 2\sqrt{10}bc \leq 144 R^2$
by Cauchy we get
$10\sqrt{2}ab + 4\sqrt{5}ac + 2\sqrt{10}bc \leq 10a^2 + 10b^2 + 10c^2$
so we need to prove that
$20a^2 + 15b^2 + 12c^2 \leq 144 R^2$
it's true because we have the inequality
$(\alpha + \beta + \gamma)^2 \geq \alpha \beta c^2 + \beta \gamma a^2 + \alpha \gamma b^2$
by Cauchy again
$6 \geq \sqrt{10} \sin A + \sqrt{5} \sin B + \sqrt{2} \sin C \geq 6 \sqrt[6]{\frac{\sqrt{10}^3}{27} \sin ^3 A \frac{5}{4} \sin^2 B \sqrt{2} \sin C} $
and we have (*)  :D

Arosisi, in your beautiful solution  can you please complete the two deductions:

1)........by Cauchy we get

10 (\sqrt2)ab+4(\sqrt5)ac+2(\sqrt10)bc <=10(a^2+b^2+c^2)

2)......it is true because we have the inequality

R^2(x+y+z)^2>=xyc^2+yza^2+xzb^2

Thank you very much.

dear manlio
for the first one
we have the equality iff

$\sin A = \frac{3}{\sqrt{10}},  \sin B = \frac{2}{\sqrt{5}} , \sin C ={1}{\sqrt{2}} $

or $\frac{a^2}{18} = \frac{b^2}{16} = \frac{c^2}{20}$

for the second
let I is the point so that 3(vector)IA + 4(vector)IB + 5(vector)IC = 0
then ${3OA^2+4OB^2+5OC^2=12OI^2+\frac{1}{12} (20a^2+15b^2+12c^2}$
or $144R^2=144OI^2+20a^2+18b^2+12c^2$
so we have $144R^2\geq 20a^2+18b^2+12c^2$
i hope this is what you need  :D

Sorry, Arosisi if I am so stupid but I don't understand

how   10 (\sqrt2)ab+4 (\sqrt5)ac+2 (\sqrt 10)bc <=10(a^2+b^2+c^2)

can be obtained by Cauchy!!!!!

Many thanks.

And there is no equality in the case you mentioned...

Pierre.


[quote="kunny"]Let $ f(x)=\sin^{2}x+\sin x \cos x$, find the values of $ p,\ q,$ and $ r$ such that $ f(p)=f(q)=f(r),\ 0\leq p<q<r\leq \pi$.[/quote]

[hide="Hint"]
We find relationship between $ x,y$ in the equation,
$ \sin^{2}x+\sin x \cos x = \sin^{2}y+\sin y \cos y$ for $ 0\leq y<x\leq \pi$.

$ 2\sin^{2}x+2\sin x 2\cos x = 2\sin^{2}y+2\sin y \cos y$

$ 1-\cos 2x+\sin 2x = 1-\cos 2y+2\sin y$

$ \sin 2x-\sin 2y = \cos 2x-\cos 2y$

$ 2\cos (x+y)\sin (x-y) =-2 \sin (x+y)\sin (x-y)$ note $ \sin (x-y)\not = 0$.

$ \cos (x+y) =-1 \sin (x+y)$

$ \tan (x+y) =-1 \implies x+y = \frac{3\pi}{4}$
[/hide]

$ \sin^{2}x+\sin x \cos x = \sin x (\sin x+\cos x)$

Clearly it has three zeroes in the domain: $ 0, \frac{3\pi}{4},\pi$. Therefore $ p=0$, $ q=\frac{3\pi}{4}$, and $ r=\pi$.

Ehhh... the answer comes quickly and without an explanation 'cuz that's how my thought process for this particular problem worked. Basically you know that $ \sin x$ has two zeroes on the domain and it is easy to see that $ \sin x+\cos x$ only has one.

It can be simplified as follows:

$ \sin^{2}x+\sin x\cos x=\frac{1}{2}*(1-\cos 2x+\sin 2x)=\frac{1}{2}*(1-\sqrt{2}\cos(2x+45))$.


easy question that an examinator asked me a few days ago to begin an examination:
Does there exist a matrix $A \in M_n(\mathbb{R})$ such that $exp(A)=-I$ ?
If it exists, comute it .

$A$ should be diagonalizable in $\mathcal M_n(\mathbb C)$, I believe. Also, it's eigenvalues must be $\pm \pi i$. It doesn't exist if $n$ is odd, and it exists and it's similar (in $\mathcal M_n(\mathbb R)$) to the matrix having $\frac n2$ blocks of the form $\begin{pmatrix}0&1\\-\pi^2&0\end{pmatrix}$ on the main diagonal.

Is it Ok?

[quote="Kent Merryfield"]I would have gone for

$\begin{bmatrix}0&-\pi\\ \pi&0\end{bmatrix}$

But it all works.

$\exp\left(t\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\right)= \begin{bmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{bmatrix}$[/quote]

Yes, I gave Kent's example during the examination. A quick proof to show that it works is to notice that:
$f:\mathbb{C} \to M_n(\mathbb{R})$,$a+ib \to \begin{bmatrix}a & -b\\ b&a\end{bmatrix}$ is a continuous morphism of algebra.

thank you alekk for the idea, but can you explain more , i hope succes in ENS  the previous years ( i think you are from Louis le grand or Saint  ... :P ) i didnt see how to use this countinios morphism , but i think that a recurence can show this diagonalization by block

oops, sorry for this late answer (hollydays...)

$f$ is a continuous morphism, so if $z=a+ib$ then
$f(1+z+z^2/2+..+z^k/k!+..)=f(1) + f(z) + f(z)^2/2! + ... + f(z)^k/k! + ... = exp(f(z))$
Hence
$f(exp(z))=exp(f(z))$
$f(cos(t) + isin(t))=exp(f(a+ib))$
(the examinator was quite happy to see this sln ..)

If you have any other question, don't hesitate :)


For how many positive integers $ m$ does there exist at least one positive integer $ n$ such that $ m\cdot n \le m \plus{} n$?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}$ infinitely many

[hide]If any of m or n is 1, anything satisfies it. E.[/hide]

[quote="mathcrazed"]Simon's factoring trick...[/quote]
I did that first, and then I was like, duh, you can sub in 1.

If n = 1, then we have
$m \le m + 1$ which is always true, so the answer is $\boxed {E}$.


limit
geometry
congruent triangles
geometry unsolved

Let $ABC$ a triangle with it's surface $S$. We denote $A_{1},B_{1},C_{1}$ the middles of the lenghts of triangle $ABC$, $A_{2},B_{2},C_{2}$ the middles ot the lenghts of triangle $A_{1}B_{1}C_{1}$ and so on until an certain $n$ and $S_{i}$ it's the surface of the triangle $A_{i}B_{i}C_{i}$.

(a) Show that $S_{1}=\frac{S}{4}$

(b) Compute $\lim_{n\to\infty}{\sum_{i=1}^n{S_{i}}}$.

well.. could this be problem be solved analiticaly?

I don't see why you'd want to do that. Just use the fact that you get four congruent triangles at each step and then the geometric series.

[quote="perfect_radio"]I don't see why you'd want to do that.[/quote]

Just wondering :roll:

[quote="perfect_radio"]... Just use the fact that you get four congruent triangles at each step and the geometric series.[/quote]

Those 4 congruent triangles have equal surfaces, right?


Prove that there is a perfect square between $ n$ and $ 2n$, inclusive, for every positive integer $ n$.

[hide="One Possible Approach?"]
The claim is equivalent to $ \sqrt{n} \le \sqrt{(n \plus{} 1)} \le \sqrt{(2n)}$ for positive integers $ n$. I tried going by induction with base case $ n \equal{} 1$. Unfortunately, I have not successfully completed the inductive step so I am a little stuck at this point. [/hide]

"The claim is equivalent to $ n^2 \le (n\plus{}1)^2 \le (2n)^2$ for positive integers n."

Why?

[hide="Hint"]Find a pattern of the intervals between perfect squares and apply what you know to the given conditions.[/hide]

I guess if we can just prove the condition that mathemagician had, we would be done.

I'm not exactly sure how to use induction on that though. But apparently, since we are dealing with positive integers, substituting $ k \plus{} 1$ for $ n$ gives
\[ (k \plus{} 1)^2 \le (k \plus{} 2)^2 \le [2(k \plus{} 1)]^2 \\
\implies k^2 \plus{} 2k \plus{} 1 \le k^2 \plus{} 4k \plus{} 4 \le 4k^2 \plus{} 8k \plus{} 4.\]
This clearly states that there is a square between $ k \plus{} 1$ and $ 2(k \plus{} 1).$

EDIT: Actually, I misunderstood the problem, so ignore this post. I am not sure how mathemagician's thing is correct either... Shouldn't it be of the form
\[ \sqrt {n} \le p \le \sqrt {2n},\]
where $ p \in \mathbb{N},$ and $ n \le p^2 \le 2n$?

Can someone explain me why mathmagician statement is equivalent to the original one?

I know that the difference between squares is the sum of the sequence $ 2n\plus{}1$, but i can't get the solution yet.

$ n$ has to be between some two consecutive squares. Suppose $ x^2 \le n \le (x\plus{}1)^2$, where $ x \ge 3$ (we can check all cases when $ x < 3$ by hand). Then $ x^2 \le n \le (x\plus{}1)^2\equal{}x^2\plus{}2x\plus{}1 \le 2x^2 \le 2n$.
If $ x<3$, then either $ 1 \le n \le 4$ or $ 4 \le n \le 9$. It's not hard to find a solution for all of these $ n$.

Sorry guys, just ignore my original post. I think I misread the problem when I first did it.

I'll start my new year with a proof for good luck :)

[hide="Solution"]Assume, for the sake of contradiction, that there exist positive integers $ m,n$ such that $ m^2 \plus{} 1 \leq n < 2n \leq (m \plus{} 1)^2 \minus{} 1$. It follows that:

$ \frac {(m \plus{} 1)^2 \minus{} 1 }{m^2 \plus{} 1} \geq 2 \implies (m \minus{} 1)^2 \plus{} 1 \leq 0$, which contradicts the trivial inequality.

Thus, there are no such $ m,n$.[/hide]


Hi everyone, anyone can help?

f : z --> z is a function that f(1)=f(-1) and for all m&n (m and n are integers)
f(m)+f(n) = f(m+2mn) + f(n-2mn)
Find all possible f(x).
thanks  ;)

First put n=1 and get (*)f(m)+f(1)=f(3m)+f(1-2m)
Then put m=-1 and get (**)f(n)+f(-1)=f(-1-2n)+f(3n)
These implie f(-1-2m)+f(3m)=f(m)+f(-1)=f(m)+f(1)=f(3m)+f(1-2m) hence f(-1-2m)=f(1-2m),put m=-m and get
f(x)=c for all odd values of x.
Next n=odd and get f(m)=f(m-2mn) for all m's and odd n's,from here it is easy to guess the general form of f(m).
Answer:$f(2^x\cdot y)=f(2^x)$,for all $x$ and $y$-odd.

[quote]Next n=odd and get f(m)=f(m-2mn)  [/quote]
Hi  hardsoul
i think it should be $f(m) = f(m+2mn)$
thanks  for your  solution ;)


vector
calculus
derivative
calculus computations

For what values of $t$ does the curve given by the parametric equations $x=t^{3}-t^{2}-1$ and $y=t^{4}+2t^{2}-8t$ have a vertical tangent?

When $\frac{dx}{dt}=0$ but $\frac{dy}{dt}\ne 0$, you have a vertical tangent. Imagine moving along the curve so that at time $t$ you are at the point $(x(t),y(t))$. If the $x$-component of your velocity is zero but the $y$-component is not, you are moving in a vertical direction.

This can also be interpreted in terms of vectors: for a curve with vector equation $\vec r(t)=(t^{3}-t^{2}-1)\vec\imath+(t^{4}+2t^{2}-8t)\vec\jmath$ the derivative $\vec r\,{}'(t)$ gives a tangent vector.

If both derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are equal to zero simultaneously, then anything can happen: there may be a tangent line, or there may not be one.


function
calculus
derivative
complex analysis
complex analysis unsolved

Let $ D$ be the unit disk and let $ \varphi: D\rightarrow \mathbb{C}$  be an analytic function with $ |\varphi(z)|<1$ for all $ z\in D$.
 If $ \varphi(z)\equal{}\sum\limits_{n\equal{}0}^{\infty}a_{n}z^{n}$, prove that

$ |a_{n}|\leq 1\minus{}|a_{0}|^{2}$,  $ n\geq 1$.

Do you know a solution for this problem didilica? Where is it from? I am interested, but I know no estimates of taylor coefficients except Cauchy, and possibly some strengtened versions.

Applying the Schwarz lemma to $ \frac{f(z)\minus{}a_0}{1\minus{}\bar a_0 f(z)}$, we find that $ |a_1|\le 1\minus{}|a_0|^2$.
For $ n>1$ consider the function $ f_n(z)\equal{}\frac{1}{n}\sum_{k\equal{}1}^{n}f(e^{2\pi i k/n} z)\equal{}\sum_{m\equal{}0}^{\infty}a_{mn}z^{mn}$ and apply the above to $ f_n(z^{1/n})$.

Yes, that is the solution, mlok has already posted it.
This is a result which is used for obtaining estimations for the higher order derivatives of bounded analytic functions.
Please note the simplicity and the elegance of the proof of it.

nice solution.
Also what is the name of such transformation? [quote="mlok"] $ f_n(z) \equal{} \frac {1}{n}\sum_{k \equal{} 1}^{n}f(e^{2\pi i k/n} z)$ [/quote]

Very neat but also very hard I think. $ a_1$ is kind of Schwarz-Pick, but the other ones are hard.

I don't know if that transformation has a name, but it's very familiar to me as the transformation that sums the terms of a generating function whose indices are a multiple of $ n$.  It's a useful tool in combinatorics, and applied here the property that $ |f(z)| < 1$ implies that $ |f_n(z)| < 1$ as well.  Very clever :)


induction
geometry
perimeter
geometric transformation
reflection
combinatorics proposed
combinatorics

A triple of different subsets $ S_{i}, S_{j}, S_{k}$ of a set with $ n$ elements is called a triangle. Define it's perimeter by:
$ |(S_{i} \cap S_{j}) \cup (S_{j} \cap S_{k}) \cup (S_{k} \cap S_{i})|$.
Prove that the number of triangles with perimeter $ n$ is $ \frac {1}{3}(2^{n\minus{}1}\minus{}1)(2^{n}\minus{}1)$.
This is from Issue 1, Math Reflections, 2008; since deadline for sending solutions is over I believe i can post problems now.

Let A be the set with n elements : $ a_{1},....,a_{n}$
Let's look at $ S_{i}, S_{j}, S_{k}$ as binary sequences: they get 0 at the i'th place iff they don't contain $ a_{i}$, 1 otherwise.
Let's place the 3 sequences one under another. If we want to perimeter to equal n, than in each column there should be: 1,1,1 or 1,1,0, ot 1,0,1 or 0,1,1 - total of 4 optional. So, there are $ 4^n$ possible sequences.

B=Set of equal sequences: only one option ($ S_{i} \equal{} S_{j} \equal{} S_{k} \equal{} A$)
C=Set of exactly 2 different sequences ($ S_{i} \equal{} S_{j} \equal{} A$, $ S_{k}$ can be anything except A : $ 2^{n} \minus{} 1$ options, each counted 3 times)
D=Set of 3 different sequences

$ 4^n \equal{} 6D \plus{} 3C \plus{} B \equal{} D \plus{} 3*(2^{n} \minus{} 1) \plus{} 1$
$ 6D \equal{} 4^n \minus{} 3*2^{n} \plus{} 2$
We are looking for D (because of what Darij wrote): $ \frac {4^n \minus{} 3*2^{n} \plus{} 2}{6} \equal{} \frac {1}{3}(2^{n \minus{} 1} \minus{} 1)(2^{n} \minus{} 1)$, Q.E.D.

[quote="darij grinberg"]1) Please don't use a topic title that spoils the solution of the problem (though in this case, I really wonder why you say induction whereas the problem is easier without).
2) Triples mean ordered triples. What [i]you[/i] mean by a triple is an un-ordered triple, i. e. an equivalence class of triples modulo permutations. If you would mean ordered triples, then the numberof triangles with perimeter $ n$ would be $ 2\left(2^{n \minus{} 1} \minus{} 1\right)\left(2^{n} \minus{} 1\right)$.

  darij[/quote]

Sorry, I won't use a topic title like that again.


is yielding a overline that is touching the top of the nabla operator.

\overline{\nabla \cdot u u}

\overline{u u} is doing the same thing.

Is this some config (or sty) problem?

where to look for? How to correct?

Thanks in advance,
luiz

PS - I am a newbie...

I just tried using overline in the Texer window and didn't get your problem.  ie: no touching of the overline with the div operator.

what sty are you using??

Thanks for your reply.

I am using a template from a brazilian journal:

try something like the following

\tenpoint {\overline{\nabla \cdot u u}}

I am in math mode, would it work?

I got the message:

./eptt2006.tex:196:Undefined control sequence. \epsilon_{\Pi} = \tenpoint

Actually, when zoom to 200% I can see a gap. But can I modify (increase) this gap? I think it is too small...

Thanks a lot for your patience...

:D

fooda(spelling in pt?) pa!

the \tenpoint just sets the font size and vertical spacing so as to alow the overstrike

try \eightpoint it maybe small


probability
set theory
superior algebra
superior algebra solved

[quote="yannick"][quote="shyong"]I wish to know the basic idea to prove these type of questions

1) Prove that $(A\cap B)\subset (A\cup B)$

2) if $A$ and $B$ are independent events , prove that $A^{c}$ and $B^{c}$ are also independent events .

thx[/quote]

For 1) you can take an [b]arbitrary[/b] element $x \in A\cap B$. The only thing you need to do then, is showing that also $x \in A \cup B$. That proves the inclusion.

For 2) I don't know what you mean with 'events'[/quote]

well , question 1 might seems to be an easy conclusion but how are we going to prove that conclusion ?

for question 2 , you can think it as two independent sets $A,B$

[quote="shyong"][quote="yannick"][quote="shyong"]

well , question 1 might seems to be an easy conclusion but how are we going to prove that conclusion ?

[/quote]

This follows from the definition:

"$X \subset Y$ if and only if for every $x \in X$, also $x \in Y$."

Since we have chosen $x \in A \cap B$ [b]arbitrary[/b], it could have been [b]every[/b] $x \in A \cap B$ actually. Hence the conclusion follows from the definition above.

[quote="shyong"]
for question 2 , you can think it as two independent sets $A,B$[/quote]
If you mean $A \cap B = \emptyset$ with independent, you can use the fact that $X \setminus Y = X \cap Y^{c}$ where $X$ is the bigger set you are working in and $Y$ a subset of it. You can easily verify this by drawing Venn-diagrams. If we now take $X=A \cup B$, assume that $A \cap B = \emptyset$ and notice that $B^{c}$ is a subset of $A \cup B$, we can say that $B^{c}= (A \cup B) \cap B^{c}= (A \cup B) \setminus B = A$. Similarly, or by taking complements, you get that $B = A^{c}$. Hence it is easy to observe that $\emptyset = A \cap B = B^{c}\cap A^{c}$ and we are done.[/quote]

ok thanks  :)[/quote]

I will assume that this is a small part of a homework assignment. If I am wrong, then accept my apologies. If you don't, then sue me. I must say I like your signature though (otherwise this topic would long have been deleted).

If you are happy with the proof above, then I would be having a great laugh if I was there when you presented the homework to your teacher.
The proof above is correct, but I don't think it proves what you need. The assumption $X=A\cup B$ is too much of a restriction for the problem text.

When you hear events, then you shouldn't think about set theory, but about probability spaces.
$A,B$ independent means $P(A)P(B)=P(A\cap B)$
$P(A^{c})P(B^{c})=(1-P(A))\cdot (1-P(B))=1-P(A)-P(B)+P(A)P(B)=1-(P(A\cap B)+P(A\cup B))+P(A\cap B)=1-P(A\cup B)=P((A\cup B)^{c})=P(A^{c}\cap B^{c})$ which means that $A^{c}$ and $B^{c}$ are independent.

[quote="amfulger"][/quote][quote="shyong"][quote="yannick"][quote="shyong"]

well , question 1 might seems to be an easy conclusion but how are we going to prove that conclusion ?

[/quote]

This follows from the definition:

"$X \subset Y$ if and only if for every $x \in X$, also $x \in Y$."

Since we have chosen $x \in A \cap B$ [b]arbitrary[/b], it could have been [b]every[/b] $x \in A \cap B$ actually. Hence the conclusion follows from the definition above.

[quote="shyong"]
for question 2 , you can think it as two independent sets $A,B$[/quote]
If you mean $A \cap B = \emptyset$ with independent, you can use the fact that $X \setminus Y = X \cap Y^{c}$ where $X$ is the bigger set you are working in and $Y$ a subset of it. You can easily verify this by drawing Venn-diagrams. If we now take $X=A \cup B$, assume that $A \cap B = \emptyset$ and notice that $B^{c}$ is a subset of $A \cup B$, we can say that $B^{c}= (A \cup B) \cap B^{c}= (A \cup B) \setminus B = A$. Similarly, or by taking complements, you get that $B = A^{c}$. Hence it is easy to observe that $\emptyset = A \cap B = B^{c}\cap A^{c}$ and we are done.[/quote]

ok thanks  :)[/quote][quote="amfulger"]

I will assume that this is a small part of a homework assignment. If I am wrong, then accept my apologies. If you don't, then sue me. I must say I like your signature though (otherwise this topic would long have been deleted).

If you are happy with the proof above, then I would be having a great laugh if I was there when you presented the homework to your teacher.
The proof above is correct, but I don't think it proves what you need. The assumption $X=A\cup B$ is too much of a restriction for the problem text.

When you hear events, then you shouldn't think about set theory, but about probability spaces.
$A,B$ independent means $P(A)P(B)=P(A\cap B)$
$P(A^{c})P(B^{c})=(1-P(A))\cdot (1-P(B))=1-P(A)-P(B)+P(A)P(B)=1-(P(A\cap B)+P(A\cup B))+P(A\cap B)=1-P(A\cup B)=P((A\cup B)^{c})=P(A^{c}\cap B^{c})$ which means that $A^{c}$ and $B^{c}$ are independent.[/quote]

Very much thanks to amfulger for your beautiful proof . Actually these questions are asked by my friend but since I dont know the exact way to prove it , so I post here to seek for help .  :)


We officially close recruiting new members.  We now have 10 members:

   [quote] [b]Andrei Frimu [/b](Moldova)
    [b]Yimin Ge [/b](Austria)
    [b]Ofir Gorodetsky[/b] (Israel)
    [b]Daniel Kohen[/b] (Argentina)
    [b]David Kotik [/b](Canada)
    [b]Hojoo Lee[/b] (Korea)
    [b]Soo-Hong Lee[/b] (Korea)
    [b]Cosmin Pohoata[/b] (Romania)
    [b]Ho Chung Siu[/b] (Hong Kong)
    [b]Peter Vandendriessche[/b] (Belgium)[/quote]
Plus, we still encourage you to join the PEN group in facebook!


linear algebra
matrix
linear algebra unsolved

A $ n\times n$ matrix which entries from $ {1,2,...,(2n\minus{}1)}$ is called "special" if for each $ i$ the union of the $ i$-th row and the $ i$-th column contains $ (2n\minus{}1)$ distinct entries. Prove that there exist no special matrix if $ n\equal{}2007$. Prove that there exists an infinite number of special matrix.

see here : http://www.mathlinks.ro/Forum/viewtopic.php?p=611#611


group theory
abstract algebra
superior algebra
superior algebra unsolved

I'm in trouble understanding the proof that

If a group $ G$ is simple and of order $ 60$, then it is isomorphic to $ A_5$.

In the proof we let $ P$ be a Sylow $ 5$-subgroup of $ G$, and $ \Omega\equal{}\{g^{\minus{}1}Pg: g\in G\}$. Now $ G$ acts in the set $ \Omega$ by conjugate action and it induces the mapping $ \rho: G\to S_6$. Let $ H\leq S_6$ be the image of this homomorphism. Then in the proof we denote $ \Gamma : \equal{}\operatorname{cos}(A_6: H)$. Now it was said that $ |\Gamma|\equal{}\frac{1}{2}6!/60\equal{}6$. But I think that $ \Gamma$ is a single number, not a set.

I think that it means that $ \Gamma$ is the set of cosets of $ H$ in $ A_{6}$.

That was what I was thinking as well. But is there a reason we should believe that $ |H|\equal{}60$? That seems rather unlikely.

Yes, $ \ker{\rho}$ must be $ 1$ or $ G$, but cannot be the latter because the action on $ \Omega$ is transitive.  So $ \rho$ is injective and $ |H| \equal{} |G| \equal{} 60$.  I do not immediately see why $ H\leq A_6$, though.

Consider a finite group of order $ 60$ and suppose that $ G$ is simple. Let $ x$ be the number of Sylow $ 5$-subgroups. Then we know that  $ x$ divides $ 24$ and is congruent to one modulo $ 5$. Thus $ x \equal{} 6$. Letting $ G$ act on its Sylow subgroups by conjugation, it follows that there is a representation $ \rho : G \rightarrow S_{5}$. Since $ G$ is simple, it follows that $ G$ is isomorphic to a subgroup of $ S_{5}$. But $ |S_{5}| \equal{} 120$ and $ |G| \equal{} 60$, What does it means? it means that $ \rho(G)$ is a subgroup of index $ 2$ so it is normal in $ S_{5}$ so we must have $ G \cong A_{5}$ as desired.

[quote="alejnaser"]it follows that there is a representation $ \rho : G \rightarrow S_{5}$. Since $ G$ is simple, it follows that $ G$ is isomorphic to a subgroup of $ S_{5}$.[/quote]
$ S_6$, you mean?

I just don't understand where your $ \rho$ came from.  What five things are you permuting?

The action of conjugation on the [i]six[/i] Sylow $ 5$-subgroups gives an injective map into $ S_6$, that much is clear to me.  Finishing that proof requires demonstrating that a simple subgroup of index $ 12$ in $ S_6$ is isomorphic to $ A_5$.

I tried to do a different proof, but came back to what you said.
Well, you said that that gives an inyective map into $ S_{6}$ and we have to show that a simple subgroup of index $ 12$ in $ S_{6}$ is isomorphic to $ A_{5}$. It is the same to prove that $ A_{5}$ is the only subgroup of index $ 12$ in $ S_{6}$. But this is easy, Let $ H$ be another subgroup of $ S_{6}$ of index $ 12$, then $ H \cap A_{5}$ is a normal subgroup of $ A_{5}$, but $ A_{5}$ is simple, so the intersection is either the trivial subgroup of the whole $ A_{5}$, In the first case we have that $ HA_{5}$ is a subgroup of $ S_{6}$ of order $ |H\parallel{}A_{5}|/|H \cap A_{5}| \equal{} 60^2 > |S_{6}|$ a contradiction!, so it has to be $ H \cap A_{5} \equal{} A_{5}$ and we are done.

[quote="alejnaser"]Let $ H$ be another subgroup of $ S_{6}$ of index $ 12$, then $ H \cap A_{5}$ is a normal subgroup of $ A_{5}$[/quote]
Why?  Be careful here.  Nobody said $ H$ was normal.  In fact, it is not hard to see that there are at least $ 6$ subgroups of $ S_6$ that are isomorphic to $ A_5$, so showing that $ H$ is equal to some particular $ A_5$ is ridiculous.

[quote="LydianRain"]I do not immediately see why $ H\leq A_6$, though.[/quote]
Actually, this is pretty straightforward.  In fact, any simple subgroup of $ S_n$ whose degree is greater than $ 2$ is contained in $ A_n$, because the restriction of the sign homomorphism $ S_n\to Z_2$ must be trivial.

The proof from here is a lot of fun, actually.  We have $ H\leq A_6$ with index $ 6$.  So consider the action of $ A_6$ on the cosets of $ H$ by left multiplication.  This gives us an injection $ A_6\to S_6$ which, since $ A_6$ is simple, restricts to an isomorphism $ A_6\to A_6$.  What is the image of $ H$ under this isomorphism?

Left multiplication by elements of $ H$ certainly fix the coset $ H$, so the image of $ H$ acts on only $ 5$ elements--in other words, it is contained in some group $ K$ which is isomorphic to $ A_5$.  By counting orders, it must equal $ K$, and it follows that $ H$ is isomorphic to $ A_5$.


calculus
derivative
function
calculus computations

Hi guys,
   I got these two problems recently, and the wording of them has me a little confused.  Im not sure what I am supposed to do with them exactly, can you point me in the right direction on how to start these guys.  Thanks again!

a. If $y = f(x) = \frac x{(x+2)}$, find the value of x, call it C, on the interval $1 < x < 4$
such that $f '(C) = \frac{[f(4) - f(1)]}{{(4 - 1)}}$.

b. Let $f(x) = \frac{x + 1}{{x - 1}}$. Show that there is no value of C such that
$f(2) - f(0) = f '(C)(2 - 0)$

Chris

a. Solve the equation.

b. I would probably study the second derivative and draw the first derivative in order to show that there are no solutions.

a)
$f(x)=\frac{x}{x+2}$
$f'(x)=\frac{(x+2)\frac{d}{dx}x-x\cdot\frac{d}{dx}(x+2)}{(x+2)^2}=\frac{2}{(x+2)^2}$
$f'(C)=\frac{f(4)-f(1)}{4-1}$
$\frac{2}{(C+2)^2}=\frac{\frac{2}{3}-\frac{1}{3}}{3}$
$(C+2)^2=18$
$C^2+4C-14=0$
$C=\frac{-4\pm\sqrt{4^2+4(14)}}{2}=-2\pm3\sqrt{2}$
$\because C\in(1,4)\therefore C=3\sqrt{2}-2$

b)
$f(x)=\frac{x+1}{x-1}$
$f'(x)=-\frac{2}{(x-1)^2}$
$f(2)-f(0)=f'(C)(2-0)$
$3-(-1)=2f'(C)$
$-\frac{2}{(x-1)^2}=2$
$(x-1)^2=-1$
contradiction

Remark: I think the aim of these two problems are to introduce the mean value theorem: If a function $f(x)$ is continuous and differentiable on the interval $(a,b)$, then there exists a point $c\in (a,b)$  such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. For question a, since $f(x)=\frac{x}{x+2}$ is continuous on the interval $(1,4)$, such $C$ exists. However, in question b, since $f(x)=\frac{x+1}{x-1}$ is not continuous at $x=1\in(2,4)$, such $C$ may not exist.

I was referring to Boousaf.  I have a strong hunch that he is my long lost best friend and roommate from the Air Force.  I've been searching for him for a long time.  Sorry for the confusion.  I just was not ABSOLUTELY sure that it was him.  I sent a private message to him but was not sure if he would check his messages.  I just wanted to post a short message in this post since it was his most recent.  I figured he'd know who I was by my handle.  I hope this is not a violation of the forum rules.


If a, b, c are positive reals such that  $a^{2}+b^{2}+c^{2}= 1$ prove that
$a+b+c+\frac{1}{abc}\geq 4\sqrt{3}$


Reply here if you are interested in participating in an unofficial State Math Contest. Anyone can participate, including Neal and Tare. The rules & regulations of the contest will be rather similar to what Tare is doing. However, I will be posting problems from the NC state math contest. You can post more than one solution for each problem, but only the last will be taken into account. You will receive no feedback as to your solutions until the end of the contest. (muahahaha, doesn't that sound evil :twisted::twisted:)

There are about 5 or 6 rounds.
Calculators are allowed.
Tare, don't feel sad, I've been meaning to do this for a long time and decided to wait until I became a moderator.

registered peeps so far: Tare, Mr. Southpaw, Neal, Martin, Mathfiend

oh boy, first Neal decides to post 20 questions right after I finish round 1, and now this...I think I'm going to go cry in the corner :cry: ...*sob*...*sniff* *sniff*...*blows nose*...Oh no! I blew my nose on my brand new AoPS 1!!! Aaaaaaaaaah! :)

Seriously though, if people hate my contest and want me to quit, then just tell me. It hurts more when you say it "in between the lines"

P.S. My prediction: we'll have another comod by next week

i don't understand why peeps can't work on 2 or more problems at the same time. whoever said they didn't like your contest? i'm just trying to pick up Mathcounts activity since it wasn't that hot over the last week.

yeah, I know...even sadder thing is that usually peeps answer Neal's or your questions, but not mine.

*sigh* :cry:

I'll register, although I doubt I'll get nething

humbug, its a learning experience. i should really answer those questions shouldnt i? meh, ill do it later

sorry guys
count me out
no time
AP's are killing me


[quote="myc"]How do you convert fractions from base to base?  for example,  how do you convert $37/97$[size=75]base 5[/size] into base 10?

Thanks[/quote]
Maybe $\frac{a}{b}_{5}=\frac{a_{5}}{b_{5}}$. But then, you could only use digits 0, 1, 2, 3, and 4.

[quote="myc"]How do you convert fractions from base to base?  for example,  how do you convert $37/97$[size=75]base 5[/size] into base 10?

Thanks[/quote]

There is no $\frac{37}{97}_{5}$

Actually, it's roughly the same as converting whole numbers?[hide=" :huh: "]Well, is it?[/hide]

o, right,i knew that.. hehe :oops: .  how about 43/32 [size=75]5[/size]  into base ten.

To convert fractions, convert them the same way you would convert integers.  $\frac{43}{32}_{5}=\frac{4\times5+3}{3\times5+2}=\frac{23}{17}_{10}$

[quote="SplashD"]So $\frac18$ would be the same in both base 9 and 10?[/quote]
Yes, since $8$ would be the same in both base $9$ and base $10$. This only works for single-digit numbers though, because, for example, $16_{9}=15_{10}$ is not equal to $16_{10}$.


In some algebra problems i've seen, the summation doesn't specify any variables. could some explain what that means? (something like $ \sum abc = 2abcd$)

It means a "cyclic sum" basically a premutation of everything you add.

So, $ \sum ab^{2}= ab^{2}+bc^{2}+ac^{2}$ get the idea?


A recent autobiography used 2808 digits to number the pages of the book.  How many numbered pages are in the book?

What I'd do for this problem is add up the number of digits for 1-999.

9 x 1 + 90 x 2 + 900 x 3 = 2889.

Then subtract 2,808 from 2,889, to get 81.

Divide this result by 3 to get 27.

Subtract 27 from 999 to get the answer of [b]972[/b].


Hmm....I think its easy but i cant think of a way to solve this

ff(x)=6x-3

find f(x)....

i tried letting f(x)=ax+b or f(x)=ax^2+b......i cant seem to think of a way that ff(x) has a power 2 in it.........

I assume by "ff(x)" you mean $ f(f(x))$?

If so, then there is no unique solution to this. More information would need to be given about the nature of the function $ f$ - like, for instance, that $ f$ is linear. If $ f$ is linear then the answer would be $ f(x) \equal{} \sqrt{6}x \minus{} \frac{3}{5}(\sqrt{6}\minus{}1)$ or $ f(x) \equal{} \minus{}\sqrt{6}x \plus{} \frac{3}{5}(\sqrt{6}\plus{}1)$.

But there are many other functions (even badly discontinuous ones) that will work.

So that means that thers no solution to this? YEa and i meant ff(x) by f(f(x))

Let $ deg f(x) \equal{} n$, then we know that $ deg f(f(x)) \equal{} n^2 \equal{} 1$, therefore $ n \equal{} 1$. So we can express $ f(x)$ as $ ax \plus{} b$ with $ a,b \in R$.

$ a(ax \plus{} b) \plus{} b \equal{} 6x \minus{} 3$

$ a^2x \plus{} ab \plus{} b \equal{} 6x \minus{} 3$

Solving this, we find $ a \equal{} \pm \sqrt{6}$ and $ b \equal{} \frac{\minus{}3}{1 \pm \sqrt{6}}$.

ah...wait....I got a correction.....its f(f(x))=6x^2-3    , How do i find f(x)

See if you've learned from the last version of the problem how to do this version. Try it out and post what you have.

[By the way, the problem is still vague in that it doesn't say what kind of function $ f$ has to be. modularmarc101 made the assumption, as was probably intended, that the function $ f$ is a polynomial.]


Suppose T:V-->V is a nilpotent linear map.Then there exists subspaces W1,W2,...Wk of V such that
1)T(Wi) is contained in Wi ,for i=1,...,k
2)Each Wi is T-cyclic subspace of V
3)V=direct sum of Wi's ,for i=1,...k.

In the proof of this question ,as T(Wi) is contained in Wi,we get a induced linear map,T':V/Wi-->V/Wi defined by T'([v])=[Tv],for all v in V.
how do i show that the map is well defined..

You need to show that if $ v \in W_i$, then $ T(v) \equal{} 0$. This is obvious since $ T(v) \in W_i$.

Why does this show well-definedness? Because if $ [v] \equal{} [w]$ then $ v\minus{}w \in W_i$, and we notice that $ T([v]) \equal{} [Tv] \equal{} [Tw \plus{} T(v\minus{}w)] \equal{} [Tw] \plus{} [T(v\minus{}w)] \equal{} [Tw] \equal{} T([w])$.


How do you evaluate $\sum_{n=1}^{\infty} \frac{n^2}{6^n}$?

Thanks.

Nice work, but on the last subtraction, that should have been
$\frac{25}{36}S=\frac{5}{36}+\frac{2}{30}=\frac{37}{180}$

So $S=\frac{37}{125}$.

Yeah, it's just slightly misleading because he wrote $\frac{1}{36}$ instead of $\frac{1}{6}$ in the last expression $(\frac{25S}{36})$


The following problem is not difficult, but its absolutely not [i]that[/i] easy as it may seem at first glance. To the more advanced problem-solvers, I leave the pleasure to discover the appropriate generalization (a much more difficult problem).

[b]Problem. [/b]  For an arbitrary sequence $\{x_i\}_{i \geq 0}$ of real numbers, define the sequence $\{y_i\}_{i \geq 0}$ by
\[
y_i := x_{i+4} + 3x_{i+3} + 7x_{i+2} + 11x_i.
\]
Prove: the sequence $\{y_i\}$ converges if and only if the sequence $\{x_i\}$ converges. What is the appropriate generalization?

Have fun!  :D
--Vesselin

The stability of $ \{x_n\}$ defined by  $ x_{n \plus{} m} \plus{} \sum_{k \equal{} 0}^{m \minus{} 1}c_kx_{n \plus{} k} \equal{} y_n$ depends on  the characteristic polynomial  $ z^m \plus{} \sum_{k \equal{} 0}^{m \minus{} 1}c_kz^k \equal{} 0$ :  the roots  have to be $ z \equal{} 1$ or $ |z| < 1$
 :cool:  Have you seen than the exposed problem doesn't work ?  :cool:


Set $a_i = +1$ and we have the sum $S = \sum_{cyc} a_1a_2a_3a_4 = n$. Changing the sign of $a_i$ affects $4$ different sums. Of these $4$ say $k$ of them are positive which become negative; then $4-k$ are negative which become positive. $S$ then changes by $-2k+2(4-k) = 8-4k = 4(2-k)$. So $S$ is invariant $\pmod{4}$ which means $n$ has to be divisible by $4$ for $S$ to possibly be $0$.

[hide="another solution"]
Let $x_{n+k}=x_k$ and denote $x_kx_{k+1}x_{k+2}x_{k+3}$ by $y_k$. Then $y_1+y_2+\cdots +y_n=0$.

Clearly each $y_k$ is $\pm 1$. The product $y_1y_2\cdots y_n$ must equal 1 because each $x_k$ appears 4 times. So an even number of the $y_k$ are -1's.

Suppose $2a$ of the $y_k$ are -1's. Since $y_1+y_2+\cdots +y_n=0$, there are also $2a$ 1's among the $y_k$ for a total of $4a$. This means $n=4a$.
[/hide]


Yes. It's easy but nice. But the extender is more diffucult.  :(

Vasc is this the best generalization?

I can't see if equality can be achieved or not ?!!(like in the first one)

maybe adding a $\frac{2^n}{n^2}$  :D

right  :wallbash:  :play_ball:
just see the case $x_1\rightarrow inf x_2=x_3=...=x_n \rightarrow 0$
 :play_ball:

Ps :
I think that it can be solved by seeing the discriminant of the expression but I'm not sure, so I'm not posting anything
 :D

[quote="Vasc"]If $n \geq 4$ and $x_i$ are positive numbers such that $x_1x_2...x_n=1$, than
$(1+x_1^2)(1+x_2^2)...(1+x_n^2) \geq (x_1+x_2+...+x_n)^2$[/quote]

Use induction.  ;)

[quote="hungkhtn"]This is very hard problem

Prove that if $a,b,c,d \ge 0$ and $abcd=1$ then
\[ (1+a^2)(1+b^2)(1+c^2)(1+d^2) \ge (a+b+c+d)^2+6(a+b+c+d-4)  \][/quote]

Any idea for it? It's very difficult problem and i hope a week for any solution.

I have an approach of mixing variables for this one
maybe with some improvement we can find a similar one for your ineq !
$(1+a^2)(1+b^2)(1+c^2)(1+d^2)\geq (a+b+c+d)^2+4(a+b+c+d-4) \leftrightarrow \sum_{sym} (1-ab)^2 + \sum f(a) \geq 0$ where $f(x)=3+\frac{1}{x^2}-4x$

we can easily  show that $f(a)+f(b)\geq 2f(\sqrt{ab})$ for $ab\le 1$

Can you complete your solution? This one is a sample of problem with 4-variable which almost can be prove by one-method. I use Stronger Mixing Variable Method (SMV Theorem).

maybe we can show that $(1-ab)^2+\frac{1}{a^2}+\frac{1}{b^2}+10-6(a+b)\geq 0$
but im not sure :(

Why havent I heard for SMV?
What are you refering to hunghtn?

Yes, Mixing Variable is one of chapter in my book about problem solving, and SMV mean Stronger Mixing Variable. It is stronger than usual but the proving is only use elementary mathematical.


i have a better problem. inspired  on you, of course

${\sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \sqrt{2 \square \ldots }}}}}}}}}}}}}=2}$

where $\square$ is $+$ or $*$ operator.
each square that appear is different from another, so we can have a combinanation of (+,*,*,*,+,*,+,*,+) ... etc.. that all possible combination convergers to $2$

PS: this is a very hard problem, not a intermediate. need to be moved or something else

[hide]
Choose $a_{0}= k > 0$ and take $a_{n+1}= \sqrt{2+\sqrt{2 a_{n}}}$
We have $a_{n}> a_{1}> 1$. When $0 < k < 2$ $a_{n+1}> a_{n}$, otherwise $a_{n+1}\leq a_{n}$
Now take $\tau = \lim_{n \to+\infty}a_{n}$
We have $\lim_{n \to+\infty}\frac{a_{n+1}}{a_{n}}= \lim_{n \to+\infty}\frac{\sqrt{2+\sqrt{2 a_{n}}}}{a_{n}}= \frac{\sqrt{2+\sqrt{2 \tau}}}{\tau}= 1$
This become $\tau = \sqrt{2+\sqrt{2 \tau}}\Longrightarrow \tau^{2}= 2+\sqrt{2 \tau}\Longrightarrow \tau^{4}-4\tau^{2}-2 \tau+4 = 0$
Now $(\tau-2)(\tau^{3}+2\tau^{2}-2) = 0$
The solutions are $\tau_{1}= 2$ and $0 < \tau_{2}< 1$, so $\lim_{n \to+\infty}a_{n}= \tau_{1}= 2$
[/hide]

[quote]${f(n+1)=\sqrt{(2+f(n))\lor (2 f(n))}}$
${f(0)=k}$

${\tau =\lim_{n\to \infty }\, f(n)}$

${\lim_{n\to \infty }\, \frac{f(n+1)}{f(n)}=\frac{\sqrt{(2+\tau )\lor (2 \tau )}}{\tau }=1}$

${\left(\frac{\sqrt{2+\tau }}{\tau }=1\right)\lor \left(\frac{\sqrt{2 \tau }}{\tau }=1\right) \Rightarrow (\tau =2)}$[/quote]

we can use it to some my general problem!?

because the solution of rbtqwt is not of the problem of "Love U"

Or you could just prove that $x = 2$ when $\sqrt{2+\sqrt{2x}}= 2.$

$\sqrt{2+\sqrt{2x}}= 2$
$\Leftrightarrow 2+\sqrt{2x}= 4$
$\Leftrightarrow \sqrt{2x}= 2$
$\Leftrightarrow x = 2. \blacksquare$

As yagaroon noted, the recursion that rbtqwt and Karth have solved doesn't give the desired expression but rather $\sqrt{2+\sqrt{2\sqrt{2+\sqrt{2\sqrt{2+\sqrt{2\sqrt{2+\ldots}}}}}}}$

[quote="Ravi B"]And even for the modified problem, Karth's solution shows, at best, that if the expression converges, then it converges to 2.  His solution, unlike that of rbtqwt, doesn't show that the expression converges.[/quote]

It's actually quite well known that this type of sum converges  :wink:

Heres my approach... seems kind of weak though

[hide]
Note that $2^{2}-2 = 2$ and $\frac{2^{2}}{2}= 2$ (uhh.. I guess we can call this lemma 1?)

Assuming the "nested radical" does equal to 2, note that, looking within the radical sign of the given problem, every number is $2$ and the only operations are addition and multiply.

That means everytime you square both sides of the equation, the left hand side is either $2+\sqrt{x}$ (where x is some number) or $2 * \sqrt{x}$, while the right hand side becomes $2^{2}$.

To isolate x after squaring both sides, you can do only two possible choice: either subtract $2$ from both sides or divide $2$ from both sides. And from lemma 1 we know that both of these operations will always have the right hand side return its original value of $2$. Therefore you can square both sides and isolate $x$ infinitely, and the right hand side will always have the same value of $2$ ($claim 1$).

Now look inside the "nested radical" and note that every single number is 2. Therefore, if you square the radical and isolate $x$ infinitely amount of times such that you take out all radical sign, the only possible "last number" (if it exist) that x can have is $2$ (since there are only $2$ in the radical and no other number) ($claim 2$).

From claim 1, we know that the right hand side will always be $2$ despite the amount of times we square both sides and isolate $x$. From claim 2, we know that squaring both sides and isolating $x$ infinitely would have the left hand side's only possible value be 2 and no other number, provided that such a number exist. This means that if you square both sides and isolate $x$ infinitely, the equation you are left with is $2 = 2$, which is a true statement leading to the fact that the assumption of the "nested radical" being equal to $2$ true.
[/hide]


function
induction
algebra proposed
algebra

Find all strictly increasing $f: N\rightarrow N$ satisfying
$f(n+f(n))=2f(n)$


Let a, b, c> 0. Find all triples (a,b,c) of positive real numbers such that

x+y+z=a+b+c
(a^2)x+(b^2)y+(c^2)z+abc=4xyz


function
calculus
derivative
algebra
functional equation
real analysis
real analysis unsolved

If $ a,b \in \mathbb{R}^*$ are such that $ a\plus{}b \not\equal{} 0$, then find all the differentiable functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(x\plus{}ay)\plus{}f(x\plus{}by)\equal{}2f(x)\plus{}ky, \forall x \in \mathbb{R}, \forall y \in \mathbb{R}^*$, where $ k$ is a real constant.

All solutions are given by functions of the form $ f(x)\equal{}\frac{kx}{a\plus{}b}\plus{}t$ with $ t\in\mathbb{R}$ arbitrary.

[i]Proof[/i]: Differentation of the functional equation by $ y,x$ yields the two equations $ k\equal{}af'(x\plus{}ay)\plus{}bf'(x\plus{}yb)$ and then $ 2af'(x)\equal{}k\plus{}(a\minus{}b)f'(x\plus{}yb)$. Substitution of $ y\equal{}0$ into the second identity gives: $ (a\plus{}b)f'(x)\equal{}k$, therefore we can write $ f(x)\equal{}\frac{kx}{a\plus{}b}\plus{}t$ with some $ t\in\mathbb{R}$. Direct calculation shows that any function of this form fulfills the functional equation.


Hoping this is problem has a trick and a nice solution... :
[img]http://x5.freeshare.us/121fs662214.gif[/img]

1. Note that $M_{(a,0)}=aI,$ and arithmetic for such matrices is just the arithmetic on $\mathbb{R}.$

2. Note that $M*N=MN+M+N=(M+I)(N+I)-I.$ This is just ordinary matrix multiplication in disguise.

How far will that take you?


We are a group of math instructors in Saudi Arabia  who would like to start a math Olympiad training for our students. We need your help in finding good places for training instructors first to be trainers for the students. Also, we would like to know is there any upcoming workshops or conferences for the Math Olympiad in any place in the world.

The World Federation of National Mathematics Competitions is having a workshop this summer in England.  Details are on the WFNMC web site at [url]http://www.amt.edu.au/wfnmc.html[/url] (the 3rd link on that page is for the conference).


group theory
abstract algebra
induction
number theory
prime factorization
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and for every subgroup $ H$ of $ G$ there is a homomorphism $ f: G\to H$ such that $ f(h) \equal{} h$ for all $ h\in H$. Prove that $ G$ is isomorphic to a direct product of cyclic groups of prime order.

Let $ P$ be a Sylow $ p$-subgroup for some prime $ p$.  Then we are given a surjective map $ f: G\to P$, so there is some  subgroup $ N$ such that $ G/N \cong P$.  Similarly, we have a surjective map $ g: G\to N$, so there is a normal subgroup Q such that $ G/Q \cong N$.  Since $ \frac{|G|}{|Q|} \equal{} |N|$ and $ \frac{|G|}{|N|}\equal{}|P|$, we have $ |Q|\equal{}|P|$, so $ Q$ is a normal Sylow $ p$-subgroup.  Hence all Sylow $ p$-subgroups are normal.

Since $ p$ was arbitrary, $ G$ is the direct product of its Sylow subgroups, so $ G$ is abelian.  Suppose that $ G$ was not the direct product of cylic groups of prime order, then $ G$ would contain a copy of $ Z_{p^2}$ for some $ p$.

But $ Z_{p^2}$ does not satisfy the condition, as if $ x$ has order $ p^2$, there any homomorphism $ \phi: \langle x\rangle\to \langle x^p\rangle$ has $ \phi(x^p)\equal{}1\neq x^p$, contradiction.

A little bit elementary:
We proceed by induction on number $ n$ of primes (not necessarily distinct) in prime factorization of $ |G|$.
Case $ n \equal{} 1$ is obvious.
Assume that for fixed $ n$ our statement is true.
If $ |G| \equal{} p_{1}\cdot \ldots\cdot p_{n}\cdot p_{n \plus{} 1}$ (where $ p_{i}$ are prime), then let $ P$ be the subgroup of $ G$ of order $ p_{n \plus{} 1}$ (which exists by Cauchy's Theorem).
We have $ f: G \to P$ s.t. $ f(h) \equal{} h, \ \forall h \in P$. Define $ K \equal{} \ker f$.
By Isomorphism Theorem we have $ G/K \simeq P$, in particular $ |G/K| \equal{} p_{n \plus{} 1}$ and hence, by Lagrange's Theorem $ |K| \equal{} p_{1}\cdot \ldots \cdot p_{n}$.
There also exists $ g: G\to K$ s.t. $ g(k) \equal{} k, \ \forall k \in K$.
Putting $ P_{1} \equal{} \ker g$ we obtain $ G \equal{} K \oplus P_{1}$, since $ K, P_{1}\triangleleft G,\ K\cap P_{1} \equal{} \{e\}$, and $ |P_{1}| \equal{} p_{n \plus{} 1}$ (again by Isomorphism Theorem and Lagrange's Theorem).
By induction hypothesis $ K$ is direct product of its cyclic subgrops of prime order, and this fact ends the induction step.

@LydianRain:
If $ G$ is the direct product of its sylowgroups it is nilpotent but neccessary abelian.

Ah, whoops.  I suppose that Sylow groups were a digression and that ifk's modification is the way to go.

If $ 0\to A\to B\to C\to 0$ is an exact sequence of (not necessary abelian) groups and if there exists a projection $ B\to A$, then $ B$ is isomorphic to $ A\times C$. Consequently, if $ G$ is a group satisfying the "projection condition" stated above, every subgroup of $ G$ must be a direct summand. Thus, by considering cylic subgroups of $ G$ and noting that a subgroup of $ G$ also satisfies the projection condition, we reduce the problem to the case when $ G$ is cyclic. In that case, we can decompose $ G$ into cylic summands of order a prime power and in that case the claim is obvious.

[quote="Hanno"]If $ 0\to A\to B\to C\to 0$ is an exact sequence of (not necessary abelian) groups and if there exists a projection $ B\to A$, then $ B$ is isomorphic to $ A\times C$[/quote]

I am always bewildered when a theorem is employed in solving a problem, that is almost equivalent to the problem to be solved.


$ a,b,c >0$ .Prove that:
$ \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} < \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$

First of all , we use Chebyshev's for 2 increasing sequences :
    $ a, b , c $ and $ \frac{1}{b+c} , \frac{1}{c+a} , \frac{1}{a+b} $
  $ => \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \leq \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}  $
  We will prove :
   $ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \leq \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{a+c}} + \sqrt{\frac{c}{b+a}} $
   We prove more general ineq :
  $ ^n\sqrt{\frac{a}{b+c}} + ^n\sqrt{\frac{b}{a+c}} + ^n\sqrt{\frac{c}{b+a}} > \frac{n}{n-1}^n\sqrt{n-1}  (1) $

    Using AM-GM ineq :
   $ \frac{\frac{(a+b)(n-1)}{c} + (n-1)}{n} \geq ^n\sqrt{\frac{(a+b)(n-1)}{c}} $
   $ => \frac{(a+b+c)(n-1)}{nc} \geq ^n\sqrt{\frac{(a+b)(n-1)}{c}} $
   $ => ^n\sqrt{\frac{c}{a+b}} \geq \frac{n}{n-1}^n\sqrt{n-1} \frac{c}{a+b+c} $
    $ ==> (1) $ is right .

   With n=2  $ => \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{a+c}} + \sqrt{\frac{c}{b+a}} > 2 $
    Moreover $ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} < 2 $
  $ => dpcm $

Interesting solution, but the statement of the problem is a little bit different. Anyway, I think the originaly posted inequality cannot be true and I congratulate you for the beautiful solution.

Hi!
[quote="nttu"]Moreover $ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} < 2 $
  $ => dpcm $[/quote]
Actually, it is wrong inequality. I think you mean
\[ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} < 2.\]

[quote="harazi"]And actually I am a bit idiot this morning. Thank's, Myth, you are right, the original inequality is true.[/quote]
It is because my local time is 13:30pm, and my morning have already ended.  :D

Oh, boy, this means that I have to wait at least 2 hours to stop being idiot  :)  :) . Please, afternoon, come also in France!  :)  :)

Oh , thank you Myth for showing a mistake . It's easy to fix it .

I don't undestand why we need so long proof for such easy problem (I mean initial problem)?  ;)
We have
\[\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\leq \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}.\]
Note that $\frac{a}{a+b}<1$, so $\sqrt{\frac{a}{a+b}}>\frac{a}{a+b}$.
That's all.


helo
Find the set $ A$ such as:
$ A\equal{}${$ x\in\mathbb{Z}/\frac{x^2\minus{}x\plus{}2}{2x\plus{}1}\in\mathbb{Z}$}

[quote="sami.d"]helo
Find the set $ A$ such as:
$ A \equal{}${$ x\in\mathbb{Z}/\frac {x^2 \minus{} x \plus{} 2}{2x \plus{} 1}\in\mathbb{Z}$}[/quote]

Im sorry is that $ \mathbb{Z} mod\left( \frac {x^2 \minus{} x \plus{} 2}{2x \plus{} 1}\right)$

or all integers in the form of $ \frac {x^2 \minus{} x \plus{} 2}{2x \plus{} 1}$.

The notation means "the set of all integers $ x$ such that the given fraction is integer."

[hide]$ {x^2 \minus{} x \plus{} 2\over 2x \plus{} 1} \equal{} n\implies 2x \minus{} 3 \plus{} {11\over 2x \plus{} 1} \equal{} 4n$

Hence $ 2x \plus{} 1\in\{\pm 1,\pm 11\}\implies x\in\{ \minus{} 6, \minus{} 1,0,5\}$

Checking shows that all the values work, so $ A \equal{} \{ \minus{} 6, \minus{} 1,0,5\}$[/hide]


together a chair a table and a lamp cost $ 562. the chair cost 4 times as much as the lamp, and the table cost$23 less than the chair. Find the cost of the table.

$ c \plus{} t \plus{} l \equal{} 562$

$ c \equal{} 4l$

$ t \equal{} c \minus{} 23$

Put the equation in terms of $ c$.

$ c \plus{} (c \minus{} 23) \plus{} \frac{1}{4} c \equal{} 562$

$ \frac{9}{4} c \minus{} 23 \equal{} 562$

$ \frac{9}{4} c \equal{} 585$

$ c \equal{} 260$

Thus, the table costs $ \boxed{237 \text{ dollars}}$.

Or if you have to use dollar signs, use this:

[code]$ \$ x $[/code]

Use $ x$ as the amount of money.


Hello evrybody

I have this matrix
\[\begin{pmatrix}0 & 4 \\ 0 & 1 \\ 1 & 3 \end{pmatrix}\]

I can't type this matrix all the time and I just replace it with M

so i have to find matrix X from equation

AX = M

from here

X = $\frac{M}{A}$ or M * $A^{-1}$

or
X = M * inverse(M)

and problem is how to inverse this matrix
please tell me if i'm in wrong way

hope understand me

The standard way to solve equations $AX = M$ is to compute the pseudoinverse $A^{+}$ of $A$. Then $X=A^{+}M$.

Note that in matrix algebra there is no such thing as $\frac{M}{A}$ and if $A$ is an invertible square matrix satisfying $AX=M$, then $X=A^{-1}M$, not $X=MA^{-1}$.


prob. 9
here is a right triangle $ ABC$ which $ AB \equal{} 20, BC \equal{} 25, CA \equal{} 15$. $ H$ is the foot of perpendicular that is put down from $ A$ to $ BC$. M is in the middle of $ AB$. find $ 5AP$ when ∠$ AHM \equal{}$ ∠$ AHP$ in respect to the point $ P$ on the side $ AC$.

It is well known that, when AH is an altitude and M and P on AB and AC respectively such as $ \angle AHM \equal{} \angle AHP$, then BP and CP meet on AH.
Obviously, $ \frac{BH}{CH} \equal{} \left ( \frac {AB}{AC} \right )^2$, then applying Ceva we get $ 5\cdot AP \equal{} 48$.

Best regards,
sunken rock


Give me an example of a continuous surjection from (-1,1) to        [-1,1].

Does there exist any differtiable surjections from (-1,1) to [-1,1]

[quote="jmerry"]$\cos(36x)$.[/quote]
Can you tell me how can you hit upon such an example. :?
What is the signifance of 36.Can we replace it with 72 or for that matter any other number.

[quote="the game"]What is the signifance of 36[/quote]
$36>\frac{\pi}{2}.$

It was just the first number that came to mind.

The problem would be more interesting if we asked for a continuous bijection, what is, of course, not possible!


Neka su $a,b,c$ strane trougla tako da $2s=a+b+c=6.$ Pokazite da
\[\frac{10}{(s-a)(s-b)(s-c)}\;+\;3\left(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right)\;+\;1\geq \frac{16}{abc}\left(4+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).\]

Jooj Dje ovo nadje Izgleda strasno al ipak morat cu bar pokusat.


search
geometry
circumcircle
trigonometry
cyclic quadrilateral
perpendicular bisector
geometry proposed

Let $ \triangle ABC$ be a triangle and $ M$ be the midpoint of $ BC$.Denote the circle with diameter $ AM$ as $ \omega$.Let $ \omega\cap AB\equal{}D$ and $ \omega\cap AC\equal{} E$.Prove that $ PB\equal{}PC$,where $ P$ is the intersection of the tangent lines to $ \omega$ at $ D$ and $ E$.

[quote="silouan"]It was posted again

http://www.mathlinks.ro/viewtopic.php?search_id=527669261&t=16779

Erken which is your solution (or the official) ?[/quote]
I can not find it in your link.My solution uses only polar theory. :wink: ...

[quote="Erken"] I can not find it in your link[/quote]

The problem dear Erken is the same  :wink:

Here is my proof:
[img]http://img76.imageshack.us/img76/5882/adbd1.png[/img]
Proof:
Let $ ME\cap AB \equal{} K$,$ MD\cap AC \equal{} L$ and $ AM\cap DE \equal{} X$.Since $ DE$ lies one the polar of $ P$ w.r.t to $ \omega$ and $ X\in DE$,thus
$ P$ lies on the polar of $ X$ w.r.t $ \omega$.It is well-known fact that $ KL$ is the polar of $ X$ w.r.t $ \omega$.Hence $ K,P,L$ are collinear.
Denote $ F\in DE\cap KL$.As we know $ BM\equal{}MC$,hence $ AF\|BC$(it is well-known fact too :lol: ).We also know that $ KDEL$ is a cyclic quadrilateral with diameter $ KL$ and $ PD \equal{} PE$(as tangent lines to $ \omega$),so $ P$ is the center of circumcirlce of $ KDEL$.Also it is well-known that $ PM\perp AF$.But as we noticed before $ AF\|BC$,hence
$ PM\|BC$,thus $ PB \equal{} PC$.

[b][u]Remark.[/u][/b] Prove easily that $ PB \equal{} \frac a4\cdot\sqrt {4 \plus{} \tan^2A}$, what is symmetrically in $ b$ , $ c$ . Thus,

 $ PB \equal{} PC$ (the positon of the point $ P$ depends only $ a$ and $ A$ ). From it results $ MP \equal{} \frac a4\cdot\tan A$ .

See and my mesage from [url=http://www.mathlinks.ro/viewtopic.php?search_id=527669261&t=16779][color=red][b]here[/b][/color][/url] .

Nice solution Erken  :wink: . See here for other solutions: http://www.mathlinks.ro/viewtopic.php?t=168866 (problem 11)

[quote="plane geometry"]why“BM=MC,hence BC‖AF ” ？
could you tell some details？thanks！[/quote]
It is a property of harmonic fours.Indeed,we know that(for example,from Ceva's and Menelaus theorems) that $ (DE,XF)\equal{}\minus{}1$.
So if $ BC\cap AF\equal{}Y$,then $ (BC,MY)\equal{}\minus{}1$,but $ BM\equal{}MC$,hence $ Y\equal{}\infty$.Thus we are done.

[quote="plane geometry"]thanks ，i dont know much about harmonic fours,but now i am  trying to understand it  ,thanks![/quote]
You can learn more about them from here:
http://www.mathlinks.ro/viewtopic.php?t=168866

[hide="Solution"]
Let the line through $ M$ and perpendicular to $ PM$ meet $ AD$ and $ AE$ at $ B'$ and $ C'$ respectively.  Furthermore, let the feet of the perpendiculars from $ P$ to $ AB$ and $ AC$ be $ X$ and $ Y$ respectively.  Then, $ \angle PMB'\plus{}\angle PYB'\equal{}90\plus{}90\equal{}180$, so $ PYB'M$ is cyclic.  Similarly is $ PMXC'$.  From http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1907178321&t=195897, we have that $ \angle PYM\equal{}\angle PXM$, so $ \angle PB'M\equal{}\angle PYM\equal{}\angle PXM\equal{}\angle PC'M$.  Thus, $ MB'\equal{}MC'$.  Now, if $ B'\ne B$ and $ C'\ne C$, then $ BC$ and $ B'C'$ bisect each other as they share midpoint $ M$, so $ BB'\parallel CC'$.  Yet, they meet at $ A$, so $ B\equal{}B'$ and $ C\equal{}C'$, so $ PM\perp BC$, so $ P$ lies on the perpendicular bisector of $ BC$, so $ PC\equal{}PB$ are desired. [/hide]


(A.Zaslavsky, 8--9) Given a circle and a point $ O$ on it. Another circle with center $ O$ meets the first one at points $ P$ and $ Q$. The point $ C$ lies on the first circle, and the lines $ CP$, $ CQ$ meet the second circle for the second time at points $ A$ and $ B$. Prove that $ AB\equal{}PQ$.

D is an arbitrary point on ⊙O1
∠OCQ=∠OCP  using sin law  we obtain sin∠CBO=sin∠CPO
So ∠CBO=∠CPO or ∠CBO+∠CPO=180
But ∠CDO+∠CPO=180  ∠CBO >∠CDO  =〉∠CBO=∠CPO   =〉CB=CP =〉AB=PQ

Draw $ D$ such that $ OD$ is a diameter of the first circle. Then $ \angle DPO$ is right so $ DP$ is tangent to the second circle. Similarly, $ DQ$ is tangent to the second circle. We have
\begin{align*}
\angle AOP &= \angle DPC \\
&= \angle DQC \\
&= \angle QOB.
\end{align*}
This means $ \angle AOB=\angle POQ$ or $ AB=PQ$ as desired.

[geogebra]31aa5346dea7f3a0b753f9a66c361e20d6e7840b[/geogebra]

From $PCQO$ cyclic we get $\widehat{OPA}=\widehat{BQO}$ $(1)$,  triangles $AOP$ and $BOQ$ are isosceles and, with $AO=BO$ they are congruent, hence $AB=PQ$, done.

Best regards,
sunken rock


integration
trigonometry
limit
calculus
real analysis
real analysis unsolved

for allpositives integers a,b and n ,suppose that:
 [tex ] I_n = \int^a/b_0 \frac 1{n!} x^n (bx-a)^n \sin x dx [tex ]
prove that [tex ] I_n [tex ] is integer and \lim_{n\rightarrow+\infty} I_n =0  [tex ]

[quote="RaduSorici"][quote="tchebytchev"]for allpositives integers $a,b$ and $n$ ,suppose that:
 $I_n = \int_{a}^{b}\frac 1{n!} x^n (bx-a)^n \sin x dx$
prove that $I_n$ is integer and $\lim_{n\rightarrow+\infty}$ $I_n =0$[/quote]

Is this the problem?[/quote]
i don't understand how the limit of an integer non-zero sequence could be $0$.

I think the integral is actually from 0 to $\frac{a}{b}$.

Seems like this is just a distorted standard proof of irrationality of $\pi$. The problem should start with "Assume $\pi=\frac ab$" (otherwise it is just false). I don't know whether it is a good idea to post a problem concerning a well-known proof in general, but it is definitely a bad thing to post it with errors and omissions :(.


Let $ k,n\in N,\geq 0$
Find$ m_{max}$ such that Existing $ B_1,B_2,...B_m \subset A \equal{} {1,2,...n}$and $ /B_i\cap B_j/ \leq k$ for all $ i\geq j$
There $ /A/$ is the number of element of $ A$


calculus
integration
linear algebra
matrix
calculus computations

Hi

"Express the integral:
$ I\equal{}\int_0^{\infty}dx\int_0^{1}dy\int_0^{x}dx$ $ xe^{Ax/y\minus{}Bxy\minus{}Cyz}$
in terms of the new variables $ \alpha \equal{} x/y, \beta\equal{}xy$ and $ \gamma\equal{}yz$. Hence show that:
$ I\equal{}\frac{1}{2A(A\plus{}B)(A\plus{}B\plus{}C)}$"

This is exactly how the question was given, but I'm not sure what it means. It looks like a product of 3 single integrals to me?

Many thanks :)

I think you have a typo. Did it perhaps actually say

$ I \equal{} \int_0^{\infty}dx\int_0^{1}dy\int_0^{x}dz\,xe^{Ax/y \minus{} Bxy \minus{} Cyz}\ ?$

Another notation for the same thing would be

$ I \equal{} \int_0^{\infty}\int_0^{1}\int_0^{x}xe^{Ax/y \minus{} Bxy \minus{} Cyz}\,dz\,dy\,dx$

Does that notation make more sense to you?

Ah yes thank you. I've been reading up on it, but I can't figure out how to change the limits of the integral after a substitution. How do I do this?

Thanks

First, we attend to the change of variables by computing a Jacobian:

$ \frac{\partial(\alpha,\beta,\gamma)}{\partial(x,y,z)}\equal{} \begin{vmatrix}\frac1y&\minus{}\frac x{y^2}&0\\y&x&0\\0&z&y\end{vmatrix}\equal{}2x.$

Hence $ 2x\,dz\,dy\,dx\equal{}d\alpha\,d\beta\,d\gamma.$

Since there is an $ x$ in a nice place, we can make this change without too much trouble.

Changing the limits of integration will be trickier. Since we have $ x,y,$ and $ z$ all positive, we also had $ \alpha,\beta,$ and $ \gamma$ all positive.

Since $ z\le x,$ we have $ yz\le xy,$ hence $ \gamma\le\beta.$ Now we note that $ \frac{\beta}{\alpha}\equal{}y^2.$ Since $ y\le 1,$ we conclude that $ \frac{\beta}{\alpha}\le 1$ or $ \beta\le\alpha.$ That pretty much uses up our restrictions, so we have $ 0\le\gamma\le\beta\le\alpha<\infty.$ One way to write what we want is:

$ I\equal{}\frac12\int_0^{\infty}\int_0^{\alpha}\int_0^{\beta} e^{A\alpha\minus{}B\beta\minus{}C\gamma}\,d\gamma\,d\beta\,d\alpha.$

Or we could try that last integral with the variables in some different order.

Now let $ u\equal{}A\alpha\minus{}B\beta\minus{}C\gamma$, so $ \frac{du}{\minus{}C}\equal{}d\gamma$.  Hence the first integral becomes
$ \int\minus{}\frac{e^{u}}{C}\,du\equal{}\minus{}\frac{e^{A\alpha\minus{}B\beta\minus{}C\gamma}}{C}\bigg|_{0}^{\beta}\equal{}\minus{}\frac{e^{A\alpha\minus{}\beta(B\plus{}C)}\plus{}e^{A\alpha\minus{}B\beta}}{C}$
This is our new integrand.  Let's move onto the second integral:
$ \int_{0}^{\alpha}\minus{}\frac{e^{A\alpha\minus{}\beta(B\plus{}C)}\plus{}e^{A\alpha\minus{}B\beta}}{C}\,d\beta$
Again, let $ u\equal{}A\alpha\minus{}\beta(B\plus{}C)$ so $ \frac{du}{\minus{}B\minus{}C}\equal{}d\beta$ so we get
$ \int\frac{e^{u}}{C(B\plus{}C)}\,du\equal{}\frac{e^{A\alpha\minus{}\beta(B\plus{}C)}}{C(B\plus{}C)}\bigg|_{0}^{\alpha}\equal{}\frac{e^{\alpha(A\minus{}B\minus{}C)}\minus{}e^{A\alpha}}{C(B\plus{}C)}$,
for the first half of the integral, and then substituting $ u\equal{}A\alpha\minus{}B\beta$, $ \frac{du}{\minus{}B}\equal{}d\beta$ for the second half,
$ \int\frac{e^{u}}{CB}\equal{}\frac{e^{A\alpha\minus{}B\beta}}{CB}\bigg|_{0}^{\alpha}\equal{}\frac{e^{\alpha(A\minus{}B)}\minus{}e^{A\alpha}}{CB}$.
Hence the whole thing is
$ \frac{e^{\alpha(A\minus{}B\minus{}C)}\minus{}e^{A\alpha}}{C(B\plus{}C)}\plus{}\frac{e^{\alpha(A\minus{}B)}\minus{}e^{A\alpha}}{CB}$.
This is now the final integrand.  You can try it.

Chances are, I botched some algebra in the preceding since I worked all of this out as I was typing it and didn't check it so before you proceed, make sure I didn't make any mistakes.


∑n=0 to ∞ ((n+1)/10^n) * (x-4)^n  I know this is a Taylor series, but i need help in finding its interval of convergence and if the end points are to be included or not.

Since $ \sum_{n\equal{}0}^{\infty}\frac{n\plus{}1}{10^n}(x\minus{}4)^n$ is a power series, you should use the Ratio Test first. The result will be of the form: convergent when $ |x\minus{}a|<R$, divergent when $ |x\minus{}a|>R$. Here $ R$ is the radius of convergence and $ a$ is the center of the series. At the endpoints $ x\equal{}a\pm R$ the series turns out to be divergent by the Test for Divergence (a.k.a. the $ n$th term test).


function
calculus
derivative
limit
analytic geometry
graphing lines
slope

A function $f$  has derivative  on $[a,b]$ and $f'(a) = f'(b) = 0$.Prove that there exists $c$ in $(a,b)$ , such that :
  $f'(c)= \frac{f(c)-f(a)}{c-a}$

[u]Babis[/u]

draw a picture and try to see geometrically where is $c$ ..

Geometrically yes , there is a tangent through A(a,f(a)) , but which is this point ? I mean a calculus approach .If you have any idea , please write.

 Babis

Could you use the definition of derivation? $f'(x)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So, $f'(c)=\lim_{c\to a}\frac{f(c)-f(a)}{c-a}$. $c$ equals(really close to?) $a$ ?

Another approach:
$y=f(c)$
$f'(c)=\frac{dy}{dx}=\frac{f(c)-f(a)}{c-a}=\frac{\Delta y}{\Delta x}$
So obviously $\Delta x\to 0$

P.S. I dont know how much of this above is correct but I think its solved by using Rolle's theorem.

[quote="fedja"][quote="stergiu"]Geometrically yes , there is a tangent through A(a,f(a)) , but which is this point ? I mean a calculus approach.
[/quote]
Once you see something geometrically, to translate it into "calculus" language should not be a problem. In this particular case, just note that the tangent line you are talking about is the secant line with an extremal slope. And the slope of the secant line is easy to express analytically. You'll need to show that either maximum or the minimum is attained in an inner point of the interval, of course, but it should not be too hard.[/quote]

  I can prove that $f$ has at least max or min in the interior of $(a,b)$ , but I'm afraid that the max or min point does not work , because in this point of $C_{f}$ the tangent has slope $0$.
   We must use the conditions ! I'll try . I have to prove that $f'$ takes all real values . If I'm lucky ...

[quote="fedja"][quote="stergiu"]
  I can prove that $f$ has at least max or min in the interior of $(a,b)$ [/quote]
You need an extremum not of $f$ but of the slope $g(x)=\frac{f(x)-f(a)}{x-a}$! :roll:[/quote]

 Yes ! This is the secret !!! (Theorem of Flett !!! I had forgotten it.)
 Thank you .
[u]    Babis[/u]

Stergiu, have you managed to solve this problem? Every time I try I hit dead ends.

[quote="ciprian"]What is theorem of Flett?[/quote]

[b]
THEOREM [T.M. Flett-(1958)][/b] [color=blue][b] Let $f$be a real-valued function differentiable in  $[a, b]$ with $f^{\prime}(a) = f^{\prime}(b).$
 Then there exists a point  $\vartheta \in (a, b)$ such that [/b][/color]
\[f(\vartheta)-f(a) = (\vartheta-a) f^{\prime}(\vartheta). \]
[u]Some references:[/u]

[1] Aczel, J.: [i]A mean value property of the derivative of quadratic polynomials-without mean values and derivatives.[/i]
 Math. Mag., 58, 42–45 (1985)
[2] Haruki, S.: [i]A property of quadratic polynomials.[/i]  Amer. Math. Monthly, 86, 577–579 (1979)
[3] Davies, R. O., Rousseau, G.: [i]A divided difference characterization of polynomials over a generalized field.[/i]   Aequationes Math., 55, 73–78 (1998)
[4] Schwaiger, J.:[i] On a characterization of polynomials by divided differences.[/i] Aequationes Math., 48, 317–323
(1994)
[5] Riedel, T., Sablik, M.: [i]On a functional equation related to a generalization of Flett’s mean value theorem.[/i]
International Journal of Mathematics and Mathematical Sciences, 23(2), 103–107 (2000)
[6] Riedel, T., Sablik, M.: [i]Characterizing polynomial functions by a mean value property,[/i]
Publicationes Mathematicae Debrecen, Tomus, 52, 597–609 (1998)
[7]T. Riedel and  Sablik M. : [i]A Different Version of Flett’s Mean Value Theorem and an Associated Functional Equation.[/i]
Acta Mathematica Sinica  29,No.6(2000) 1073–1078.
[8] Sahoo, P, K., Riedel, T.:[i] Mean Value Theorems and Functional   Equations,[/i] World Scientific Publishing Co.,NJ, 1998
[b][9] Flett, T. M.: [i]A mean value theorem.[/i] Math. Gazette, 42, 38–39 (1958).[/b]
[10] Sablik, M.:[i] Taylor’s theorem and functional equations.[/i] Aequationes Math., 60, 257–267 (2000)
[11] Pawlikowska, I.:[i] Charakteryzacja odwzorowa´n poprzez twierdzenia o warto´sci ´sredniej,[/i] (in Polish), PhD
Dissertation, Uniwersytet ´Sl¸aski Katowice, 2002
[12] Aczel, J., Dhombres, J.:[i] Functional equations in several variables,[/i] Cambridge Univ. Press, Cambridge,1989


im doing the 10A tomorrow :)
good luck all :D

I'm taking the 10A in approximately 43 minutes.

Good luck!

I just finished the AMC 10A this morning, before school. Yeah...pretty painful to wake up at 5:30 in the morning and get to school at 6:30.

[quote="suma_milli"]The school I'm taking it at doesn't offer the A date. (Kind of weird..)  I would say good luck,  but as it's over I hope you did well! :D  When do you get your results?  I'll be in the math jam tomorrow if I remember though.[/quote]

I am in the same situation >_>. I would really be great if I had two chances to qualify for the AIME.

We were allowed to look at the solutions during math team today.  But we weren't actually given enough time to read them.  But I know my score...

Darn for one problem I spent quite some time and finally solved it.
And then I did

89 - 72 = 15

and that "quite some time" was wasted.  :(

Oh gosh, such strikingly similar stupidity...

Except I did 89-72=7, apparently I'm thinking in mod 10...

(Sorry if this is considered spam, but I just got answers from someone else and felt very stupid right now...)


Find the maximum number of increasing arithmetic progressions that can have a finite sequence of real numbers $a_1<a_2<\cdots<a_n$ of $n\ge 3$ real numbers.

The Problem is:
Find the maximum number of increasing arithmetic progressions "with three terms from" a sequence of real numbers $a_1 < a_2 < \cdots < a_n$?

And this is the problem 2 of 1980 USAMO.


modular arithmetic
number theory unsolved
number theory

[quote="thcong1345"]i don't know how to solve it :

let $ x \equal{} 2k \plus{} 1,x \in \mathbb{Z}$ prove that exist $ y\in Z$ such  that
$ x\equiv {y^{11}} (\ mod {2^{2009}})$
thanks.[/quote]

Let $ y_1 , y_2$ be two distinct odd integers.

Then $ y_1^{11} \equiv y_2^{11} \pmod{2^{2009}}$
$ \Leftrightarrow 2^{2009}|(y_1\minus{}y_2)(y_1^{10}\plus{}y_1^9y_2\plus{}...\plus{}y_1y_2^9\plus{}y_2^{10})$
$ \Leftrightarrow 2^{2009}|y_1\minus{}y_2$ (because $ y_1^{10}\plus{}y_1^9y_2\plus{}...\plus{}y_1y_2^9\plus{}y_2^{10}$ is odd)

So there are $ 2^{2008}$ different values of $ x$ such that $ x \equiv y^{11} \pmod{2^{2009}}$ and $ y \in \{1,3,5,...,2^{2009}\minus{}1 \}$ and $ 0 \le x \le 2^{2009}\minus{}1$ .  It's easy to see that $ x \in \{1,3,5,...,2^{2009}\minus{}1 \}$ and in the set there are only $ 2^{2008}$ elements.

Thus for all $ x \in \{1,3,5,...,2^{2009}\minus{}1\}$ ,
$ \exists y$ such that $ x\equiv {y^{11}} (\ mod {2^{2009}})$ . The result follows.

i think you have a mistake in thinking $ x\in \{{1,3,5,...,2^{2009}-1\}}$
in fact it 's true every $ x \in \mathbb{Z}$
thanks.

[quote="thcong1345"]i think you have a mistake in thinking $ x\in \{{1,3,5,...,2^{2009}\}}$
in fact it 's true every $ x \in \mathbb{Z}$
thanks.[/quote]we only need to consider the case that $ x\in \{{1,3,5,...,2^{2009}\}}$


You are all invited to our next Math Jam during which we will walk through the solutions to this year's AMC B exams.  The discussions will be focused on conveying not simply the solutions, but more importantly on how students can think creatively to solve even problem types they have never seen before.

The format will go as follows: First, Richard and I will go through each of the last 5 problems on both of the B exams.  All of the instructors will then be open to requests regarding any of the problems from either exam.

The Math Jam will take place this Saturday at 7 pm EST/4 pm PST.  All information about math jams can be found here:

http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php

This is going to get de-stickified.  Right now.

Yes, sorry, go ahead.  If I ever post announcements and forget to pull them down after their effective date, feel free to knock them down the priority chain.


function
floor function
number theory proposed
number theory

For an arbitrary real number $ x$, $ \lfloor x\rfloor$ denotes the greatest integer not exceeding $ x$. Prove that there is exactly one integer $ m$ which satisfy $ \displaystyle m\minus{}\left\lfloor \frac{m}{2005}\right\rfloor\equal{}2005$.

m-[m/2005]=2005
2006 is the only possible solution.........

for m>2006
m-[m/2005]>2005          (you can observe the difference is constantly increasing  )

and for m<2006             (you can also that the difference is constantly increasing)
m-[m/2005]<2005


inequalities quadratics calculus integration algebra polynomial

[hide="1"]We know that $(2x-1)+(2x-1)>5x-7$ and $5x-7\ge1.$

Simplifying the first inequality gives $x<5.$ Simplifying the second gives $x\ge\frac{8}{5}.$ We are trying to find integer solutions, so the second inequality is $x>1.$

$\therefore x=\{2,\,3,\,4\}$[/hide]

[quote="Ceva Theorem"]The following 2 problems are from the UNC Mathematics contest in Colorado (First Round).

1. The length of each leg of an isoceles triange is $2x-1$. If the base is $5x-7$ determine all possible integer values of $x$.

2. Let $f(x)\;=\;5x^{4}-6x^{3}-3x^{2}+8x+2$. Determine coefficients$\;a,\;b,\;c,\;d,\;and\;e\;$ so that $f(x)\;=\;a+b(x-2)+c(x-2)^{2}+d(x-2)^{3}+e(x-2)^{4}$

[hide="hint"]
Hint for 1: Pythagreon Triples
Hint for 2: expand $f(x)$ and combine like powers[/hide][/quote]

[hide]
I hope my idea for number one is correct.

The sum of the two legs must be greater than the base.

$4x-2 > 5x-7$
$5 > x$
$x < 5$

However, the sum of one leg and the base must be greater than the other leg, as well.

$2x-1+5x-7 > 2x-1$
$5x-7 > 0$
$x > \frac{7}{5}$

Therefore,

$\frac{7}{5}< x < 5$

So, for integral values of x, x can be 2, 3, or 4.[/hide]

[quote="Ceva Theorem"]2. Let $f(x)\;=\;5x^{4}-6x^{3}-3x^{2}+8x+2$. Determine coefficients$\;a,\;b,\;c,\;d,\;and\;e\;$ so that $f(x)\;=\;a+b(x-2)+c(x-2)^{2}+d(x-2)^{3}+e(x-2)^{4}$[/quote]

[hide="The worst way to do it"] Expand that huge expression to solve for $a, b, c, d, e$ as a system of five linear equations in five variables. [/hide] [hide="The better way to do it"] Let $t = x-2$.  Then $f(x) = f(t+2)$.  Expand that to get $a, b, c, d, e$ more directly. [/hide] [hide="The best way to do it"] Repeated synthetic division. ;) [/hide]

i guess synthetic division is basically the same idea as considering numbers in base x, x-2?
also what can you say about my solution, t0r?

[quote="me@home"]i guess synthetic division is basically the same idea as considering numbers in base x, x-2?[/quote]

Yes.  It's equivalent to the corresponding method of transforming from, say, base $10$ to base $8$.  (A surprising number of operations on integers make sense when applied to polynomials!  Consider, for example, the Euclidean Algorithm.)

[quote="me@home"]also what can you say about my solution, t0r?[/quote]

It looks wrong to me.  For one thing, $d$ can't possibly equal $-6$.  It looks to be $34$.

[b]Remark:[/b]  Can you show that the method I gave as "best" requires the least number of operations?  (As in, addition and multiplication)


[quote="Interval"]On the first six tests in her mathematics course, Jerelyn’s scores were
92, 78, 86, 92, 95, and 91. Determine the median and the mode of
her scores. If Jerelyn took a seventh test and raised the mean of her
scores exactly 1 point, what was her score on the seventh test?[/quote]

Her current average is 89, and the mode is 92. Her score on the seventh test must be 96% in order to raise her average score to a 90.

i_like_pie:

Why did you equate the same numerator using different denominators?

Sorry...I did NOT see the x term on the left side numerator.


percent
ratio
geometric series
real analysis
real analysis unsolved

Okay, so here's the problem, and I'm not exactly sure how to go about solving it so any help would be appreciated:

Economic planners wish to estimate the impact of bringing a major league franchise to a certain city.  They estimate that crowds coming to games would spend 50 million dollars over the course of a season.  Some of that 50 million dollars would go into savings, some of it would be spent somewhere else, but they believe that about 60% of the money spent will be spent again in the city.  That's another 30 million dollars for a total of 80 million.  But that's not all.  Sixty percent of the extra 30 million will be spent in the city, that's 18 million, and then 60% of 18 million, and then....

How much should they estimate will be spent altogether?

The geometric series with the initial term $a=50\cdot 10^{6}$ and common ratio $r=0.6$ has the sum $\frac{a}{1-r}=\dots$

I hope that economic planners don't actually get their estimates in that way. :ninja:


When you test the hypothesis, in the final step it doesn't matter if you have equality or not. The normal distribution is continuous, so P(X<=a)=P(X<a).
I read the reason is that "the parameter( for example, the population mean ) has already some specific value". But this doesn't convince me.
It makes sense to me to always put the claim( either it has equality or not) on the null hypothesis and not on alternative hypothesis.

Since it doesn't matter, you shouldn't care about it.


[hide="Further Explanation"]
The problem was just

$ \frac{a^2\minus{}b^2}{a\plus{}b}$ so $ a\minus{}b$ is the answer as 777-66 = 711. [/hide]


Can you find the explanation for the following?

Snoop and Watson had just enjoyed an excellent meal (chips in a basket) in their favourite restaurant (the Goat and Compasses). Having partaken liberally, not to say magnificently, of beers, wines and spirits with their gourmet fare (or fayre, as the establishment had it), they decided to round off the occasion with a cup of black coffee each.
"I say, Snoop," said Watson when Snoop, after stirring it, began to sip his coffee with the air of one unused to enjoying such sobering refreshments, "isn't that a pair of bluebottles in your cup?"
Snoop spluttered, discharging a fine mist of coffee into various parts of the neighbourhood. "Not a pair, Watson," he then pronounced, "but a single - not to say singularly large specimen of bluebottlia restauratis. [i]Waiter![/i]"
Snoop explained the situation to the waiter and (having got through the obligatory rigmarole about everyone wanting one) his cup was removed and a new one brought, minus fly.
It was definitely a different cup, and certainly free of fauna, but on taking a sip, Snoop spluttered again. "Pah!" he said, "this is the same coffee I had before - only the cup has been changed!"
"How do you know?" asked Watson, "is it cold?"
"It doesn't seem any colder than it was before," replied Snoop, "but it's definitely the same coffee."
"How can you be so sure?" Watson goggled, "after all, cheap coffee's cheap coffee."
"Ah, but there is one thing you've forgotten..." began Snoop.

What had Watson forgotten, and how could Snoop be so sure it was the same coffee in a different cup?

Yeah, you're right. I didn't know it was that easy... though the original problem didn't have that part about him stirring it before he drinks it the first time.


Can someone suggest a good book, Ebook, or a site, for learning number theory, i am completly new to it, i have no idea what it is, what mod is +... i just think it is to do with numbers, bcoz of its name.

Well an obvious answer is to get Intro to NT, from AoPS. it sets a nice foundation for your NT skills.

Another is to just catch on. when i was first learning NT, that's what i did...