What is the median of the prime numbers between 20 and 50?

List the primes numbers between 20 and 50:
23,29,31,37,41,43,47
Median means the middle number. It is clear that the middle number is 37.

The prime numbers are 23,29,31,37,41,43, and 47.

The median is the middle of them, which is 37.


Let P(x) and Q(x) be two polynomials in (R-Z)[x] having the following property for all real numbers like x:
P(x) and Q(x) are the same in being integer or not.
prove that: |P(x)|=|Q(x)|

I think I misunderstood the problem. Can you explain why $ Q(x) \equal{} P(x)\plus{}c$ isn't a counter example for $ c\neq 0$?


1.benzene to m-nitro acetophenone
2. b3enzaldehyde to benzophenone  in not more than 2 steps

[hide="Answer"]1. Friedel-Crafts acylation with CH3COCl/AlCl3, followed by nitration with HNO3/H2SO4.
2. Reactions with C6H5MgBr, followed by oxidation.
[/hide]


probability
combinatorics proposed
combinatorics

An instructor gives her class a set of 10 problems with the information that the final exam will conssit of a random selection of 5 of them. if a student has figured out how to do 7 of the problems, what is the rpobability that he or she will answer correctly
(a) all 5 problems;
(b) @ least 4 of the problems.

Thanks in advance

Probablity that he will answer all five correctly is
7[b]C[/b]5/10[b]C[/b]5

Hence probability of 5 correct is 21/252=0.0833...

Probablity taht he will answer [u]exatcly 4[/u] correctly is
(7[b]C[/b]4)*(3[b]C[/b]1)/10[b]C[/b]5

Hence probability of [u]exactly 4[/u] correct is 35*3/252=0.4166...

Hence probablity of atleast 4 correct is 21/252+105/252=126/252=0.5


let p(x),q(x) be two nonconstant poly. with rational coeficients.if every value of q(t) for t rational is also value of p(s) for some rational s,show there exists rational poly.  s.t.
q(x)=p(f(x)) :D


induction
combinatorics solved
combinatorics

Two players A and B play the following game:
Suppose on the table is the positive integer n. Player A chooses an positive integer x and player B will replace n by either n+x or n-x on the table. Find all values of k such that A can always ensure that at some time, a power of k will appear on the table. (Here we consider 1 to be a power of any integer).

I think I can find strategies for the case $k=2^t+1$ and I think this might be the answer, but I'm still trying to show that these are the only numbers that work (maybe that's not true and I'm trying in vain, but this seems to be the case).

Here's a proof that it doesn't work if $a | (k-1)$ for some odd $a$.
Player B's strategy: Choose a starting $n$ with $n$ not congruent to 1 mod $a$. At each stage, choose whichever of $n-x$ or $n+x$ is not congruent to 1 mod $a$. This is possible because if both were congruent to 1 mod $a$, $2x$ would be divisible by $a$ and $a$ would divide $x$, which forces $x$ to be 1 mod $a$. Since all powers of $k$ are 1 mod $a$, this strategy avoids them.

[quote="lhvietbao"]Two players A and B play the following game:
Suppose on the table is the positive integer n. Player A chooses an positive integer x and player B will replace n by either n+x or n-x on the table. [/quote]

Does A then do the same as B, and they continue alternating, or is it a one round game?

For each round, the process is repeated, with no role reversal.

Since grobber hasn't posted the details of his strategy, I'll finish this one.
Let $k=2^t+1$. Let $S$ be the set of all starting values for which Player A has a winning strategy. Obviously, the powers of $k$ are in $S$. Also, if $n \in S$, $kn \in S$; we can just multiply the $x$ used in the strategy by $k$. Finally, if $a$ and $b$ are in $S$ and $a+b$ is even, $\frac{a+b}{2} \in S$ since we can just take $x=|\frac{a-b}{2}|$ to reach one of them. Now, by averaging $1$ and $k$, we can reach all integers between $1$ and $k$ in at most $t$ steps; $a = \frac{a-1}{2^t}k+\frac{k-a}{2^t}\cdot 1$.
We finish the argument by induction. Assume $n>k$ and $a \in S$ for all $a<n$. If $k|n$, then $\frac{n}{k} \in S$ and $n \in S$. Otherwise, let $k|(n+b)$ with $0<b<k$. Since $\frac{n+b}{k} <  \frac{2n}{k} < n$, $\frac{n+b}{k} \in S$ and $n+b \in S$. Also $n-b \in S$, so $n \in S$. This forces $S$ to be the full set of natural numbers, and we are done.

The complete answer: The desired values of $k$ are the integers of the form $2^t+1$.

Oh, I had forgotten about this one. I think there is more than one strategy. Here's one for which I don't have a proof:

First of all, it's easy to show that if $A$ has a winning strategy when n is in the interval $[2,k-1]$ then we can show that $A$ has a strategy in the other cases as well (we write the numbers on the table in base $k$, and each time $A$ can choose the complement of the last digit in base $k$ of the number on the table. If $B$ subtratcts the digit from the number then we're left with a smaller number, and if $B$ adds it to the number on the table then the number on the table will have fewer digits than in the previous step, so we can use some sort of induction on the size of $n$). All we need to do is find a strategy for numbers smaller than $k$, and I noticed that if when $n$ is on the table $A$ chooses $x$ to be the smallest of $n-1$ and $k-n$ then $A$ wins.

Another strategy, for which I have a proof (I think :)) would be  doing pretty much the same thing as above, only that we work in the unique interval $[2^t+1,2^{t+1}+1)$ which contains $n$ (at each step $A$ chooses $x$ to be the smallest of $n-(2^t+1)$ and $2^{t+1}+1-n$).

However, the proof is rather nasty, and I don't think I'll be posting it now that there's such a nice proof here :).


vector
rotation
geometry
geometric transformation

Find the direction of the magnetic field at point P in the following situations.

[img]http://i19.tinypic.com/2nrmq6h.jpg[/img]

I done it like this but keep getting them wrong

(going from left to right)
a)Using right hand palm rule - I put my thumb down the page following the arrow, force is to the right of the page and magnetic field into the page

b) consider the top arrowing i.e. < ----. Thumb to the left, force down the page magnetic field into the page  ANSWER:magnetic field out of page"

c) current out of paper,
Is this right hand palm rule or right hand grip rule? ANSWER: right of page

d) current into paper,
Is this right hand palm rule or right hand grip rule?ANSWER: down the page

please help
thanks

The magnetic field created by a current in a point is proportional to $ \Delta \vec{l} \times \vec{r}$, where the first is a vector with the direction of the current, and the second is the positional vector of the point relative to the current. To find the direction of the magnetic field we can use the right hand rule. Try to do this:

1) Put the origin of the vector r in the extreme of vector l.

2) Now imagine that the first vector (l) can rotate about its origin, and that the two vectors are equivalent to two connected sticks.

3) Push the second vector (r) along its direction. What were then be the sense of rotation of both sticks?

4) Curve and rotate the four biggest fingers of your right hand in the same direction of rotation of the vectors, always keeping the thumb perpendicular to the other rotating fingers.

5) The direction of the magnetic field is the same as the direction to which the thumb is pointing.

You can also do this analitically, by choosing a system of three axes, see what are the directions of vectors l and r, and then make the vectorial product.

thanks carcul

But this is much too complex for me.
I learnt to find magnetic field use right hand palm rule where palm=force
fingers=magnetic field
thumb=current

So in the first example the arrow means current right?
So i put  4fingers into point P(into page)
hence thumb is pointing down and force is to the right.
Is this correct?

for the third one (with the x)
current into page.
I placed thumb into the page at point P. So force is going left and magnetic field down the page.
THis is the dilemna. If I put finger into page and rotate my wrist 180^0 then thumb is still into page but force is to the right and magnetic field up the page :S

thanks for the help above

I am very courious  to know how boss did the picture


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

I. Let $p(x)$ be an irreducible polynomial of degree $n$ with coefficients in a field $K$. Let $q(x) = d^{n}p\left(\frac{x}{d}\right)$ for some nonzero $d \in K$.
Show that $K[x]/(p(x))$ and $K[x]/(q(x))$ are field isomorphic.

II. Let $p(x)$ be a monic irreducible polynomial of degree $n$ with coefficients in the field $\mathbb{Z}$. Let $V_{r}\subset \mathbb{Z}[z]$ denote the $\mathbb{Z}$-submodule with basis $1, x, \ldots, x^{r}$. There is a ring homomorphism $\phi$ defined as the composite $\mathbb{Z}[x] \subset \mathbb{Q}[x] \rightarrow \mathbb{Q}[x]/(p(x))$.
Show that $\phi$ restricted to $V_{n-1}$ is injective.

I am having trouble constructing the homomorphisms. In problem II, is $\phi$ the inclusion map in both mappings or just for $\mathbb{Z}[x] \hookrightarrow \mathbb{Q}[x]$?

1. we have $K[x]/p \cong K(a)$, where $a$ is a root of $p$. but then $d a$ is a root of $q$, so $K(a)=K(da) \cong K[x]/q$. explicitely, the isomorphism of the factor rings is given by $x \to dx$.
2. the field $\mathbb{Z}$? do you mean $p \in \mathbb{Q}[x]$ or $p \in \mathbb{Z}[x]$?
$\mathbb{Q}[x] \to \mathbb{Q}[x]/p$ is no inclusion, but the canonical projection.


p(x) is a polynomial of odd degree which satisfies p(x 2 -1) = p(x) 2  - 1 for all x. Show that p(x) = x.

from the given condition, we easily get that:

$p(x)^2 = p(-x)^2$ but then since p(x) is of odd degree, as x approaches infinity, p(x) tends to $\pm \infty$ while p(-x) tends to $\mp \infty$ so for infinitely many real numbers t, we have that p(t) = -p(-t), and we must have p(x) = - p(-x) implying p(x) is an odd function.

Therefore, we get that p(0) = 0, and p(-1) = p(0)^2 - 1 = -1, implying that p(1) = 1.

now define the sequence $a_n$ as follows: $a_1 = 1$, and $a_{n+1} = \sqrt{a_n+1}$, note that $a_n \ge 1$ for all n.

now, i claim that $p(a_n) = a_n$ for all n.  We already have the base case, so assume that $p(a_k) = a_k$ for some k.

then $p(a_{k+1}^2-1) = p(a_k) = a_k = p(a_{k+1}^2)-1$

which implies that $p(a_{k+1}) = \pm a_{k+1}$ now, if $p(a_{k+1}) = -a_{k+1}$ then repeating the calculation for $a_{k+2}$ reveals: $p(a_{k+2})^2 = -a_{k+1}+1 < 0$.  So indeed p(a_{k+1}) = a_{k+1} as required.

therefore since p(x) = x for infinitely many values of x, p(x) must be x, and checking we see that p(x) = x satisfies the given relation.


number theory
greatest common divisor
algebra proposed
algebra

[quote="April"]The denominators of two irreducible fractions are 600 and 700. Find the minimum value of the denominator of their sum (written as an irreducible fraction).[/quote]

$ gcd(q_1,600)\equal{}1, gcd(q_2,700)\equal{}1$

$ \frac{q_1}{600}\plus{}\frac{q_2}{700}\equal{}\frac{7q_1\plus{}6q_2}{6 \times 700}$

It's easy to see that $ gcd(6, q_1)\equal{}1, gcd(7, q_2)\equal{}1$

So $ gcd (6 \times 7 \times 4, 7q_1\plus{}6q_2)\equal{}1$

To minimize the denominator , we try to find $ q_1,q_2$ such that $ 25|7q_1\plus{}6q_2$

$ (q_1,q_2)\equal{}(1,3)$ is a choice.

$ \frac{1}{600}\plus{}\frac{3}{700}\equal{}\frac{1}{168}$

Min. value of denominator is $ 168$

Let the two fractions be $\frac{a}{600}$ and $\frac{b}{700}$ where $\gcd(a,600)=1$ and $\gcd(b,700)=1$. Then, $\frac{a}{600}+\frac{b}{700}=\frac{700a+600b}{420000}=\frac{7a+6b}{4200}$. Now, we want the numerator and denominator to share as many factors as possible. First, note that $4200=2^3 \times 3 \times 5^2 \times 7$. By Bezout's Lemma, all integers can be formed by $7a+6b$. Let $7a+6b=n$, where $n$ is an integer. It is easy to see that $a=n, b=-n$ is a solution, so the solution set for $7a+6b=n$ is $a=n+6t, b=-n-7t$ for some integer $t$. Since $\gcd(a,600)=1$ and $\gcd(b,700)=1$, $\gcd(n,6)$ and $\gcd(n,7)$ must both equal 1. Hence, 2,3,7 cannot divide $n$. This means that the maximum possible gcd of 4200 and $n$ is 25, so the minimum denominator of $\frac{7a+6b}{4200}=\frac{n}{4200}$ is $4200/25=168.$ A denominator of 168 is satisfied by $a=25$ and $b=-25$.


summer program
MathPath
percent
absolute value

[quote="curl"]What are your pet peeves (excluding forum-related)?

I can't stand loud crunchers.[/quote]
neither can I! Another one is when people say they're they know something when  they sortof know they really don't.(bad explanation)

I absolutely cannot stand the phrase "I told you so."

A lot of people who knew me at MathPath probably know a couple dozen other things that piss me off :P .

[quote="star99"]or when people say 'I knew that!'[/quote]

omg that drives me insane!!! especially when it's so totally obvious that they had no idea.

my list: ( in no particular order )

"i knew that"
ppl that click their pens (although i always do it)
ppl that take bites of food that are WAY too big
being interrupted when reading
being interrupted when doing a math problem!!!
people who think they know everything
PREPS!!!
teachers who can't stand being corrected even when wrong
....................................

it keeps going tho, i promise. lol

hmm...when people ask why you are mad when you really aren't, that really ticks me off and actually makes me mad..

also when people eat with their mouth open, or make annoying noises in class, really loud barking dogs, and...lots more..

Ditto, even though I'm guilty of most of those things on occasion. The only new one I can add is: my own ineptitude: I hate it when I stumble over myself. ...

Definitely loud crunchers is up there.

I think the worst for me iz ppl wh0 Typ liK dIs. I guess that's forum related...um people who sing loudly off-key drive me insane.

[quote="MathFiend"]
teachers who can't stand being corrected even when wrong[/quote]
Absolutely!!! (Also everyone else who think that they know everything but actually they don't have a clue what they're talking about.)
And people who moan about their food all the time but they don't realise that there are people who have nothing to eat.

[quote]Also everyone else who think that they know everything but actually they don't have a clue what they're talking about
[/quote]
Amen! that is my biggest pet peeve.[/quote]

spammers....and i don't care that it is forum-related.

[quote="elston"]hmm...when people ask why you are mad when you really aren't, that really ticks me off and actually makes me mad..

[/quote]

Exactly.

My dad always asks me "Why did you have such an attitude" when I'm doing fine and this ticks me off and I say "I don't!" and he's like "See, I told you." and I get in a bad mood then. Grr makes me mad thinking about it. :P

[quote="dts"]When people treat, for example, "3 times more" as being the same as "3 times as much".

It gets on my nerves because when they say, for example, "3 times more", I don't know if they mean "3 times as much" (how most people would interpret it) or "4 times as much" (what it actually means).

The most irritated with this thing is when I encountered it on the ACT, and the two interpretations lead to different answers, both of which were choices.[/quote]

I think that "3 times more" always means "3 times as much", despite the fact that this is counterintuitive. Kind of like when people say "I could care less" when they really mean "I couldn't care less", but people still know what they mean. Most people don't give second thought to things like these (why second thought? do they give it first thought?). I've wondered about the "3 times more" issue too though.

I don't think I could think of pet peeves off the top of my head, but here are two rather trivial ones that came up today:
1) When people say "when x increases, y decreases" and then add "when x decreases, y increases". Comes up a lot on multiple choice tests.
2) When people are using a variable x which ALWAYS has a negative value, but they want to make it positive, so instead of just negating it, they use the absolute value. This would be fine if they also used the absolute value for a corresponding variable which is always positive, but they don't, and the inconsistency bothers me.

bad manners *poke poke*
people who talk way too much also fit in

[quote="aznmonkey1992"]people who talk way too much also fit in[/quote]

along the same lines, people who try to be cool. either you're cool or you're not, so don't try to be something you're not. and who really cares if you're "cool" or not anyways? why should you let that affect you?

People who stand and socialize in the middle of hallways and, my favorite, right in front of doorways.

[quote="jmadsen"][quote="dts"]When people treat, for example, "3 times more" as being the same as "3 times as much".

It gets on my nerves because when they say, for example, "3 times more", I don't know if they mean "3 times as much" (how most people would interpret it) or "4 times as much" (what it actually means).

The most irritated with this thing is when I encountered it on the ACT, and the two interpretations lead to different answers, both of which were choices.[/quote]

I think that "3 times more" always means "3 times as much", despite the fact that this is counterintuitive. Kind of like when people say "I could care less" when they really mean "I couldn't care less", but people still know what they mean. Most people don't give second thought to things like these (why second thought? do they give it first thought?). I've wondered about the "3 times more" issue too though.[/quote]

No, think about it.  50% more means 1.5 times as much.  100% more means twice as much.  Therefore, 200% more means three times as much.  Converting out of percent, we get two times more means three times as much, and so on.


I ordered the book like a week ago and I set the book on priority mail so it should have arrived at my house last Thursday. But it didn't. What is up with that?

Please PM me or email me at crawford@artofproblemsolving.com about the particulars of your order and I will check and see if everything was processed correctly.


The number of persons ($ p$) needed to do a job varies directly as the amount ($ w$) of work to be done and inversely as the time ($ t$) in which the job must be done. If 2 men can erect 500 feet of fence in 10 hours, how long will it take 5 men to erect 875 feet of fence?

[hide]We have
\[ p\equal{}k\frac{w}{t}.\]
Plugging in $ p\equal{}2$, $ w\equal{}500$, and $ t\equal{}10$, we find that $ k\equal{}\frac{1}{25}$.
Then for $ p\equal{}5$, $ w\equal{}857$, and $ k\equal{}\frac{1}{25}$, we find that
\[ t\equal{}\boxed{7 \text{ hours}}.\][/hide]


calculus
derivative
function
real analysis
real analysis solved

Construct a continuous monotonic $f$ on $\mathbb{R}^1$ so that $f$ is not constant on any segment although $f'(x)$ = 0 a.e

We have been talking about this over on the "Gallery of Examples" topic:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22230&postorder=asc&start=45]This topic[/url]

See posts #25, 32, and 46 through 51.

The $g(x)$ in that example maps $[0,1]$ onto $[0,1]$, but all we would need to do to get an example for the whole line is to define $f(x)=n+g(x-n)$ for $x\in[n,n+1).$

I realize that soon after I posted. But I would appreciate very much any solution involving some constructing work. It would fit more to the context in the textbook!! ( We can't just give students the Bilingsley funtion as a solution :D  )

Let $f(x)$ be the Cantor-Lebesgue function (also mentioned in the "gallery of examples" topic.) Consider it to be equal to $0$ for $x<0$ and $1$ for $x>1.$

Let $g(x)=\sum_{k=0}^{\infty}\frac1{4^k}\sum_{j=0}^{3^k-1}f(3^kx-3j-1).$

That should work on $[0,1].$

Well. it looks as well like the Weierstrass function ( Which is also in the gallery  ;) )
 Is it the idea for you to find that function??
 But I have to think more of it, can't understand right now!! :(


inequalities
parameterization
real analysis
real analysis unsolved

Hi,
I have the following econ problem. I calculated two equations for utilities. Now I wand to compare the utilities to know which one is "the better one".
My Utilities are:
U_1=(((A(6+α)+2c(1-α)-w(8-α))(A(2-α)-2c(5-α)+w(8-α)))/(96))
U_2=(((2-α)(A-2c+w)²α)/(24))
I know that 0<=a<=1
A-c>0
A-w>0
I hope that the firs utility is larger than the second, but I am not sure. I substracted both terms, tried to show that this term is larger than 0 but I failed.

Does anyone have an idea?

Thanks a lot,
all the best
Anna

You should know more about your parameters to get the desired inequality. Plugging in $\alpha=0$, we see that we need to compare $(6A+2c-8w)(2A-10c+8w)/96$ with $0$. The problem is that the constraits $A-c>0$ and $A-w>0$ are insufficient to guarantee either of the 2 inequalities. Indeed, when $A=1$, $c=w=0$, the first quantity is positive and when $A=1$, $c=0.5$, $w=0$, it is negative.


How many integers less than 10000 can be made with the eight digits 0, 1, 2, 3, 4, 5, 6, 7?

[hide="Can digits be used more than once?"]
Complementary counting $ 10000-\left\lfloor \frac{10000}{9}\right\rfloor-\left\lfloor \frac{10000}{8}\right\rfloor = 7639$.
[/hide]

If not , then the answer xpmath gave.

Didn't you post a couple of these questions already?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159964[/url]

[quote="xpmath"][hide]It must be a 4 digit number, and the first digit can't be 0, so $ 7*8^{3}$?[/hide][/quote]
[hide]Umm, it can also be a 3-dig. 2-dig. or 1-dig., so it should be $ 8^{4}= \boxed{4096}$
Also, in the other thread that mountain.dew posted, 0 should count, since it's an integer too, right?, so you should add on to 4095, I think. [/hide]

Whoops, I thought it said 4-digit. I agree with you then


Every June 1, an ecologist takes a census of the number of wrens in a state park. She noticed that the number is decreasing by $ 40\%$ each year. If this trend continues, in what year will be the census show that the number of wrens is less than $ 10\%$ of what it was on June 1, 2004?

Each year, it becomes $ 1\minus{}0.4\equal{}0.6$ of what it was the previous year. We want $ 0.6^{y\minus{}2004}<0.1\implies y\minus{}2004\equal{}5\implies y\equal{}\boxed{2009}$.


inequalities
function
inequalities proposed
algebra
highschoolmath

I'll prove:
  $\sum\frac{1}{2c^{2}+2ac+2bc+3ab}\leq\frac{1}{\sum ab}$
  $\leftrightarrow \sum\frac{2c(a+b-2c)}{2c^{2}+2ac+2bc+3ab}\leq 0$
  $\leftrightarrow \sum\frac{a+b-2c}{2\sum a+3\frac{ab}{c}}\leq 0$(*)
Finally,use chebyshev's inequality to prove (*)

[quote="silver1989"]I'll prove:
  $\sum\frac{1}{2c^{2}+2ac+2bc+3ab}\leq\frac{1}{\sum ab}$
  [/quote]
Maybe $\sum\frac{1}{4c^{2}+2ac+2bc+ab}\leq\frac{1}{\sum ab}$ $?$ :wink:

Nice Problem! My solution:

see that $\frac{ab+bc+ca}{(2a+b)(2a+c)}= \frac{b(2a+c)+c(2a+b)}{2(2a+b)(2a+c)}$

$= \frac{b}{2(2a+b)}+\frac{c}{2(2a+c)}$

now, do in this same way with the other fractions.

let S our sum!

by cauchy, we get:
$(b(2a+b)+c(2a+c)+a(2b+a)+c(2b+c)+a(2c+a)+b(2c+b)) . S$

$\geq (\frac{2a+2b+2c}{\sqrt{2}})^{2}$

$\Longrightarrow S \geq 1$

$\frac{ab}{ab+ac+bc}\leq\frac{ab+ac+bc}{ab+2c^{2}+2c}$ (which is easy to show) and the rest is easy.

[quote="Megus"]Observe that
$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$
is equivalent to (algebraic manipulations only, us condition $a+b+c=1$ to homogenize firstly)

$2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$
which is easy  :)[/quote]

[b]NOTE[/b]

Observe now that

$\frac{1}{2+x}+\frac{1}{2+\frac{1}{x}}\leq \frac{2}{3}$
  Use this for
  $x = \frac{a}{b}, x = \frac{b}{c},x = \frac{c}{a}$
  and we have a simple solution , as Megus pointed out , too !

 Babis

Or (more general form) consider the function $f(x)=\frac{1}{2+x}\iff f(x-2)=\frac{1}{x}$ which is clearly concave and thus by Jensen:
\[\sum^{n}\frac{1}{2+x_{i}}\leq \frac{n}{2+\frac{\sum^{n}x_{1}}{n}}\leq \frac{n}{2+\prod^{n}\sqrt[n]{x_{i}}}\] where the last is by Am-Gm. Our ineq. is the case of $n=6, \prod x_{i}=1$ when \[\frac{n}{2+\prod^{n}\sqrt[n]{x_{i}}}=\frac{6}{2+1}=2\]

[quote="me@home"]Or (more general form) consider the function $f(x)=\frac{1}{2+x}\iff f(x-2)=\frac{1}{x}$ which is clearly concave ...
[/quote]

 Sorry  :( , but is this funtion concave ? I see it is convex on $(0,+\infty)$.
 ( I find $f''(x) =+\frac{2}{(x+2)^{3}}$). But I'll try it again !

[u]  Babis[/u]

[quote]Let $ a, b, c$  be positive reals such that their sum is $ 1$. Prove that
\[ \frac {1}{ab \plus{} 2c^{2} \plus{} 2c} \plus{} \frac {1}{bc \plus{} 2a^{2} \plus{} 2a} \plus{} \frac {1}{ac \plus{} 2b^{2} \plus{} 2b}\geq \frac {1}{ab \plus{} bc \plus{} ac}.\]
[/quote]

By $ AM\minus{}GM$, we obtain: $ a^2c^2\plus{}b^2c^2\ge 2abc^2$. Which is equivalent to:
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}2abc\ge a^2b^2\plus{}2abc^2\plus{}2abc$ or
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}2abc(a\plus{}b\plus{}c)\ge ab(ab\plus{}2c^2\plus{}2c)$
$ \Leftrightarrow$ $ (ab\plus{}ca\plus{}bc)^2\ge ab(ab\plus{}2c^2\plus{}2c)$
$ \Leftrightarrow$ $ \frac {1}{ab\plus{}2c^2\plus{}2c}\ge \frac {ab}{(ab\plus{}bc\plus{}ca)^2}$.
Having the same argument, we conclude that:
$ LHS$ $ \ge \frac {ab\plus{}bc\plus{}ca}{(ab\plus{}ca\plus{}bc)^2}\equal{}\frac {1}{ab\plus{}bc\plus{}ca}$ $ \equal{}RHS$. $ \square$
Our proof is completed then.

This problem has actually been posted before. See [url=http://www.mathlinks.ro/viewtopic.php?t=293234]here[/url].

$ \sum_{cyc}\frac {1}{2c^{2} + 2c + ab}\geq\frac {1}{ab + bc + ac}$

$ 2c^{2} + 2c + ab = 2c^{2} + 2(a + b + c)c + ab = (2c + a)(2c + b)$

$ \implies$  $ \sum_{cyc}\frac {2}{(2c + a)(2c + b)}\geq \frac {2}{ab + bc + ac}$

$ \implies$  $ \sum_{cyc}\frac {2(ab + ac + bc)}{(2c + a)(2c + b)}\geq 2$

$ 2(ab + ac + bc) = b(2c + a) + a(2c + b)$

$ \implies$  $ \sum_{cyc}\frac {2(ab + ac + bc)}{(2c + a)(2c + b)} = \sum_{cyc}(\frac {b}{2c + b} + \frac {a}{2c + a})$

$ \implies$ $ \sum_{cyc}(\frac {b}{2c + b} + \frac {a}{2c + a})(\sum_{cyc}{b(2c + b) + a(2c + a))\geq 4(a + b + c)^{2}}$

$ (\sum_{cyc}{b(2c + b) + a(2c + a))} = 2(a + b + c)^{2}$

$ \implies$   $ \sum_{cyc}\frac {2(ab + ac + bc)}{(2c + a)(2c + b)}\geq 2$ Q.E.D

Secondly we can use muirhead inequality,but it is very long solution

Using $uvw$ method write the terms in a function with $w^3$ and fix u and v. Suppose that the function will be like that \[\sum\frac{1}{4c^{2}+2ac+2bc+ab}\ge\frac{1}{\sum ab} \Longleftrightarrow g(w^3)=Aw^6+Bw^3+C\ge0\]
it is easy to see $A$ is negative so $g$ will be concave hence we must look for the minimum and the max of $w^3$ by $uvw$ theorem we must look the cases $a=b$ and $c=0$ which is obvius.

Consider the lagrange function $f(a,b,c,\lambda)=\sum\frac {1}{ab+2c^2+2c}+\frac {1}{\sum ab}-\lambda(\sum a)$
Now $f'(a)=-\sum\frac {b}{(ab+2c^2+2c)^2}-\frac {3a+2}{(bc+2a^2+2a)^2}+\frac {b+c}{(ab+bc+ac)^2}-\lambda a$.
So basically now we've like $\lambda(a+t)+\frac {a^2(3a+2)}{(m+2a^3+2a^2)^2}=g$ where $t,g,m$ are constant.
So $a>b>c$ implies $g(x)$ is decreasing , where $g(x)=\frac {x^2(3x+2)}{(m+2x^3+2x^2)^2}$.
Now $g'(x)<0$ gives $(2a^3+2a)(9a^2+4)<(m+2a^3+2a)(9a^2+4)<2a^2(3a+2)(6a^2+2)$ now that implies $a^2+a>\frac {2}{3}\implies a>\frac {1}{2}$. Now as $g$ is decreasing in the interval $(\frac {1}{2},\inf )$ so by the condition $a,b,c\in\(\frac {1}{2},\inf )$ and that implies $\sum a>1$, absurd, similarly we can do this for if two of the are equal, now so rest and last case is we must have $a=b=c$ to obtain minimum value and that is indeed zero.

Excuse me for opening this old topic, but this inequality has another short proof that I found and want to share. Using Am-gm we have:

$(2a+b)(2a+c)=\frac{1}{bc}(2ac+bc)(2ab+bc)\le \frac{1}{bc}(ab+ac+bc)^2$

Therefore:

$\sum \frac{1}{ab+2c^2+2c}=\sum \frac{1}{(2a+b)(2a+c)}\ge \sum \frac{ab}{(ab+ac+bc)^2} =\frac{1}{ab+ac+bc}$ . $\blacksquare$

Here is an elementary proof :)
..hope that I haven't made any mistake
First bring the inequality in the form
$\sum\frac{(a+b)^2}{(a+b)^{2}(4c^2+2cb+2ca +ab)}\geq \frac {1}{\sum ab}$
From Cauchy-Schwartz the inequality reduces to
$4(\sum ab)(\sum a^2) +8(\sum ab )^2\geq \sum (a+b)^{2}(4c^2+2cb+2ca +ab) $ or equivalently $\sum a^{3}b +\sum ab^{3} \geq 2 \sum( ab)^2$ ... Which is true from AM-GM

Let $a, b, c$  be positive reals such that their sum is $1$. Prove that \[\frac{a}{bc+2a^{2}+2a}+\frac{b}{ca+2b^{2}+2b}+\frac{c}{ab+2c^{2}+2c}\geq \frac{1}{3(bc+ca+ab)}.\] [url]http://www.artofproblemsolving.com/community/c6h610084p3626533[/url]
Let $a,b,c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c1068820h2200767p16573227]Prove that[/url]
$$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}}  \leq 1$$$$\leq \sqrt{\frac{bc}{(2a+b)(2a+c)}}+\sqrt{\frac{ca}{(2b+c)(2b+a)}}+\sqrt{\frac{ab}{(2c+a)(2c+b)}}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}} $$$$\sqrt{\frac{a^2+bc}{(2a+b)(2a+c)}}+\sqrt{\frac{b^2+ca}{(2b+c)(2b+a)}}+\sqrt{\frac{c^2+ab}{(2c+a)(2c+b)}}\ge \sqrt{2}$$

[quote=sqing]Let $a, b, c$  be positive reals such that their sum is $1$. Prove that \[\frac{a}{bc+2a^{2}+2a}+\frac{b}{ca+2b^{2}+2b}+\frac{c}{ab+2c^{2}+2c}\geq \frac{1}{3(bc+ca+ab)}.\] [/quote]

It's easy S.O.S :lol:

Clearing the denominator, we get

$4(\sum_{cyc}a^4b^2+\sum_{cyc}a^4c^2)+4(\sum_{cyc}a^4bc)+4(\sum_{cyc}a^3b^3)\geq (\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2)+42a^2b^2c^2$

This holds by Muirhead inequality or AM-GM inequality :D

Notice that
$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$
is equivalent to $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ after homogenizing.
However $\frac{1}{2+x}+\frac{1}{2+\frac{1}{x}}=\frac{4+x+\frac{1}{x}}{5+2(x+\frac{1}{x})}=\frac{4+y}{5+2y}$. Yet $\frac{4+y}{5+2y}$ is montonically decreasing and $y \geq 2$, so $\frac{4+y}{5+2y} \leq \frac{2}{3}$. The result follows.

[quote=Litlle 1000t]Let $a, b, c$  be positive reals such that their sum is $1$. Prove that \[\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}.\][/quote]
[quote=sqing]Let $a, b, c$  be positive reals such that their sum is $1$. Prove that \[\frac{a}{bc+2a^{2}+2a}+\frac{b}{ca+2b^{2}+2b}+\frac{c}{ab+2c^{2}+2c}\geq \frac{1}{3(bc+ca+ab)}.\] http://www.artofproblemsolving.com/community/c6h610084p3626533[/quote]
$$ 2a^{2} + 2a + bc = 2a^{2} + 2(a + b + c)a + bc = (2a + b)(2a + c)$$
 $$\sum \frac{1}{bc+2a^{2}+2a}=\sum \frac{1}{(2a+b)(2a+c)},\sum \frac{a}{bc+2a^{2}+2a}=\sum \frac{a}{(2a+b)(2a+c)}$$

Let $a,b,c $  be positive reals . [url=https://artofproblemsolving.com/community/c6h1692708p10832736]Prove that :[/url]
$$\frac{1}{ab+bc+ca}\le \frac{1}{(2a+b)(2a+c)}+\frac{1}{(2b+c)(2b+a)}+\frac{1}{(2c+a)(2c+b)}\le \frac{1}{3\sqrt[3]{a^2b^2c^2}}$$
$$\frac{a}{(2a+b)(2a+c)}+\frac{b}{(2b+c)(2b+a)}+\frac{c}{(2c+a)(2c+b)}\le \frac{1}{a+b+c}$$

[quote=sqing]
$$\frac{a}{(2a+b)(2a+c)}+\frac{b}{(2b+c)(2b+a)}+\frac{c}{(2c+a)(2c+b)}\le \frac{1}{a+b+c}$$[/quote]

It is equivalent to $\sum (a-b)(a-c)\cdot \frac{1}{(2a+b)(2a+c)} \ge 0$.
Wlog $a \ge b\ge c$. We have $\frac{1}{(2a+b)(2a+c)}\le \frac{1}{(2b+c)(2b+a)}  \le \frac{1}{(2c+a)(2c+b)}$ which implies the result from generalization of Schur Inequality.

It is equivalent to
$$\frac{2(ab+bc+ca)}{ab+2c^2+2c}+\frac{2(ab+bc+ca)}{bc+2a^2+2a}+\frac{2(ab+bc+ca)}{ac+2b^2+2b}\ge 2$$
We have
$$\frac{2(ab+bc+ca)}{ab+2c^2+2c}=\frac{2(ab+bc+ca)}{ab+2c^2+2c(a+b+c)}=\frac{b(2c+a)+a(2c+b)}{(2c+a)(2c+b)}=\frac b{2c+b}+\frac a{2c+a}$$
Hence, we need to prove that
$$\frac a{2c+a}+\frac b{2c+b}+\frac b{2a+b}+\frac c{2a+c}+\frac a{2b+a}+\frac c{2b+c}\ge 2$$
Now, just see that
$$\frac a{2b+a}+\frac b{2a+b}=\frac {a^2}{2ab+a^2}+\frac {b^2}{2ab+b^2}\ge \frac{(a+b)^2}{a^2+b^2+4ab}\ge \frac 23$$
The rest follows.

Let $a,b,c$ positive reels such that $a+b+c=\frac{2}{3}$. Then prove the following

$$\frac{1}{3ab+4c^2+4c}+\frac{1}{3bc+4a^2+4a}+\frac{1}{3ca+4b^2+4b}\geq \frac{1}{3(ab+bc+ca)}$$

Let $a,b,c$ positive reels such that $a+b+c=\frac{7}{12}$. Then prove the following

$$\frac{1}{6ab+7c^2+7c}+\frac{1}{6bc+7a^2+7a}+\frac{1}{6ca+7b^2+7b}\geq \frac{1}{6(ab+bc+ca)}$$

Let $a,b,c$ positive reels such that $a+b+c=\frac{5}{8}$. Then prove the following

$$\frac{1}{4ab+5c^2+5c}+\frac{1}{4bc+5a^2+5a}+\frac{1}{4ca+5b^2+5b}\geq \frac{1}{4(ab+bc+ca)}$$

Let $a,b,c$ positive reels such that $a+b+c=\frac{3}{4}$. Then prove the following

$$\frac{1}{2ab+3c^2+3c}+\frac{1}{2bc+3a^2+3a}+\frac{1}{2ca+3b^2+3b}\geq \frac{1}{2(ab+bc+ca)}$$

[b]Generalization 1[/b]
Let $a,b,c$ positive reels and $k\geq 1$ such that $a+b+c=\frac{k+1}{2k}$. Then prove the following

$$\frac{1}{kab+(k+1)c^2+(k+1)c}+\frac{1}{kbc+(k+1)a^2+(k+1)a}+\frac{1}{kca+(k+1)b^2+(k+1)b}\geq \frac{1}{k(ab+bc+ca)}$$

[hide=understandable]
[b][color=red]1-)[/color] Solution[/b]

$$\sum_{cyc}{\dfrac{k(ab+bc+ca)}{kab+(k+1)c^2+(k+1)c}}\geq 1$$

$$\dfrac{k(ab+bc+ca)}{kab+(k+1)c^2+(k+1)c}\geq \dfrac{ab}{ab+bc+ca}$$
We should proof the below inequality

$$k(ab+bc+ca)^2=ka^2b^2+kb^2c^2+kc^2a^2+2kabc(a+b+c)\geq ka^2b^2+(k+1)abc^2+(k+1)abc$$
$$kb^2c^2+kc^2a^2+2kabc(a+b+c)\geq (k+1)(abc^2+abc)$$
Remember $a+b+c=\dfrac{k+1}{2k}$ and replace it
$$kb^2c^2+kc^2a^2+2kabc(a+b+c)\geq kb^2c^2+kc^2a^2+(k+1)(a+b+c)\geq (k+1)(abc^2+abc)$$
$$\rightarrow kb^2c^2+kc^2a^2\overbrace{\geq}^{AGO} 2kabc^2\geq (k+1)(abc^2)$$
$$2k\geq k+1$$
$k\geq 1$ which had given in the question. Proof completed

[quote=Megus]Observe that
$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$
is equivalent to (algebraic manipulations only, us condition $a+b+c=1$ to homogenize firstly)

$2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$
which is easy  :)[/quote]

[quote=arqady]Very nice proof, Megus![/quote]

Indeed


The [url=http://www.imo-official.org/year_reg_team.aspx?code=USA]USA IMO team[/url] got the following result at the 50th International Mathematical Olympiad (IMO) in Bremen, Germany. Students had to solve six problems each worth seven points giving a cumulative maxium of 42 points. The scores as follows:

[b]Name:[/b] [i]P1-P2-P3-P4-P5-P6[/i]:[b]Sum[/b]
[b]John Berman:[/b] 7-7-7-7-7-0:[b]35[/b]
[b]Wenyu Cao:[/b] 7-7-1-3-7-0:[b]25[/b]
[b]Eric Larson:[/b] 7-7-3-7-7-3:[b]34[/b]
[b]Delong Meng:[/b] 7-7-1-7-7-0:[b]29[/b]
[b]Evan O'Dorney:[/b] 7-7-7-2-7-0:[b]30[/b]
[b]Qinxuan Pan:[/b] 7-7-1-7-7-0:[b]29[/b]

The cut-offs for the respective medal ranks are due to get computed soon. There are rumours that the gold cut-off is at 29-30 points which implies three to five gold medals for the US team. Congrats everybody. Well done!  :)

Congratulations to everyone! I wonder what the cutoffs will be  :)


[b] A man turned off the light and went to bed. He woke up in the middle of the night, and he watched the news, discovering that 2 men had died. He went upstairs and he hung himself. why?? :maybe: [/b]

I've heard this one before.

None of it is a figure of speech. Wether he knew the 2 men that died is irrelevant, but he was responsible for their deaths.

Did he attempt to kill, or greatly weaken the men?

It seems like this one has many,many possible answers.
Besides, we know nothing of the temperament of the man, and some people would commit suicide over seeing two deaths on t.v. from something they feel they caused (like a factory owner lowering wages to watch his workers go on strike, in the strike a couple are killed, and the owner feels really guilty).

And my guess is that it has something to do with politics...

[quote="tjhance"][hide="guess"]
did he own a lighthouse, and by turning off the light, two men at sea were lost and could not get to safety?
[/hide][/quote]

I think that is the correct answer, but wait till klutzycherry14 agrees.


inequalities
calculus
algebra proposed
algebra

I know that it's an easy problem but I want to see another solution, thx  ;)

Find all positive integer $n$ s.t. for every $x\geq n$ the inequality holds:
$\sqrt {x-1}+\sqrt {x-2}+...+\sqrt {x-n} <x$

Hi,

For every $x \geq n$ ?, but for x = n LHS is $\geq 1+ \sqrt{2} + n/2 * \sqrt{n/2}$ (for $n \geq 4$) which is $\geq n$.
n=1,2 are checked easily and works.
n=3 need probably be checked with some calculus ?


I was pondering various things, as I am often wont to do, and somewhere in that series of random events came the question: why do the more 'complex' biological organisms tend to have clustered sensory organs?  Is there a reason for locating the eyes in such proximity to the brain?  The ears? The tongue?

I began to hypothesize about the data transmission, in regard to the eyes, but I've yet to do any real research.  I have my ideas about why the organs are clustered, but what are yours?  Do you know of any studies that attempt to answer this question?

Any ideas, however strange or underdeveloped, are welcome.

I have a simple idea. For faster data processing and quicker reactions based on input (you see a ball coming at you, so you catch it), the sensory organs are loacted near the brain. If we heard with our feet, the signals might take a longer time to travel to our brain (plus it would be very inconvenient).

[quote="i_like_pie"](plus it would be very inconvenient).[/quote]
Define this term please in an objective way ;)

Yes, signal speed is a good reason. Also, mouth and nose are of very similar origin, so they are close. I'm not sure, but this could also hold for eyes and ears since they need "holes", and thats something rather difficult to develop.

Organs are clustered because the need to transmit stuff (former meals for example) I think.

Though this is a late response, you guys pretty much hit the nail on the head insofar as my initial analysis goes.  Data transfer was the first thing I thought of, too.

While I don't believe in evolution, I think you have missed the boat.  The eyes and ears largest advantage being on the head may well be this is the highest point.  (Same for nose for smelling prey, food or danger earlier might be of some benefit.)  One might suggest that the real question is why the brain might have been designed to be in the head.  (Crabs have eyes on stalks, far from their brains.)  (The Greeks were convinced that reasoning was in the heart and the brain was a cooler for the blood.)

Part of the design for the location of the brain might well be heat loss.  (Think of a runner, who has muscles low to the ground and in the middle and has 1/3 of his blood flow going to the brain).  Where would you design the brain?

Think of the design for someone feeding on the beach, wading to eat food from the surf and trying to keep warm.  Where would you design the brain with 1/3 the blood flow to conserve heat?

God did a great job designing us.  We can even see some of the benefits.

Have a great day.


$S$ is a set with $2007$ elements of positive integers. Called 'not friendly' set such that  if it  contains $x,y$  it doesn't contain $x+y$. Prove that there is 'not friendly' set $X$ which is subset of $S$such that $\mid{X}\mid\ge{670}$

Posted before : http://www.mathlinks.ro/Forum/viewtopic.php?t=77010


Find the sum of all integers $ k$ such that $ \binom{23}{4} \plus{} \binom{23}{5} \equal{} \binom{24}{k}$.

We use $ \binom{n}{r} \equal{} \binom{n\minus{}1}{r} \plus{} \binom{n\minus{}1}{r\minus{}1}$.
Thus, $ k\equal{}\boxed{5}$.

[quote="Kouichi Nakagawa"]We use $ \binom{n}{r} \equal{} \binom{n \minus{} 1}{r} \plus{} \binom{n \minus{} 1}{r \minus{} 1}$.
Thus, $ k \equal{} \boxed{5}$.[/quote]

However, 19 also works.

$ \binom{24}{n}\equal{}\binom{24}{24\minus{}n}$, so the sum is 24.


function
number theory unsolved
number theory

Given any positive integer $ k$, prove that there exist infinitely many integers $ n$ such that
$ \mu(n\plus{}1) \equal{} \mu(n\plus{}2) \equal{} ... \equal{} \mu(n\plus{}k)$.

[hide="Solution"]
Set $ n \plus{} i \equiv 0\bmod p_i^2$, where the $ p_i$'s are the primes in increasing order (for example, $ p_1 \equal{} 2$, $ p_2 \equal{} 3$, etcetera).  Then, by the Chinese Remainder Theorem, we may find that $ n\equiv x\bmod (\prod_{i \equal{} 1}^k {p_i})^2$ which gives that\[ \mu(n \plus{} 1) \equal{} \mu(n \plus{} 2) \equal{} \cdots \equal{} \mu(n \plus{} k) \equal{} 0\]Clearly, there are infinitely many values of $ n$ that satisfy $ n\equiv x\bmod (\prod_{i \equal{} 1}^k {p_i})^2$, so we are done. [/hide]

[quote="The QuattoMaster 6000"][hide="Solution"]
Set $ n \plus{} i \equiv 0\bmod p_i^2$, where the $ p_i$'s are the primes in increasing order (for example, $ p_1 \equal{} 2$, $ p_2 \equal{} 3$, etcetera).  Then, by the Chinese Remainder Theorem, we may find that $ n\equiv x\bmod (\prod_{i \equal{} 1}^k {p_i})^2$ which gives that
$ \mu(n \plus{} 1) \equal{} \mu(n \plus{} 2) \equal{} \cdots \equal{} \mu(n \plus{} k) \equal{} 0$
Clearly, there are infinitely many values of $ n$ that satisfy $ n\equiv x\bmod (\prod_{i \equal{} 1}^k {p_i})^2$, so we are done. [/hide][/quote]


I could not solve the following problem from the Newman seminars. Does anyone here have the book/know a solution?

[b]Problem. [/b] Two players alternatively choose a binary digit $b_i$. The digits are used to construct a real number $0.b_1b_2b_3\ldots_{(2)}$. The first player wins if and only if $b$ is transcendental. Who has a winning strategy?

--Vesselin

I think the first player has a winning strategy.

There are only a countable number of algebraic numbers.  So the first player can use a diagonalization strategy to avoid them.  Namely, on his first turn, avoid the first algebraic number.  On his second turn, avoid the second algebraic number, etc.

One minor problem is that some numbers have two representations, such as 0.1 = 0.0111..., but that is easy to deal with.  Instead of listing the algebraic numbers, we can list all the representations of algebraic numbers, which is still countable.

[Edited to reverse first and second players.]

What's wrong with a diagonalization argument?

The first player can make an enumeration of the algebraics, which are countably many (written in binary), and, at step $n$, put a digit which does not coincide with the digit on the same position of the $n$'s number on the list.

I bet there's a huge mistake in there, but I'm tired, so be forgiving... :)

:blush:  :blush:  :blush:

Shame on me: it's a triviality; there's nothing wrong with this diagonalization argument.

On that note, I have the following conjecture that I believe is [i]much[/i] more difficult (the initial problem is a trivial consequence with $p=2$):

[b]Conjecture. [/b] Consider an integer $p \geq 2$ and a real number $x \in (0,1)$ with binary expansion $x=0.a_1a_2a_3\ldots_{(2)}$. If the number $0.a_{p}a_{2p}a_{3p}\ldots_{(2)}$ is transcendental, then $x$ is also transcendental.

--Vesselin


Find all functions $ f: \mathbb Z\to \mathbb Z$ such that
\[ f(x\minus{}y\plus{}f(y))\equal{}f(x)\plus{}f(y),\] for all $x, y \in \mathbb Z$.

After $ x \equal{} y$ we will have $ f(f(x)) \equal{} 2f(x)$.
After $ y\rightarrow f(y)$ we will have $ f(x \plus{} f(y)) \equal{} f(x) \plus{} 2f(y)$.
For $ x \equal{} 0$: $ f(f(y)) \equal{} f(0) \plus{} 2f(y)$, but $ f(f(y)) \equal{} 2f(y)$, so $ f(0) \equal{} 0$.

Now, we have two cases:
1. Suppose, there are exist $ a\neq b\in Z$ such that $ f(a) \equal{} f(b)$. So, we have:
$ f(x \minus{} a \plus{} f(a)) \equal{} f(x \minus{} b \plus{} f(b))$
So, $ f$ is periodic. $ f: Z\rightarrow Z$, and $ f$ - periodic, so there exist $ f_{max}$ and $ f_{min}$.
By taking $ x \equal{} y \equal{} u$ with $ f_{max} \equal{} f(u)$, we will have:
$ f(f(u)) \equal{} 2f_{max}$, so $ f_{max}\leq 0$.
By taking $ x \equal{} y \equal{} v$ with $ f_{min} \equal{} f(v)$, we will have:
$ f(f(v)) \equal{} 2f_{min}$, so $ f_{min}\geq 0$.
From $ f_{max}\leq 0$ and $ f_{min}\geq 0$ we have $ f(x) \equal{} 0$ for all $ x$ - it's solution.

2. From $ f(a) \equal{} f(b)$ we will have $ a \equal{} b$.
Put $ x \equal{} 0:$ $ f(f(y) \minus{} y) \equal{} f(y)$, so $ f(y) \minus{} y \equal{} y$, and $ f(x) \equal{} 2x$ is solution - it's easy to check.

[quote="Kirill"]lgmco, thanks to you. It's very nice Functional Equation.
Is there any mistakes in my solution (I hope, no)?[/quote]
Quite OK and nice solution!  :)
Congrats.

by simmetry : $ f(x\minus{}y\plus{}f(y))\equal{}f(y\minus{}x\plus{}f(x))$

$ y\equal{}0$ $ \implies$ $ f(x)\equal{}f(f(x)\minus{}x)$

so $ x,y\rightarrow  f(y)\minus{}y$ $ \implies$ $ f(x\minus{}f(y)\plus{}y\plus{}f(y))\equal{}f(x)\plus{}f(f(y)\minus{}y)\equal{}f(x)\plus{}f(y)$

so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$

and it is cauchy equation $ \forall x,y \in \mathbb{Z}$ so $ f(x)\equal{}ax$

replace in the original equation :

$ a(x\minus{}y\plus{}ay)\equal{}a(x\plus{}y)\implies a\equal{}2$

 :)


AMC
AIME
geometry
ARML
AMC 12
AMC 10
number theory
\/closed

We have tentative plans to run the following problem series that will straddle the new year:

AMC-10 on Wednesdays starting in November and ending in late January.
AMC-12 on Mondays starting in November and ending in late January.
AIME on Tuesdays starting in December and ending in early March.

I would like feedback on the following concerns:

Christmas and New Years day are on Saturdays.
Would students prefer to have class on Dec 20, 21, 27, and 28, or to skip those days (or just the later days), or have class during those days and begin the classes earlier?

I would mention that it IS okay to skip classes and review the transcripts on your own.  Some students do this from time to time anyway.  I am mainly concerned with students who feel uncomfortable missing those days due to family vacation and feel the need to attend every class.

[quote="joml88"]I'm not sure what my schedule will be for Christmas time but that looks fine for me.[/quote]

What looks fine, holding class those days or skipping them?

Oh sorry,  I meant to clarify that having classes those days was fine with me.

[quote="mathnovice"]What will the schedules be for the subject classes?

Also, are all the classes that will be run next year already listed?

I was hoping that there would be AMC/AIME Geometry and AMC/AIME Number Theory classes.[/quote]

Most of the Fall course schedules will be posted soon in the Enrollment area of the site.

We plan to offer a MATHCOUNTS/AMC Geometry course starting around the beginning of October.

Unfortunately we will not be offering a next level Number Theory course until at earliest some time in 2005.  How quickly that class gets written depends on a number of things that simply cannot yet be determined.

We stay in town for the holidays so any schedule at that time of the year works for us. I appreciate the excellent line-up of classes on days of the week other than Thursdays, which once again will be A+MATH's day for his curricular math course, which meets in the late afternoon to early evening at the University of Minnesota. He will be out of town for summer programs beginning tomorrow, so we won't be able to sign him up for AoPS classes until late August or so. The other participants here have certainly said many things about the AoPS classes that make us eager to try them at our first opportunity.

[quote="MithsApprentice"]This is like slightly off topic but could I request a ARML problem series class?[/quote]

That's a future possibility, but not a high priority.  I think students could prepare well for ARML through our subject courses and the AMC12 and AIME problem series.

We have at least a dozen courses that we think would be immediately more valuable.  I personally see the subject courses are better options for students and the problem series as "problem practice" and "filling gaps in knowledge".  The subject courses give more solid, in-depth coverage of problem solving methods.

Is the Olymp Geom going to be offered this summer?

Not this summer, but we will offer it this fall.

We will eventually, but unfortunately not this fall.

[quote="mathnovice"]Yeah, I think an AMC/AIME Geometry class would be more useful than a MC/AMC Geometry class because most people are already familiar with the basics. Reaching the next step is what is the most difficult.[/quote]

I think our MC/AMC Geometry class will cover many concepts useful all the way up through the AIME.  We do plan to offer a Geometry class between that class and the Olympiad class in the not too distant future, but it probably won't be offered before sometime in 2005.


A tank can be filled by a tap in exactly 5 hours ,A pipe at the bottom of the tank can fully empty the tank of water in exactly 10 hours .How much time would the tank take to fill to the Brim if both the taps were opened simultaneously. :)

These are fun problems :D

[hide]
Rate of fill = $\frac{1}{5}$
Rate of empty = $\frac{1}{10}$

Rate of fill-empty = $\frac{1}{5} - \frac{1}{10} = \frac{1}{10}$

Total time = $10$ hours.[/hide]

[quote="Devendra Patel"]buzzer11 ,this is the most commonest mistake we tend to make ,I did it myself[/quote]

how is he wrong???

Hi sheepwarrior
Since the pressure for water to drain out would be different at different levels of water in the tank ,we would not be able to find out the exact time for the tank to fill ,atleast mathematically .

if only you had given the hieght of the tank, i would be able to calculate the time it takes--with physics :lol: .....(assuming that the tank is not an irregular shaped object)


there is a G (intersection of a countable collection of open
sets)with $E \subset G$ ,$m^{*}(G-E)=0$prove that $E$is measurable

Since the outer measure of $G\setminus E$ is 0, it is measurable. Also $G$ is measurable. Thus $E=G\setminus(G\setminus E)$ is mesurable too.

[quote="lebesgue"]let $A$ be measurable set and outer measure of $A,B$ be equal then $B$ is measurable too?[/quote]
No. You know for every set $E$, there exists $G_\delta$ set $G$ containing $E$ such that $m^\ast(E) = mG$.


[quote="Farenhajt"]Solve in reals:

$9^{x}-6^{x}=4^{x+{1\over 2}}$[/quote]

[hide]$3^{2x}-3^{x}2^{x}-2^{2x+1}=0$
$(3^{x}+2^{x})(3^{x}-2^{x+1})=0$

The first factor will yield no solution as powers are always postive.

The second factor will yield $x=\frac{\ln{2}}{\ln{\frac32}}$[/hide]

[quote="SplashD"][quote="Farenhajt"]Solve in reals:

$9^{x}-6^{x}=4^{x+{1\over 2}}$[/quote]

[hide]$3^{2x}-3^{x}2^{x}-2^{2x+1}=0$
$(3^{x}+2^{x})(3^{x}-2^{x}+1)=$

The first factor will yield no solution as powers are always postive.

The second factor will yield $x=\frac{\ln{2}}{\ln{\frac32}}$[/hide][/quote]

There seems to be a typo in the second factor.


In a perpendicular triangle the perimeter is 60 and the altitude on the hypotenuse is 12. Then, the length of the hypotenuse is

$ \text{(A)}\ 24 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 28$


Find all functions $ f : \mathbb{R} \rightarrow \mathbb{R}$ such that
$ f(x)f(yf(x) \minus{}1) \equal{} x^2f(y) \minus{} f(x)$
for all real numbers $ x$ and $ y$.

Assume that $ f$ is not identically zero.

- For $ x\equal{}0$, the given equation leads to $ f(0)\equal{}0$ or $ f(yf(0)\minus{}1) \equal{} \minus{}1$. If $ f(0) \neq 0$ then, since $ yf(0)\minus{}1$ is an affine expression, we deduce that $ f(x)\equal{}\minus{}1$ for all $ x$. Back to the initial equation, we get a contradiction. Thus $ f(0)\equal{}0$.

- For $ y\equal{}0$ we then have $ f(x)f(\minus{}1)\equal{}\minus{}f(x)$ for all $ x$. Since $ f$ is not identically zero, it forces $ f(\minus{}1)\equal{}\minus{}1$.

- For $ y \equal{} f(b)$, we have $ f(x)f(f(b)f(x)\minus{}1) \equal{} x^2f(f(b))\minus{}f(x)$, so that $ f(b)f(x)f(f(b)f(x)\minus{}1) \equal{} x^2f(b)f(f(b))\minus{}f(x)f(b)$.
The LHS is symmetric in $ x$ and $ b$ then so must be the RHS. It follows that $ x^2f(b)f(f(b)) \equal{} b^2f(x)f(f(x)).$
For $ b \equal{} \minus{}1$, it follows that $ f(x)f(f(x)) \equal{} x^2$ for all $ x$.  (1)
Note that it forces $ f(x) \neq 0$ for all $ x \neq 0$.

- Back to the initial equation, let $ x\equal{}y\equal{}f(a)$ then $ f(f(a))f(f(a)f(f(a))\minus{}1)\equal{}f^2(a)f(f(a))\minus{}f(f(a))$ for all $ a$.
For $ a \neq 0$ we may divide by $ f(f(a))$ and using (1), it gives $ f(a^2 \minus{}1 ) \equal{} f^2(a)\minus{}1$. (2)
Note that (2) still holds for $ a\equal{}0$.

- Back to the initial equation, for $ y\equal{}f(x)$ we then have $ f(x)f(f^2(x)\minus{}1) \equal{} x^2f(f(x))\minus{}f(x)$
Therefore, using (2), we have $ f(x)(f^2(x)\minus{}1) \equal{} x^2f(f(x))\minus{}f(x)$
thus $ f^3(x) \equal{} x^2f(f(x))$.
Then, using (1), we have $ f^4(x) \equal{} x^2f(x)f(f(x)) \equal{} x^4$, which leads to $ f(x)\equal{}x$ or $ f(x)\equal{}\minus{}x$.

- Assume that $ x,y \neq 0$ with $ f(x)\equal{}x$ and $ f(y)\equal{}\minus{}y$ :
Using the initial equation, we get $ xf(xy\minus{}1)\equal{}\minus{}x^2y\minus{}x$
which is $ f(xy\minus{}1)\equal{}\minus{}xy\minus{}1$.
If $ f(xy\minus{}1)\equal{}xy\minus{}1$ then $ xy\equal{}0$, a contradiction.
If $ f(xy\minus{}1) \equal{} \minus{}xy\plus{}1$, then $ 1\equal{}\minus{}1$, which is absurd.

Therefore $ f(x)\equal{}x$ for all $ x$ or $ f(x)\equal{}\minus{}x$ for all $ x$. Since $ f(\minus{}1)\equal{}\minus{}1$ the latter case is impossible. The first case gives obviously a solution.

Then, the solutions are $ f(x)\equal{}0$ for all $ x$, or $ f(x)\equal{}x$ for all $ x$.

Pierre.

I realize that there is a mistake in the proof.

Pierre had
"- Back to the initial equation, for  we then have f(x)f(f^2(x)-1) = x^2f(f(x))-f(x)
Therefore, using (2), we have f(x)(f^2(x)-1) = x^2f(f(x))-f(x)
thus . "

But you can only conclude the following
"- Back to the initial equation, for  we then have f(x)f(f^2(x)-1) = x^2f(f(x))-f(x)
Therefore, using (2), we have f(x)((f(f(x)))^2-1) = x^2f(f(x))-f(x)
thus . "

I have a solution to the problem. I'll post it in a day or two.

[quote="nicetry007"]
Therefore, using (2), we have f(x)((f(f(x)))^2-1) = x^2f(f(x))-f(x)
thus . "

[/quote]

It seems to have to be $ f(x)(f^3(x)\minus{}1) \equal{} x^2f(f(x))\minus{}f(x)$ ?

You are right, there is a mistake...too much $ f$ for my head...

Let's give a correct solution (I hope) :

Assume that $ f$ is not identically zero.

- For $ x \equal{} 0$, the given equation leads to $ f(0) \equal{} 0$ or $ f(yf(0) \minus{} 1) \equal{} \minus{} 1$. If $ f(0) \neq 0$ then, since $ yf(0) \minus{} 1$ is an affine expression, we deduce that $ f(x) \equal{} \minus{} 1$ for all $ x$. Back to the initial equation, we get a contradiction. Thus $ f(0) \equal{} 0$.

- For $ y \equal{} 0$ we then have $ f(x)f( \minus{} 1) \equal{} \minus{} f(x)$ for all $ x$. Since $ f$ is not identically zero, it forces $ f( \minus{} 1) \equal{} \minus{} 1$.

- For $ y \equal{} f(b)$, we have $ f(x)f(f(b)f(x) \minus{} 1) \equal{} x^2f(f(b)) \minus{} f(x)$, so that $ f(b)f(x)f(f(b)f(x) \minus{} 1) \equal{} x^2f(b)f(f(b)) \minus{} f(x)f(b)$.
The LHS is symmetric in $ x$ and $ b$ then so must be the RHS. It follows that $ x^2f(b)f(f(b)) \equal{} b^2f(x)f(f(x)).$
For $ b \equal{} \minus{} 1$, it follows that $ f(x)f(f(x)) \equal{} x^2$ for all $ x$.  (1)
Note that it forces $ f(x) \neq 0$ for all $ x \neq 0$.

- For $ x\equal{}y\equal{}1$ in the initial equation, we have $ f(1)f(f(1)\minus{}1)\equal{}0$, with $ f(1) \neq 0$. Thus $ f(f(1)\minus{}1)\equal{}0$ which gives $ f(1)\equal{}1$.

- For $ x\equal{}1$, the initial equation gives $ f(y\minus{}1)\equal{}f(y)\minus{}1$ for all $ y$.  (2)

- From (2) and (1), we easily deduce that for all $ x$, we have
$ (a\plus{}1)^2 \equal{} f(a\plus{}1)f(f(a\plus{}1))$
$ \equal{}(f(a)\plus{}1)(f(f(a))\plus{}1)$
$ \equal{} f(a)f(f(a))\plus{}f(a)\plus{}f(f(a))\plus{}1$
$ \equal{} a^2\plus{}1\plus{}f(a)\plus{}f(f(a))$.
Thus $ f(a)f(f(a)) \equal{} a^2$ and $ f(a)\plus{}f(f(a)) \equal{} 2a$, which forces $ f(a)\equal{}a$.

Conversely, the solutions are $ f$ identically zero or $ f(x)\equal{}x$ for all $ x$.

Pierre.

Find f(x) have f'(x) continuous in R satisfying:
      $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}2xy \forall x;y\in R$

Pierre,

        Thats a nice solution!! much better than mine.

[quote="karis"]Find f(x) have f'(x) continuous in R satisfying:
      $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \plus{} 2xy \forall x;y\in R$[/quote]
Setting $ g(x)\equal{}f(x)\minus{}x^2$ we have $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)\,\, \forall x,y\in\mathbb{R}$.

[quote="karis"]Find f(x) have f'(x) continuous in R satisfying:
      $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \plus{} 2xy \forall x;y\in R$[/quote]

What does it here? :wink:

Pierre.

N T Tuan has solved it. Nice one N T Tuan!! Since $ f'(x)$ is continuous, $ f(x)$ is continuous. This implies $ g(x)$ is continuous. The fact that $ g(x)$ satisfies the cauchy equation in turn implies that $ g(x) \equal{} g(1)x \equal{} cx$.

Hence, the solution to the problem is $ f(x) \equal{} x^2 \plus{} cx$.

[quote="pbornsztein"]You are right, there is a mistake...too much $ f$ for my head...

Let's give a correct solution (I hope) :

Assume that $ f$ is not identically zero.

- For $ x \equal{} 0$, the given equation leads to $ f(0) \equal{} 0$ or $ f(yf(0) \minus{} 1) \equal{} \minus{} 1$. If $ f(0) \neq 0$ then, since $ yf(0) \minus{} 1$ is an affine expression, we deduce that $ f(x) \equal{} \minus{} 1$ for all $ x$. Back to the initial equation, we get a contradiction. Thus $ f(0) \equal{} 0$.

- For $ y \equal{} 0$ we then have $ f(x)f( \minus{} 1) \equal{} \minus{} f(x)$ for all $ x$. Since $ f$ is not identically zero, it forces $ f( \minus{} 1) \equal{} \minus{} 1$.

- For $ y \equal{} f(b)$, we have $ f(x)f(f(b)f(x) \minus{} 1) \equal{} x^2f(f(b)) \minus{} f(x)$, so that $ f(b)f(x)f(f(b)f(x) \minus{} 1) \equal{} x^2f(b)f(f(b)) \minus{} f(x)f(b)$.
The LHS is symmetric in $ x$ and $ b$ then so must be the RHS. It follows that $ x^2f(b)f(f(b)) \equal{} b^2f(x)f(f(x)).$
For $ b \equal{} \minus{} 1$, it follows that $ f(x)f(f(x)) \equal{} x^2$ for all $ x$.  (1)
Note that it forces $ f(x) \neq 0$ for all $ x \neq 0$.

- For $ x \equal{} y \equal{} 1$ in the initial equation, we have $ f(1)f(f(1) \minus{} 1) \equal{} 0$, with $ f(1) \neq 0$. Thus $ f(f(1) \minus{} 1) \equal{} 0$ which gives $ f(1) \equal{} 1$.

- For $ x \equal{} 1$, the initial equation gives $ f(y \minus{} 1) \equal{} f(y) \minus{} 1$ for all $ y$.  (2)

- From (2) and (1), we easily deduce that for all $ x$, we have
$ (a \plus{} 1)^2 \equal{} f(a \plus{} 1)f(f(a \plus{} 1))$

$ \boxed{ \equal{} (f(a) \plus{} 1)(f(f(a)) \plus{} 1) \equal{} f(a)f(f(a)) \plus{} f(a) \plus{} f(f(a)) \plus{} 1}$

$ \equal{} a^2 \plus{} 1 \plus{} f(a) \plus{} f(f(a))$.
Thus $ f(a)f(f(a)) \equal{} a^2$ and $ f(a) \plus{} f(f(a)) \equal{} 2a$, which forces $ f(a) \equal{} a$.

Conversely, the solutions are $ f$ identically zero or $ f(x) \equal{} x$ for all $ x$.

Pierre.[/quote]

 [i]Sorry for my stupidity but  Pierre , Can you explain how you get $ f(f(a \plus{} 1)) \ \equal{} \ f(f(a)) \plus{} 1$ :lol: ?[/i]

From Pierre's solution, we have $ f(y\minus{}1) \equal{} f(y) \minus{}1 \Rightarrow f(y) \equal{} f(y\minus{}1) \plus{} 1$.
Using this repeatedly, we get $ f(f(a\plus{}1)) \equal{} f(f(a) \plus{} 1) \equal{} f(f(a)) \plus{} 1$


geometry
rectangle
perimeter
algebra
polynomial
3D geometry
trigonometry

tokenadult or any of the other moderators feel free to move this wherever you deem fit....these questions are pretty simple

post solutions in spoiler please

1. The lengths of two sides of a right triangle are 3 and 4. What is the smallest possible value of the third side?

2. How many real solutions does the equation 2003^x=2004^(2x) have?

3. Kate drives to work at an average speed of 50 miles per hour. On her way home, she drives at an average speed of 60 miles per hour. What was her average speed for the round trip to work and back, to the nearest tenth?

4. A rectangle has area 9 square inches. The diagonals of the rectangle intersect at a 60 degree angle. What is the length of the longer side of the rectangle?

5. What is the ones digit of 7^2004?

6. If sin(A)=2cos(A), then what is the value of sin(A)cos(A)?

7. If f(x)=ax+b and f(f(f(x)))=8x+21, then what is the value of a+b?

8. Given that a+b=12 and a :^2: +b :^2: =250, what is the value of a :^3: +b :^3: ?

9. A regular pentagon has perimeter 50 in. What is its area?

10. What is the sum of the reciprocals of the roots of the polynomial x :^4:-11x :^3: +28x :^2: +36x-144?

11. Two circles are concentric. The area between the two circles is 75 :pi: . If the radius of the large circle is twice the radius of the small circle, what is the radius of the large circle?

12. Anne's sock drawer contains 13 blue socks, 6 red socks, and 8 white socks. In the dar, Anne cannot see the colors of the socks. How many socks would she have to take from the drawer to guarantee that she had a matched pair of any color?

13. Let f(x)=x/(x+1). If f^n(x) is the result of composing f with itself n times, compute f^2004(2). (Note: f^1(x)=f(x).)

14. For how many integers n is the expression (n :^2: +n)/(n+4) an integer?

15. Evaluate: 2004 :^2: -2003 :^2: +2002 :^2: -2001 :^2: +...+4 :^2: -3 :^2: +2 :^2: -1 :^2:

16. The numbers 1, 2, 3,..., 2004 are placed in order around a circle. We go around the circle marking every 15th number (starting with 1, 16, 31, etc.) until we mark some number twice. What is the smallest unmarked number once this procedure is complete?

17. What is the largest prime factor of 1 :^3: +2 :^3: +3 :^3: +...+2004 :^3: ?

18. How many subsets does the set {{1, 2, 3,} 4, 5} have?

19. What is the real part of e^(5i :pi: /6)?

20. Four pieces of wire 12 inches long are bent into a square, a circle, an equilateral triangle, and a regular hexagon. Which of these shapes has the largest area?

21. The number 2004! ends in how many zeroes?

22. What is arcsin(cos(17 :pi: /6))?

23. You visit Fibber Island, an island filled with inhabitants that always answer question with "yes" or "no", and always lie if possible. If they cannot lie, the inhabitants simply do not answer. To which of the following questions must an inhabitant always answer "yes"?

A. Do you live on Fibber Island?
B. Is the answer to this question "yes"?
C. Can you ever say "I am lying"?
D. Are there questions that you cannot answer?
E. None of the above

24. Sam and Frodo are at the foot of Mount Doom. The volcano is shaped like a circular cone with height 4000 feet and the diameter of the base is 6000 feet. Sam idly wonders what the surface area of Mount Doom is. Help him out.

I'll only post my answers for some of the questions to give others a chance.  I also don't have time to do all of them and some of them might take me a while. I might post some more later though after other ppl have had a chance.

[hide]1. If you let 4 be the hypotenuse you get x :^2: +9=16

x :^2: =7   so the other side is equal to  :sqrt: 7

2.  It has one solution when x is equal to 0.  I don't believe that there would be any when x is positive because 2003^x will never "catch up to" 2004^2x.  I believe that there will be one negative solution so that adds up to 2 solutions.

15.  If you factor every two as the difference of two squares you get (2004-2003)(2004+2003)+(2002-2001)(2002+2001)...(2+1)(2-1)

The x-y part of every factorization equals one so your left with 2004+2003+2002...2+1  (2005)(2004)/2=2009010

22.  If you subtract 2 :pi: from 17 :pi: /6 you get arcsin cos5 :pi: /6 because they are coterminal angles.  cos 5 :pi: /6=- :rt3: /2  Since I think that arcsin is restricted to the first 360 :^o:   the answers are 240 :^o:  and 300 :^o: I am not very sure about this one though.  [/hide]

Most of the questions sure look like Getting Started level to me. Have at it, anyone who is a beginner in problem-solving.

Quote: 20. Four pieces of wire 12 inches long are bent into a square, a circle, an equilateral triangle, and a regular hexagon. Which of these shapes has the largest area?
[hide] Square: Each side would be 3 inches long; the area would be 3x3 = 9
Circle: The circumference is 12, the radius is 12/2:pi: =6/:pi:, so the area is :pi:(6/:pi:):^2: = 36/:pi: :approx: 11.46
Triangle: Each side is 4 inches, the area is .5(4)(2:rt3:)=4:rt3: :approx: 6.93
Hexagon: Each side is 2 inches, the area is .5(2)(:rt3:)(6) = 6 :rt3: :approx: 10.39
The one with the largest area is the circle. [/hide]

Quote:23. You visit Fibber Island, an island filled with inhabitants that always answer question with "yes" or "no", and always lie if possible. If they cannot lie, the inhabitants simply do not answer. To which of the following questions must an inhabitant always answer "yes"?

A. Do you live on Fibber Island?
B. Is the answer to this question "yes"?
C. Can you ever say "I am lying"?
D. Are there questions that you cannot answer?
E. None of the above

Ooh, let me try this one! Knowing me, I got it wrong, but hey.
[hide]A. The inhabitants would answer this question "no", since they have to lie.
B. The inhabitants can't answer this question; answering "yes" or "no" would mean telling the truth
C. I think I interpreted this wrong, but the inhabitants can't say they're lying because that would be telling the truth, so the truthful answer to this question would be "no"; therefore, they have to lie and answer "yes".
D. They would answer "no" to this question.
So I think they have to answer "yes" to C, but I'm not sure. I kept getting mixed up. [/hide]
 [/hide][/quote]

[quote]
19. What is the real part of e^(5i  /6)? [/quote]

[spoiler]I'm not really sure where to go with this one.  I know that e^pi*i is equal to -1.  so I changed the expression to (e^ :pi:*i)^5/6 So that is (-1)^5/6  I believe this means the real is 0 but someone help me please.

joml88: [hide] use the fact that e^i :theta: =cos :theta: +i(sin :theta: )

[/hide]

[hide] 12. Wouldn't this just be 4 socks?  In a worse case scenario, she picks 1 blue sock, 1 red sock, and 1 green sock; if she picks 1 more sock, she'll have a pair. [/hide]

Thanks for the tip.

[hide]so that would make it cos(5 :pi: /6) + i*sin(5 :pi: /6). The real part is the cos (5 :pi: /6)  I like degrees better than radians. So 5 :pi: /6 is equal to 150 :^o:  150 :^o: is equal to - :rt3: /2 (- because it is in the 2nd quadrant)[/hide]

BHorseMath wrote:[hide] number six is 0.  Not sure how I got it...[/hide]

How can you not be sure how you got it? You must have had some reasoning  But if you think about what the question is, you can see that it must be the wrong answer..

I'm not that great at math.  I can't ever tell how I get anything unless it's really simple.  Yea my answers wrong I have no idea how to fix it but gosh how dumb can I get???? :oops:

[quote="s0mp"]6. If sin(A)=2cos(A), then what is the value of sin(A)cos(A)?[/quote]

[tex]\displaystyle{\sin{\theta}=2\cos{\theta}}[/tex]
[tex]\displaystyle{\sin^2{\theta}-4\cos^2=0{\theta}}[/tex]
[tex]\displaystyle{\sin^2{\theta}+\cos^2{\theta}-5\cos^2{/theta}=0}[/tex]
[tex]\displaystyle{1=5\cos^2{\theta}}[/tex]
[tex]\displaystyle{\frac{1}{\sqrt{5}}=\cos{\theta}}[/tex]

From the original question, we see
[tex]\displaystyle{\frac{\sin{\theta}}{2}=\cos{\theta}}[/tex]
[tex]\displaystyle{\frac{\sin{\theta}*\cos{\theta}}{2}=\cos^2{\theta}=1/5}[/tex]

Therefore, [tex]\displaystyle{\sin{\theta}*\cos{\theta}=2/5}[/tex]
P.S.  My first time using Tex!  Yay.  Man, what a pain in the arse.

s0mp wrote:
10. What is the sum of the reciprocals of the roots of the polynomial x :^4:-11x :^3: +28x :^2: +36x-144?

[hide]Let's call the roots a and b. We are trying to find 1/x  +  1/y.  We can combine that into (x+y)/xy.  The sum of the roots of any polynomial is    -b/a  where a is the coefficient of the highest power term and b is the coefficient of the 2nd highest.  In this case it is 11.  The product is c/a c is the constant.  So it is -144 here.  Therefore the answer is -11/144[/hide]

[quote="ComplexZeta"]But you did some work to get that answer. I presume answers to problems don't just come to you from a supernatural source (hm that might be convenient), so you should explain what you did to get that answer, regardless of how crazy it may seem.[/quote]

Agreed. Either 1) you were sitting looking at the equation and suddenly the number 0 came into your head, or 2) you wrote down something that helped give you the idea of 0. I'm guessing you are under number 2, so you could just write that in here :)

joml88 wrote:10.  [hide]Therefore the answer is -11/144[/hide]

No.  Your method is on the right track, but somewhat misappplied.

How could I forget that it was a fourth degree? :-x

3) I've done this problem before, but they sort of gave the time, which made it really easy.  Let's see, I know it's definitely not 55. So...

[hide] Distance remains constant, since she's driving the same route and everything. So using d = rt,

d = 50t  on the way there and d = 60 (5/6)t

Set them equal to solve for t: 50t = 60 (5/6)t, t = 1 (did I need to do this step or could I have plugged in a random distance for d?)

Since t = 1, d = 50 miles.  Her total time was 1 + (5/6) = 11/6 hrs, total distance was 100 miles (50 there, 50 back).

So, 100 = r (11/6), and r = 600/11  :approx: 54. 5 mph.

I think you can just plug in a random distance for d, but I didn't realize that until I was doing it. *shrugs* [/hide]

trig_is_fun88 wrote:[hide]
I think you can just plug in a random distance for d, but I didn't realize that until I was doing it. *shrugs*[/hide]

[hide]Yeah, I think it's a little more respectable or whatever to do it algebraically, but for me it's a little quicker and easier to choose a nice value for d to work with.  For this problem, I'd choose something like 300.  This quickly yields times of 6 and 5 hours, with a total distance of 600 miles, so 600/11=54.5, as you found.[/hide]

Try closing your spoiler tags.  [hide]Like this.[/hide]

10. What is the sum of the reciprocals of the roots of the polynomial x^4 - 11x^3 + 28x^2 + 36x - 144?

If this was not already correctly answered, here's my solution:

[hide]f(x) = x^4 - 11x^3 + 28x^2 + 36x - 144.  Let g(x) = f(1/x).  The roots of g(x) are the inverses of the roots of f(x).

g(x) = f(1/x) = 1/x^4 - 11/x^3 + 28/x^2 + 36/x - 144.

The roots don't change if we multiply through with x^4, so it becomes

-144x^4 + 36x^3 + 28x^2 - 11x + 1

Divide through by -144 to get

x^4 - x^3/4 - 7x^2/36 + 11x/144 - 1/144

When expanding (x-a)(x-b)(x-c)(x-d), the x^3 coefficient is -(a+b+c+d).  a, b, c, and d are obviously roots of their expanded polynomial, so the sum of the roots is the -1 times the x^3 coefficient for any 4th degree polynomial.  This 4th degree polynomial's coefficient is -1/4, and -1 times that is 1/4.  Therefore the sum of the roots of g(x) is 1/4.  The roots of g(x) are the reciprocals of the roots of f(x), so it follows that 1/4 is the sum of the inverses of the roots of f(x).[/hide]

A simpler solution to #10.

[tex]\displaystyle{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{acd+abd+abc+bcd}{abcd}}[/tex]

According Newton's laws, we see that the numerator is equal to 36 and the denominator is equal to 144.

[tex]\displaystyle{36/144=1/4}[/tex]

[quote="Magnara"]
According Newton's laws, we see that the numerator is equal to 36 and the denominator is equal to 144.
[/quote]

What is Newton's law?

It's a fancy way of saying something you already know:  if you write the polynomial as a product of linear terms, you get expressions for the coefficients in terms of the roots.  That's all he's using.

How exactly do you get that abc+abd+acd+bcd=36?

Look at the product (x - a)(x - b)(x - c)(x - d).  Since we have a monic polynomial, this will represent our polynomial.  Now, when we multiply it out, abc + abd + acd + bcd will be multipled by -x -- so it's the negative of the coefficient of the x term of the polynomial, or in this case -36.

It says that for a degree 4 equation with roots abcd, you get:
x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd

Btw, it's definately something you should learn for math contests.  It comes up a lot.  [url]http://mathworld.wolfram.com/VietasFormulas.html[/url].  They accredit it to Vieta here, but I've always seen Newton.  Same idea though.[/url]

s0mp wrote:
7. If f(x)=ax+b and f(f(f(x)))=8x+21, then what is the value of a+b?

[hide]Plug ax+b in for f(x). We get f(f(ax+b)).  f(ax+b)=a(ax+b)+b=a :^2: x +ab  +b

Put this in the function again a(a :^2: x +ab+b) + b = 8x+21

a :^3: x + a :^2: b +ab+b = 8x + 21

a :^3: = 8 so a = 2

a :^2: b + ab+b =21

Plug a=2 in and you get 4b+2b+b=21

a=2

b=3

a+b=5[/hide]

s0mp wrote:5. What is the ones digit of 7^2004?

[hide]Look at the powers of 7 mod 10.

7

9

3

1

7

9

Ok, well, that was easy, it started repeating nice and early.  So, we do 2004/4=501 with no remainder, which means that 7^2004  :equiv: 1 mod 10, so the last digit is [tex]\displaystyle{1}[/tex][/hide]

s0mp wrote:4. A rectangle has area 9 square inches. The diagonals of the rectangle intersect at a 60 degree angle. What is the length of the longer side of the rectangle?

hmm... I'm not sure of this one.  The answer I got is weird.

[hide]If you draw the diagonals you get an equilateral triangle with one side being the small side of the rectangle.  The area of this is 9/4 or x :^2: * :rt3: /4.

 So 9/4=x :^2: *:rt3: /4.

9=x :^2: :rt3:

x= :sqrt: 9/ :rt3:

this times the longer side equals 9

the longer side equals 3* :sqrt:  :sqrt: 3[/hide]

Joml88, when you say [hide]:sqrt::sqrt:3[/hide], do you mean 3^(1/4)? Sorry, notation was a bit confusing.

If so, I got the same answer as you, albeit written a bit differently: [hide]3^(5/4)[/hide].  The only part I wasn't sure about was if the diagonals divided the rectangle into triangles of equal area; I did a few trials and they seemed to, though.

Try drawing heights of the triangles from the center (where the diagonals meet; the point that is in all 4 triangles) and finding their areas in terms of the area of the whole rectangle, and see what you get :)

Oh, I get it now! :oops:  I'm so stupid.

joml88 wrote:s0mp wrote:17. What is the largest prime factor of 1 :^3: +2 :^3: +3 :^3: +...+2004 :^3: ?

[hide]In another thread they talked about how 1 :^3: +2 :^3: ... + n :^3: =(1+2+...+n) :^2:
So if we apply that we get (1+2+3+...+2004) :^2:

The sum of 1 to 2004 is 2005(2004)/2
Which equals 2005*1002
1002 breaks into 167*3*2
2005 into 5*401
401 is the largest.[/hide]

Oooh, nice.


geometry
circumcircle
function
national olympiad

The problems have been posted in ML 2 months ago, but I still can't find the problems in resource.  :D

[b]DAY ONE.[/b]

Problem 1. Given triangle $ ABC$. Points $ D,E,F$ outside triangle $ ABC$ are chosen such that triangle $ ABD$, $ BCE$, and $ CAF$ are equilateral triangles. Prove that the circumcircles of these three triangles are concurrent.

Problem 2. Prove that for $ x$ and $ y$ positive reals,
$ \frac {1}{(1 \plus{} \sqrt {x})^2} \plus{} \frac {1}{(1 \plus{} \sqrt {y})^2} \ge \frac {2}{x \plus{} y \plus{} 2}$.

Problem 3. Find all natural number which can be expressed in
$ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b}$
Where $ a,b,c$ are positive integers satisfying
$ gcd(a,b) \equal{} gcd(b,c) \equal{} gcd(c,a) \equal{} 1$.

Problem 4. Let $ A \equal{} {1,2,. . .,2008}$
(a) Find the number of subset of $ A$ which satisfy: the product of its elements is divisible by $ 7$.
(b) Let $ N(i)$ denotes the number of subsets of $ A$ which sum of its elements remains $ i$ when divided by $ 7$. Prove that $ N(0) \minus{} N(1) \plus{} N(2) \minus{} N(3) \plus{} N(4) \minus{} N(5) \plus{} N(6) \minus{} N(7) \equal{} 0$.

[b]DAY TWO.[/b]

Problem 5. Let $ m,n > 1$ are integers which satisfy $ n|(4^m \minus{} 1)$ and $ 2^m|(n \minus{} 1)$. Is it a must that $ n \equal{} 2^m \plus{} 1$?

Problem 6. In a group of $ 21$ persons, every two person communicate with different radio frequency. It's possible for two person to not communicate (means there's no frequency occupied to connect them). Only one frequency used by each couple, and it's unique for every couple. In every $ 3$ persons, exactly two of them is not communicating to each other. Determine the maximum number of frequency required for this group. Explain your answer.

Problem 7. Given triangle $ ABC$ with sidelengths $ a,b,c$. Tangents to incircle of $ ABC$ that parallel with triangle's sides form three small triangle (each small triangle has 1 vertex of $ ABC$). Prove that the sum of area of incircles of these three small triangles and the area of incircle of triangle $ ABC$is equal to
$ \frac {\pi(a^2 \plus{} b^2 \plus{} c^2)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}{(a \plus{} b \plus{} c)^3}$.

Problem 8. Find all function $ f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying
$ f(mn) \plus{} f(m \plus{} n) \equal{} f(m)f(n) \plus{} 1$
for all $ m,n \in \mathbb{N}$.

I don't understand how to attach the file  :blush: , so I just uploaded it in Rapidshare.  :lol:

http://rapidshare.com/files/149346417/OSN_2008.pdf.html

Bye.

Edited: my grammar  :blush: . sorry for some mistakes.  :lol:

If you or anybody else wants to have added problems to the resources section please pm a moderator like me. Here is an example of what you should be writing us:

Problem 1: URL
Problem 2: URL
Problem 3: URL
Problem 4: URL
Problem 5: URL
Problem 6: URL


If you remember the formula to compute the number of divisors :huh: , then:
[hide]60 = (2^2)(3^1)(5^1)
Number of divisors= (2+1)(1+1)(1+1) = 12 divisors.
You can pair up each divisor with another divisor to equal 60:
1x60=60
2x30=60
3x20=60
4x15=60
5x12=60
6x10=60
If you multiply everything on the LH side, you get the product of the divisors, and the right hand side = 60^6.
The answer is 60^6. [/hide]

[hide]The formula for finding the product of a number's divisors is $n^{\frac{t(n)}{2}}$, where $t(n)$ denotes the number of divisors the number has.

$60=2^{2}\cdot 3 \cdot 5$, so it has $(2+1)(1+1)^{2}$ divisors. $60^{\frac{12}{2}}=\boxed{60^{6}}$.[/hide]

[quote="ch1n353ch3s54a1l"]What is the product of the divisors of 60?[/quote]
46656000000 :D

[hide]
There are 6 ways to get 60 with two numbers's product.
So, $60^{6}=\boxed{46656000000}$.[/hide]

[hide][quote="ch1n353ch3s54a1l"]What is the product of the divisors of 60?[/quote]

1*2*3*4*5*6*10*12*15*20*30*60
7200*300*1440
=some big number[/hide]

[quote="jackdillon"]Its 60^6

The message is too small. Please make the message longer before submitting.[/quote]

Dude, post solutions.  :wink:

[quote="jackdillon"]Its 60^6

The message is too small. Please make the message longer before submitting.[/quote]
1) Don't revive old topics.
2) Post solutions explaining how you solved the problem.
3) Hide your answers/solutions.


function
topology
real analysis
real analysis unsolved

Suppose that $f$ is a real function such that for a certain $n$ the function $f^n$ (that is the n-th iterate) is a contraction. Does it follow that f has a fixed point?

$f^n$ has a unique fixed point $x_0$.
$f^n(f(x_0))=f(f^n(x_0))=f(x_0)$. So $f(x_0)$ is a fixed point of $f^n$. But this fixed point is unique.
So $f(x_0)=x_0$ ?

Theorem: Let $X$ be a complete metric space, let $J$ be a positive integer, choose $0<M<1$, and let $T$ be a self-map of $X$ satisfying $\min_{1\le k\le J}(d(T^kx,T^ky))\le Md(x,y)$. Then $T$ has a unique fixed point, and for any $x$, some subsequence of $T^nx$ converges to the fixed point.

[url=http://www.ams.org/proc/2002-130-04/S0002-9939-01-06169-X/home.html]Reference[/url]:
An application of Ramsey's Theorem to the Banach Contraction Principle

Author(s): James Merryfield; Bruce Rothschild; James D. Stein Jr.
Journal: Proc. Amer. Math. Soc. 130 (2002), 927-933.

(This reference has the additional hypothesis that $T$ is continuous. The full version is published in a more obscure journal, and I can't find it online)


Does anyone know how to make custom images in mathematica? Can you make say, a circle with radius 2 with a square inscribed inside?

In Mathematica?  You can graph functions and use the "show" command to stick them all into one image, but that can get tiresome pretty quickly.


f(x1,x2)=3/11*(5*x1+x2) for x1>0,x2>0,x1+2*x2<0
f(x1,x2)=0 elsewhere
how to find the distribution function?
please showthe steps....Thanks

How can $x_1+2*x_2<0$ if $x_1>0$ and $x_2>0$?  Or am I misreading this problem...


The aim of this game is to form the fractional part of a positive real number.
Successively, I and You , give decimal figures : I give a decimal figure, You give a decimal figure, I give a decimal figure , and so on …
I win if the limit number is rational else you are the winner .
Could You win against me ?!  :D  :D
.........

depends, how long does the game last?

also what is a fractional number exactly :?

It's an infernal game, so you have to spend infinite time !!

If I give the decimals of a known irrational number (pi or simpler  \sum 10^-n! (or an other Liouville's), ...) the result number must be irrational because the decimal can't be periodical ??

Looks a bit too obvious as solution but I don't see the flaw.


[geogebra]fc88c82aa049ac34f0a6d1106899555f6279d3c7[/geogebra]

Sorry for the bad picture...
Ok, so there is a pivoting arm attached to the ground
and there is this cable, one end attached to the ground at the left and the other end attached to a weight.
When the arm moves left, it raises the weight; when it moves right it lowers it.

I don't know them right now but eventually I will know what the weight, distance between the pivot point and the point where the cable touches the ground, and the length of the arm are, so we can just leave them as variables for now.
What I want to know is the force on the cable left of the arm when given the angle of the arm.

You should try the physics forum. But there is a the weight is exerting force (N) on the cable due to gravity, 9.8x weight(grams). And the angle, you will have to multiply by I think cos(angle).. Not sure though.. Again, try the physics forum.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hello, I found an exercise in my notes (which can be used for the problem I posted in the topic "reducible poly") which asks something about how many squares are in $ \mathbb F_{p^n}$ as $ p>2$.

Say $ p^n\equal{}q$. I considered the application $ \Phi: F_q^* \rightarrow F_q^*$ defined as $ \Phi(a) \equal{} a^2$. It is well defined and is a group homomorphism from and to the multiplicative group of the field.
We have $ K\equal{}\ker \Phi \equal{} \{square\ roots \ of\ 1\ in\ F_q\} \equal{} \{\pm 1\}$.

Say $ Im(\Phi)\equal{}: S \leq F_q^*$ the subgroups of the squares of something, we have for homomorphism theorem \[ F_q^*/K \simeq S\] and considering orders we get \[ \frac{q\minus{}1}{2} \equal{} |F_q^*/K| \equal{} |S|\]

But the exercise asks to prove that $ |S| \equal{} \frac{q\plus{}1}{2}$. Is this a mistake or something's wrong in my proof?

Probably the author of the exercise counts $ 0$ to $ S$ as well (after all, $ 0$ is a square, just not a square of a unit).

  darij

yes you're right because I defined S as a sub[i]group[/i] of $ \mathbb F_q^*$, so I must add the zero and $ \frac {q \minus{} 1}{2} \plus{} 1 \equal{} \frac {q \plus{} 1}2$
thanks anyway :)


analytic geometry
calculus
integration
combinatorics unsolved
combinatorics

n points in the plane are such that for any three points we can find a cartesian coordinate system in which the points have integral coordinates. Show that there is a cartesian coordinate system in which all n points have integral coordinates. :D

Don't you read the other posts in the forum?  Two days ago: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199142

Sorry, in fact I've seen that one. It is not solved there...


[i]In how many ways can one select 3 different integers ranging from 1 to 30 inclusive so that their sm is a multiple of 3 (order of seclection does not matter)?

(A) 3160
(B) 1360
(C) 1240
(D) 1353
(E) 3240

[hide="Answer"]Answer is B[/hide][/i]
This appeared on the 1985 Australian Maths Competition Intermediate Section. Please tell me how to do it and try not to use advanced theorems because it was assumed that this could be done with normal grade 9 syllabus. Thanks.

Divide the numbers up into $ 3$ sets:

$ \{1,4,7,........,28\}$
$ \{2,5,8,........,29\}$
$ \{3,6,9,........,30\}$

Where the first set is congruent to $ 1$ modulo $ 3$ and the second is congruent to $ 2$ and the third $ 0$. So we can pick $ 3$ numbers from one set, $ 2$ from one set and a third from another or pick $ 1$ number from each set. Obviously the second case never give us a number divisible by $ 3$. So the total number of ways are:

$ \binom{10}{3}\times 3+[\binom{10}{1}]^{3}= 1360$

Answer : $ \boxed{B}$.


ratio
number theory proposed
number theory

For a real $\theta$, let $\| \theta \|$ denote the distance of $\theta$ to the nearest integer.
Prove that \[{\begin{eqnarray*}\sum_{n = 1}^{\infty}\| \varphi^{n}\| & = & \varphi\\ \sum_{n = 1}^{\infty}\| \varphi^{n}\|^{2}& = & \varphi-1 \end{eqnarray*}\] Notice that $-(\varphi-1)$ is a (Galois) conjugate of $\varphi$ !
EDIT: For the ones who didn't read the title (e.g Rust) by $\varphi$ I mean the Golden ratio.

[quote="{x}"]For a real $\theta$, let $\| \theta \|$ denote the distance of $\theta$ to the nearest integer.
Prove that \[{\begin{eqnarray*}\sum_{n = 1}^{\infty}\| \varphi^{n}\| & = & \varphi\\ \sum_{n = 1}^{\infty}\| \varphi^{n}\|^{2}& = & \varphi-1 \end{eqnarray*}\] Notice that $-(\varphi-1)$ is a (Galois) conjugate of $\varphi$ ![/quote]
It is not true. For example $\varphi =\sqrt 2$.
Must be \[\sum_{n=1}^{\infty }\parallel \varphi \parallel ^{n}=\varphi \\ \sum_{n=1}^{\infty }\parallel \varphi \parallel ^{2n}=\varphi-1\] It is obviosly.

NO. By $\varphi$ I meant the Golden ratio  :mad: ... It's in the title !!!
Moreover I believe that $\varphi$ is the unique real with the above proprety (first formula).
Actually the first (and second) sum converge if and only if the real in it is a Pisot number (or an integer)
Anyway, I hope some mathlinker will seriously consider, and solve this problem  :|

Edit: Some rudeness erased [Excuse me !].

I am sorry. I don't undestand your problem.
[quote="{x}"]
Notice that $-(\varphi-1)$ is a (Galois) conjugate of $\varphi$ !
EDIT: For the ones who didn't read the title (e.g Rust) by $\varphi$ I mean the Golden ratio.[/quote]
Let $x,y$ are real roots of equation $z^{2}-az\pm 1=0$,  were a is integer and $|x^{2}|>2$. Then $x^{n}+y^{n}$ is integer, therefore $\parallel x^{n}\parallel =|y|^{n}$, when n>1.
It give \[\sum_{n}\parallel x^{2n}\parallel =\frac{1}{1-y^{2}}\]  \[\sum_{n}\parallel x^{n}\parallel =\frac{1+\parallel y\parallel }{1-y^{2}}.\] It give solution your problem.

Exactly. There is an shorter solution using that $\parallel \varphi^{n}\parallel = \varphi^{-n}$ (which is anyway a corollary of what you have shown).  It should follow (I guess) in the same paths as yours.


$ a,b>0.$     $ a\plus{}b\plus{}1\equal{}3ab$ prove that   $ \frac{1}{a}\plus{}\frac{1}{b}\geqslant2$

hello, this problem is solved allready.
Sonnhard.

$ 3ab\equal{}a\plus{}b\plus{}1 \ge 3 \sqrt[3]{ab} \Rightarrow (ab)^3 \ge (ab) \Rightarrow (ab) \ge 1 \Rightarrow \frac{1}{ab} \le 1$

$ \frac{1}{a}\plus{}\frac{1}{b}\equal{}3 \minus{}\frac{1}{ab} \ge 3 \minus{}1\equal{}2$


$ F_7\equal{}\left\{0,1,2....,6\right\}$

$ I\equal{} \left(\begin{array}{cc}1 & 0 \\
           0&1 \end{array}\right)$

$ J\equal{} \left(\begin{array}{cc}0 & 6\\
           1&0 \end{array}\right)$

$ F^* \equal{}\left\{aI\plus{}bJ : a,b\in F_7\right\}$

All operations  are under mod 7

(a)Prove that $ F^*$ has totally 49 elements.

(b)If $ A,B \in F^*$ , then $ A\plus{}B ,AB \in F^*$  and $ AB\equal{}BA$

(c)If $ A\in F^*$
 and  $ A \not \equal{} 0$
  then   $ A^{48 }\equal{} I$

(d) If $ A\in F^*$
      there exists a unique  $ B \in F^*$   such that
    $ AB\equal{}I$

I'd rather call those two matrices "$ 1$" and "$ i$". If you've ever seen the representation of complex numbers as $ 2\times 2$ matrices, this is the same structure. The essential algebraic facts here are that $ 7$ is prime making the integers mod $ 7$ a field, and that $ \minus{}1$ has no square root mod $ 7$.

The inverse of $ a\plus{}bi$ is, of course, $ \frac{a\minus{}bi}{a^2\plus{}b^2}$. Dividing by $ a^2\plus{}b^2$ works mod $ 7$ as long as $ a^2\plus{}b^2$ is not zero mod $ 7$, which happens whenever at least one of $ a,b$ is nonzero.


search
geometry
geometric transformation

good question. but before that, let's establish this... bharata (the person after whom india was given its original name) was a king, and he was no ordinary king. he was virtuous, brave, just and a great king in every way like Rama. the reason we don't know as much about him is probably because, 1, he didn't actually kill any demons like Rama did, and if he did, he didn't have an epic written after him, and 2, Rama was an avatar of Vishnu, popularizing him more. he was definitely worthy of such an honor. but why him? good question. i'll search on google for more info.

so many different sources

o yea, i know about mahabharat and ramayan
but wat about krishna
thats a whole epic (get it on dvd today. jk.)

krishna... what about

the story associated with krishna is called krishnalila and is his story. Krishna is an avatar of Vishnu like Rama but in a different time period. Krishna is also associated with the Mahabharata. Arujuna is considered a disciple of Krishna... Bhagavad Gita... anymore questions?

well... i know the Gita was written by Veda Vyasa, spoken by Krishna to Arjuna, but I believe it was written after the Mahabharata but it was spoken during the Mahabharata... not sure... though

According to Kalidas - legend has it that Bharat was named after the  the son of Dushyanta and Shakuntala.

The  love story  became famous because Kalidas (5th century  CE ?)  wrote a beautiful  play Abhijnanashakuntalam

(The play is beautiful and  if you do not know Sanskrit, its worth to learn it, just to enjoy the play in original - Of course, you could read it in translation  ... One of my favorite translation was by M.S Gupta's Shakuntala (in Hindi - But I also read it in Sanskrit too)

(I think old name(s) used  for India were Aryavrat (Really old used in Veda's etc..) ) ...Hindustan ( In Persian - after sindh river )  name is also a few thousand years old.

my dad emailed me this so i didn't write this.

Bharat was the forefather of Kurus, the family name of Pandavas and Kauravas. He was was even before Santanu,
the father of Bhisma. Bharat was son of Dusyanta and Shakuntala. Dusyanta was a very powerful king and Shakuntla
was a beautiful daughter of  Menaka ( the most beautiful heavenly damsel ) and Vishwamitra. Once Indra sent Menaka to
get destroy the tapasya of Vishwamitra and get married with him. He did so because he was scared that after the success
in tapasya Vishwamitra will take of the reign of Swarga from him. Later  Menaka  got married with Vishwamitra and had a beautiful
daugher  named  Shakuntla . Bharata was a very powerful king. Behind him the Aryavarta became Bharatvarsha or Bharata.

so india was originally after mahabharat

interesing...

well... not really :wink: i think ur confusing mahabharata and bharata. mahabharata was an epic and bharata was a king who ruled a couple generations before the mahabharata. it is named after the person.

[quote="Iversonfan2007"][quote="akalra1"]hi

i was wondering, instead of a small topic for all countries, one for india! lets begin

if  it was named after someone, why not vishnu, who was in both scriputres[/quote]

well, technically Brahma is known as the "creator" so I would wonder why it was not named after him. Although my last name came from Vishnu so I'm kinda ticked off that it wasn't named after him too :P[/quote]

brahma isn't a main character in the scriptures

hower, Ramayan is named after Ram, and Ram is an avatar of Vishnu

[quote="Iversonfan2007"]true, along with krishna. But Brahma is still known as the "creator".[/quote]

hindu trinity:

brahma - creator
vishnu - preserver
shiva - destroyer

vishnu came in to preserve the earth, so it is vishnu who had to be there

r u from india?

[quote="akalra1"]

r u from india?[/quote]

maybe :wink:

[quote="Iversonfan2007"][quote="akalra1"]r u?

becuz if u r, u r cool
becuz if u rn't, u r really smart[/quote]

what if i'm half indian, half non indian? cool and smart?

if so, then i'm half indian

[quote="Iversonfan2007"] Although my last name came from Vishnu [/quote]

as u can tell, i'm from Siberia[/quote]

i am from india
(i am half indian)

so r u half indian or not- make ur psot clear

Repeated warning from other thread:  akalra, if you continue talking in chatspeak, I will begin deleting your posts.  If you want to ask a particular question to a particular forum member, use private messages.

[quote="akalra1"]bharat ruled during mahabharat

maha[b]bharat[/b][/quote]
this is NOT true. the bharat in mahabharat comes from the country bharat, which was named after the king bharat. why is it named after the king? because be4 he was king the whole Aryavarta(India) was divided into separate kingdoms. he combined all kingdoms to create one "nation" and thus it was named after him. The Mahabharata came long after him. this answers your question. QED

EDIT: BTW, the king Dhritirashtra is associated with the Mahabharat.

For those, who are interested in all the genealogy :)
Yes  Don't remember all the names but ... Dushyanta ->  (Son Bharat ->  (adopted grandson) Bharadwja ) ... many generation .. Hasti - Riksha ...many generation ..Kuru ---many generations .....Shantanu ... Pandu...  ....
Arjun/Yudhistir/et all (Time of Mhahabharata) -- grandson
  Pariixit  ..27 more generation then the dynasty died.

(According to legend in Mahabharata -of course!)


Given two positive integers $ n,k$. Find the number of natural roots of the following equation: $ \sum_{i \equal{} 1}^n x_i\equal{} k$

If you mean by natural roots with each xi in natural numbers [1,inf] it is  C (k-1,k-n)

if you want for [0,inf] it is C(n+k-1,k) you can derive the former from latter because in each box you have to place atleast 1.

The latter is got since if you have k pebbles and you want to seperate them into n parts, you can place n-1 sticks between k pebbles.


[quote="Kouichi Nakagawa"]$ (2\times3^2 \plus{} 4\times(3\times4)) \plus{} 4\times(2\times6) \equal{} (18 \plus{} 48) \plus{} 48 \equal{} \boxed{114}$[/quote]

Actually there are $ 64$ blocks, so the rectangular solid of least surface area is a cube with sides $ 4$. The surface area for the cube is $ 4*4*6\equal{}\boxed{96}$.


limit
inequalities
algebra unsolved
algebra

Let $ a>0$, sequence $ (x_i)$ satisfied: $ x_i > a \forall i \in N$.
Prove that: $ \displaystyle\lim_{n\to\infty}\ \displaystyle\sum_{i\equal{}1}^{n}\frac{x_i}{\displaystyle\sum_{j\equal{}1}^{i}\ x_j} \equal{} \infty$.

There are two possible cases :

a) [b]X_i is bounded above[/b]

So a < X_i < b
=> Its easy to see the each term of the sum is > (a/b)*(1/n)

From the fact that the harmonic series is divergent we reach the required conclusion

b) [b]X_i is not bounded above[/b]

We will invoke the pringsheim's theorem (necessary but not sufficient condition for convergence )that if the sum of a sequence of +ve numbers is convergent, then
 n*X_n  -> 0 as n -> infinity

So we will prove that the necessary condition does not hold here and thus the sum is divergent...

As the sequence is not bounded above, we can always find arbitrarily large 'n' such that X_n is greater than all of its preceding members(X_i such that i < n)

=> the 'n'th term of the given sum is > 1. (easy inequality manipulation)

So lim (n-> infinity) n*X_n does not tend to zero..

Hence the sum is divergent...


1.
[img]http://i6.photobucket.com/albums/y208/outoflimitz/math02.jpg[/img]
2.

[img]http://i6.photobucket.com/albums/y208/outoflimitz/math04.jpg[/img]

This is even worse than homework :roll:  Do not post [b]too many [/b]questions like these,search through your text books first.

For the first one,substitute vaklue of y in terms of x and put it in the second equation

In the second one,draw two lines with the given equations and the point where they intersect is the answer.To draw the lines,put any value of x in each of the given equations and then obtain the respective y ordinate and mark those points on the graph and join them.

[quote="HostGFX"]pleace show me man i will never ask anything no more:([/quote]

No hard feelings,sorry if I hurt you or anything.Its my fault then :( All I meant to say is that questions like these should maybe be asked in some other forum(like mathcount,MOEMS,etc)  because it will be treated like homework here in G.S

In the 1º equation of the first system, we got that $y=2x-10$. Plug this result in the second equation and solve it.....