trigonometry
quadratics
Vieta
algebra
quadratic formula

Determine the sum of the values of $ \tan{\theta}$ for which $ 0\le x<\pi$ and $ 1\equal{}2004\cos{\theta}(\sin{\theta}\minus{}\cos{\theta})$.

Determine the smallest positive angle $ \alpha$ for which $ 8\cos{\alpha}\cos{3\alpha}\cos{5\alpha}\cos{7\alpha}\plus{}\cos{12\alpha}\equal{}\frac{1}{2}$, leaving your answer in radians.

[quote="Smartguy"]Determine the sum of the values of $ \tan{\theta}$ for which $ 0\le x < \pi$ and $ 1 \equal{} 2004\cos{\theta}(\sin{\theta} \minus{} \cos{\theta})$.[/quote]
Beautiful problem! Where did you get this problem, my friend??
I hope you like my solution:
We see that $ cos{\theta} > \equal{} 0$ to satisfy the equation(since $ 0 \equal{} < {\theta} \equal{} < pi$) and $ cos{\theta} \equal{} 0$ is not a solution to this equation!
So, we just need to care about $ cos{\theta} > 0$
Now, dividing both sides by $ cos{\theta}$ we get: $ \frac {1}{cos^2{\theta}} \equal{} 2004(tan{\theta} \minus{} 1)$.
This equation is equivalent to: $ 1 \plus{} tan^2{\theta} \equal{} 2004tan{\theta} \minus{} 2004$.
This is the quadratic equation, so we easily get: $ tan{\theta} \equal{} 2005$ or $ tan{\theta} \equal{} \minus{} 1$. Since $ sin{\theta}$ and $ cos{\theta} > 0$, we eliminate all negative values of $ tan{\theta}$. Finally, the sum of $ tan{\theta} \equal{} 2005$(Q.E.D)
Hmm... The 2nd one seems to be tougher than this one even though it looks familiar to me.I'll post the solution to it later.

whoops, forgot to post  :huh:

ghjk: In the first question, $ 0 \le \theta < \pi$, so $ \tan \theta$ can take any real.  :wink:
[hide="1"]
Dividing both sides by $ \cos^2 \theta$ (clearly $ \cos \theta \neq 0$), we have $ \sec^2 \theta = 1 + \tan^2 \theta = 2004 \tan \theta - 2004$, and by vieta's [each value occurs once] the sum is $ 2004$.
[/hide]
[hide="2"]
Repeatedly applying the product to sum formula yields
\begin{align*}8\cos \alpha \cos 3\alpha \cos 5\alpha \cos 7\alpha & = 2 \left[\cos 8 \alpha + \cos 2\alpha\right]\left[\cos 8\alpha + \cos 6 \alpha \right] \\
& = 2\left[\cos^2 8\alpha + \cos 8\alpha \cos 6\alpha + \cos 8\alpha \cos 2\alpha + \cos 6\alpha \cos 2\alpha\right] \\
& = \left[\cos 2\alpha + \cos 4\alpha + \cdots + \cos 14\alpha \right] - \cos 12\alpha.\end{align*}
Thus, we have $ \frac {1}{2} = \sum_{n = 1}^{7} \cos 2n\alpha = \text{Re}\left(\sum_{n = 1}^7 e^{i2n\alpha}\right) = \text{Re}\left(\frac {e^{i16 \alpha} - 1}{e^{i2\alpha} - 1}\right) = \frac {\sin 7\alpha \cos 8\alpha}{\sin \alpha}$. Then $ \sin \alpha = 2\sin 7\alpha \cos 8\alpha \cdot \frac {\sin 8\alpha}{\sin 8\alpha} = \frac {\sin 7 \alpha \sin 16\alpha}{\sin 8 \alpha}$ $ \Longrightarrow 2\cos 9\alpha = \cos 23\alpha + \cos 7\alpha$ $ \Longrightarrow \cos 9\alpha = \cos 15\alpha \cos 8 \alpha$ .. shouldn't be too much harder from here ..
[/hide]

[quote="azjps"]whoops, forgot to post  :huh:

ghjk: In the first question, $ 0 \le \theta < \pi$, so $ \tan \theta$ can take any real.  :wink:
[/quote]
Hmm... Did you read my argument above? I already showed that $ cos{\theta}>0$(I leave its proof for you to think of it first!). Therefore, $ tan{\theta}>0$ with $ 0 \le\theta <\pi$ :)

[quote="azjps"]Edit: Actually, I'm right  :lol:  its $ \tan^2 \theta \minus{} 2004\tan \theta \plus{} 2005 \equal{} 0$, the roots are both positive.[/quote]
My equation is right, but I made a dumb factorization. :o  Since the quadratic equation gives two positive roots for tan, the sum is indeed 2004(you're lucky since it gives both positive roots, aren't u?) :oops:

[quote="ghjk"]Beautiful problem! Where did you get this problem, my friend??
I hope you like my solution.[/quote]

 :D I really like your solution!

[quote="azjps"]
[hide="2"]
Repeatedly applying the product to sum formula yields
\begin{align*}8\cos \alpha \cos 3\alpha \cos 5\alpha \cos 7\alpha & = 2 \left[\cos 8 \alpha + \cos 2\alpha\right]\left[\cos 8\alpha + \cos 6 \alpha \right] \\
& = 2\left[\cos^2 8\alpha + \cos 8\alpha \cos 6\alpha + \cos 8\alpha \cos 2\alpha + \cos 6\alpha \cos 2\alpha\right] \\
& = \left[\cos 2\alpha + \cos 4\alpha + \cdots + \cos 14\alpha \right] - \cos 12\alpha.\end{align*}
Thus, we have $ \frac {1}{2} = \sum_{n = 1}^{7} \cos 2n\alpha = \text{Re}\left(\sum_{n = 1}^7 e^{i2n\alpha}\right) = \text{Re}\left(\frac {e^{i16 \alpha} - 1}{e^{i2\alpha} - 1}\right) = \frac {\sin 7\alpha \cos 8\alpha}{\sin \alpha}$. Then $ \sin \alpha = 2\sin 7\alpha \cos 8\alpha \cdot \frac {\sin 8\alpha}{\sin 8\alpha} = \frac {\sin 7 \alpha \sin 16\alpha}{\sin 8 \alpha}$ $ \Longrightarrow 2\cos 9\alpha = \cos 23\alpha + \cos 7\alpha$ $ \Longrightarrow \cos 9\alpha = \cos 15\alpha \cos 8 \alpha$ .. shouldn't be too much harder from here ..
[/hide][/quote]

Is there a better solution for this?

[quote="Altheman"]Lets solve the system $ a \equal{} x(y \minus{} x)$ $ x^2 \plus{} y^2 \equal{} 1$ where $ a$ is given.

Naturally, we homogenize in $ x,y$ to get $ a(x^2 \plus{} y^2) \equal{} x (y \minus{} x)$ and we easily solve for $ x: y$ by the quadratic formula and we are done.

This motivates the other solution that was given...

This works for any system in the form $ A \equal{} (Bx \plus{} Cy)(Dx \plus{} Ey)$ and $ x^2 \plus{} y^2 \equal{} 1$...[/quote]

What in the world does this have to do with trig?

He was just explaining the motivation for the solution to #1, namely homogenizing. (there should be a comma between $ a \equal{} x(y\minus{}x)$ and $ x^2 \plus{} y^2 \equal{} 1$, and now look at the substitution $ x \equal{} \cos \theta, y \equal{} \sin \theta$).


LaTeX
advanced fields
advanced fields theorems

What is it?
How to prove things with it?
For example, how to proof that (in this method)
A -> B |-- (C V A) -> (C V B)

(I can prove it with true-table only...)

Thank you

Please $ \text{\LaTeX}$ your formula. Is it $ (A \rightarrow B) \wedge (C \vee A) \rightarrow C \vee B$ ?

I believe it's: $ A \Rightarrow B \vdash (C \vee A) \Rightarrow (C \vee B)$

The symbol "$ p, q, r, \ldots \vdash c$" means "given $ p, q, r, \ldots$, prove $ c$".

$ A \Rightarrow B \vdash (C \vee A) \Rightarrow (C \vee B)$   Yes.

(I am sorry, I relatively new user and I am not so good in Latex)


group theory
abstract algebra
geometry
geometric transformation
rotation
superior algebra
superior algebra solved

Without any computation show $S_{4}$ has a subgroup isomorphic to $Z_{4}$ and a subgroup isomorphic to $K_{4}$ (Klein four group). Extend this for $S_{n}.$

[quote="ZetaX"]$S_{n}$ always contains an $n$-cycle $(1,2,3,...,n)$ clearly having order $n$.
And $K_{4}$ is the group of permuations of two stacks of two elements without perturbation, thus also contained.
$S_{n}$ also always contains the dihedral group $D_{n}$ since it is a group of permutations on $n$ objects. By this, the rotations give another way of getting an element of order $n$.[/quote]

You are right, but note that I said that without any computation!! What you say about this problem:

Without any computation show $S_{8}$ has subgroups isomorphic to $Z_{8}, Z_{2}\times Z_{4},Z_{2}\times Z_{2}\times Z_{2}, Q$ and $D$ where $Q$ is the quaternion group of order 8 and $D$ dihedral group of order 8.

Where did I use computations (even for the cycle I gave a pure combi-algebraical proof).
And the generalisation is simply that $S_{n}$ contains every group that is a permutation group on $\leq n$ element set, so especially contains any group of order $\leq n$ (Cayley's theorem).

Ok, this is my solution exactly.

Thanks


Jest dany trojkat $ABC$ z  $\angle ABC = 3\angle CAB$. Punkty $M$ i $N$ sa wybrane na boku $AC$ z $N$ pomiedzy $A$ i $M$ tak, ze $\angle CBM =\angle MBN = \angle NBA$. Niech $L$ bedzie dowolnym punktem wewnetrznym odcinka $BN$, zas  $K$ punktem na $BM$ takim, ze  $LK || AC$. Udowodnij, ze proste $AL$,$NK$ i $BC$ przecinaja sie w jednym punkcie.

P.S. reanimacja trwa  :P

Najwyzszy czas podac rozwiazanie :P Niech $ NK\cap BC\equal{}P$, $ LK\cap AC\equal{}Q$ oraz $ LK\cap AB\equal{}R$. Z tw. Menelausa mamy $ \frac{BP\cdot QK\cdot CN}{PQ\cdot KL\cdot BN}\equal{}1\iff \frac{BP\cdot BC\cdot QC}{PQ\cdot BN\cdot BC}\equal{}1\iff \frac{BP}{PQ}\equal{}\frac{BN}{QC}$. Teraz mamy $ \frac{BN\cdot QL\cdot AR}{PQ\cdot RL\cdot AB}\equal{}\frac{BN\cdot CN\cdot CQ}{CQ\cdot AN\cdot BL}\equal{}1$ qed

Całkiem nieźle: po blisko 3 latach rozwiązanie (nie sprawdzałem czy poprawne...)

 :)

Sa literowki. Ma byc
$ LK\cap BC \equal{} Q$ i w ostatniej rownosci$ \frac{BP\cdot QL\cdot AR}{PQ\cdot RL\cdot AB}\equal{}\frac{BN\cdot CN\cdot CQ}{CQ\cdot AN\cdot BC}\equal{}1$


algebra
polynomial
calculus
derivative
induction
algebra proposed

If $b_{1}, b_{2}, \ldots, b_{n}$ are non-negative reals not all zero, then prove that the polynomial \[x^{n}-b_{1}x^{n-1}-b_{2}x^{n-2}-\ldots-b_{n}=0\] has only one positive root $p$, which is simple. Moreover prove that any root of the polynomial does not exceed $p$ in absolute value.

[quote="iura"]If $b_{1}, b_{2}, \ldots, b_{n}$ are non-negative reals not all zero, then prove that the polynomial
\[x^{n}-b_{1}x^{n-1}-b_{2}x^{n-2}-\ldots-b_{n}=0 \]
has only one positive root $p$, which is simple. [/quote]
It is obviosly, consider
(*) $1-b_{1}y-b_{2}y^{2}-...-b_{n}y^{n}, \ y=1/x$.
[quote]Moreover prove that any root of the polynomial does not exceed $p$ in absolute value.[/quote]
It followed from (*).

Part 1: Use Descartes' Sign Rule .
Part 2: If $x_{0}$ is a real root, then $|x_{0}|^{n}\leq b_{1}|x_{0}|^{n-1}+...+b_{n}$. And we're done.

[quote="Goblin"]In book "Polynomials" by Prasolov, this is stated as a theorem (Cauchy's)[/quote]
Yes, theorem 1.3, page 11. But prove in that not as of mine :P

[quote="iura"]If $b_{1}, b_{2}, \ldots, b_{n}$ are non-negative reals not all zero, then prove that the polynomial
\[f(x)=x^{n}-b_{1}x^{n-1}-b_{2}x^{n-2}-\ldots-b_{n}=0 \]
has only one positive root $p$, which is simple. Moreover prove that any root of the polynomial does not exceed $p$ in absolute value.[/quote]
This is prove in that book.

Put $F(x)=-\frac{f(x)}{x^{n}}=\frac{b_{1}}{x}+\frac{b_{2}}{x^{2}}+...+\frac{b_{n}}{x^{n}}-1.$ If $x\not = 0$ then $f(x)=0$ is equivalent to $F(x)=0$. Now if $x\to+\infty$ then $F(x)\to-1$ and if $x\to 0^{+}$ then $F(x)\to+\infty$, by $F$ is continuous and monotone on $(0,+\infty)$ we have $f(x)$ has only a positive root, denote it by $p$. We also have $-\frac{f'(p)}{p^{n}}=F'(p)=-\frac{b_{1}}{p^{2}}-...-\frac{nb_{n}}{p^{n+1}}<0$, so $p$ is simple.

Assume that $x_{0}$ is a root of $f$. We have  $q=|x_{0}|\leq p$. In fact, if $q>p$ we have $F(q)<0$ and so $f(q)>0$. But from $x_{0}^{n}=b_{1}x_{0}^{n-1}+...+b_{n}$ we have $q^{n}\leq b_{1}q^{n-1}+...+b_{n}$ or $f(q)\leq 0$. Contradiction!

You could use derivatives from the beginning:
Use induction on $n$. If $f$ is the polynomial in the problem, then $f^\prime$ satisfies more or less the induction hypothesis (you should watch out for the special case $x^{n}-\alpha$). Let $x_{0}$ be the positive root of $f^\prime$. Then, we have that $f$ is strictly decreasing on $[0,x_{0}]$ and strictly increasing on $\left[ x_{0}, \infty \right)$. But $f(0) \leq 0$, so $f$ has a unique positive root.

You must remember that we need prove $p$ is SIMPLE!

I thought that was clear. We have that $f \left( x_{0}\right) < 0$, so the positive root $u$ of $f$ must satisfy $u > x_{0}\, \Longrightarrow \, f^\prime (u) > 0$.


I cant seem to comprehend this type of questions. Thorough explanation would help me a lot.

2. A flare is dropped from a plane flying over level ground at a velocity of $70 m/s$ in the
horizontal direction. At the instant the flare is released, the plane begins to accelerate horizontally
at $0.75 m/s^{2}$. The flare takes $4.0 s$ to reach the ground. Assume air resistance is negligible.
Relative to a spot directly under the flare at release, where does the flare land?

I don't see why the acceleration of the plane is important... The time the flare needs to reach the ground is $t$, so the distance it covers in the horizontal direction is $vt$, since there are no forces acting on it in that direction. That should be the answer.


find the last two digits of
$2^{2004},3^{2004},4^{2004}$
how would you find the first two digits of the same numbers

SM4RT, Euler's totient theorem is only valid when the base is relatively prime to the modulus.

But it gets the correct answer for each problem, lol.

[hide="$2^{2004}$"]
We have that
\[16^{ \varphi(25)}=16^{20}\equiv 1 \bmod 25 \ \ \wedge \ \ 16^{20}\equiv 0 \bmod 4 \stackrel{\text{CRT}}{\implies }16^{20}\equiv 76 \bmod 100 \]
so
\[2^{2004}=16^{501}\equiv 16 \cdot (76)^{25}\bmod 100 \]
But, notice that
\[76 \bmod 100 \equiv 76^{2}\bmod 100 \equiv 76^{3}\mod 100 \equiv \ldots \]
so our expression is equivalent to:
\[16 \cdot (76) \bmod 100 \equiv \boxed{16 \bmod 100}\]
[/hide]

[hide="$3^{2004}$"]
We have
\[(3,100)=1 \implies 3^{ \varphi(100)}=3^{40}\equiv 1 \bmod 100 \]

\[\implies 3^{2004}\equiv 3^{4}\equiv \boxed{81 \bmod 100}\]
[/hide]

[hide="$4^{2004}$"]
From the first problem, we know that $2^{2004}\equiv 16 \mod 100$. But,
\[4^{2004}=\left( 2^{2004}\right)^{2}\equiv 16^{2}\bmod 100 \equiv \boxed{ 56 \bmod 100}\]
[/hide]
(Click on each expression to see the solution.  :D )


How would you produce formic acid from acetic acid?

[hide="Is this OK?"]$ \ce{CH3COOH}\plus{}\ce{NaOH}\longrightarrow\ce{CH3COONa}\plus{}\ce{H2O}\\
\ce{CH3COONa}\plus{}\ce{NaOH}\,(\ce{CaO},\Delta )\longrightarrow\ce{CH4}\plus{}\ce{Na2CO3}\\
\ce{CH4}\plus{}\ce{I2}\longrightarrow\ce{CH3I}\plus{}\ce{HI}\\
\ce{CH3I}\plus{}\ce{NaOH}\,(\text{aq})\longrightarrow\ce{CH3OH}\plus{}\ce{NaI}\\
\ce{CH3OH}\plus{}2[\ce{O}]\longrightarrow\ce{HCOOH}\plus{}\ce{H2O}$[/hide]

2 mistakes in step 3 itself.

[hide="1"]$ \ce {HI}$ is a reducing agent and you will get back the alkane, meaning to say, the reaction is not completely towards the right. Hence one must use an oxidising agent also, say $ \ce {HIO_3}$ or $ \ce {HNO_3}$[/hide]

[hide="2"]Even still, you cannot gaurantee a 100% pure product. You will get a mixture of products. ($ \ce {CH_2I_2}$, $ \ce {CHI_3}$ etc will also be formed.)[/hide]

[b]Side question:[/b]

Well I was working on Schmidt reaction but I cant proceed further:

+$ N_3H$ (Schmidt reaction) to give $ CH_3NH_2$

How can we go on to produce acid from an amine.

amine to amide
then hydrolysis of amide formed

Amine -> Amide how? Note that I want to form HCOOH

Thanks all for your help. A friend of mine suggested creating $ \ce{Ca}$ salt and decomposing it to get propanone. Then create $ \ce{HCOONa}$ and iodoform. But I can't understand the last step. Could anyone please help?

@hell_ever: Yes I thought about the second mistake, but I couldn't find any other way of doing it. Could you please help fix the mistake or find another way of doing this?

[quote="Carcul"]May I ask what in God's name might be the purpose of preparing formic acid from acetic acid? If you want to have practice in organic reactions there are much more instructive examples to work with.

Here's the method I would use:

1) Sell a bottle of acetic acid.

2) With the money earned in 1), buy a bottle of formic acid.[/quote]

Yes I'm pretty amateur in organic chemistry and I've been trying to learn organic reactions. But from what you say above, I don't think I should learn anymore about these, I can go to a store and buy whatever I need. You helped me lose my interest in this subject and now I think it's nothing but memorizing a stupid bunch of reactions. Thank you very much for making me realize this.

[quote="nayel"]i think it's nothing but memorizing a stupid bunch of reactions[/quote]
atleast organic is a lot more than this

wat if suddenly a fault appears and the starting product used by most of industries disappear....sm1 will hav 2 divise new method :oops:

[hide][quote="nayel"]i think it's nothing but memorizing a stupid bunch of reactions[/quote]
actually i also think so :rotfl:  :P [/hide]

Nayel, I apologize if my post above looked a bit harsh, but, as you said, you are "pretty amateur in organic chemistry". Therefore, according to my experience, this doesn't look like a problem with pedagogic value. Even for starters, the problem BanishedTraitor has posted about protoanemonin is an excellent one for learning several important reactions, their mechanisms, and to start thinking like a synthetic organic chemist. I advise you to look at it.

Also, there was nothing im my post above that implies Organic Chemistry "it's nothing but memorizing a stupid bunch of reactions". It's much more than that, specially in the laboratory.

Finally, don't forget you can always prove me wrong by providing a nice synthesis of formic acid from acetic acid.

[hide="@ Carcul"]I understand; I also apologize for my comments. But as a friend once told me: we are here to help one another, so please make sure no one gets upset. :) [/hide]

What if we reduce the acetic acid (with for example $ \ce{LiAlH4}$) to ethanol, and dehydrate it to ethylene. Then ozonolysis of the ethylene in presence of water should produce formic acid.

Shouldn't it work? :maybe:

Actually, ozonolysis in the presence of water should produce formaldehyde. In the presence of hydrogen peroxide (or direct treatment with hot potassium permanganate) would, in principle, yield formic acid.


If you consider the problem:

How many base-4 numbers with $ k$ digits have at least one digit a "2" AND have no "0"s before that "2"? (Note the number doesn't have to have a "0")?

This is the answer written recursively:

[hide]
$ k_1 \equal{} 1$
$ k_n \equal{} 4^{n\minus{}1} \plus{} 2*k_{n\minus{}1}$
So, for $ k\equal{}1,2,3,4$, you get $ k_n \equal{} 1, 6, 28, 120$
[/hide]

My question is

[hide]
Why are the solutions perfect numbers? What is the relationship (an algebraic one would be fine, but if anyone knows an intuitive relationship, that would be better)
[/hide]

[quote="magixter"]Why are the solutions perfect numbers?[/quote]
They are not all perfect numbers. For example, $ k_6 \equal{} 2016$ is not a perfect number. :lol:

There is a theorem that states that an integer is an even perfect number if and only if it is of the form $ 2^{p \minus{} 1}M_p$ for some prime $ p$ and Mersenne prime $ M_p \equal{} 2^p \minus{} 1.$ From the recursive definition, it is easy to easy why some of the $ k_i$ should be of this form and hence are even perfect numbers. ;)

[quote="ValentineA"][quote="magixter"]Why are the solutions perfect numbers?[/quote]
They are not all perfect numbers. For example, $ k_6 \equal{} 2016$ is not a perfect number. :lol:[/quote]  Or, uh, $ k_4 \equal{} 120$ ;)  To slightly restate what ValentineA wrote, you can solve that recurrence to get $ k_n \equal{} 2^{n \minus{} 1} \cdot (2^n \minus{} 1)$, so it follows from the theorem ValentineA quoted that every even number is of the form $ k_p$ for some $ p$.

By the way, a LaTeX suggestion for the O.P.: \cdot makes a much nicer multiplication dot than the asterix.  If you want a multiplication cross, you can use \times.


Prove that for all positive integers $n$,\[0<\sum^n_{k=1}\frac{g(k)}{k}-\frac{2n}{3}<\frac{2}{3}\]where $g(k)$ denotes the greatest odd divisor of $k$

are you sur that this inégalité it's true ?? :huh:

Well indeed is not dificult at all...first let´s see that $\displaystyle 0<\sum^n_{k=1}\frac{g(k)}{k}-\frac{2n}{3}<\frac{2}{3}$ is iff ${\displaystyle \frac{2n}{3}<\sum^n_{k=1}\frac{1}{p(k)}<\frac{2n}{3}+\frac{2}{3}}$ where $p(k)$ is the largest power of 2 which divides k.  Now lets call $Q(n)=\sum^n_{k=1}\frac{1}{p(k)}$ then lets see that if n fit in the equality then 2n also...that is because:

$Q(2n)=\sum^{n}_{x=1}\frac{1}{p(2x-1)}+\sum^{n}_{x=1}\frac{1}{p(2x)}$ now see that in the first sume there are just 1´s...

$Q(2n)=n+\sum^{n}_{x=1}\frac{1}{p(2x)}=n+\frac{1}{2}\sum^{n}_{x=1}\frac{1}{p(x)}=n+\frac{1}{2}Q(n)$
and if n fit in the equality then $\frac{2(2n)}{3}+\frac{2}{3}>n+\frac{n}{3}+\frac{1}{3}>n+\frac{1}{2}Q(n)>n+\frac{n}{3}=\frac{2(2n)}{3}$
 so 2n fits...

also y $2n$ fits then $2n+1$ and that is because $Q(2n+1)=Q(n)+\frac{1}{p(2x+1)}=Q(2n)+1$
and that make 2n+1 fits in the equality...just check i have to leave...but lets see that $Q(1)=1$ and it fits so every other natural fits...just multiply by 2 and sum 1


:rotfl:  :)  :o
please prove this inequlity

I think it maybe classify number theory.

See this
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=605111911&t=6233


Problem. Give $ A,B \in M_n(k)$. Prove that:
1.
$ rank(A \plus{} B) \leq rank(A) \plus{} rank(B)$
2.
$ rank(A) \plus{} rank(B) \minus{} n \leq rank (AB) \leq min(rank(A),rank(B))$

1. More generally, let $ A,B \in M_{m \times n} (\mathbb{K})$, and let $ T: V \to W$, $ L: V \to W$ (with $ \dim V \equal{} n$, $ \dim W \equal{} m$) be linear transformations represented by matrices $ A, B$, respectively. Then, $ A \plus{} B$ represents the linear transformation $ T \plus{} L$. Now, since $ Im(T\plus{}L), Im(T), Im(L)$ are subspaces of $ W$ and $ Im(T\plus{}L) \subseteq Im(T) \plus{} Im(L)$, then $ Im(T\plus{}L)$ is a subspace of $ Im(T) \plus{} Im(L)$, and so

$ \dim Im(T\plus{}L) \leq \dim [Im(T) \plus{} Im(L)]$

$ \equal{} \dim Im(T) \plus{} \dim Im(L) \minus{} \dim [Im(T) \cap Im(L)]$

$ \leq \dim Im(T) \plus{} \dim Im(L)$.

Since for any linear transformation F, the dimension of its image equals the rank of any matrix that represents it, we conclude from the previous inequality that

$ rank(A\plus{}B) \leq rank(A) \plus{} rank(B)$. $ \Diamond$

2. Let A,B be matrices such that the product AB is defined. Since the columns of AB are linear combinations of the columns of A, we have $ rank(AB) \leq rank (A)$. On the other hand, the rows of AB are linear combinations of the rows of B, and so $ rank(AB) \leq rank B$. Hence, $ rank(AB) \leq min\,\,[rank(A),rank(B)]$. The other inequality, $ rank(A) \plus{} rank(B) \leq rank(AB) \plus{} n$, is Sylvester's inequality, which can be proved directly or as a special case of the Frobenius inequality. In any case, you can also use the concept of linear transformation, and I will leave that to you.

[quote="vipCD_A1"]Problem. Give $ A,B \in M_n(k)$. Prove that:

2.
$ rank(A) \plus{} rank(B) \minus{} n \leq rank (AB)$[/quote]

Can you detail solve?
I have thought about it but I can't solve it. Please, carcul!

For instance, start by assuming that $ A \equal{} diag \{I_{r(A)}, O\}$, where $ r(A)$ is the rank of A. Then show that $ r(AB) \geq r(A) \plus{} r(B) \minus{} n$. Next, if A does not have the form above, there are invertible matrices P,Q such that $ D \equal{} diag \{I_{r(A)}, O\} \equal{} PAQ$. Using the previous result we have $ r(DB) \equal{} r(PAQB) \geq r(D) \plus{} r(B) \minus{} n \equal{} r(A) \plus{} r(B) \minus{} n$. Now show that $ r(PAQB) \equal{} r(AB)$.


Im new to the pigeonhole prinicple and i think i understand the theory but i cant solve tough problems...e.g.

(1)If {a1,a2,a3...a1995} (numbers are subscripts) be a sequence of positive integers whose sum is 3989, show that there is a block of r sucessive ai's (i is subscripts here) whose sum is 95.

(2)A Chessmaster who has 11 weeks to prepare for a tournament decides to play at least one game everyday, but no more than 12 games during a week.Show that there exists a succession of days during which the chessmaster played exactly 21 games.

Can someone please give me advice on how to attack such problems.....also it would be nice if someone posts the solution to the to probs.......

i was going to post something one this or might have already. but yeah


A,b,c,d are real nunbers .a^2+b^2+c^2+d^2=1
Prove that ab+ac+ad+bc+bd+cd<=4abcd+5/4

[quote="stevenwang1992"]A,b,c,d are real nunbers .a^2+b^2+c^2+d^2=1
Prove that $ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\leq4abcd\plus{}\frac{5}{4}$[/quote]
Let $ a\equal{}\min\{a,b,c,d\}.$
$ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\leq4abcd\plus{}\frac{5}{4}\Leftrightarrow$
$ 5(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\plus{}16abcd\geq4(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)(a^2\plus{}b^2\plus{}c^2\plus{}d^2),$
which is easy to check after substitution  $ b\equal{}a\plus{}x,$ $ c\equal{}a\plus{}y$ and $ d\equal{}a\plus{}z.$ :wink:


Taken a point A in a regular polygon A1...An. Prove that there exists at least one angle A_i O A_j verifying the inequalities:

                                   pi(1-1/n)  <  A_i O A_j  <  pi
[/hide]

it is pigeonhole on the n wells $(\frac{k}{n}, \frac{k+1}{n})$ for $1\leq k\leq n$


$1+2+3+...+n=\sum k$
$1\cdot 2+2\cdot 3+...+n\cdot (n+1)=\sum k(k+1)=\sum k^{2}+\sum k$
$1\cdot 2\cdot 3+2\cdot 3\cdot 4+...+n\cdot (n+1) \cdot (n+2)=\sum k(k+1)(k+2)=\sum k^{3}+3\sum k^{2}+2\sum k$
All the sums from the $RHS$ are known, and divisibility should give easy.

hmm
sorry, my representation is somewhat wrong.

exactly writting is following
3^{2} | 1*2+1*3+...+2*3+...+(3^{3}-3)*(3^{3}-1) (just sum of 2-element combination of S)

actually problem can be written like this...(hmm above rep was somewhat stupid :blush: )
3^{3} | (x+1)(x+2)...(x+(3^{3}-1))'s 2th-coeff
3^{2} | (x+1)(x+2)...(x+(3^{3}-1))'s 3th-coeff
3^{1} | (x+1)(x+2)...(x+(3^{3}-1))'s 4th-coeff


Given that the largest triangle is equilateral and all the smaller triangles are formed by connecting the midpoints of the sides of the larger triangle that contains, them, how many equilateral triangles are there?
[asy]import olympiad; import geometry_dev; import graph; size(100); defaultpen(linewidth(0.8));
  void drawEqTriangle(real side, pair blCorner, real degrees = 0){
    pair c1 = blCorner;
    pair c2 = c1 + side*dir(degrees);
    pair c3 = c2 + side*dir(120 + degrees);
    draw(c1--c2--c3--cycle);
  }
  drawEqTriangle(1,origin); drawEqTriangle(1,(1,0)); drawEqTriangle(1,(0.5,sqrt(3)/2));
  for(int i = 1; i <=3; ++i){
    for(int j = 0; j <i; ++j){
      drawEqTriangle(1/4, (1 - i/8 + 1/4*j, sqrt(3)/8*i));
    }
  }[/asy]

From smallest triangles to largest: 16+7+3+4+1=31


function
real analysis
real analysis unsolved

Show that if $ f: R \rightarrow R$ is a monotone increasing function, then the set where $ f$ is discontinuous has measure zero.

The set where f is dicontinuous is countable, because they are all jump discontinuities. Uncountably many jump discontinuities of positive size on some interval $ (a,b)$ would make $ f$ increase more than $ f(b) \minus{} f(a)$ on the interval

in fact, I saw what you said, but I don't know how to prove that [i]uncountably many jump discontinuities of positive size on some interval $ (a,b)$  would make  increase $ f$ more than  on the interval $ (f(a),f(b))$.[/i]

[quote="Kalle"]I was just loosely saying that an "uncountable sum" of positive numbers is infinite. Index all the discontinuities in $ (a,b)$ as $ x_j$ where $ j$ runs through some index, and let the corresponding jump size at $ x_j$ be $ K_j$. Then the set of $ x_j$ such that $ K_j \geq 1/n$ is finite. The set of all $ x_j$ is the countable union of the former sets as $ n$ runs through the positive integers. The countable union of finite sets is countable, so the number of $ x_j$ is countable.[/quote]

Thank you Kalle !

[quote="Kalle"]I was just loosely saying that an "uncountable sum" of positive numbers is infinite. Index all the discontinuities in $ (a,b)$ as $ x_j$ where $ j$ runs through some index, and let the corresponding jump size at $ x_j$ be $ K_j$. Then the set of $ x_j$ such that $ K_j \geq 1/n$ is finite. The set of all $ x_j$ is the countable union of the former sets as $ n$ runs through the positive integers. The countable union of finite sets is countable, so the number of $ x_j$ is countable.[/quote]

why is the set of $ x_j$ s.t. $ K_j \geq \frac{1}{n}$  finite?


For natural numbers $m, n$, set $a = (n+1)^{m}-n$ and $b = (n+1)^{m+3}-n$.

(a) Prove that $a$ and $b$ are coprime if $3 \nmid m$.

(b) Find all numbers $m, n$ for which $a$ and $b$ are not coprime.

This is solve at
http://www.mathlinks.ro/Forum/viewtopic.php?t=116060


geometry
3D geometry
sphere
Pythagorean Theorem

Given two balls. The larger ball of radius 9 leans against the wall of a room, and the smaller ball, or radius 4, rests upon the larger ball. Each ball is tangent to the wall, and the larger ball is tangent to the floor. What is the distance from the floor to the point of tangency of the smaller ball?

The solution given is 21, but I'm getting 22.

The solution uses the picture.

[hide="Agree with the book"]By the Pythagorean Theorem, $ GE = 12$. $ 9 = DA = EB$. Hence, the length is $ 12+9 = 21$[/hide]

the correct answer is $ 21$.

[hide="solution"]
[asy]size(200); draw((0,0)--(100,0)); draw((0,0)--(0,100)); pair A,B,C,D,R,F,M,P,Q,O; O=origin; A=(0,27); B=(27,0); C=(54,27); D=(0,63); R=(24,63); F=(12,51); draw(circumcircle(A,B,C)); draw(circumcircle(D,R,F)); label("D",D,W); label("A",A,W); P=(27,27); Q=(12,63); label("P",P,E); label("Q",Q,E); dot(P); dot(Q); draw(P--Q); M=(12,27); label("M",M,S); dot(M); draw(A--P); draw(Q--M); label("O",O,W);[/asy]
from the diagram, we get using pythagorean theorem, $ PQ=9+4=13=\sqrt{QM^{2}+(9-4)^{2}}$ thus $ QM=AD=12$.
so $ OD=AD+AO=12+9=21$.
[/hide]

EDIT: sorry vishalarul, didn't see your post there.

Thanks for your help, I see my error!  :lol:


trigonometry geometry function probability AMC

This is a stupid question and can be deleted at soon as there is one accurate response.  How much time do you get for the AMC 10/12?  I've heard 75 and I've heard 90.

Edit: Perhaps I'll substantiate this post slightly by asking, what tips/strategies people have for maximizing productivity under limited and straining time conditions.  Also, how can one reach the later level questions quickly (as they require more time than the earlier questions), without making silly careless errors?

AMC 10/12 has a 75 minute time limit.  The old AHSME had a 90 minute time limit (AHSME turned into AMC).

Tips:

Don't waste too much time on a problem.  If, for example, there is a trig problem but you just learned trig last week in your precalc class, skip it (unless you are confident you can do it in a timely fashion).

Try to stay organized.  You don't want to be unable to correctly read what you wrote.  It also helps to have organized work in case you have the chance to go back and check your work.

There aren't really any tips I can give for you to get through the 'easier' problems quicker.  That mostly comes with practice (the more times you've seen a certain type of problem, the faster you'll recognize how to approach, and so forth).

Why should you skip a problem that you just learned how to do?  That seems like it would be one of the problems you would want to make sure you solve.

He means skip it, since you were just recently introduced to the concept, so that you don't spend a disproportionate amount of time on it. Obviously, if you've finished all the problems that you can do easily, go back to that one and try your hand at it, but don't go after it when you get to it (unless it's at the end of the test, obviously :P). As joml said, if you're sure you can do it quickly, go for it, but make sure you assess your capabilities in that subject realistically.

Wow, I just used a whole lot of big words. :what?:

More tips:

1. Always check as you go. Of course it's nice to check back at the end, especially if you've organized your work well, but often you don't have time anyway. Also, you might have forgotten some special cases that you noticed the first time. Being meticulous the first time is usually the most effective method.

2. Don't be too hasty. Actually I'm not really sure if this helps, but a lot of times I dive into a problem because I immediately see a way, when in fact there's an easier way that I could have used to save up to a minute or so if I had just spent a few seconds longer before diving in.

After 3 minutes and you've gotten no where on a problem, it's not the end of the world, just skip it and come back to it after doing like 2 or 3 other problems.

i would do all the problems that i am good at (geometry, nt, functions,etc) and skip all but the easiest probability (probability and combinatorics make me mad :mad: ), then if i have time, go back to the problems i am bad at, also most of the early amc problems can be done with 2 sentences of explaination if you do them correctly, so don't attack a tedious solution, find a better one, or move on, if you find the best solution, very little work is needed, and don't hesitate to use lots of paper, writing big helps you check better

I'm certainly no expert but if you look at a problem and it looks hard to you dont waste time, do the easy problems first, and then investigate those problems..

I think the key is to focus correctly.

Focus on reading every word of the problem, and taking in the meaning.
Then just get everything clear in your mind exactly what you need to do.
Organize your work and write out as much as you can (without going too slowly). Continue to keep your mind clear as to what you're doing and what you need to do. Especially, don't go through sloppy calculations or thoughts in your head; write things down.
Once you have the answer just write it in and go on to the next problem.
Still remember not to take too much time on each problem.

One example is that in a math competition, I had a factoring/geometry problem. I got three solutions, 0, 8, and 60. I quickly thought through in my head that 0 didn't work, and didn't bother to check 8 and 60, thinking they worked because they were positive. As it turned out, 8 didn't work. Only 60 did.

Although that particular problem might not apply to AMC 12 directly, the idea behind it does. Rather than quickly thinking through in my head, I should have paid attention to each answer individually, doing the full process of checking each answer.
Anyway, checking the answers should have only taken a few seconds apiece. One can still focus in this way and do a problem quickly.


Joyce wants to share six identical pieces of licorice equally among four people. What is the mnimum number of pieces that must be cut?

2 is correct, but if you don't see why, I will explain.

Sharing 6 pieces amoung 4 people means each person gets $\frac64=1+\frac12$.  Give each person a stick, and you have 2 sticks remaining.  Now, cut them in half so there are 4 halfs to give to the people.  You only have to cut 2 of the sticks.


Find all $ x,y\in Z^\plus{}$ such that $ 3^x\equal{}2^x.y\plus{}1$
If post before,please give links  :)

Also, refer to the exponent lifting lemma


$ n\neq{1}$and$ m\neq{1}$
Suppose by absurd $ \frac{m^{n}-1}{n-1}\notin{Z}$ and $ \frac{n^{m}-1}{m-1}\notin{Z}$ then
$ (m-1)\frac{(m^{n}-1)}{n-1}\in{Z}$ and $ (n-1)\frac{n^{m}-1}{m-1}\in{Z}$
$ \Leftrightarrow \frac{m-1}{n-1}\in{Z}$ and $ \frac{n-1}{m-1}\in{Z}\Leftrightarrow{m=n}$absurd


If $ \sin x\plus{}a \cos x\equal{}b$, express $ |a \sin x \minus{}\cos x|$ in terms of $ a,b$.

Let's calculate $ (\sin x \plus{} a \cos x)^2\plus{}|a \sin x \minus{} \cos x|^2.$


superior algebra
superior algebra unsolved

Show that $C$ has infinitely many field automorphisms. What is the cardinality of the set of these automorphisms?

There are as many as possible: $2^{2^{\aleph_{0}}}$. This is because any permutation of a transcendence basis of $\mathbb C$ over $\mathbb Q$ (and all such bases have cardinality $2^{\aleph_{0}}$) extends to an automorphism of $\mathbb C$.


calculus integration limit logarithms geometry rectangle inequalities

Let $ n$ be natural number. Find the limit value of ${ \lim_{n\to\infty} \frac{1}{n}(\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{5}}}+\cdots\cdots +\frac{n}{\sqrt{n^2+1}}).$

hello kunny, we get
$ \lim_{n\to \infty}\frac{1}{n}\left(\sum_{i\equal{}1}^{n}\frac{i}{\sqrt{i^2\plus{}1}}\right)\equal{}1$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello kunny, we get
$ \lim_{n\to \infty}\frac {1}{n}\left(\sum_{i \equal{} 1}^{n}\frac {i}{\sqrt {i^2 \plus{} 1}}\right) \equal{} 1$
Sonnhard.[/quote]
Your answer is circular answer.
You are the same as answering by feeling.

[b]My Solution[/b]
Let $ f(n) \equal{} \frac{n}{\sqrt{n^2\plus{}1}}$.
So,
$ f(1) \plus{} \int_1^n f(x)dx \le \sum_{i \equal{} 1}^{n} f(i) \le \int_1^n f(x)dx \plus{} f(n)$
$ \iff \frac{1}{\sqrt{2}} \plus{} \sqrt{n^2\plus{}1} \minus{} \sqrt{2} \le \sum_{i \equal{} 1}^{n} f(i) \le \sqrt{1\plus{}n^2} \minus{} \sqrt{2} \plus{} \frac {n}{\sqrt {n^2\plus{}1}}$
$ \iff \sqrt{n^2\plus{}1} \minus{} \frac{1}{\sqrt{2}} \le \sum_{i \equal{} 1}^{n} f(i) \le \frac {n^2\plus{}n\plus{}1}{\sqrt {n^2\plus{}1}} \minus{} \sqrt{2}$
$ \iff \sqrt{1\plus{}\frac{1}{n^2}} \minus{} \frac{1}{n\sqrt{2}} \le \frac{1}{n}\left(\sum_{i \equal{} 1}^{n} f(i)\right) \le \frac {1\plus{}\frac{1}{n}\plus{}\frac{1}{n^2}}{\sqrt {1\plus{}\frac{1}{n^2}}} \minus{} \frac{\sqrt{2}}{n}$
$ \Rightarrow 1 \le \frac{1}{n}\left(\sum_{i \equal{} 1}^{n} f(i)\right) \le 1 \ \ \ \ \ (n \to \infty)$

Thus, $ \lim_{n \to \infty} \frac{1}{n}\left(\sum_{i \equal{} 1}^{n} f(i)\right) \equal{} \boxed{1}$

We have:
$ 1 \geq \frac{1}{n}\sum_{i\equal{}1}^{n}\frac{i}{\sqrt{i^2}} \geq \frac{1}{n}\sum_{i\equal{}1}^{n}\frac{i}{\sqrt{i^2 \plus{} 1}} \geq \left(\prod_{i\equal{}1}^{n} \frac{i}{\sqrt{i^2 \plus{} 1}}\right)^{\frac{1}{n}} \geq \left(\prod_{i\equal{}1}^{n} \frac{i}{\sqrt{\left(i \plus{} 1\right)^2}}\right)^{\frac{1}{n}} \equal{} \left(\prod_{i\equal{}1}^{n} \frac{i}{i \plus{} 1}\right)^{\frac{1}{n}} \equal{} \left(1 \plus{} n\right)^{\frac{1}{n}}$

Applying the sandwich method we get the limit to be 1.

[quote="akech"]We have:
$ 1 \geq \frac {1}{n}\sum_{i \equal{} 1}^{n}\frac {i}{\sqrt {i^2}} \geq \frac {1}{n}\sum_{i \equal{} 1}^{n}\frac {i}{\sqrt {i^2 \plus{} 1}} \geq \left(\prod_{i \equal{} 1}^{n} \frac {i}{\sqrt {i^2 \plus{} 1}}\right)^{\frac {1}{n}} \geq \left(\prod_{i \equal{} 1}^{n} \frac {i}{\sqrt {\left(i \plus{} 1\right)^2}}\right)^{\frac {1}{n}} \equal{} \left(\prod_{i \equal{} 1}^{n} \frac {i}{i \plus{} 1}\right)^{\frac {1}{n}} \equal{} \left(1 \plus{} n\right)^{\frac {1}{n}}$

Applying the sandwich method we get the limit to be 1.[/quote]

your solution and Kouichi Nakagawa's both are excellent. I do agree with Kouichi Nakagawa that Dr Sonnhard Graubner's answer is probably based on answering by feeling.

I think instead of $ \left(1 \plus{} n\right)^{\frac {1}{n}}$
write $ \frac{1}{\left(1 \plus{} n\right)^{\frac {1}{n}}}$

and $ \lim_{n\to\infty}\left(1 \plus{} n\right)^{\frac {1}{n}}\equal{}1$

[color=blue]
My solution as follows:

As $ \ \frac {r}{\sqrt {(r\plus{}1)^2}}<\frac {r}{\sqrt {r^2\plus{}1}}<\frac {r}{\sqrt {r^2}}\ \forall \  r \equal{} 1,\ 2,\ 3,\ \dots$

$ \implies\frac {r}{r\plus{}1}<\frac {r}{\sqrt {r^2\plus{}1}}<\frac {r}{r}$

$ \implies1\minus{}\frac {1}{r\plus{}1}<\frac {r}{\sqrt {r^2\plus{}1}}<1$

$ \implies \sum_{r \equal{} 1}^{n}\left(1\minus{}\frac {1}{r\plus{}1}\right)<\sum_{r \equal{} 1}^{n}\frac {r}{\sqrt {r^2\plus{}1}}<\sum_{r \equal{} 1}^{n}1$

$ \implies n\minus{}\sum_{r \equal{} 2}^{n\plus{}1}\frac {1}{r}<\sum_{r \equal{} 1}^{n}\frac {r}{\sqrt {r^2\plus{}1}}< n$

As $ \ \int_{1}^{n\plus{}1}\frac{1}{x}\ dx \plus{} \frac {1}{n\plus{}1}\ge \sum_{r \equal{} 2}^{n\plus{}1}\frac {1}{r}$

$ \implies n\minus{}\int_{1}^{n\plus{}1}\frac{1}{x}\ dx \minus{} \frac {1}{n\plus{}1}<\sum_{r \equal{} 1}^{n}\frac {r}{\sqrt {r^2\plus{}1}}< n$

$ \implies 1\minus{}\frac{\ln (n\plus{}1)}{n} \minus{} \frac {1}{n(n\plus{}1)}<\frac{1}{n}\sum_{r \equal{} 1}^{n}\frac {r}{\sqrt {r^2\plus{}1}}< 1$

As $ \ \lim_{n\to\infty}\frac{\ln (n\plus{}1)}{n}\equal{}0$  and  $ \ \lim_{n\to\infty}\frac{1}{n(n\plus{}1)}\equal{}0, $

By applying sandwich theorem we get,

$ \boxed {\boxed {\lim_{n\to\infty}\frac{1}{n}\sum_{r \equal{} 1}^{n}\frac {r}{\sqrt {r^2\plus{}1}}\equal{}1}}$

[/color]

Another approach: let $ a_n \equal{} \frac{n}{\sqrt{n^2 \plus{} 1}}$.  As $ n$ approaches infinity, $ a_n$ approaches 1.  So by Cesaro's theorem, $ \frac{1}{n} \sum_{i\equal{}1}^n a_i$ approaches 1 too.

[quote="Ravi B"]Another approach: let $ a_n \equal{} \frac {n}{\sqrt {n^2 \plus{} 1}}$.  As $ n$ approaches infinity, $ a_n$ approaches 1.  So by Cesaro's theorem, $ \frac {1}{n} \sum_{i \equal{} 1}^n a_i$ approaches 1 too.[/quote]

Can u explain Cesaro's theorem in detail

If you have a sequence of numbers $ a_n$ that converges to $ L$, then the sequence of partial averages $ \frac{1}{n} \sum_{i\equal{}1}^n a_i$ also converges to $ L$.

Note that the sum is equivalent to the area of the rectangles above the curve $ y = \frac {x}{\sqrt {x^2 + 1}}$ on $ [0,n]$, where each rectangle has width $ 1$. Note also that each term of the sum is less than $ 1$, so these rectangles all lie below $ y=1$. Hence, we get the inequality

$ \int_0^n \! \frac {x}{\sqrt {x^2 + 1}} \, dx \le \sum_{i = 1}^{n}\frac {i}{\sqrt {i^{2} + 1}} \le \int_0^n \! dx$

and subsequently,

$ \frac {\sqrt {n^2 + 1} - 1}{n} \le \frac {1}{n} \sum_{i = 1}^{n}\frac {i}{\sqrt {i^{2} + 1}} \le 1$

Since $ \lim_{n\to\infty}\!\frac {\sqrt {n^2 + 1} - 1}{n}\, = 1$ and $ \lim_{n\to\infty}1 = 1$, by the Squeeze Theorem, we get

$ {\lim_{n\to\infty}\frac {1}{n}\sum_{r = 1}^{n}\frac {r}{\sqrt {r^{2} + 1}} = 1}$

Thank you for your replies! :lol:

Kouichi Nalagawa and Elixir Solution are a standard for Japanese high school students.

My solution is almost same as  Elixir 's, but I write $ \int_0^n f(x)\ dx < \sum_{k \equal{} 1}^n \frac {k}{\sqrt {k^2 \plus{} 1}} < \boxed{\int_1^{n \plus{} 1}}\ f(x)\ dx$

Cesaro's theorem by Ravi B is very useful, but it usually appeares in University mathematics in Japan.

I am rather pleased to see that Japanese high school* students are taught some interesting theory and asked questions that actually require some thought - a "hard" question in the UK would be probably be off a STEP paper;

http://www.admissionstests.cambridgeassessment.org.uk/adt/step

I just remmbered this is a medical entrance exam - in the UK a medical student would also hopelessly struggle as mathematics is not a requirement for entrance into medical schools.


inequalities
linear algebra
matrix
linear algebra unsolved

Let $A_i \in {\mathbb C}^{n \times n},i=1,2,...,n$. If there is at least a $k \in \{1,2,...,n\}$ such that $A_k = O_n$ then show that the following inequality holds: $\sum_{i=1}^{n} rank(A_i) \leq n(n-1)$.
 :ninja:

:huh:
Is $O_n$ the zero matrix (then it's trivial), or is it an orthogonal matrix (then it's false) ?


probability
AMC
AIME
counting
distinguishability
summer program
Mathcamp

What do ppl mean when they say u need to know the balls-and-urns argument in order to solve many of the probability problems in AIME? Can someone tell me what that argument is.  :D

If you have $b$ indistinguishable balls and $u$ urns, the number of ways you can put $b$ balls in $u$ urns is $\dbinom{b+u-1}{u-1}$. If there is the additional restriction that there must be at least one ball in each urn, the number of ways is $\dbinom{b-1}{u-1}$.

What if u have different types of balls? Then does the same argument still hold or is it somewhat different?

Oh, I forgot, sorry, they have to be indistinguishable balls. Distinguishable balls have different methods.

A specific question: If you have $6$ indistinguishable balls that you need to pout into $4$ distinguishable urns, how many ways can you put the balls into the urns?

Answer:
We make a bijection. Instead of putting balls into urns, let us imagine that the balls are arranged in a straight line and that we are putting dividers between the balls. There will be 3 dividers.

For example, oo|o||ooo means put two balls in the first urn, one ball in the second urn, zero balls in the third urn, and 3 balls into the fourth urn.

Each permutation of oooooo||| represents a different way to put the balls into the urns, and each way the balls can be put into urns can be represented as a permutation of oooooo|||. However, it is easy to count the number of permutations of oooooo|||, which is $\binom{9}{3}$.

In general with b balls and u urns, you get what davidyko has.

[quote="turtlecloud"]woah.. i was about to ask the same question... So where did everyone learn the balls and urns argument from? Was it in Aops?1 or 2?[/quote]
It's in intro to C&P and AoPS vol. 2.

[quote="randomdragoon"]I learned in MOP, I think it's in AoPS, and the teacher in my school's honors math class even talked about it! This concept is amazingly useful. For example:

How many 4-digit numbers have the property that no digit is less than the digit to its left?[/quote]

[hide="hint"]

I interpret it as, "How many way are there to pick 4 digits from 9 if we allow repetition?"

[/hide]

[quote="Kamil Witek"]Where is it found in aops 2? i looked through the book but i still cant find it.[/quote]
Eh, in the counting in the twighlight zone section I think.

Not really sure.

[quote="davidyko"]Meh, I don't think it is in there. Oh well, I learned it from these forums.  :P[/quote]
I'm almost positive it's in there.
It was a problem about a person buying 8 german dogs with three breeds to choose from.

Never mind, you're right.  :roll:
Chapter 17.2, Clever Correspondences, the first problem about 8 dogs and three breeds, just in case anyone wanted to know.

[quote="123s"]What about indistinguishable urns? How would you do that?[/quote]

I believe, correct me if I'm wrong, that that's counting the number of partitions, which has an extremely extremely complicated formula (Bell numbers) that I think was on the back of some mathcamp shirts and went all the way across.

[quote="13375P34K43V312"]For that you could be 1337 and use generating functions, then plug into mathematica...

coefficient of $x^{n}$ in $(1+x+x^{2}+\cdots)^{\infty}$ is the number of partitions of $n$.[/quote]

Heh Heh no.

That expression you gave is always $\infty$. (look at the expansion for the coefficient of x).

The correct generating function would be
$(1+x+x^{2}+\cdots)(1+x^{2}+x^{4}+\cdots)(1+x^{3}+x^{6}+\cdots)(\cdots)$

which is equal to

$\prod_{n\in\mathbb{N}}\frac{1}{1-x^{n}}$

[quote="machack"][quote="randomdragoon"]I learned in MOP, I think it's in AoPS, and the teacher in my school's honors math class even talked about it! This concept is amazingly useful. For example:

How many 4-digit numbers have the property that no digit is less than the digit to its left?[/quote]

is the answer 12 choose 4? (which is 495)[/quote]
12 choose 3 is the answer. Remember that if you have n urns you will only have n-1 divider bars!

EDIT: This post is wrong, see below

I'm feeling incredibly stupid after posting about balls-and-urns...
Will someone enlighten me why it's [b]12[/b] choose 3?

I get 12C4 as well...isn't it analogous to the situation of having to give out 4 apples to 9 kids (we label them from 1 to 9) with no restrictions?

In that case, it would be 12C3.
But why 12? Why 9 kids? I think I'm missing something big here.

Perhaps what you are missing is that the first digit can't be zero. Since no digit is less than the one on its left, this means that no digit can be a zero. :roll:

i really think its 12 c 4 because it isn't exactly the concept of ball and urn (the numbers don't have to add up to a ceertain sum. and the "dividing" concept don't work in this example.
Anyone agree with me?

[quote="randomdragoon"]

For example, oo|o||ooo means put two balls in the first urn, one ball in the second urn, zero balls in the third urn, and 3 balls into the fourth urn.

Each permutation of oooooo||| represents a different way to put the balls into the urns, and each way the balls can be put into urns can be represented as a permutation of oooooo|||. However, it is easy to count the number of permutations of oooooo|||, which is $\binom{9}{3}$.
[/quote]

i thought that was the method behind hockey-stick identity? :huh:

Never mind, got it. Yep, it's $\dbinom{12}{3}$.
A good night's sleep can work wonders.

I don't see the conection between the 4-digit number problem and the balls-in-urns formula

Sorry it is $\binom{12}{4}$. I needed a night's sleep as well.  :oops:

Explanation: There aren't 4 urns, there are 9 urns, and 4 balls.


calculus
integration
algebra
function
domain
superior algebra
superior algebra unsolved

If $R$ is an integral domain which has only finitely many ideals, then show that $R$ is a field.

  Darij

Solution without thinking: finitely many ideals => artinian => zero-dimensional, and zero-dimensional + domain => field. owk

Is it the Krull dimension you are talking about? In fact, I am not that far in Atiyah-Macdonald...

[hide="Solution without looking up"][i]Solution.[/i] Let $ a$ be a nonzero element of $ R$. Denote the (finitely many) ideals of $ R$ (including both $ R$ itself and $ 0$) by $ I_{1}$, $ I_{2}$, ..., $ I_{n}$ such that $ I_{1} \equal{} R$ and such that $ I_{i}\neq I_{j}$ for any two integers $ i$ and $ j$ from the set $ \left\{1,2,...,n\right\}$ satisfying $ i\neq j$. Then, $ aI_{1}$, $ aI_{2}$, ..., $ aI_{n}$ are also ideals of $ R$.

Assume that $ aI_{i} \equal{} aI_{j}$ holds for two integers $ i$ and $ j$ from the set $ \left\{1,2,...,n\right\}$. Then, for every $ s_{i}\in I_{i}$, there exists some $ s_{j}\in I_{j}$ satisfying $ as_{i} \equal{} as_{j}$ (because $ as_{i}\in aI_{i} \equal{} aI_{j}$); this becomes $ a\left(s_{i} \minus{} s_{j}\right) \equal{} 0$, what yields $ s_{i} \minus{} s_{j} \equal{} 0$ because $ a$ is nonzero (and $ R$ is an integral domain). Hence, $ s_{i} \equal{} s_{j}$, so that $ s_{i}\in I_{j}$. Since this holds for every $ s_{i}\in I_{i}$, we thus get $ I_{i}\subseteq I_{j}$. Similarly, $ I_{j}\subseteq I_{i}$. Thus, $ I_{i} \equal{} I_{j}$, what yields $ i \equal{} j$ (since we else would have $ I_{i}\neq I_{j}$).

So we see that if $ i$ and $ j$ are two integers from the set $ \left\{1,2,...,n\right\}$, then $ aI_{i} \equal{} aI_{j}$ holds only if $ i \equal{} j$. On the other hand, $ aI_{1}$, $ aI_{2}$, ..., $ aI_{n}$ are ideals of $ R$ and thus elements of the set $ \left\{I_{1},I_{2},...,I_{n}\right\}$. Thus, $ aI_{1}$, $ aI_{2}$, ..., $ aI_{n}$ are $ n$ pairwise distinct elements of the set $ \left\{I_{1},I_{2},...,I_{n}\right\}$. This yields that the set $ \left\{aI_{1},aI_{2},...,aI_{n}\right\}$ is a permutation of the set $ \left\{I_{1},I_{2},...,I_{n}\right\}$. Thus, in particular, there exists some $ k\in\left\{1,2,...,n\right\}$ such that $ I_{1} \equal{} aI_{k}$. Since $ 1\in R \equal{} I_{1}$, this yields $ 1\in aI_{k}$, so that there exists some $ b\in I_{k}$ satisfying $ 1 \equal{} ab$. Hence, $ a$ has a multiplicative inverse in $ R$. Since $ a$ can be any arbitrary nonzero element of $ R$, we have thus shown that every nonzero element of $ R$ has a multiplicative inverse, and thus $ R$ is a field, qed.[/hide]

  darij

just consider $(x) \supseteq (x^{2}) \supseteq ...$, and you're done. this also shows domain + artinian = field.

[quote="darij grinberg"]Is it the Krull dimension you are talking about? In fact, I am not that far in Atiyah Macdonald...
[/quote]

Yes, but zero-dimensional just means that all prime ideals are maximal. owk

[quote="darij grinberg"]If $ R$ is an integral domain which has only finitely many ideals, then show that $ R$ is a field. [/quote]

The nonzero principal ideals form a group, being a finite cancellative monoid.


Can anyone suggest me some good books for preparing for the RMO . i am currently in 12th and i need some advice on this

The book Excursion in Mathematics cover all the topics that could ever come in RMO.For the depth on those topics you can try books on a particular topic like
Burton's Number Theory,Balakrishnan's Combinatorics and Sharygin's Geometry(Though out of print).If you are looking for a problem book then arthur engel is a unique collection of problems.But most problems are very difficult.There are other olympiad problem books like titu's Mathematical Olympiad Challenges but their coverage is incomplete.

Is  Challenge and Thrill of Pre-College Mathematics by  V. Krishnamurthy, C. R. Pranesachar, K. N. Ranganathan and B. J. Venkatachala(New Age International Publishers) good?Well;I find the questions pretty tough!

i had to bear much difficulties before ensuring myself a copy of'challenge & thrills.......'
the book is a nice one............

I want it to! Asking a friend in Delhi to mail it :P

u may also contact the publication office...

My parents went to Delhi for 2 days, so they got it for me :)

I bought my copy at the Kolkata Book fair

lol
u guys shud be in pune , people complete CAT in their 8th :P

I am happier in kgp, :)
over there,I would have suffered from a worse inferiority complex! :(

[quote="Scary math"]Are u from kol? :o[/quote]
Lived there from 4-10 years of age.


abstract algebra
calculus
integration
group theory
Ring Theory
superior algebra
superior algebra unsolved

Assume R is commutative. Prove that the set of prime ideals in R has a minimal element with respect to inclusion. (by Zorn's lemma)

the cut of a chain in the set of prime ideals is a obviously a lower bound for it, so zorn's lemma (applied to the "dual" inclusion relation) yields the result.

"the cut of a chain in the set of prime ideals is a obviously a lower bound"

...can you explain a little bit more? I have no idea on how to construct a chain in it.....

zorn's lemma advises you to take an arbitrary chain (totally ordered subset) - it must not be constructed - and find a lower bound ..

related to this question..
me and friend were wondering: is the proposition still true if you ask for [i]non-zero[/i] minimal prime ideals?
of course, forget the case of $R$ field, or maybe (just in order to let the statement have a sense) consider also the whole field as a prime ideal.
or maybe reformulate the statement for ideals which are non-zero primes or the whole field :P

the question reduces to this: does it exist a (non-noetherian, integral) ring with an infinite descending chain of non-zero prime ideals with trivial intersection?
actually, i have some problem in finding a ring with an infinite descending chain of prime ideals :blush: :maybe: ...

Look what happened in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125503]this[/url] thread: the algebraic question was transferred, via valuation rings, to a question on totally ordered abelian groups, and then the problem consisted simply of finding a totally ordered abelian group with certain properties. We can do the same thing here.

Let $\nu: K^{*}\to\Gamma$ be, just as in the other thread, a valuation of the field $K$ onto the totally ordered abelian group $\Gamma$, and let $A$ be the valuation ring corresponding to $\nu$, i.e. $A=\{a\in K\ |\ \nu(a)\ge 0\}$. A subgroup $\Delta\le\Gamma$ is called [i]isolated[/i] if whenever $0\le\beta\le\alpha$ and $\alpha\in\Delta$, we also have $\beta\in\Delta$. If $\frak p$ is a prime ideal of $A$, then it can be shown that the set $\nu(A\setminus\frak p)$ is the set of non-negative elements of an isolated subgroup of $\Gamma$, and that this gives a bijection between prime ideals and isolated subgroups of $\Gamma$. Now, the question of finding infinite descending chains of prime ideals in $A$ becomes, through this correspondence, equivalent to finding infinite ascending chains of isolated subgroups of $\Gamma$.

We can choose $\Gamma$ to be any totally ordered abelian group we want. Take it to be the set of infinite sequences $(\alpha_{n})_{n\in\mathbb Z}$ (infinite in both directions) of integers such that $\alpha_{n}$ is zero for small enough $n$, with componentwise addition. We order this set lexicographically, i.e. $(\alpha_{n})_{n}<(\beta_{n})_{n}$ if $\alpha_{n}<\beta_{n}$ for the smallest $n$ where the sequences differ. Now define $\Delta_{n},\ n\le 0$ to be the set of elements in $\Gamma$ whose components with index smaller than $n$ is zero. It is easy to check that $\Delta_{n}$ for an infinite ascending chain of isolated subgroups of $\Gamma$, so we’re done. Moreover, this example can be modified to give descending chains of non-zero prime ideals of any cardinality we want.


MATHCOUNTS 2004-2005 Warm-Up 13

A two-pan balance scale comes with a collection of weights.  Each weight weighs a whole number of grams.  Weights can be put in either or both pans during a weighing.  To ensure any whole number of grams up to 100 grams can be measured, what is the minimum number of weights needed in the collection?

[hide="Answer"]Here is how I did and wondering if there is a better way:

We start with 1 and 3 to cover number 1-4.
To get 5, we need 4+5=9, so 1, 3 and 9 can cover number 5-13.
To get 14, we need 13+14=27, so 1, 3, 9 and 27 can cover number 14-40.
To get 41, we need 40+41=81, so 1, 3, 9, 27 and 81 can cover number 41-121.

So, the answer is [b]5[/b].[/hide]

If you think of the famous base 2 numbers magic trick, you'll realize that you just need the powers of two, including 1. The answer is the number of powers of 2 less than 100, which are 1,2,4,8,16,32, and 64. That gives you the answer of 7. Also you could do log(2)64, which also equals 7.

[quote]
If you think of the famous base 2 numbers magic trick, you'll realize that you just need the powers of two, including 1.[/quote]

But that doesn't give you the [b]minimum[/b] number of weights, because you are allowed to put weights on both sides of the scale.

Yakko


logarithms
function
Euler
complex analysis

[b]
Determine is this Conv. Or Div. and Is there any effect if $ z\in \mathbb {C}$ OR $ z\in \mathbb{R}$.

$ \prod\limits_{n = 1}^\infty {\left( {1 - \frac {z}{n}} \right)} e^{\frac {z}{n}}$

I need details enough to understand why?

My thinking , it's not Con. , since no any factor of it's equal to 0 yes Or not ?

[/b]

If one of the factors is $ 0$, it certainly converges to $ 0$. Otherwise, take logarithms and reduce the problem to that about convergence of a series (you need to know that $ \log(1 \minus{} w) \equal{} \minus{} w \plus{} O(|w|^2)$ for small $ w$).

By the way, [b]entire posts in bold are slightly irritating[/b], so, please, use normal fonts switching to [b]bold[/b], [i]italic[/i], [color=red]color[/color], etc. only when you need to emphasize something :).

Also, please spell the words in the topic title according to English standards!

I'm sorry  :blush:  about Bold and my spelling . I'll use Standard English In titles and bold in something need to emphasize about it.

Thank u about Your Polite way. :)

Anyway,

Could you do more steps about my question please ? I think I still have warping about it.

Thank you.

Let's see. There are two possibilities for an infinite product $ \prod_n a_n$ to converge.
1) One of the factors is $ 0$. Then the entire product obviously converges to $ 0$.
2) All $ a_n\ne 0$, and $ \sum_n \log a_n$ converges. Then the limit of the partial products is just the value of the exponential function at the limit of the partial sums of the logarithms.

So, assume that none of the factors in your product is $ 0$. When $ n$ is much larger than $ |z|$, you can write
\[ \log(1\minus{}\tfrac zn)\plus{}\frac zn\equal{}\minus{}\frac zn\plus{}O(n^{\minus{}2})\plus{}\frac zn\equal{}O(n^{\minus{}2})\]
so the series of logarithms is dominated by the positive series $ \sum_N \frac 1{n^2}$, which converges. I hope that the rest is clear :)

That's clear dude.

I'm So thankful.

 :D  :wink:

[quote="fedja"]
1) One of the factors is $ 0$. Then the entire product obviously converges to $ 0$.
[/quote]
I'd just like to point out that typically the product is not considered to converge just because one of the factors is 0. This is because if you allow this for convergence, then you cannot say anything about the terms. As an example, you'd often like to know if the product converges to a continuous or differentiable functions, and in such discussion it becomes apparent that when factors are 0 you need to throw out the zero factors and look at the rest of the factors.

(fedja knows this, others might not  :wink: )

I'm going to go ahead and move this topic into the complex analysis forum. The standards for convergence in Kalle's post are those typically elaborated in a complex analysis course. And by those standards, this product converges as nicely as we could possibly ask for: uniformly on each compact subset of $ \mathbb{C}.$ Hence, the function defined by this product is an entire function, and it has a simple zero at each positive integer.

Now let me add a question: what does it converge to? Can we recognize and identify this entire function? Can we at least find some useful algebraic properties of this entire function?

Well, applying the logarithm to the function and doing some basic manipulation, we can get th expression:

$ \prod^\infty_{n\equal{}1}\left(1\minus{}\frac{z}{n}\right)e^{\frac{z}{n}}\equal{}\frac{e^{z\, \gamma}}{\Gamma(1\minus{}z)}$

where $ \gamma$ is the Euler-Mascheroni constant.


Write an equation for the tangent line at (c,f(c)).

f(x) =sqrt(x); c=4

Since I can flip open my calculus textbook and find many homework exercises (in a fairly early chapter) that look exactly like this, I'll have to ask: what have you done so far?


trigonometry
inequalities
geometry proposed
geometry

Prove that

$ a^4\plus{}b^4\plus{}c^4\geq16s(s\minus{}a)(s\minus{}b)(s\minus{}c)(\cot A\cot B \plus{} \cot B \cot C \plus{} \cot C \cot A)$

Same form like previous problems
Use $ 4R\equal{}\frac{abc}{S}, 2R\equal{}\frac{a}{\sin A}, \cos A \equal{}\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{2bc}$
Then it goes to obvious.

[quote="Heebeen, Yang"]Same form like previous problems
Use $ 4R \equal{} \frac {abc}{S}, 2R \equal{} \frac {a}{\sin A}, \cos A \equal{} \frac {b^{2} \plus{} c^{2} \minus{} a^{2}}{2bc}$
Then it goes to obvious.[/quote]

Dear [color=red][b]Yang[/b][/color] can you describe better your solution? Thanks

If we use them,
$ \cot A\equal{}\frac{2R(b^{2}\plus{}c^{2}\minus{}a^{2})}{2abc}$and $ \cot B, \cot C$ defined similar.
Then, above inequality is equivalent to
$ a^{4}\plus{}b^{4}\plus{}c^{4} \geq \sum_{cyc}(b^{2}\plus{}c^{2}\minus{}a^{2})(c^{2}\plus{}a^{2}\minus{}b^{2})$
<=>
$ a^{4}\plus{}b^{4}\plus{}c^{4} \geq a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}$

$ 16S^2\equal{}16s(s \minus{} a)(s \minus{} b)(s \minus{} c) \equal{} \sum (a^2\plus{}b^2\minus{}c^2)(a^2\plus{}c^2\minus{}b^2)\equal{} 2\sum b^2c^2 \minus{} \sum a^4$

and $ \sum\cot B\cot C \equal{} 1$ . [b]Simply ![/b] The proposed inequality becomes $ \sum a^4\ge \sum b^2c^2$ what is evidently.

Well.... That's right wow.
I've forgottend that property.
Thank you for your advice ^^

[quote="Virgil Nicula"]$ 16S^2 \equal{} 16s(s \minus{} a)(s \minus{} b)(s \minus{} c) \equal{} \sum (a^2 \plus{} b^2 \minus{} c^2)(a^2 \plus{} c^2 \minus{} b^2) \equal{} 2\sum b^2c^2 \minus{} \sum a^4$

and $ \sum\cot B\cot C \equal{} 1$ . [b]Simply ![/b] The proposed inequality becomes $ \sum a^4\ge \sum b^2c^2$ what is evidently.[/quote]

Very Nice my Dear Friend.. Thanks  :)

[quote="Heebeen, Yang"]If we use them,
$ \cot A \equal{} \frac {2R(b^{2} \plus{} c^{2} \minus{} a^{2})}{2abc}$and $ \cot B, \cot C$ defined similar.
Then, above inequality is equivalent to
$ a^{4} \plus{} b^{4} \plus{} c^{4} \geq \sum_{cyc}(b^{2} \plus{} c^{2} \minus{} a^{2})(c^{2} \plus{} a^{2} \minus{} b^{2})$
<=>
$ a^{4} \plus{} b^{4} \plus{} c^{4} \geq a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2}$[/quote]

Thanks Dear [color=red][b]Yang[/b][/color]....


quadratics
LaTeX
trigonometry
limit
algebra unsolved
algebra

Solve:
$ e^{x\minus{}y}\equal{}\frac{sinx}{siny}$
$ 10\sqrt{x^6\plus{}1}\equal{}3(y^4\plus{}2)$
$ \pi<x;y<\frac{5\pi}{4}$

[quote="karis"]Solve:
$ e^{x \minus{} y} \equal{} \frac {sinx}{siny}$
$ 10\sqrt {x^6 \plus{} 1} \equal{} 3(y^4 \plus{} 2)$
$ \pi < x;y < \frac {5\pi}{4}$[/quote]

$ sinx\equal{}\frac{e^{ix}\minus{}e^{\minus{}ix}}{2i}$

$ e^{x\minus{}y}\equal{}\frac{sinx}{siny}$ <-> $ e^{x}siny\equal{}e^{y}sinx$ ----> $ e^{x}*\frac{e^{iy}\minus{}e^{\minus{}iy}}{2i}\equal{}e^{y}*\frac{e^{ix}\minus{}e^{\minus{}ix}}{2i}$

$ e^{x\plus{}iy}\minus{}e^{x\minus{}iy}\equal{}e^{y\plus{}ix}\minus{}e^{y\minus{}ix}$-----> $ \minus{}\minus{}e^{x\minus{}iy}\equal{}\minus{}e^{y\minus{}ix}$-----> $ x\minus{}iy\equal{}y\minus{}ix$----> $ (1\plus{}i)(x\minus{}y)\equal{}0$-----> $ x\equal{}y$

---> $ 10\sqrt {x^6 \plus{} 1} \equal{} 3(x^4 \plus{} 2)$ -----> $ 9x^{8}\minus{}100x^{6}\plus{}36x^{4}\minus{}64\equal{}0$----> $ x\equal{}?$

[quote="zaya_yc"]$ 9x^{8} \minus{} 100x^{6} \plus{} 36x^{4} \minus{} 64 \equal{} 0\implies x \equal{} ?$[/quote]
It can be factorized as:
$ (9x^{4} \minus{} 10x^{2} \plus{} 8)(x^{4} \minus{} 10x^2 \minus{} 8) \equal{} 0$
Now we can use the quadratic formula. :wink:

[hide="Latex Tips"]The latex code of "$ \implies$" is "\implies "
And "\iff " for "$ \iff$"
The code "\sin " gives "$ \sin$" which looks better than "$ sin$", and similarly for $ \cos$, $ \lim$ etc.
For multiplication "\cdot " gives "$ \cdot$", it looks better than "$ *$"
Sometimes "\cdots " is needed for "$ \cdots$"[/hide]

Let $ f(x)\equal{}\frac{\sin x}{e^x}$. $ e^{x\minus{}y}\equal{}\frac{sinx}{siny} \iff f(x)\equal{}f(y)$
$ f'(x)\equal{}\frac{\cos x \minus{}\sin x}{e^x}<0$ for $ \pi < x < \frac{5\pi}{4} \implies x\equal{}y$


hi!!
Let f a function such that: $ f(x) \equal{} x^{n} \minus{} \sum_{k \equal{} 0}^{n \minus{} 1}x^{k}$ with $ n\ge2$ is even.
prove that f has two solutions $ a$ and $ b$ such as: $ a < 0 < b$

thanx.

why $ f(0) \equal{} 0$ and $ f(1) \equal{} 1 \minus{} n$?
I think we have $ f(0) \equal{} \minus{} 1$.

I forgot that $ n\ge2$.


calculus
integration
inequalities
Euler
function
real analysis
real analysis unsolved

Try to find max $C$
for wich we have
if $0<{a}\leq{f(x)}\leq{b}$ and  $0<{c}\leq{g(x)}\leq{d}$ and ${1/p}+{1/q}=1$ then
${C}({\int^{l}_{k}{f^{p}(x)}})^{1/p}({\int^{l}_{k}{g^{q}(x)}})^{1/q}\leq{\int^{l}_{k}{f(x)}{g(x)}}\leq({\int^{l}_{k}{f^{p}(x)}})^{1/p}({\int^{l}_{k}{g^{q}(x)}})^{1/q}$
where $f(x)$ and $g(x)$ in interval$[k,l]$ is continious derivetiv function.
When we have equal?

[quote="Extremal"]Try to find max $C$
for wich we have
if ${a}\leq{f(x)}\leq{b}$ and  ${c}\leq{g(x)}\leq{d}$ and ${1/p}+{1/q}=1$ then
${C}({\int^{l}_{k}{f^{p}(x)}})^{1/p}({\int^{l}_{k}{g^{q}(x)}})^{1/q}\leq{\int^{l}_{k}{f(x)}{g(x)}}\leq({\int^{l}_{k}{f^{p}(x)}})^{1/p}({\int^{l}_{k}{g^{q}(x)}})^{1/q}$
where $f(x)$ and $g(x)$ in interval$[k,l]$ is continious derivetiv function.
When we have equal?[/quote]
Few comment: right inequality is known It's calld Holders integral inequality  :lol: , but it's know everyone.But you don't know leftintegral inequality, it's very nice and interesting. it's open many interesting things in real anaysis and not only here.It's  can bu used to solve difficult problems posted in the extremal theories.Every problems wich solved by holders integral inequality now we can use this this and give problems different face. I don't know is it known or not!! But I will post Sulution of this problem after Conferenc Eulers in Saint Petersburg.
The sulution is hard.  Good luck.
 :|

your way is simple, but it's not best constant. I have more big than your  :wink:

[quote="Kent Merryfield"]Very simple, very crude estimates give us that $\int fg\ge ac(k-l)$ while $\|f\|_{p}\|g\|_{q}\le bd(k-l).$ Hence, $C=\frac{ac}{bd}$ would work.

I would guess that that's not best possible and that one can find $C$ larger than that.

Incidentally, the condition for equality in Hölder's inequality - the right hand inequality - (assuming positive functions) is that $\frac{f^{p}}{g^{q}}$ is constant.[/quote]

Kent Merryfield,  $\int fg\ge ac(k-l)$ it is true and equl when f=a,g=c,
then  $\|f\|_{p}\|g\|_{q}\le bd(k-l).$ it's true and equal when f=b, g=d.
then you write
$(\int fg)/{ac}\ge{(k-l)}\ge{(\|f\|_{p}\|g\|_{q})/{bd}}$ here can't be achievd first equal and seccond equal togather/.  therefore this constant is wrong.

[quote="Kent Merryfield"]I knew all of that already, and I was already assuming this wasn't best possible. I was just putting a floor under it.[/quote]
OK. I have a question: Do you anyvere heardabout this inequality? I mean left  :lol:

[quote="flip2004"][quote][b] Assume:
1) $p>1\; ,\; \frac{1}{p}+\frac{1}{q}=1$ ,
2) $x\to f(x)^{p}\; ,\; x\to g(x)^{q}$ are integrable functions  on $[a,b] ,$
3)  $0<m_{1}\leq{f(x)}\leq M_{1}$ ,  $0<{m_{2}}\leq{g(x)}\leq{M_{2}}\; ,\; \forall x\in [a,b].$
Find the best constant $C$ in inequality[/b]
\[{{{C}(\int_{a}^{b}}{f^{p}(x)\; dx}})^{1/p}({\int_{a}^{b}{g^{q}(x)\; dx}})^{1/q}\leq{\int^{b}_{a}{f(x)}{g(x)\; dx}}\]
[/quote]
[b]
[u]Answer:[/u]
\[C=C(p,m_{1},m_{2},M_{1},M_{2}): =\frac{(pm_{2}M_{2}(M_{1}M_{2}^{q-1}-m_{1}m_{2}^{q-1}))^{\frac{1}{p}}(qm_{1}M_{1}(M_{2}M_{1}^{p-1}-m_{2}m_{1}^{p-1}))^{\frac{1}{q}}}{M_{1}^{p}M_{2}^{q}-m_{1}^{p}m_{2}^{q}}\]
[/b]

Possible references:
[1]E.Beck ,[i]Komplementaere Ungleichungen bei vergleichbaren Mittelwerten,[/i] Monatsh.Math., 73(1969) 289--308.
[2] J.B.Diaz, A.J.Goldman and F.T.Metcalf,[i]Equivalence of certain inequalities complementary to those of Cauchy-Schwarz and Hoelder,[/i]
J.Res.Nat.Bur.Standards Sect.B, 68 B(1964)147--149.
[3]Z.Nehari ,[i]Inverse  Hoelder inequalities,[/i] J.Math.Anal.Appl., 21(1968)405-420.[/quote]
 :(  The answer correct. I have deceived: the sulution is not hard ( i mean my sulution)


With \alpha, \beta \in I; m,n \in Z+. Show that:
[m\alpha]\not= [n\beta ]

Do you mean the following: Let $ \alpha$, $ \beta \in \mathbb{I}$ (some interval, i guess $ \mathbb{I} \subseteq \mathbb{R}$) and $ m$, $ n \in \mathbb{Z^ \plus{} }$. Prove that $ \lfloor m \alpha \rfloor \ne \lfloor n \beta \rfloor$. (without any resrictions on $ \alpha$, $ \beta$ that's clearly false)

[quote="smiley4t"][quote="FelixD"]Do you mean the following: Let $ \alpha$, $ \beta \in \mathbb{I}$ (some interval, i guess $ \mathbb{I} \subseteq \mathbb{R}$) and $ m$, $ n \in \mathbb{Z^ \plus{} }$. Prove that $ \lfloor m \alpha \rfloor \ne \lfloor n \beta \rfloor$.[/quote]
Yes, It's right. I am sorry because i don't know how to write right. Can you help me??Thanks.[/quote]

As FelixD said, it's wrong without more constraint. For example, choose $ \alpha,\beta\in\mathbb N$ and $ m\equal{}\beta$ and $ n\equal{}\alpha$  :)

You should check the exact statement of the problem if you want to get some useful help.


Prove that the polynomial $ (x\minus{}a_{1})(x\minus{}a_{2})...(x\minus{}a_{n})$-1 where $ a_{1}$,$ a_{2}$,...$ a_{n}$ are distinct integers cannot be written as the product of two non constant polynomials with integer coefficients,i.e.,it is irreducible

This particular problem is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150955]here.[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150957]A related one.[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66699]Even more.[/url]


Pozdrav!
A je šel kdo letos na logiko?  :nhl:

Vsa leta so iste naloge tukej  :cleaning:

js sm tud bla... Kak so se vam pa zdele naloge? men se zdi da je bla 1. res sam za motivacijo  :)

Ja, se strinjam... In še razlago so hotl LoL

no ja, rešitev bi pr tej res z lahkoto prepisu... zadnjo sta rešla?

Jz sm biu tud na logiki pa se mi ni zdel prou težko.
Sam mal mi je časa zmankal za tretjo, tko k vedno :(

Z 3.?  pa sej ta ni tak problem, sam če si nisi si probi recimo izpisvat ugotovitve pa vedet morš, kje začet...  Men je zmankal pol bol na koncu pr 4... Aja, a mate rajši logiko al "navadno" matematiko?  :)

Četrta mi je bla bl všeč k tretja pa sem jo šou prej rešvt.
Vsazga mal - mislm oboje skupi :)

bol ti je bla všeč 4.?  :)  js najprej sploh nism cele prebrala in tak sm najprej misla da so mal čudni vsi da jih 3x eno in isto praša... sam pol pa ok, no  :)  kolk si pa dobu pol točk pr tej?

:sleep2: tale logika je (vsaj zame, k ne vem nobenih special trikov :roll: ) precej obrtniška zadeva, ker moraš v kratkem času obdelat precej podatkov, tko da mam raje matematično tekmovanje   :lol:

ja, specialni triki... skor bi blo bolj objektivno, če takih trikov pri logiki sploh ne bi blo, ker se mi zdi namen da sam iztuhtaš način reševanja... In kak ste se kej odrezal?  :)

Kolk točk sm dubu ne vem, pa najbrž tudi ne bom zvedel :|
Vem samo da jih je bilo dovolj za uvrstitev na državno  :lol:
Kakšni nej bi pa bli teli "specialni" triki?

pol mogoče sploh niso pregledal?  ;)  ej, kolk pa vas lahko gre iz tvojga letnika pač iz vaše gimnazije na državno?

aja, triki... ja recimo pr 2. nalogi je "predpisan" postopek, da vsakemu imenu prirediš zaporedno številko itd. sej mogoče tak poznaš to.
najbol zanimiv bi blo, če bi ble vsako leto čist drgačne naloge  :)  sam bi blo pa težko za sestavlat...

Una druga naloga je čist butasto drevo, kjer iščeš dolžine poti do najbolj oddaljenih točk.
Drugač pa sm mislu s "special triki" razne optimizacije, da prhranš na času pa probleme poenostavš  :oops:

Jz nimam nč proti, tud če niso pregledal :) Zdi se mi da greva lahko dva, nisem pa prepričan.

Tista druga naloga bi bla pa mal bl zanimiva če bi se kšni cikli pojavl.

Ja, to se strinjam... Ej Nino, pa sej to je glih največji užitek če kako nalogo zlo na preprost rešiš in je glih najbolš, če se morš tega sam spomnt  :) Recimo če pogruntaš hitrejši način za dolgovezno 2. nalogo ti bom hvaležna, če mi ga zaupaš  :)

Tom@ž, kira šola si ti?  :blush:

Ravnokar sem prebral da gremo lahko iz tretjih letnikov Gimnazije Vič trije.
Za drugo nalogo pa je edini hitrejši način tak, da uganeš osebo. :)

[quote="majaa"]Recimo če pogruntaš hitrejši način za dolgovezno 2. nalogo ti bom hvaležna, če mi ga zaupaš  :) [/quote]

Za začetek probi narisat točke, ki predstavljajo imena ljudi in potem povezat s črtami tiste, ki imajo cifre drug od drugega med sabo :-) Če slika grda jih probi prerazporedit in dobit nekaj tazga, da se črte ne križajo -> dobiš ravninski graf in to celo drevo. Potem pa gruntaj kaj naprej... (a je to dost hitro?)

Btw: Od kod si pa ti?

No ja, ta rešitev je precej podobna "predpisani" sam sej po moje res ni krajšega načina (razen Tom@ževega  ;)  )

pr nas lahko gre pa mnda iz vsakega letnika sam po eden... drgač je pa to I. gimnazija v Celju   :)

[quote="majaa"]No ja, ta rešitev je precej podobna [color=blue]"predpisani"[/color] sam sej po moje res ni krajšega načina (razen Tom@ževega  ;)  )[/quote]

hehe, sm si pa že skor ustvaru iluzijo, da sm kej novga proguntu...  :oops:

no sej če si to sam pogruntu je vseen zlo v redu ker si se očitno spomnu najboljše možne poti  :)


Who's actually going to own up to IE? Seriously

What about the multibrowser option? I use Safari and Firefox. On the Windows machine, that's Firefox for everything, while it's Safari for almost everything (except AOPS) on my Mac.

(I know Safari is now on the list of supported browsers here, but there were problems a year or two ago)

[quote="K81o7"]Apparently IE is by far the most used.  Surprising...I now use Firefox, but personally I don't prefer either over the other.  I'm just more used to clicking on the firefox icon on my desktop...and it's useful sometimes to have two different web browsers.[/quote]

IE is most used because anyone who doesn't know what a browser is uses it.

No  :wink: . I use IE7 whenever my Firefox doesn't support a certain plugin (too lazy to install).

[quote="SimonM"][quote="7h3.D3m0n.117"]I use Firefox, but I keep IE handy just in case.[/quote]

Are you sure thats not because MS have embedded it so deeply in the OS its too hard to get rid of?[/quote]

Actually I got rid of it... you just remove it, but I reinstalled it again lol

I have a Mac. I use Safari more often, and use Firefox for AoPS classes and Java applets. If I have to, I go to Parallels and open Windows to go on IE or Firefox.

FIREFOX!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I use 10 extensions and 5 themes. Firefox is definitely the best.

Current theme: Vista-Aero

Well IE7 is, I admit, cool, but
1. Vista-Aero can change Firefox to that identical look
2. CSS can make it even cooler
3. Vimperator
4. 1337key
5. AiOS
6. A lot of things only people really deep into HTML would know.

FF for all eternity!

[quote="math_explorer"]4. 1337key[/quote]
I see, the ultimative reason to use FireFox  :wink:

I use FireFox only, I never had big problems with it. Well, the pdf-plugin sometimes crashes...
The only real disadvantage is the big amount of RAM it occupies; especially, it's memory usage sometimes grows and grows without any reason.

But, 1337key is broken, AiOS feels granted, and Vimperator just isn't that useful. Firebug is useful. Download Statusbar, Gmail Manager, YSlow, Talkback, Google Notebook. That's all the extensions I have.

Yippee!

just wondering, what's the 1337key?

i like firefox. just that my computer doesn't have it b/c my parents are like, "why do you need firefox? you have IE."

i hope to get a mac soon and get firefox on there and all my awesome programs that i'm not allowed to download.

Firefox has better security...i think...

Yeah, and it supports pseudo-classes for everything. There are standards for how the web pages are supposed to be written and how they should then look. IE fails to follow a lot of these standards.

Something like that...

Plus, if you don't like the default font (I despise it!), there's this wonderful part of Firefox where you get to change it. Just that you need to know some CSS and be good at Googling.

And so on.

[quote="7h3.D3m0n.117"][quote="kimmystar94"]just wondering, what's the 1337key?[/quote]

Plugin that turns everything into L337 so you don't have to manually translate and type as you go.

[quote="kimmystar94"]i like firefox. just that my computer doesn't have it b/c my parents are like, "why do you need firefox? you have IE."[/quote]

Your parents aren't very knowledgeable of computers and Internet, I see.[/quote]

thanks 7h3.D3m0n.117... and my parents are not very knowledgeable of computers and internet.
however, i want to be very knowledgeable.. i asked them and they're like, "no" you can always use IE for testing web layouts and programming...
and i want a mac!!!

How about links? You can view some images, and the benefits of browsing without Javascript and Flash and Java are myriad. But I don't see links/lynx on the poll...
.
[hide="."]/troll[/hide]

I often use firefox.  I sometimes use safari if I'm using a website that I know works well in safari.

I use firefox on the family computer, but IE on my mom's laptop.


figure out as much as u could

[img]http://img252.imageshack.us/img252/7154/untitledyy5.png[/img]

Which are the ones which are on the row? it isn't clear.
   Bharath

my guess is that its the horizontal rows which are in a 1-4-2-4-1 formation :maybe:

This is impossible because you cant have 2 different numbers on the first/last row that are the same.

Maybe its the diagonal rows/column/whatever.

i think its the sides of the two triangles...  :maybe:


How do I convert a mixed number into a proper fraction? Like for example 35 and 4/6th to a proper fraction?

Thanks.

Dude, read your book ...

Multiply it by 6, then divide by 6.

If you multiply that number by 6, you get $ 35 \times 6 \plus{} 4 \equal{} 214$

Now divide by 6 to get $ \frac {214}{6}$

You can simplify it by dividing the top and bottom by $ 2$, to get $ \frac {107}{3}$

Actually that would be an improper fraction.
It is impossible to convert to a proper fraction.

[quote="Walk Around The River"][quote="leroyjenkens"][quote="Ryuk"]I don't think he really meant proper.

As Zuton shows, simply multiply the whole number by the denominator in the fraction, and then add this product to the numerator in the fraction.[/quote]
I meant proper, because on this website I saw it done, but it didn't make sense to me.
http://www.purplemath.com/modules/percents.htm

Halfway down the page, under "another example", for 33 and 1/3 he converted that to 1/3.[/quote]

Oh, that was for percentages.  If you meant percentages, just put your number over 100, and reduce.[/quote]
Oh, I didn't think that mattered. Alright thanks. How do I reduce the percentage part of it? Like for example 24 and 2/4ths. How do I reduce the 2/4ths along with the 24?
Isn't it still considered a mixed number if it's a percent?

A percent just means that the number is a numerator in a fraction where 100 is the denominator.

e.g.

$ a$% = $ \frac{a}{100}$

Here's an example of converting fractions:

$ 35\frac{4}{6}$

$ \equal{}35\frac{2}{3}$

$ \equal{}\frac{(35)(3)}{3}\plus{}\frac{2}{3}$

$ \equal{}\frac{105}{3}\plus{}\frac{2}{3}$

$ \equal{}\frac{105\plus{}2}{3}$

$ \equal{}\frac{107}{3}$

BTW: As people have said, this is converting to an improper fraction. A proper fraction is one where the numerator is less than the denominator.

[quote="fishythefish"]A percent just means that the number is a numerator in a fraction where 100 is the denominator.

e.g.

$ a$% = $ \frac {a}{100}$

Here's an example of converting fractions:

$ 35\frac {4}{6}$

$ \equal{} 35\frac {2}{3}$

$ \equal{} \frac {(35)(3)}{3} \plus{} \frac {2}{3}$

$ \equal{} \frac {105}{3} \plus{} \frac {2}{3}$

$ \equal{} \frac {105 \plus{} 2}{3}$

$ \equal{} \frac {107}{3}$

BTW: As people have said, this is converting to an improper fraction. A proper fraction is one where the numerator is less than the denominator.[/quote]
Yeah, the reason I asked was because in the link I gave, the guy converts an improper fraction, which is a percentage (33 and 1/3 percent) into a proper fraction. He converted it to 1/3rd. I was wondering how he did that.
Thanks for the responses.

It's because $ 33\frac{1}{3}$%

$ \equal{}\frac{33\frac{1}{3}}{100}$

$ \equal{}\frac{1}{3}$

That's all he did.

[quote="fishythefish"]It's because $ 33\frac {1}{3}$%

$ \equal{} \frac {33\frac {1}{3}}{100}$

$ \equal{} \frac {1}{3}$

That's all he did.[/quote]
For that one you can see that it's 1/3rd just by looking at it, but I don't know the method of turning it into a proper fraction. For example, one that's more complex, like 44 and 2/6ths percent. How do I turn that into a proper fraction?
Thanks.

[quote="Ihatepie"]multiply the top and bottom by the denominator of the fraction.

(44 2/6)/100
(44 1/3)/100
133/300

can't be simplified.[/quote]
Oh, well that's simple enough. Thanks. But just so I know, this can't be done with just a mixed number? It has to be a fraction? Thanks.


Consider a $25 \times 25$ grid of unit squares. Draw with a red pen contours of squares of any size on the grid. What is the minimal number of squares we must draw in order to colour all the lines of the grid?

It can be done trivially with 48 squares, just connect $(i,i)$ to $\left(25\left[\frac{25-i}{25}\right],25\left[\frac{25-i}{25}\right]\right)$ and $(i,-i)$ to $\left(25\left[\frac{25-i}{25}\right],-25\left[\frac{25-i}{25}\right]\right)$, where $[...]$ denotes round(...).

Now can it be done with 47? No, consider the 24 inner horizontal lines and 24 inner vertical lines. None of them can be covered by just 1 square, so we need at least 2 squares per line, and since a square has 4 sides we need at the very least $\frac{2.24.2}{2}=48$ lines.


LaTeX
calculus
integration
linear algebra
matrix
function

Hello All,
 I need to get a large integral sign,
say if i have to put an integral sign in front of a 3X3 matrix...

thnx in advance
Pparakh

ok, here's what I found.

Two packages can vary the size of the integral sign.  [b]mtpro[/b] and [b]bigint[/b].  None of them are in the miktex distribution.  I tried [b]bigint[/b] but it didn't work, something about using the cmex10 font but I don't know.  Look for bigint.sty in google.

[b]mtpro[/b] seems much better but I couldn't find it on ctan (didn't try really hard though).

If you find a way, tell me.

Ledurt

P.S.: Never seen an integral of a matrix but I like the idea;)

That would be some funny-looking notation, and I'm not sure I'd do it if I could. Of course you can take the integral of a matrix - the integral is linear, after all. But I'd probably use more than one line.

$A(t) = $(matrix of functions, in detail), followed by

$\int_a^bA(t)\,dt.$

Part of the charm and power of matrix notation is its abilitity to compress notation into a small space, if you'll let it.

great solution Kent.  Plus it's less scary for the readers to see
$\int A(t) dt$
than the integral of a matrix.

good call

ledurt


1.  Find, as a function of $n$, the sum of the digits of  $9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right)$, where each factor has twice as many digits as the previous one.


integration
calculus
algebra
polynomial
real analysis
real analysis unsolved

QUESTION: Following proposition  is correct ?
[color=green] [b] If $f , w \in C [a,b], w \ge 0$ on  $[a,b], \int\limits_{a}^{b}w(x) \; dx =1 ,$    then there exists $c , c\in (a,b)$  such that  $\int\limits_{a}^{b}f(x)w(x)\; dx = f(c) .$
 [/b] [/color]  If not, why ? If it is false ,  please a counterexample and  give the correct assertion

it's true , because if we put m=min(f) and M=max(f) then
   for all x in [a,b] we have  m.w(x)<=f(x).w(x)<=M.w(x)  and by integral we get m<=integral(f(x).w(x)dx,a,b)<=M , f is continuous , the result follows (intermidiate values  )

[quote="yassinus"]it's true , because ..... [/quote]  [b]The mean-value theorem, posted  by proposer, [u] is'nt correct[/u] !   Why ?[/b]

i dont undertand what you want !! what i have write doesn't convaincu you ?

[quote="zhaobin"]but by what you wrote,we can only obtain $c \in [a,b]$  not $c \in (a,b)$,right?[/quote]
[color=blue] [b] [u]Yes[/u] , but  please  give a counterexample ! [/b] [/color] Thanks,for interest/proposer .

Sorry,I can't give a couterexample,Because I think the proposition  is correct ;)
[b]Solution[/b]
Let $m=\min\{f|x \in[a,b]\},M=\max\{f|x \in[a,b]\}$
We can easily get:\[ m \le \int_{a}^{b}f(x)w(x)dx \le M \]
(We may assume $M>m$,if not then f is constant on [a,b],the result follow quickly)
It is easy to prove the problem,if \[ m<\int_{a}^{b}f(x)w(x)dx<M \]
Then we should do it only   \[ m=\int_{a}^{b}f(x)w(x)dx, M=\int_{a}^{b}f(x)w(x)dx \]
For  $m=\int_{a}^{b}f(x)w(x)dx$.
because,\[ \int_{a}^{b}w(x) dx =1  \],we can get there exist $\xi \in [a,b]$ such that $f(\xi)>0$
then because $w \in C[a,b]$,we can get a intervuale $[d,e] \subseteqq [a,b]$,such that
\[ f(x)>\frac{f(\xi)}{2},x \in [d,e] \]
then we get
\[ 0=\int_{a}^{b}\left(f(x)-m\right)w(x)dx \ge \int_{d}^{e}\left(f(x)-m\right)w(x)dx \ge \frac{f(\xi)}{2} \int_{d}^{e}\left(f(x)-m\right)dx \ge 0 \]
That means $f(x)=m,x \in [d,e]$,then we can choose every $c \in [d,e]$ as desired.
and similar when $M=\int_{a}^{b}f(x)w(x)dx$

[b]More generality:[/b]If $f \in C[a,b]$,$w \in R[a,b]$ and g doesn't change sign on $[a,b]$,then there exist $c \in (a,b)$ such that:
\[ \int_a^b f(x)w(x)dx=f(c)\int_a^b w(x)dx \] :)

[quote="zhaobin"]... I think the proposition  is correct ...
[b]If $f \in C[a,b]$,$w \in R[a,b]$ and g doesn't change sign on $[a,b]$,then there exist $\theta \in (a,b)$ such that: [/b]  \[ \int_a^b f(x)w(x)dx=f(\theta)\int_a^b w(x)dx  \] :)[/quote]
OK, but $c =a$or $c=b$ is also possible !
Let $f \in C[a,b]$  and consider the mean-value theorem
$(\pounds)\; \; \; \; \frac{1}{b-a}\int\limits_{a}^{b}f(x) \; dx= f(\theta)\; ,\theta \in [a,b] .$  (That is $w(x)=\frac{1}{b-a}$ ).

Let $[a,b]: =[0,1]$ , i.e. for the mean  value theorem $\int\limits_{0}^{1}f(x)\; dx =f(\theta) \; :$
Take  $f_*(x)= x-\frac{3}{4}x^2 ;$  In this case equation   $\int\limits_{0}^{1}f_*(x)\; dx =f_*(\theta) \; ,$  has solution $\theta=1$  but  also  $\theta=\frac{1}{2}$ ( which belongs to  $[0,1]$ ) verifies the equation.

[b] Other remarks:[/b]
[color=green][b]1) [/b][/color] If there exists $h\in C^1[a,b]$  such that $F=h^{\prime}$ on $[a,b]$,   then (according to Lagrange mean value theorem of differential calculus) we have $\theta \in (a,b)$  (see  $\pounds$).

[color=green][b]2)[/b] [/color]  In ($\pounds$) it's clear that [b] it's  possible to have $\theta= a$ or/and  $\theta=b .$  [/b]
Another example , (see also above), in case $[a,b]=[-1,2]$  may be :  $f_1(x)=\left\{\begin{array}{lcl} P_n(x) &, & x\in [-1,1] \\ \frac{4}{5}(1-x) +1 &, & x\in (1,2] \; ,\\ \end{array}\right.$
where $P_n(x)=\frac{1}{2^n n!}\left((x^2-1)^n\right)^{(n)}$  is the Legendre polynomial of degree $n$ . Remind that
$\left|P_n(x) \right| \le 1$ for $x\in [-1,1]$ , $\int\limits_{-1}^{1}P_n(x)\; dx = 0$ , $(n\ge 1),\;$ $P_n(-x)=(-1)^nP_n(x)\; ,\; P_n(1)=1 .$
 If we  consider  the corresponding mean value
$(\pounds ')\; \; \; \int\limits_{-1}^{2}f_1(x)\; dx = 3f_1(\theta)$  , then  $\theta=2$  is a possiblity (a solution) .
 [b] But $\theta$ may be also any other solution , from $[-1,1]$ [/b],  of equation $P_n(\theta)-\frac{1}{5}=0 .$ Because ,e.g. in case   $n=2\; ,\; P_2(x) =\frac{3}{2}x^2-\frac{1}{2} ,$ and  we conclude that  $\theta \in \left\{-\sqrt{\frac{7}{15}},\; \sqrt{\frac{7}{15}},\; 2\right\}\limits_{n=2} .$

Thanks,nice.(But I didn't say that It is impossible that $\theta=a$ or $b$) :)  :?


geometry
3D geometry
calculus
calculus computations

2. A right cylinder is inscribed in a right circular cone with the radius 3 cm and height 5 cm. Find the dimensions of the cylinder of maximum volume.

$ V\equal{}\pi r^2 h$, we can make r depend on h since we are inscribed in the cone

$ r(h)\equal{}3\minus{}\frac{3}{5}h$ so $ V(h)\equal{}9\pi h (1\minus{}\frac{h}{5})^2$

so $ V'(h)\equal{}9\pi[ (1\minus{}\frac{h}{5})^2\minus{}2\frac{h}{5}(1\minus{}\frac{h}{5})]\equal{}9\pi [1\minus{}\frac{3h}{5}](1\minus{}\frac{h}{5})$

has max at $ h\equal{}\frac{5}{3}$, which is $ V(\frac{5}{3})\equal{}9\pi \frac{5}{3} (1\minus{} \frac{1}{3})^2\equal{}\frac{20\pi}{3}$


Given $k$ be a positive real;$x,y,z >0$ satisfying $xy+yz+zx=3$.Prove that:
                 $(x+y)^{k}+(y+z)^{k}+(x+z)^{k}\geq 3.2^{k}$

hint
[hide]
$(x+y)(y+z)(z+x) \ge \frac{8}{9}(x+y+z)(xy+yz+zx)$[/hide]


factorial function MATHCOUNTS calculus integration

In the strictest sense, you can't take a factorial of a negative number.

However, in advanced math there is a thing called the gamma function, which is basically a generalized factorial. But I'd better stop.

Woot, 1200th post!

[quote="mathnerd314"]In the strictest sense, you can't take a factorial of a negative number.
[/quote]

Thats what I thought but I kept having nightmares that it will appear in the Mathcounts Chapters tommorow  :maybe:

Gratz mathnerd314  :lol:

[quote="7h3.D3m0n.117"][quote="mathnerd314"]In the strictest sense, you can't take a factorial of a negative number.
[/quote]

Thats what I thought but I kept having nightmares that it will appear in the Mathcounts Chapters tommorow  :maybe: [/quote]
Factorials don't appear very often, so I wouldn't worry about negative or non-integral factorials.

If they start putting negative factorials on mc I'm not going

Factorial of any negative number would be undefined.
I find it interesting, though, that 0!=1. I understand 1!=1 and, therefore, 0!=1/1=1. But some how it just doesn't make sense to me because n! is defined as 1*2*3*...*n.

[quote="myspacesucks"]If you actually tried to do this, (-1)! wouldn't be defined because it would end up as 1/0, same thing with all other negative factorials.   The Gamma Theorem makes people believe it is possible, but the math greats have yet to determine negative factorials as something other than the number/0.  If they put this on tests for mathcounts I would stop going to practices.[/quote]

It's the gamma function, not the gamma theorem. Its relationship to the factorial is:

$\Gamma(x+1)=x!$

if the factorial is defined. If it's not, you use calculus to get the answer.

If they put this on tests for Mathcounts, I would keep going to practices. I would just complain to the people who write the tests.

sorry, I meant the gamma function not theorem, but calculus is not at mc level.  That's why I didn't talk about that.

[quote="7h3.D3m0n.117"]What happens when you have

$-5!$?

is it

$-(5 \times 4 \times 3 \times 2 \times 1) =-120$?

or

$-5 \times-4 \times-3 \times-2\times-1$?[/quote]

$-5!=-(5 \times 4 \times 3 \times 2 \times 1) =-120$

but $(-5)!\neq-5 \times-4 \times-3 \times-2\times-1$

[quote="7h3.D3m0n.117"]What happens when you have

$-5!$?[/quote]

This isn't a question about negative factorials as much as it is a question of [b]order-of-operations.[/b]

$-5! =-(5!) =-120$

That's all there is to it.

uuhh...

I thought that

$(n-1)!=n!/n$

But, with n equal to 0, it will be undefined.

No. You get $\frac{1}{1}$ which is clearly $1$.

I don't think we need to worry about this. This stuff definitely won't be at chapter, or state, probably not even nationals. Unless they explain to you how they want you to do it, which would make it very, very easy.

I guess they'd probly say $-(5!)$ if visharul is to be right

why r we still on this? i mean just forget about it


$ \ a,b,c>0$ và  $ \ a\plus{}b\plus{}c\equal{}1$ .

Prove that :

$ \frac{a^2\plus{}b^3}{b\plus{}3c^2}\plus{}\frac{b^2\plus{}c^3}{c\plus{}3a^2}\plus{}
\frac{c^2\plus{}a^3}{a\plus{}3b^2}\geq\frac{2}{3}$.

You had a mistake in type, RHS is $ \frac {3}{2}$, not  $ \frac{2}{3}$
Indeed, it's a weak ineq  :)
We have: $ \frac {b^3}{b \plus{} 3c^2} \plus{} \frac {c^3}{c \plus{} 3a^2} \plus{} \frac {a^3}{a \plus{} 3b^2}\geq\frac {3}{4}$
So, we only need to prove: $ \frac {a^2}{b \plus{} 3c^2} \geq\ \frac {3}{4}$
It's weaker than my old result:
Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \frac {a}{\sqrt {4b \plus{} 4c^2 \plus{} 1}} \plus{} \frac {b}{\sqrt {4c \plus{} 4a^2 \plus{} 1}} \plus{} \frac {c}{\sqrt {4a \plus{} 4b^2 \plus{} 1}} \geq\ 1$
$ k \equal{} \frac {1}{4}$ is also the best constant for this ineq trues  :)

[quote="nguoivn"]You had a mistake in type, RHS is $ \frac {3}{2}$, not  $ \frac {2}{3}$
Indeed, it's a weak ineq  :)
We have: $ \frac {b^3}{b \plus{} 3c^2} \plus{} \frac {c^3}{c \plus{} 3a^2} \plus{} \frac {a^3}{a \plus{} 3b^2}\geq\frac {3}{4}$
So, we only need to prove: $ \frac {a^2}{b \plus{} 3c^2} \geq\ \frac {3}{4}$
It's weaker than my old result:
Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \frac {a}{\sqrt {4b \plus{} 4c^2 \plus{} 1}} \plus{} \frac {b}{\sqrt {4c \plus{} 4a^2 \plus{} 1}} \plus{} \frac {c}{\sqrt {4a \plus{} 4b^2 \plus{} 1}} \geq\ 1$
$ k \equal{} \frac {1}{4}$ is also the best constant for this ineq trues  :)[/quote]
I can solve this inequality with $ c \ge\ b \ge\ a$  :D


function
calculus
integration
search
number theory
complex numbers
complex analysis

hi, i'm trying to do some basic number theory , and the riemann function turns out to be very important

sum(1/n^z, n=1..infinity) can easily be shown to be convergent for every z with real number greater than one

1)but for which z in general does this series converge?

if z is real then it is quite clear that is must be greater than 1, but how about imaginary numbers

2) i often hear about extending this function to the entire plane except the the line with real part 1, by analytic continuation

what is analytic continuation,
it has something to do with power summands, but is it easy how exactly you must extend the function?

You have surely already got your question answered. There are several ways to make an analytic continuation of this function to the entire complex plane by making it a mermomorphic function with a pole of residue 1 at  z = 1 and no other singularities. I believe Ahlfors covers this issue in his Complex Analysis book, but I don't have that book anymore. Anyways, I remember rewriting the "normal" sum-definition as a sum of a convergent sum for Re s > 0 or mabye it was >= 0 and an integral then you repeadily can use the functional equation etc etc. I think this should work, don't have time to check it. But a search on "analytic continuation of zeta function" on google would give answers.

For a proof, check out http://www.math.harvard.edu/~elkies/M259.02/zeta1.pdf

Andrei


geometry
geometric transformation
reflection
MATHCOUNTS
\/closed

Some students have noted that our classes start too early for them to participate.  We are considering changing our schedule for the next set of classes we offer.  Please only answer this poll if you would consider taking one of our classes at the times mentioned in the poll.  Also, please make any comments about timing you might have.

I voted for 7-8:30 est, because this gives me an early enuff time that I'm not absolutely dead (waking up at 6:15 every morning doesnt agree with me) and enuff time to finish whatever afterward. I woudl like to make one appeal, though. Please stop scheduling classes that i want to take on Tuesdays. Actually, just reschedule all classes that have olympiad in front of them currently on tuesdays. Please? Really, pretty please? With a giant cherry and proportioned whip cream and your favorite food (whatever it may be) on top?

Heh... I just read Anthem (Ayn Rand) the other day, thus all mention of 'we' makes me think again.

[quote="tokenadult"]I just had A+MATH click on his preferred time, which reflects our schedule in the Central Time Zone. My only other comment would be that making sure whatever was on Thursday one semester is on some other day of the week the following semester would help us, as his evening accelerated math class meets on Thursdays (and an occasional additional Tuesday) this year. We are looking forward to him getting heavily into this semester's online classes.[/quote]

We intend to move class days around in most cases although there might be instances when we do not.  A small part of our scheduling is dictated by circumstances, though we're fairly flexible.

Anything but 7:00-8:30 ET is good for me. Some of my other extra-curricular activities run until 7:00, so by the time I'm home it's 7:15, so I can't make it to the 7:00 ET class.

I agree with EFuzzy.  Extra curriculars force me to lean toward later times.  I usually stay up to at least 11 anyways and get up at 5:30 so any of those times wouldn't have much of an effect except for the earlier times which might interfere with other stuff.

[quote="RC-7th"]Interesting how people either like the earliest time or the latest time. Seems like someones going to be unhappy no matter what time is chosen.[/quote]

That all depends on how the future of the school develops.  At some point we could run back to back classes, but we can only do that if enrollment grows large enough.  Perhaps next year we will offer some classes that way.

[quote="cutiepi"]i like 7:30-9, because by then all my other activities are over, also its not tooo late, and its not too early for west coast people.[/quote]

AH HA!
"i didnt take any AoPS classes" she says.
its like ur ashamed to admit it or something.  ;)

[quote="cutiepi"]i like 7:30-9, because by then all my other activities are over, also its not tooo late, and its not too early for west coast people.[/quote]

I agree.

Sometimes at the later times, after a long day, I'm just not up for doing problem solving. I just feel like going to bed and get frustrated a lot easier.

I voted for 9:00 to 10:30 time because in winter, I don't have any sport going on and I rather want to finish my homeworks, tests, etc before classes so I can simply attend class and go to sleep after it.

I might stay some after to review some school stuffs but still, I don't want class to be starting at too early like 7:30.  :D

[quote="white_horse_king88"]Yeah... I wouldn't want later than to finish at 9 EST, because I also have to get up around 5:45 every morning. Thanks for the poll![/quote]

I would add that especially for the middle school level classes (MATHCOUNTS and Intro classes) ending after 9 would be difficult for many people.

It's quite difficult, I'm sure, to accomodate schedules in multiple timezones!  I can see how starting at 4 or 4:30 could be difficult for those on the west coast.

Good luck working it all out!


How would one invert the left-right layout of the wiki as you did?

[quote="#H34N1"]Use CSS[/quote]
Oh, so there isn't a php variable you can edit or anything?

There is, but its easier to just use CSS.

I don't know; you have to look into the monobook template.