inequalities function calculus derivative blogs AMC USA(J)MO

Lagrange Theorem:
Let $ f$ be a continuous and differentiable function of several variables in an interval. Also let $ g_{i}\equal{} 0\;(i \equal{} 1,2,\cdots,k)$ be conditions that must be satisfied. Then the maximum or minimum values of $ f$ with the conditions $ g_{i}\equal{} 0\;(i \equal{} 1,2,\cdots,k)$, occur in the bounds of the interval or occur in the points which the derivative of $ L \equal{} f\minus{}\sum^{k}_{i \equal{} 1}\mu_{i}g_{i}$ ($ \mu_{i}$'s are real numbers) in relation to the variables of $ f$, gets zero.

Now let's solve $ 2$ problems here so we can understand the theorem completely! :)

$ \textbf{Problem1: }$For real and not necessarily positive numbers $ a,b,c$ where $ a\plus{}b\plus{}c > 0$, Prove that:
$ \boxed{a^{3}\plus{}b^{3}\plus{}c^{3}\leq\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)^{\frac{3}{2}}\plus{}3abc}$
$ \textbf{Problem2: }$We have $ a\plus{}b\plus{}c\plus{}d \equal{} 1$ where $ a,b,c,d$ are positive reals, Prove that:
$ \boxed{abc\plus{}bcd\plus{}cda\plus{}dab\leq\frac{1}{27}\plus{}\frac{176}{27}abcd}$

$ \textbf{Solution of Problem1: }$Its easy to check that if we define:
$ A \equal{}\frac{a}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}$
$ B \equal{}\frac{b}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}$
$ C \equal{}\frac{c}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}$
Then the main inequality will be:
$ A^{3}\plus{}B^{3}\plus{}C^{3}\leq\left(A^{2}\plus{}B^{2}\plus{}C^{2}\right)^{\frac{3}{2}}\plus{}3ABC$
Where obviously $ A^{2}\plus{}B^{2}\plus{}C^{2}\equal{} 1$, so now its sufficient to solve the following inequality with only one condition that is $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$:
$ a^{3}\plus{}b^{3}\plus{}c^{3}\leq1\plus{}3abc$
Here we use the Lagrange Theorem, let:
$ f \equal{} a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc$
$ g \equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}1 \equal{} 0$
(Notice that in all the conditions $ g_{i}$ the LHS must be zero.)
We have to prove that the maximum value of $ f$ is $ 1$ or we always have: $ f\leq1$
Let: $ L \equal{} f\minus{}\mu g \equal{} a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc\minus{}\mu\left(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}1\right)\Longrightarrow$
$ \frac{\partial L}{\partial a}\equal{} 3a^{2}\minus{}3bc\minus{}2\mu a \equal{} 0\Longrightarrow\mu \equal{}\frac{3\left(a^{2}\minus{}bc\right)}{2a}$
$ \frac{\partial L}{\partial b}\equal{} 3b^{2}\minus{}3ca\minus{}2\mu b \equal{} 0\Longrightarrow\mu \equal{}\frac{3\left(b^{2}\minus{}ca\right)}{2b}$
$ \frac{\partial L}{\partial c}\equal{} 3c^{2}\minus{}3ab\minus{}2\mu c \equal{} 0\Longrightarrow\mu \equal{}\frac{3\left(c^{2}\minus{}ab\right)}{2c}$
$ \Longrightarrow\frac{2\mu}{3}\equal{}\frac{a^{2}\minus{}bc}{a}\equal{}\frac{b^{2}\minus{}ca}{b}\equal{}\frac{c^{2}\minus{}ab}{c}$
$ \Longrightarrow\frac{a^{2}\minus{}bc}{a}\equal{}\frac{b^{2}\minus{}ca}{b}\rightarrow a^{2}b\minus{}b^{2}c \equal{} ab^{2}\minus{}a^{2}c\rightarrow(a\minus{}b)(ab\plus{}bc\plus{}ca) \equal{} 0$
In a same way we have $ (b\minus{}c)(ab\plus{}bc\plus{}ca) \equal{} 0$ and $ (c\minus{}a)(ab\plus{}bc\plus{}ca) \equal{} 0$. By solving these three equations we must have $ ab\plus{}bc\plus{}ca \equal{} 0$ or $ a \equal{} b \equal{} c$, but we also have $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$ so:
$ (i)$If $ a \equal{} b \equal{} c$ then $ a \equal{} b \equal{} c \equal{}\plus{}\frac{1}{\sqrt{3}}\rightarrow f(a,a,a) \equal{} 0$

$ (ii)$If $ ab\plus{}bc\plus{}ca \equal{} 0$ then $ (a\plus{}b\plus{}c)^{2}\equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}2(ab\plus{}bc\plus{}ca) \equal{} 1$
And because $ a\plus{}b\plus{}c > 0$ we'll have $ a\plus{}b\plus{}c \equal{} 1$ and:
$ f \equal{} a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{} (a\plus{}b\plus{}c)\left(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca\right) \equal{}\plus{}1$
As a consequence the minimum is $ 0$ and the maximum of $ f$ is $ \plus{}1$ and we're done.  :D

$ \textbf{Solution of Problem2: }$ Here similarly let:
$ f \equal{} abc\plus{}bcd\plus{}cda\plus{}dab\minus{}\frac{176}{27}abcd$
$ g \equal{} a\plus{}b\plus{}c\plus{}d\minus{}1 \equal{} 0$
We have to prove that always we have: $ f\leq\frac{1}{27}$
Let: $ L \equal{} f\minus{}\mu g \equal{} abc\plus{}bcd\plus{}cda\plus{}dab\minus{}\frac{176}{27}abcd\minus{}\mu a\minus{}\mu b\minus{}\mu c\minus{}\mu d\plus{}\mu$
$ \frac{\partial L}{\partial a}\equal{} bc\plus{}cd\plus{}db\minus{}\frac{176}{27}bcd\minus{}\mu \equal{} 0$
$ \frac{\partial L}{\partial b}\equal{} ac\plus{}cd\plus{}da\minus{}\frac{176}{27}acd\minus{}\mu \equal{} 0$
$ \frac{\partial L}{\partial c}\equal{} ab\plus{}bd\plus{}da\minus{}\frac{176}{27}bad\minus{}\mu \equal{} 0$
$ \frac{\partial L}{\partial d}\equal{} bc\plus{}ca\plus{}ab\minus{}\frac{176}{27}bca\minus{}\mu \equal{} 0$
$ \Longrightarrow\mu \equal{} bc\plus{}cd\plus{}db\minus{}\frac{176}{27}bcd \equal{} ac\plus{}cd\plus{}da\minus{}\frac{176}{27}acd \equal{} ab\plus{}bd\plus{}da\minus{}\frac{176}{27}bad \equal{} bc\plus{}ca\plus{}ab\minus{}\frac{176}{27}bca$
Now notice that:
$ bc\plus{}cd\plus{}db\minus{}\frac{176}{27}bcd \equal{} ac\plus{}cd\plus{}da\minus{}\frac{176}{27}acd\Longrightarrow(b\minus{}a)\left(c\plus{}d\minus{}\frac{176}{27}cd\right) \equal{} 0$
by the same way we'll have the following equations too:
$ (b\minus{}c)\left(a\plus{}d\minus{}\frac{176}{27}ad\right) \equal{} 0$
$ (b\minus{}d)\left(a\plus{}c\minus{}\frac{176}{27}ca\right) \equal{} 0$
$ (a\minus{}c)\left(b\plus{}d\minus{}\frac{176}{27}bd\right) \equal{} 0$
$ (a\minus{}d)\left(c\plus{}b\minus{}\frac{176}{27}cb\right) \equal{} 0$
$ (c\minus{}d)\left(a\plus{}b\minus{}\frac{176}{27}ab\right) \equal{} 0$
By solving these six equations we must have $ a \equal{} b \equal{} c \equal{} d(\textbf{Why?})$ and so by the condition $ a\plus{}b\plus{}c\plus{}d \equal{} 1$ we'll have $ a \equal{} b \equal{} c \equal{} d \equal{}\frac{1}{4}$ and:

$ f(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}) \equal{}\frac{1}{4^{3}}\plus{}\frac{1}{4^{3}}\plus{}\frac{1}{4^{3}}\plus{}\frac{1}{4^{3}}\minus{}\frac{176}{27}\cdot\frac{1}{4^{4}}\equal{}\frac{1}{27}$

And we're done!  :lol:

It is a very good explanation. Let's see if I got it right.

Here's a problem from [b]sinajackson[/b]'s blog:

We have  $ x_1+x_2+...+x_n=1$. Where $ x_i$'s are positive reals.
Prove that:
$ \boxed{ \sum x_i(1-x_i)\leq \frac{n-1}{n}}$

[hide="Solution"]Let:
$ f= x_1(1-x_1) + x_2(1-x_2-1) + ... +x_n(1-x_n)$
$ g=x_1+x_2+...+x_n-1=0$

$ L= x_1-x_1^2+x_2-x_2^2...+x_n-x_n^2-\mu(x_1+x_2+...+x_n-1)$

$ \frac {\partial L}{\partial x_i}=1-2x_i-\mu=0$
$ \mu=1-2x_i$      $ \forall x_i$
And so:
$ x_1=x_2=...=x_n=\frac{1}{n}$
And:
${ f\leq \frac{n-1}{n}}$

Am I doing it right?[/hide]

Thats Right [b]skuge[/b], nice solution, some other cool and easy ones: :)

$ \text{(USAMO 77/5)  }$ If $ a,b,c,d,e$ are positive reals bounded by $ p$ and $ q$ with $ 0<p\leq q$, prove that:

$ \left(a\plus{}b\plus{}c\plus{}d\plus{}e\right)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\plus{}\frac{1}{e}\right)\leq25\plus{}6\left(\sqrt{\frac{p}{q}}\minus{}\sqrt{\frac{q}{p}}\right)^{2}$

$ \text{(USAMO 80/5)  }$Show that for all non-negative reals $ a,b,c\leq1$,

$ \frac{a}{b\plus{}c\plus{}1}\plus{}\frac{b}{c\plus{}a\plus{}1}\plus{}\frac{c}{a\plus{}b\plus{}1}\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c)\leq1$

Just a note, those two problems can actually be solved very easily using a convex/concave argument. Sorry that's kind of irrelevant. :oops:

[quote="Altheman"]Computationally, the solutions here are correct.

However, you would get 0 points for not including certain conditions.

You need to show that grad $ g_i\ne 0$(zero vector)

If your domain isn't only the constraint, you have to (like if you have x\ge 0, then you have to check that too). If your domain is open (or partially open) then I don't know what you have to do to be rigirous.

The gradient of f and g_i must be continuous.[/quote]

Well in fact I was worried about that. I would like to know what other conditions must be included to be completely rigorous. [b]Altheman[/b], can you give us an example of a complete solution?

Well here is the most basic situation: we have f a function and g=k a restraint. We must show
i) grad g is not zero
ii) grad f and grad g are continuous in the domain
iii) the independent variables are a closed and bounded set
The the absolute max/min occur when grad f = K* grad g.

Now we can make the problem a bit more complicated by putting simple nonstrict inequality restraints on the variables (ex. $ x\ge 0$).

The theorem then says that the max/min may occur on one of these boundaries (when the strict inequality becomes equality). Hence we must systematically check these boundaries.

Notice that iii is necessary for the situation, say, maximize x^2+y^2+z^2 given that x+y+z=1 where x,y,z are real numbers.

If we have a strict inequality, I do not know what to do exactly.

This theorem might be of use as well:

Extreme value theorem: let f(x_1,x_2,...,x_n) be a continuous function on a closed and bounded subset of R^n, then f attains maximum and minimum values.


The Maxwell's relation in electromagnetic theory states that refractive index is equal to the square root over the di-electric constant.

This relation holds for some simple gases.But more generally,it does not.My question is why this is the case.
Eugene Hecht says that this is because n (refractive index) is a function of frequency.But I do not understand.After all, di-electric constant is also frequency dependent quantity.

What I think is that for di-electric substances,due to the very structure of the substances,the frequency is only measurable at low frequencies.So,that when we try to match the calculated n (observed in visual range of EM spectrum) with the di-electric constant =Limit (frequency--->0) [K(w)]   (where K is the di-electric constant and w is the frequency),the two values do not match.

Please confirm if I am correct.

http://en.wikipedia.org/wiki/Refractive_index
http://en.wikipedia.org/wiki/Phase_velocity

the formula is $n=\sqrt{\epsilon_{r}}\sqrt{\mu_{r}}$

thats why, read the "Definition" in the first link

I know that.But in visible range we may safely use the definition I wrote.

you make a approximation and after you say "the result is wrong"

dont make the subsituition $\mu_{r}= 1$ and everybody will be happy

yagaron,you may be contented with that arguement.But the reality is more subtle.What I wrotethere was correct.


Six is the coefficient of x, and seven is the constant.  To isolate x, subtract the constant from both sides, then devide both sides by the coefficient of x.

$6x+7 = 0$
$6x=-7$

from here to get x alone,
we must divide 6 to both sides.

$6x/6 =-7/6$
and then 6x/6 would be x so

$x=-7/6$

[quote="jenny221"]Solve 6x + 7 = 0[/quote]

[hide]
6x=-7
x=-7/6
[/hide]

Could you seriously not figure this out :?: [hide]6x=-7, x=-7/6[/hide]

Goto http://www.TheMathWebSite.com and print out as many practice sheets as you need until you have mastered factoring quadratics.  All the sheets have solutions and answers.
Good Luck! :D

@mathalete: Look. You post a problem here for help.

Secondly, another thing is that pre-algebra, standard US math, is, I believe, 6-7th grade. Seriously. I have a looot of friends who are in a different school than me, and they are the most advanced in their math class, and they are in 6th grade doing prealgebra. If you don't want to answer easyy questions, then don't come here. No need for contumely.

@kstan013: you've been to mc nat's so you probably think its easy, but please don't post comments like this.

@jrsjrs: that's high school website...and this isn't quadratics

@inom: same goes for you

obviously the answer if -7/6. But some people in our school are not even in the [i]prealgebra[/i] class.  :( our school sucks at math. We're planning to join the mathcounts comp (county) yet we havent even hired the coach :(. basicly; some schools suck and we can't help it if there are people who are "not as cappable" in subjects in math.

[quote="Eli"][quote="Mathalete"]Yea, why would you even need to post such a thing? That's 4th grade math! You need to stop posting this junk you're doing.[/quote]
3 things:
1. This is very rude
2. This is still classroom math
3. The original poster might be in 4th grade, for all you know.[/quote]I agree that was quite rude. As for #c, have you looked at where else the author has posted? I saw many High School Basics/Intermediate Topics :wink:

[quote="ragnarok23"][quote="Mathalete"]Yea, why would you even need to post such a thing? That's 4th grade math! You need to stop posting this junk you're doing.[/quote]
if I remember you could barely multiply fractions in 4th grade[/quote]

Actually, I could multiply fractions in 2nd grade, quit math in 4th grade. If I can remember correctly, you couldn't even add fractiosn in 4th grade.

EVERYONE PLEASE!!! The person who posted this must have seriously had a question! As for all of you who think you know the answer, please just post an answer and leave this topic alone. If you all think you know a lot, go to the higher topics  to exhibit your knowledge. We're here to help and learn from our mistakes! thank you!  :wink:


Does the sequence of squares contain an infinite arithmetic sequence?

possibly not. because $f(n)=(n+1)^{2}-n^{2}=2n+1$ is strictly increasing.

[hide="Solution"]
No. It doesn't.

Assume there is.

Say the common difference between terms is $d$.

Then, for consecutive terms $a_{k}^{2}$ and $a_{k+1}^{2}$ in the sequence, we will have:
\[a_{k+1}^{2}-a_{k}^{2}= (a_{k+1}+a_{k})(a_{k+1}-a_{k})=d \]
Now, since both $a_{k}$ and $a_{k+1}$ are positive integers, for each distinct factorization of $d$ into two integers, we will obtain one distinct, solvable system (not necessarily in the integers, though) of two equations in these two variables.

But, since there must be only a finite number of distinct factorizations of $d$, there can only be a finite number of these systems of equations, and thus, only a finite number of terms in the arithmetic sequence. Contradiction. [/hide]

I interpreted it to mean whether or not there is an arithmetic subsequence of the sequence of squares that has an infinite length.

[hide="Solution"]
Consider an arithmetic subsequence of $S_{n}=n^{2}$ with common difference is $d$. The difference $S_{n+1}-S_{n}=(n+1)^{2}-n^{2}=2n+1$, so the shortest distance from $S_{n}$ to a higher term is $2n+1$. However, $\forall d \in \mathbb{R}, \; \exists n \; : \; 2n+1>d$ $\implies \forall d \in \mathbb{R}, \; \exists n \; : \; S_{n+1}-S_{n}> d$.

In other words, the differences between consecutive squares will eventually become larger than the common difference, and such an arithmetic sequence would have to terminate.

[b]Therefore, any arithmetic subsequence of the sequence of squares must terminate.[/b]

This is in fact true for any sequence defined by a function that is eventually concave up. On the other hand, many concave-down functions that increase without bound (such as $\log_{n}$ and $\sqrt[k]{n}, \; k \in \mathbb{Z}$) have such subsequences. [/hide]

[quote="The Zuton Force"]I interpreted it to mean whether or not there is an arithmetic subsequence of the sequence of squares that has an infinite length.
...
This is in fact true for any sequence defined by a function that is eventually concave up.[/quote]

Your first sentence is a non-sequitur -- you've interpreted it the same way the previous two posters did, and your method is essentially the same as the one suggested by nayel.  The second sentence I've quoted doesn't really make sense -- what you were trying to say is that a monotonically increasing sequence has an infinite subsequence which is arithmetic implies that its first differences are bounded.

Oops, I didn't see nayel's. I wasn't sure after looking at cincodemayo's, which talks about factorizations...

Thanks for clarifying the second.

[hide="WOP"]Note that the difference between two squares is in the form $2n+1$ where $n$ is increasing. Assume that such a sequence existed. If we chose an index $i$ and another index $i+k_{1},$ the other distinct indexes $i+k_{2}, i+k_{3}, \cdots$ are determined where $k_{i}> 0.$ Now, since the sequence $k_{1}, k_{2}, k_{3}, \cdots$ is monotonically decreasing and $k_{i}\in \mathbb{Z}^{+},$ there must be a smallest value $k_{m}$ in the sequence. But, note that $k_{m+1}< k_{m}.$ A contradiction exists, so there is no such infinite sequence. $\blacksquare$[/hide]

Edit: WOP = Well Ordering Principle

[b]Solution.[/b] Assume to the contrary that there exists an infinite arithmetic progression $0\leq a_{1}^{2}<a_{2}^{2}<a_{3}^{2}<\hdots$. We may assume that $a_{i}\geq 0$.  It follows that
\[a_{2}^{2}-a_{1}^{2}=a_{3}^{2}-a_{2}^{2}\implies (a_{2}-a_{1})(a_{1}+a_{2})=(a_{3}-a_{2})(a_{2}+a_{3}).\]
Since $a_{1}+a_{2}<a_{2}+a_{3}$, $a_{2}-a_{1}>a_{3}-a_{2}$.
Similarly, we obtain an infinite sequence of decreasing positive integers $a_{2}-a_{1}>a_{3}-a_{2}>a_{4}-a_{3}>\hdots$.  Contradiction! $\Box$

[quote="LordoftheMorons"]Uh... can't you find a sequence of some arbitrary length, though?[/quote]

Perhaps surprisingly, no.  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111997
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82742

Also interesting is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74178


inequalities
geometry
inradius
circumcircle
geometry proposed

Here are two problems on geo ineqs.
I admit that they are not hard.
However, give a try and ...have fun!  :)

Generalizations, comments, further examples and problems are left for [i]Virgil[/i] (and not only...)  :D

Well, here  goes:

[color=darkred]A)[/color]
Show that in each triangle having sidelengths $a, b, c$, inradius $r$
circumradius $R$ and semiperimeter $p$ the following inequality holds:
 $\sqrt{1+4\frac{R}{r}}\geq 1+\sqrt{\frac{p-a}{p-b}}+\sqrt{\frac{p-b}{p-a}}$ .

[color=darkred]B)[/color]
Prove that in any triangle the following ineq is true:
$(a+b+c)^3>4[a(a^2+bc)+b(b^2+ac)+c(c^2+ab)]$  .

 :)

[hide="A"]Plausible[/hide]
[hide="B"]$(1+1+1)^3> 4\cdot 3\cdot [1(1^2+1\cdot 1)]$. Plausible and this. [Corrected][/hide]

[quote="Virgil Nicula"]B: $(1+1+1)^3\ne 4\cdot 3\cdot [1(1^2+1\cdot 1)]$. Isn't plausible.[/quote]

Why?
Are you reffering to the degree??

 :)

Using the relations
$4R+r=r_a+r_b+r_c$ (ex-circle radii)
and
$\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=\frac{1}{r}$

we need to prove that

$\sqrt{(r_a+r_b+r_b)(\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c})}\geq 1+\sqrt{\frac{p-a}{p-b}}+\sqrt{\frac{p-b}{p-a}}=1+\sqrt{\frac{r_b}{r_a}}+\sqrt{\frac{r_a}{r_b}}$

which is true owing to cauchy schwarz ineq.

$\sqrt{(r_b+r_a+r_c)(\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c})}\geq 1+\sqrt{\frac{r_b}{r_a}}+\sqrt{\frac{r_a}{r_b}}$

  :)

I have another solution for B.
$(a+b+c)^3-4[a(a^2+bc)+b(b^2+ac)+c(c^2+ab)]=3(b+c-a)(c+a-b)(a+b-c)>0$
:D  :D

[quote="Virgil Nicula"][hide="A - It is O.K."] I shall use the wellknown relations $(2p=a+b+c):$
$(p-a)+(p-b)+(p-c)=p;\ (p-a)(p-b)(p-c)=pr^2;$
$(p-a)(p-b)$ $+(p-b)(p-c)+(p-c)(p-a)=r(4R+r).$
Denote: $x=p-a,\ y=p-b,\ z=p-c$, where $x>0,\ y>0,\ z>0$. Therefore,
$1+\sqrt{\frac{p-a}{p-b}}+\sqrt{\frac{p-b}{p-a}}\le \sqrt{1+\frac{4R}{r}}\Longleftrightarrow$ $\left( 1+\sqrt{\frac xy}+\sqrt{\frac yx}\right)^2\le \frac{(x+y+z)(xy+yz+zx)}{xyz}\Longleftrightarrow$
$1+2+2\left(\sqrt{\frac xy}+\sqrt{\frac yx}\right)+\frac xy+\frac yx\le 3+\sum \frac{x+y}{z}\Longleftrightarrow$ $2\left(\sqrt{\frac xy}+\sqrt{\frac yx}\right)\le \frac{x+y}{z}+\frac{z(x+y)}{xy}\ \ (1).$
From the A.M.-G.M. inequality $(n=2)$ results:
$\frac{x+y}{z}+\frac{z(x+y)}{xy}\ge 2\sqrt{\frac{x+y}{z}\cdot \frac{z(x+y)}{xy}}=\frac{2(x+y)}{\sqrt{xy}}=2\left(\sqrt{\frac xy}+\sqrt{\frac yx}\right)$,
i.e. the relation $(1).$
==================================================================[/hide] [hide="B - It is O.K."] You are true, Socrates ! I will prove that
$\boxed {\ (a+b+c)^3> (a+b+c)^3-3(b+c-a)(c+a-b)(a+b-c)=4\sum [a(a^2+bc)]\ }\ \ (1).$
[b]A solution.[/b]
$E\equiv (a+b+c)^3-4\sum [a(a^2+bc)]=s^3_1-4S_3-12s_3=$
$=s^3_1-4(s^3_1-3s_1s_2+3s_3)-12s_3=-3s^3_1+12s_1s_2-24s_3=$
$=-3(s^3_1-4s_1s_2+8s_3)=$ $-3[8p^3-8p(p^2+r^2+4Rr)+32Rrp]=$
$=24pr^2=24(p-a)(p-b)(p-c)=3(b+c-a)(c+a-b)(a+b-c)>0\Longrightarrow (1)$
[u]Remark.[/u] I used the following notations and wellknown properties:
$s_1\equiv a+b+c=2p;\ \ s_2\equiv ab+bc+ca=p^2+r^2+4Rr;\ \ s_3\equiv abc=12Rrp.$
$S_2\equiv a^2+b^2+c^2=s^2_1-2s_2;\ \ S_3\equiv a^3+b^3+c^3=s^3_1-3s_1s_2+3s_3.$
$S\equiv [ABC]=pr;\ S^2=p(p-a)(p-b)(p-c);\ abc=4RS$$;\ (p-a)(p-b)(p-c)=pr^2.$[/hide] I fill with joy by the .... Greek's presents ![/quote]

Thanks! I am happy you liked them!! :lol:


Suppose a refrigerator has a COP of 5.5.  a) How much energy does it consume in removing 1850 cal from its interior?
b) What power rating will this refrigerator have if it can remove 1850 cal from its interior each minute?

Any help would be appreciated.  I do not even know how to set the problem up.

Thanks.

Sorry for the late answer...

A refrigerator is an example of a cooling machine -- its job is to remove the heat $Q_2$ from the interior by using work $A$ and to release the heat $Q_1$ into the surroundings. In your problem, you have the heat $Q_2$. I'm not sure what COP stands for -- in words -- but it's got to be the cooling coefficient defined by $k=\frac{Q_2}{A}$. The 'consumed energy" is the work $A$, so $A=\frac{Q_2}{k}$. The power of the engine is $P = \frac{A}{t} = \frac{Q_2}{kt}$.

Thank you so much!  I think that I understand it, but does it matter that the work units are calories?  Do they have to be in Joules?  So, in the end, my power is measured in J/s or Watts?  Or can it be cal./s?

J/s are Watts, of course :) You do need to convert the work from calories to Joules in order to obtain the answer in Watts; if you keep calories, you'll get cal/s.


modular arithmetic
number theory unsolved
number theory

$ k,m,n \in \mathbb{N}$

Prove that
$ 1^m\plus{}2^m\plus{}3^m\plus{}\ldots \plus{} (n^k\minus{}1)^m\plus{}(n^k)^m| n^{k\minus{}1}$

$ 1^{m} \plus{} 2^{m} \plus{} 3^{m} \plus{} \ldots \plus{} (n^{k} \minus{} 1)^{m} \plus{} (n^{k})^{m}| n^{k \minus{} 1}$ $ \implies$
$ n^{k \minus{} 1} > (n^{k})^{m}$ it is false
i think  this question is wrongly

It is obliged to be so

$ (1^m\plus{}2^m\plus{}3^m\plus{}\ldots \plus{} (n^k\minus{}1)^m\plus{}(n^k)^m)| n^{k\minus{}1}$

[quote="aktyw19"]$ k,m,n \in \mathbb{N}$

Prove that
$ 1^m \plus{} 2^m \plus{} 3^m \plus{} \ldots \plus{} (n^k \minus{} 1)^m \plus{} (n^k)^m| n^{k \minus{} 1}$[/quote]

Is it
$ n^{k\minus{}1}|(1^m\plus{}2^m\plus{}3^m\plus{}\ldots \plus{} (n^k\minus{}1)^m\plus{}(n^k)^m)$

ok

So prove that

$ n^{k \minus{} 1}|(1^m \plus{} 2^m \plus{} 3^m \plus{} \ldots \plus{} (n^k \minus{} 1)^m \plus{} (n^k)^m)$

$ i)2\nmid m$ $ \implies$  $ 1^{m} \plus{} 2^{m} \plus{} .. \plus{} (n^{k})^{m}$
$ 2|n$ $ \implies$  $ n \equal{} 2l$ $ \implies$  $ 1^{m} \plus{} 2^{m} \plus{} .. \plus{} l^{m} \plus{} ... \plus{} (n^{k})^{m}\equiv 0\pmod{n^{k \minus{} 1}}$
similarly $ 2\nmid n$  but I didn't solved for $ 2|m$


In the preamble (before \begin{document}) put [code]\usepackage{makeidx}
\makeindex[/code]then each word that you want to index you put \index{}. So if you want [i]mathematics[/i] to appear in the index you type [code]\index{mathematics}[/code] Finally, put [code]\printindex[/code]where you want the index to appear. You will need to compile the file at last twice before the index appears, even if you're just updating the document.

You can find more detail in section 4.3 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to Latex2e[/url]. Other language versions can be found [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]here[/url].

For example, this index

[img]http://img403.imageshack.us/img403/6695/inha6.png[/img]

I add in the preamble

\usepackage{makeidx}
\makeindex

But I didn't get where put \index{mathematics} and \printindex

I hope you can explain me anyway how did you do that index, which looks like very good.

Thanks for everything

That's not an index but a table of contents. In LaTeX language the index is at the end of the document where you look up words in the text. For example you might look up [i]mathematics[/i] and find it's on pages 1, 6, 200. But the list chapters and sections at the beginning of a document, like your example, is called a table of contents.

That's very simple to do because all you do is use \tableofcontents wherever you want it to appear and it will automatically list your chapters and sections. You will have to compile twice to get it.

Thanks, but I'm a complete novice at LaTeX

Then where do I put \tableofcontents?

[quote="stevem"]use \tableofcontents wherever you want it to appear[/quote]

You typically put the table of contents near the beginning of your book or article.

Thanks, but.. could you give me a little example please?

You normally put \tableofcontents just after \begin{document} but it can come later if you have a title page or something else you want first. So you have
\documentclass {...}
\usepackage{...}
...
\begin{document}
This is a title or a first page
\tableofcontents
....
rest of document
\end{document}

Thank you very much man, I really appreciate all of this.


I know this has been brought up somewhere, but I can't find it anymore!  How can we send in images if we are using Latex without drawing in Latex (it seems really hard to learn and I don't really want to do it).  Can we fax you the drawings or attach in JPEG(I dont know any virus that can lie in pictures)?

You may attach jpeg files and email them to us.

In any event, please don't email your solutions but fax your pictures -- send your entire submission via one method.


function
trigonometry
integration
logarithms
real analysis
real analysis unsolved

I am sorry but what is $ q$ here? is it a function or just conjugate of $ p$?

It was just the abuse of notations  :) I edited the post.

[quote="Yustas"]Suppose that for the sequence of positive functions $ f_n\in L_p(\mu), \ 1 < p < \infty$ such that $ \log(f_n)\in L_r(\mu)$, $ r < \infty$, we know $ f_n\to f$ weakly, $ \log(f_n)\to q$ weakly. Does this imply $ f \equal{} e^q$ a.e. for $ r$ big enough? $ \mu$ is a finite measure on $ R$.[/quote]No, this conclusion does not hold. A simple counterexample: Take $ f_n(x) \equal{} e^{\sin (2n \pi x)}$ for $ x \in (0,1)$.

Then $ f_n \to c \equal{} \int_0^1 f_1(x) dx \approx 1.26606587775384...$ weakly in $ L^p(0,1)$, but $ \log f_n \to 0$ weakly in $ L^p$, for all $ p< \infty$. (To show the weak convergence, just use smooth periodic test functions and change variables.)


The first one. Now let 17091982=n Now we should prove that (n!)^2 more than n^n. Now n! = n(n-1)....1 And (n!)^2 = (1.n)(2.(n-1))... (n-k)(k+1) for k = 0,1,2.... nk+n-k^2 -k is bigger than n because n ise more than k-1. So we're done. [/u][/code]

We claim that $17091982!^2$ is larger than $17091982^{17091982}$. To do so, we generalize this for all integers $n>2$.
We rewrite $n!^2$ as $(n\cdot 1) \cdot (n--1)2 \cdot (n-2)3 \cdot ... \cdot 2(n-1)\cdot 1(n)$ (has $n$ amount of terms). It is simple to see that all of the terms are greater than $n$ except for the first and last, which are equal to $n$.
Therefore, $n!^2>n^n$ for all integers $n>2$, proving that $17091982!^2$ is greater than $17091982^{17091982}$. $\square$

EDIT: 200th post :D


Given $ a, b, c$ satisfy $ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4.$
Prove that: $ (2\minus{}a)(2\minus{}b)(2\minus{}c) \geq\ abc$
PS:Am-Gm can killed it  :)

I dont think it must be used AMGM
$ Let a\equal{}2cos A.b\equal{}2cosB.c\equal{}2cosC$
So this is a very well-know ineq $ (1\minus{}cosA)(1\minus{}cosB)(1\minus{}cosC)\ge cosAcosBcosC$


function
advanced fields
advanced fields unsolved

Let $f: \mathbb{Z}\longrightarrow \{0,1\}$ and $n\in \mathbb{N}$, $n\geq 1$, such that the number of the words

$f(x)f(x+1)\ldots f(x+n-1) , x\in \mathbb{Z}$

are $\leq n$. Prove that the function $f$ is periodic, namely, there is $0\neq T \in \mathbb{Z}$ such that $f(x+T)=f(x), \forall x\in \mathbb{Z}$.

Alexandros

What about this question?

Have you any ideas?

Alexandros

You can easily modify the proof in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90063]this[/url] topic. In your terminology, that problem deals with functions $f$ defined on $\mathbb N$instead of $\mathbb Z$, and asks for a proof that in the same conditions as in your problem, such a function is [i]eventually[/i] periodic, so it's slightly different, but a practiaclly identical technique takes care of this one as well.

P.S.

Notice that the "alphabet" you're using doesn't matter; it
s not necessary for $f$ to take only two values.

P.P.S.

This problem belongs in the Combinatorics section.

[quote="grobber"]P.S.

Notice that the "alphabet" you're using doesn't matter; it
s not necessary for $f$ to take only two values.
[/quote]

Yes I know, but the problem becomes more attractive if the function $f$ takes only two values instead of taking the abstract values $a_{1},a_{2},\ldots,a_{n}$

Alexandros

P.S. Thanks for your answer


[quote="2006"]Let $\ a,b,c$ be positive real number such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=ab+bc+ca.$
Prove that $\ a+b+c\leq3$
        good luck  :D[/quote]
This problem is wrong!!!!! :P
The sign in the ineq. should be reversed! :P
Let's prove the folowing ineq.
$(a+b+c)^{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}\geq 27(ab+bc+ac)^{2}$ for any positive real numbers.
We can assume that $a+b+c=3$,then we need to prove that
$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ac$<=>
$a^{2}+b^{2}+c^{2}+2(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$
As you can easy see $a^{2}+2\sqrt{a}=a^{2}+\sqrt{a}+\sqrt{a}\geq 3a$(AM-GM),so summing the last ineq. for all tree varibles a,b,c we will have that
\[a^{2}+b^{2}+c^{2}+2(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 3(a+b+c)=9\]
so we proved that
\[(a+b+c)^{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}\geq 27(ab+bc+ac)^{2}\]
and if $\sqrt{a}+\sqrt{b}+\sqrt{c}=ab+bc+ca$=>
\[(a+b+c)^{3}\geq 27\]
<=>
\[a+b+c\geq 3\]
 :wink:


Let {a[n]}_n be a sequence in R satisfying the condition :
a[1]>0 , a[2]>0 , a[n+2]=(2004+a[n+1])/(2004+a[n]) for every n in N.
Prove that {a[n]}_n has a limit and find it .

I solved the same problem but with 2 instead of 2004. Check it in analisys section.


calculus
integration
real analysis
real analysis unsolved

Find:
1. $ \int e^{x}\frac {tgx}{lnx} \,dt$
2. $ \int \frac{dx}{dx}$
3. Is it true that: $ \int e^{x} \equal{} f(un)$?

Nothing is not wrong. These are not serious integrals. Their intention is just to improve your mood/humour and to provocate your thinking.
--
math10.com

Third one asks if sex = fun ;-) - it have no common with real math of all. But the other two are far more interesting.


6.Find all nonnegative integer solutions (x,y,z,w) of the equation

(2^x).(3^y) - (5^z).(7^w) = 1.

It was already solved on the forum, of course. You don't need to post it again. Just look at the Current Olympiads pane on the main page http://www.mathlinks.ro/Forum/portal.php and choose CMO 2005.


I need to use De Moivre's theorem to prove trigonometrical identites, can anyone help me with this?

Given that

             cos x		=	[u]ejx  +  e–jx[/u]
			                                             2

	sin x		=	[u]ejx  –  e–jx[/u]
				          2j

Prove that Sin 2 X+ Cos2 x = 1

When you square and add you get,

[(e^ix + e^-ix)/2]^2 + [(e^ix - e^-ix)/2i]^2

just expand and you get

1/4 . [e^2ix + e^-2ix +2 -(e^2ix + e^-2ix -2)]

= 1/4 .(4)
= 1
as required.
Note that I've used i instead of j(as per engineer notation)


Compute in terms of x.
$ \sum_{n \equal{} 1}^\infty 3^{n \minus{} 1}\sin^3\left(\frac {x}{3^n}\right).$

Cf.
[quote="Sugaku Koshiki II (Mathematical Formulas II) (Iwanami Shoten, Tokyo, 1987) [in Japanese]"]
$ \sum_{n \equal{} 1}^\infty 3^n \sin^3 \frac {x}{3^n} \equal{} \frac {3}{4}(x \minus{} \sin x)$
$ \sum_{n \equal{} 1}^\infty 2^{2n} \sin^4 \frac {x}{2^n} \equal{} x^2 \minus{} \sin^2 x$[/quote]


[hide]You multiply 5 and 7 to get 35, so $\log_3 (35)$.[/hide]

[quote="easyas3.14159..."][hide]Log (base a)b+Log (base a)c=Log (base a)bc, so Log (base 3)35.[/hide]

Sorry, I don't have tex... :blush:[/quote]
You don't need to "have" $\LaTeX$... Just put dollar signs ($\$$) before+after the expression ;)

For example...
[code]\log_a b + \log_a c=\log_a bc[/code]
If you put dollar signs to the code above, you'll see something like $\log_a b + \log_a c=\log_a bc$ :)


Let $ N_m$ be the set of all integers not less than a given integer $ m$. Find all functions $ f: N_m\rightarrow N_m$ such that

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

for all $ x,y\in N_m$


How many different ways can nickels, dimes, quarters or half-dollars be combined to equal one dollar if no more than four coins of any type may be used?


geometry
trapezoid
USAMTS
AMC
USA(J)MO
USAMO
congruent triangles

Why must we prove that the trapezoid is isosceles. Isn't it trivial enough?
I want to know why this is in the grading criteria so I can be more rigorous in the future.

How is it trivial? Obvious by looking at the diagram, maybe, but not trivial.

Well, if something is obvious from the diagram its might not be trivial... if we were allowed to do that geometry problems on olympiads would be much easier...

You should prove that they are equal unless its stated in the problem.

[quote="worthawholebean"][quote="jtw"]Well, In the diagram I had I put the lengths of the radii that were on the equal sides. Did I really need to show that they were equal?[/quote]No, but you did have to show that it was a trapezoid.[/quote]
Not necessarily; I didn't show that the centers form a trapezoid, and I didn't lose any points for it.

[quote="jtw"]Well, In the diagram I had I put the lengths of the radii that were on the equal sides. Did I really need to show that they were equal?[/quote]

Thats exactly the issue; the radii below are corresponding and equal here .. and .. it's not a trapezoid.

[quote="azjps"]
[asy]pen f = fontsize(10);pair C1 = MP("1",(0,0),NE,f), C2 = (2,0), C4 = MP("3",4*expi(100*pi/180),SE,f), C3=IP(CR(C2,4),CR(C4,6));D(CR(C1,1));D(CR(C2,1));D(CR(C3,3));D(CR(C4,3));D(D(C1)--D(C2)--D(C3)--D(C4)--cycle,linewidth(0.7));[/asy]
[/quote]

I showed rigorously that it is a trapezoid; for some reason, I was worried that I assumed that the center of the 5th circle was in the interior of the 4 circles  :blush:

Clearly usamts grading isn't as strict as usamo grading...

Well, you can use some similar (or congruent) triangles to show that its an isoceles trapezoid.

OK, I didn't realize that equal sides don't mean isosoles trapezoid.
It's just frustrating because that pretty much blew my gold chances.


Prove that For all positive integers $m$ and $n$ , one has $| n \sqrt{2005} - m | > \frac{1}{90n}$

If $\vert n\sqrt{2005} - m \vert < \frac{1}{90n}$ then $\vert n\sqrt{2005}+m \vert < 2(\sqrt{2005}+1/90)n < 90n$.  But we have $\vert 2005n^2 - m^2 \vert \ge 1$.  So $\vert n\sqrt{2005}-m \vert \cdot \vert n\sqrt{2005} + m \vert \ge 1$.  Contradiction.


AMC
AMC 10
AMC 10 B
AMC 12
AMC 12 A
AIME
AMC 8

My AMC Profile:

AMC 8 2006: 23
AMC 10A 2007: 114
AMC 10B 2007: XXX Shooting for a 132  :blush:

I just happen to have not taken the AMC before this year, so I'm sure others have much longer history

Oh and one more thing. No complaining about how you could have gotten blah blah points more.  :rotfl:

[quote="nat mc"]8th grade: 2005 AMC 10A: 144

9th grade: 2006 AMC 12A: 143
[color=white]9th grade: [/color]2006 AMC 12B: 139.5

10th grade: 2007 AMC 12A: 144[/quote]
Dude, how'd you get 139.5?

@iamagenius, trajirou is in 11th grade IIRC

8th Grade:
2006 AMC10B: 146.5 (Yes, I'm a wuss. Get over it)

9th grade:
2007 AMC12A: 138

Mine's a rather short history, eh?

Age: 11; 6th grade: AMC10: 111

Age: 12; 7th grade: AMC 10: 138

Age: 13;8th grade: 2007 AMC 10A: 132

6th grade (age: 9) : AMC 10 A: 129
7th grade (age: 10) : AMC 10 A: 137
8th grade (age: 11) : 2007 AMC 10 A: 150! :lol:  :D

[quote="somebody"] Dude, how'd you get 139.5?[/quote]

Shame on somebody.

He got 22 correct and left 3 blank.

I have a less than impressive one compared to some of you guys, but whatever:

7th Grade:
2003 AMC 8 - 22

8th Grade:
2004 AMC 8 - 24

9th Grade:
2005 AMC 10A - 122.5
2005 AMC 12B - 111

10th Grade:
2006 AMC 12A - 126.5
2006 AMC 12B - 137

11th Grade:
2007 AMC 12A - 150

AMC 8: Never

AMC 10: Never (Planning to take it this fall)

AMC 12: Never (Planning to take it this fall)

Yeah, haha funny. 98.5 isn't possible. Nor is 117.5 GG

All right, I'll play, just for the anachronism of it all:

1969 AHSME: 72
1970 AHSME: about 88?
1971 AHSME: about 85?

Three-time school winner, Sooner HS, Bartlesville, OK.

Notes:

1. The scoring system was different - a completely blank test would have been a zero (not 37.5 or 62.5 or anything like that) and negative scores were possible. But the maximum possible was still 150.

2. There was no AIME.

3. I vaguely remember seeing a list on which that 1970 score was the 3rd highest score in Oklahoma.

4. That high school no longer exists, although the building is still in use, as a "mid-high" school for 9th and 10th graders. (Or so it was the last I heard, several years ago.)

[quote="nealth"][quote="gighiuhui"][quote="nealth"]AMC10:117.5
AMC12:98.5
Hope to make AIME (or usamo possibly) next year[/quote]

Dude delong you're funny except WHOA a 97.5 this year on the 12 would have ALREADY qualified you for AIME.

but nice try.[/quote]
huh, bubula, why did u disguise yourself?
and who is delong?[/quote]

Actually, I was [b]not[/b] the author of that post.

As probability1.01 says, everything written by Delong (in the form of nealth or trigonometry1.01 or whatever) must be taken with a grain of salt.

[quote="bubala"]As probability1.01 says, everything written by Delong (in the form of nealth or trigonometry1.01 or whatever) must be taken with a grain of salt.[/quote]
A grain?

I bid two or three bucketfuls at [i]least[/i]. ;)

now updated

7th grade:

Amc 8: 21
Amc 10A: 117(what luck)
AIME 2: 2  :(

My History:

That's it. :blush: Compounded, of course, by the fact that I'll be in 11th grade next year. Yea, I'm a late starter, but at least I'm working on it!

(Never heard of all these contests before I joined this forum actually; I blame the math teachers)

[quote="Kent Merryfield"]All right, I'll play, just for the anachronism of it all:

1969 AHSME: 72
1970 AHSME: about 88?
1971 AHSME: about 85?

Three-time school winner, Sooner HS, Bartlesville, OK.

Notes:

1. The scoring system was different - a completely blank test would have been a zero (not 37.5 or 62.5 or anything like that) and negative scores were possible. But the maximum possible was still 150.[/quote]

Kent, that's interesting to see how different it is scored now and then.  It seems to change every few years.  And hey, congratulations on doing so well.

I never did math contests in high school.  I was the ballet dancer.  My senior year, though, a student who was scheduled to take a competition had to drop out.  Someone suggested me as the replacement.  I had no idea what a math competition was, but I took it.  I placed in the top ten, which surprised a lot of people, including myself.

In college, once, when I was a dance major, I saw a flier about a test the math department was giving.  So for the heck of it, I showed up and took the test.  Or tried.  I had never done a proof, except the ones in high school geometry.  I didn't even recognize a lot of the notation on the test, like the binomial coefficients.  I did pretty poorly.  But it got me curious about math, and it was one of many factors that contributed to my decision to change from a dance major to math and sciences, and eventually get a doctorate in theoretical physics.

The name of the test was the Putnam.  :-)

Best,
Catherine

I remember those AHSME scores better than I remember my Putnam scores. I think I took the Putnam twice; I think my scores were positive - that's all I can say. In the immediate family (me, my wife, my son, my daughter) we'll claim 9 cases of being a school winner on the AHSME or AMC. Of course, jmerry's record is on a whole different level than the rest of us.

[quote="jmerry"]Since my record has been invoked, here it is:

1997: 135 AHSME, 6 AIME, USAMO qualifier (decent score, not MOP).
1998: 130 AHSME, 7 AIME, USAMO HM, MOP.
1999: 145 AHSME, 14 AIME, USAMO HM, MOP.
2000: 138 AMC-12, 13 AIME, USAMO HM.

That 14 AIME was amazing even to me. Among southern CA students, only Tedrick has matched it.[/quote]

That's impressive.  Congratulations.  :-)  It was even harder to get those scores when there was no A or B tests.

One thing I've noticed, when students take both the A and the B AMC tests, is that errors can crop up when they report their scores in other places, such as summer applications, ARML practices, or college applications.  I think it's a good idea students can take it twice, for those who normally would do well and might have had a bad day.  However, it's possible for someone looking up those scores to see the lower one and not realize the student has two scores.

This is especially important on applications.  Students should always note if they took both the AMC A and AMC B, and say the score they are reporting is the higher of the two.  Certainly the higher score should be reported first, but students should at least mention the existence of the other one, even give the number.  Otherwise, if an administrator or coach goes to check on the score and sees the lower one first, they may think that student misrepresented his or her score.

For the medals, I've heard students outside schools wonder "How could that person have been the school winner with such a low score?"  It's always useful to double-check the published scores for both the A and B to make sure.

One thing I like, as a coach, is that when students do it twice, two students in the school have a chance to be the school winner.  I don't think I've ever had a case where the same student won the AMC-12A and AMC-12B.  It happened only once with the AMC-10.

Best,
Catherine

8th - 2006

AMC 8 = 19
AMC 10A = 86.5
AMC 12B = 67.5

9th - 2007

AMC 10A = 112.5
AMC 12B = 82.5

Yeah I know I'm an idiot.

I'm purchasing the Art of Problem Solving Book 1 soon from this website.


vector
function
topology
advanced fields
advanced fields unsolved

in this book , lemma 2.4 said if a trivial n-bundle has a metric，then there are  n orthonormal  sections。I can't quite
understand the proof：
n-bundle is trivial means there are n linearly independent sections，This part is OK。
At each base point, using Gram-Schmit process，make the n vectors orthonormal。
then these n new sections “clearly” are continuous，

I can't see why “clearly” they are continuous: using Gram-Schmit process point-wise，
how to insure continuous？did I miss something obvious?

[quote="Third Edition"]Can you see that if $ X$ is a continuous section of the vector bundle and $ \varphi$ is a continuous function on the base manifold, that $ \varphi X$ is a continuous section?[/quote]

 :blush: so easy


Here are some ARML practice problems that I meant to distribute on Saturday for my geometry talk, but never got around to. I want everyone on the South Carolina Team to try these at some point and post your solutions in "Hide" mode.

If you've already seem them I'll post some new problems for you to try.

SC Assistant Team Coach
Mr. Neal

EDIT: I thought these were gonna be geometry problems.... :|

[hide="I3"]Notice that they're infinite series

so it will be $(\frac{1}{1-a})(\frac{1}{1-b})=\frac{1}{1-(a+b)+ab}=\frac{1}{1-(\frac{4}{11}+(\frac{-2}{11}})=\boxed{\frac{11}{5}}$[/hide]

Wow, why does no one else want to do these :huuh:  :fool:

I'll just post the solutions to the ones i know how to do:

[hide="I1"]Since she was paid $k$ dollars a year for $17$ years and then was paid another million for each of the $x$ years her team made the playoffs, we have $17k+x=72$, since $x$ needs to be less than $17$, than $x=\boxed{4}$ is the only solution that works out. [/hide]
[hide="I2"]
We have to use power of point (for those who don't know it, check out my "Circles Lesson"). $PB \cdot PA=PC \cdot PD$, where $D$ and $C$ are the points of intersection of the circle concurrent w/ line $OP$. Since $OD=OB=PD$, without loss of generality, let us call each of them, or in other words, the radius of the circle $1$, thus $PB=\sqrt{5}$, now $\sqrt{5} \cdot PA=3 \Longrightarrow PA=\frac{3}{\sqrt{5}}$, thus $\frac{PA}{AB}=\boxed{\frac{3}{2}}$. [/hide]
[hide="I3"]
Notice that they're infinite series, so it will be $(\frac{1}{1-a})(\frac{1}{1-b})=\frac{1}{1-(a+b)+ab}=\frac{1}{1-(\frac{4}{11}+(\frac{-2}{11}})=\boxed{\frac{11}{5}}$[/hide]
[hide="I4"]Look at my diagram. Use the law of cosines. $PQ^2=4^2+2^2-2(2)(4)(\cos{60}) \Longrightarrow PQ=\boxed{2\sqrt{3}}$[/hide]
[hide="I5"]Using the change of base formula, we can simplify the expression to $\frac{\log{c}}{\log{b}}=\frac{19}{99}$, since this is $\log_{b}{c}=\frac{19}{99} \Longrightarrow b^{\frac{19}{99}}=c$, $\frac{b}{b^\frac{19}{99}}=b^{\frac{80}{99}}=c^{\frac{80}{19}}$, so $k=\boxed{\frac{80}{19}}$[/hide]

The rest of these problems are beyond my level...so good luck doing them yourself!

Bumping lots of ARML problems for the Carolinians to try.


Drops of water fall continually from height $ h$ (each drop is identical). After some time a pattern of a circle will form in the ground (that is perfect flat, etc etc).

How can we calculate Theoretically the radius of this circle?

[hide="My Approach"]Each drop has energy $ E_0\equal{}m g h$. Wen it collide with the ground the drop is (for simple purpose) subdivided in $ n$ smaller drops of mass $ \frac{m}{n}$. The energy is not conserved. But we can assume that for some subdivision the energy is $ \frac{E_0}{n}$ because it don't make contact with the ground, just with the water behind it (it's this valid?).
This sub-drop will fly upward forming a angle with the normal of the ground. Since the probability of each angle exist, to find the maximum distance we make theta equal to 45 degrees.
The problem is $ n$. There is no way to find $ n$ (assuming this Semi-conservation of energy holds)[/hide]

The answer should depend on the strength of [url=http://en.wikipedia.org/wiki/Surface_tension]surface tension[/url] - that's the only thing bounding the radius of the circle.  The answer should also depend on the properties of the ground (dirt vs. linoleum, for example).  Energy (in the form you've used it) doesn't enter into the computation - a puddle is an essentially static situation.


A 2-inch elastic band is fastened to the wall at one end, and there's a bug at the other end. Every minute (beginning at time $0$), the band is instantaneously and uniformly stretched by 1 inch, and then the bug walks 1 inch toward the fastened end. Will the bug ever reach the wall?

I think I must have misinterpreted the question because in my solution, the bug does reach the wall, and the number of steps is just a messy and uninteresting calculation.

Did you have the rubber band stretch instantaneously once per minute as in the problem, or did you have the rubber band stretch at a constant rate of 1 inch/minute?

That's the only misinterpretation I can think of.

I'm getting:

t = 0: bug is 3 from the wall, elastic is length 3
t -> 1: bug is 2 from the wall, elastic is length 3

t = 1: bug is 2*4/3 =8/3 from the wall, elastic is length 4
t -> 2: bug is 5/3 from the wall, elastic is length 4

t = 2: bug is 5/3*5/4=25/12 from the wall, elastic is length 5
t -> 3: bug is 13/12 from the wall, elastic is length 5

t = 3: bug is 13/12*6/5=13/10 from the wall, elastic is length 6
t -> 4: bug is 3/10 from the wall, elastic is length 6

t = 5: bug is 3/10*7/6 = 7/20 from the wall, elastic is length 7
t -> (5+7/20) bug reaches wall.

So if I interpreted it correctly, it takes the bug $5.35$ time units to reach the wall. But I agree, it's not a terribly interesting calculation.

That's exactly what I got. If someone knows what the problem should be, please speak up.

[quote="nateharman1234"]How about if the rubber band stretches a mile each time, does the bug still reach the end?

Still not that hard of a problem:
[hide]
Let k be the number of inches in a mile after one minute the bug is 1/k of the way there, after two minutes it goes another inch but this time its out of 2k so thats a total of 1/k + 1/2k.  This pattern continues and after n minutes it is 1/k + 1/2k + ... 1/nk of the way. Which is the same as 1/k ( 1 + 1/2 + 1/3 ... 1/n), and sinc ethe harmonic series diverges it will eventually be above k (and I stress eventually) and the bug will reach the end.

As I said, not much harder but a certainly more interesting result[/hide][/quote]

And since the harmonic series diverges, it will be true for any stretch distance. That bug just needs to keep on truckin'.  :P

([i]Another Question[/i]: If I told you how long the bug took to finally reach the wall, could you figure out how long the rubber band was stretched each time?)


[quote="THECHAMP"]Triangle ABC has vertices of A(0,0), B(2,-4) and C(x,y). C is in quadrant 3. What coordinates for C would make ABC a right triangle.[/quote]
[hide]There are several options, we can have either $ A$, $ B$, or $ C$ as the right angle. I'll do the first one and you can do the rest. Let's make $ A$ that right angle. If the line segments, $ AB$ and $ AC$ make a right angle, they are perpendicular and the product of their slopes is $ \minus{}1$. The slope of $ AB$ is $ \minus{}2$. The slope of $ AC$ is $ \frac{y}{x}$. Thus $ \frac{y}{x}\equal{}\frac{1}{2}$. Since $ C$ is in the 3rd quadrant, we have that it can be $ (\minus{}1,\minus{}2), (\minus{}2,\minus{}4), (\minus{}3, \minus{}6)...(\minus{}n,\minus{}2n)$ where $ n$ is a positive integer. [/hide]


I am unable to find the solution to this following question:-

a:b=5:3[color=blue](corrected from 3:5)[/color]
c: d=4:7
a+2c:b+d=3:2

Find a-2c:2b-d

Can anybody help me out?

$12a+5c=0$ and $63b+25d=0$ from here you can find it!!!

im so stupid, i meant to type a:b=5:3! sorry bout the mistake, can you work out with the correction please? and possibly show how to get the answer?


Hello, I have problems with prooving this one:

2^(2n+3)  -  1 > n(n+1(2n+1)

While solving this, I am geting to certain "point" where I am unsure if that is proof or not.

Help :)

[hide]

Notice that the LHS is equal to $ 2\plus{}2^2\plus{}2^3\plus{}\cdots \plus{} 2^{2n\plus{}1}\plus{}2^{2n\plus{}2}$ and the RHS $ 6(1\plus{}2^2\plus{}3^2\plus{}\cdots \plus{} n^2)$. Rewrite the LHS as $ 2(1\plus{}2)\plus{}2^3(1\plus{}2)\plus{}2^5(1\plus{}2)\plus{}\cdots \plus{} 2^{2n\plus{}1}(1\plus{}2)\equal{}3(2^1\plus{}2^3\plus{}\cdots \plus{} 2^{2n\plus{}1})$.

$ 3(2^1\plus{}2^3\plus{}\cdots \plus{} 2^{2n\plus{}1}) > 6(1\plus{}2^2\plus{}3^2\plus{}\cdots \plus{} n^2)$

$ (\frac{1}{2})(2^1\plus{}2^3\plus{}\cdots \plus{} 2^{2n\plus{}1}) > 1\plus{}2^2\plus{}3^2\plus{}\cdots \plus{} n^2$

$ 1\plus{}2^2\plus{}2^4\plus{}\cdots \plus{} 2^{2n} > 1 \plus{}2^2\plus{}3^2\plus{}\cdots \plus{} n^2$

Now just show $ 2^{2n}>n^2$ for $ n\ge 1$ by induction.

Base case. k=1

$ 2^2 \equal{} 4 > 1^2\equal{}1$.

Inductive step. Assume $ 2^{2k}> k^2$

We want to prove $ 2^{2k\plus{}2}>(k\plus{}1)^2$

When n increases by 1, the LHS is multiplied by 4, while we add to 2n+1 to right side. Since $ n\ge 1$, multiplying by 4 increases the number more than adding 2n+1, so therefore $ 2^{2n}>n^2$, and we are done.

[/hide]


function
real analysis
real analysis unsolved

let $f$ be an integrable function then  $\{x|f(x)\neq 0\}$ is $\sigma$ finite measure

Let $E_{n}=\{x : |f(x)|>1/n\}.$ (Or whatever your favorite sequence that goes to zero is.) What do you know about the measure of $E_{n}?$


My brother will be in 6th grade next year, and I'm thinking that he'll want to do MathCounts.  So normally, he would just take the school test, and if he makes to team, he can go on to later rounds.

However, my school district is arranged a little differently - we have a middle school, which is for 5th and 6th graders, and a junior high, for 7th and 8th graders.  The junior high has a MathCounts team this year, but not the middle school.  So I'm just wondering whether he should just try to get a team started at his middle school, or try to join the team at the junior high, or maybe just go on his own (I don't know if that's possible though).  Because I'm not sure if it would be a good idea to have a team of just 6th graders go to MathCounts, but having him try to join to junior high team doesn't really seem fair to them, especially if he gets in.

[quote="kstan013"]I would suggest going to the junior high. A team of 6th graders probably wouldn't do so well. Besides it might be too late. Did your region have chapters yet?[/quote]
He said next year ;)
EDIT: OK, apparantley I had no clue about what I was talking about when I said this.
@h_s_potter, I suggest that you ask the junior high. Is your brother very talented at math competitions? If so tell them that they should give him an aptitude test if they don't agree.
They especially shouldn't mind if he goes as just an individual.

[quote="h_s_potter2002"]However, my school district is arranged a little differently - we have a middle school, which is for 5th and 6th graders, and a junior high, for 7th and 8th graders.  The junior high has a MathCounts team this year, but not the middle school.  So I'm just wondering whether he should just try to get a team started at his middle school, or try to join the team at the junior high, or maybe just go on his own (I don't know if that's possible though).  Because I'm not sure if it would be a good idea to have a team of just 6th graders go to MathCounts, but having him try to join to junior high team doesn't really seem fair to them, especially if he gets in.[/quote]

My understanding of Mathcounts rules is that a student from the middle school could not compete on the junior high's team, unless both schools qualify as "small schools" (fewer than 8 kids in each grade, 6th 7th and 8th).  Mathcounts teams must consist of only students from a single school.  Mathcounts also does not allow students to go on their own (without a school).  His options are to get his school to start a team, or to get his school to register him as an individual from the school (similar to going on his own, but he is officially representing his school).  He'd need a teacher or volunteer from the school who was willing to be the "coach" for the middle school, and get him to competitions.  (That could be one of your parents, if the school and your parents agreed to it.  I am the parent-volunteer coach for my son's school.)

In VT, we have had both individuals and teams of just 6th graders from K-6 schools attend Mathcounts.  Last year such a team came first in our chapter!  (VT is not the most competitive state, but that's still quite the accomplishment!)  Usually the teams do not do so well, but the kids still get the experience and are that much better prepared for the team they may be on in their 7th grade year at their junior high.  If he goes with a team of 6th graders, even if the team does poorly, he may qualify to move on to State as an individual, depending on your state's rules.  We have had 6th graders qualify to State as individuals in VT too.   Finally, if he can't get a team together but his school is still willing to sign him (and maybe a friend or two, if there are others interested) up as individual(s) from the middle school, again he could still qualify to state as an individual.

Good luck to your brother!

From the handbook thingie (http://www.mathcounts.org/webarticles/articlefiles/132-0607_HB_Intro.pdf):

School Enrollment Status: [b]A student may compete only for his/her official school of record.[/b] A student’s school of record is the student’s base or main school. A student taking limited course work at a second school or educational center may not register or compete for that second school or center, even if the student is not competing for his/her school of record. MATHCOUNTS registration is not determined by where a student takes his/her math course.

Small Schools: Schools with eight or fewer students in each of the 6th, 7th and 8th grades are permitted to combine to form a MATHCOUNTS team. Only schools from the same or adjacent chapters within a state may combine to form a team. The combined team will compete in the chapter where the coach’s school is located.

Team Registration: Only one team (of up to four students) per school is eligible to compete. Members of a school team will participate in the Sprint, Target and Team Rounds. Members of a school team also will be eligible to qualify for the Countdown Round (where conducted). Team members will be eligible for team awards, individual awards and progression to the state and national levels based on their individual and/or team performance. It is recommended that your strongest four Mathletes form your school team. [b]Teams of fewer than four will be allowed to compete; however, the team score will be computed by dividing the sum of the team members’ scores by four (see “Scoring” on page 15 for details). Consequently, teams of fewer than four students will be at a disadvantage.[/b]

I guess that makes sense to make sure the teams are legit.  And it does not make it unfair for individuals.  So I guess he could just try to get a program started and if he can't, he could try to get himself registered. This seems to be what mathmom suggests...

At our chapter this year we definitely had a kid compete on his own without a school.  He had been on his school's team in 7th grade and then he transferred to a private school that didn't have mathcounts.  So he was at chapter completely on his own.

The registration guidelines in the handbook say that up to 4 individuals may be registered in addition to or in LIEU of the team.  So a school can register just one kid in lieu of a team.

Frost

[quote="frost13"]At our chapter this year we definitely had a kid compete on his own without a school.  He had been on his school's team in 7th grade and then he transferred to a private school that didn't have mathcounts.  So he was at chapter completely on his own.

The registration guidelines in the handbook say that up to 4 individuals may be registered in addition to or in LIEU of the team.  So a school can register just one kid in lieu of a team.[/quote]

Presumably, that kid wasn't officially there [i]without [/i]a school, but rather was registered as an individual (in lieu of a team) from his school.  Which I think is what you were saying, but saying he was "without a school" might be somewhat confusing.

Well, he didn't have a coach with him other than a parent.  My point is that if a kid goes to a school without a MATHCOUNTS team and he wants to compete, he can.  MATHCOUNTS is not going to keep a kid from participating.  If you don't make your team you can't then go and register individually of course.

[quote]My brother will be in 6th grade next year, and I'm thinking that he'll want to do MathCounts. So normally, he would just take the school test, and if he makes to team, he can go on to later rounds. [/quote]

I dunno about your brother-- at my district, 6th graders do Math Olympiad, not MATHCOUNTS. :|

Last year my son was the only mathlete from our school, and I was his coach, so same situation.  But he was still officially representing his school, and doing so with the knowledge and support of the school.  (And after three years, he has put his school "on the map" so to speak -- they have even finally learned how to spell the name our our school.)

[quote="13375P34K43V312"]Hypothetically, could you go around and get 3 homeless kids from the street that don't go to school and form an official team with them?[/quote]

No.  Unless they are eligible to and do register to attend your school.

I [i]believe [/i]that 4 homeschoolers can get together and form a homeschool association, and compete as a team representing that homeschool association.  (Some coaches here who work with homeschoolers can probably comment more on the homeschool association rules.  I don't see anything about homeschool "associations" in the rulebook.)

I volunteer for a "small school" (according to the official definition quoted above) and we got a ruling at our State level that we could form a combined team with another "small school" OR with a "small" homeschool association.  (But so far, we've had no takers from the other small school in town, or the local homeschoolers.)  If your state interprets the small school rules the same way, then if your school were a small school, and the 3 homeless kids were homeschooled, you might be in luck. ;-)   But except for the "small schools" and "homeschool association" exceptions, everyone on a team must actually be registered for and attend the same school.


Find all $n$ in set $\mathbb{N}$ such that
$3^{n}-1$ divisble by $2^{n}-1$

It is interesting problem. I check: if $\frac{3^{n}-1}{2^{n}-1}$ is integer, than $(n,2*3*5*7*11)=1.$

[quote="justdudxit"]I don't understand the notation with 3 vertical dots...[/quote]
$a\vdots b\iff b|a$.

[quote="tanlsth"]$3^{n}-1 \vdots 2^{n}-1$ so $n \no \vdots 2$
So $3^{n+1}-3 \vdots 2^{n}-1$
Give $p \in P,2^{n}-1 \vdots p$
So $(\frac{3}{p})=1$
So that $p=12t+1$ or $p=12t-1$
So $2^{n}-1=12k+1$ or $2^{n}-1=12k-1$
This is fail[/quote]
It is not solution, because it is easy to show, that $(n,2*3*5*7*11)=1$. It give, that n is odd.

Rust, see the topic "Find all n" (just below) , I posted a solution.

Yes it is correct, but I don't undestand what is mean your first post:
[quote="tanlsth"]$3^{n}-1 \vdots 2^{n}-1$ so $n \no \vdots 2$
So $3^{n+1}-3 \vdots 2^{n}-1$
[/quote]


function
trigonometry
real analysis
calculus
calculus computations

Given that $ f: S\rightarrow \mathbb{R}^m$ is uniformly continuous on $ S$ and $ \{x_k\}$ is a Cauchy sequence in $ S$, show that $ \{f(x_k)\}$ is also a Cauchy sequence.

whats an example of  a cauchy sequence $ \{x_k\}$ in $ (0,\infty)$ and a continuous function $ f: (0,\infty)\rightarrow \mathbb{R}$ (not necessarily continuous), such that $ \{f(x_k)\}$ is not cauchy?

$ f(x)\equal{}\cos\left(\frac{\pi}{x}\right)$ is continuous on $ (0,\infty).$

$ x_n\equal{}\frac{1}{n}$ is a Cauchy sequence in $ (0,\infty)$ because it is convergent in $ \mathbb{R}.$ However it is not a convergent sequence in $ (0,\infty)$ because it has no limit inside that set.

And $ f(x_n)\equal{}(\minus{}1)^n$ is clearly not Cauchy.

This can be taken as one way to prove that $ f$ is not uniformly continuous.


Let $A \in \mathcal M_{2}\left( \mathbb R \right)$ be a matrix such that $A \neq A^\star$ and $A^{3}= \left( A^\star \right)^{3}$.
Prove that $\det A = \text{tr}^{2}A$.

[i]Remark.[/i] [url=http://planetmath.org/encyclopedia/ClassicalAdjoint.html]$A^\star$ is the classical adjoint / adjugate of $A$.[/url]

[quote="perfect_radio"]Let $A \in \mathcal M_{2}\left( \mathbb R \right)$ be a matrix such that $A \neq A^\star$ and $A^{3}= \left( A^\star \right)^{3}$.
Prove that $\det A = \text{tr}^{2}A$.

[/quote]

Let $A=\left(\begin{matrix}{a}&b\\ c&{d\cr}\end{matrix}\right)$ then $A^{*}=\left(\begin{matrix}{d}&-c\\-b&{a\cr}\end{matrix}\right),A\neq A^{*}\Rightarrow a\neq d, b\neq-c$.

From $A^{3}= \left( A^\star \right)^{3}$ we got $a^{3}+abc=d^{3}+dbc$ so $(a-d)(a^{2}+ad+d^{2}+bc)=0$ but $a\neq d \Rightarrow a^{2}+ad+d^{2}+bc=0$ so $(a+d)^{2}=ad-bc\Leftrightarrow detA=tr^{2}A$


Show that there is a unique sequence of positive integers $ (a_{n})$ satisfying the following conditions:

$ a_{1}\equal{}1, a_{2}\equal{}2, a_{3}\equal{}12,$

$ a_{n\plus{}1}a_{n\minus{}1}\equal{}a{n}^{2} \pm 1$ for $ n\equal{}2, 3, 4, ...$ .

Sorry: too late to edit.

The last line should be:

$ a_{n\plus{}1}a_{n\minus{}1} \equal{} a_{n}^{2} \pm 1$ for $ n\equal{}2, 3, 4, ...$


calculus
integration
limit
Asymptote
geometry
rectangle
calculus computations

Our teacher gave us the problem $\int_1^4{\frac{dx}{x-2}}$ and then went on to show that it diverges.

But this makes no logical sense to me.  I know you obviously can't just say things like $\infty-\infty = 0$, but because it is perfectly symmetric, why is this not true?:

$\int_1^4{\frac{dx}{x-2}} = \int_3^4{\frac{dx}{x-2}} = \text{ln}2$

That's because $\int_1^4{\frac{dx}{x-2}}=\lim_{a\rightarrow 2^-}\int_1^a{\frac{dx}{x-2}}+\lim_{b\rightarrow 2^+}\int_b^4{\frac{dx}{x-2}}=\infty-\infty$ is an indeterminate form. $\infty-\infty$ has no meaning because $\infty$ is not a number. How would you know how big is $\infty$?

indeed, but they are symmetric.  The fact that they are exactly the same shape.. I just don't get why they wouldn't theoretically cancel.  I mean I wouldn't think just any old integral across an asymptote would cancel, but take like any dx rectangle positioned at 2+x and it would cancel with the one at 2-x.  But sigh, I guess it is one of those oddities about infinity.  Before learning all these techniques, I woulda expected $\int_{-\infty}^{\infty}} {\frac{1}{x}dx= 0$  :oops: / :| / :huh: / :ninja:

Oh also then can "diverge" describe any indeterminate form?  I only thought of it as increasing or decreasing without bounds or oscillating or things of that nature.

You can take 'principal values' and then the portions that go to infinity will cancel out, which works out well sometimes.  However, you need to note when you are doing this.

Hmm, are you sure this term is correct?  I can't seem to find it in Hardy anywhere.  If so, could you direct me somewhere where I could get a more detailed explanation of this?  I'm not really sure what you mean.

What about the argument that:
$\int_1^4 \frac{dx}{x-2}=\int_{-1}^2\frac{dx}{x}=\int_{-1}^0\frac{dx}{x}+\int_0^1\frac{dx}{x}+\int_1^2\frac{dx}{x}$
Then shift the first integral by x=-t
$=\int_{1}^0\frac{d(-t)}{-t}+\int_0^1\frac{dx}{x}+\int_1^2\frac{dx}{x}$
$=-\int_{0}^1\frac{dx}{x}+\int_0^1\frac{dx}{x}+\int_1^2\frac{dx}{x}=\int_1^2\frac{dx}{x}$

Neither $\int_{-1}^0\frac{dx}{x}$, nor $\int_{0}^1\frac{dx}{x}$ have any particular meanings as numbers, so you can't work with them like that...

It is a definition (at least when generalizing one variable riemann integrals) that each improper integral has to be convergent for an integral that is improper in several points to be convergent. You could make another definition.


Find all function $f: R^{+*} \to R$ satisfy:
$f(x+f(y))=\frac{y}{1-xy}$

what if I take $x=\frac{1}{y}$?Since there are infinite many $x,y$ such that $xy=1$ I believe that something is fishy :lol:

All right, so the statement is missing a line. How about this:
$f(x+f(y))=\frac{y}{1-xy}$ whenever $x+f(y)>0$

Now we can easily show that $f(x)\le-\frac1x$ for all $x$, and $f(x)=-\frac1x$ is a solution.

[quote="Hong Quy"] Are you sure $f: R+\to R$[/quote]
Yes,I'm sure.
[quote="Lovasz"]or may like this...
[quote]Find all function $f: R^{+} \to R$ satisfy:
$f(x+f(y))=\frac{y}{1+xy}$[/quote][/quote]
No!With this problem,it's so easy !(Note :We need prove $f(x)\geq 0$ first ,and answer is$f(x)=\frac{1}{x}$)
With my problem,we have to understand that :$f(x+f(y))=\frac{y}{1-xy}$ with all $x;y$ satisfy $xy\neq 1$.
@jmerry:
Please make details! :idea:

[quote="Hong Quy"]Take $xy>1$ we have $f(x+f(y))<0$. If $f(y)<0$ we can shoose $x$ such that $x+f(y)<0$ and it's impossible because $f: R^+\to R$. There fore there are no such functions[/quote]
I disagree with you!
When the problem is writen like that,we must understand :The equation happen [u][b]with and only with[/b][/u]:
 $x+f(y)> 0$ and $xy \neq1$
 :!:

$f(x+f(y))=\frac{y}{1-xy}$ so $f(x)=\frac{y}{1-xy+yf(y)}=\frac{1}{1-x+f(1)}$ so with our results of polynomials(variable x) we get $f(y)y=f(1)$ and $f(1)=-1$
so the only solution is $f(x)=-\frac1x$


What is the 1,000th digit of  :sqrt: .999999999...?

Next time, post in normal font please.

About your problem- does the decimal expand out to infinity? Because then it would be 1, and $\sqrt{1}=1 or -1$

Steeler suck. Pats '05. Just like Red Sox '05.

The square root of 1 does [size=200]not[/size] equal 1 or -1. It equals 1.

Oops, sorry. Forgot about that.  :sqrt: 1=1, not -1.

[quote="JBL"]I disagree, MMM.   :sqrt: 1 does not equal -1, but "the square root of 1" is really "the square roots of 1."[/quote]

OK, I concede that fact :) but you know what I mean.

The question asks for the 1000th digit; so the answer is 9,not 1. It doesn't ask what the series converges to.

since .9999999999.... = 1

we have to square root 1

the funny thing is if you type it in your calcualtor, it's just 1

1 still equals to .9999999999999999999.....

so the 1000th digit has to be 9

[quote]Steeler suck. Pats '05. Just like Red Sox '05[/quote]

Steelers was amazing this year.... but lost to Patriots.....

It's all Ben Rothlesburgers' fault.( the rookie quaterback...)
omg...
you guys should have seen one of STEELERS players Heinz Ward.. He was crying because he couldn't make it to the super bowl.... I feel bad for him

i think the digit would be 9

i hate the patriots more that all of you do!  :mad:  i really wanted the steelers to win.  tom brady is such a pretty boy... blech  they're just really lucky... i mean they lost to the miami dolphins! aaaaaaaaaaaaaaaauuuuuuuugh   i cant stand another patriot super bowl

It asks for the [size=150]1000th digit[/size] of the square root not the value of the square root.

[size=200]GO SEAGULLS!!![/size]

Next year, The Steelers are going to beat the Patsies IN New England IN the championship to end "the dynasty".  If Roethlisburger wasnt a rookie , the Patsies would have gotten killed.  Iv heard enough of all this dynasty junk.

[size=200]GO SEAGULLS!!![/size]

Holy ****, so this is how it feels to want to abuse power. Just one click on the edit button...  ;)

But hey, I'm not gonna argue with you guys. Go to the Super Bowl topic in Games and Fun. For now, I'll just say this. Results speak for themselves.

ln(dx/dy), theaznkungfu your both wrong.  Shinwoo it's not the rookie quarterback's fault.  It was no one on the steelers team fault.  Its just the Patriots are so much better  :lol:  The patriots will own the eagles.  sigh... just watch the superbowl game and just see your proof. sigh...

Oh yeah I also think the 1000th digit is 9. :D

elvenchamp777

No, that's not true.  The function $\sqrt{}$, when we're not dealing with complex numbers, is defined as $\sqrt{} : \mathbb{R}^{\geq 0} \to \mathbb{R}^{\geq 0}$.  (That is, it always takes non-negative real input and always gives non-negative real output.)  The symbol $\sqrt$ represents the positive square root function, always.  This is equivalent to the distinction between $\text{Arcsin}$ and $\arcsin$, one of which is a function from $[-1, 1]$ to $[-\frac{\pi}{2},\frac{\pi}{2}]$ and the other of which is a function from $[-1, 1]$ to the powerset of the reals.

Are you guys saying that

square root of .99999999  isn't 9?

hm...

but since .9999999999.... equals to one and square root of 1 equlas to 1, it would have to be .9999999999..... and It  would have to be 9

I guess what I'm trying to say is that the question could have a couple of answers. You could argue that .999999... is 1 and say there is no answer, because 1 is just one digit. You could just as well argue that since it goes on forever, the 1000th digit of .9999999... is 9. But the thing is whether you define .99999999... as 1 or not.

Steelers did have an amazing season. I'd be the first to say Roethlesburger was amazing in the regular season. But don't worry, he'll improve a ton next year. If he was this successful in his first season, I'm pretty scared to think about what he'll do next year.  :)

Actually, Ward wasn't crying mainly because they didn't make it. He was more saddened by the fact that Jerome Bettis might not be back next year and they wanted to get a Super Bowl ring for him before he retired. But we'll see next year.

This really isn't a good question, because 1 has two decimal expansions.  If we assume the question is asking about the 10000th digit after the decimal place, we have that 1.00...(10000times)...0000 = 0.999...(10000times)...999....

when you said we had to define the thing, I just remebered the argument I had with some people ...

it was one of the problem from getting started forum proving that .999999... = 1

this is how he defined it...

$\displaystyle .999\ldots=\sum^{\infty}_{k=1}\frac{9}{10^k}=\lim_{n\to\infty}\sum^{n}_{k=1}\frac{9}{10^k}$

well,,

people thought Ward looked really cute ( I strongly disagree since I'm a guy.. ) when he was crying... so

most of the girls in my school wore the Steelers t-shirt just for Ward....

In case anybody wanted to see the topic discussing $\overline{.999}$, here it is.

http://www.artofproblemsolving.com/Forum/topic-19502.html

I know that this is a math site, but  :mad: i hate the patriots!  and btw elvenchamp, patriots are not soooo much better, if they were, they would have smushed the dolphins.  steelers were just having a bad day.  and patriots were not having a bad day with the dolphins, because "dynasties" dont got bad days.... erg.   sooo mad at pats.... too bad chargers didnt go, just local pride  :D


LaTeX
inequalities
inequalities unsolved

[quote="Hinagiku"]I can't prove this inequality.

Let a,b,c > 0 and a ≥ b ≥ c   Prove that

2(a+b√(b/a)+c∛(c^2/ab))  ≥ √(a^2+bc)+√(b^2+ca)+√(c^2+ab)

 :)  :)

sorry I can't use latex .[/quote]

I can't understand your inequality means...

[quote="Hinagiku"]I can only write it in Microsoft Word.

 :)  :)[/quote]

I help you to write down your problem here:

Let $ a, b, c > 0$ and $ a\ge b\ge c$, Prove that
\[ 2\left( a \plus{} b\sqrt {\frac {b}{a}} \plus{} c\sqrt [3]{\frac {c^2}{ab}} \right) \ge \sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ca} \plus{} \sqrt {c^2 \plus{} ab}.\]

ummm...
that's hard for me too...
I just know that inequality is equivalent to
$ k,t\ge1$
\[ 2\left( {kt \plus{} k\sqrt {\frac{1}{t}}  \plus{} \sqrt[3]{{\frac{1}{{k^2 t}}}}} \right) \ge \sqrt {k^2 t^2  \plus{} k}  \plus{} \sqrt {k \plus{} kt}  \plus{} \sqrt {1 \plus{} k^2 t}\]
...
I don't know how to prove it...

[quote="Dr Sonnhard Graubner"]hello, let $ a,b,c > 0$ and $ a\geq b\geq c$. Prove that
$ 2\left(a \plus{} b\sqrt {\frac {b}{a}} \plus{} c\sqrt [3]{\frac {c^2}{ab}}\right)\geq\sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ca} \plus{} \sqrt {c^2 \plus{} ab}$.
Is this your problem?
Sonnhard.[/quote]
It is known that
$ \frac {3}{2}(a \plus{} b \plus{} c) \ge \sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ca} \plus{} \sqrt {c^2 \plus{} ab}.$
Therefore, in order to prove the desired inequality, we only need to prove that
$ 4\left( a \plus{} b\sqrt {\frac {b}{a}} \plus{} c\sqrt [3]{\frac {c^2}{ab}}\right) \ge 3(a \plus{} b \plus{} c).$
This inequality is equivalent to
$ a \plus{} 4b\sqrt {\frac {b}{a}} \plus{} 4c\sqrt [3]{\frac {c^2}{ab}} \ge 3b \plus{} 3c.$
By the AM-GM Inequality, we have
$ \frac {a}{2} \plus{} 2b\sqrt {\frac {b}{a}} \plus{} 2b\sqrt {\frac {b}{a}} \ge 3\sqrt [3]{2}b > 3b,$
and
$ \sqrt [3]{\frac {c^2}{ab}} \equal{} \frac {c}{\sqrt [3]{abc}} \ge \frac {3c}{a \plus{} b \plus{} c} \ge \frac {3c}{2a \plus{} c}.$
so it is enough to check the following inequality
$ \frac {a}{2} \plus{} \frac {12c^2}{2a \plus{} c}\ge 3c.$
This is equivalent to
$ \frac {2a \plus{} c}{4} \plus{} \frac {12c^2}{2a \plus{} c} \ge \frac {13}{4}c,$
which is obviously true according to the AM-GM Inequality
$ \frac {2a \plus{} c}{4} \plus{} \frac {12c^2}{2a \plus{} c} \ge 2\sqrt {3}c > \frac{13}{4}c.$
The proof is completed. Note that equality cannot occur.

Can you show that

$ \frac{3}{2}(a\plus{}b\plus{}c)\ge\sqrt{a^{2}\plus{}bc}\plus{}\sqrt{b^{2}\plus{}ca}\plus{}\sqrt{c^{2}\plus{}ab}.$

 :)


calculus
derivative
inequalities
inequalities solved

:blush: is my english so bad, or did I miss something?

could you give a little more explanation please??

Is not so tough as the original one. Let
f(x) = 2x - sqrt(3x^2+1), we have to show that if xyz=1 then
f(x) + f(y) + f(z) >= 0 = 3f(1).
It is enough to show that  f(x) + f(y) >= 2f(sqrt(xy)) (1) for all x, y.

Let g(x) = f(exp(x)), we have

g"(x) = 2exp(x) + 9(exp(x))^4/T^3 - 6(exp(x))^2/T

where T = sqrt(3x^2+1).

Apply AM-GM for  exp(x), exp(x) and 9(exp(x))^4/T^3 we get

2exp(x) + 9(exp(x))^4/T^3 >= 3.(9)^(1/3)(exp(x))^2/T > 6(exp(x))^2/T.

So g"(x) > 0 and we have (1). Thus, the prb is solved.

Namdung

Yes, but the problem was intended for the 9-th form, where they do not know derivatives! And yes, it is far less hard than the original one.

Prove that if a,b,c>0 verify abc=1, then sum sqrt(3a^2+1)<=2 sum a.

so you get a,b and c positive reals, with product one

then you get sum sqrt(3a^2+1)<=2 sum a.
a formula in which b and c do not come back?? and what is the sum about, is a ascending or so???

harazi why not use \sum instead of plain sum ? :)

[quote="harazi"]sum a=a+b+c, Peter![/quote]

:? how am I supposed to know that?

so it comes down to:
Prove that if ( a,b,c>0 AND abc=1 ), then \sqrt(3a^2+1) + \sqrt(3b^2+1)+ \sqrt (3c^2+1) \leq 2a+2b+2c

right?

Ok then that sounds more intelligible, will try.

Harazi .could u pleeease post ur proof for this inequality
i have found very weak inequality relating the lhs and the best constant i could get on the  \sum x  term was 7/4
so please give me ur proof for this beautiful ineq. which is nice and hard enough for me :D
thanx
best regards

My solution is quite long, but I like it. First of all, if ab+bc+ca<=a+b+c then ( \sum  \sqrt 3a 2 +1) 2  \leq ( \sum a)(3 \sum a+ \sum ab) by Cauchy and this is at most 4( \sum a) 2 .
       Now let the other case, that is (a-1)(b-1)(c-1)<0.We can assume that a<1 and b,c>=1. Then ( \sqrt 3b 2 +1+ \sqrt 3c 2 +1)  2  \leq (b+c)(3b+1/b+3c+1/c)=(b+c) 2 (3+a). Thus, it is enough to prove that (b+c)(2- \sqrt 3+a) \geq  \sqrt 3a 2 2+1-2a. This is equivalent to b+c \geq (1+a)(2+ \sqrt 3+a)/[2a+ \sqrt 3a 2 2+1]. Since 2+ \sqrt 3+a \leq 4 and b+c \geq 2 \sqrt 1/a, it is enough to prove that 2a+ \sqrt 3a 2 2+1 \geq 2a \sqrt a+2 \sqrt a. If a<=1/3, it follows from 3a 2 2+1 \geq 4a. Suppose that 1/3<a<1. It becomes (3a-1)/[2a+ \sqrt 3a 2 +1] \leq 2a/[1+ \sqrt a] and it is true.
   Now it's time to see if I have learned to write using smilies.


For any positive integer $k$, let $f_1(k)$ denote the square of the sum of the digits of $k$.  For $n \ge 2$, let $f_n(k) = f_1(f_{n - 1}(k))$.  Find $f_{1988}(11)$.

[hide]Seeing that $f_n(k)$ is the same as $f_1(f_1(...f_1(k)$ with $n$ occurrences of $f_1$, we can say that $f_{1988}(11) = f_1(f_1(...f_1(11)$, with 1988 occurrences of $f_1$. Let the notation {$a,b$} show the number of occurrences of $f_1$ there are in the expression as $a$ and the number inside the function as $b$. Then the first few pairs of numbers are:
{$1988,11$},{$1987,4$},{$1986,16$},{$1985,49$},{$1984,169$},{$1983,256$},{$1982,169$},{$1981,256$}....
It is clear that the sequence will continue with {$odd,256$} and {$even, 169$}. Thus when $a=0$, $b=\boxed{169}$.[/hide]
(edit: forgot to hide :P)


geometry
parallelogram
geometric transformation
reflection
circumcircle
trigonometry
geometry proposed

Two circles with different radii are tangent to a line $ l$ in points $ A,B$ and intersect each other at $ C,D$. Let $ H_1,H_2$ be the orthocenters of $ \triangle ABC,ABD$. Prove that $ H_1CH_2D$ is a parallelogram.

[hide="Solution"]
Reflect $ H_1$ and $ H_2$ over $ AB$ to form $ X$ and $ Y$ respectively.  Now, $ \angle H_2AB\equal{}90\minus{}\angle ABD$ and $ \angle H_2BA\equal{}90\minus{}\angle DAB$, so \[ \angle AYB\equal{}\angle AH_2B\equal{}180\minus{}\angle H_2AB\minus{}\angle H_2BA\equal{}\angle DAB\plus{}\angle DBA\equal{}180\minus{}\angle BDA\] and thus, $ DAYB$ is cyclic.  Similarly, $ ACBX$ is cyclic.  Now, \[ \angle ACB\equal{}180\minus{}\angle CAB\minus{}\angle DBA\equal{}180\minus{}\angle ADC\minus{}\angle BDC\equal{}180\minus{}\angle BDA\equal{}\angle AH_2B\] and it follows that $ ACH_2B$ is cyclic.  Similarly, $ AH_1DB$ is cyclic.  Now, the circumradius of $ \triangle ADB$ is equal to that of $ \triangle AH_1B$, which is equal to that of $ \triangle AXB$, which is equal to that of $ \triangle ACB$, which is equal to that of $ \triangle AH_2B$ from the cyclic quadrilaterals and the reflections.  Let this common radius be $ R$.  It follows that $ CH_1\equal{}|2R\cos \angle ACB|\equal{}|2R\cos (180\minus{}\angle ACB)|\equal{}|2R\cos \angle ADB|\equal{}DH_2$.  Now, we see that $ CH_1\perp AB$ and $ DH_2\perp AB$, so $ CH_1\parallel DH_2$.  Finally, we conclude that $ DH_2CH_1$ is a parallelogram, as desired. [/hide]


Alright so, I just took AMC12B and I got a score of 99. Considering the difficulty of the test relative to AMC12A, what do you guys think will be my chances of making the cutoff for AIME? What do you guys think the cut off will be?

I thought the 12B exam was harder than 12A, so the cutoff shouldn't be any higher (97.5).
I made AIME last year with a 99 on 12B
You should be able to make it.

In my opinion, you will almost certainly make it. The cutoff for the AMC 12A, which most AoPSers considered easy, was already 97.5.

You'll probably make it; in the past few years, the cutoff has almost always been lowered. In addition, this was not an easy test in most people's view.

I think it will be 97.5 for the 12B. Most people I talked to thought this was harder than the 12A and harder than past years.

thanks guys,
do you know when the cutoff will be released?

Just be patient; wait another week or two.

The official cutoff for AMC 12B was 100. Sorry dude

missed it by one :(
did u make it on AMC 10 though?


linear algebra
matrix
inequalities
vector
symmetry
linear algebra unsolved

Let $ A\in M_{n\times m}(F),B\in M_{m\times n}(F)$,where $ F$ is a field.If $ AB$ is Idempotent.(Which means $ (AB)^{2}\equal{} AB$)
Prove:$ rank(AB) \equal{} rank(BA)$

We will first show that under the assumptions the inequality $ rk(BA)\geq rk(AB)$ is fullfilled:

Define the sub-vector space $ Ker(B)\leq V=\{\bar{v}\in\F^{n}\,|\,B\bar{v}\in Ker(A)\}$ and extend to a basis of the complete vector space, i.e. assume $ F^{n}=V\oplus W$.

For any $ \bar{w}\in W$ compute $ AB\bar{w}=\bar{w'}+\bar{v}$ with $ \bar{w'}\in W\,,\,\bar{v}\in V$. Then $ AB(\bar{w}-\bar{w'})=AB(\bar{w}-\bar{w'}-\bar{v})=AB(\bar{w}-AB\bar{w})=\bar{0}$. Since $ \bar{w}-\bar{w'}\in W\cap V$, this implies $ \bar{w}=\bar{w'}$ and as a consequence $ W$ contains all eigenvectors to the eigenvalue $ 1$ of $ AB$ and $ rk(AB)=dim(W)$.

Now choose arbitrary $ \bar{w}\in W$ and set $ \bar{u}=B\bar{w}$. This yields $ BA\bar{u}=B\bar{w}=\bar{u}$ and $ rk(BA)\geq dim(W)=rk(AB)$.

For any $ \bar{x}\in Im(BA)\leq F^{m}$, i.e. we can write $ \bar{x}=BA\bar{y}$, then $ [(BA)^{2}-BA]\bar{x}=[(BA)^{3}-(BA)^{2}]\bar{y}=[B(AB)^{2}A-(AB)^{2}]\bar{y}=\bar{0}$ and we conclude that the restriction of $ BA$ to $ Im(BA)$ is idempotent. Let $ U$ be the image of $ Im(BA)$ in $ F^{n}$ under transformation by $ A$. Note that $ W\leq U$ and consider the restrictions $ A_{(r)}=A|_{Im(BA)}: Im(BA)\rightarrow U\,,\,B_{(r)}=B|_{U}: U\rightarrow Im(BA)$.

By a symmetry argument we get $ rk(A_{(r)}B_{(r)})\geq rk(B_{(r)}A_{(r)})$ and since by construction the rank of the two respective products is preserved, the proof is complete.

[quote="perfect_radio"]In other words, if $ AB \equal{} 0$, then $ BA \equal{} 0$. Maybe you meant something else?[/quote]
Hei,thank you,you are right.This problem is wrong :blush:
And so is hjbrasch's proof.
But Thank you all the same.

to clarify, the second part of the proof showed the opposite inequality as well:

$ rk(AB)\equal{}rk(A_{(r)}B_{(r)})\geq rk(B_{(r)}A_{(r)})\equal{}rk(BA)$

[quote="hjbrasch"]to clarify, the second part of the proof showed the opposite inequality as well:

$ rk(AB) \equal{} rk(A_{(r)}B_{(r)})\geq rk(B_{(r)}A_{(r)}) \equal{} rk(BA)$[/quote]

I have not read your post but $ A\equal{}\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $ B\equal{}\left(\begin{array}{cc}0&1\end{array}\right)$ yields a counterexample (and you can easily get a counterexample with two $ 2\times 2$ matrices from this one, if you want $ n\equal{}m$).

  Darij

in your counterexample the product $ AB$ is not idempotent, since $ AB\not \equal{} 0\equal{}(AB)^{2}$, but if you feel there is sth incorrect, pls let me know

[quote="hjbrasch"]in your counterexample the product $ AB$ is not idempotent, since $ AB\not \equal{} 0 \equal{} (AB)^{2}$, but if you feel there is sth incorrect, pls let me know[/quote]

Oops, read $ A$ for $ B$ and vice versa. ;)

To your proof: How do you explain the two equality signs in $ rk(AB)\equal{}rk(A_{(r)}B_{(r)})\geq rk(B_{(r)}A_{(r)})\equal{}rk(BA)$ ?

  darij

[quote="darij grinberg"]To those who rated Hjbrasch's post spam: He proved the weaker (and correct!) assertion that $ \text{rank}\left(BA\right)\geq\text{rank}\left(AB\right)$.

  Darij[/quote]
 :) This can deduced obviously by $ rank(BA)\ge rank(ABAB)\equal{}rank(AB)$ :)


logarithms
calculus
calculus computations

For each of the series below, decide which of the descriptions:

(i) Does not converge;
(ii) Converges, but not absolutely;
(iii) Converges absolutely

applies, and prove your answer.

1. [sum] from n=1 to oo ((-1)^{n+1} / (n)^{1+(1/n)})

2. [sum] from n=1 to oo (e^{1/n} - 1)

3. [sum] from n=1 to oo (1 - cos(1 / n^{2/3}))

1. convergent by alternating series test, although you'll have to work to verify monotonicity. Does not converge absolutely, by comparison to $\frac{1}{n\log n}$ (use logarithms).

2. divergent by limit comparison to $\frac{1}{n}$. Use Maclauren series.

3. convergent by limit comparison to $\frac{1}{n^{4/3}}$. Use Maclauren series. Also abs. convergent (positive terms).


geometry
geometric transformation
reflection
trigonometry

I am curious to know, how many prooves are there for the Butterfly Theorem??
And if you have a good proof, can you post it here?
Thanks.

[quote="djimenez"][quote="shobber"]I am curious to know, how many solutions are there for the Butterfly Theorem??
And if you have a good proof, can you post it here?
Thanks.[/quote]

What do you mean by solutions? do you mean proofs? I mean, it is a theorem and it does not stablish nothing about solutions of equations or something like that.
Any way, you can find a proof [url=http://mathworld.wolfram.com/ButterflyTheorem.html]here[/url].

Best,[/quote]

edited, thanks

[quote="jpark"]Or you can reflect one of the "Wings" of the "butterfly" to the other side, which proves the result nicely.[/quote]
I know this proof. It is a very simple method.

And i have anther two. One is by using the Law of Sine and the other one is by $Area$ ratios. But i know there must be a lot of other good ways which i don't know.  That's why i post the first topic.

[quote="mecrazywong"]In fact, there is a generalization of butterfly theorem ;)[/quote]
Yes......but i am only asking for the proof of the most original one, the one djimenez has posted.


superior algebra
superior algebra unsolved

what is the difference between the normaliser and center of a group.
Explain this theorem
If G is a group and  'a' belongs to Z if and only if N(a)=G

In this particular instance, I would prefer to use the term “centralizer” to “normalizer”. Given $ a\in G,$ the centralizer of $ a$ in $ G$ is $ C_G(a)\equal{}\{x\in G: xa\equal{}ax\}.$

The centre of the group $ G$ is defined as $ Z(G)\equal{}\{g\in G: xg\equal{}gx\ \mbox{for all}\ x\in G\}.$

Hence $ a\in Z(G)$ $ \iff$ $ ax\equal{}xa\ \mbox{for all}\ x\in G$ $ \iff$ $ C_G(a)\equal{}G.$


abstract algebra
algebra
function
domain
Ring Theory

Έστω $ \Lambda$ μία περιοχή κυρίων ιδεωδών και $ A,B,C$ πεπερασμένα παραγόμενοι Λ-μόδιοι.
α)Αν $ A \oplus C \simeq B \oplus C$ να δειχτεί ότι $ A \simeq B$.
Να βρεθούν αντιπαραγείγματα που δεν ισχύει αυτό αν
β)τα $ A,B,C$ δεν είναι πεπερασμένα παραγόμενα
γ)το $ \Lambda$ δεν είναι περιοχή κυρίων ιδεωδών.
Χρειάζομαι βοήθεια please για τα α,γ :( .Το β νομίζω το έχω κάνει(Αν $ V$ ένας διανυσματικός χώρος αριθμήσιμης διάστασης τότε $ V \simeq V \oplus V$).

Για το (α) χρειάζεσαι το [url=http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain]Structure theorem for finitely generated modules over a principal ideal domain[/url].

Ίσως βοηθάει να αποδείξεις το θεώρημα για $ \Lambda \equal{} \mathbb{Z}$ χρησιμοποιώντας ότι κάθε πεπερασμένα παραγόμενο $ \mathbb{Z}$-μόδιο (= πεπερασμένα παραγόμενη αβελιανή ομάδα) μπορεί να γραφτεί με μοναδικό τρόπο σαν $ \mathbb{Z}^n \oplus \mathbb{Z}_{q_1} \oplus \cdots \oplus \mathbb{Z}_{q_r}$, όπου τα $ q_1,\ldots,q_r$ είναι δυνάμεις πρώτων. (Αυτό είναι συνέπεια του κινέζικου θεωρήματος.)

Η γενική απόδειξη είναι ουσιαστικά η ίδια.

Δημήτρη ευχαριστώ πολύ :) Το έκανα το α. και το κατάλαβα.
Μήπως έχεις και κάποιο αντιπαράδειγμα για το γ? :lol: Δεν μπορώ να σκεφτώ κάποιο :(

[quote="giannis18"]
Μήπως έχεις και κάποιο αντιπαράδειγμα για το γ? :lol: Δεν μπορώ να σκεφτώ κάποιο :([/quote]

Ούτε και εγώ.  :( Θα το ξαναδώ αργότερα.

αν πάρουμε το σύνολο $ T = \{(x_1,x_2,\dots)|x_i\in \mathbb{Z}\}$ ειναι αβελιανη ομαδα(με προσθεση).
παίρνοντας σαν $ \Lambda =$ $ end(T)$ με εξωτερικο πολ/σμο την συνθεση και προσθεση κανονικα. θεωρώντας
το $ \Lambda,$ $ \Lambda$-μοδιο ειναι ελεύθερο με βαση το μονοσυνολο $ {I_{\Lambda}}$ την  ταυτοτικη απεικονιση.
$ e_1,e_2,\dots\in T$ με $ e_1 = (1,0,\dots)e_2 = (0,1,0\dots)$ κτλ.. η απεικονισεις $ f,g\in T$
που ορίζονται    $ f(e_i) = \left\{ \begin{array}{ c l } e_n, & i = 2n \\
0, & i = 2n + 1 \end{array} \right. g(e_i) = \left\{ \begin{array}{ c l } 0, & i = 2n \\
e_n, & i = 2n - 1 \end{array} \right.$
ευκολα ελεγχουμε οτι ειναι και αυτη μια βαση με δυο στοιχεια αρα $ \Lambda\cong\Lambda\oplus\Lambda$
edit:τυπογραφικο

Juki, ευχαριστούμε για την λύση. Μόλις τώρα την πρόσεξα.

Ευχαριστώ πολύ για την απάντηση αν και λίγο καθυστερημένα


the probability is $ \frac {2(\sum n)\minus{}n}{^{2n\plus{}1}C_3}\equal{}\frac {n^2}{^{2n\plus{}1}C_3}$


ratio
arithmetic series
arithmetic sequence

The sums of N terms of two arithmetic progressions are in the ratio (7N+2)/(N+4). Find the ratio of their 5th term.

Thanks!

[quote="Virgil Nicula"]$\frac{a_n}{b_n}=\frac{7n+2}{n+4}$[/quote]

Virgil, this should read $\frac{\sum{a_n}}{\sum{b_n}}=\frac{7n+2}{n+4}$

[quote="xixi"][quote="Virgil Nicula"]$\left(\exists \right)r,\ q$ so that $\left(\forall\right)n\in N^*,\ a_{n+1}=a_n+r,\ b_{n+1}{=b_n+q}$ and $\frac{a_n}{b_n}=\frac{7n+2}{n+4}\Longrightarrow \frac{a_5}{b_5}=\ ?$
[hide="Answer : "]$5.$[/hide][/quote]

Thanks. the answer's 5.
I don't really follow your line of reasoning though. Could you explain a little more?[/quote]
$\frac{\sum a_n}{\sum b_n}=\frac{7n+2}{n+4}$

$\sum a_n=\frac{n(2a_1+(n-1)d_a)}2$

$\sum b_n=\frac{n(2b_1+(n-1)d_b)}2$

$\frac{\sum a_n}{\sum b_n}=\frac{2a_1+(n-1)d_a}{2b_1+(n-1)d_b}=\frac{7n+2}{n+4}$

You are looking for $\frac{a_5}{b_5}$, so

$\frac{a_5}{b_5}=\frac{a_1+4d_a}{b_1+4d_b}=\frac{2a_1+(9-1)d_a}{2b_1+(9-1)d_b}$

Compare above and below and you can figure that $n=9$ and there you get your answer :)

Virgil didn't include any reasoning; he only rewrote the question in symbols.

[hide]The formula for an arithmetic series is $na_1+\frac{(n-1)n}{2}d_a$, where $a_1$ is the first term, $n$ is the number of terms and $d_a$ is the common difference.
So we have $\frac{na_1+\frac{(n-1)n}{2}d_a}{nb_1+\frac{(n-1)n}{2}d_b}=\frac{7n+2}{n+4}$
$\frac{a_1+\frac{n-1}{2}d_a}{b_1+\frac{n-1}{2}d_b}=\frac{7n+2}{n+4}$
Note that the ratio of the fifth terms is $\frac{a_1+4d_a}{b_1+4d_b}$. So substitute $n=9$ to get:
$\frac{a_1+4d_a}{b_1+4d_b}=\frac{65}{13}=5$
I think this implies that the sequences must be of length 9 in order for the ratio of the 5th terms to be unique. Interesting...[/hide]

Denote the ratios are $x$, $y$ respectively.
Thus, $\left(\forall\right) n\in N^*,\ \frac{\frac{a_1+a_n}{2}\cdot n}{\frac{b_1+b_n}{2}\cdot n}=\frac{7n+2}{n+4}\Longrightarrow$
$(n+4)\left[2a_1+(n-1)x\right]=(7n+2)\left[2b_1+(n-1)y\right],\ \left(\forall\right) n\in N^*\ .$
$n: =1\Longrightarrow 5a_1=9b_1$. Suppose w.l.o.g. that $y=2$.
Therefore, $a_1=9,\ b_1=5,\ x=14\Longrightarrow \frac{a_5}{b_5}=\frac{a_1+4x}{b_1+4y}=5\ .$

[quote="Bictor717"]I think this implies that the sequences must be of length 9 in order for the ratio of the 5th terms to be unique. Interesting...[/quote]The above implication is wrong. I can still have 20 terms, 50 terms, and the ratio is still the same. The right implication is that the sequences of length 9 allowed you to find the ratio of 5th terms because of the formula (7N+2)/(N+4), which is given by the problem.

[quote="Bictor717"]I think this implies that the sequences must be of length 9 in order for the ratio of the 5th terms to be unique. Interesting...[/quote]

It doesn't. Take your fraction

[quote="Bictor717"]$\frac{a_1+\frac{n-1}{2}d_a}{b_1+\frac{n-1}{2}d_b}=\frac{7n+2}{n+4}$[/quote]

and rewrite it thus:

$\frac{2a_1-d_a+nd_a}{2b_1-d_b+nd_b}=\frac{7n+2}{n+4}$

Now compare the corresponding coefficients in the two fractions: they must be proportional, hence there exists a real number $t$ such that

$2a_1-d_a=2t$
$d_a=7t$
$2b_1-d_b=4t$
$d_b=t$

Therefore, you can define both sequences in the terms of $t$ and $n$, and consequently the ratio $a_n/b_n$ won't depend on $t$ and will be uniquely determined for any $n$, no matter how long is the sequence.

EDIT: 10000th_user's note on 9-term sequence allowing to find the ratio of 5th terms is also wrong - all we have to find are the initial terms and the differences in the terms of $t$.

could someone please explain the symobls that Virgil used.

i have no idea what they stand for.  :(

\begin{tabular}{c|c} \implies & \text {``implies''}\\\exists & \text {``there exists''}\\ \forall & \text {``for all''}\\ \in & \text {``is an element of''}\\ \mathbb{N}^* & \text {the set of natural numbers} \end{tabular}

I can't remember if $0\in\mathbb{N}^*$ or not...

[quote="10000th User"]$\frac{a_5}{b_5}=\frac{a_1+4d_a}{b_1+4d_b}=\frac{2a_1+(9-1)d_a}{2b_1+(9-1)d_b}$

Compare above and below and you can figure that $n=9$ and there you get your answer :)[/quote]

Why did you write that? How do you figure out that n=9?

Choosing $n=9$ makes the value of $(\frac{n-1}{2})d=4d$, which is what we're looking for since the problem is asking about the 5th terms.[/quote]

[quote="Farenhajt"]EDIT: 10000th_user's note on 9-term sequence allowing to find the ratio of 5th terms is also wrong - all we have to find are the initial terms and the differences in the terms of $t$.[/quote]Yes is wrong because instead of "ratio of sums of 9 terms" I wrote "sequence of length 9". I wanted to say something else but did not go back to revise it.

PS: Let's not waste time arguing. To test if you understood, below is mini problem, everything the same except the question:

Find the smallest positive integer $n$ for which $\frac{a_n}{b_n}>6$ :D

[hide="For Chess64."]$N^*$, $R^*$,  $(-1,1)^*$ a.s.o. means "without $0$" ! The number $0$ belongs to $N$.[/hide]

Denote $S_n=\sum a_i$ , $P_n=\sum b_i$ , given $\frac{S_n}{P_n}=\frac{7n+2}{n+4}$

Notice that by definition of airthmetric series , $a_n=\frac{a_1+a_{2n-1}}{2}$ , and similarly $b_n=\frac{b_1+b_{2n-1}}{2}$

Notice also $S_{2n-1}=\frac{2n-1}{2}\cdot\frac{a_1+a_{2n-1}}{2}=\frac{2n-1}{2}\cdot a_n$ and similar for $P_{2n-1}=\frac{2n-1}{2}\cdot b_n$

So we have $\frac{S_{2n-1}}{P_{2n-1}}=\frac{7(2n-1)+2}{(2n-1)+4}=\frac{14n-5}{2n+3}$

which is equivalent to $\frac{a_n}{b_n}=\frac{14n-5}{2n+3}$ :)

To answer 10000th User's question , we have $\frac{14n-5}{2n+3}>6$ $\implies n>11.5$

in fact , a similar question is posted here , but without solution  ;)
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76281[/url]

Since the given ratio of the sum of n terms is true for all n, in particular it is true for n=9, when the ratio is (7x9+2)/(9+4) = 65/13 = 5.
Since the sum of  the first 9 terms of an arithmetic sequence is 9 times the 5th term, the ratio of 9 times the 5th terms is therefore 5.  So the ratio of the 5th terms is 5.
Further, and unnecessary to solve the given problem, analysis yields examples of how this could occur: the first sequence could be 9, 23, 37, 51, 65, etc, and the second could be 5, 7, 9, 11, 13, etc.


It was All Soviet Union Olympiad 1986 problem (Problem Number 999 in Kvant). Also the constant 2 can not be repalced by any smaller constant.


[quote="AIME15"]The expression $ (81)^{ \minus{} 2^{ \minus{} 2}}$ has the same value as:

$ \textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B)}\ \frac {1}{3} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 81^4$[/quote]
[hide]-2^-2=-1/4, 81^-1/4=B... 1/3[/hide]


The ABC string has n positions.
How many different strings can we make if the number of A's, B's and C's must have the same parity as n (odd-odd, even-even).
kweenie

we'll do the even case. the odd case can then be gotten similarly or via a bijection with the even case.

we need only count the number of ways to place an even number of both A's and B's among $ n$ positions, as the remaining empty positions will be even in number too and will have to be filled by C's. for each $ i$, $ i \equal{} 0,1,...k$, the number of strings containing a total of $ 2i$ A's and B's is $ \binom{2k}{2i}\sum_{j \equal{} 1}^{i}\binom{2i}{2j} \equal{} 2^{2i \minus{} 1}\binom{2k}{2i}$ if $ i\neq 0$ and 1 if $ i \equal{} 0$. the total number of strings of length $ n \equal{} 2k$ is then $ 1 \plus{} \frac {1}{2}\sum_{i \equal{} 1}^{k}2^{2i}\binom{2k}{2i} \equal{} 1 \plus{} \frac {1}{2}\left(\frac {3^{2k} \plus{} 1}{2} \minus{} 1\right) \equal{} \frac {3^n \plus{} 3}{4}$.

Thanks pleurestique, that was quick and short.

For the even and the odd case we have then: (3^n + 3*(-1)^n)/4.
Is there a way to get rid of the (-1)^n, in other words to have 1 simple formula for the two cases?

kweenie


Could you help me on this please?
Prove f(x) is convex.

f(x)=(SUM from k=0 to k=n) C(n,k)^2 * x^(2k) * (1-x)^(2n-2k)

where C(n,k) the binomial coefficient of n and k.
Thank you so much...

hello, do you mean $ \sum_{k\equal{}0}^n \binom{n}{k}^2x^{2k}(1\minus{}x)^{2n\minus{}2k}$?
Sonnhard.