calculus
integration
logarithms
calculus computations

Without explicitly solving either of the 2 integrals below, show:

$\int_{0}^{1}\;\frac{\ln{x}}{1+(x+1)^{2}}\;dx\; =\;\alpha\;\int_{0}^{1}\frac{\ln{x}}{1+x^{2}}\;dx\qquad\;{(}\;where\quad\alpha\;\in\;\mathbb{R}\;{)}\;$

What is special about the constant $\alpha$?

Are you sure the integral on the left is correct as written?

Yes sir.  :|

In fact, it can [b]also [/b]be shown, again without explicitly solving either of the 2 integrals below, that

$\;\int_{0}^{1}\;\frac{\ln{x}}{x^{2}+(x+1)^{2}}\;dx\;=\;\beta\;\int_{0}^{1}\;\frac{\ln{x}}{1+x^{2}}\;dx\; \quad \; ( \;where\quad\beta\;\in\;\mathbb{R}\; )$

A rather beautiful relationship exists between $\alpha$ and $\beta$.

[ [u]PS:[/u] S. Ramanujan was the first to discover this unexpected relationship. ]


Prove: For every positive number d, there is a polynomial P(x) with integer coefficients such that for all x in [0.01, 0.99] it is true that |P(x) - 0.5| < d.


Note that

$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=1$ iff $(x+y)(x+z)(y+z)=0$.

wlog assume $x+y=0$, then $xy+yz+zx=z(x+y)+xy=xy=-x^{2}<0$, since equality can't hold because otherwise $\frac{1}{x}$ would undefine.


Is there a continuous and injection function from $ R^n$ onto $ R^2$?($ n\neq 2$)

There's certainly no homeomorphism, although that's not quite as strong as what you're asking.

I'm thinking of something similar though.
Take some compact subset of $ \mathbb{R}^n$, look at $ f$ restricted to it. Then we have a homeomorphism, and we should be able to get to a contradiction by removing say a circle from the image in $ \mathbb{R}^2$, which would make the image not connected anymore.

@Tyl:
I don't understand what's your point. Removing something from the image can make it disconnected, but what does this show?

Here is a complicated argument. Perhaps it can be done much easier...

Look at the images of $ \overline{B_n(0)}$. They are compact and cover the whole space. By baire's theorem for any sequence of closed subspaces $ F_n$ with $ \mathbb{R}^2\equal{}\bigcup_{n} F_n$ there is one $ n$ s.t. $ F_n$ has non-empty interior.
Because $ F_n\equal{}f(\overline{B_n(0)})$ and $ f$ is continuous + injective, $ f$ is a homeomorphism $ \overline{B_n(0)}\to F_n$. Because $ F_n$ contains a non empty open set (open in $ \mathbb{R}^2$ !) it restricts to an homeomorphism of non empty open sets of $ \mathbb{R}^n$ and $ \mathbb{R}^2$.
One can show that non empty open sets of $ \mathbb{R}^k$ and $ \mathbb{R}^m$ are homeomorphic iff $ k\equal{}m$ (looking at the (co)homology groups for example). So we get $ n\equal{}2$.

I was thinking of a fancier version of the 'remove-a-point'-trick. You also remove the pre-image of the circle from the other space, but that one is still connected then.

Similarly one can show $ S^1$ and $ [a,b]$ are not homeomorphic. If you would have a homeomorphism, remove the point $ c \equal{} \frac{a\plus{}b}2$ from $ [a,b]$, and remove the point $ p \in S^1$ such that $ f(p) \equal{} c$ from $ S^1$. Then $ f$ restricted to this set is still a homeomorphism. But $ [a,b] \setminus \{c\}$ is not connected, while $ S^1 \setminus \{p\}$ still is.

You could almost certainly remove a point from the image in $ \mathbb{R}^2$ and then note that it has nontrivial
fundamental group, where a closed ball in $ \mathbb{R}^n$ ($ n > 2$) remains simply connected when you remove
a point.

nukular, it seems you're assuming that the map is an homeomorphism with its image, in your last post: this needs not to be true, even for more regular maps (of course, with inequality on dimensions reversed).

i was thinking of another approach: you can approximate any continuous map between differentiable manifolds with a differentiable map, which can't (at least, shouldn't) be the approximation of any injective map. maybe in all of this you need to further perturb your map and use sard's theorem, but mine wants to be just a sketch :)

I was working off Tyl's comment that you can restrict $ f$ to a homemorphism out from a closed ball in $ \mathbb{R}^n$ and its image.

[quote="Gockel"]

Because $ F_n \equal{} f(\overline{B_n(0)})$ and $ f$ is continuous + injective, $ f$ is a homeomorphism $ \overline{B_n(0)}\to F_n$. Because $ F_n$ contains a non empty open set (open in $ \mathbb{R}^2$ !) it restricts to an homeomorphism of non empty open sets of $ \mathbb{R}^n$ and $ \mathbb{R}^2$.[/quote]

how we can construct a homeomorphism of non empty open sets of $ \mathbb{R}^n$ to non empty open sets in $ \mathbb{R}^2$.
If above statment is true ,by a similar way we can show that there is no continuous ,injective function from $ R^n$ onto $ R^m$ .
I proved it( there is no continuous ,injective function from $ R^n$ onto $ R^2$ with $ n > 2$ ) by elementary methods but thats too long and complicated.

Thanks for your attention. :)


vector
topology
real analysis
real analysis unsolved

Let $X$ be the space of real sequences such that ${\forall_{(x_{n}) \subset X}}\exists_{n_{0}}\forall_{n > n_{0}}\ x_{n}= 0$. Let $\| \cdot \|$ be any norm on $X$.

Proof that $(X,\| \cdot \|)$ is not complete.

I would say that this is false.
Maybe you meant to say "Let $X$ be THE space ...".
Then I think the statement follows from the category theorem.

If you insist on "Let $X$ be A space ..." then consider the space
$\{(x_{n}): \forall n\ge n_{0}: x_{n}=0\}$ for fixed $n_{0}$.
It satifies the hypothesis and is a finite dimensional real vector space,
hence complete.

Ok, so now consider $X_{k}=\{(x_{n}): \forall n\ge k: x_{n}=0\}$ for $k=1,2,\ldots$.
Every $X_{k}$ is finite dimensional, hence closed in $X$.
For fixed $k$, every neighbourhood of $0$ and hence every open set in $X$ contains elements which are not in $X_{k}$.
(Take some $x\in X\setminus X_{k}$. Then $t\cdot x\to0$ in $X$ for $t\to 0$ in $\mathbb{R}$, but $t\cdot x\notin X_{k}$.)
Therefore $X_{k}$ has empty interior. If $X$ were complete, the category theorem would state that $\cup X_{k}$ also has empty interior, contradicting the fact  $X=\cup X_{k}$.

I know. When you said "category theorem" in your first post I found exactly the same proof :). Thanks one more time.


I can't find this anywhere (I tried to look hard)!

What is the symbol for arc?

For instance, how do you write arc $AB$ in Latex?

There's probably a better way, but \frown\above AB seems to work:

$\frown\above AB$

This question is also discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18587]here[/url] in the bottom half of the thread; there's not really a nice way to do it.

Yeah, it's really strange that nobody that designs $\text{\TeX}$ packages didn't thought about that yet :)

[quote="solafidefarms"]How about \widehat{}? $\widehat{AB}$[/quote]

In the thread that jmerry linked to, that was discussed as a possible solution.  As a summary of that thread, here are the possible solutions offered so far:

1.  \stackrel{\frown}{}

2.  \widehat{}

3.  When the tipa package is used, \textroundcap is the command.

4.  With the phonetic package, you may use \ohill

5. I'm going to add another option to that list (one which I just found and like better than the others).

First of all, you need to install the yhmath package and put \usepackage{yhmath} in the preamble of your document.  Then use \wideparen{ABC}.


function
percent
calculus
derivative
analytic geometry
graphing lines
slope

[quote="linj"]A computer software company claims that the following function models the "learning curve" for their software.
P(t) = 100 / ( 1 + 48.2e^(-0.52t) )
where t is measured in months and P(t) is the average percent of the software program's capabilities mastered after t months.
1. Roughly, when can a person expect to "learn the most in the least amount of time"?
2. If the company's claim is true, how many months will it take to have completely mastered's program?[/quote]
Is at 50% a person expect to "learn the most in the least amount of time"?

[quote="LightEntropy"]is there a way to do this without graphing it on a calculator and finding the maximum? Is there a way to do this without the calculator all together?[/quote]
What is mean "maximum"? If it mean maximum value, then it obvious 1000 as time goes infinite.

[quote="pi^2/6"]For part 1 you would just maximize the derivative.

For part 2 if you mean when does P(t) = 100, the answer is never.[/quote]
What is derivative?

[quote="pi^2/6"]the derivative is the slope (rate of change) of a function. Here it would be the percentage of the program learned per month for a given time t.

so finding the time when we "learn the most in the least amount of time" is finding the largest value for "percentage of program learned per month."

is this problem from a calculus book? (I don't know how to do it without calculus.)[/quote]
This question from the algebra book.

[quote="linj"][quote="pi^2/6"]oh. forget about derivatives then.
I guess you can look at the graph and see where the slope looks steepest.[/quote]
I have saw the graph. Seems when in the 50% the graph is steepest, but could you use derivative to confirm or disprove it? Please![/quote]
Or Is there any way to do it without graph or derivative?

yes the maximum slope is at P(t)=50.
The function P(t) is the solution to a logistic equation with carrying capacity of 100. The derivative is maximized when P(t) is half the carrying capacity.

so you don't need a calculator or graphs if you know about logistic equations (but that still involves calulus).


factorial
ratio
limit
calculus
calculus computations

What is the interval of convergence of the series ${ \sum_{n = 1}^{\infty}\{\prod_{i = 1}^n(9i^2 - 3i - 1)(3i - 2)}\frac{x^{3n}}{(3n)!}\} + \sum_{n = 1}^{\infty}\{{\prod_{i = 1}^n(9i^2 + 3i + 1)(3i - 1)}\frac{3^{3n+1}}{(3n+1)!}\}$?

Can someone please help. Thank you very much. :P

[edited]

[quote="JBL"]If you want an interval of convergence, there probably should be some free variable available to you.  Are those two sums just supposed to have an $ x^n$ before the product?  If so, the growth of the coefficients is faster than geometric, so the interval of convergence is just 0.  If something else, you need to tell us what you meant.[/quote]

Thank you for replying JBL, :lol: I have indeed mis-typed the free variable, it should be:

${ \sum_{n = 1}^{\infty}\{\prod_{i = 1}^n(9i^2 - 3i - 1)(3i - 2)}\frac{x^{3n}}{(3n)!}\} + \sum_{n = 1}^{\infty}\{{\prod_{i = 1}^n(9i^2 + 3i + 1)(3i - 1)}\frac{3^{3n+1}}{(3n+1)!}\}$

Tactical observation: given a power series whose coefficients include either factorials or factorial-like parts, I want to use the ratio test.

Only the first series is a power series. (Unless you haven't finished fixing it.) Apply the ratio test to that:

$ \lim_{n\to\infty}\frac {\prod_{i \equal{} 1}^{n \plus{} 1}(9i^2 \minus{} 3i \minus{} 1)(3i \minus{} 2)|x|^{3n \plus{} 3}} {(3n \plus{} 3)!}\cdot\frac {(3n)!}{\prod_{i \equal{} 1}^n(9i^2 \minus{} 3i \minus{} 1)(3i \minus{} 2)|x|^{3n}}$

$ \equal{} \lim_{n\to\infty}\frac {(9(n \plus{} 1)^2 \minus{} 3(n \plus{} 1) \minus{} 1)(3n \plus{} 1)|x|^3} {(3n \plus{} 1)(3n \plus{} 2)(3n \plus{} 3)} \equal{} |x|^3.$

With the test being whether this is less than $ 1,$ we get that the radius of convergence of this power series is $ 1.$

The second series, the numerical series, is now problematical. (Unless you really meant for the powers in the numerator to be $ x^{3n \plus{} 1}$ rather than $ 3^{3n \plus{} 1}.$) If that's an $ x,$ a very similar calculation will also show a radius of convergence of $ 1.$ If that's a $ 3,$ then the same calculation shows that the series diverges badly. At face value the question makes no sense because that second series always diverges. If that's really $ x^{3n \plus{} 1}$ in the second series, then we have a radius of convergence of $ 1.$

[quote="Kent Merryfield"]Only the first series is a power series. (Unless you haven't finished fixing it.) [/quote]

Yes, you are right Kent Merryfield, it should be $ x^{3n\plus{}1}$ instead of $ 3^{3n\plus{}1}$ in the second series.
Thank you very much. :P


[quote="math92"][hide](0+2+4+6)/2=$\boxed{3}$[/hide][/quote]

$0$ is not positive.  ;)

[hide]$\frac{2+4+6+8}{4}=5 \implies \boxed{C}$[/hide]

[hide]
You want them to we as small as possible.
$2+4+6+8=20$
Thus the average is 5[/hide]

[hide]The four smallest positive even integers are 2,4,6,8 and their average is 5, so answer C.[/hide]

[hide]The smallest 4 positive even integers are 2,4,6, and 8

So the average is $(2+4+6+8)/4$

$5$[/hide]

[hide]Smallest possible even intergers are $2, 4, 6$ and $8$. Thus $(2+4+6+8)/4 =5$.[/hide]

[quote="236factorial"]What is the smallest possible average of four distinct positive even integers?

a) 3
b) 4
c) 5
d) 6
e) 7

Source: AMC 8 2002 #3[/quote]

not that hard...

[hide]

2+4+6+8=20/4=5

c [/hide]

2+4+6+8 are the integers.

when added it equals 20
 and 20/4=[b]5[/b]

The smallest possible numbers with the smallest differences ar 8,4,6, and 2. The average of those digits would equal a total 5. The answer is C.


function
number theory unsolved
number theory

Find all bijective functions $f:N^+->N^+$ satisfying: $f(m)$ divides $f(n) \Leftrightarrow m$ divides $n$
(Where $N^+$ be set of positive integer numbers)

This is a well-known problem from Turkey TST. But the solution is too long for me to write it here.

It's not hard to see that f(1) = 1 and f takes primes to primes.  From here, I get the following: Let p_1, p_2, ... be the primes, and let \pi be an arbitrary permutation of the positive integers.  Then the function f defined by
f(p_1^{e_1} p_2^{e_2} ...) = p_1^{e_{\pi(1)}} p_2^{e_{\pi(2)}} ...
is a solution, and I would guess that all solutions are of this form.

Note: The TeX option is currently not working.

TeX works now:
[quote="jhaussmann5"]It's not hard to see that $f(1) = 1$ and $f$ takes primes to primes.  From here, I get the following: Let $p_1, p_2, ...$ be the primes, and let $\pi$ be an arbitrary permutation of the positive integers.  Then the function $f$ defined by
$f(p_1^{e_1} p_2^{e_2} ...) = p_1^{e_{\pi(1)}} p_2^{e_{\pi(2)}} ...$
is a solution, and I would guess that all solutions are of this form.[/quote]

hxtung, do you know an arrow can simply be done using \righarrow? so $f:\mathbb{N}\rightarrow\mathbb{N}$
Also, why $\mathbb{N}^+$? naturals are always positive... ;)


Find all positive integers such that:
\[ \frac{n}
{{\varphi (n)}} \in N\]

It is [url=http://www.research.att.com/~njas/sequences/A007694]A007694[/url] at OEIS. (Seems like multiples of 2s and 3s, but I don't have a proof right now /now there's one proof under  :) /)

Suppose $ n\equal{}\prod_{i}p_{i}^{\alpha_i}$, where $ p_i$ are the prime divisors of $ n$, and the $ \alpha_i$ are the maximum power of each $ p_i$. Then it is a standard result that $ \phi(n) \equal{} \prod_i(p_i^{\alpha_i}\minus{}p_i^{\alpha_i\minus{}1})$, then

\[ \frac{n}{\phi(n)}\equal{}\frac{\prod_i p_i}{\prod_i p_i\minus{}1} \]

If n has more than one odd prime divisor than in the denominator is divisible by 4 but the numerator can be divisible by 2 at most. Therefore n has at most one odd prime divisor,. If n has one odd prime divisor then the numerator must be divisible by 2 and
\[ \frac{n}{\phi(n)} \equal{} \frac{2p}{p\minus{}1}\]
$ (p\minus{}1,p)\equal{} \text{if}$ $ p> 3$ so $ p\equal{}3$.

Therefore $ \frac{n}{\phi(n)}$ $ \in$ $ \mathrm{Z}$ iff the prime divisors of $ n$ are 2,3.


Given $ a, b, c > 0$ satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that: $ \frac{ab}{c} \plus{} \frac{bc}{a} \plus{} \frac{ca}{b} \plus{} \frac{9abc}{4} \geq\ \frac{21}{4}$

My dear friend, I hope you'll like it  :)

[quote="nguoivn"]Given $ a, b, c > 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {9abc}{4} \geq\ \frac {21}{4}$
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ t\geq1$ and
$ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {9abc}{4} \geq\ \frac {21}{4}\Leftrightarrow w^6 \minus{} 5u^3w^3 \plus{} 4u^2v^4\geq0\Leftrightarrow$
$ \Leftrightarrow w^3\leq\frac {5u^3 \minus{} \sqrt {25u^6 \minus{} 16u^2v^4}}{2}.$
$ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Id est, we need to prove that
$ 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq\frac {5u^3 \minus{} \sqrt {25u^6 \minus{} 16u^2v^4}}{2}.$
But $ 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq\frac {5u^3 \minus{} \sqrt {25u^6 \minus{} 16u^2v^4}}{2}\Leftrightarrow$
$ \Leftrightarrow9t \minus{} 6 \minus{} 4\sqrt {\frac {(t \minus{} 1)^3}{t}}\geq\sqrt {25t^2 \minus{} 16}\Leftrightarrow$
$ \Leftrightarrow14t^2 \minus{} 27t \plus{} 13 \plus{} \frac {4(t \minus{} 1)^3}{t}\geq6(3t \minus{} 2)\sqrt {\frac {(t \minus{} 1)^3}{t}}\Leftrightarrow$
$ \Leftrightarrow14t \minus{} 13 \plus{} \frac {4(t \minus{} 1)^2}{t}\geq6(3t \minus{} 2)\sqrt {\frac {t \minus{} 1}{t}}\Leftrightarrow$
$ \Leftrightarrow18t^2 \minus{} 21t \plus{} 4\geq6(3t \minus{} 2)\sqrt {t(t \minus{} 1)}\Leftrightarrow(3t \minus{} 4)^2\geq0.$ :)

[quote="nguoivn"]Nice work, arqady. I wait for another proof of phamhuuduc.
Equality occurs when $ a \equal{} b \equal{} c$ or $ a \equal{} 4b \equal{} 4c$ and the cyclics.
PS: Am-Gm solved it easily  :)[/quote]

Dear nguoivn, thank you very much for your gift of this nice problem.

Below is my first attempt, which is indirect but fairly short: Rewrite the inequality as
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\plus{}\frac{9}{4}\ge\frac{21}{4abc}$
Put $ x\equal{}a^{\minus{}1}$, $ y\equal{}b^{\minus{}1}$ and $ z\equal{}c^{\minus{}1}$. Then $ xy\plus{}yz\plus{}zx\equal{}3xyz$ and the inequality becomes
$ 4(x^2\plus{}y^2\plus{}z^2)\plus{}9 \ge 7(xy\plus{}yz\plus{}zx)$
or after homogenizing
$ 4(x\plus{}y\plus{}z)^2\plus{}\frac{81(xyz)^2}{(xy\plus{}yz\plus{}zx)^2} \ge 15(xy\plus{}yz\plus{}zx)$
Without loss of generality, assume $ x\plus{}y\plus{}z\equal{}1$. Put $ xy\plus{}yz\plus{}zx\equal{}(1\minus{}q^2)/3$. Then as in
  [url]http://reflections.awesomemath.org/2007_2/3variable.pdf[/url], we have
$ xyz\ge \frac{(1\plus{}q)^2(1\minus{}2q)}{27}$
It remains to show that
$ 4\left(\frac{(1\minus{}q^2)}{3}\right)^2\plus{}81\left(\frac{(1\plus{}q)^2(1\minus{}2q)}{27}\right)^2\ge 15\left(\frac{1\minus{}q^2}{3}\right)^3$
or
$ 4(1\minus{}q^2)^2\plus{}(1\plus{}q)^4(1\minus{}2q)^2\minus{}5(1\minus{}q^2)^3\ge 0$
But this is reduced to
$ q^2(1\plus{}q)^2(1\minus{}3q)^2 \ge 0$
which is obvious.

And here is my proof, my friends. I hope you'll like it  :)


¡¡¡S37ñä NVW>Vd ´as!mhue ´+nq ´a&h+ o+ pjey oo+ s! asejy& s!y+ +ey+ ×u!y+ I

wuq^cxz.l×fybJpse&o!nh+jamÞ

From the title, you can tell it means Decipher This Message, so i=d, a =e, etc.

but the first character isn't an i, it's an upside-down exclamation point.

[quote="1=2"][hide="giveaway"]turn upside down[/hide][/quote]
YES
[quote="33286"]·ɹəʇʇəq ṣı ɓụıʇ̣ıɹʍ uʍop əp̣ısdn ʎɯ
Turning the message upside down, we get someting like this:

I +ʎ¡n× +ʎə+ +ʎ¡s [size=150]₈[/size]ʎſ̣əsɐ ¡s +oo ʎəſ̣d +o +ɥ[size=150]₈[/size]ɐˏ ənɥɯ¡sɐˏ PΛ<MΛN ɐ̤ṵ[i]L[/i][size=150]ɛ[/size]S!!![/quote]
don't be so literal!!!

Example: 3 upside down isn't [size=150]ɛ[/size], is E.

and, it makes it harder, no? :wink:

and, another clue is the thing behind the message.


modular arithmetic
number theory proposed
number theory

Prove that there exist infinitely many positive integers $n$such that $n~|~2^{n}+1$. Find all such prime numbers!

[quote="Oskar"]Prove that there exist infinitely many positive integers $n$such that $n~|~2^{n}+1$.[/quote]
I'm quite sure it's posted before, but anyway, take $n = 3^{k}, \ \ k \in \mathbb{N}.$

[quote="Oskar"]Find all such prime numbers![/quote]
Ok, so, let's call them $p$.
We want $p|2^{p}+1$, but since $p$ is prime, we know that (FLT) : $2^{p}\equiv 2 \pmod p$. Hence $p|(2^{p}+1)-(2^{p}-2) = 3$, so the only such prime can be $p = 3$, and it clearly works, so we're done.

if we make your  problem more dificult:
prove that existe infinitly $n$:
1: $n$ has exactly 2007 deferent prime divisor
2:$n|2^{n}+1$.

lets try to prove this result:
theorem:let $a$ and $b$ tow natural numbers such $a+b$ is not power of $2$ then: exists infinitly many $n$ such:
1:$n$ has exactly $s$ deferent prime divisor
2:$n|a^{n}+b^{n}$.
$preuve$:
we use this lemma: let $p$ a
prime divisor of $a+b$ then:
$p^{k+1}|a^{p^{k}}+b^{p^{k}}$.
you can use
hensel lemma tow prove this.
now lets conseder $x_{k}=a^{p^{k}}+b^{p^{k}}$ and $y_{k}=\frac{x_{k+1}}{x_{k}}$.

we complet the preuve: the first result wich we can find easly  $\frac{y_{k}}{p}$ is a positive integer. and $(\frac{x_{k}}{p},\frac{y_{k}}{p})=1$. and by lemma we have also :$(\frac{y_{k}}{p},p^{k})=1$.
then we can chosse an  odd prime divisor    $q_{k}$ of $\frac{y_{k}}{p}$.
we now have a sequense of odd prime divisor satisfying:
1:$(x_{k},q_{k})=1$
2:$(p,q_{k})=1$
3:$q_{k}|x_{k+1}$.
4: the eequense consists of distintc prime.

now we set $n_{0}=p^{s}q_{1}.....q_{s-1}$ and $n_{k+1}=pn_{k}$.
we can easly prove that $n_{k}$ is streacly increasing. and $n_{k}|a^{n_{k}}+b^{n_{k}}$. and we are done.
now your problem is easy with this theorem.
NB: we can use this theorem to solve iran2005.

                                                                      reference: excalibur v  9 :wink:


Prove that \[ x^3+y^3+z^3+3xyz\geq\frac{2}{3}(x^2+y^2+z^2)\sqrt{3(x^2+y^2+z^2)}  \] for positive real numbers $x,y,z.$

EDIT: This isn't true

[quote="Davron"]What do you mean :-)?[/quote]

About what exactly  :?:

[quote="shyong"]Counter example :

pick $a=b=1$ , $c\rightarrow 0^+$ , LHS$\rightarrow 2$ while RHS$\rightarrow \frac{4\sqrt{6}}{3}$

and $2<\frac{4\sqrt{6}}{3}$[/quote]

I know, I also found a mistake.  :wallbash_red:


Let $ a,b,c,d \in [1; 2]$. Prove that: $ (a\plus{}b\plus{}c\plus{}d)\displaystyle \left ( \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}  \right )$ $ \le 18$.

[quote="thanhnam2902"]Let $ a,b,c,d \in [1; 2]$. Prove that: $ (a \plus{} b \plus{} c \plus{} d)\displaystyle \left ( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d} \right )$ $ \le 18$.[/quote]
Let $ f(a,b,c,d)\equal{}(a \plus{} b \plus{} c \plus{} d)\left ( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d} \right ).$ Then
$ \frac{\partial^2f}{\partial a^2}>0,$ $ \frac{\partial^2f}{\partial b^2}>0,$ $ \frac{\partial^2f}{\partial c^2}>0,$ and $ \frac{\partial^2f}{\partial d^2}>0.$ Id est,
$ \max_{\{a,b,c,d\}\subset[1,2]}f\equal{}\max\{f(1,1,1,1), f(1,1,1,2), f(1,1,2,2), f(1,2,2,2), f(2,2,2,2)\}.$

[quote="thanhnam2902"]Let $ a,b,c,d \in [1; 2]$. Prove that: $ (a \plus{} b \plus{} c \plus{} d)\displaystyle \left ( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d} \right )$ $ \le 18$.[/quote]
Follows from Shweizer inequality:
If $ m \equal{} min\{a_i\}$ and $ M \equal{} max\{a_i\}$,then
$ (\sum_{i \equal{} 1}^{n}\frac {1}{a_i})(\sum_{i \equal{} 1}^{n}a_i)\leq \frac {(M \plus{} m)^2}{4Mm}n^2$


Given $ a, b, c$ are three sides of triangle. Prove that: $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq\ 3 (\frac{a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^{\frac{2}{3}}$

I know it's trues with $ a, b, c > 0$ and it's a hard problem but with $ a, b, c$ are three sides of triangle, we have a nice solution   :)

[quote="nguoivn"]Given $ a, b, c$ are three sides of triangle. Prove that: $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq\ 3 (\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^{\frac {2}{3}}$

I know it's trues with $ a, b, c > 0$ and it's a hard problem but with $ a, b, c$ are three sides of triangle, we have a nice solution   :)[/quote]
It is already posted here by me before. ;)


Well, i've read the chapter and tried a few other online resources only to get more confused what should probably be a simple matter.

Let T: R^2 -> R^3 be the linear operator given by:
T(x,y) = (2x-y, -8x+4y)

and we're asked to determine whether some vectors are in the range, R(T).

Well, my question is, how would i determine if a vector (u1,u2) is part of the range? One given vector to try is (5,0) and, well i'm stuck  :oops: Any help is appreciated, as always.

notice that $-8x+4y=-4(2x-y)$, so $(u,v)$ is in the range if and only if $v=-4u$, since of course any real number can be written in the form $2x-y$.

That helped a great deal. I had a friend show me something similar using bases vectors that helped me breeze through similar problems. Thanks again.


geometry
geometric transformation
reflection
rotation
analytic geometry

Let $ P \equal{} (2,3)$.First, translate $ P$ one unit to the right and call the image $ R$.Next, reflect $ R$ over the y-axis and call the image $ S$. Finally rotate $ S$ 90 degrees clockwise about the origin and call the image $ T$.What are the coordinates of $ T$ ?

Well a movement to the right means that $ R\equal{}(3,3)$.  Reflecting about the $ y$-axis negates the $ x$-coordinate, so $ S\equal{}(\minus{}3,3)$.  A rotation of 90 degrees negates the $ y$-coordinate and switches them, so $ T\equal{}(\minus{}3,\minus{}3)$.

If you want to get better at those problems then play FTW.
Or just ask people like me to make up problems for you.

[quote="JRav"]Well a movement to the right means that $ R \equal{} (3,3)$.  Reflecting about the $ y$-axis negates the $ x$-coordinate, so $ S \equal{} ( \minus{} 3,3)$.  A rotation of 90 degrees negates the $ y$-coordinate and switches them, so $ T \equal{} ( \minus{} 3, \minus{} 3)$.[/quote] It's clockwise, so wouldn't it be $ (3,3)$?


percent
invariant
group theory
abstract algebra
algorithm
superior algebra
superior algebra unsolved

For a finite group $G$ define $g_G(n)$ to be the cardinal of elements having order $n$. Prove that if the applications $ g_{G_1}, g_{G_2}$ are identical, then the two groups are isomorphic.

There are already counterexamples among groups of order 16 or 32 if I remember.
There is a principle of Lovasz that says $| Hom (G, H)|$ determines $G$ if $H$ varies in some large family of finite groups (and the same for graphs and other finite structures).  There is an exercise about this in his book on combinatorics, extracted from one of his papers circa 1970.

What does the notation you used stand for, fleeting_guest? :?

Edit: Thanks, I see you've edited it :).

I really don't understand anything from what you say, man. Is the problem wrong or not?  :?

The problem is wrong, there are nonisomorphic groups of order 32 (or 16, or 64; some small power of 2) with, for each $k$, the same number of elements of order $k$.

The notation means "number of homomorphisms from $G$ to $H$".   Sorry for any unclarity.

Could you please give such explicit examples of groups?

I certainly do not remember any explicit example, but I remember that the example I saw was attributed to Polya.  This was in some standard textbook on finite groups, such as J. Dixon problem book or Philip Hall book on $p$-groups.   If I am at the library I will see if I can find a better reference.

by the way, the point of the $Hom(G,H)$ remarks is that the problem claimed that $G$ is determined by $|Hom(\mathbb{Z}/n,G)|$ for all $n$, and something like this is true, but with a different family of homomorphisms.

Too bad we cannot see some counter-examples. And I was wondering why after so much struggle I found nothing. Thanks anyway.

Counterexample: $G_1=Q_8\times C_2$, $G_2=$ semidirect product of $C_4$ with $C_4$.

[quote="harazi"]Too bad we cannot see some counter-examples. [/quote]

I think that any non-zero amount of information (even if only a hint or a wild idea) is a reasonable thing to post and need not elicit sarcasm.

The example with the quaternion group looks very much like the one I was trying to remember.  As far as I know, there is no "idea" for finding it; it just follows from the enumeration of $2$-groups, and similar examples are expected (and known) for $p > 2$.  So it is just good fortune, that the simplest example can be described in terms of well-known groups and constructions.

[quote="grobber"][quote="fleeting_guest"]
by the way, the point of the $Hom(G,H)$ remarks is that the problem claimed that $G$ is determined by $|Hom(\mathbb{Z}/n,G)|$ for all $n$, and something like this is true, but with a different family of homomorphisms.[/quote]

Which one is it then? $Hom(G,\mathbb Z_n)$ for all $n$, or $Hom(\mathbb Z_n,G)$ for all $n$? And harazi, where did you see the problem? Was it stated exactly as you wrote it (i.e. wrong)?[/quote]

The one that Lovasz proved involved $|Hom(G,H)|$ for finite $H$, and the family $H = Z_n$ does not suffice.     The statement at the top of this thread is equivalent to one involving $a_n = |Hom(Z_n,G)|$ (exercise: knowledge of all $a_n$ is equivalent to knowledge of all $b_n$ = number of elements of order $n$).

To fleeting-guest: I apologise, I really did not want to offend you.
  To grobber: actually, now I realized that the problem was a question and not an assertion. Sorry.  :oops:

harazi - I was not offended, but I noticed that there are many pejorative remarks here about postings (or their authors...) that do not give a 100-percent solution, complete and correct with details.  I think that quasi-solutions, wild ideas, references, statements without proof, reminiscences about the literature, or really anything that promotes discussion ---- all these are relevant contributions and should not be discouraged.  This is an open bulletin board, not a scientific journal.

About the original problem, almost [i]any[/i] obvious invariants of a group that can be written down using a small amount of information compared to the specification of the group, will not characterize the group up to isomorphism.   This includes the cardinality, the number of elements of each order, the lattice structure of the subgroups, and many other such things.    It is, so far, a computationally difficult problem (I think at least as difficult as graph isomorphism) to detect isomorphism of groups, and if one could make a simple list of invariants it is likely that it would imply a good deterministic algorithm for isomorphism testing.   The situation may be different for probabilistic algorithms.    If you are interested in this problem, there are papers by Laszlo Babai with a lot of information.

Wow, thanks, I will try to find these papers of Laszlo Babai. And about the comments, you can easily see that these comments were only in case someone wanted to close a certain discussion, I never said anything about someone who gave even the worst amount of information about a certain problem or discussion, as long as the information had not the purpose to close a certain topic. But I think that posting a result is clearly against any principle of the forum, which is a discussion about the problems and not the sign that we are so smart to discover a certain obvious answer and then not to make any comentary.

I don't know what are considered "the" principles of the forum.
I understand the forum as a place to discuss mathematics, mathematical problems, and anything else that arises for discussion including politics, money, sex, religion, sports, the weather, etc.
For the mathematical part, sometimes I have a full result with proof in my head and can share it, other times I just remember (or mis-remember) a relevant theorem.  The second situation is less smart, not a claim to be more smart.

how about this example

let p be prime

consider the elementary abelian group : Cp*Cp*Cp

one element is trivial, p^3-1 others have order p

now consider the group of upper diagonal matrices with diagonal entries 1 over the field with p elements

there are p^3 elements, and all non trivial elements have order p

the two groups are not isomorphic because the last one is not abelian

I remember a previous thread discussing this exact problem (with examples), but I can't find it. There are two pairs of examples among the groups of order 16.


In  hexagon $ABCDEF$, diagonals $AD$,$BE$, and $CF$ all bisect the area of the hexagon. Show that the three diagonals are concurrent.

[quote="spiral191"]that's not actually too bad...there's actually a theorem such that either AD,CE,BF are either parallel or concurent. Clearly if they bisect the area they're not parralel so they therefore must be concurrent.[/quote]

Could you state that theorem in detail?

um yeah sure it's called Brianchon's theorem: in any hexagon (convex or concave) the diagonals of the hexagon i.e. AD,BE,CF must be either  parallel or concurent. For more try  one of tjusamo geometry sections(sorry no link here!)

Brianchon's Theorem only works when the hexagon has an inscribed circle...

[quote="spiral191"]oh yeah thanks for reminding me of that aspect of Brianchon's theoem, that we first must prove that the hexagon can have an inscribed circle. however, is it possible to prove that just because the dagonals all bisect the area that an inscribed circle exist? i'll work on it! although i've found another solution exists through the wonder of complex geometry![/quote]

How is everything going? Any progress??

no sorry no progress, however as a hint try complex and analytical geometry

Is this a general hexagon, or a convex one?

$X \equiv AD \cap BE$, and since $S = [ABCDEF], \frac{S}{2}=[ABCD]=[BCDE]=[ADEF]=[EFAB]$ we get $[ABX]=[DXE]$ i.e. $BD \parallel AE$ analogously $AC \parallel DF$ and $CE \parallel BF$ hence $\triangle ACE$ is homothetic to $\triangle DFB \Rightarrow AD,BE$ and $CF$ are concurrent.

[quote=Gryphos]In fact, Brianchon´s theorem works for hexagons that have an inscribed ellipse  :)[/quote]

Brianchon's is true for hexagons that have [b]inconic[/b], because is the duality of Pascal's Theorem, and this is true if only if have 6 points on a conic.


Well this is how I remember it off the top of my head. The wordings different so theres no copyright thing or anything is there?

Let A be a subset of the numbers 1 to 1000000 (positive integers that is) with 101 elements. Prove that you can find 100 numbers from 1 to 1000000 called t_1, t_2, .. , t_100 so that the sets
Aj = {x+t_j for all x in A} are pairwise disjoint.

When you see the solution this looks easy, but this is quite difficult under exam conditions - especially since I'm bad at combinatorics. So I got a big zero ... :( They could have given me 1 point I think, I was even hoping for 2 points back then since I had the main idea (assume alle x(i)'s till x(k) are chosen, and prove that x(k + 1) can also be chosen - but the realisation of this idea was not entirely correct.) I prefer geometry (I got my H.M. at the geometry problem 4), number theory (the number theory question 2 was the only other question for which I got points, 3/7) and algebra (damn the algebra question 5 was pretty tricky this year ... 0)


Put $x=2^{n}$. Then the expression becomes $\frac{x^{3}-1}{8x-1}={x^{2}\over 8}+\frac{x^{2}-8}{8(8x-1)}$. For $n=1, n=2$ we have no solution. For $n\geqslant 3$, ${x^{2}\over 8}=2^{2n-3}$ is always integer, hence we must have either $\frac{x^{2}-8}{8(8x-1)}=0$ which is impossible, or $\frac{x^{2}-8}{8(8x-1)}\geqslant 1\iff x\geqslant 64$. Further division yields $\frac{x^{3}-1}{8x-1}={x^{2}\over 8}+{x\over 64}+\frac{x-64}{64(8x-1)}$. For $x\geqslant 64$ the first two terms are integer, hence for the third we must have either $\frac{x-64}{64(8x-1)}=0$ or $\frac{x-64}{64(8x-1)}\geqslant 1\iff x\leqslant 0$ which is impossible. Therefore the only possibility is $x=64\iff n=6$.

[quote="kunny"]Find positive integers $n$ for which $\frac{8^{n}-1}{2^{n+3}-1}$ is a integer.[/quote]

What about $\frac{8^{n}-1}{2^{n+3}-1}=\frac{2^{3n}-1}{2^{n+3}-1}$

So we know $n+3$ must divide $3n$ This easily gives 6 as the only answer.

this might help:

$\gcd(2^{a}-1,2^{b}-1)=2^{\gcd(a,b)}-1$

for positive integers $a$ and $b$


Traveling in the direction of Hogwarts at a constant speed for the entire trip, Larry was one-eighth of the way to Hogwarts at 5:00 a.m. At 8:00 a.m. on the same day he was three-fourths of the way to Hogwarts. Knowing Larry's starting location and Hogwarts are in the same time zone, what time did Larry arrive at Hogwarts?

He completed $ \frac34\minus{}\frac18\equal{}\frac58$ of the trip in three hours. He still needs to complete $ \frac28$ of the trip, so he will drive another $ \frac25(180)\equal{}72$ minutes. The time then  is $ \boxed{9: 12}$.

The easiest way to solve this:

I agree that he completes $5/8$ of the journey in $3$ hours. Now, however, i would use a proportion. $\frac{5}{8(3)} = \frac{2}{8x}$. This was arrived by saying that he does $\frac{5}{8}$ of the journey in $3$ hours so how much time will it take to travel $\frac{2}{8}$ of the distance?

Remember he has to travel $\frac{2}{8}=\frac{1}{4}$ of the distance because he has traveled $\frac{3}{4}$ of the distance by $8:00$. So solving the proportion gives $x=\frac{6}{5}$ of an hour or $72$ minutes so he arrived at Hogwarts at $9:12$.
Hope it helped! :wink:  :)

Edit: $LaTeXified$ it

[quote="CoolCatCrash"]I realize that 180 is the minute equivalent of 3 hours, but can someone explain why you multiply the remainder of the trip (2/5) by 180? I don't quite understand. Thanks,[/quote]

The remainder of the trip is $\frac{2}{8}$. Since he completes $\frac{5}{8}$ of the trip in 3 hours, he will complete $\frac{2}{8}=\frac{2}{5} \cdot \frac{5}{8}=\frac{2}{5} \cdot 180=72$ of the trip in 72 minutes. The $\frac{2}{5}$ is just a shortcut.


induction
number theory unsolved
number theory

The Lucas Sequence $1,3,4,7,11,18...$ is defined by:
$a_1=1$
$a_2=3$
$a_n=a_{n-1} + a_{n-2}$ for $n \geq 3$
Prove by mathematical induction, that $a_n<(1.75)^n$

I suppose it's another trivial problem but ... its hard for me :(  :blush:


Let m,n \in N and let K(m,n) the smallest natural number exceeding the number ( \sqrt (m+3) +  \sqrt (m+2))2n.

a.) Determine in which cases K(m,n) is even and in which cases it is odd.
b.) Evaluate K(6,4).
c.) Solve for m such that K(m,3) = 2702.
d.) Solve for m such K(m,3) = 64m^3 + 702046.

We have (\sqrt(m + 2) + \sqrt(m + 3))2n = (2m + 5 + 2\sqrt((m + 2)(m + 3)))n.
Also, (2m + 5 + 2\sqrt((m + 2)(m + 3)))n + (2m + 5 - 2\sqrt((m + 2)(m + 3)))n is a positive integer.
So a) becomes easy.

Brrr. Guess me may put it in Solved Problems Section :)


[quote="Alibek"]skolkimi sposobami mozhno predstavit naturalnoe chislo n v vide k naturalnyh slagaemyh (predstavlenia ,otlichaushiesa poradkom slagaemyh ,schitautsa razlichnimy).
   zaranie Sposibo!!!))) :wink:[/quote]

answer: $ \binom{n\minus{}1}{k\minus{}1}$
Write $ n$ ones and $ n\minus{}1$ dot between every two 1.
$ 1\cdot 1 \cdot 1 \cdot .....1 \cdot 1 \cdot 1$
Let's choose $ k\minus{}1$ dots among $ n\minus{}1$. This $ k\minus{}1$ dots will divide $ n$ ones into $ k$ groups. For every dividing of $ k$ summand we can determine exactly one such choosing of $ k\minus{}1$ dots and for every choosing of $ k\minus{}1$ dots we can determine exactly one dividing into $ k$ summond. So quantity of both is equal.
So quantity is equal $ \binom {n\minus{}1}{k\minus{}1}$

Then, you only have to translate the question...

Pierre.

How many ways are there to write a given natural number $ n$ as a sum of $ k$ natural numbers (two ways are considered different even if they only differ in the order of the addends)?

SaYaT's solution assumes that natural numbers mean $ 1$, $ 2$, $ 3$, ...
If natural numbers mean $ 0$, $ 1$, $ 2$, $ 3$, ..., then the answer would be $ \binom{n\plus{}k\minus{}1}{n}$.

  darij


function
calculus
derivative
integration
calculus computations

Let $f: [1,\infty)$$\rightarrow R$ be a bounded differenatiable function with continuous derivative

Prove that the improper integral $\int_{1}^{\infty}\frac{f'(x)}{x}dx$ $converges$

It follows from Dirichlet's Test:
$\int_1^X f'(x)dx = f(X) - f(1)$ is bounded.

Use integration by parts on $[1,a]$ and take $a \rightarrow +\infty$.

Could u explain more specific by using the integration by parts on $[1,a]$ and take $a\rightarrow +\infty$ Thx

$\int_1^a \frac{f'(x)}{x}dx = \left.\frac{f(x)}{x}\right|_a^1 + \int_1^a \frac{f(x)}{x^2}dx$

Since $f(x)$ is bounded, $\int_1^\infty \frac{f(x)}{x^2}dx$ converges by comparing with $\int_1^\infty \frac 1{x^2}dx$

Could anyone explain more about the comparing with $\int_1^\infty \frac 1{x^2}dx$ of the last post by "liyi"?

And what Theorem are we using in the following $\int_1^a \frac{f'(x)}{x}dx = \left.\frac{f(x)}{x}\right|_a^1 + \int_1^a \frac{f(x)}{x^2}dx$

Thx

[quote]Could anyone explain more about the comparing with $\int_1^\infty \frac 1{x^2}dx$ of the last post by "liyi"?[/quote]
This is the Comparison Test, only for integrals instead of series.

$|f(x)|\le M$ for some $M.$ Hence,

$\left|\int_1^{\infty}\frac{f(x)}{x^2}\,dx\right|\le \int_1^{\infty}\left|\frac{f(x)}{x^2}\right|\,dx\le \int_1^{\infty}\frac M{x^2}\,dx=M.$

That makes the improper integral absolutely convergent. Now we argue (and the proof is similar to the analogous one for series) that an absolutely convergent improper integral is convergent.

[quote]And what Theorem are we using in the following $\int_1^a \frac{f'(x)}{x}dx = \left.\frac{f(x)}{x}\right|_a^1 + \int_1^a \frac{f(x)}{x^2}dx$ [/quote]
Of coure, liyi meant to say

$\int_1^a \frac{f'(x)}{x}dx = \left.\frac{f(x)}{x}\right|_1^a + \int_1^a \frac{f(x)}{x^2}dx$

And that is simply integration by parts. Integration by parts (along with the related summation by parts) is a powerful method for proving convergence in the presence of oscillation.


Prove that in anytime in world cup.We always can find two groups have played an equality of matches. :winner_first:

Math is my life :!:

[hide]
Heh, very nice. Let $n$ be the number of teams. If they have all played a different number of matches, they must have played $0, 1, \ldots, n-1$ matches, respectively, since $n-1$ is the max. But that means one team (the one who has played $n-1$ matches) has played everyone else. However, this is impossible since one team has to have played $0$ matches. Contradiction.[/hide]

Yes,that's right. And this is Problem2:
 The captain of team now had the ball and he noticed that if he transmit the ball to one of his friends and one of members of team B stand near his friend a distance $\leq5$ then after transmit the ball to his friend,his friend will be lost the ball.Prove that his friend always can find a place which is safe for him :lol:

Math is my life :!:


probability
expected value
arithmetic sequence
geometric sequence

52 cards are dealt evenly to 4 ppl.  whats the probability that one person gets both ace of hearts and 3 of clubs?

belenos wrote:Combinatorics. Ouch

[hide]
Is it 1/4 ?[/hide]

er. i don't think so...mebbe i'm making it more complicated than it is, but[hide] we know one person has the ace of hearts. Call this guy person A. We want to figure out the probability person A also gets the 3 of clubs. Ok, now, let us say that the cards are dealt in "rounds", where each round consists one of one card going to the first person, one card going to the second person, one card going to the third person, and one card going to the fourth person. Now, if the three of clubs is not dealt in the same round as the ace of hearts, the probability that A gets the 3 of clubs will be 1/4. If the three of clubs is dealt in the same round as the ace of hearts, the probability A gets the 3 of clubs is 0. Thus, we now need to find the probability the 3 of clubs is dealt in the same round as the ace of hearts. That will be 3/51=1/17, since the 3 of clubs can be any other one of the other 51 cards, and there are 3 places it can be to put it in the same round as the ace of hearts. So, the probability will be 1/17*0+16/17*1/4=4/17. How's that?[/hide]

I agree with Mystic's answer.  I would suggest an alternate way of coming by it:  [hide]there are 52 cards total, divided into 4 piles of 13.  2 of them are important.  This gives us C(52, 2) arrangements.  If one of the piles has both cards, there are C(13, 2) arrangements for those cards within the pile, but there are four piles, so our final answer is

C(4, 1)*C(13, 2)/C(52, 2) = 4*13*12/(52*51) = 4/17.[/hide]

JBL, did AoPS teach you that kinda stuff?

Probably just experience, doing many many many many probability problems.

Wish i started learning this kind of stuff like 4 years ago, rather than now. Arg.

[quote]Wish i started learning this kind of stuff like 4 years ago, rather than now. [/quote]

I feel you.

[color=#A0FFA0]I think you mean "I feel for you."  But anyhow, I mostly have learned it from doing lots of problems and seeing lots of solutions.  Eventually, you begin to get a feel for how to approach a problem.  One simplifying step, for example, is to notice that it doesn't matter which particular cards we're considering.  Thus, we don't need to consider those two as distinct from each other, nor do we need to consider the other 50 cards as distinct.  This is nice because it lets us use just plain old nCr.

A more complicated problem that can be solved similarly:
Four players deal out a standard deck of cards among themselves.  The rules of the game are simple:  every player starts with 1 point, and for every ace they hold their score is doubled.  Thus, if I hold all four aces, I get 16 points while the other players all get 1.  Compute the expected number of points any player will get.[/color]

JBL wrote:A more complicated problem that can be solved similarly:
Four players deal out a standard deck of cards among themselves.  The rules of the game are simple:  every player starts with 1 point, and for every ace they hold their score is doubled.  Thus, if I hold all four aces, I get 16 points while the other players all get 1.  Compute the expected number of points any player will get.

[hide]alright let's consider player A. The probability he has k aces in his hand is C(4,k)*C(48,13-k)/C(52,13). Therefore, the expected value will be the sum from k = 0 to 4 of 2^k*C(4,k)*C(48,13-k)/C(52,13), which I don't feel like computing. Correct?[/hide]

Um, call me stupid. Because I am, definitely, comparitively, stupid. But is it [hide]2[/hide]?

Well, the question it asks is what is the expected number of points each player will get. By intuition, thing thats most likely to happen is for [hide]the aces to get evenly distributed, which probability, I think, is 6/13 (could be wrong about this too)[/hide] Andif this happens, then each player will get [hide]2[/hide].

I think you can "average" the number of aces a player will get if each ace added one point, in other words if the number of points a playa has is an arithmetic progression of the number of aces the guy has. In this case, it is a geometric progression, so I don't think you can do that...

for the original question, the 'correct' answer is 12/51.  ace of hearts must goto someone, and that person have 12/51 chance of getting the other card=4/17


limit
trigonometry
calculus
calculus computations

Without using L'Hospital's Theorem, calculate the value of $ \lim_{x \to 0} \frac{\tan ax \minus{} ax}{\tan bx \minus{} bx}$, where $ a$ and $ b$ are different constants.

Use Taylor series for $ \tan(x)\equal{}x\plus{}\frac{x^3}{3}\plus{}o(x^3)$.

...which gives you immediately $ \frac {a^3}{b^3}$  :)

Actually, that problem can be solved only by using limit theorem $ \lim_{x\to 0}\frac {\sin x}{x} \equal{} \lim_{x\to 0}\frac {\tan x}{x} \equal{} 1$ and trigonometric identities.

And I come up with this..
[hide="Solution"]
I noticed that $ \tan ax \equal{} \frac {\sin ax}{\cos ax}\times \frac {2\sin ax}{2\sin ax} \equal{} \frac {2\sin^2 ax}{2\cos ax \sin ax} \equal{} \frac {2\sin^2 ax}{\sin 2ax}$. In the same way, $ \tan bx \equal{} \frac {2\sin^2 bx}{\sin 2bx}$
So, I substitute it unto the problem..
$ \lim_{x\to 0}\frac {\tan ax \minus{} ax}{\tan bx \minus{} bx}$
$ \equal{} \lim_{x\to 0}\frac {\left(\frac {2\sin^2ax}{\sin 2ax} \right) \minus{} ax}{\left( \frac {2\sin^2 bx}{\sin 2bx}\right) \minus{} bx}\times \frac {(\sin 2ax)(\sin 2bx)}{(\sin 2ax)(\sin 2bx)}$
$ \equal{} \lim_{x\to 0}\frac {(2\sin^2 ax \minus{} ax\sin 2ax)(\sin 2bx)}{(2\sin^2 bx \minus{} bx\sin 2bx)(\sin 2ax)}$
Since $ \sin 2x \equal{} 2\sin x \cos x$, then $ \sin 2ax \equal{} 2\sin ax \cos ax$, similarly $ \sin 2bx \equal{} 2\sin bx \cos bx$.
$ \equal{} \lim_{x\to 0}\frac {(2\sin^2 ax \minus{} ax.2\sin ax\cos ax)(\sin 2bx)}{(2\sin^2 bx \minus{} bx.2\sin bx\cos bx)(\sin 2ax)}$
$ \equal{} \lim_{x\to 0}\frac {2\sin ax (\sin ax \minus{} ax\cos ax)}{2\sin bx (\sin bx \minus{} bx\cos bx)}\times \lim_{x\to 0}\frac {\sin 2bx}{\sin 2ax}$
$ \equal{} \lim_{x\to 0}\frac {\sin ax}{\sin bx}\times \lim_{x\to 0} \left( \frac {\sin ax \minus{} ax\cos ax}{\sin bx \minus{} bx\cos bx} \times \frac {\frac {1}{x}}{\frac {1}{x}} \right) \times \lim_{x\to 0}\frac {\sin 2bx}{\sin 2ax}$
$ \equal{} \lim_{x\to 0}\left( \frac {\sin ax}{ax}\times\frac {bx}{\sin bx}\times\frac {a}{b}\right)\times\lim_{x\to 0}\frac {(a\times\frac {\sin ax}{ax} \minus{} a\cos ax)}{b\times\frac {\sin bx}{bx} \minus{} b\cos bx}\times \lim_{x\to 0}\left( \frac {\sin 2bx}{2bx}\frac {2ax}{\sin 2ax}\times\frac {b}{a}\right)$
$ \equal{} \frac {a}{b}\times\frac {b}{a}\times\lim_{x \to 0}\frac {a \minus{} a\cos ax}{b \minus{} b\cos bx}$
$ \equal{} \lim_{x\to 0}\frac {a(1 \minus{} \cos ax)}{b(1 \minus{} \cos bx)}$
By using the identity that $ \tan \frac {1}{2}x \equal{} \frac {1 \minus{} \cos x}{\sin x}$, $ x\neq 0$, I obtain $ 1 \minus{} \cos ax \equal{} \sin ax \tan \frac {1}{2}ax$ and $ 1 \minus{} \cos bx \equal{} \sin bx \tan \frac {1}{2}bx$.
$ \equal{} \frac {a}{b}\times\lim_{x\to 0}\frac {\sin ax\tan \frac {1}{2}ax}{\sin bx\tan\frac {1}{2}bx}$
$ \equal{} \frac {a}{b}\times\lim_{x\to 0}\left( \frac{\sin ax}{ax}\times \frac {bx}{\sin bx}\times\frac {\tan\frac {1}{2}ax}{\frac {1}{2}ax}\times\frac {\frac {1}{2}bx}{\tan\frac {1}{2}bx}\times\frac {a^2}{b^2}\right)$
$ \equal{} \frac {a}{b}\times\frac {a^2}{b^2}$
$ \equal{} \frac {a^3}{b^3}$
[hide]
A long but nice solution..
 :D  :D  :D
[/hide]
[/hide]

[quote="silvergrasshopper"]Actually, that problem can be solved only by using limit theorem $ \lim_{x\to 0}\frac {\sin x}{x} \equal{} \lim_{x\to 0}\frac {\tan x}{x} \equal{} 1$ and trigonometric identities.

And I come up with this..

$ \equal{} \lim_{x\to 0}\left( \frac {\sin ax}{ax}\times\frac {bx}{\sin bx}\times\frac {a}{b}\right)\times\lim_{x\to 0}\frac {(a\times\frac {\sin ax}{ax} \minus{} a\cos ax)}{b\times\frac {\sin bx}{bx} \minus{} b\cos bx}\times \lim_{x\to 0}\left( \frac {\sin 2bx}{2bx}\frac {2ax}{\sin 2ax}\times\frac {b}{a}\right)$
$ \equal{} \frac {a}{b}\times\frac {b}{a}\times\lim_{x \to 0}\frac {a \minus{} a\cos ax}{b \minus{} b\cos bx}$
$ \equal{} \lim_{x\to 0}\frac {a(1 \minus{} \cos ax)}{b(1 \minus{} \cos bx)}$
By using the identity that $ \tan \frac {1}{2}x \equal{} \frac {1 \minus{} \cos x}{\sin x}$, $ x\neq 0$, I obtain $ 1 \minus{} \cos ax \equal{} \sin ax \tan \frac {1}{2}ax$ and $ 1 \minus{} \cos bx \equal{} \sin bx \tan \frac {1}{2}bx$.
$ \equal{} \frac {a}{b}\times\lim_{x\to 0}\frac {\sin ax\tan \frac {1}{2}ax}{\sin bx\tan\frac {1}{2}bx}$
$ \equal{} \frac {a}{b}\times\lim_{x\to 0}\left( \frac {\sin ax}{ax}\times \frac {bx}{\sin bx}\times\frac {\tan\frac {1}{2}ax}{\frac {1}{2}ax}\times\frac {\frac {1}{2}bx}{\tan\frac {1}{2}bx}\times\frac {a^2}{b^2}\right)$
$ \equal{} \frac {a}{b}\times\frac {a^2}{b^2}$
$ \equal{} \frac {a^3}{b^3}$
[/quote]

There appears to be a big error in your solution.Please remember that in any problem on limits you can use the formula or standard results only once in the entire question.The moment  where you have written

 $ \equal{} \frac {a}{b}\times\frac {b}{a}\times\lim_{x \to 0}\frac {a \minus{} a\cos ax}{b \minus{} b\cos bx}$

Now this is totally wrong.You should have written the value of $ \lim_{x \to 0}\frac {a \minus{} a\cos ax}{b \minus{} b\cos bx}$ here itself.

Even in the previous step you have used the standard results in between and then have come to this step.
I am afraid your solution is wrong despite the answer being correct.

[quote="JBL"][quote="pankajsinha"]Please remember that in any problem on limits you can use the formula or standard results only once in the entire question.[/quote]  What are you talking about?[/quote]

he means that silvergrasshopper substituted $ \lim_{x\to\ 0} \frac {\sin ax}{x}$, but didn't substitute $ \lim_{x\to\ 0}a\cos ax$ and
$ \lim_{x\to\ 0} b\cos bx$ which would have ended the expression up as $ \frac {0}{0}$. so his solution is wrong.

First we'll prove $ \lim_{x\to 0}{\tan x-x\over x^3}={1\over 3}$

\begin{eqnarray*}{\tan 2x-2x\over 8x^3} &=& \frac{{2\tan x\over {1-\tan^2x}}-2x}{8x^3}\\ &=& \frac{2\tan x-2x+2x\tan^2x}{8x^3(1-\tan^2x)}\\ &=& {1\over 1-\tan^2x}\left({1\over 4}\cdot{{\tan x-x}\over x^3}+{\tan^2 x\over 4x^2}\right)\end{eqnarray*}

Driving the limit through this, and using $ \lim_{x\to 0}{\tan x\over x}=1$, we get

$ L=1\cdot \left({1\over 4}L+{1\over 4}\right)\iff L={1\over 3}$

Now rewrite the initial limit as $ \lim_{x\to 0}\frac{\tan ax-ax}{a^3x^3}\cdot {a^3x^3\over b^3x^3}\cdot\frac{b^3x^3}{\tan bx-bx}={a^3\over b^3}\cdot{1\over 3}\cdot 3={a^3\over b^3}$


The following problems require constants to be found for the inequalities :

[1] Find the largest $ y$ such that $ \frac{1}{1\plus{}x^2}\ge \frac{y\minus{}x}{y\plus{}x}$.

[2] Find the least real number $ k$ with the following property: if the real  numbers $ x,y,z$ are not all positive, then
$ k(x^{2}\minus{}x\plus{}1)(y^{2}\minus{}y\plus{}1)(z^{2}\minus{}z\plus{}1) \geq (xyz)^{2}\minus{}xyz\plus{}1.$

[3] Find the maximum value of real number $ k$ such that
$ \frac{(b\minus{}c)^2(b\plus{}c)}{a}\plus{}\frac{(c\minus{}a)^2(c\plus{}a)}{b}\plus{}\frac{(a\minus{}b)^2(a\plus{}b)}{c} \ge k(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca).$

[4] Find $ A$ and $ B$ where,
$ A \ge\frac{x\plus{}1}{xy\plus{}x\plus{}1}\plus{}\frac{y\plus{}1}{yz\plus{}y\plus{}1}\plus{}\frac{z\plus{}1}{zx\plus{}z\plus{}1}\ge B$

[5] If  $ \prod_{n\equal{}1}^{n}{x_i}\equal{}a^n$ and $ x_i>0$ for $ i\equal{}1,2,...,n$, find $ m$ where $ \prod_{n\equal{}1}^{n}{(x_i\plus{}k)}\ge m, k$ being any positive real number.

I want mathlinkers' help.

[quote="Shanku"]The following problems require constants to be found for the inequalities :

[1] Find the largest $ y$ such that $ \frac {1}{1 \plus{} x^2}\ge \frac {y \minus{} x}{y \plus{} x}$.

[2] Find the least real number $ k$ with the following property: if the real  numbers $ x,y,z$ are not all positive, then
$ k(x^{2} \minus{} x \plus{} 1)(y^{2} \minus{} y \plus{} 1)(z^{2} \minus{} z \plus{} 1) \geq (xyz)^{2} \minus{} xyz \plus{} 1.$

[3] Find the maximum value of real number $ k$ such that
$ \frac {(b \minus{} c)^2(b \plus{} c)}{a} \plus{} \frac {(c \minus{} a)^2(c \plus{} a)}{b} \plus{} \frac {(a \minus{} b)^2(a \plus{} b)}{c} \ge k(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca).$

[4] Find $ A$ and $ B$ where,
$ A \ge\frac {x \plus{} 1}{xy \plus{} x \plus{} 1} \plus{} \frac {y \plus{} 1}{yz \plus{} y \plus{} 1} \plus{} \frac {z \plus{} 1}{zx \plus{} z \plus{} 1}\ge B$

[5] If  $ \prod_{n \equal{} 1}^{n}{x_i} \equal{} a^n$ and $ x_i > 0$ for $ i \equal{} 1,2,...,n$, find $ m$ where $ \prod_{n \equal{} 1}^{n}{(x_i \plus{} k)}\ge m, k$ being any positive real number.

I want mathlinkers' help.[/quote]

$ x\geq 0$????

$ a,b,c\geq 0$????

 :?:

[quote="Shanku"]The following problems require constants to be found for the inequalities :

[2] Find the least real number $ k$ with the following property: if the real  numbers $ x,y,z$ are not all positive, then
$ k(x^{2} \minus{} x \plus{} 1)(y^{2} \minus{} y \plus{} 1)(z^{2} \minus{} z \plus{} 1) \geq (xyz)^{2} \minus{} xyz \plus{} 1.$

[3] Find the maximum value of real number $ k$ such that
$ \frac {(b \minus{} c)^2(b \plus{} c)}{a} \plus{} \frac {(c \minus{} a)^2(c \plus{} a)}{b} \plus{} \frac {(a \minus{} b)^2(a \plus{} b)}{c} \ge k(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca).$

[/quote]

For $ [2], k_{min}\equal{}\frac{3\plus{}2\sqrt{3}}{3},$

for $ [3],$ if $ (a,b,c)\in{R^3_{\plus{}}},$ then $ k_{max}\equal{}2,$

            if $ (a,b,c)\in{R^3},$ then $ k_{max}\equal{}\minus{}\infty.$

[quote="fjwxcsl"][quote="Shanku"]The following problems require constants to be found for the inequalities :

[2] Find the least real number $ k$ with the following property: if the real  numbers $ x,y,z$ are not all positive, then
$ k(x^{2} \minus{} x \plus{} 1)(y^{2} \minus{} y \plus{} 1)(z^{2} \minus{} z \plus{} 1) \geq (xyz)^{2} \minus{} xyz \plus{} 1.$

[3] Find the maximum value of real number $ k$ such that
$ \frac {(b \minus{} c)^2(b \plus{} c)}{a} \plus{} \frac {(c \minus{} a)^2(c \plus{} a)}{b} \plus{} \frac {(a \minus{} b)^2(a \plus{} b)}{c} \ge k(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca).$

[/quote]

For $ [2], k_{min} \equal{} \frac {3 \plus{} 2\sqrt {3}}{3},$

for $ [3],$ if $ (a,b,c)\in{R^3_{ \plus{} }},$ then $ k_{max} \equal{} 2,$

            if $ (a,b,c)\in{R^3},$ then $ k_{max} \equal{} \minus{} \infty.$[/quote]

Will you kindly post your solutions ?

Please help. I need the solutions.

The answer for [4] is $ A\equal{}B\equal{}2$. The quantity is constant. Is that correct ?


calculus integration function quadratics AMC

I'll be gone on thursday and friday, so I'm going to crank out a lot of problems...

Source: Redwood Empire Mathematics Tournament, 2002C (Mathematical Manipulation)
Difficulty Level: Very Easy (good for basic AMC prep, though!)
[Note: Some problems reworded]

i) What is the product of the integral divisors of 81?

ii) The function f is a quadratic function.  Given that f(0) = 2, f(1) = 6, and f(2) = 12, what is f(3)?  What is f(x)?

iii) Simplify secx - (sinx)(tanx).

iv) Solve for x if log(x+8) = log(x-10) + 1.

Show work for credit.

i) What is the product of the integral divisors of 81?

ii) The function f is a quadratic function. Given that f(0) = 2, f(1) = 6, and f(2) = 12, what is f(3)? What is f(x)?

iii) Simplify secx - (sinx)(tanx).

iv) Solve for x if log(x+8) = log(x-10) + 1.

i) It is known that the product of the factors of any number is given by :sqrt:n^f where n is the number and f is the number of factors.  81 has 5 factors so 81^2.5 = 9^5 = 59049.

ii)  ax:^2:+bx+c=f(x)
c=2
a+b=4
4a+2b=10
a=1
b=3
(a,b,c) = (1,3,2)
x:^2:+3x+2=(x+1)(x+2)=f(x)
f(3)=(4)(5)=20

iii) secx - (sinx)(tanx) = 1/cosx - sin:^2:x/cosx = 1-sin:^2:x/cosx = cos:^2:x/cosx = cosx

iv) log(x+8) = log(x-10) + 1.
log (x+8) - log(x-10) = 1
log (x+8)/(x-10) = 1
10=(x+8)/(x-10)
10x-100=x+8
9x=108
x=12

Right, right, right, and right.  Although it's okay if you don't know the formula for i); it works just as well.

Overall, 28/28.  Yay.

[quote]WarpedKlown said:
 81 has 5 factors so 81^2.5=9^5=6561[/quote]

Actually 9 :^5: would be 59,049

[quote="ml2"]what is secx?[/quote]

1/sinx

[quote="MysticTerminator"]um no. actually sec x = 1/cos x[/quote]
LOL. whoops, always do that.


function
real analysis
real analysis unsolved

Ok so it's been a long time since I've done this

Let f:A-->B and g:B-->C be functions

B) Show that if g(f(x)) is surjective, then g is surjective

This is what I've attempted:

=> g(f(x)):A->C and g(f(x))=C

g(F(A))=C

I get stuck here, I can't assume F(A)=B can I?

So from

G(F(A))= C

F(A)= G$ ^{\minus{}1}(C)$

F(A)=B

plug that in and get G(B)=C which states that g is surjective, does that work?


calculus
integration
quadratics
algebra
polynomial
Euler
modular arithmetic

[quote]I came up with the following idea:
Why not having a formulary where everything of interest is collected¿
It should be of interest for all, for some to look them up again, for others to see them the first time ever and also referring to it here on MathLinks and in contests should be of use.[/quote]

Thus:
This topic will be there to collect (nice, useful) formulas and theorems (when possible with their names) in Number Theory.
These then will be (re)posted by moderators (I would do it regulary) in some other thread ( http://www.mathlinks.ro/Forum/viewtopic.php?t=76610 ).
Note that it shall [u]not[/u] contain proofs (except they are just one line of text), but links to such proofs when possible (but not necessarily).
When you want to see or present your own proof, I suggest starting a new topic for that.

So I recommend everyone to [u]post theorems and formulas[/u].

Note: Please don't edit your formulas/theorems when the post is older than one day, since there is a good chance that it will not be recognized. To keep the thread small, the posts will be deleted after being implemented, so don't be scared when they vanish.
When you have corrections or further generalizations etc., please post them at the end of the topic instead of editing; they will be added then (so double-posting will not considered bad in this case).
And it would be helpful when posting at the beginning of additional stuff where it should go (e.g. "this belongs to the theorem of [...]" or "this belongs to elementary properties on [...]") and what new symbols are introduced (e.g. "the symbol [...] representing [...] will be used").

All pairwise prime triples of integers satisfying $x^2+y^2=z^2$ are given by $x=|u^2+v^2|$, $y=2uv$, $z=u^2+v^2$ for relatively prime integers $u, v$ of distinct parity.

It is not provable and not refutable from Peano Arithmetic that $\#\{x | x < a\}$ is finite for any $a$ :P

[hide="done"]
Fermat's little theorem:
For any positive integer $a$ and every prime $p$ it is $a^p \equiv a \mod p$

For any polynom $f$ with integral coefficients and any integers $a, b$ with $a \equiv b \mod m$ for some integer $m$ it is $f(a) \equiv f(b) \mod m$.
[/hide]

- unique factorization (not only in $\mathbb{Z}$, but especially there).

- primitive roots: http://www.mathlinks.ro/Forum/viewtopic.php?p=344792#p344792

- order-lemma (uses $ord_n(a)$ for the smallest number $m$ with $a^m \equiv 1 \mod n$; thus $\gcd(a,n)$ is requested for that definition): Let $a,n \in \mathbb{Z}$ be coprime and $k \in \mathbb{Z}$ be such that $a^k \equiv 1 \mod n$, then $ord_n(a) | k$.

- a special case of Dirichlet's theorem on units: the Pell equation $x^2-dy^2=1$ and it's generalisation $x^2-dy^2=c$.

[hide="done"]
- Dirichlets theorems on primes in arithmetic progression: for any coprime $a,n \in \mathbb{N}$ there exist infinetely many primes $p \equiv a \mod n$.
- valuations: $v_p(xy)=v_p(x)+v_p(y) ; v_p(x+y) \geq \min (v_p(x) , v_p(y))$ with equality in the last one everytime when $v_p(x) \neq v_p(y)$ (but not only then).
- I also suggest the content of http://www.mathlinks.ro/Forum/viewtopic.php?p=336145#p336145 could be of interest.
- Quadratic residues: http://www.mathlinks.ro/Forum/viewtopic.php?t=34664
- The theorem of Hasse-Minkowski (or more exactly the local-to-global-principle for quadratic forms):
Assume you have a rational quadratic form $q(x_1, x_2 , ... , x_n)=q(x)= \sum_{1 \leq i,j \leq n} a_{i,j} x_i x_j$ (a homogenous rational polynomial of degree two), then there is an $x \in \mathbb{Q}^n, x\neq 0$ with $q(x)=0$ iff there are such $x_K \in K^n , x \neq 0$ for $K=\mathbb{R}$ and $K=\mathbb{Q}_p$.
[/hide]

[hide="done"][b]Fermat-Euler Theorem[/b]
If $gcd(a, m) = 1$ then $a^{\phi(m)} \equiv 1 \mod m$.[/hide]

[b]Chinese Remainder Theorem[/b]
let $n=(n_1,\dots,n_k)$ a tupel of pairwise coprime positive integers and $m=(m_1,\dots,m_k)$ a tupel of positive integers, then $\exists a$ so that:
\[ \begin{array}{c} \displaystyle a\equiv m_1 \pmod{n_1}\\ \vdots\\ \displaystyle a\equiv m_k \pmod{n_k} \end{array} \]
Also there is excactly one $m \in \mathbb Z / \mathbb Z_ {{\scriptscriptstyle \prod} n_i}$ so that:
\[ a \equiv m \pmod{{\textstyle \prod } n_i} \]

I would have expected more people posting here  :(
Why have only two people posted here (I even had to ask them to do)¿

[hide="done"][b]Gauss' theorem :[/b]
If $a|bc$ and $(a,b)=1$, then $a|c$.

The Gauss' theorem comes from :

[b]Bezout's identity :[/b]
The set $\{ax+by|x,y \in \mathbb{Z}\}$ is the set of all the multiples of $(a,b)$, that is to say :
\[ a\mathbb{Z} + b\mathbb{Z} = (a,b)\mathbb{Z}  \].[/hide]

[hide="done"]I've had some free time, so ...  :D
[b]On dealing with the floor function:[/b]
1. Let $n, m \in \mathbb{N}$, then
  \[ m \mbox{ mod } n = m - n \cdot \left \lfloor \frac{m}{n} \right \rfloor  \]

[b]Remark:[/b] Perhaps this could work with $n, m \in \mathbb{R}$ but who would use it ?

2. Let $m \in \mathbb{N}, n \in \mathbb{Z}, x \in \mathbb{R}$, then
  \[ \sum_{k = 0}^{m - 1} \left \lfloor \frac{nk + x}{m} \right \rfloor = ( m, n ) \cdot \left \lfloor \frac{x}{d} \right \rfloor + \frac{m - 1}{2} \cdot n + \frac{( m, n ) - m}{2}  \]
3. Let $m \in \mathbb{N}, x \in \mathbb{R}$, then
  \[ \left \lfloor m \cdot x \right \rfloor = \sum_{k = 0}^{m - 1} \left \lfloor x + \frac{k}{m} \right \rfloor  \]

[b]Some number theorethic sum[/b]:

1. Let $n \in \mathbb{N}$
  \[ \sum_{j = 1}^n \varphi ( j ) = \frac{3}{\pi^2} n^2 + O \left( n \log n \right)  \]
  \[ \sum_{j = 1}^n \varphi ( j ) = \frac{3}{\pi^2} n^2 + O \left( n \left( \log n \right)^{2 / 3} ( \log \log n )^{4 / 3} \right)  \]
2. Let $n \in \mathbb{N}$
  \[ \sum_{j = 1}^n d \left( j \right) = n \log n + ( 2 \gamma - 1 ) n + O \left( \sqrt{n} \right)  \]
3. Let $n, k \in \mathbb{N}$
  \[ \sum_{j = 1}^n \sigma_k ( j ) = \left( \frac{1}{k + 1} \sum_{j = 1}^{\infty} \frac{1}{j^{1 + k}} \right) n^{1 + k} + R_k ( n )  \]
  where
  \[ R_k ( n ) = \left\{\begin{array}{l} O ( n ), \mbox{ when } 0 < k < 1\\ O ( n \log n ), \mbox{ when } k = 1\\ O ( n^k ), \mbox{ when } k > 1 \end{array}\right.  \]
4. Let $n \geqslant 2$. Let $Q ( n )$ denote the number of squarefree integers less than $n$. Then
  \[ Q ( n ) = \sum_{j = 1}^n \mu^2 ( j ) = \frac{6}{\pi^2} n^2 + O \left( \sqrt{n} \right)  \]
5. Let $f$ be a multiplicative function, if
  \[ S = \sum^{\infty}_{n = 1} f ( n )  \]
  converges absolutely, then
  \[ \prod_p \left( \sum_{k = 0}^{\infty} f \left( p^k \right) \right) = \sum_{n = 1}^{\infty} f ( n )  \]
  where $p$ runs through primes.

6. If $f$ is completely multiplicative then
  \[ \sum_{n = 1}^{\infty} f ( n ) = \prod_p \frac{1}{1 - f ( p )}  \]
  where $p$ runs through primes.

7. Let $f$ be a multiplicative function, then
  \[ \sum_{d|n} \mu ( d ) f ( d ) = \prod_{p|n} \left( 1 - f ( p ) \right)  \]
  \[ \sum_{d|n} \mu^2 ( d ) f ( d ) = \prod_{p|n} ( 1 + f ( p ) )  \]
  where $p$ is prime.

8. Let $n \in \mathbb{N}$, then
  \[ \sum_{d|n} \mu ( d ) = \left\{\begin{array}{l} 1, \mbox{ if } n = 1\\ 0, \mbox{ if } n > 1 \end{array}\right.  \]
9. Let $n \in \mathbb{N}$, then
  \[ \sum_{j = 1}^n \frac{1}{\varphi ( j )} = C_1 \log n + C_2 + O \left( \frac{\log n}{n} \right)  \]
  where $C_1 > 0$ and $C_2$ are real constants.

10. Let $n \in \mathbb{N}$, then
  \[ \sum_{j = 1}^n \omega ( j ) = n \log \log n + Bn + O \left( \frac{n}{\log n} \right)  \]
  \[ \sum_{j = 1}^n \Omega ( j ) = n \log \log n + ( B + C ) n + O \left( \frac{n}{\log n} \right)  \]
  \[ \sum_{j = 1}^n \omega^2 ( j ) = n \left( \log \log n \right)^2 + O \left( n \log \log n \right)  \]
  where $B, C$ are constants.

11. Let $n \in \mathbb{N}$, then
  \[ \left( \log \log x \right) - 1 \leqslant \sum_{p \leq x} \frac{1}{p} = \log \log x + B + O \left( \frac{1}{\log x} \right)  \]
  where $p$ runs through primes and $B$ is a constant.
  \[ \sum_{k \geq 2, p^k \leq x} \frac{1}{p^k} = C + O \left( \frac{1}{\log x} \right)  \]
  where $p$ runs through primes and $C$ is a constant.

12. Let $n \in \mathbb{N}$, then
  \[ \sum_{n \leq x} r_2 ( n ) = \pi x + O \left( x^{1 / 3} \log x \right)  \]
  \item Let $n \geqslant 2$ be a positive integer, then
  \[ \sum_{p \leq n} \frac{\log p}{p} = \log n + O ( 1 )  \]
  where $p$ runs through primes.

13. Let $z \in \mathbb{C}$, and $n \in \mathbb{N}$, then
  \[ \prod_{p \leq n} \left( 1 + \frac{z}{p} \right) = A ( z ) \left( \log^{} n \right)^z \cdot \left( 1 + O \left( \frac{1}{\log n} \right) \right)  \]
  for $A ( z )$ a constant depending on $z$.

14. Let $n \in \mathbb{N}$, then
  \[ \sum_{d|n} \frac{1}{d} \geqslant \frac{n}{2 \varphi ( n )}  \]
15. Let $k, l$ be two positive integers with $( k, l ) = 1$, then
  \[ \sum_{%Error. "tmscript" is a bad command.
{\begin{array}{c} p \leq x\\ \left. p \equiv l ( %Error. "tmop" is a bad command.
{mod} k \right) \end{array}}} \frac{1}{p} = \frac{1}{\varphi ( k )} \log \log x + O ( 1 )  \]
  where $p$ runs through primes.

16. Let $f$ be an additive function and $n$ a positive integer, then
  \[ \sum_{m \leq n} \left( | f ( m ) - \sum_{p \leq n} \frac{f ( p )}{p} | \right)^2 \leqslant Cn \sum_{p^k \leq n} \frac{|f \left( p^k \right) |^2}{p^k}  \]
  where $p$ runs through primes, and $C$ is a constant $( C \leqslant 32 )$.

17. Let $f$ be a strongly additive function, and $n$ a positive integer. Then
  \[ \sum_{m \leq n} \left( | f ( m ) - \sum_{p \leq n} \frac{f ( p )}{p} | \right)^2 \leqslant 2 Cn \sum_{p \leq n} \frac{|f ( p ) |^2}{p}  \]
  where $p$ runs through primes and $C$ is a constant $( C \leqslant 32 )$.

[b]Arithmetic functions[/b]

1. Let $n \in \mathbb{N}$, then
  \[ \varphi ( n ) = n \prod_{p|n} \left( 1 - \frac{1}{p} \right) = \sum_{d|n} \mu ( d ) \frac{n}{d}  \]
2. Let $n \in \mathbb{N}$, then
  \[ \frac{1}{\varphi ( n )} = \frac{1}{n} \sum_{d|n} \frac{\mu^2 ( d )}{\varphi ( d )}  \]
3. Let $n \in \mathbb{N}$, then
  \[ 0, 92129 \cdot \frac{n}{\log n} < \pi ( n ) < 1, 1055 \cdot \frac{n}{\log n}  \]
4. Let $n \in \mathbb{N}$, then
  \[ r_4 ( n ) = \left\{\begin{array}{l} 8 \sigma ( n ), \mbox{ if } n \mbox{odd}\\ 24 \sigma ( n ), \mbox{ if } n \mbox{even} \end{array}\right.  \]
5. Let $n \in \mathbb{ N }$, then
  \[ \sigma ( n ) \varphi ( n ) \geqslant \frac{6}{\pi^2 } n^2  \]
6. Let $n \geqslant 2$ be a positive integer, then
  \[ \varphi ( n ) \geqslant \frac{cn}{\log \log n}  \]
  for some positive constant $c > 0$.

7. Let $n \geqslant 2$ be a positive integer, then
  \[ \varphi ( n ) \geqslant n - \sqrt{n}  \]
8. Let $p_n$ be the $n$-th prime number, then
  \[ an \log n \leqslant p_n \leqslant bn \log n  \]
  for two constants $0 < a < b$.

9. Let $n$ be a positive integer, then
  \[ \omega ( n ) \leqslant \lg_2 n  \]
10. Let $n$ be a positive integer, then
  \[ d ( n ) \leqslant 2 \sqrt{n}  \]

What follows hardly qualifies as number theory.
Homever I believe it is sufficently important to be included in the formulary.

[b]Some sums[/b]
1. [b]Abelian summation[/b]

Let $\left( a_j \right)_{j = 1}^n, ( b_j )_{j = 1}^n$ be a finite sequence
  of complex numbers. Then
  \[ \sum_{i = 1}^n a_i b_i = \left( \sum_{i = 1}^n a_i \right) b_n - \sum^{n - 1}_{m = 1} \left( \left( \sum_{i = 1}^m a_i \right) \left( b_{m + 1} - b_m \right) \right)  \]
2. Let $( a_m )_{m = 1}^n, ( b_m )_{m = 1}^n_{}$ be two finite sequence of real numbers. Then
  \[ \left( \sum_{k = 1}^n a_k \right) \cdot \left( \sum_{k = 1}^n b_k \right) = n \sum_{k = 1}^n a_k b_k - \sum_{k = 2}^n \sum_{j = 1}^{k - 1} \left( a_k - a_j \right) \cdot \left( b_k - b_j \right)  \]
  Or equivalently
  \[ \left( \sum_{k = 1}^n a_k \right) \cdot \left( \sum_{k = 1}^n b_k \right) = n \sum_{k = 1}^n a_k b_k - \sum_{1 \leq j < k \leq n} ( a_k - a_j ) \cdot ( b_k - b_j )  \]
3. Let $n \in \mathbb{N}$
  \[ \ln n + \gamma + \frac{1}{2 n} \leqslant \sum_{j = 1}^n \frac{1}{j} = \ln n + \gamma + \frac{1}{2 n} + O \left( \frac{1}{n^2} \right)  \]
  Where $\gamma = \underset{n \rightarrow \infty}{\lim} \left( \sum_{j = 1}^n \frac{1}{j} - \ln n \right)$ is the gamma constant.[/hide]

I've made a mistake :roll:. You'll find it in the formulary post #8, equation 6. The corrected version is:
For all composite number $n$, $\varphi(n) \leq n - \sqrt{n}$ holds.
I'll try to go through my post again, to see if there is no further mistakes...

Corrected. I've also added something to number $4$ of that post and corrected a mistake in number $10$.

[quote="ZetaX"][b]On dealing with the floor function:[/b]
1. Let $n, m \in \mathbb{N}$, then
  \[ m \mbox{ mod } n = m - n \cdot \left \lfloor \frac{m}{n} \right \rfloor  \]

[b]Remark:[/b] Perhaps this could work with $n, m \in \mathbb{R}$ but who would use it ?

2. Let $m \in \mathbb{N}, n \in \mathbb{Z}, x \in \mathbb{R}$, then
  \[ \sum_{k = 0}^{m - 1} \left \lfloor \frac{nk + x}{m} \right \rfloor = ( m, n ) \cdot \left \lfloor \frac{x}{d} \right \rfloor + \frac{m - 1}{2} \cdot n + \frac{( m, n ) - m}{2}  \]
3. Let $m \in \mathbb{N}, x \in \mathbb{R}$, then
  \[ \left \lfloor m \cdot x \right \rfloor = \sum_{k = 0}^{m - 1} \left \lfloor x + \frac{k}{m} \right \rfloor  \][/quote]

also include the formula:
$\sum_{k = 0}^{n} \left \lfloor \frac{n}{2} \right \rfloor = \left \lfloor \frac{n^2}{2} \right \rfloor$

[quote="me@home"]also include the formula:
$\sum_{k = 0}^{n} \left \lfloor \frac{n}{2} \right \rfloor = \left \lfloor \frac{n^2}{2} \right \rfloor$[/quote]
What is the correct form¿ It seems not to be correct...

[b]van der Waerden's theorem (with other equivalent formulations possible):[/b]

If $\mathbb{N}=\mathbb{A}\cup\mathbb{B}$ then at least one of $\mathbb{A}$ and $\mathbb{B}$ contains an arbitrary long arithmetic progression.

[hide="done"]
[b]Dirichlet's theorem:[/b]

In every arithmetic progression $ax+b$ with $(a,b)=1$ there are infinitely many primes.
(And Dirichlet's density as well, alas, I'm not sure if I remember it correctly.)

[b]Hensel's lemma:[/b]

$p^s||a-1,\ p^k||b,\ s\ge 1\ \Rightarrow\ p^{s+k}||a^b-1$ with exception of the case $p=2,\ s=1$. (I hope I've stated it correctly.)

[b]Zeckendorf's (base Fibonacci) representation:[/b]

Every $n\in\mathbb{N}$ can be uniquely expressed as a sum of Fibonacci numbers no two of which are consecutive.

[b]Lucas' theorem:[/b]

$\binom{a}{b}\equiv\prod_{i=0}^{k}\binom{a_i}{b_i}\mod p$ where $a_i$'s and $b_i$'s are the digits of $a$ and $b$ expressed in base $p$ ($p$ is a prime) with leading zeros allowed.

[b]Kummer's theorem:[/b]

If $p^s|\binom{n}{n-k}$ then $s$ does not exceed the number of carries needed when the numbers $n-k$ and $k$ are added when expressed in base $p$ ($p$ is again a prime number).
[/hide]

OK, I attached the beta-version of the pdf (I hadn't time till now to look through it); please tell me if you see some errors etc..  :)

PS: I think it should be continued even with the Wiki...

Well this theorem: \[ gcd(a, b)=1 \Rightarrow gcd(a^n-b^n, a^m-b^m)=a^{gcd(n, m)}-b^{gcd(n, m)}  \] is very useful for problem solving (and I think I didn't see it in this pdf Forumulary), but I don't know its author, and maybe it's more strict
than you wont to be...

I don't know what you speak about with "maybe it's more strict than you wont to be..."  :?
But it will be added (also to the other part) :)

PS: I thought all (co)authors are mentioned at the introduction¿ *looks again* (Edit: they are there.)

What I tried to say is, that maybe you are looking for some more general theorems, like let's say Euclidian algorithm or Fermat's theorem, something like this, but when I look little better I see that this is not the case :)

bye

ZetaX: Can you tell me what is this Wiki stuff ? (I'm sorry if the question is silly ...  :oops: )

The admins will start a Wiki here on MathLinks sonn, see http://www.mathlinks.ro/Forum/index.php?f=416 .
And it seems like it would be better to continue there...  :?

This one is good too, it has been very helpful in many cases:

For a prime number $p>2$ and a natural number $a$ s.t. $a>1$ and $p|a-1$ it stands: \[ v_p(a-1)=\alpha \Rightarrow p^{\alpha + \beta}|a^n-1 \Leftrightarrow p^\beta|n  \]

And for $p=2$: \[ v_2(a^2-1)=\alpha \Rightarrow 2^{\alpha + \beta}|a^n-1 \Leftrightarrow 2^{\beta +1 }|n  \]

cao!

[b]Theorem[/b]: If there exist primitive root of number $m$ there are excatly $\phi (\phi(m))$ primitive roots of number $m$.

[b][color=red]Fact 1.[/color][/b]
$p_{1}, p_{2}, ... p_{n}$ are primes and $n = \prod_{i=1}^{n}p_{i}^{a_{i}}$, let $d(n)$ be number of positive divisors of integer $n$. then
\[d(n) = \prod_{i=1}^{n}(a_{i}+1) \]
[b][color=red]Fact 2.[/color][/b]
If $p$ is prime number, then for every integer $a$ congruency $x_{1}^{n}+x_{2}^{n}+...+x_{s}^{n}\equiv a (mod \ p)$ have solution if $s \geq n$

[quote="Schoppenhauer"]All pairwise prime triples of integers satisfying $ x^2 \plus{} y^2 \equal{} z^2$ are given by $ x \equal{} |u^2 \plus{} v^2|$, $ y \equal{} 2uv$, $ z \equal{} u^2 \plus{} v^2$ for relatively prime integers $ u, v$ of distinct parity.

It is not provable and not refutable from Peano Arithmetic that $ \#\{x | x < a\}$ is finite for any $ a$ :P

[/hide][/quote]

Proof: Well I could do it only for primitive Pythagorean Triplets. Anyways, assume $ x \text{ and } y$ are $ \text{even}$. Then it follows that $ z$ will be $ \text{even}$, but this cannot be since the triplets are primitive.
Assume $ x \text{ and } y$ are $ \text{ odd }$. Then $ z^2$ will be of the form $ 4k \text{ or } 4k\plus{}2$ which is impossible for a square.

Proof: Three cases
Case 1: $ (4k)^2 \equal{} 16k^2 \equal{} 4(4k^2) \equal{}4q$
Case 2: $ (4k\plus{}1)^2 \equal{} 16k^2\plus{}8k\plus{}1 \equal{} 4(4k^2\plus{}2k)\plus{}1 \equal{} 4q\plus{}1$
Case 3: $ (4k^2)^2 \equal{} 16k^2\plus{}16k\plus{}4 \equal{} 4q$
Case 4: $ (4k\plus{}3)^2 \equal{} 16k^2\plus{}24k\plus{}9 \equal{} 4(4k^2\plus{}6k\plus{}2)\plus{}1 \equal{} 4q\plus{}1$
Thus $ x \text{ and } y$ cannot be $ \text{odd}$

WLOG, let $ x$ be $ \text{even}$ and $ y$ be $ \text{odd}$. Therefore $ z$ will be $ \text{odd}$.
$ x^2 \equal{} z^2\minus{}y^2 \equal{} (z\minus{}y)(z\plus{}y).$ Since $ x$ is $ \text{even}$, both $ z\minus{}y$ and $ z\plus{}y$ must be even. (Look at cases 1 and 3)
Let $ z\minus{}y \equal{} 2u$ and $ z\plus{}y \equal{} 2v$. Now $ \text{gcd}(u,v)\equal{}1$ (otherwise their $ \text{gcd}$ will divide $ z$ and $ y$).
Also as we assumed $ x$ is even, $ \frac{x}{2} \in \mathbb{Z}$.

Thus  $ (\frac{x}{2})^2 \equal{} (\frac{z\minus{}y}{2})(\frac{z\plus{}y}{2}) \equal{} uv$

Since $ uv \equal{} (\frac{x}{2})^2$ and $ \text{gcd}(u,v)\equal{}1$, both must be perfect squares themselves. So let $ u\equal{}s^2 \text{ and } v \equal{} t^2$ and they are of different parity $ (s,t)$.

Now we may look back and conclude
$ x\equal{}2st$
$ y \equal{} s^2\minus{}t^2$
$ z\equal{}s^2\plus{}t^2$
:)

[list]

[hide="Theorem 1"][*]We can represent a prime $p$ as sum of two integer squares if and only if $p \equiv 1 \pmod 4$.

[b]Proof.[/b]

[b]Theorem 1.[/b] If $p$ be a prime of the form $4k+1 , k \in \mathbb N$ prove that the equation $a^2 \equiv -1 \pmod p$ has integer solution for $a.$

[b]Proof.[/b] Let $p=4n+1$ be a prime. Assume $p-1$ numbers $1,2,\dots,2n,p-2n,p-2n+1,\dots,p-1.$ Now we have
\[ 1 \equiv 1 \pmod p\]\[2 \equiv 2 \pmod p\]\[\vdots\]\[2n \equiv 2n \pmod p\]\[p-2n \equiv -2n \pmod p\]\[p-2n+1 \equiv -(2n-1) \pmod p\]\[\vdots\]\[p-1 \equiv -1 \pmod p\]
Multiplying, we get
\[(p-1)! \equiv (-1)^{2n} \left( (2n) ! \right)^2=\left( (2n) ! \right)^2 \pmod p\]
By [url=http://en.wikipedia.org/wiki/Wilson%27s_theorem]Wilson's theorem[/url], $(p-1)! \equiv -1 \pmod p .$ So
\[\left( (2n) ! \right)^2 \equiv (p-1)! \equiv -1 \pmod p\]
This means if $p $ be a prime of the form $4k+1$, then the equation $a^2 \equiv -1 \pmod p$ has integer solution$.\square$

[b]Theorem 2.[/b] Let $n>1$ be a positive integer and $b \in \mathbb Z$. $e,f$ are positive reals such that $e,f>1$ and $ef>n.$ Prove that there exist integers $x_0,y_0$ such that $(x_0,y_0) \neq (0,0), 0\leq x_0 \leq e , |y_0| \leq f$ and
\[x_0 \equiv by_0 \pmod n\]

[b]Proof.[/b] Suppose the set
\[A=\{ by -x | x,y \in \mathbb Z , 0 \leq y \leq \lfloor f \rfloor , 0 \leq x \leq \lfloor e \rfloor, (x,y) \neq (0,0) \}\]
So, if we denote $|A|$ as number of members of the set $A$, then
\[|A|= (\lfloor e \rfloor +1 )  (\lfloor f \rfloor +1) -1 > ef-1 \geq n-1 \]
This implies $A \geq n$ (because $\lfloor e \rfloor +1 >e, \lfloor f \rfloor +1 >f, |A| \in \mathbb N$).

So, by pigeonhole principle, one of members of $A$ is divisible by $n$ or there exist two members $s,t$ in $A$ such that $s \equiv t \pmod n.$

[b]First case.[/b] One of members of $A$ is divisible by $n$. Let this number be $bu-v$ with $u,v \in \mathbb Z$. So
\[v \equiv bu \pmod n , \qquad (u,v) \neq (0,0) , \qquad 0 \leq u \leq \lfloor f \rfloor \leq f , \qquad 0 \leq v \leq \lfloor e \rfloor \leq e\]
And the theorem is proved in this case.

[b]Second case.[/b] there exist two members $s,t$ in $A$ such that $s \equiv t \pmod n.$ Let $s=bu_1-v_1 , t=bu_2-v_2$ and WLOG $v_1>v_2$. So
\[bu_1-v_1  \equiv bu_2-v_2 \pmod n\]
\[ \implies b(u_1-u_2)-(v_1-v_2) \equiv 0 \pmod n\]
Let $x_0=v_1-v_2$ and $y_0=u_1-u_2.$ We have
\[x_0 \equiv by_0 \pmod n , \qquad 0 \leq x_0 \leq e , \qquad |y_0| \leq f\]
So the problem proved in both cases$.\square$

Now we prove the original theorem,

[b]Theorem.[/b] Let $p$ be a prime and $p \equiv 1 \pmod 4.$ Prove that the equation
\[x^2+y^2=p\]
has integer solution for $x,y.$

[b]Proof.[/b] Let $p=4n+1$ and $b=(2n)!.$  Using [b]Theorem 1[/b], we see that $b^2 \equiv -1 \pmod p.$ Also by [b]Theorem 2[/b], there exist integers $x,y$ such that $(x,y) \neq (0,0)$ and
\[x \equiv by \pmod p, \qquad |x|,|y| \leq \sqrt p\]
Note that $\sqrt p \not \in \mathbb Z$, so $|x|,|y| < \sqrt p.$ This implies
\[x^2+y^2 \equiv b^2y^2+y^2 =y^2(b^2+1)\]
\[\stackrel{b^2 \equiv -1 \pmod p}{\implies} x^2+y^2 \equiv y^2(-1+1) \equiv 0 \pmod p\]
And since
\[ 1 \leq x^2+y^2 \leq p+p =2p\]
We have $x^2+y^2=p. \square$[/hide]

[hide="Theorem 2"][*]If $p \equiv 1 \pmod 4$ and $a,b \in \mathbb Z : p=a^2 +b^2$ then $ \nexists c,d \in \mathbb Z : p=c^2+d^2, (a,b) \neq (c,d).$[/hide]

[hide="Lemma 1"][*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=60&t=324597&start=0][size=150]Lifting Exponent Lemma[/size][/url] [/list]

[/hide]

$\sum_{p\in \mathbb P}\dfrac{1}{p}=\infty$ :roll:

Maybe it is a good idea to select all good theorems and lemma's which are useful on IMO:

-Bezout, euclide,bertrand, order,quadratic residues, LTE, lots of simple theorems ( I mean we don't have to add: p|ab if $p|a$ ôr $p|b$ and so/ Gauss I see here)

But things like Wolstenholme’s Theorem, is that usefull?

It is because I want collect the useful ones and think there are only like 10.

Yet interesting: Vietajumping, extremal priciple

Let $\sigma_{even}(n)$ be the number of $i$ such that $i$ is less than or equal to $n$ and $\sigma(i)$ is even. Then \[\sigma_{even}(n)=n-\lfloor\sqrt{\frac n2}\rfloor-\lfloor \sqrt n\rfloor\]
This theorem can be very easily proved using the following lemma:
[b]Lemma:[/b] If $x$ is a perfect square then $\sigma(x)$ is odd.
[i]Proof:[/i] Since the divisors with even parity does not change the parity of $\sigma(x)$, we should count only odd divisors. Since there are an odd number of odd divisors, the lemma follws.
Let's define the parity function $\mathcal P$ as \[\mathcal P(n)=1\]
if $n$ odd, else \[\mathcal P(n)=0\]
Note that, \[\mathcal P(\sigma(2^kn)=\mathcal P(\sigma(n))\]
So, if $\mathcal P(\sigma(n))=1$ then so is $\mathcal P(\sigma(2n))$. Thus, we need to just count for how many numbers less than $n$ we have $\mathcal P(\sigma(n))=1$. Note that we have $\mathcal P(\sigma(n))=1$ if $n$ is of the form $a^2$ or $2a^2$. There are such $\lfloor\sqrt{\frac n2}\rfloor+\lfloor \sqrt n\rfloor$ numbers. Hence conclusion.

I would like add another theorem:
\[\sum_{a_1+a_2+\ldots+a_m=n}\binom n {a_1}\binom{n}{a_2}\cdots\binom{n}{a_m}=\binom{mn}n\]
where $a_1,a_2,...,a_m$ are non-negative integers.
Partition $mn$ into $m$ equal parts of size $n$, say gathering of sticks. We can choose $a_1$ sticks from the first pool in $\binom n {a_1}$ ways, $a_2$ sticks from the second in $\binom {n} {a_2}$ ways,... Likewise, we can conclude the theorem since there


probability
geometry
analytic geometry
trigonometry
calculus

This question seems to be easy, but I just can't figure it out...

Consider a tabletop with a number of parallel lines drawn on it, which are equally spaced and a distance D apart. A needle of length L is dropped on the table surface ruled with parallel horizontal lines where D ≥ L. If you drop the needle on the table, you will find that one of two things happens: (1) The needle crosses or touches one of the lines, or (2) the needle crosses no lines.

Show that the (theoretical) probability that the needle comes to rest crossing a
line is

$=\frac{2L}{\pi D}$

I was thinking something along the lines of the area of the circle using L as the diameter and multiplying it by D, but I still can't work it out, so my approach is probably wrong. I'll probably kick myself after seeing the answer, though.

A little help, guys? Thanks.

[hide]
First simplify the probability to this:

Choose a random point $P$ between $(0,0)$ and $(0,D)$, then choose a random direction, corresponding to a point $Q$ on the circle of length $L$ with that point as the center. What is the probability that $\overline{PQ}$ intersects the lines $y=0$ or $y=D$?

We must choose the $y$ coordinate of $P$, and the direction of the line segment. Let the $y$ coordinate of $P$ be denoted $h$, and let the direction be denoted $\theta$. $0\leq h\leq D$, and $0\leq \theta < 2\pi$, so the total "area" of choices is $2\pi D$.

If the coordinates of $P$ are $(0,h)$, then the coordinates of $Q$ are $(L\cos \theta , h + L\sin \theta)$. So clearly, $\overline{PQ}$ crosses one of the lines if $h + L\sin \theta \leq 0$ or $h+L\sin \theta \geq D$. So we write

$h \leq -L\sin \theta$ or $h \geq D - L\sin \theta$
where $0\leq h \leq D$ and $0\leq \theta < 2\pi$.

Now, you could find the area of this region using Calculus, but since this is the Intermediate board we'll just "recall" that the area under $L \sin \theta$ and above the $x$ axis for a single period is $2L$. Examining a graph we see that in this case, our desired area is twice that, or $4L$.

So our desired probability is $\frac{4L}{2\pi D} = \frac{2L}{\pi D}$.
[/hide]

Thanks a lot for the answer. In terms of P, $h - L\sin\theta < 0$ and so on would also work, right?

Just wondering, are there alternative ways of doing this? I'm asking this because the course I'm doing doesn't deal with trigonometry much (although it may well be assumed knowledge). But I guess it's unavoidable.

Well, it is just the famous [url=http://en.wikipedia.org/wiki/Buffon's_needle]Buffon's needle problem[/url]. :)


Show that if $ m,n$ are distinct integer, then $ \gcd(a^{2^m}\plus{}1,a^{2^n}\plus{}1)\equal{}1$ if $ a$ is even, or $ 2$ if $ a$ is odd.
[color=red]Latexed by moderator[/color]

[quote="MagnusZ"]Show that if $ m,n$ are distinct integer, then $ \gcd(a^{2^m} \plus{} 1,a^{2^n} \plus{} 1) \equal{} 1$ if $ a$ is even, or $ 2$ if $ a$ is odd.
[color=red]Latexed by moderator[/color][/quote]
Assume WLOG that $ 0 \le m < n$.
Assume there exists a prime $ p > 2$ such that:
$ p \mid a^{2^m} \plus{} 1$ and $ a^{2^n} \plus{} 1$. Then we know that $ p \mid a^{2^m\plus{}1} \minus{} 1$ and thus we know $ ord_p(a) \mid gcd(2^{m\plus{}1}, p\minus{}1)$, but since $ ord_p(a) > 2^m$ (Since $ 1 \neq \minus{}1 \bmod p$) we know that $ ord_p(a) \equal{} 2^{m\plus{}1}$. But then $ p \mid a^{2^n} \plus{} 1$ and thus $ p \mid 2$. Contradiction, and our assumption was wrong: Hence there exists now prime $ p > 2$.
But $ a^{2^n}\plus{}1 \equiv 1,2 \bmod 4$ so either $ \gcd(a^{2^m} \plus{} 1,a^{2^n} \plus{} 1) \equal{} 1$ (when $ a$ is even) or $ 2$ (when $ a$ is odd).


Show that there exists a point inside every triangle such that when lines are drawn from the point to the vertices, the form 120 degree angles

[hide="Tidbit"]This is the [[Brocard point]] of the triangle, and its construction is quite annoying.[/hide]

[hide]
 I'm pretty sure this is actually the fermat point, not the brocard point.

construct equilateral triangles on two of the sides. Draw the circumcircles of these two equilateral triangles, they intersect once at a vertex and again at another point.  Now some basic circle geometry shows that the two angles from the point of intersection to the two sides of the triangle are both 120, which of course makes the third angle 120 as well.

Whats more interesting is that this is the point that minimizes the sum of the distances from the point to the vertices but I don't know a quick proof of that fact.[/hide]

Oops.  :oops:  Sorry, I get those two confused a lot. (similar concepts, really)

As nateharman noted, the Fermat Point is the point where the circumcircles of the three equilateral triangles concur. It can easily be shown from there that the three segments connecting the vertices of the triangle to vertex of the equilateral triangle opposite it, are concurrent.

nateharman also mentioned that the Fermat Point (also called the Toricelli Point and Steiner Point) minimizes AP+BP+CP. The general idea of the proof is to rotate PAB 60 degrees about B, and apply the Triangle Inequality.

[quote="CatalystOfNostalgia"]As nateharman noted, the Fermat Point is the point where the circumcircles of the three equilateral triangles concur. It can easily be shown from there that the three segments connecting the vertices of the triangle to vertex of the equilateral triangle opposite it, are concurrent.

nateharman also mentioned that the [b]Fermat Point (also called the Toricelli Point and Steiner Point)[/b] minimizes AP+BP+CP. The general idea of the proof is to rotate PAB 60 degrees about B, and apply the Triangle Inequality.[/quote]

um, yeah

Hmm.... These circles concur on an obtuse triangle with an angle greater than 120 degrees, but on the outside, not inside.  So is there a point that is on the inside of such a triangle where the angles created make 120 degree angles?  My guess is no.


Solve the trig eq
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/2b12a03703452d39344af5ce475b6995.gif[/img]

I got this: $sin(9x)=-8cosxcos(2x)sin(3x)cos(7x)$

well it would be too long to write all the process, but i used:

$2sen^{2}x=1-cos2x$
$2cos^{2}x=1+cos2x$
$2senxseny=cos(x-y)-cos(x+y)$
$2cosxcosy=cos(x+y)+cos(x-y)$


Hi all! I want to show you an interesting problem. It was proposed in the Bulgarian Math Olympiad 1997 - Second Round - district Vratza. It is Emanuil Yasinovui's /Arad Israel/ problem. Official solution were published in "Math and Informatics" bulgarian journal by Ivan Tonov. I found an error in solution and published this problem in "Math Forum" another bulgarian journal. It wasn't absolutely correct again. Now I know the problem have infinitely many solutions (I've proved by using Pell). I just want to ask you for a simple formula describing all of them.

The problem: [b]Find all integers m for which 7m^2+10m-8 is square of an integer.[/b]

7m^2+10m-8=(7m-4)(m+2)

Case 1. (7m-4)=p^2.(m+2)
Case 2. (m+2)=q^2.(7m-4)
Case 3. m+2=p.s^2, 7m-4=p.t^2

Case 1 and 2 leads to limited number of solutions.
But Case 3 is interesting after eliminating m and p it leads to few Pell's equations. Some of them don't have solutions, but some - have solutions (There is a common method for solving such equations - math excalibur 7.1 you may use). My question is: "Is there a simple formula for common solutions to the problem?"

I think there is not common method for solving this equation. It is necessary consider each Pell's equation.
m has the form
1. $ m \equal{} 2k$
2. $ m \equal{} 3k\plus{}1$
3. $ m \equal{} 6k\plus{}4$
4. $ m \equal{} 9k\plus{}7$
5. $ m \equal{} 18k\plus{}16$
For example, all solutions of even m are $ 6\cdot\left (\left (\frac{1}{2}\minus{}\frac{3}{\sqrt{63}}\right )\cdot (8\plus{}\sqrt{63})^{n}\plus{}\left (\frac{1}{2}\plus{}\frac{3}{\sqrt{63}}\right )\cdot(8\minus{}\sqrt{63})^{n}\right )^{2}\minus{}2$

And all solutions are
$ X_{0}\equal{}\minus{}2$
$ Y_{0}\equal{} 0$

and also:
$ X_{0}\equal{} 1$
$ Y_{0}\equal{}\minus{}3$

and also:
$ X_{0}\equal{} 1$
$ Y_{0}\equal{} 3$

and also:
$ X_{0}\equal{}\minus{}3$
$ Y_{0}\equal{}\minus{}5$

and also:
$ X_{0}\equal{}\minus{}3$
$ Y_{0}\equal{} 5$

and also:
$ X_{0}\equal{}\minus{}4$
$ Y_{0}\equal{}\minus{}8$

and also:
$ X_{0}\equal{}\minus{}4$
$ Y_{0}\equal{} 8$

and also:
$ X_{0}\equal{} 4$
$ Y_{0}\equal{}\minus{}12$

and also:
$ X_{0}\equal{} 4$
$ Y_{0}\equal{} 12$

and also:
$ X_{0}\equal{}\minus{}11$
$ Y_{0}\equal{}\minus{}27$

and also:
$ X_{0}\equal{}\minus{}11$
$ Y_{0}\equal{} 27$

and also:
$ X_{0}\equal{} 22$
$ Y_{0}\equal{}\minus{}60$

and also:
$ X_{0}\equal{} 22$
$ Y_{0}\equal{} 60$

and also:
$ X_{0}\equal{}\minus{}34$
$ Y_{0}\equal{}\minus{}88$

and also:
$ X_{0}\equal{}\minus{}34$
$ Y_{0}\equal{} 88$

and also:
$ X_{0}\equal{}\minus{}51$
$ Y_{0}\equal{}\minus{}133$

and also:
$ X_{0}\equal{}\minus{}51$
$ Y_{0}\equal{} 133$

$ X_{n\plus{}1}\equal{} P X_{n}\plus{}Q Y_{n}\plus{}K$
$ Y_{n\plus{}1}\equal{} R X_{n}\plus{}S Y_{n}\plus{}L$

$ P \equal{} 8$
$ Q \equal{} 3$
$ K \equal{} 5$
$ R \equal{} 21$
$ S \equal{} 8$
$ L \equal{} 15$
P.S. They are given by http://www.alpertron.com.ar/QUAD.HTM  :)

Thank you! It's a very interesting site I've never visited this one before. I'm not familiar with the Number Theory - I never took a course. But I think it was an interesting (and not easy) problem.


Samson derives utility according the relation $ \text{Utility} \equal{} \text{hours of math done} \times \text{hours of frisbee played}$.  On Monday he plays $ t$ hours of frisbee and spends $ 8 \minus{} t$ hours doing math.  On Tuesday, he ends up getting the same amount of utility as Monday while spending $ 2\minus{}t$ hours playing frisbee and $ t\plus{}3$ hours on math.  Find $ t$.  Answer as a fraction in lowest terms.

$ t(8\minus{}t)\equal{}(2\minus{}t)(t\plus{}3)$
$ 8t\minus{}t^2\equal{}\minus{}t\minus{}t^2\plus{}6$
$ t\equal{}\frac 23$


Quadrilateral $ABCD$ is inscribed in a circle. $AB=6, BC=7, CD=4, DA=8,$ and $AC=8.$ Find $BD$.

[hide]10, Ptolomey's Theorm-8*7+6*4=8x, wehre x is BD, solve to get 10[/hide]

[hide="ptolomey"]In a cyclic quadrilateral the sum of the products of the oppsite sides is the product of the diagonals.[/hide]


I reading up on NT the other day, and I was wondering if there was an algorithm for calculating the $ nth$ magic square, assuming that 1 is the first magic square. Where magic squares are numbers that are both perfect squares and sums of consecutive integers starting from one.

Please hurry and answer cuz I just HAVE to know if an algorithm exists, or else this problem will keep me up at night.

Thanks.

[quote="bigboy82892"]magic squares are numbers that are both perfect squares and sums of consecutive integers starting from one.[/quote]

This is not what people usually mean by a [url=http://en.wikipedia.org/wiki/Magic_square]magic square[/url]; you're talking about a [url=http://mathworld.wolfram.com/SquareTriangularNumber.html]square triangular number[/url], whose closed form is given by

$ S_n \equal{} \frac{1}{32} \left( (1 \plus{} \sqrt{2})^{2n} \minus{} (1 \minus{} \sqrt{2})^{2n} \right)^2$


[quote="ckck"]Find the constant term in the expansion $(2x^{3}-\frac{1}{4x^{2}})^{5}$.[/quote]

what does this have to do with logs?  anyways, its when 2x^3 is to the 2nd power and the other is to the third power

so

5C2*2^2*(-1/4)^3=40/64[b]=-5/8[/b]


geometry
3D geometry
number theory
relatively prime
Diophantine Equations

Suppose that $A=1,2,$ or $3$. Let $a$ and $b$ be relatively prime integers such that $a^{2}+Ab^2 =s^3$ for some integer $s$. Then, there are integers $u$ and $v$ such that $s=u^2 +Av^2$, $a =u^3 - 3Avu^2$, and $b=3u^{2}v -Av^3$.

I think it was meant to be $ a\equal{}u^3\minus{}3Auv^2$, not $ u^3\minus{}3Avu^2$

$ s^3\equal{}a^2\plus{}Ab^2$
$ s^3\equal{}(a\plus{}b\sqrt{\minus{}A})(a\minus{}b\sqrt{\minus{}A})$

Note $ 2$ cannot divide $ s$, since then $ a$ and $ b$ must both be even (look $ \mod 8$), but they are relatively prime. So $ 2$ cannot divide either $ a\plus{}b\sqrt{\minus{}A}$ or $ a\minus{}b\sqrt{\minus{}A}$.

$ \gcd(a\plus{}b\sqrt{\minus{}A},a\minus{}b\sqrt{\minus{}A})\equal{}\\
\gcd(2a,a\minus{}b\sqrt{\minus{}A})\equal{}\\
\gcd(a,a\minus{}b\sqrt{\minus{}A})\equal{}\\
\gcd(a,\minus{}b\sqrt{\minus{}A})\equal{}1$

since $ a$ and $ b$ are relatively prime. Thus, $ a\plus{}b\sqrt{\minus{}A}$ and $ a\minus{}b\sqrt{\minus{}A}$ are relatively prime, and since their product is a cube, each must be a cube. So let

$ (a\plus{}b\sqrt{\minus{}A})\equal{}(u\plus{}v\sqrt{\minus{}A})^3$
$ (a\plus{}b\sqrt{\minus{}A})\equal{}(u^3\minus{}3Auv^2\plus{}(3u^2 v\minus{}Av^3)\sqrt{\minus{}A})$

So

$ a\equal{}u^3\minus{}3Auv^2$ and
$ b\equal{}3u^2 v\minus{}Av^3$

Also note that $ s\equal{}(u\plus{}v\sqrt{\minus{}A})(u\minus{}v\sqrt{\minus{}A})\equal{}u^2\plus{}Av^2$, as desired.


Solve

$ x(x\plus{}y)\equal{}9$

$ y(x\plus{}y)\equal{}16$.

hello, dividing both equations we get
$ \frac {x}{y} \equal{} \frac {9}{16}$
from here we get by inserting $ x \equal{} \frac {9}{16}y$ in the first equation
$ \frac {9}{16}y(\frac {9}{16}y \plus{} y) \equal{} 9$
solving this equation we get
$ y_1 \equal{} \frac {16}{5}$ or $ y_2 \equal{} \minus{} \frac {16}{5}$.
Sonnhard.

Hi Sonnhard.

Alternatively, adding the two equations, I get

$ (x\plus{}y)^{2}\equal{}25$

so

$ x\plus{}y\equal{}\pm5$.

Substituting these for $ x\plus{}y$ in the given equations, I get

$ x\equal{}\frac{9}{5}, y\equal{}\frac{16}{5}$ and $ x\equal{}\frac{\minus{}9}{5}, y\equal{}\frac{\minus{}16}{5}$,

as you did.


Given x>0 find minimum value of
$ P \equal{} \frac {(x \plus{} \frac {1}{x})^6 \minus{} x^6 \minus{} \frac {1}{x^6} \minus{} 2}{(x \plus{} \frac {1}{x})^3 \plus{} x^3 \plus{} \frac {1}{x^3}}$

An easy problem ?  :D i want to see the simple sollution

[hide="answer"]The numerator factors as $ \frac {3(1 \plus{} x^2)^2(2 \plus{} x^2 \plus{} 2x^4)}{x^4}$ while the denominator is $ \frac {(1 \plus{} x^2)(2 \plus{} x^2 \plus{} 2x^4)}{x^3}$ so we are minimizing
\[ P \equal{} \frac {3(1 \plus{} x^2)}{x} \equal{} 3\left(x \plus{} \frac {1}{x}\right)
\]
which is clearly $ 6$.[/hide]

[quote="JoeBlow"]The numerator factors as $ \frac {3(1 \plus{} x^2)^2(2 \plus{} x^2 \plus{} 2x^4)}{x^4}$ while the denominator is $ \frac {(1 \plus{} x^2)(2 \plus{} x^2 \plus{} 2x^4)}{x^3}$ so we are minimizing
\[ P \equal{} \frac {3(1 \plus{} x^2)}{x}
\]
[/quote]

I am not yet checking  but if you right ; you have nice and simple sollution  :)  thanks ! but how to factor the numerator and denominator ; using Maple ?! :D

what's about other ?  :D

Let $ x\plus{}\frac{1}{x}\equal{}a$ then $ x^3\plus{}\frac{1}{x^3}\equal{}a^3\minus{}3a$ and $ x^6\plus{}\frac{1}{x^6}\equal{}(a^3\minus{}3a)^2\minus{}2$ so
$ P\equal{}\frac{a^6\minus{}(a^3\minus{}3a)^2}{2a^3\minus{}3a}\equal{}\frac{6a^4\minus{}9a^2}{2a^3\minus{}3a}\equal{}3a\ge6$ when $ x\equal{}1$ $ P\equal{}6$

hello, $ p(x)\equal{}\frac{3(x^2\plus{}1)}{x}\geq6 \Leftrightarrow (x\minus{}1)^2\geq0$ which is obviously.
Sonnhard.