inequalities
Euler
inequalities proposed

prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\le\frac {\sqrt {3}}{2r}$

[quote="xzlbq"][quote="Ligouras"]prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {\sqrt {3}}{2r}$[/quote]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\le\frac {\sqrt {3}}{2r}$[/quote]

$ \Longleftrightarrow \frac {3}{4r^2}\geq (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2.........(*)$

substitution  $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$

$ (*)\Longleftrightarrow \frac {3(x \plus{} y \plus{} z)}{xyz}\geq \frac {(x^2 \plus{} y^2 \plus{} z^2 \plus{} 3xy \plus{} 3yz \plus{} 3xz)^2}{(x \plus{} y)^2(y \plus{} z)^2(z \plus{} x)^2}$

$ \Longleftrightarrow \sum_{sym}(6x^5y^2\plus{}5x^5yz\plus{}18x^4y^3\plus{}42x^4y^2z\plus{}31x^3y^3z\plus{}42x^3y^2z^2)\geq 0$

obvious true.

[quote="xzlbq"][quote="Ligouras"]prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {\sqrt {3}}{2r}$[/quote]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\le\frac {\sqrt {3}}{2r}$[/quote]

Yes!!!!

[quote="xzlbq"][quote="Ligouras"]prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {\sqrt {3}}{2r}$[/quote]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\le\frac {\sqrt {3}}{2r}$[/quote]

$ \iff \frac{{s^2  \plus{} 4Rr \plus{} r^2 }}{{4sRr}} \le \frac{{\sqrt 3 }}{{2r}}$

$ \iff s^2  \plus{} 4Rr \plus{} r^2  \le 2\sqrt 3 sR$

$ \iff \left( {s^2  \plus{} 4Rr \plus{} r^2 } \right)^2  \le 12s^2 R^2$

$ \iff f(s^2)\equal{} \minus{} s^4  \plus{} \left( {12R^2  \minus{} 8Rr \minus{} 2r^2 } \right)s^2  \minus{} 16R^2 r^2  \minus{} 8Rr^3  \minus{} r^4  \ge 0,$

by Gerretsen Inequality , just need to prove $ f(4R^2\plus{}4Rr\plus{}3r^2) \ge 0$ and $ f(16Rr\minus{}5r^2) \ge 0$, by Euler Inequality, we have

$ f(4R^2\plus{}4Rr\plus{}3r^2)\equal{}4(R\minus{}2r)(8R^3\plus{}12R^2r\plus{}9Rr^2\plus{}2r^3) \ge 0$

and

$ f(16Rr\minus{}5r^2) \equal{} 4r(R\minus{}2r)(48R^2\minus{}19Rr\plus{}2r^2) \ge 0$,

done.


inequalities
trigonometry
function
calculus
derivative
integration
real analysis

Prove the following inequality without using [b]Taylor expansion[/b].

\[\frac{100}{120}<\sin 1<\frac{101}{120}\]

Again, can anyone solve this problem? :)

Well, I'm not sure about what exactly you are driving at when saying "solve without using Taylor expansion". The following argument formally avoids it but I doubt it will be acceptable for you. Just take the inequality $\sin x\le x$ for $x\ge 0$, integrate it 2 times to get $\sin x\ge x-\frac{x^3}{6}$ and integrate 2 more times to get $\sin x\le x-\frac{x^3}{6}+\frac{x^5}{120}$. Formally here there is no Taylor expansion involved (and even no differentiation, integration only) but, surely, it is just cheating: the whole thing is still based on  the fact that the first few derivatives of $\sin x$  are what they are. :( The point of this post is that I do not  understand clearly what the rules of this game are and I doubt that anybody else does. :? Probably, that's why nobody answered.

My solution is exactly same as yours.Japanese high school students study usually calculus such as differentiation and integral at the age of 17,18, but they are not allowed to use l'Hopital's theorem and Taylor expansion, which means, this problem is the very difficult. I have been posting the Japanese university entrance exam, the replies for these problem are the content which Japanese university student usually learn.

kunny


Let $ n$ be a positive integer, and let $ M\equal{}\left\{  1,2,...,n\right\}$. Two players take turns at the following game: Each player, at his turn, has to select an element of $ M$ and remove all divisors of this element (including this element itself) from the set $ M$.
[b]a)[/b] Assume that the player who cannot move anymore (because the set $ M$ is empty when it's his move) wins. For which values of $ n$ does the first player have a winning strategy?
[b]b)[/b] Assume that the player who cannot move anymore (because the set $ M$ is empty when it's his move) loses. For which values of $ n$ does the first player have a winning strategy?

For $ a)$ the answer is all $ n\ge2$, and for $ b)$ - any positive integer.

For $ n \equal{} 1$ the first player clearly loses.
Assume for some $ n\ge2$ the first player hasn't a winning strategy. This means no matter how he plays, he loses (the game always ends). But then, taking at his first move the number $ 1$, he transmits the move to the second player, because the number $ 1$ is removed anyway from $ M$ at the first move. So, now the second player is to move, and he is found in a situation identical to the one of the first player at the beginning of the game. According to our assumption - he hasn't a winning strategy - that is the first player wins.
Note that a [i]winning strategy[/i] is not explicitly defined, only its existence is proved.

The reasoning for $ b)$ is completely similar.

Yes, this is a correct solution. I believe such kind of proof is called "strategy stealing", and I haven't heard of any explicite winning strategy for this game.

  Darij

A very much related problem was given at [i]All-Soviet Union Olympiad[/i] in [i]1987[/i] for [i]8th grade[/i], as last problem in the second day. Here's the statement of this problem:

[i]Two players, one after another write on a blackboard positive integers not exceeding $ p$. The rules of the game do not allow writing on the blackboard a number that is a divisor of a previosly written number. The player who cannot move loses.
a) Determine which player has a winning strategy for $ p \equal{} 10$.
b) Determine which player has a winning strategy for $ p \equal{} 1000$.[/i]


Solve for $ x \in \mathbb{R}$,

$ \sqrt{2x^{2}\plus{}3x\plus{}5} \plus{} \sqrt{2x^2 \minus{}3x\plus{}5} \equal{} 3x$.

[quote="AndrewTom"]Solve for $ x \in \mathbb{R}$,

$ \sqrt {2x^{2} \plus{} 3x \plus{} 5} \plus{} \sqrt {2x^2 \minus{} 3x \plus{} 5} \equal{} 3x$.[/quote]

[hide="solution"]

$ \sqrt {2x^{2} \plus{} 3x \plus{} 5} \plus{} \sqrt {2x^2 \minus{} 3x \plus{} 5} \equal{} 3x$

$ \frac{(2x^2\plus{}3x\plus{}5)\minus{}(2x^2\minus{}3x\plus{}5)}{\sqrt {2x^{2} \plus{} 3x \plus{} 5} \minus{} \sqrt {2x^2 \minus{} 3x \plus{} 5}}\equal{}3x$
$ \frac{2}{\sqrt {2x^{2} \plus{} 3x \plus{} 5} \minus{} \sqrt {2x^2 \minus{} 3x \plus{} 5}}\equal{}1$

$ \sqrt {2x^{2} \plus{} 3x \plus{} 5} \minus{} \sqrt {2x^2 \minus{} 3x \plus{} 5} \equal{}2$
$ \sqrt{2x^2\plus{}3x\plus{}5}\equal{}\frac32x\plus{}1$
the rest is easy

[/hide]

Thanks.

Alternatively, multiplying both sides by $ (\sqrt{2x^{2}\plus{}3x\plus{}5} \minus{} \sqrt{2x^{2} \minus{}3x\plus{}5})$ we get

$ 2x\equal{} x(\sqrt{2x^{2} \plus{}3x\plus{}5} \minus{} \sqrt{2x^{2} \minus{}3x\plus{}5})$.

Since $ x\equal{}0$ is not a solution, we can divide both sids by $ x$ and get the pair of simultaneous equations:

$ \sqrt{2x^{2} \plus{} 3x \plus{}5} \plus{} \sqrt{2x^{2} \minus{} 3x \plus{} 5}  \equal{} 3x$,

$ \sqrt{2x^{2} \plus{} 3x \plus{} 5} \minus{} \sqrt{2x^{2} \minus{} 3x \plus{} 5} \equal{} 2$.

Adding these two,

$ 2\sqrt{2x^{2} \plus{}3x\plus{}5} \equal{} 3x\plus{}2$

Squaring both sides and simplifying, we get

$ x^{2} \equal{} 16$.

Checking, we see that $ x\equal{}\minus{}4$ cannot be a root. Therefore the only solution is $ x\equal{}4$.


HI I have a few questions and I was wondering if you know how to solve them?

to start: $sqrt(2-x)+2=sqrt(3x-2)$

$6x^3-11x^2+6x-1$ if one of the zeros is 1 what are the others?

$x^3-4x^2-x+4$

Thanks, also, if you solve one of them the same way as the other just tell me and I will solve, thanks for your help,
lyra

For finding roots of polynomials,just apply the rational root theorem.What ever the last constant term is,find its prime factors and substitute the value of each factor as $x$.For the factor for which the resultant comes $0$ after substitution,then that is a root of the equation.Hence $x-a$(where a is the factor) is a facctor of the polynomial.Then divide the plynomial by $x-a$ to get another polynomial,in the above cases,a quadratic which is then easy to solve.Keeoing this in mind,i think u should be able to solve the problems lyra.

FYI-We find the prime factors of the constant term because the general form of a polynomial(lets take cubic) is:-

$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$.So constant term is basically the product of the roots.(where roots are a,b,c)

For the second one, I think synthetic division is easier than dividing by $x-a$. You can also use synthetic division, on the third one, to check the possible roots after using the rational root theorem.


Given 8 light bulbs in a box. A man checked at random 4 bulbs and found that there are 3 good light bulbs and a defective one. If the man continues to check 3 more bulbs, what's the probability that there will be 1 good bulb and 2 defective ones?

What are [i]a priori[/i] probabilities that bulbs are good/defective?

After answering this question you can use Bayes' theorem to determine a posteriori probability $ p$ for a lamp to be good and then, as usual, $ P(\{\text{1 good bulb of 3}/\{\text{3 good bulbs of 4}\}\} \equal{} \binom{3}{1} p^1(1\minus{}p)^2$ (binomial law).


There is a satellite of mass $m$ which is moving in the alliptic path around a planet of math $M$.The half axis of ellipse are $a$ and $b$.At some distance $r$ from the planet,a rock hits the satellite in radial direction so that it changes its path into circular motion around the planet.But the mass of satellite doesn't change.Find this distance $r$ in terms of parameters $a$ and $b$.

That;s correct. :) Can you post your solution.

Total energy of the satellite on the elliptic path equals:
\[E=-\frac{GMm}{2a}\]
In perigee the distance of the satellite from the planet equals
\[r_{p}=a-\sqrt{a^{2}-b^{2}}\]
its angular momentum equals:
\[L=mv_{p}r_{p}\]
and its energy equals:
\[E=-\frac{GMm}{r_{p}}+\frac{mv_{p}^{2}}{2}=-\frac{GMm}{r_{p}}+\frac{L^{2}}{2mr_{p}^{2}}\]
Putting $r_{p}$ and $E$ we get the angular momentum of the satellite:
\[L=mb\sqrt{\frac{GM}{a}}\]
The rock hits the satellite in radial direction so angular momentum is conserved
\[L=mb\sqrt{\frac{GM}{a}}=mvr\]
For a circular path we get:
\[\frac{GMm}{r^{2}}=\frac{mv^{2}}{r}\]
Elliminating $v$ from the last two equations we get
\[r=\frac{b^{2}}{a}\]


function
trigonometry
calculus
calculus computations

Find $ \mathcal{L}\{ x(t)\}$ where:

$ x(t) \equal{} 1\plus{}\frac{1}{1!}t\plus{}\frac{1}{2!}t^{2}\plus{}\cdots\plus{}\frac{1}{n!}t^{n}$

Assume the convention that when we give a formula, we're talking about functions defined on (or perhaps supported on) $ [0,\infty).$

Note that $ \mathcal{L}(t^{n})\equal{}\frac{n!}{s^{n\plus{}1}}.$ From this,

$ \mathcal{L}(x)(s)\equal{}\sum_{k\equal{}0}^{n}\frac{1}{s^{k\plus{}1}}\equal{}\frac{\frac{1}s\minus{}\frac{1}{s^{n\plus{}2}}}{1\minus{}\frac{1}s}\equal{}\frac{1\minus{}\frac{1}{s^{n\plus{}1}}}{s\minus{}1}$

$ \equal{}\frac{1}{s\minus{}1}\minus{}\frac{1}{s^{n\plus{}1}(s\minus{}1)}$

In some sense - exploring in exactly what sense is a delicate business - this tends to $ \frac{1}{s\minus{}1}$ as $ n\to\infty.$ Hence (well, maybe not "hence" but something on a formal level), in some sense, $ x(t)$ tends to $ e^{t}.$

Another:

Obtain $ \mathcal{L}\{f(t)\}$ where:

$ f(t) \equal{} e^{\minus{}t}(t\minus{}2)\cos{(t\minus{}2)}H(t\minus{}2)$

H is the "Heaviside" function.

We know that $ \mathcal{L}\{e^{at}f(t)\}\equal{}\mathcal{L}\{f(t)\}_{s\to s\minus{}a}$ and $ \mathcal{L}\{H(t\minus{}b)f(t\minus{}b)\}\equal{}e^{\minus{}bs}\mathcal{L}\{f(t)\}$. Making $ a\equal{}\minus{}1$ and $ b\equal{}2$ we obtain
$ \mathcal{L}\{e^{\minus{}t}(t\minus{}2)cos(t\minus{}2)H(t\minus{}2)\}\equal{}\mathcal{L}\{(t\minus{}2)cos(t\minus{}2)H(t\minus{}2)\}_{s\to s\plus{}1}\equal{}\left[ e^{\minus{}2s}\mathcal{L}\{tcos(t)\}\right]_{s\to s\plus{}1}\equal{}\left[ e^{\minus{}2s}\frac{s^{2}\minus{}1}{(s^{2}\plus{}1)^{2}}\right]_{s\to s\plus{}1}\equal{} e^{\minus{}2(s\plus{}1)}\frac{(s\plus{}1)^{2}\minus{}1}{((s\plus{}1)^{2}\plus{}1)^{2}}$.


En un cuadrilátero bicéntrico $ABCD$ con incírculo $(I)$ y circuncírculo $(O)$. Los puntos de tangencia del incírculo $(I)$ con los lados $AB, BC, CD, DA$ son $K, M, L, N$ respectivamente. Si desde el incentro $I$ trazamos la perpendicular $IQ$ sobre la diagonal $BD$, probar que:
$[ABD].[QML] = [BCD].[KQN]$. Donde $[XYZ]$ = Area $XYZ$.

Maestro ya tengo una solucion para el problema pero como ya no tengo internet en casa no tengo el timepo suficiente para postearla pero prometo acerlo cuanto antes

 :!:

Saludos

Solo tenemos que aplicar el teorema de Euler :
$\frac{[ABD]}{[KQN]}=\frac{[BCD]}{[QML]}=\frac{R^{2}-d^{2}}{4R^{2}}$
Debido a que $KQN$ y $QML$ son triangulos pedales de $I$ con respecto a los triangulos $ABD$ y $BCD$ respectivamente. Además $IO=d$ y $R$= cincunradio de $ABCD$


calculus
integration
parameterization
function
derivative
real analysis
real analysis unsolved

I am practicing on old exams, and I would like some advice on a past problem. The problem is more like a calculation problem; it is very boring and in my opinion a very bad exam problem (still I would like to know how to solve it). The problem is:

Let $1<p<2$ and $f \in L^{p}(\mathbb{R})$. Show that the function
$h(y) = \int_{\mathbb{R}}\frac{f(x)}{\sqrt{|x|+|y|+1}}\, dx$
is well defined and continuous. Where is it differentiable?

Now, to show well-definedness I just use Hölder. For continuity and differentiability for $y\neq 0$ I use bounded convergence yada yada. However, I don't know what to do differentiability at $y=0$. Obviously $h$ can be differentiable there, but it seems likely that often it is not. Just how would you go about sorting this out in a reasonably quick manner?

With $f(x)=\chi_{[0,1]}$ you have $h(y)=2(\sqrt{|y|+2}-\sqrt{|y|+1})$, which is not differentiable at $y=0$. (Reason: $h$ is an even function with nonzero one-sided derivative at $0$).

Actually, I forgot that the mean value theorem applies even if the function is not differentiable at the endpoints, so the argument with bounded convergence does show that we have differentiability at $y=0$ iff
$\int \frac{f(x)}{(|x|+1)^{3/2}}\, dx = 0$
Stupid mistake of me


calculus
derivative
logarithms
complex analysis

[quote="Davron"]Find derivative of: e^x^x ?

D>L[/quote]

Is it (e^x)^x or it is e^(x^x)?

If it is the first one we get $ e^{x^{2}}$ with derivative $ (x^{2})'e^{x^{2}}\equal{} 2x e^{x^{2}}$ by chain rule. If it is the second one we get as derivative $ (x^{x})'e^{x^{x}}$. Now to find the derivative of $ x^{x}$ note that $ x^{x}\equal{} e^{x\ln x}$ now continue ...


Call a positive integer $N$ a $\textit{7-10 double}$ if the digits of the base-7 representation of $N$ form a base-10 number that is twice $N.$  For example, 51 is a 7-10 double because its base-7 representation is 102.  What is the largest 7-10 double?

[hide="Answer"]We have the restrictions for the base-10 number that we can never have a $9$, we cannot have two or more consecutive $8$'s, and we cannot have any $4$'s, as each of those will cause numbers greater than $6$ in the base-7 number. Finally, no numbers $\ge 343$ will work, as this will give a value in the $7^3$ place for the base-7 number, and from there the $7^2$ place will increase too rapidly for the base-10 number to ever again reach the right amount. From there, we test decreasing numbers to find that the answer is $\boxed{315}$.[/hide]

Sorry, but I think 315 is wrong. I don't know how to use the solution tags, so I'll post the solution further down.
Edit: Oh wait i forgot the fact that base 7 numbers cannot have any digits higher than 6, making 315 the right answer. never mind.
Edit2: Also just realized than 4everwise posted the exact same solution as I did, with better notation, making my post completely useless.

For the base 7 representation of the number, denote a as the 1's digit, b as the 7's digit, c as the 49's digit, d as the 343's digit, and so on.  (For example, 51's base 7 representation 102, so a = 2, b = 0, and c = 1.)  So we know that a + 7b + 49c +343d... etc. = the number in base 10, which I will denote here as n.  The base 10 representation has to be 2 times as great as the base 7 representation, so n = 2a + 14b + 98c + 686d... etc.  n is a base 10 number, and has the digits ...dcba (the ellipsis before dcba is used to indicate that there may be digits before d), and therefore can equivalently be written as a + 10b + 100c + 1000d... etc. So you have a + 10b + 100c + 1000d... = 2a + 14b + 98c + 686d... Now, combine like terms so that only positive terms remain.  The only positive terms on the right hand side will be a + 4b.  The positive terms on the left hand side will be 2c + 314d... and so on.  a,b,c,d... are all positive single digit numbers because they represent digits.  From this, it can be seen that the number n you are looking for can be no greater than 3 digits (No single digit, positive substitution for a,b,c,d can satisfy a + 4b = 2c + 316d...), which leaves you with a + 4b = 2c.  Because c is the first digit, you want to maximize the number, and therefore maximize c, so you plug in 9 for c, and now you want the maximum value of b satisfying a + 4b = 18, so you plug in 4 for b, leaving you with 2 for a.  So a = 2, b = 4, c = 9, leaving you with n = 941.  941/2 gives you the base 7 representation, making the answer 471.  (You can double check that 9*(49) + 7*(7) + 2 does indeed equal 471.)  Answer: [b][u]471[/u][/b]

315 is correct:

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2001-aa/aimeans32701.html

$ m\equal{}49s_1\plus{}7s_2\plus{}s_3$
$ 2m\equal{}100s_1\plus{}10s_2\plus{}s_3\equal{}m\plus{}51s_1\plus{}3s_2$
$ m\equal{}51s_1\plus{}3s_2$

plugging this back into the first line one obtains \[ 2s_1\equal{}4s_2\plus{}s_3\] with positive integers $ s_1,s_2,s_3$ it's quite clear that under no circumstance may $ s_1\equal{}0$. However, setting $ s_2$ and $ s_3$ equal to zero we obtain the respective identities \[ s_3\equal{}2s_1\Rightarrow m_{10}\equal{}51s_1\] \[ s_1\equal{}2s_2\Rightarrow m_{10}\equal{}105s_2\] Likewise we simply double the coefficients to assure $ m_7$ is proper (no digit exceeds 6). This results in $ \max{m_7}\equal{}630\Rightarrow m_{10}\equal{}315$

Well,I also get 471...And it turns out to be right.Mmm,I'm really puzzled :o

[quote="4everwise"][hide="Solution"]Let the largest 7-10 double in base $10$ be $d_{n+1}d_{n}\ldots d_{1}d_{0}.$ We're given that
\[2(d_{n+1}d_{n}\ldots d_{1}d_{0})_{7}=(d_{n+1}d_{n}\ldots d_{1}d_{0})_{10}\]
Expanding,
\[2(d_{n+1}\cdot 7^{n}+d_{n}\cdot 7^{n-1}+\cdots+7\cdot d_{1}+d_{0})=d_{n+1}\cdot 10^{n}+d_{n}\cdot 10^{n-1}+\cdots+10\cdot d_{1}+d_{0}. \]
Simplifying,
\[4d_{1}+d_{0}=2d_{2}+314d_{3}+\cdots+(10^{n}-2\cdot 7^{n})d_{n+1}. \]
However, we're working in base $7,$ so the largest possible value of each digit is $6.$ Hence, the largest possible value of the left hand side is $30,$ implying that we can't have a number larger than $3$ digits. (which was obvious because of how the AIME works :P)

From here, one sees that using $d_{2}=6$ yields a maximum value of $N$ of $630_{7}=315_{10}.$
[/hide][/quote]

I don't think 315 is correct answer.. As far as I am concerned, 420 is greater than 315 .. and base 7 representation of 420 is 840... Even 471 is the correct answer..

Read the whole thread?
[quote="JBL"]Read the whole thread? [quote="username718"]Edit: Oh wait i forgot the fact that base 7 numbers cannot have any digits higher than 6, making 315 the right answer. never mind.[/quote][/quote]

@Vickie both 471/420 don't work because in base 7 you cannot have 7's, 8's or 9's. Only digits allowed are 0-6.

Let N = 49a+7b+c. So 100a+10b+c is twice that. 98a+14b+2c=100a+10b+c. That becomes c+4b=2a. Maximize digits so get 630. Since 6 is highest is base 7. In base 10 thats 315.

Denote the base $7$ representation by $abc$. Then we have $100a + 10b + c = 98a + 14b + 2c$, or equivalently $2a-4b-c = 0$. Keeping in mind that $a,b,c<7$, the biggest possible solution is $a = 6, b = 3, c = 0$ which gives an answer of $\boxed{315}$.

bruh this was in my intermediate nt homework a few weeks ago

I just checked and it is, but it was also in my homework

[quote=joshualiu315]huh i thought this is aime problem also why 12 year bump[/quote]

@OlympusHero had a 12 yeah bump

[b]Solution:[/b] Realize that if $N \ge 500$, then $2N$ in base $7$ will always have a decimal equivalent less than $N$. Clearly, by definition of base $7$, any $334 \le N \le 500$ does work either. Now we can test all values of $N \le 333$, and see that $N = \boxed{315}$ is the largest one that works. Alternatively, we can let $\overline{abc}$ be the representation of $N$ in base $7$. Then we have that $100a + 10b + c = 49(2a) + 7(2b) + 2c \Rightarrow a=6, b=3, c=0$ is the largest solution. So, either way $\boxed{315}$ is our answer.


trigonometry
AMC
AIME
geometry
USA(J)MO
USAMO
AMC 10
\/closed

So here's where I'm at: My average AMC 10 is like 144 and my average amc 12 is about 120. I've nearly (not quite) mastered the C&P and algebra sections (I can get 4/5 on average of these problems in the last 5 right) but i'm only average at NT (i can only get like 1/2 on average of this in the last 5 right), while my knowledge of trig is limited to right angle triangles. Would the AMC 12 and/or AIME classes help?

Also, since the AMC 12 and AIME classes overlap, how would I be able to access both?

I'd recommend the relevant subject courses in your weaker areas instead of the AMC Problem Series, and going ahead and moving up to AIME Prep rather than taking the AMC 12 Problem Series.  Probably the best course for you would be the Precalculus course, which will take you through trig, complex numbers, and get you a start on linear algebra.  It will also cover all the trig and complex numbers you'll need for the AMC 12 and AIME, and give you a fighting shot on some relevant USAMO problems.

If you don't know modular arithmetic, go through the Intro Number Theory book before taking on Intermediate Number Theory (you probably don't need the Intro Number Theory class).


real analysis
calculus
integration
limit
function
real analysis unsolved

Hello!

Assume $ \mu$ denote Lebesgue measure on $ \Delta \equal{} [0,1]$ and $ \varphi_n: [0,1]\rightarrow \mathbb{R}$ be a sequence of nonnegative integrable(with respect to $ \mu$) functions. Assume that
\[ \displaystyle\lim_{n\to\infty}\displaystyle\int_{\Delta}\varphi_{n}d\mu \equal{} 0.\]
Is it true that $ \mu\{x|\displaystyle\lim_{n\to\infty}\varphi_n(x)\neq 0\} \equal{} 0?$

Thanks!

Nope! Enumerate the dyadic intervals $ E_n$ and let $ \phi_n$ be the characteristic function of $ E_n$. Then $ \phi_n$ converges in no point. However, $ \phi_n \to 0$ in measure, so there is a subsequence which converges almost everywhere to zero.

can you answer more simpler, I don't understand well. Can you give explicitly counterexample.

Consider the intervals $ E_{j,k} \equal{} [\frac{j}{2^k}, \frac{j\plus{}1}{2^k}]$, $ 0 \leq j \leq 2^k\minus{}1$. There are countably many, so we index them $ E_n$. Now let $ \phi_n(x)$ be 1 if $ x \in E_n$ and 0 otherwise. Then $ \int \phi_n \to 0$, while $ \phi_n(x)$ does not converge for any $ x$.

[quote="Kalle"]Consider the intervals $ E_{j,k} \equal{} [\frac {j}{2^k}, \frac {j \plus{} 1}{2^k}]$, $ 0 \leq j \leq 2^k \minus{} 1$. There are countably many, so we index them $ E_n$. Now let $ \phi_n(x)$ be 1 if $ x \in E_n$ and 0 otherwise. Then $ \int \phi_n \to 0$, while $ \phi_n(x)$ does not converge for any $ x$.[/quote]
Dear Kalle,
In your example, the set $ \{x: \lim\limits_{n\to\infty}\varphi_n(x)\ne 0\}\equal{}\emptyset$ then its measure is 0. And i think that the answer of this exercise is yes because by the Fatou's lemma, we have
\[ 0\leq\int\limits_{\Delta}\liminf\limits_{n\to\infty}\varphi_nd\mu\leq \liminf\limits_{n\to\infty}\int\limits_{\Delta}\varphi_nd\mu\equal{}0\]
thus
\[ \int\limits_{\Delta}\liminf\limits_{n\to\infty}\varphi_nd\mu\equal{}0\]
This follow that $ \mu(\{x: \liminf\limits_{n\to\infty}\varphi_n(x)\ne 0\})\equal{}0$. And, by $ \{x: \lim\limits_{n\to\infty}\varphi_n(x)\ne 0\}\subset \{x: \liminf\limits_{n\to\infty}\varphi_n(x)\ne 0\}$ then its measure is 0.

Kalle is right. And the example - I usually call it the "rotating tooth" example - is [i]the[/i] example both for this (that convergence in norm doesn't imply convergence [i]a.e.[/i]) and also the best-known example that shows that convergence in measure doesn't imply convergence [i]a.e.[/i] (In other words, Kalle wasn't being unusually inventive there - he was just rattling a well-known example off the top of his head.)

Yes, it is true that $ \liminf\varphi_n(x) \equal{} 0$ everywhere. But it's also true that $ \limsup\varphi_n(x) \equal{} 1$ everywhere. The limit exists nowhere. You seem to be operating on the mistaken assumption that not converging to zero is the same as converging to something nonzero - an assumption that leaves out the possibility of divergence.

It [i]is[/i] true that there exists a subsequence $ \varphi_{n_k}$ that converges to $ 0$ almost everywhere. That can be proved in general, but can you see such a subsequence in this particular example?

Dear all, thanks.
Yes Kalle is right, I understand his solution.And we can prove that (by the Fatou's lemma)there exists a subsequens of $ \varphi_{n}$ that converges to 0 almost everywhere.

So,
\[ \int\limits_{\Delta}\liminf\limits_{n\to\infty}\varphi_{n}d\mu \equal{} 0.\]

Does it follows from this that there is subsequence $ \{\varphi_{n_k}\}$ such that
\[ \int\limits_{\Delta}\lim\limits_{k\to\infty}\varphi_{n_k}d\mu \equal{} 0?\]


trigonometry
number theory
relatively prime

Given that $\sum_{k=1}^{35}\sin5k=\tan\frac{m}{n}$, where angles are measured in degrees, and $m$ and $n$ are relatively prime positive integers that satisfy $\frac{m}{n}<90$, find $m+n$.

Is there a solution that does not involve DeMoivre's?


Show that a triangle with side lengths  $a\ge b,c$  is right-angled if and only if  $(\sqrt{a+b}+\sqrt{a-b})\cdot(\sqrt{a+c}+\sqrt{a-c})=\sqrt{2}\cdot (a+b+c)$.

i dont know latex but here goes:

first we square the whole equation to get:

2(a+sqrt(a2-b2)(a+sqrt(a2-c2)=(a+b+c)2 (the twos are exponents) now we have two cases:
when A is obtuse we  have a2>b2+c2 hence rhs> 2(a+b)(a+c)=2a2+2ab+2ac+2bc>a2+b2+c2+2ab+2ac+2bc=(a+b+c)2
when A  is acute we have a2<b2+c2 hence rhs<2(a+b)(a+c)=2a2+2ab+2ac+2bc<a2+b2+c2+2ab+2bc+2ac=(a+b+c)2 it is very easy to see thet when when A is right we have: 2(a+b)(a+c)=(a+b+c)2. hence we are done.

hardsoul моё решение на отборочном :) :
 $\left((\sqrt{a+b}+\sqrt{a-b})(\sqrt{a+c}+\sqrt{a-c})\right)^2-2(a+b+c)^2=$

$2\left[a^2-b^2-c^2+2a\left(\sqrt{a^2-b^2}-c+\sqrt{a^2-c^2}-b\right)+2(\sqrt{a^2-b^2}\sqrt{a^2-b^2}-bc) \right]=2(a^2-b^2-c^2)\left(1+\frac{2a}{\sqrt{a^2-b^2}+c}+\frac{2a}{\sqrt{a^2-c^2}+b}+\frac{2a^2}{\sqrt{a^2-b^2}\sqrt{a^2-c^2}+bc}\right)$
 Then my solution can be easily completed

[quote="Mamat"]hardsoul моё решение на отборочном :)[/quote]

????????????what is that???????????
but a nice solution though.

Wow. That just killed the problem. Very nice way.

xxreddevilzxx
   that mean "solution which i wrote in the exam"  in russian

[quote="fuzzylogic"][quote="pkothari13"]a less time consuming way to prove that is by saying:

$a = m^2 + n^2$
$b = 2mn$
$c = m^2 - n^2$

where $m,n$ are integers and $m > n$

when you plug them in, the radicals turn out nice and voila![/quote]

But $a,b,c$ may not be integers.[/quote]

I don't think $m$ and $n$ need to be integers, the only restriction I think is $m>n$ and of course they are both positive

[quote="fuzzylogic"][quote="pkothari13"]a less time consuming way to prove that is by saying:

$a = m^2 + n^2$
$b = 2mn$
$c = m^2 - n^2$

where $m,n$ are integers and $m > n$

when you plug them in, the radicals turn out nice and voila![/quote]

But $a,b,c$ may not be integers.[/quote]
they don't have to be.
pkothari13's solution is correct
.

[quote="pkothari13"]oops sorry, when he said right angled triangle, i somehow assumed that all the sides had to be integers. but it still works if m and n they are not integers[/quote]

Your solution is not complete, you need to prove the converse as well.
If the equation is true then a,b,c must be the sides of a right triangle.

[color=darkred][b][u]PP[/u].[/b] Show that a triangle with side lengths  $a\ge b$ and $a\ge c$  is $A$-right-angled $\iff$

$E\equiv \left(\sqrt{a+b}+\sqrt{a-b}\right)\cdot\left(\sqrt{a+c}+\sqrt{a-c}\right)=2s\sqrt{2}$ , where $2s=a+b+c$ . [/color]

[b][u]Proof[/u].[/b] Denote $\left\{\begin{array}{c}
D=b^2+c^2-a^2\\\\
X=(a+b)(a+c)\ ;\ Y=(a-b)(a-c)\\\\
Z=(a-b)(a+c)\ ;\ T=(a+b)(a-c)\\\\
E=\sqrt X+\sqrt Y+\sqrt Z+\sqrt T\end{array}\right|$ . Define the equivalence relation

$m\ .s.s.\ n\ \iff mn>0\ \vee\ m=n=0$ $\iff$ $\mathrm{sign}(m)=\mathrm{sign}(n)$ ($m$ and $n$ have [size=120][b]s[/b][/size]ame [size=120][b]s[/b][/size]ignature).

Prove easily that $2s^2-X=2(s-a)^2-Y=$ $2(s-b)^2-Z=2(s-c)^2-T=D$ and

$\left(s\sqrt 2-\sqrt X\right)\ .s.s.\ \left[(s-a)\sqrt 2-\sqrt Y\right]\ .s.s.\ \left[(s-b)\sqrt 2-\sqrt Z\right]\ .s.s\ \left[(s-c)\sqrt 2-\sqrt T\right]\ .s.s.\ D$ .

Since $\left(s\sqrt 2-\sqrt X\right)+\left[(s-a)\sqrt 2-\sqrt Y\right]+$ $\left[(s-b)\sqrt 2-\sqrt Z\right]+\left[(s-c)\sqrt 2-\sqrt T\right]=2s\sqrt 2-E$ obtain that

$\left(2s\sqrt 2-E\right)\ .s.s\ D\ .s.s.\ \left(90^{\circ}-A\right)$ , i.e. $\left\{\begin{array}{cccc}
(1) & E<2s\sqrt 2& \iff & A<90^{\circ}\\\\
(2) & E=2s\sqrt 2& \iff & A=90^{\circ}\\\\
(3) & E>2s\sqrt 2& \iff & A>90^{\circ}\end{array}\right|$ . Our [b]PP[/b] is the particular case $(2)$ .


In each $ 2007^{2}$ unit squares on chess board whose size is $ 2007\times 2007$,
there lies one coin each square such that their "heads" face upward.
Consider the process that flips four consecutive coins on the same row, or flips four consecutive coins on the same column.
Doing this process finite times, we want to make the "tails" of all of coins face upward, except one that lies in the $ i$th row and $ j$th column.
Show that this is possible if and only if both of $ i$ and $ j$ are divisible by $ 4$.

Some slight variation of K81o7's trick from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162039 should work, I think.

Let $ r_{1}, r_{2}, r_{3}, r_{4}$ be the number of heads in rows that are 1 mod 4, 2 mod 4, 3 mod 4, and 0 mod 4 respectively.  Then $ r_{1}\equal{} r_{2}\equal{} r_{3}$, and each flip either increases or decreases each of the $ r_{i}$'s by 1, or it increases or decreases a single $ r_{i}$ by 0, 2, or 4.  Thus after any sequence of flips, we have $ r_{1}\equal{} r_{2}\equal{} r_{3}(\mod 2)$.  Thus it cannot be the case we have a lone head in one of the rows that are 1,2, or 3 mod 4.  Similarly, we cannot have a lone head in a column that is 1,2, or 3 mod 4.

For the converse, let's start by making all the coins tails except for  (4,4).  Flip all the rows from 4 to 2007 using vertical flips, then for the first three rows, flip all the columns from 4 to 2007 using horizontal flips. This leaves a 3x3 square of heads in the upper left corner.  Now use four vertical flips to leave three heads at (4,1), (4,2) and (4,3), then one horizontal flip will leave (4,4).  From here, we can move the lone head a distance of four using two adjacent flips, allowing us to move to any $ (i,j)$ with $ i \equal{} j \equal{} 0 (\mod 4)$.

Well,I actually had the same solution as above.
In fact ,I first found the construction and then the if condition


I have the molecule (2S, 3R)-3-bromo-2,3-diphenylpentane, but drawn with a fischer projection with the chain of carbons on the vertical axis, I am confused with assigning priorities for the 3rd Carbon. I know that Br is 1st, but I'm not sure if the phenyl is second or the C2 (which has a phenyl, a H, and a methyl). I know the ethyl group is last, but I can't decide between those two. Can anyone help?

-Br > -Ph > -CH(CH3)Ph > -CH2CH3

The first carbon of -Ph is unsaturated, while the first carbon of -CH(CH3)Ph is saturated. Therefore, the former has higher priority.


Let x= SV in right triangle SVE.  If the sum of the lengths of the legs is 24,
write a function A(x) which gives the area of the triangle and find the maximum value of A(x).

SV is a leg, VE is a leg, and SE is the hypotenuse

Let $ SV\equal{}x, \ VE\equal{}y$.
Then $ x\plus{}y\equal{}24$ and the area is $ \displaystyle\frac{xy}{2}$.
We have $ y\equal{}24\minus{}x\Rightarrow A(x)\equal{}\frac{x(24\minus{}x)}{2}\equal{}\frac{1}{2}x^2\plus{}12x$.
The maximum of $ A(x)$ is $ \minus{}\frac{\Delta}{4a}\equal{}72$


calculus
derivative
function
real analysis
real analysis unsolved

If the graph of a derivative of a function is strictly monotonically increasing, does that mean the function itself is strictly monotonically increasing?

No. Consider $f(x)=x^{2}$ whose derivative is $f'(x)=2x$.

Ah, didn't think of that since the problem I've been focussed on only has a derivative and graph defined for x>0. So then, would it be possible?


geometry
incenter
circumcircle
Euler
angle bisector
perpendicular bisector
geometry proposed

The perpendicular bisectors of the sides AB and BC of a triangle ABC meet the lines BC and AB at the points X and Z, respectively. The angle bisectors of the angles XAC and ZCA intersect at a point B'. Similarly, define two points C' and A'. Prove that the points A', B', C' lie on one line through the incenter I of triangle ABC.

[i]Extension:[/i] Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC.

  Darij

[quote="darij grinberg"]
[i]Extension:[/i] Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC.

  Darij[/quote]

Actually this is a big hint  :) .

For the moment I will give only a very concise sketch.

a) Is enough to prove that B',O and I are collinear.

b) One can prove (is not very difficult, indeed) that the bisectors (of both angles and sides) meet on c(ABC).

c) Then using Pascal theorem one can prove the claim.

This is highly interesting. Could you tell me some details of steps b) and c)? I cannot complete your solution. On the other hand, my own solution is very long.

Thanks!
  Darij

What sprmnt is saying is, I think, that $OX$ and $AB'$ meet on $(ABC)$, and this is a simple angle chase (the same holds, of course, for $OZ$ and $B'C$). All we need to do is compute the angle between $AB'$ and the perp. bisector of $AB$. On the other hand, we know that the bisectors of $\angle A,\angle C$ meet the perpendicular bisectors of $BC,AB$ (respectively) on $(ABC)$.

mmm i don't know if this is ok or not, because it seems to be quite easy... what i had thought is exactly what grobber said: $OX$ and $AB'$ meet on $(ABC)$ (let $M$ be that intersection), and the same holds for $OZ$ and $CB'$ (they meet in a point $N$, in $(ABC)$... let $P$ and $Q$ be the second intersections of $OX$ and $OZ$ with $(ABC)$ (i.e., the intersection of the perpendicular bisectors of $AB$ and $BC$ with the circumcircle), respectively. Now you just have to notice that in the cyclic hexagon $PCNQAM$ the intersections of the opposite sides lie on a line...and those intersections are $PC\cap AQ = I$, $CN\cap AM = B'$ and $NQ\cap MP = O$, so B' lies on the line OI and we are done  :)

[quote="darij grinberg"]This is highly interesting. Could you tell me some details of steps b) and c)?
Darij[/quote]

Tanks to Grobber and RamLaf, I sould have only to attach a picture.

BTW, here follow the arguments which prove the claims.

If K (opposite to A) and H are the common points to c(ABC) and the axis of BC, then AI pass through K. furtermore, as BCZ is isosceles, the angle bisector of <ZBA pass through H.

Similarly for all the other bisectors.

PS

Darij, what contest is 239MO?

Wow, this was a cook. Probably I should have thought about the problem for a longer time, and not at midnight. Anyway, all of your solutions are correct and nice. My own solution used nine-point centers, Kosnita points and isogonal conjugates and was so monstrous that I don't want even to mention it. A note to your post, Sprmnt21:

[quote="sprmnt21"]If K (opposite to A) and H are the common points to c(ABC) and the axis of BC, then AI pass through K. furtermore, as BCZ is isosceles, the angle bisector of <ZBA pass through H.[/quote]

I guess <ZBA should be < ZCA. Again, your proofs are very nice. Now I have another synthetic proof, based on your ideas:

Let the lines AX and AB' meet the circumcircle of triangle ABC at the points U and M (apart from A). Since the line AB' bisects the angle XAC, we have < MAC = < UAM, and thus, replacing chordal angles by arcs on the circumcircle of triangle ABC, we have arc MC = arc UM. On the other hand, the point X lies on the perpendicular bisector of the segment AB, and thus < XBA = < BAX, or, in other words, < CBA = < BAU. In terms of arcs, this yields arc CA = arc BU. Together with arc MC = arc UM, this implies arc MA = arc MC + arc CA = arc UM + arc BU = arc BM, what yields that the point M is the midpoint of the arc ACB on the circumcircle of triangle ABC. In other words, the point M is the midpoint of the segment $I_aI_b$, where $I_a$, $I_b$, $I_c$ are the excenters of triangle ABC. Similarly, the point N where the line CB' meets the circumcircle of triangle ABC (apart from C) is the midpoint of the segment $I_bI_c$. Now, applying the Pappos theorem to the collinear point triplets (C, M, $I_a$) and (A, N, $I_c$), we see that the points $CN \cap AM$, $MI_c \cap NI_a$ and $I_aA \cap I_cC$ are collinear. But $CN \cap AM = B^{\prime}$. Further, $MI_c \cap NI_a$ is the point of intersection of the lines joining the vertices $I_c$ and $I_a$ of triangle $I_aI_bI_c$ with the midpoints M and N of the opposite sides $I_aI_b$ and $I_bI_c$; in other words, the point $MI_c \cap NI_a$ is the centroid of triangle $I_aI_bI_c$. Finally, the point $I_aA \cap I_cC$ is the point I, the orthocenter of triangle $I_aI_bI_c$. Hence, we have obtained that the point B' is collinear with the centroid and the orthocenter of triangle $I_aI_bI_c$. In other words, the point B' lies on the Euler line of triangle $I_aI_bI_c$. But this Euler line is the line OI. Hence, the point B' lies on the line OI. Similarly, we can show that the points C' and A' lie on the line OI, and the problem is solved.

[quote="sprmnt21"]Darij, what contest is 239MO?[/quote]

The Open Olympiad of the 239th Mathematical Lyceum of St. Petersburg, Russia. This is a very special olympiad, organized by enthusiasts, and containing very hard problems.

  Darij

The perpendicular bisectors of the sides $AB$ and $BC$ of a triangle $ABC$ meet the lines $BC$ and $AB$ at the points $X$ and $Z$, respectively. The angle bisectors of the angles $XAC$ and $ZCA$ intersect at a point $B'$. Similarly, define two points $C'$ and $A'$. Prove that the points $A'$, $B'$, $C'$ lie on one line through the incenter $I$ of triangle $ABC$.


Given a segment $AB$, find the midpoint of $AB$ only with a compass.

First we can have two point $C,D$ such that $\triangle ABC,ABD$ are equilateral triangle.(only need to construct the circle with center $A$ and radius $AB$ and the circle with center $B$ and radius $AB$,$C,D$ are their intersection points.)
Then we can similarity get $E$ such that $CDE$ is a equilateral triangle,then $EB=AB$ and $E$ is on the straight line $AB$,we just need to construct the 反演 of $E$ by the circle with center $A$ and radius $AB$(sorry I don't know how to say it in English).And we can construct a circle with center $E$ and radius $EA$,the circle intersect the first circle(反演圆) at $F,G$, then we can construct two circle with radius $FA(GA)$ and center $F,G$,respectively.The other intersection of them(not A) is the midpoint of $AB$.


[hide]For n = 7 or higher, n! will end in a 0 in base 7, so n*n! will also end in a 0 in base 7. Every factor of 7, another 0 is added. Summing n*n! for n ranging from 1 through 6 inclusive, we get 5039, which is equivilent to 6 (mod 7). So d = 6.

Your hunch is nearly correct. Every factor of 7, n! gets another 0. But multiples of 49 have 2 factors of 7 and multiples of 343 have 3 factors of 7, so that would also have an effect on it. So my guess is that the number would end in [2006/7]+[2006/49]+[2006/343] = 286+40+5 = 331 6's[/hide]

I understand that multiple factors of the base will cause more zeros to occur at the end of each term (beyond a certain point) in such a sum.   But exactly how does your "mod" observation justify the REPEATED  9's (for base 10) or 6's (in base 7).  What is causing the repetition?

[quote="SJS"]I would like ideas for non-electronic solutions to the following problem:

The numerical (base 7) representation of summation (x*x!, x =1 to x =2006) ends in a digit d repeated M times.  Find d and M.

Regarding a base 10 solution, I noticed the fourth partial sum ends in 9; an additional 9 is appended at each fifth partial sum thereafter.  I'm puzzled by this behavior, however.  Any ideas?

Using a base 7 converter, I noticed the sixth partial sum ends in 6 and the thirteenth partial sum ends in 66; my hunch is that an extra 6 is appended after every 7 terms (after the sixth term).   Is this correct?   Same rationale?[/quote]

would this happen to be on your arml homework  :P
also, there is a more rigirous solution out there

Yes.  I thought I'd try the problems (my arml students were assigned) myself.  Got the others; just need to finish this one!   If you're a Chicago team member, congratulations on making the team.

$\sum_{k=1}^{2006}k*k!=\sum_{k=1}^{2006}(k+1)!-\sum_{k=1}^{2006}k!=2007!-1$, then $\sum_{1}^\infty \left\lfloor\frac{2007}{7^k}\right\rfloor=331$ is the number of factors of 7 in that factorial, each of these $7$'s correspond to a $0$ on the end in base $7$, then taking off $1$, it is $331$ $6$'s, so $(6,331)$

yes i did make the team  :), i would like to discuss #12 since it i believe that it has no solution, and #10 i am not sure how to prove the range for $k$, i got $1<k\le e$, but i don't really know about rigirous convergence proofs


:idea:  :idea:  :idea: There should be a separate thread on the forum home page for the Mandelbrot Competition. It's such an awesome, well-known competition. I really think it would be cool to have a separate space to discuss these problems, concepts, and topics. :rotfl:

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205683]here[/url]. Any new threads should be created in that forum.


This is a coincidence topic. You can only post coincidences that happened to you or near you.

example: U.S.A.
John Phillip SoUSA, a patriotic composer

I can't start because I don't have a coincidence...

[quote="1=2"]I can't start because I don't have a coincidence...[/quote]
What a coincidence! Neither do I
:D

Once I had an English Test on some story, and there were 50 problems, and I knew only 39 of them since I didn't read it and only looked at Sparknotes. I was like I am SOO going to get an A on this no matter what it takes. I guessed and got 8 out of 11 right.

I have a lot more. too.  :lol:

If you search Lincoln Kennedy on Google, you can find lots of coincidences among them.

A numerical coincidence from my past:
As a high school student, I took math classes at the local state university. I was listed as a math major, code 6666 in the system.
When I went to college full time, the last four digits of my student ID number were 6666.

there are a lot of vocab words in my planner... and one time, I needed to find out the meaning of a word but i couldn't, and then i just happened to see it inside my planner right when i was looking for it.

[quote="h_s_potter2002"]One time I watched Jeopardy, and I wanted this guy to win.  And he did.  And then I watched Wheel of Fortune, and wanted another guy to win.  And he did.

And one time, when we were presenting our science projects, I wanted to be the last one to go.  And I was.

And I have like a whole bunch more.[/quote]

=O what was the probability of that happening =O

what a sick math problem

1/9 the reciprocal of the # of kids in his class. duh.

[quote="bubka"]I don't know whether this fits here, but the google search reminded me:

If you google searched "French Military Victories", it returned: "Sorry no results found. Do you mean "French Military Defeats"?

I put this here, because this is actually an example of a coincidence if you think carefully, as clearly it wasn't an intended joke.[/quote]
uh what?
that's not true...it's like a joke on albinoblacksheep?


A two-inch by six-inch board costs 24 cents per linear foot. Jennifer needs to purchase 18 boards that are each 10 feet long. What is the total cost of the lumber?

please see the figure to the following pdf  warm up -2 problem no 10

http://sch.ci.lexington.ma.us/~jfrost/MC/2004%20-%202005%20Warmups.pdf?FCItemID=S0133B27B

I'm pretty sure that is just $ 18\cdot 10\cdot 24\equal{} \boxed{\$43.20}$ since I'm pretty sure they would ask for dollars not cents.

Do not put answers online, repeat DO NOT PUT ANSWERS ONLINE!!!

The rule is quite flexible. We generally post the solutions anyway.

Why don't you go PM AIME15 regarding this rule?

In that case, why don't you ask him in person to clarify this rule, and to clarify why he hasn't banned us all?

[quote="mathlearner2012"]wow AIME15 is my friend, he like tells me everything what to do on AOPS at school.[/quote]

I PM'ed you yesterday about this, but apparently you did not read it.  Additionally, you don't talk to me about AoPS at school.

It is OK to post answers and solutions.


modular arithmetic
number theory
Diophantine equation

$ 6x + 4y = 5$

I had some difficulty dividing in mod 4.

[hide="My Work"]

$ \begin{eqnarray*} 6x + 4y & = & 5 \\
6x & = & 5 - 4y \\
6x & \equiv & 5 - 4y \pmod {4} \\
6x & \equiv & 5 \pmod{4} \\
6x & \equiv & 9 \pmod{4} \\
2x & \equiv & 3 \pmod{4} \end{eqnarry*}$

And I'm stuck.
[/hide]

[quote="xpmath"][hide]The left side is always even, the right side is odd, so no solution.[/hide][/quote]

That's what I thought at first, so there's simply no way to put it in terms of $ k$?

[b](a consequence of) Bezout's Lemma:[/b]  For given integers $ a, b, d$, the Diophantine equation $ ax \plus{} by \equal{} d$ is solvable iff $ \gcd(a, b) | d$.


function
limit
real analysis
real analysis unsolved

Given a differentiable function $ f(x)$ such that $ f(x\plus{}y)\equal{}f(x)f(y)$ for all real $ x$ and $ y$.
Prove that $ f(x)\equal{}p^x$ where $ p\equal{}f(1)$.

$ f(x \plus{} 0) \equal{} f(x) f(0)$ gives $ f(0) \equal{} 1$ or $ f(x) \equal{} 0 \forall x$.  In the former case,

$ f'(a) \equal{} \lim_{h \to 0} \frac{f(a \plus{} h) \minus{} f(a)}{h} \equal{} f(a) \lim_{h \to 0} \frac{f(h) \minus{} 1}{h} \equal{} f(a) f'(0)$

and the rest follows.


Gasiti toate perechile de numere intregi pozitive (x;y) ce satisfac ecuatzia:

x ^ y = y ^ (x-y)

[hide="solutie"]pentru inceput solutia $(1,1)$.
deci $x,y\geq 2$, si deci si $x>y$. de aici rezulta $x-y>y$, (altfel $x^{y}>y^{x-y})$.
apoi, fie $p$ prim care sa il divida pe $y$. atunci $p|x$. fie $a$ exponentul lui $p$ in $x$ si $b$ exponentul lui $p$ in $y$. atunci exponentul lui $p$ in $x^{y}$ este $ay$, si exponentul lui $p$ in $y^{x-y}$ este $b(x-y)$. cum $x^{y}=y^{x-y}$, rezulta ca $ay=b(x-y)$. dar $x-y>y$, deci $a>b$, adica $\exp_{p}(y)<\exp_{p}(x)$ cum $p$ a fost ales arbitrar dintre divizorii primi ai lui $y$, rezulta ca pt orice divizor prim al lui $y$ este adevarat faptul demonstrat mai sus. deci $y|x$. atunci
$a=\frac{b(x-y)}{y}=b\Bigg(\frac{x}{y}-1\Bigg)$. cum $x-y>y$ implica $x>2y$, si $y|x$, rezulta ca $b|a$. din nou aceasta proprietate este generala, iar factorul $s=\frac{x}{y}-1$ este acelasi, pentru fiecare $p|y$, cu $p$ prim. atunci rezulta ca $x=y^{s}$. $s>1$, deoarece $x>2y$.
$y^{sy}=y^{y^{s}-y}$ implica $sy=y^{s}-y$, sau $s=y^{s-1}-1$.
de aici iese banal...
pentru $y=2$, rezulta $s=3$.
pentru $y=3$, rezulta $s=2$.
pentru $y\geq 4$ nu avem solutii
deci solutiile sunt $(1,1),(8,2),(9,3)$.[/hide]


Proof by contradiction should suffice.

Bomb


[b]a)[/b] We have $f(n)=\sum_{d|n}d\phi(\frac nd)$, so $\frac{f(n)}n=\sum_{d|n}\frac{\phi(d)}d$. It's a well-known fact that if $g$ is multiplicative, then so is $G(n)=\sum_{d|n}g(d)$, and since $g(n)=\frac{\phi(n)}n$ is multiplicative, then so is $G(n)=\frac{f(n)}n$, Q.E.D.

(by multiplicative I mean $f(mn)=f(m)f(n)$ whenever $(m,n)=1$)

[b]b)[/b] Since we have established the multiplicativity of $f$, it's enough to show the conclusion holds for $n=p^k$, with $p$ prime. $f(p^i),\ i\ge 1$ is, however, equal to $(i+1)p^i-ip^{i-1}$. Now the conclusion follows easily (a very simple computation, since we have a telescopic sum).

[quote=sam-n]We define $ f: \mathbb{N} \rightarrow \mathbb{N}$, $ f(n) \equal{} \sum_{k \equal{} 1}^{n}(k,n)$.

a) Show that if $ \gcd(m,n)\equal{}1$ then we have $ f(mn)\equal{}f(m)\cdot f(n)$;

b) Show that $ \sum_{d|n}f(d) \equal{} nd(n)$.[/quote]

Well we can see that any divisor $d$ of $n$ is repeated $\phi(\frac{n}{d})$ times , so
$$f(n)=\sum_{d|n} id \times \phi (\frac{n}{d})=id \bigstar \phi$$ where $\bigstar$ denotes the Dirichlet convolution.
After this everything follows serially


Answer

There is no solutions if $ n$ is odd as this would imply that $ z-1 = z$

However, if $ n$ is even then:

$ z-1 = \pm \,z$

We have already established that there is no solution for the $ +$ case, but the $ -$ case yields $ z = \frac{1}{2}$

So solution is $ z = \frac{1}{2}$ when $ n$ is even, or every $ z$ is a solution if $ n = 0$

[quote="mdk"]Find all solutions of the equation (z-1)^n = z^n.[/quote]
[hide]$ z=\tfrac12\quad\text{if}\quad n\equiv0\bmod2$

$ z\in\mathbb{R}\quad\text{if}\quad n=0$[/hide]

None of you are considering complex solutions ;)

[quote="sludgethrower"]I tried to think of complex solutions, but then I realised that if we took the simple case where $ n$ is a positive integer, then we would have a polynomial in degree $ n\minus{}1$, which would have $ n\minus{}1$ complex roots. And I wouldn't fancy attempting to factorise a general polynomial. Also the case where both $ z$ and $ n$ are complex would be even worse. The mathematics needed to deal with such problems wouldn't be regarded as "high school basics". But if you have a solution in the complex numbers, then please post it.[/quote]

Although the problem is not very clear, the letters used let us think that z can be a complex number and n an integer. Then :

$ (z\minus{}1)^{n}\equal{}z^{n}$ $ \Leftrightarrow$  $ (1\minus{}\frac{1}{z})^{n}\equal{}1$

So $ 1\minus{}\frac{1}{z}$ is any n-th root of 1, except 1 itself.

So we have $ n\minus{}1$ complex solutions $ z_{k}\equal{}\frac{1}{1\minus{}e^{\frac{2k\pi}{n}i}}$ $ \forall k\in[1,n\minus{}1]$

Ah...brilliant. I never had the inspriation to divide through like that. Out of curiosity, how would you treat the problem if n were complex?


inequalities function vector algebra polynomial

Does anyone know a non-inductive proof that $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=...=\frac{x_{n}}{y_{n}}$ is the only case in which equality holds in the Cauchy=Schwarz Inequality? It seems like such a proof should exist, but I can't seem to come up with it.  (Don't ask why I'm doing this now...it's a long story...)

Consider the function $\sum (a_{i}x-b_{i})^{2}\geq 0$.  This is a nonnegative quadratic.

What is its discriminant/what can you conclude about the sign of the discriminant?

When does the discriminant equal zero?

Ahhh...Ok, thanks! I was beating around the bush, as usual.

[quote="WindingFunction"]Oh, OK.  The inner product space thing was what I was trying to do in the first place.  We didn't spend much time on inner products.  Only one lecture at the end.  So I figured I'd better try to learn it on my own.  In class we were too busy fiddling around with all those normal forms and polynomial ideals all semester...[/quote]

For a general inner product space, the proof is less trivial, and IMO, isn't really motivated by anything.  You just kinda expand, substitute, and it works.  You can find it on Wikipedia: [url]http://en.wikipedia.org/wiki/Cauchy-Schwarz[/url].

[quote="Elyot"]Just use linear algebra... let $A = (a1, a2, ..., an)$ and $B = (b1, b2, ..., bn)$ be vectors in n-space.  Then Cauchy-Schwarz becomes the inequality $(A \centerdot B)^{2}\leq (|A|*|B|)^{2}$.  But of course, $A \centerdot B = |A|*|B|*cos(\theta)$, where $\theta$ is the angle between $A$ and $B$.  This gives an easy proof, since $cos(\theta)$ is always less than $1$.  Then the equality condition only holds iff $cos(\theta)=1$, which is when one vector is a scalar multiple of the other.

This actually holds for inner product spaces in general, but is slightly less trivial to prove.  I'm actually pretty surprised you didn't see this in first year linear algebra.[/quote]

You usually use the scalar product to define a notion of angle in $\mathbb{R}^{n}$ for $n\geq 4$.


function
calculus
derivative
algebra proposed
algebra

Determine all real numbers x for which 3x + 41/x = 11.

x=2 is a solution of course.. let's check if there are more. :)

classic ln/log method failed

I tried rewriting them as  \sqrt3^x + 2^(1/x) =  \sqrt(11-4\sqrt3) but that didn't really solve anything

I think the solution will be starting with 12^x - 11.4^x + 4^(x+1/x) = 0
say x=2+a

then we get 144.12^a-176.4^a+16.4^((a+1)²/(2+a)) = 0  (where a=0 is a solution indeed)
this should be solvable in some way, too bad I don't know how.

Interesting problem though!

Let f(x) be the function f(x) = 3 x  + 4 (1/x) . It's trivial that f(x)=11 has only positive solutions.

The derivative f'(x) = ln(3)*3 x* - ln(4)*(1/x 2)*4 (1/x) is  increasing on [0, +\infty[, so f'(x)=0 has only one root, so the equation f(x)=11 has  at most 2 solutions.

It's easy to see that if u is a solution then v = ln(4 1/u )/ln(3) is also a solution.

Manifestly u=2 is a solution, so the oher one is v= ln(2)/ln(3) .


[quote="Rushil"]If $a,b,c$ are positie reals such that $abc = 1$ , prove  that
\[ \frac{1}{\sqrt{ b + \frac{1}{a} + \frac{1}{2}}} + \frac{1}{\sqrt{ c+ \frac{1}{b} + \frac{1}{2}}} + \frac{1}{\sqrt{ a + \frac{1}{c} + \frac{1}{2}}} \geq \sqrt{2} \][/quote]

Nice one. :)

From $\sqrt{1+2x} \le 1+x$ we get: \[\sum^3_{cyc} \frac{ \sqrt{2} }{ \sqrt{ 2\left(b + \frac{1}{a}\right) + 1}} \geq \sqrt{2} \sum^3_{cyc}\frac{1}{b + \frac{1}{a} + 1}\]

Now we substitute $a=y/x, b=z/y, c=x/z$ for our constraint, and we get: \[\sum^3_{cyc}\frac{1}{b + \frac{1}{a} + 1} = \sum^3_{cyc}\frac{y}{x+y+z} = 1\] Q.e.D.

I just have a question( by the way your solution is good!!!)... What is the guiding facor in making the sunstitutions for $a$ as you have done?!!!

This was posted before.

Peter, maybe you should rather give a link than post the solution again...


quadratics
probability
probability and stats

Let $ M_t$ be a continuous martingale that satisfies for all $ t \geq 0$: $ E(M_t^2) < \infty$. Show that the quadratic variation $ V_t(\omega) \equal{} \sup \{ \sum_{t_1 < t_2 < \ldots < t_n < t} |M_{t_i}(\omega) \minus{} M_{t_{i_1}}(\omega)|^2\}$ is almost finite everywhere for a fixed time $ t$. Then show that $ M_t^2 \minus{} V_t$ is a martingale too !

Fix the partition of the interval; then $ EV_t \equal{} E \sum_{t_1 < t_2 < \ldots < t_n < t} |M_{t_i} \minus{} M_{t_{i \minus{} 1}}|^2 \equal{} EM_{t_n}^2 \minus{} EM_{t_1}^2$, which can be proved by opening the brackets and using conditional expectation. We know $ EM_i M_{i \plus{} 1} \equal{} EE(M_iM_{i \plus{} 1}\parallel{}F_i) \equal{} EM_i^2$, so everything in the sum will disappear besides $ M_n^2$ and $ M_1^2$.

what is a martingale and also what branch of mathematics is this(so complex :o )?

just to say that what I wrote is completely false  :D
for example, take a brownian motion: almost every path has an infinite quadratic variation on $ [0;1]$ ..

Well, i'll try to find a mistake in my calculations :)
Of course, i didnt pay attention to $ \sup$ :blush: ; under fixed partition my proof works.

Hi Alekk do you know where I can find a proof that shows that almost surely the quadratic variaition of a Brownian path is infinite ?

thx

I think that it has something to do with Levy modulus of continuity of brownian paths, if I remember well..

Well Alekk, I found an elementary proff ofthis fact at exercise 1.13 page 40 with correction at page 315 of this document which seems to be a good account on Brownian motion :

http://www.stat.berkeley.edu/~peres/bmbook.pdf

see ya

After I took a look at the correction page 315

I find the three following points very unclear :

First in (a) there is a use of Fubini's theorem which I don't understant to justify the fact that almost surely $ N(s)<\infty$ for almost every $ s \in [0,1]$

Second I don't know how this can be used to find a partition every even intervals of which have total length above 1/2 and satisfy equation (*).

Third in (b) I simply don't get why from (a) and the fact that "finite partitions formed with dyadic numbers  are countable" we can deduce that there is a partition  with the property that with probability superior to 1-1/n we have that the quadratic spreads are superior to n.

Any help appreciated


Pretty sure this is the right place to post this (as the forum description mentions ARML.)

I'd like a hint on this problem ([i]not[/i] a solution!)

Given that the sum of all positive integers with exactly two proper divisors, each of which is less than 30, is 2397, find the sum of all positive integers with exactly three proper divisors, each of which is less than 30 (a proper divisor of n is a positive integer that divides but is not equal to n).

Thanks!

Is this an individual or a team round question?

Regardless, think about the formula of the number of divisors of an integer. Then you should realize what form an integer with exactly two proper divisors and an integer with exactly three proper divisors would take. Then you do a little algebraic manipulation and arithmetic to find your answer.

[quote="darkmage2009"]Is this an individual or a team round question?
[/quote]

Not sure. Someone showed it to me and said it was from a mock ARML.

Thanks for the hint!

I know an integer with exactly two proper divisors is in the form $ p_1^2$ and an integer with exactly three proper divisors is in the form $ p_1^3$ or $ p_1 \cdot p_2$, where $ p$ is a prime factor, but I'm still not sure where to go from there.

Bah, I'll have to give this some more thought.

I remember this was from an Altheman mock ARML. Not sure which one, though.

[hide="hint"]
$ 2\plus{}3\plus{}5\plus{}\cdots \plus{}29\equal{}129$. Now think symmetric sums.
And yes, it requires a lot of computation...[/hide]

Let me state a fact that would perhaps help some of the unexperienced readers:

Let $ n\equal{}p_1^{a_1}p_2^{a_2}\cdots p_j^{a_j}$ be the prime factorization of $ n>0$. Then, the number of positive divisors of $ n$ is $ \prod_{k\equal{}1}^j(a_k\plus{}1)\equal{}(a_1\plus{}1)(a_2\plus{}1)\cdots(a_j\plus{}1)$.

[quote="isabella2296"]I know an integer with exactly two proper divisors is in the form $ p_1^2$ and an integer with exactly three proper divisors is in the form $ p_1^3$ or $ p_1 \cdot p_2$, where $ p$ is a prime factor, but I'm still not sure where to go from there.

Bah, I'll have to give this some more thought.[/quote]You're on the right path. Adding the numbers of the first form shouldn't be a problem for you.
[hide="For those of the second form"]
look at the multinomial expansion $ \left(\sum_{sym}a\right)^2\equal{}\sum_{sym}a^2\plus{}2\sum_{sym}ab$. Solve for $ \sum_{sym}ab$. Set up a few bounds on the primes and the rest will be easy. I'm sure you get my idea by now. :)[/hide]

[quote="isabella2296"]Yay, I think I got it!

Is the answer 7282?[/quote]That's right :) Here is your prize:  :first:


1. Find the equivalent resistance of the following bridge network:

2. Obtain the equivalent resistance at the terminals a-b in the following circuit:

3. Obtain the equivalent resistance at the terminals a-b in the following circuit:

[hide="2"]
the 2nd part of this circuit is clearly a balanced whetstone's bridge.
so the $ 10\Omega$ resistor can safely be removed .
rest the two $ 20\Omega s$ and $ 10\Omega s$ are in series and respectively parallel with each other and the $ 30\Omega$ resistance.
so $ R_{eq}\equal{}\frac{1}{\frac{1}{20}\plus{}\frac{1}{40}\plus{}\frac{1}{30}}\equal{}\frac{240}{13}\approx 18.46$
[/hide]

4. Find the equivalent resistance for the circuit shown bellow. All resistors are 1 Ohm.

5. Obtain the equivalent resistance between a-b in the following circuit (all resistors are 30 Ohm):

All the resistences are in cojuction with the resistene $ 25ohm$ and all the other resistences are independent to one of each other so they complete the condition to use $ \frac {1}{R} \equal{} \frac {1}{R_1} \plus{} ..... \plus{} \frac {1}{R_n}$ then we can find the eaquevalent $ R_1$ and finally we can find the resistence of the circuit $ R \equal{} R_1 \plus{} 25ohm$

$ R_1 \equal{} 12ohm$
$ R \equal{} (12 \plus{} 25)ohm$
$ R \equal{} 37ohm$ :maybe:

Three questions:

1) What problem are you answering?

2) What are resistances in conjunction?

3) And what are resistances in disjunction?

1. I aswered problem  3

2. With conjuction "^" fom logical mathematic . In a circuit  the resistences who are depend to each other.

3.  with disjuction . In  a circuit the resistences who are indepedent to each other

Can you tell me how to expres these in another mode.

I have some knowledge of circuit analysis and as far as I know two or more resistances can only be: 1) in parallel; 2) in series; 3) neither in series or in parallel. I have never heard of conjunction and disjunction. Also, I think two resistances in a given circuit can never be independent of each other (unless some special configurations), in the sense that the current in one is always dependent of the presence or absence of another. Finally, your answer to problem 3 is wrong (although very close).

I don't know very well the english so i expres series and parallelles in another mode . Can you give the exact reponse to the problem. Apropos des conjction and disjucion they are terms of the logical math . With them You can write also different kind of circuits .Is like the Pythagorian teorem if you  see  $ a^{2}\plus{}b^{2}\equal{}c^{2}$ you can imagine the triangle. If you undertand what i mean .

[quote="Elio (n)"]Can you give the exact reponse to the problem.[/quote]

The numerical answer will not be of any good for you. Explain your reasoning first.

[quote="Elio (n)"]conjction and disjucion they are terms of the logical math[/quote]

Which are not used (as far as I know) in the elementary description of circuits.

[quote="Elio (n)"]3. with disjuction . In a circuit the resistences who are indepedent to each other[/quote]

First explain what do you mean by "independent" (whenever we use a terminology different from the one commonly accepted, we should at least explain it). Then give me an example of a circuit with "independent" resistances.

[hide="2"] Actually it is $ \; {120 \over 13}\;\approx\;9.23\;\Omega\;.$ [/hide]

3

There is also an balanced Wheatston bridge

 [hide]  \[ R_{eq}\;\equal{}\;{145\over 4}\;\equal{}\;36.25\;\Omega\;.\] [/hide]

Elio (n), could you tell us where are you from?

Yes Immanuel Bonfils, your answers are correct.

So, seems your name has an Italian origin, am I right? Do you live on Tirana?

If Italian is easier to you, may be would be good to have a look at:

http://it.wikipedia.org/wiki/Resistenza_elettrica

 and after that at:

http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Resistors

Yeah  that's true. Where are you from? Bonfils has a french origin , hasn't it ? Are you  french?
Thank you for interesting internet  address .

Just to mention that the discussion is going quite off topic. There are certain types of information that should be exchanged by PMs.

Sorry, but I was just trying to help, having in mind your dialogue with Elio...

About your 4th e 5th problems,it's a quite tedious work (the 5th is a little better)

to solve, requiring either delta-wye convertion Or Kirchoff's laws (at least it's principles).

[hide]4

 \[ R_{eq} \;\equal{}\;{169 \over 55} \; \Omega\;\]

      5

 \[ R_{eq} \;\equal{}\;{100 \over 3} \; \Omega\;\][/hide]

 What about a slight modification,like that?

[img]http://img100.imageshack.us/img100/181/reistances1ch1.th.jpg[/img]

[img]http://img406.imageshack.us/img406/931/reistances2il0.th.jpg[/img]

Your answers are correct: the purpose was to use delta-wye transformations.

I think problem 5. is not a normally electic circuit  I think it is an electronic circuit, am I right?

What Elio is talking about is Logic Circuits. A Logic Circuit is a device that allows us to simplify logical propositions, by assigning to each proposition p,q,r,... a switch, that can be activated if the proposition is true, or not activated if it is false. The proposition simplyfies, if we assign certain proposition to be false and some to be true. Then if the corrient reaches the end of the circuit, the preposition is true for these values assigned to prepositions.

We often think as conjunction as a serie-circuit, conjunction is similar to "and". Like "the sun is high and the sky is blue and it is warm" is like p and q and r. If any of p,q,r is false, the whole preposition is false (the corrient doesn't pass).

For disjunction, we think it as a parallel circuit. Disjunction is like "or". Like "the sun is high or the sky is blue, or it is warm", can be represented as p or q or r. If a single proposition p,q,r is true, the whole proposition is true, so the corrient passes.

Now, having talked about logic circuits, I'll tell why Elio can't apply logical laws to solve a resistance circuit problem:

Logican Preposition Circuit is based on YES/NO switches. Resistances are not YES/NO switches. They have a numerical value, but corrient passes on them and this can not be determined by laws that apply only for Logical Proposition, I mean, YES/NO circuits. We can have a logical proposition to be false or true, not to be 60 ohm, 10 ohm, 3, ohm,... you get it?

On the other hand, resistances can be combined in such a way that they are not in series nor in parallel.

Yes that is very true what Pul de Algodoncito sad. In fact this kind of terms are used for the keys in the circuit.
I know, I have not find the right words to express (series ^ paralel)  :D .But I hoped you understood what I mean.

Hey, I think it can be solved in an easier way, by the method of untrue junction. In the picture it would make no difference in the equivalent circuit if the "pedal triangle " resistor delta was not connected to the arm  joining a and b. this can be done because the resistances on the connected arms are proportional.
After that its just parallel, and the answer is 100/3 ohms.
Try this method on prob 4 too.
And please contact me. tittu1990@yahoo.com


inequalities
linear algebra
linear algebra unsolved

Let $ A,B \in M_{n} \mathbb{(R)}$ such that [b]rank(B)=1[/b]
prove that :
1) $ \det \left[\begin {array} {LL} A & B \\
B & A \end {array} \right] \leq (\det(A))^{2} \leq \det \left[\begin {array} {LL} A & B \\
- B & A \end {array} \right]$.
2) $ \det \left[\begin {array} {LL} A & B \\
B & A \end {array} \right] \det \left[\begin {array} {LL} A & B \\
-B & A \end {array} \right] \leq (\det(A))^{4} \leq \frac {1}{2} ((\det \left[\begin {array} {LL} A & B \\
B & A \end {array} \right])^{2} + (\det \left[\begin {array} {LL} A & B \\
- B & A \end {array} \right])^{2} )$

[quote="Roosta"]First, let $ X = \det (A),Y = \det (B)$, and X and Y are both real numbers. We note that:

$ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| = X^2 - Y^2$
$ \left|\begin {array} {LL} A & B \\
- B & A \end {array} \right| = X^2 + Y^2$

[/quote]

Why can we conclude this ?

I thought that was how determinants of block matrices worked. I am not as experienced with block matrices as I am with the rest of linear algebra, so I guess I was wrong. I will work on this problem some more to see what I can come up with.

[hide="Hint"]Using the assumption $ rg(B)\equal{}1$ show that WLOG every entry of $ B$ can be assumed to be 0 but the left upper element.[/hide]

[quote="Roosta"]What follows, as of now, is wrong, but is still here because I don't know how to delete this post (there is no button for it as there is for others).

First, let $ X = \det (A),Y = \det (B)$, and X and Y are both real numbers. We note that:

$ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| = X^2 - Y^2$
$ \left|\begin {array} {LL} A & B \\
- B & A \end {array} \right| = X^2 + Y^2$

It is clear that $ X^2 - Y^2\leq X^2\leq X^2 + Y^2$, so that:

$ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| = X^2 - Y^2\leq X^2 = (\det(A))^2\leq X^2 + Y^2 = \left|\begin {array} {LL} A & B \\
- B & A \end {array} \right|$

Next, note that:

$ (X^2 - Y^2)(X^2 + Y^2) = X^4 - Y^4\leq X^4\leq X^4 + Y^4 = \frac {1}{2}((X^2 - Y^2)^2 + (X^2 + Y^2)^2)$

Substituting, this becomes:

$ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| \left|\begin {array} {LL} A & B \\
- B & A \end {array} \right| \leq (\det(A))^4\leq \frac {1}{2} \left( \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| ^2 + \left|\begin {array} {LL} A & B \\
- B & A \end {array} \right| ^2\right)$

Equality holds when $ \det (B) = 0$, which should be the case for $ n\geq 2$, because of $ B$'s rank equaling 1.[/quote]
       We have : $ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| = \det(A - B)\det(A + B)$;$ \left|\begin {array} {LL} A & B \\
- B & A \end {array} \right| = \det(A - iB)\det(A + iB)$
you can prove that : $ \det(A - xB)\det(A + xB) = (\det(A))^{2} - x^{2}(\delta)^{2}$
from $ x = 1$ ,$ x = i$
$ i^{2} = - 1$
from here :$ \det(A - B)\det(A + B) = (\det(A))^{2} - (\delta)^{2}$,$ \det(A - xB)\det(A + xB) = (\det(A))^{2} +(\delta)^{2}$ and we have inequality

[quote="QuyBac"]We have : $ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right| = \det(A - B)\det(A + B)$[/quote]Why is this? I am wondering if you used the assumption that $ B$ has rank 1 here?

$ \left|\begin {array} {LL} A & B \\
B & A \end {array} \right|=\left|\begin {array} {LL} A-B & B-A \\
B & A \end {array} \right|=\left|\begin {array} {LL} A-B & 0 \\
B & A+B \end {array} \right|=\det(A - B)\det(A + B)$
 :roll:

[quote="Roosta"]What do you mean when you say "$ \delta$"?[/quote]
You need use [b]Rank(B)=1[/b] then the exist $ P,Q \in GL_n(R)$ such that $ B=Q^{-1}J_{n,1}P$ here $ J_{n,1} \in M_{n}(R)$ have $ J_{11}=1,J_{ij}=0$ if $ (i,j) \neq (1,1)$ Let $ A_1=QAP^{-1}=(a^'_{ij})_{ij}$
and $ \det(A+xB) \det(A-xB)=\det^{2}(Q^{-1}) \det^2(P)\det(A_1+xJ_{n,1})\det(A_1-xJ_{n,1})$ from here you can see $ \delta$
 :roll:


limit
function
algebra
polynomial
real analysis
real analysis unsolved

Let $f: [0,1]\to \Re$ be a continous function. Prove that
      $\lim_{n\to\infty}\frac{1}{2^n}\sum_{k=0}^{n}(-1)^kC_n^kf(\frac{k}{n})=0$

I think I solved it using the fact that the function is uniformly continous and that $C_n^k=C_{n-1}^k+C_{n-1}^{k-1}$...  How can I prove that there a sequence of polynomials that converges to continous function?

[quote="xirti"][hide="hint"]Prove this for polinomyals and use the fact that for every continuous function there exists a sequence of polynomials which converges uniformly to that function. There might be a simpler solution.  :D  [/hide][/quote]
 :blush: I know the fact,but I still don't how to prove this one,can you detail?


Let $ f$ be a polynomial having only 0-1 coefficients such that $ f(1)$ is a prime number and any non-zero root of $f$ is on the unit circle. What are the possible forms of f? Additional question: find the TST problem related to this one... :D  :D

I believe the problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13981]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13981[/url]. It can be reduced to a problem weaker than the one you've posted here, namely to this:

If $1+x+\ldots+x^{m-1}=fg$ for some $m$, where $f$ is as you have described it and $g$ also has $0,1$ coefficients, then the exponents of $f$ form an arithemtic progression.

Am I right? :)

And are you sure the answer is known? I mean, I suppose you don't have a solution, since the problem is in the "Unsolved" section, but since it's not in the "Open Questions" forum, I'm guessing that you know the answer or at least know that the answer is known? :?

Oopps, sorry, I did not posted it in the right forum. It should have been open questions. Unfortunately, I have no idea about the answer.


Support
advanced fields
advanced fields unsolved

The data comprises hours devoted to patient care per week for practitioners in conventional and alternative medical practices.

[b]Part 1:[/b]

Q. Can we infer that the mean amount of time devoted to patient care per week by doctors using conventional methods exceeds that of doctors using alternative methods?  Test at the 10% level of significance.

[b]Results:[/b]
t-Test: Two-Sample Assuming Unequal Variances

	Alternative	Conventional
Mean	49.34615385	53.80769231
Variance	41.43538462	46.80153846
Observations	26	26
Hypothesized Mean Difference	0
df	50
t Stat	-2.421843846
P(T<=t) one-tail	0.009555236
t Critical one-tail	1.298712959
P(T<=t) two-tail	0.019110472
t Critical two-tail	1.675905423

[b]Part 2:[/b]
Q. Imagine the data was organised in a different way.  In this scenario there are 26 medical practices.   Each medical practice has a doctor practicing conventional medicine and a doctor practicing alternative medicine.  Test at the 5% level of significance to see if a difference exists between alternative and conventional methods with respect to the time spent on patient care.

[b]Result:[/b]

t-Test: Paired Two Sample for Means

	Variable 1	Variable 2
Mean	49.34615385	53.80769231
Variance	41.43538462	46.80153846
Observations	26	26
Pearson Correlation	-0.155569189
Hypothesized Mean Difference	0
df	25
t Stat	-2.253213153
P(T<=t) one-tail	0.016633158
t Critical one-tail	1.708140189
P(T<=t) two-tail	0.033266316
t Critical two-tail	2.05953711

Need to know the conclusions for the hypothesis tests condeucted.
View the attachment for the excel files.

For [b]Part 1[/b]: Put

$H_{0}: \mu_{c}\leq\mu_{a}$
Claim=$H_{1}: \mu_{c}>\mu_{a}$ [right-tailed test]

I get the same sample means you do, I skipped the sample standard deviation calculations, but using your numbers I get the

t-stat: $t=\frac{\bar{x_{c}}-\bar{x_{a}}}{\sqrt{\frac{s_{c}^{2}}{n_{c}}+\frac{s_{a}^{2}}{n_{a}}}}=+0.36394379583697833$

but you have "t-stat = -2.42184384601166" on your spreadsheet!

Anyhow, using a t-stat = +0.36394379583697833 which is less than your "t Critical one-tail +1.29871295939665", we should [u]Fail to reject[/u] $H_{0}$, as the [u]data does not support the claim[/u] that [i]"the mean amount of time devoted to patient care per week by doctors using conventional methods exceeds that of doctors using alternative methods".[/i]

[/i]

Promblem is that was calculated in excel and since i have nt done that manually i cannot verify that for you.
which value would be the true value for t-stat


A right circular cone has for its base a circle having the same radius as a given sphere.  The volume of the cone is one-half that of the sphere.  The ratio of the altitude of the cone to the radius of its base is:

$ \textbf{(A)}\ \frac{1}{1} \qquad
\textbf{(B)}\ \frac{1}{2} \qquad
\textbf{(C)}\ \frac{2}{3} \qquad
\textbf{(D)}\ \frac{2}{1} \qquad
\textbf{(E)}\ \sqrt{\frac{5}{4}}$

[hide="Click for solution"]
By the given information we have $ \frac {1}{3} \pi r^2h \equal{} \frac {1}{2} \cdot \frac {4}{3} \pi r^3 \implies \frac {h}{r} \equal{} \frac {2}{1}$ or $ \boxed{\textbf{(D)}}$.
[/hide]

I believe you mean $ \frac{1}{2} \cdot \frac{4}{3} \pi r^3$


as you know , lim $\sqrt{{}{2+\sqrt{{}{2+\sqrt{{}{2+...}}}}}}$
is 2 . it's very easy!
but I want to generalize it
that's : calculate lim $\sqrt{{}{a+\sqrt{{}{a+\sqrt{{}{a+...}}}}}}$
// a :real positive
I know lim is $(\frac tion{1+\sqrt{{}{1+4a}}})/2$ but I want to know more way to solve it
and the more I can't solve
lim $\sqrt[k]{t+\sqrt[k]{t+\sqrt[k]{t+...}}}$
with k:integer and t:real positive
what will it be if k is real positive , or in case of t<0

[quote="nttjuvwamsncc"]$\lim\sqrt[k]{t+\sqrt[k]{t+\sqrt[k]{t+\ldots}}}$
with k:integer and t:real positive
what will it be if k is real positive , or in case of t<0[/quote]
The limit $x$ satisfies equation $\sqrt[k]{t+x}=x\iff x^{k}-x-t=0$. I think that for $k\geqslant 5$ there's no efficient way to find roots of such equations. However, for $k\leqslant 4$ it can be solved by general formulas for roots of polynomial equations of low degree. For example, in the case $\sqrt{a+\sqrt{a+\ldots}}$ you mentioned we have $x^{2}-x-a=0$, which yields $\frac{1\pm\sqrt{1+4a}}{2}$. The solution $\frac{1-\sqrt{1+4a}}{2}$ is negative and can't be the limit of a sequence consisting of nonnegative numbers, so the limit is in fact $\frac{1+\sqrt{1+4a}}{2}$.

[quote="nttjuvwamsncc"][quote="Rzeszut"]
The limit $x$ satisfies equation $\sqrt[k]{t+x}=x\iff x^{k}-x-t=0$.[/quote]
I know that but I want to know WHAT THEORY IS IT BASED ON ?
I only foungd it BASED ON MY SENSE
thank you very much Rzeszut[/quote]

It's based on the fact that increasing bounded sequences converge.  It turns out that solving these essentially reduces to solving polynomials; thus we can't expect an exact solution as the degrees increase.


[hide]outerspace :lol: i think or a tv[/hide]

A record book? a museum?  a tv?  :lol:  :rotfl:

[hide]Mount Everest, Sahara Desert, and Niagara Falls?[/hide]

[quote="mathnerd314"]Where in the world could you go to see the world's tallest mountain, largest desert, and highest waterfall?[/quote]

[hide]how about on the internet?[/hide]

[quote="Walk Around The River"][quote="mathnerd314"][quote="klutzycherry14"]wait...does the question mean like, where u can go to see all of them? like, they are all there at that same place? :maybe:[/quote]

[hide]No. The point of the puzzle is that you're supposed to assume they're all in one place, which would be contradictory. However, nowhere in the problem does it say that. With this in mind, the answer is easy:

Tallest mountain--Everest, in Nepal.
Largest desert--Sahara, in Africa.
Highest waterfall--I thought it was Niagara but apparently I am wrong.[/hide][/quote]

isnt it that one is venezuela? like angel falls?  i think we learned that last year.[/quote]

yes it is..it think it is like 3212 feet or something...

I think I've got it!!!

[hide="stupid answer..."]A library. :P The highest mountain, the largest waterfall, and the largest desert should be in a guiniess world records book.....[/hide]

[hide]The Guinness Book of World Records[/hide]

[quote="mustafa"][quote="probability1.01"]Errr... Niagra Falls is not the world's highest waterfall. Also outer space is not exactly "in the world." But the other answers work.[/quote]

I wasn't sure, so I guessed.

[quote="I know someone is going to say this, so I'll say it for you"]Wikipedia is your friend[/quote][/quote]
No, wiki is your friend and also for this question there are as many opinions as there are men and tow tongued men

I'm just wondering, but who started the "Wikipedia is your friend" thing? Everyone knows this and it's getting pointless. Besides occasionally wikipedia even has mistakes

[quote="Xantos C. Guin"]Sorry, I didn't read your post before posting.[/quote]

isn't the solution already given?

Yes, moogra, so I will start a new "Where in the world" off of this one. I am thinking of a city. This is lateral thinking.

Is it fictional?

Is it the capital of a state/country?

Is it currently a city? (For example an enormous Anasazi civilization in the southwest US collapsed when they ruined the habitat)

klutzycherry14,

you can be consistent.

Don't state the problem is "easy" and refer to yourself "...apparently I'm wrong"
in the same post.   :)


Some  dvi  files  can  be   found  [url=http://terrytao.wordpress.com/books/additive-combinatorics/]here[/url]!

I  want  to convert   them   to  pdf?  I  am  using   Miktex+Texniccenter.

How  to  get  that?

many  thanks!

LaTeX=>PS=>PDF does that for tex files. If you look in Build, Define Output Profiles, Postprocessor you can see how it is done and how you need to do it wthout the original tex file ie dvips (to get PS) and ps2pdf (to get PDF). The command parameters are given in PostProcessor.
Use a batch file (lots of tutorials about batch files on the net) somethng like:
[code]C:\path to\miktex\bin\dvips.exe -P pdf "%1.dvi"
C:\path to\bin\gswin32c.exe -sPAPERSIZE=a4 -dSAFER -dBATCH -dNOPAUSE -sDEVICE=pdfwrite -sOutputFile="%1.pdf" -c save pop -f "%1.ps"[/code]

Thank you very much!

Could  you explain  in  detail?
for  example, you have  a  dvi  file, say, the attachment,  how do you  convert   it  to  pdf?
1,  first......
2, then........
3,..........
4
.......

finally, the pdf   shows  up!

I have just said what to do. For just the one file additive_sample.dvi  here are the DOS commands. You have to put in your own paths of course. Run the command prompt and type:
[code]cd \path to additive_sample.dvi
C:\path to\miktex\bin\dvips.exe -P pdf additive_sample.dvi
C:\path to\bin\gswin32c.exe -sPAPERSIZE=a4 -dSAFER -dBATCH -dNOPAUSE -sDEVICE=pdfwrite -sOutputFile="%1.pdf" -c save pop -f additive_sample.ps[/code]
The batch file I suggested would allow you to do this automatically.

save the file  filename.dvi  in  d:\

start--->run--->cmd--->enter--->dvipdfmx -o d:\filename.pdf d:\filename.dvi--->enter

pdf shows up!


I have read three of Neil's novels, and all of them were among the best books I've ever read. Has anyone else read any of Gaiman's works, and what do you think of them?

If not check it out: [url=http://www.neilgaiman.com]www.neilgaiman.com[/url]

Unfortunately, the only Neil Gaiman book I've read was his biography of Douglas Adams (I mean to read more, but I've never found the time.)  However, I have heard that he is quite good, and the one book I have read supports those reports.

I've read a couple of his books ( Good omens and Neverwhere...)....They're simply amazing....... :)

[i]Good Omens[/i] was excellent; [i]American Gods[/i] less so.
At least, in my opinion.

I was also reading a collection of his short stories in the bookstore a while back and came across a poem/story thing about Santa Claus-and how he was a convict forced to give toys to children as his punishment.  It was exactly 100 words long.
Awesome.

He writes poems as well ???  :?

Hmmm....... Could someone give me the names of any other poems of his ??  :lol:

i read good omens i though it was fairly good, I'm doing a project on postmodernism so I've been reading weirder stuff but the way he described the end of the world was very funny.  I definately plan to read more are the other books by him that you would recommend.

I would recommend anything you can get your hands on. Of the three that I've read (American Gods, Neverwhere, Stardust) and the one I'm currently reading (Anansi Boys), I thought all of them were excellent. However, I would probably recommend Neverwhere the most, but this might be because it is still fresh in my mind.


If A and B are symmetric matrices, then I think, by using several examples, that $ ABA$ is certainly symmetric. But I can't find a way to prove this.

Can anyone prove this?

Thanx

$ (ABA)^T\equal{}A^TB^TA^T\equal{}ABA$. Transposes reverse the order of multiplication, but that works with what we have.

Slightly more generally:  If $ B$ is symmetric then $ ABA^T$ is symmetric (by the same argument).  In other words, symmetry is invariant under [url=http://en.wikipedia.org/wiki/Matrix_congruence]congruence[/url].


inequalities
search
inequalities proposed

Let $z;b;c>0$.prove that:
$\frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3}\leq\sqrt[3]{a.\frac{a+b}{2}.\frac{a+b+c}{3}}$

[quote="N.T.TUAN"]Trivial by Kedlaya's theorem  :wink:[/quote]

whats that?

Search for Keyword ''Proof of a Mixed Arithmetic-Mean, Geometric-Mean Inequality ''  :wink:


here are some notes to those for which these are helpful(mostly projective
geometry) i foud them on net.


Diophantine equation
number theory unsolved
number theory

Prove that there exist infinitely many integers $ n$ such that
$ n,n\plus{}1,n\plus{}2$ are each the sum of the squares of two integers.

Posted before may time.
Here is an easy solution.
Consider $ n\equal{}m^2\minus{}1$
$ n\plus{}1\equal{}m^2\plus{}0^2$
$ n\plus{}2\equal{}m^2\plus{}1^2$
we prove that exist infinite $ m\in N$ such that:$ m^2\minus{}1\equal{}2y^2\equal{}y^2\plus{}y^2$
It is the Pell equation :
$ m_0\equal{}1,m_1\equal{}3$
$ y_0\equal{}1,y_1\equal{}2$
And $ m_{k\plus{}2}\equal{}6a_{k\plus{}1}\minus{}a_k$
       $ y_{k\plus{}2}\equal{}6y_{k\plus{}1}\minus{}y_k$
is also solution of this equation.
So has many n satisfy condition.

Or a bit shorter:
$ 8,9,10$ are all sums of two squares. If $ n \minus{} 1, n ,n \plus{} 1$ are sums of two squares, then so is $ n^2 \minus{} 1 \equal{} (n \minus{} 1)(n \plus{} 1), n^2, n^2 \plus{} 1$. Thus there are infinitely many.

An easy problem :$ k\equal{} 3$ is the largest integer such that each term of the sequence $ x,x \plus{} 1,x \plus{} 2,x \plus{} k$ can represent as sum of two square.

Obviosly $ k < 3$, because else one of numbers $ n \plus{} i \equal{} 3\mod 4$ and for one prime divisor $ p \equal{} 3\mod 4$ $ ord_p(n \plus{} i)$ is odd.
Therefore may be only $ 4n,4n \plus{} 1,4n \plus{} 2$.


[b] Problem 1.[/b] Consider the set $A=\{1,2,3,4,\dots ,2^n\}$, $n\geq 2$. Find the number of its subsets $B \subseteq A$, such that if $x$ and $y$ are different elements of $A$, with sum a power of two, then exactly one of $x$ and $y$ is in $B$.

[i]Aleksandar Ivanov[/i]


Two cylinders of equal radius are intersected at right angles, exact at their perfect middles. Find the volume of the intersection between the two cylinders, without using calculus.

Note: Solved by the mathematician Archimedes (287 B.C. - 212 B.C.), waaay before calculus came around!

Sorry for upping this old topic, but I really need help on this. I didn't figure out how to solve this yet   :(


floor function
number theory unsolved
number theory

Determin value of
$ [2\sqrt{a_n}]$
as
$ a_n\equal{}10^{2n}\minus{}10^n\plus{}1$
for all natural n

For $ n\equal{}0$ it's $ 2$. For $ n\geq 1$ one has $ 4a_n \equal{} (2\cdot 10^n \minus{}1)^2 \plus{} 3$, so $ \lfloor 2\sqrt{a_n} \rfloor \equal{} 2 \cdot 10^n \minus{}1$.

[quote="NikolayKaz"]Determin value of
$ [2\sqrt {a_n}]$
as
$ a_n \equal{} 10^{2n} \minus{} 10^n \plus{} 1$
for all natural n[/quote]

$ 4a_n\equal{}(2\cdot 10^n\minus{}1)^2\plus{}3$ and so $ (2\cdot 10^n\minus{}1)^2<4a_n<(2\cdot 10^n)^2$ and so $ [2\sqrt {a_n}]\equal{}[\sqrt{4a_n}]\equal{}2\cdot 10^n\minus{}1$


Define a sequence  of integers $ (a_n)_n\ge 1$ by
$ \sum_(d dividing n ) (a_d) \equal{} 2^n$ where the sum is taken over the psoitive divisors of $ n$. Prove that $ n$ divides $ a_n$ for all $ n\ge 1$.

Not necessary though it is easy to do it using mobius inversion.
Could you please psot it {Ur mobius inversion soln} or someone else please post the mobius inversion solution as I have never applied it in any problems.
So please help.

It's a PEN problem-
http://www.mathlinks.ro/viewtopic.php?t=150834
some more solutions-
http://www.mathlinks.ro/viewtopic.php?t=219847

I do not understand Zetax's solution which is the only one using mobius inversion.
Could someone please elaborately write the solution using mobius inversion formula? Please. It is very urgent and please help.


superior algebra
superior algebra unsolved

a. Prove that $ R[x]/(x)$ is isomorphic to $ R[x]/(x\plus{}1)$
b. Is $ Z[x]/(x)$ is isomorphic to $ Z[x]/(x\plus{}1)$
Thanks in advance..
Note : $ R$ is the set of real numbers and $ Z[x]$ is the set of integers.

In general, for any ring $ A$, you have $ A[x] / (x\minus{}a) \cong A$.

The specific automorphism is that for each constant $ c$, the class of $ c$ goes to $ c$.


10 integers are given. We know that there is no way of choosing some of them whose sum is divisible by 10. What can be the remainders of the 10 integers when divided by 10?

from the first condition, we get that no number is divisible by 10.
so, the set of possible residues modulo 10 is {1,2,3,4,5,6,7,8,9}.

all the residues cannot be distinct. there will have to be more than one number with the same residue.

I cannot get farther than this :(

if you exclude all doubles which sum to 10, all triples which sum to 10, all quadruples which sum to 10, etc, it will become exceedingly tedious.
appears a very tough problem.

the set of residues {9,9,9,9,9,9}, {8,8,8,8,8,8,8}, {7,7,7,7,7,7,7,},
{6,6,6,6,6,6,6,} work. maybe you can do something with a computer, but I can't see any combinatoric way of doing this.

[quote="dogcat"]10 integers are given. We know that there is no way of choosing some of them whose sum is divisible by 10.[/quote]

I'm not sure this is possible.

Let the ten integers be $ a_1, a_2, \ldots , a_{10}$. Consider the ten partial sums of the form $ \sum^{n}_{i \equal{} 1} a_i$, with $ 1 \le n \le 10$. They can't all be distinct modulo $ 10$; otherwise one of the sums would be divisible by $ 10$. If any two of them are the same modulo $ 10$, say $ a_1 \plus{} \ldots \plus{} a_x$ and $ a_1 \plus{} \ldots \plus{} a_y$, with $ x < y$, then the sum $ a_{x \plus{} 1} \plus{} \ldots \plus{} a_y$ must be divisible by $ 10$, being the difference of two numbers that are congruent modulo $ 10$.

Did I mess up?  :maybe:

[quote="Arrange your tan"][quote="dogcat"] We know that there is no way of choosing some of them whose sum is divisible by 10. [/quote]
What is meant by this sentence?

Is it:

"There is no way of choosing any of the numbers, in pairs,
in triples, in quadruples, and so on, such that their sums
are not divisible by 10."[/quote]
yeah as per given, it means that, and it is impossible as far as i can see
1, 2, 3,4, .., 9
atleast one has to be more than 1 and thereafter its all easy to prove we can get a sum of 10

[quote="mathwizarddude"]I don't think I understood the question; let's say the ten integers are the successive integers from 1 to 10, but then the sum of the pairs 1+9 = 2+8 = 3+7 = ... = 10 are divisible by 10.  Is there any restriction on the ten integers?[/quote]
Yeah: There is no way of choosing any of the numbers, in pairs, in triples, in quadruples, and so on, such that their sums are divisible by 10.

I think leoxnlin is right. (It's not possible.)


geometry
rectangle
geometric transformation
rotation
reflection
counting
distinguishability

1) In how many ways can two rooks by placed on an $8 \times 8$ chessboard such that they do not attack eachother?

2) In how many ways can four squares, not all in the same column or row, be selected from an $8 \times 8$ chessboard to form a rectangle?

Just a couple clarifications:

For (1), do rotations and reflections count as the same?

And for (2), by squares forming rectangles do you mean that if you connect the centers of the squares that it will form a rectangle? And again, do rotations and reflections count as the same?

the chess board has columns a-h and rows 1-8 to form a grid, so rotations/reflections count
[hide="1"]
there are 64 place to put rook#1. the rook attacks an entire column and an entire row, leaving 7 rows and 7 columns (or 49 possible squares) for rook#2.
but since the rooks are indistinguishable, the answer is the numerical value of $\frac{64\times49}{2}$
[/hide]

Rotations are distinct. For 2), go by the centers but you cannot connect them diagonally .

[hide="2"]For 2, let's create an 8 by 8 array of points and attach coordinates to them for ease - each point is $(x,y)$ where $x$ and $y$ belong to $\{ 1,2,3,4,5,6,7,8 \}$, with $(1,1)$ on the lower left.

We will consider only the lower left and upper right corners, since these two will uniquely define the rectangle.

For a point with coordinates $(a,b)$ of the lower left coordinate, the upper right coordinate can be anything from $(a+1,b+1)$ through $(8,8)$, so there are a total of $(8-a)$ possibilities for the horizontal coordinate and $(8-b)$ possibilities for the vertical cordinate.

We now simply want the sum of $(8-a)(8-b)$ as $a$ and $b$ range from $1$ through $7$. Note that this is equivilent to the sum $ab$ as $a$ and $b$ range from 1 through 7, which factors to $(7+6+5+4+3+2+1)(7+6+5+4+3+2+1)=\boxed{784}$[/hide]

[hide="2 tricky 4 you"] Pick two rows and two columns and take the four intersection squares. $\binom{8}{2}^{2}= 28^{2}= 784$.[/hide]

[quote="The Zuton Force"]My bad  :lol:[/quote]

Oh, your solution was good, too, I wasn't directing the hide tag comment at you lol.


Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages $ 40$ miles per hour, he arrives at his workplace three minutes late. When he averages $ 60$ miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

$ \textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$

[hide]IT's the harmonic mean of 40 and 60, which is 2*40*60/(40+60)=48,B.[/hide]

Lets say that he has to get to work in $ a$ hours. Say that the distance that he travels is $ d$ miles. From the equations, we have
\[ d\equal{}\left(a\plus{}\frac{1}{20}\right)\cdot 40\]
\[ d\equal{}\left(a\minus{}\frac{1}{20}\right)\cdot 60\]

We want to find the value of $ \frac{d}{a}$. Divide both sides by the speeds and add both sides. The extra 3 minutes cancel

\[ \frac{d}{40}\plus{}\frac{d}{60}\equal{}2a\implies \frac{d}{a}\equal{}\frac{2}{\frac{1}{40}\plus{}\frac{1}{60}}\equal{}48\] Hence the answer is $ \boxed{B}$.


From a pile of 100 pennies (P), 100 nickels (N), and 100 dimes (D), select 21 coins which have a total value of exactly 1 dollar. In your selection you must use at least one coin of each type. How many coins of each of the three types (P,N,D) should be selected?

How would you do this problem?

[hide="Solution"]
Let the number of pennies be $P$, the number of nickels be $N$, and the number of dimes be $D$.

Now, we have two equations:  $P+N+D=21$ and $P+5N+10D=100$.

Subtracting the first from the second we obtain: $4N+9D=79 \iff N= \frac{79-9D}{4}$

Now, $N$ must be an integer, so we have $79-9D = 4(19-2D)+(3-D)$ is divisible by $4$, thus we must have that $3-D$ is divisible by $4$, which gives $D=3,7,11, \ldots$.

Surely, we can only have either $D=3$ or $D=7$. Testing, both values of $D$ give solutions:
\[\boxed{ (P,N,D) = (10,4,7) \text{ or }(P,N,D) = (5,13,3) }\]
[/hide]

[hide="solution"]Since you have to have one coin of each, we have to have $84$ cents with 18 coins.

We also need 4 more pennies because a penny is the only coin worth $1\bmod 5$.  That leaves $80$ cents and 14 coins.

Notice that if the rest of the coins were nickels, that would be $70$ cents so we need $10$ more cents.  This can be done by replacing 2 nickels with 2 dimes.

Therefore, there are 5 pennies, 13 nickels, and 3 dimes.[/hide]

(there are more than one solution by the way.)

[quote="jli"](there are more than one solution by the way.)[/quote]
Then, in your solution, you should probably find all of the solutions...  :P


Hi, I have these two problems that neither I nor the teacher could get; I've just totally forgotten how to.
Any help would be greatly appreciated.

$e^x+e^{-x}=3$

and

$2^x+2^{-x}=5$

Wow I never even thought of that. Thanks so much!


symmetry
inequalities unsolved
inequalities

Tiks , post  this in the inequalty section and they will take care of it for you.

It 's[tex] \frac ba[/tex] or[tex] \frac{b}{a+b}[/tex]

non of them, its $\frac{a-b}{a+b}$
i think , i solved this before and there is  no need of the exact power $11$ , it should be just  an odd number.

Not a bad problem... Obviously, it suffices to prove that $\sum{(\frac{1-x}{1+x})^{11}}\leq 1$ if $xyz=1$.  We have two cases (because of the symmetry):
  Case 1: $x$ is the only number among the three which is smaller than 1. Then it is clear, since the terms corresponding to y,z in the LHS are negative and the term corresponding to x is at most 1.
  Case 2: $z,y$ are smaller than 1. We must prove that $1+(\frac{x-1}{x+1})^{11}\geq (\frac{1-y}{1+y})^{11}+(\frac{1-z}{1+z})^{11}$.  But it is clear that it is enough to prove that $\frac{x-1}{x+1}\geq\frac{1-y}{1+y}$ , which comes down to $ xy\geq 1$ or $z\leq 1$, clearly true.


Find all function satisfying the relation  $ R\minus{}\minus{}\minus{}> R$ : $ f(x\plus{}1)\minus{}3f(x)\plus{}1\equal{}0$

The function $ g(x) \equal{}f(x) \minus{} 1/2$ satisfies $ g(x\plus{}1) \equal{} 3g(x)$, which means $ g$ is uniquely determined by its (arbitrary) values on any left-closed right-opened interval of length $ 1$, e.g. $ [0,1)$.


calculus
integration
logarithms
function
trigonometry
calculus computations

$ y\equal{} \ln x$ as a real function is only defined for $ x>0$. Then $ \frac{dy}{dx} \equal{} \frac{1}{x}$ for $ x>0$.

So $ \int \frac{1}{x} \ dx \equal{} \ln x \plus{} c$ for $ x>0$.

How do we deal with $ \int \frac{1}{x} \ dx$ for $ x<0$?

Or $ \int \cot x\ dx\equal{} \ln \mid \sin x\mid \plus{}c$?

Thanks JRav. I realise that. What I was wondering is how to integrate $ \ln x$ when $ x<0$.

The function $ \ln x$ isn't defined when $ x < 0$, so you certainly can't integrate it there.  But that's not the question you actually appear to have asked.  The answer to the question you did ask is that over $ ( \minus{} \infty, 0)$ we have $ \int \frac {1}{x} \, dx \equal{} \ln( \minus{} x) \plus{} C ( \equal{} \ln |x| \plus{} C)$.  (This is what JRav's answer means.)  Similarly, the other formula you gave is valid on any interval in which $ \cot x$ is defined and nonzero (and in those intervals, $ \sin x$ is of constant sign).

I prefer to think of the integral of $ \frac1x$ as two separate formulas for two separate domains. Any convergent integral will confine itself to just one of those domains. That is, $ \int_2^5\frac1x\,dx$ and $ \int_{\minus{}4}^{\minus{}3}\frac1x\,dx$ both make sense but $ \int_{\minus{}2}^6\frac1x\,dx$ is divergent.

Suppose we know that $ x<0.$ Make the substitution $ u\equal{}\minus{}x,du\equal{}\minus{}dx,$
\[ \int\frac1x\,dx\equal{}\int\frac1{\minus{}u}\,(\minus{}du)\equal{}\int\frac1u\,du\equal{}\ln u\plus{}C\equal{}\ln(\minus{}x)\plus{}C.\]
So the two formulas for this integral are
\[ \int\frac1x\,dx\equal{}\begin{cases}\ln x\plus{}C&\text{if }x>0\\ \ln(\minus{}x)\plus{}C&\text{if }x<0\end{cases}\]
Writing that as $ \ln|x|\plus{}C$ is simply a notational convenience to get it all onto one line, nothing more. It's still basically two formulas.