calculus
derivative
function
algebra
polynomial
Asymptote
calculus computations

for what interval is f(x)=xe^(-x^2) concave down?

What do you have for $ f'(x)$ and $ f''(x)?$ Let's see your calculations. Don't forget to use the product rule (when appropriate) as well as the chain rule.

Those derivatives are incorrect.

If $ g(x) \equal{} e^{ \minus{} \frac {x^2}2},$ then $ g'(x) \equal{} \minus{} 2xe^{ \minus{} \frac {x^2}2}.$

But you said that $ f(x) \equal{} xe^{ \minus{} \frac {x^2}2},$ so that has a different derivative.

Are you correctly using the product rule? $ \frac{d}{dx}[f(x)g(x)]\equal{}f'(x)g(x)\plus{}f(x)g'(x).$ You shouldn't ever be multiplying a derivative times a derivative.

----

Here's something that can be used as a partial check: every function here is either even or odd. In particular, $ e^{ \minus{} \frac {x^2}2}$ is an even function, and even and odd polynomials are easily recognized. An even polynomial has only even powers of $ x.$ (Note that a constant term is an even term.) An odd polynomial has only odd powers of $ x.$ An odd function times and even function is odd. And so on.

The derivative of an odd function must be even and the derivative of an even function must be odd. If you find yourself violating these symmetries, then you're making some mistake.

I think Kent used the wrong function.  I think $ f''(x)\equal{}\minus{}e^{\minus{}x^{2}}\minus{}2xe^{\minus{}x^{2}}\plus{}4x^{2}e^{\minus{}x^{2}}$.  Factor out $ e^{\minus{}x^{2}}$ and then you have a nice quadratic.  $ e^{\minus{}x^{2}}\neq 0$ but the remaining part yields $ x\equal{}\frac{1\pm\sqrt{5}}{8}$ so these are the two critical points.  Now it's just a matter of testing them.

Not so much the "wrong" function as a different function, as an example. (Note that I called it by a different name, to emphasize that point.)

However, JRav, your answer cannot be correct. Please read what I said at the bottom of my previous post about even functions and odd functions - and why I can tell that you're wrong without replicating your work.

My mistake.  $ f'(x)\equal{}e^{\minus{}x^{2}}\minus{}2x^{2}e^{\minus{}x^{2}}$.  So $ f''(x)\equal{}e^{\minus{}x^{2}}\minus{}2x^{2}e^{\minus{}x^{2}}\plus{}4x^{3}e^{\minus{}x^{2}}\minus{}4xe^{\minus{}x^{2}}$.  I believe this one is correct.  That yields $ x\approx\minus{}.9068$ and $ x\approx.2353$.  I hope that is right.

That's still wrong. $ f''(x)$ must be an odd function, and the inflection points must include zero and otherwise be symmetric about the origin.

Draw a quick sketch of the graph of the function $ f(x)=xe^{-x^2}.$

(Ah, now I see what JRav meant about me using the wrong function - I was misreading $ e^{-x^2}$ as $ e^{-\frac{x^2}2}.$ All the more reason to learn $ \text{\LaTeX}.$)

Anyway, that's an odd function, so the graph is symmetric about the origin. It is positive for $ x>0$ and negative for $ x<0,$ and it tends (rapidly) to zero as either $ x\to\infty$ or $ x\to-\infty.$ Other than throwing in some extra ornate wiggles that this function probably doesn't have, the graph pretty much has to have certain features. There has to be a maximum at some $ x=a>0,$ and a matching minimum at $ x=-a.$ Then the function must be concave down at the maximum, but for significantly larger $ x,$ it must be concave up to have the horizontal asymptote that it does have. Hence there has to be a point $ x=b>a$ that is a point of inflection. And again, we have to match all of that on the other side.

Visually, that means that the graph of $ f$ will be concave down on the interval $ (-\infty,-b)$ and on the interval $ (0,b).$ And it will be concave up on $ (-b,0)$ and $ (b,\infty).$

(Either that, or it has a much more complicated shape - but trust me, this one's shape is not more complicated.)

We still need to (correctly!) find $ f'(x)$ and $ f''(x)$ just to find out what the values of $ a$ and $ b$ are, but we know what we're looking for.

You just apply the chain rule and the product rule a few times.

Given: $ y \equal{} e^{(\minus{}x^2)}$
We have:
$ \ln(y) \equal{} \minus{}x^{2}$
$ \frac{1}{y}dy \equal{} \minus{} 2x dx \implies y' \equal{} \frac{dy}{dx} \equal{} \minus{} 2xy$
$ y'' \equal{} \frac{d^{2}y}{dx^{2}} \equal{} \minus{}2(y \plus{} xy') \equal{} \minus{}2[e^{\minus{}x^2} \minus{} 2x^{2}e^{\minus{}x^2}]$.

[quote="cdymdcool"]How can you calculate y''(x) of y(x)=e^(-x^2) ?
Please leave the answer.[/quote]
[url=http://calc101.com/webMathematica/derivatives.jsp]WebMathematica[/url] can give you not just the answer, but also step-by-step solution, which you can then submit as your own work.

$ \left[f(x)\cdot e^{g(x)}\right]'\equal{}$ $ \left[f'(x)\plus{}f(x)\cdot g'(x)\right]\cdot e^{g(x)}\mathrm {\ .s.s.\ }\left[f'(x)\plus{}f(x)\cdot g'(x)\right]\ !$

[b][u]Notation.[/u][/b] $ A\ .s.s.\ B\Longleftrightarrow A\equal{}B\equal{}0\ \ \vee\ \  AB>0\ .$


analytic geometry
graphing lines
slope
absolute value

Is there an easy way to graph an absolute value equation in the form

y=a|bx+c|+d?

For proabolas the equation is in relatively the same form and you can find the vertex in a flash and then use a little trick to graph (has to do with coeficents and odd numbers)...

Does anything exist for these?

[hide]In absolute value equations, there are 2 rays that meet with both of their vertices at a certain point.
y=a|bx+c| + d
The graph gets shifted -c/b units from the y axis. For example, if b = 2 and c = 1, then the graph gets shifted 1/2 units to the left of the y axis. d is how many units the graph goes up or down. For example, if d = 3, then the graph (where the 2 rays meet) will start at 3. If b is 2, c is 1, and d is 3, then the graph will start at (-.5, 3). a is the "slope" if the graph. If a is 3, then the "slope" of the graph will be 3. However, if b is 2, then the "slope" of the graph is ab = 6. Also, which way the rays go (up or down) depends on whether a is negative. If a is negative, then the graph will go in a downward direction. If a is positive, then it goes in an upward direction. So if y = 3|2 + 1| + 3, then the graph will look like 2 rays that meet and stop at (-.5, 3) with slopes of 6 and -6. If y = -3|3x + 2| - 8, then the graph will like 2 rays that are going down (towards the negative points of the y axis) with slopes of 9 and -9 that meet and stop at (-2/3, -8).[/hide]

Thanks...I learned that  a long time ago and I completely forgot about it.

[hide]Translationg of coordinate axes and the points[/hide]


When I try to build a latex file on TeXnic Center it sometimes runs and then freezes for about 10 minutes before resuming its build. Stopping the build and starting over doesn't seem to fix the problem. Other times, it works fine, but I'd rather not wait 10 minutes every other time I build for the computer to unstick itself.

The output is attached.

Thanks!

edit: The tex on the bottom is what TeXnic Center gives me when it freezes, and then it completes the rest of the output after it unfreezes itself.

From the log it looks like it might be a problem with the geometry package. What options are you using for it? In other words what goes in the square brackets in the line \usepackage[ [i]options[/i] ]{geometry}?
Some tests you can do:
1) comment out the geometry line using % and compile? Do you still get a freeze (ignore errors about undefined commands or text size or whatever);
2) try a simple document
[code]\documentclass{article}
\usepackage{geometry}
\begin{document}
Hello world
\end{document}[/code]
Does that freeze and if so does it freeze without the \usepackage{geometry} line?

If these documents work OK then cut down your weightedaverage.tex ([b]but back it up first![/b]) until you isolate the cause of the error and hence are able to produce a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] which we can analyse.

By the way, the log shown in TexnicCenter's bottom window is output to weightedaverage.log as text so making it easier to copy here.


There is a number that is 6 times the sum of its digits. What is this number

[hide=" click to reveal hidden content"]

$ 0$

$ 54$

[/hide]

[hide="Algebraic Solution"]You can say x and y are two, one-digit, integers. We can then set up an equation.
where $ 10x\plus{}y$ is the number and $ x\plus{}y$ is the sum of its digits. We then have $ 10x\plus{}y\equal{}6(x\plus{}y)$. We can make this $ 10x\plus{}y\equal{}6x\plus{}6y$ and then, $ 4x\equal{}5y$, we know that x and y are both one-digit integers so we can easily say $ (x,y)\equal{}(5,4)$ So the answer is $ \boxed {54}$[/hide]
I believe this is more of a solution than the previous post, hope this is what you were looking for  :lol:

But you didn't post how you got it, so rksmash didn't just know how you got it, so I posted the solution.

[hide] $ 10x\plus{}y\equal{}6(x\plus{}y)$
 $ 4x\equal{}5y\rightarrow (x,y)\equal{}(5,4)$ [/hide]

Answer of: "find the number" --> 54  :lol:

Hey Guys, Please do not retrieve a topic that is over a month old.

no need to write Answer is 54 if you do not post a solution or if you repeat the same solkution that have been posted before.Use your common sense please.


The tens digit of a two-digit number exceeds the units digit by 1. If the number is divided by the sum of its digits, the quotient is 6. What is the number?

We can write the number in the form 10(x+1)+x. When it is divided by the sum of its digits, the quotient is 6. Thus, we get the equation:
(10(x+1)+x)/(x+1+x)=6, which simplifies to
(11x+10)/(2x+1)=6. We can then multiply each side by 2x+1 and solve for x, which equals 4. The answer is [b]54[/b].


geometry
email
Davidson Institute
\/closed

We plan to build a page in our resources area listing math scholarships and we are counting on AoPSers to help us identify any such scholarships and websites.  There are many scholarships out there, so this project will be an ongoing effort.

[quote="siuhochung"]will it include those that can be applied by non-US students?[/quote]

The list will mostly contain scholarships for U.S. students, but I do plan to include a section for scholarships outside the U.S. as well.

However, I don't plan to do 100 hours of research, so I'm not likely going to discover many non-U.S. scholarships on my own.  The list is going to be good and thorough only if AoPSers make a community effort from which all can benefit.

[quote="Silverfalcon"]So how far is the research gone yet?

I know there's Siemens-Westing House.. But that's REALLY hard. :|[/quote]

I have a list of some scholarships, but the universe of scholarships is enormous.  We'll put a list up soon and encourage AoPSers to suggest additions.

I'll be sure to post to that page. The [url=http://www.ditdservices.org/Articles.aspx?ArticleID=24&NavID=0_0]Davidson Institute Young Scholars Program[/url] is a program all of the more conspicuously "into-it" middle-school-AoPSers in the United States ought to be applying to.

Hopefully I will have an initial list completed soon.  I'm mostly spamming here to bring attention to this project once again as I'm going to take the time to polish off the initial listing soon and post it and I hope anyone who knows about any good resources will email me about them.  Thanks in advance!

[quote="h_s_potter2002"]I think there's a scholarship in Ohio for the person with the highest AMC 12 score.[/quote]

Do you have a link for it?

No.  I just remeber that it was mentioned in the book with all the Ohio top scorers in the AMC's.  I lost it, but there might be some people who have it with more information.  And I e-mailed the AMC director for Ohio about it.

Our [url=http://www.artofproblemsolving.com/Resources/AoPS_R_MathScholarships.php]Math Scholarships[/url] page is now online.


abstract algebra
number theory
greatest common divisor
group theory
superior algebra
superior algebra unsolved

1. Let $ R$ be a commutative ring. Show that $ R^m\otimes R^n \cong R^{mn}$.
2. For cyclic $ \mathbb{Z}$-module $ \mathbb{Z}_m$ and $ \mathbb{Z}_n$ with generator $ a$ and $ b$, respectively, show that $ \mathbb{Z}_{m} \otimes \mathbb{Z}_{n} \cong \mathbb{Z}_{(m,n)}$ with generator $ a\otimes b$, where $ (m,n)$ is the gcd of $ m$ and $ n$.
3. If $ A$ is a finite abelian group, show that $ A\otimes_{\mathbb{Z}}\mathbb{Q}\equal{}0$.

Thanks in advance

You should start by finding a suitable bilinear map.

Can you guys give me 1 example to start from? I'm totally blank at the moment :(

First, try to understand why you're not getting responses.  You've posted three problems at once, and haven't said where they come from--when I see this I usually assume that someone is posting homework.  It also suggests that you have not even tried to solve 1% of them yourself and this is a frustrating situation to work with.

That said, virtually every tensor product problem like #1 and #2 this can be solved using bilinear maps.  Specifically, if we want to construct a map $ F: A\otimes B \to C$, we first find a bilinear map $ f: A\times B \to C$, then use the universal property of tensor products.

For example, say we want to show $ R\otimes R^n \cong R^n$.  First we write down the obvious bilinear map $ (r,(r_1,r_2,\ldots r_n))\mapsto (rr_1,rr_2,\ldots rr_n)$, giving us a map sending $ r\otimes (r_1,r_2,\ldots r_n)$ to $ (rr_1,rr_2,\ldots rr_n)$.  Now we look at the map in the other direction, sending $ (r_1,r_2,\ldots r_n)$ to $ 1\otimes (r_1,r_2,\ldots r_n)$.  It is a simple calculation to verify that these maps are inverses.

There's a slightly easier way to do #1, if you've seen how tensor product and direct sum interact.

As for #3, you really should be able to do this yourself.  Take a simple tensor and... well, play with it.  This is mathematics, sometimes you have to really get your hands dirty.

Just a quick note, doesn't tensor product in general commute with direct products, not direct sums? (Since the tensor product is a right adjoint, so it commutes with limits, but not necessarily colimits)

Edit: Nevermind, I was mixing them up

1. the tensor product commutes with direct sums, this follows immediately from the adjointness or the universal property. besides, $ R \otimes_R M\equal{} M$ also follows from the universal property ($ r \otimes m \mapsto rm$ and $ m \mapsto 1 \otimes m$ define inverse maps). from this the claim follows. note that this also implies $ rank(A \otimes_R B) \equal{} rank(A) * rank(B)$ for free $ R$-modules $ A,B$.

2. $ M/U \otimes_R N/V \cong (M \otimes N)/ \langle M \otimes V , U \otimes N \rangle$ follows from the universal property. in particular, $ R/I \otimes_R R/J \equal{} R/(I \plus{} J)$ for ideals $ I,J \subseteq R$. for $ R \equal{} \mathbb{Z}$ you get your result.

3. if $ A$ is a torsion abelian group and $ B$ is a divisible abelian group, then $ A \otimes_\mathbb{Z} B \equal{} 0$. this generalization shows how the proof must(!) look like.

It seems to me that to show isomorphism between these tensor products, you either use the universal property (some other object that makes the diagram commutes) or gives explicit maps between themselves. However, I'm battling with
[quote="-oo-"]2. $ M/U \otimes_R N/V \cong (M \otimes N)/ \langle M \otimes V , U \otimes N \rangle$ follows from the universal property. in particular, $ R/I \otimes_R R/J \equal{} R/(I \plus{} J)$ for ideals $ I,J \subseteq R$. for $ R \equal{} \mathbb{Z}$ you get your result.[/quote]

How exactly do you use the universal property for these examples? I just got back to this so I'm slow and lost at the moment :(

$ Hom(M/U \otimes N,T) \equal{} Hom(M/U,Hom(N,T))$ $ \equal{} \{f \in Hom(M,Hom(N,T)) : U \subseteq \ker(f) \}$
$ \equal{} \{f \in Hom(M \otimes N,T) : \langle U \otimes N \rangle \subseteq \ker(f)\}$ shows $ M/U \otimes N \equal{} (M \otimes N) / \langle U \otimes N \rangle$, where $ \langle U \otimes N \rangle$ means the image of $ U \otimes N \to M \otimes N$. now do this for $ N/V$ to get your result.

usually, when you start with tensor products, you are advised to show that $ \overline{m} \otimes n \mapsto \overline{m \otimes n}$ defines the desired isomorphism, but the verification of this is best written like above. otherwise you reprove general facts (homomorphism theorem, yoneda-lemma).

another proof (essentially with the same arguments): since $ \minus{} \otimes N$ is a left-adjoint, it is right exact. in particular, $ U \otimes N \to M \otimes N \to M/U \otimes N \to 0$ is exact, which is exactly the claim.


It has been rewritten.

Do I use A = pir^2?

If so, what numbers do I plug into the formula?

[quote]How many cans will they need to paint the
whole GEODESTIC GLOBE?[/quote]

you need the formula for SA

[quote="sharkman"]The GEODESTIC EPCOT CENTER has a diameter of 265 feet. They need to paint it. One paint covers 400 square feet. How many cans will they need to paint the
whole GEODESTIC GLOBE?[/quote]

radius- 132.5

so

$4 \pi (132.5)^{2}$

$4 \pi 17556.25$

$70225 \pi$

$220506.5$ square feet.

$\frac{220506.5}{400}= 551.2$ so therefore 552 cans of paint?

well okay we can be sure that we need at the least 511 cans, right?
well we're going to need another can. You can't buy 1/5 of a can.

[quote="sharkman"]So, 551.2 rounded to the nearest unit is 552?

I thought it was 551.[/quote]

yeah if you only bought 551 cans then you would have like a little part of the sphere to paint. so therefore you need 552 to cover ALL of the sphere. a general rule is to think it logically  :wink:


vector
linear algebra
linear algebra unsolved

[size=134]1. Suppose U, V ⊂ Rⁿ are subspaces and U ⊂ V. Prove that V┴ ⊂ U┴.

2. Let V ⊂ Rⁿ be a subspace. Prove that V ⊂ (V┴)┴. Do you think more is true?

3. Suppose V = Span(v_1, ... , v_k) ⊂ Rⁿ. Prove that there are vectors w_1, ..., w_k ∈ V that are mutually orthogonal that also span V.

4. Suppose U and V are subspaces of Rⁿ. Prove that (U + V)┴ = U┴ ⋂ V┴.[/size]

1) If a vector is orthogonal to everything in $ V$, then it is certainly orthogonal to everything in $ U$.

2) A vector in $ V$ is orthogonal to any vector that is orthogonal to any vector in $ V$.

3) Use the Gram-Schmidt process.  Let $ w_1 \equal{} v_1$.  Let $ w_2 \equal{} v_2 \minus{} \frac{< v_2, w_1 >}{|w_1|^2}w_1$.  And so on.


inequalities
algebra
polynomial
inequalities unsolved

For positive x,y,z prove
\[\frac{xy}{x^2+xy+yz}+\frac{yz}{y^2+yz+zx}+\frac{zx}{z^2+zx+xy}\le \frac{x}{z+2x}+\frac{y}{x+2y}+\frac{z}{y+2z}.\]

It can be easily rewritten as symmetric inequality with respect to $\frac{x}{y}$, $\frac{y}{z}$, $\frac{z}{x}$, but I don't know how to prove it.
Obviously, there exists straightfowvard solution (multiply by product of denominators and then prove by weight AM-GM). I'm interested in proof not using the conversion to polynomial inequality. Any ideas?

You should rather post it in the Inequality Unsolved problems...
Anyway mods will move it in a couple of hours (at last I hope  ;) )

Sorry for my inattention, I forgot about existance of such division  :blush:

The left-hand-side of the inequality is equal to $1$. Then the problem can be solved using [b]Cauchy[/b].

By putting $a=\frac{x}{y}$ , $b=\frac{y}{z}$ and $c=\frac{z}{x}$ the ineq becomes :
$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\leq \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$
with $abc=1$
and I remember that I saw something like this on the forum, but I don't know where exactly  :?

Bingo I found it  :D
Here it is : [url]http://www.mathlinks.ro/Forum/post-115579.html[/url]

I meant exactly this. What do you propose next? ;)
PS Thanks!
PPS
[quote="darij grinberg"]Both solutions are almost the same: Brute force. The inequality doesn't seem to have a better proof.[/quote]
It's a pity...

Yeeeh... It's a pity. I don't think there is another way  :lol:

[quote="Feuerbach"][quote="Andrei"]The left-hand-side of the inequality is equal to $1$.[/quote]
Actually, it does not :) Or do you mean some conversion?[/quote]

Yeah, sorry :(. Guess I had made a terrible calculation mistake :(


A fair, standard six-faced die is tossed eight times.  The
sequence of eight results is recorded to form an eight-digit
number.  What is the probability that the number formed is a
multiple of eight?  Express your answer as a common fraction.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=254317]Click because I'm lazy.[/url]


Construct a sequence $ a_1,a_2,...,a_N$ of positive reals such that $ n_1a_{n_0}\plus{}n_2a_{n_1}\plus{}...\plus{}n_ka_{n_{k\minus{}1}}>2.7(a_1\plus{}a_2\plus{}...\plus{}a_N)$ for arbitrary integers $ 1\equal{}n_0<n_1<...<n_k\equal{}N$.


Suppose that each distinct letter in the
equation $ MATH \equal{} COU \plus{} NTS$ is replaced
by a different digit chosen from 1 through 9
in such a way that the resulting equation is
true.  If $ H \equal{} 4$, what is the value of the greater of $ C$ and $ N$?
\begin{tabular}[t]{cccc}
&C&O&U\\
+&N&T&S\\\hline
M&A&T&H\\
\end{tabular}

[quote="Mewto55555"][quote="GameBot"]Suppose that each distinct letter in the
equation $ MATH \equal{} COU \plus{} NTS$ is replaced
by a different digit chosen from 1 through 9
in such a way that the resulting equation is
true.  If $ H \equal{} 4$, what is the value of the greater of $ C$ and $ N$?
$ \begin{tabular}[t]{cccc}
&C&O&U\\
\plus{}&N&T&S\\\hline
M&A&T&H\\
\end{tabular}$[/quote][/quote]
Playing around a little, we see that.

$M=1$

$A=3$

$T=2$

$H=4$

$C=7$

$O=9$

$U=8$

$N=5$

$T=2$

$S=6$

So it is:

$ \begin{tabular}[t]{cccc}&7&9&8\\ +&5&2&6\\\hline 1&3&2&4\\ \end{tabular} $

So the greater of $C$ and $N$ is $\boxed{7}$.


(1) Suppose $ A$ and $ B$ are positive real numbers for which $ \log_{A}B \equal{} \log_{B}A$.  If neither $ A$ nor $ B$ is $ 1$ and $ A\neq B$. find the value of $ AB$.

(2) The formula $ N \equal{} 8\cdot10^{8}\cdot x^{ \minus{} 3/2}$ gives, for a certain group, the number of individuals whose income exceeds $ x$ dollars.  What is the smallest possible value of the lowest income of the wealthiest $ 800$ individuals?

(3) Given that $ \log_{3}2 \equal{} 0.631$, find the smallest positive integer $ a$ such that $ 3^{a} > 2^{102}$.

[hide]Hint: Show that $ \log_{3}2^{202} \equal{} 102\log_{3}2$.[/hide]

(4) Show that $ \log_{6}2 \plus{} \log_{6}3 \equal{} 1$

Thanks in advance!

yay i can do #4
ok adding logarithims in the same base is the same as multiplyings the 2 *the two things that are being logarithmed* and taking the same logarithmic base
in otherwords,
$ \log_m n\plus{}\log_m o\equal{}\log_m no$
so we can change the problem into show that
$ \log_6 6\equal{}1$
or $ 6^1\equal{}6$ since this is true, i guess im done... heh

1: This is just $ \log_AB\equal{}1/\log_AB$
$ \log_A^2B\equal{}1$
Since $ A\ne B$,
$ \log_AB\equal{}\minus{}1$.
Follows that AB=1.

2:We have $ 800\equal{}8\cdot 10^8 \cdot x^{\minus{}3/2}$.
$ 10^{\minus{}6}\equal{}x^{\minus{}3/2}$
Taking logb10,
$ \minus{}6\equal{}\minus{}3/2 \log x$
$ \log x \equal{}4$
x=10000 *small salary for #800 :D

3: taking logs base 2 give $ a \log_2 3 > 102$
$ a\cdot .631>102$
$ a>161.648$
a=162.

[hide="Quick Q for elcric"]
Are these from Algebra for College Students, by any chance?[/hide]

[quote="someperson01"]
Are these from Algebra for College Students, by any chance?[/quote]

Actually, it is from the Logarithms chapter from AoPS Volume 1.

[quote="elcric"]
(4) Show that $ \log_{6}2 \plus{} \log_{6}3 \equal{} 1$[/quote]
[quote="stevenmeow"]
$ \log_m n \plus{} \log_m o \equal{} \log_m no$[/quote]

The motivation behind this problem, if I remember correctly, was to prove that this is true, because it was not proven in the text of AoPS Volume 1, but rather developed through this problem. So they would want you to do something like this:

[hide="Solution."]
$ \log_62 \plus{} \log_63 \equal{} 1$
$ 6^{\log_62 \plus{} \log_63} \equal{} 6^1$
$ 6^{\log_62}\cdot6^{\log_63} \equal{} 6$
$ 2\cdot 3\equal{}6$
$ 6 \equal{} 6$, and we're done.
[/hide]

For number 1, I don't get how this was gotten: $ \log_AB \equal{} 1/\log_AB$.

For number 3, why would you take the $ \log_{2}$ of both sides?  Also, how would you do this?

Thanks!

[quote="jvenezuela716"][quote="elcric"]
For number 3, why would you take the $ \log_{2}$ of both sides?  Also, how would you do this?[/quote]

You wouldn't; his answer is incorrect.

[hide="Solution."]
First, I will prove that $ n\log_ba \equal{} \log_ba^n$.
Let $ x \equal{} n\log_ba$. Taking $ b$ to each side, we have $ b^x \equal{} b^{n\log_ba} \implies b^x \equal{} \left(b^{\log_ba}\right)^n \implies b^x \equal{} a^n$. Taking the logarithm base $ b$ of both sides, $ x \equal{} \log_ba^n$, as desired.

Now apply this to the problem:
$ 3^a > 2^{102}$
Remember that we want $ a$ alone on that left side, so why not take the logarithm base $ 3$ of both sides to clear the $ 3$? We can do this operation because logarithms are strictly increasing.

$ \log_33^a > \log_32^{102}$, so we use our property above:
$ a > 102\log_32$, and we are given $ \log_32$, so:
$ a > 102\cdot0.631$
$ a > 64.36 \implies a \equal{} \boxed{65}$.
[/hide][/quote]

How did you get from $ \log_33^a > \log_32^{102}$ to $ a > 102\log_32$?  Thanks.

[quote="jvenezuela716"]First, I will prove that $ n\log_ba \equal{} \log_ba^n$.
Let $ x \equal{} n\log_ba$. Taking $ b$ to each side, we have $ b^x \equal{} b^{n\log_ba} \implies b^x \equal{} \left(b^{\log_ba}\right)^n \implies b^x \equal{} a^n$. Taking the logarithm base $ b$ of both sides, $ x \equal{} \log_ba^n$, as desired.[/quote]

Study this proof very carefully, as it will help you get used to logarithms. This property of logarithms basically says that when i take the logarithm of a power, i can "pull" the exponent outside of the logarithm.

So when I have $ \log_33^a$, this is the same as $ a\log_33$. [i]Make sure you see why our property allows us to say this.[/i] $ \log_33$ is just $ 1$, so the left-hand side becomes $ a\cdot1\equal{}a$. The right side just becomes $ \log_32^{102}\equal{}102\log_32$, again by the property I proved above.

Putting it all together, $ \log_33^a>\log_32^{102} \implies a>102\log_32$. If anything above is at all unclear, I advise you to reread this chapter in the AoPS Volume 1 book, and if you have the solutions manual for it, read the solution there, as well.

[quote="jvenezuela716"][quote="jvenezuela716"]First, I will prove that $ n\log_ba \equal{} \log_ba^n$.
Let $ x \equal{} n\log_ba$. Taking $ b$ to each side, we have $ b^x \equal{} b^{n\log_ba} \implies b^x \equal{} \left(b^{\log_ba}\right)^n \implies b^x \equal{} a^n$. Taking the logarithm base $ b$ of both sides, $ x \equal{} \log_ba^n$, as desired.[/quote]

Study this proof very carefully, as it will help you get used to logarithms. This property of logarithms basically says that when i take the logarithm of a power, i can "pull" the exponent outside of the logarithm.

So when I have $ \log_33^a$, this is the same as $ a\log_33$. [i]Make sure you see why our property allows us to say this.[/i] $ \log_33$ is just $ 1$, so the left-hand side becomes $ a\cdot1 \equal{} a$. The right side just becomes $ \log_32^{102} \equal{} 102\log_32$, again by the property I proved above.

Putting it all together, $ \log_33^a > \log_32^{102} \implies a > 102\log_32$. If anything above is at all unclear, I advise you to reread this chapter in the AoPS Volume 1 book, and if you have the solutions manual for it, read the solution there, as well.[/quote]

Thanks, I got it now!

Do you know how $ \log_AB \equal{} 1/\log_AB$ was gotten for problem 1?


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ (K,+,*)$ be a finite field,with 81 elements.
Prove that the multiplicative group $ (K^{*},*)$ is $ cyclic$

Let $ (A,+,0)$ be a finite abelian group having the property that for every $ n>0$, the equation $ na=0$ has at most $ n$ distinct solutions in $ A$. Show that $ A$ is cyclic. owk

Even more generally (though straying from the question I suppose):

Let $ (K,+, *)$ be a finite field. Show that $ (K^{*}, *)$ is cyclic.

This is a specific case.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=root+primitive&t=27050]Posted Before[/url]

(In my proof I didnt realize that $ \mathbb{F}$ is infinite, but it doesnt matter because it was proven for a finite case which is exactly what you are asking.)


I have a method of using AM-GM to prove a kind of inequality like this:


algebra
polynomial
number theory proposed
number theory

It is obviosly $m|(f(a+m)-f(a) \ \forall \ f(x)\in Z[x]$.

We must have that f bijects Z mod 101 to Z mod 101. If such an f existed, then f(f(x)) would be a degree 100 polynomial having the same property. However, it is well-known that no such polynomial exists...


How many unlabeled families of subsets of a $ 3$-set are there?

So we know that there are $ 2^{2^3} \equal{} 256$ labeled families. How do we get the number of unlabeled?

What do you mean by "unlabeled"?  The elements of each set are already unordered, and so is the collection.

The only sense I could make of this is that perhaps it wants two-element subsets of the original three-element set to be indistinguishable?  But then I don't know if I'm supposed to be able to take up to three of them (but I don't really think this makes sense), up to infinitely many of them (in which case the question is really silly, so presumably not this) or only up to one of them (in which case I'm just looking at subsets of a four-element set).


Given nonnegative real numbers have $a+b+c+d=4$. Show that: \[(1+a^{3})(1+b^{3})(1+c^{3})(1+d^{3}) \ge 16.\]


vector
analytic geometry
inequalities
topology
probability and stats

Suppose that $ E_1, E_2,\dots , E_k$ are linear sets of PD's on a finite set $ X$ with $ E \equal{} \cap_{i \equal{} 1}^{k}E_i \neq \emptyset$
Can someone show how $ E$ is closed?

I'm not sure I understand what is going on here.
For the special case $ k\equal{}1$ to be true, the given sets must be assumed closed (whatever that means). If your concept of a set being closed is like mine, then the intersection of any closed sets is closed.

Yes, the question boils down to the following: If $ E$ is a linear set of PD's then $ E$ is closed. The metric I am talking about here is the variational distance $ \parallel P\minus{}Q\parallel : \equal{}\sup_{B\subset\mathbb{X}}|P(B)\minus{}Q(B)|\equal{}\frac {1}{2} \sum_{x\in X}|P(x) \minus{} Q(x)|$
$ E$ can be looked as an intersection of a linear subset of $ \mathbb{R}^{|X|}$ with the simplex representing the PD's on $ X$. I don't know how to show $ E$ is closed from here.

1) Any linear subspace $ V$ of $ \mathbb R^n$ is a closed set. This can be proved by choosing an orthonormal basis for $ V$. The limit of any convergent subsequence in $ V$ is a linear combination of these basis vectors (because the limit can be taken coordinate-wise).

2) The simplex of all PDs is a closed subset of $ \mathbb R^{|X|}$. Indeed, it's determined by non-strict inequalities and one equality. Each of these conditions determines a closed set (why?) so the set where all of them are fulfilled at once is also closed.

[quote="mlok"]1) Any linear subspace $ V$ of $ \mathbb R^n$ is a closed set. [/quote]
Doesn't the subspace also have to be finite dimensional?

Since $ \mathbb R^n$ has dimension $ n$, any linear subspace of it will have dimension at most $ n$... (I suppose we are talking about linear spaces over $ \mathbb R$)


for $x,y \in \mathbb{R^{+}}$ prove that $x^{x}+y^{y}\ge x^{y}+y^{x}$ in as many ways as possible.
does the generalization is right? prove it. it reminds me a bit the rearrangement inequality, can we use it?

[hide="Method 1"]
Assume wlog that $x \geq y$.
The inequality is equivalent to:
$x^{x}-x^{y}\geq y^{x}-y^{y}$
$x^{y}(x^{x-y}-1) \geq y^{y}(y^{x-y}-1)$

This is true since $x^{y}\geq y^{y}$ and $x^{x-y}\geq y^{x-y}$ (and both are $\geq 1$).[/hide]


[hide="Method 1"]Substitute $ u \equal{} x \plus{} \frac 52$.

We obtain the following:

$ (u \minus{} 3/2)(u \minus{} 1/2)(u \plus{} 1/2)(u \plus{} 3/2) \equal{} \minus{} 1 \implies (u^2 \minus{} 9/4)(u^2 \minus{} 1/4) \equal{} \minus{} 1$.

Substitute $ v \equal{} u^2$ and we have a quadratic equation:

$ (v \minus{} 9/4)(v \minus{} 1/4) \equal{} \minus{} 1 \implies v^2 \minus{} 5/2v \plus{} 9/16 \equal{} \minus{} 1 \implies v^2 \minus{} 5/2v \plus{} 25/16 \equal{} 0$, which is just $ (v \minus{} 5/4)^2 \equal{} 0$, so $ v \equal{} 5/4$.

Therefore, $ u \equal{} \pm\sqrt {5/4} \equal{} \pm\frac {\sqrt 5}2$, and $ x \equal{} u \minus{} \frac 52 \equal{} \frac { \minus{} 5\pm\sqrt 5}2$.[/hide]

[hide="Method 2"]
\[ (x \plus{} 1)(x \plus{} 2)(x \plus{} 3)(x \plus{} 4) \plus{} 1 \\
\equal{} (x^4 \plus{} 10 x^3 \plus{} 35 x^2 \plus{} 50 x \plus{} 24) \plus{} 1 \\
\equal{} x^4 \plus{} 10 x^3 \plus{} 35 x^2 \plus{} 50 x \plus{} 25 \\
\equal{} (x^2 \plus{} 5x \plus{} 5)^2 \equal{} 0
\]

The roots are $ \frac { \minus{} 5\pm\sqrt 5}2$.[/hide]

I got a different answer, where did I make my mistake?

EDIT: Fixed. Thanks Yongyi.

[hide]
We have

$ (x \plus{} 1)(x \plus{} 2)(x \plus{} 3)(x \plus{} 4) \equal{} (x \plus{} 1)(x \plus{} 4)(x \plus{} 2)(x \plus{} 3) \equal{} (x^2 \plus{} 5x \plus{} 4) (x^2 \plus{} 5x \plus{} 6) \equal{} \minus{} 1.$

Letting $ t \equal{} x^2 \plus{} 5x \plus{} 5$, we have

$ (t \minus{} 1)(t \plus{} 1) \equal{} \minus{} 1\implies t^2 \equal{} 0\implies x^2 \plus{} 5x \plus{} 5 \equal{} 0\implies x \equal{} \frac { \minus{} 5 \pm \sqrt {25 \minus{} 4(5)}}{2} \equal{} \boxed{\frac { \minus{} 5 \pm \sqrt {5}}{2}}.$
[/hide]

[quote="sunehra"]
Letting $ t \equal{} x^2 \plus{} 5x \color{red}{\plus{}} \color{black}5$, we have...

...$ (t \minus{} 1)(t \plus{} 1) \equal{} \minus{} 1\implies t^2 \equal{} 0\implies x^2 \plus{} 5x \color{red}{\minus{}} \color{black}5 \equal{} 0\implies x \equal{} \frac { \minus{} 5 \pm \sqrt {25 \minus{} 4( \minus{} 5)}}{2} \equal{} \boxed{\frac { \minus{} 5 \pm 3\sqrt {5}}{2}}.$
[/hide][/quote]

Therein lies your error.


Considering that $ x \in \mathbb{R}$, determine all values of $ x$ for wich $ \sqrt{4x \minus{} 8} \plus{} \sqrt[3]{14x \minus{} 20} \equal{} \sqrt{24 \minus{} 4x} \plus{} \sqrt[3]{2092 \minus{} 182x}$.

For the square roots to be real we need $ x \in [2,6]$
but for $ x \le 6$ in this range $ y \equal{} \sqrt {4x \minus{} 8} \plus{} \sqrt [3]{14x \minus{} 20} \minus{} \sqrt {24 \minus{} 4x} \minus{} \sqrt [3]{2092 \minus{} 182x} \le 4 \plus{} 4 \minus{} 0 \minus{} 10 \equal{} \minus{}2$

So $ y$ is always negative and there are no roots.

Note that even if we allow imaginary roots then at most one of the square roots is imaginary at a time and the rest of the expression is real so there can still be no roots.


A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

$ \textbf{(A) } \frac {1}{2187} \qquad \textbf{(B) } \frac {1}{729} \qquad \textbf{(C) } \frac {2}{243} \qquad \textbf{(D) } \frac {1}{81} \qquad \textbf{(E) } \frac {5}{243}$

[hide="solution1"]
notice that the number ways the bug can go on all 7 corners in 7 moves, must be EVEN,
becuase the bug has 3 choices for its first move and only 2 for the second move (since he can't go back to the first corner) and the denomiator WILL be a power of 3 anyways...so C

[/hide]
[hide="OR"]
do what i did on the test...
actually find the total number of ways it can go on all corners after 7 moves. which is 18 so C

this was discussed in a thread in Pre-Olympiad (or maybe Intermediate) forum, like a month ago (whcih was about in how many ways you can go on each corner after 7 moves)...[/hide]

Amir's solution 1 is what i used on the test. It was #25 on the AMC 10.

[hide]Obviously, you can never return to a point. Because of symmetric connections, the first two moves don't matter (3x2 = 6 valid sets of first two moves). From there, notice that if you go in one direction, you have two possible paths and if you go in the other direction, you have only one possible path, meaning 3 possiblities given the first two moves. This means the bug has 6x3 = 18 ways to reach every vertex, and the probability is 18/3^7 = 2/243[/hide]

[quote="Sly Si"]Whoever posted on another thread that this question has poorly chosen distractors (wrong answers) is absolutely right--the probability that it won't return to the first vertex on move #2 is 2/3, and that 2 in the numerator will not cancel out no matter what the other probabilities are.  So if you're really on top of things, this is about a 10 second problem.[/quote]

I'm beating myself because I was intimidated by the fact it appeared to require geometry (which I lack) and that it was #25 on the 10  :oops:

Oh well.

[quote="chess64"]Why would a bug be moving along the edges of a cube?[/quote]To avoid being squashed :)

This was also number 25 of the AMC10s... so basically...
[hide]I took this problem case by case. For the first move the probability is 1 that he "stays on track". Then 2/3. [b]Then 1/3[/b] (b/c the bug gets "stuck" if it goes down. A lot of people missed this and got 8/729 or something like that) Then the bug can go down or continue the square at the top. If the bug goes down, there's one combination to what he can do, and if the bug finishes the top square, then there are two combinations. So.... I got
$1*2/3*1/3*(1/3*1/3*1/3*1/3+1/3*1/3*2/3*1/3)$

That should come out to C; $2/243$
[/hide]

Hmm the choices are just horrible, once you get it must be a multiple of 3*2 you immediately eliminate A, B, D, E.

way too easy for a #20 on the 12
[hide=easy \$\$]
Due to parity, note that the bug can only end at either the other corner of the cube or exactly $1$ unit from it's original position. For the first case, we can work backwards to get $3 \cdot 2$ ways. For the second case, thee are $3$ subcases which each give the same number of ways due to symmetry. We can brute-force here and find there are only $4$ ways per sub-case, so there are $3 \cdot 4$ total ways in this case. The final answer is $\frac{3 \cdot 2 + 3 \cdot 4}{3^7}=\boxed{\textbf{(C) } \frac {2}{243}}$

Anyone here like me instantly knew the problem the moment they saw the title?

I have always hated these types of problems, and have only been able to resort to counting the number of paths. Thanks, guys for showing me the better way.

we can use a lot of symetry here

every possible way to choose 3 vertices are symetric so there are 3*2 to choose the next 2. then by simple counting, we get 3 for each of those

so there are 3^2*2 ways. cancel this with the 3^7 to get 2/(3^5) so answer is C

thanks for coming to my ted talk :)

[b]Solution: [/b]
3 ways for the first move.
2 ways for the second move. Casework from here, shown in the picture. 1) 2 way 2) 1 way.
Total number of valid ways: $3\cdot 2\cdot (2+1)=3^2\cdot 2$.

P(every vertex in 7 moves): $\frac{3^2\cdot 2}{3^7}=\frac{2}{3^5}=\boxed{\frac{2}{243}}$.

[b]Short Solution: [/b]
The total number of valid ways must be 3x2xk for some positive integer k. In other words, the numerator is divisible by 6.
A. 1/2187 X
B. 1/729=3/2187 X
C. 2/243=18/2187 YES
D. 1/81=27/2187 X
E. 5/243=45/2187 X
So the answer is (C).

[quote=Trex4days]FYI dont think abt why bugs are moving along edges of cubes or why someone buys 82 watermelons
its just cuz they need an object and they want to use common objects not like aliens or stuff
they could just as easily say that an arrow is moving along the edges of a cube[/quote]

Good point.

Or it could be someone painting an arrow along a cube.

I was trying to practice cheesing and this is a great problem for that.

We try to find the total number of paths
Notice that this number must be divisible by $3$ because each of the first three choices are symmetric. Consider where we are now. We have exactly two choices which are both symmetric so the answer must be divisible by $6$. $\boxed{C}$


topology
advanced fields
advanced fields unsolved

Show that if $X$ is a Hausdorff space that is locally compact at the point $x$, then for each neighbourhood $U$ of $x$, there is a neighbourhood $V$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subset U$.

First we prove a lemma: If $X$ is Hausdorff and compact, $U$ is an
open set in $X$, $x\in U$, then there exists an open set $V$
containing $x$ such that $\overline V \subseteq U$. The proof is
easy. $U^c$ is closed in $X$, $x\notin U^c$, then there exist open
neighbourhoods of $x$ and $U^c$, say $O_1$ and $O_2$. Then $O_1$
satisfies the conditions.

Suppose $K$ is a compact neighbourhood of $x$. $K$ is Hausdorff and
compact. $K\cap U$ is open in $K$, hence there exists open set $W$
in $K$, $x\in W$, such that $\overline W|_K \subseteq K\cap U\subseteq U$. Since $K$ is closed $X$, we have $\overline W|_K = \overline W \cap K = \overline W$ (because $\overline W\subseteq \overline K = K$). Let $V=K^\circ\cap W$, it is open in $X$ and
contains $x$. $\overline V$ is what we desired, as $\overline V\subseteq \overline W\subseteq U$.

What is $K^\circ$? And is $\bar{V}$ compact?

$K^\circ$ is interior of $K$
$\overline V$ is closed in $K$, since $K$ is compact, $\overline V$ is compact.


real analysis
integration
geometry unsolved
geometry

We have given $n$-dimensional hypersphere $S$ of radius 1. Pick a point $p$ on the surface of $S$ and draw a $n$-dimensional hypersphere $S'$ with radius $r$ and center $p$. Determine $r$ if $\frac{m(S \cap S')}{m(S)}=\frac{1}{2}$, where $m$ stands for $n$-dimensional Lebesgue measure.

The answer can be found by solving the equation
$\int_{r^{2}/2}^{r}(r^{2}-x^{2})^{(n-1)/2}dx+\int_{1-r^{2}/2}^{1}(1-x^{2})^{(n-1)/2}dx=\int_{0}^{1}(1-x^{2})^{(n-1)/2}dx$,
with respect to $r$ where $1<r<2$. This is possible only numerically.


counting
derangement
induction
linear algebra
combinatorics proposed
combinatorics

Find the number of even permutations of $ \{1,2,\ldots,n\}$ with no fixed points.

It is known that an even permutation of this kind is called an even [i]derrangement[/i]. In the following, for sake of clarity, denote by $ O_{n}$, $ E_{n}$, and $ D_{n}$ the numbers of odd, even, and all [i]derangements[/i] of the $ n$ numbers.  Now, I shall figure out a connection between $ O_{n}$ and $ E_{n}$, the conclusion following then from the obvious fact that $ O_{n} \plus{} E_{n} \equal{} D_{n}$.

First, it is clear that $ E_{1} \equal{} O_{1} \equal{} 0$, $ E_{2} \equal{} 0$, $ O_{2} \equal{} 1$. From the initial ordering of the $ n$ numbers, interchange the first and the $ j$-th numbers. There will be $ E_{n \minus{} 2}$ derangements of the remaining $ n \minus{} 2$ letters which lead to odd derangements of the entire set. Since $ j$ may be any one of the integers $ 2$, $ 3$, $ \ldots$ , $ n$, there are $ (n \minus{} 1)E_{n \minus{} 2}$ odd derangements of this type. Again, starting with the original arrangement of the $ n$ integers, consider the first one followed by any one of the $ E_{n \minus{} 1}$ even derangements of the last $ n \minus{} 1$ numbers. Now interchange the two numbers from the first and $ j$-th positions, and we obtain an odd derangement of all $ n$ numbers. There are $ (n \minus{} 1)E_{n \minus{} 1}$ odd derangements of this type. Thus, $ O_{n} \equal{} (n \minus{} 1)(E_{n \minus{} 1} \plus{} E_{n \minus{} 1})$, and likewise, $ E_{n} \equal{} (n \minus{} 1)(O_{n \minus{} 1} \plus{} O_{n \minus{} 2})$.
In conclusion, by a small induction, one can deduce that $ O_{n} \minus{} E_{n} \equal{} ( \minus{} 1)^{n}(n \minus{} 1)$, and since $ O_{n} \plus{} E_{n} \equal{} D_{n}$, we have that $ E_{n} \equal{} \frac {1}{2}(D_{n} \plus{} ( \minus{} 1)^{n\minus{}1}(n \minus{} 1))$, where $ D_{n} \equal{} n!\left(1 \minus{} \frac {1}{1!} \plus{} \ldots \plus{} \frac {( \minus{} 1)^{n}}{n!}\right)$, as a consequence, for example, of the Inclusion-Exclusion Principle.

I suspect that my approach is far from beeing new, but this is what I've found. I am also aware of a solution using some linear algebra, from "Recounting the Odds of an Even Derangement", [i]Mathematics Magazine[/i].


The rate of change of acceleration is sometimes known as "jerk." Find the direction and magnitude of jerk for a particle moving in a circle of radius $ R$ at angular velocity $ \omega$. Draw a vector diagram showing the instantaneous position, velocity, acceleration, and jerk.

So the particle has velocity $ \bold{v} \equal{} v_{r} \bold{\hat{r}} \plus{} v_{\theta} \bold{\hat{\theta}}$ (a radial and angular component). So would I just differentiate this twice? How would I incorporate the $ R$ and $ \omega$? How would the diagram look like?

[quote="arbelos08"]is there any application for jerk?[/quote]

There's more applications for jerk than snap, crackle and pop

[quote]"Many other suggestions have been made, such as jilt, jouse, jolt, and delta jerk. As more distinct terms that start with letters other than "j", the term snap has been proposed for the 4th derivative of position, with "crackle" and "pop" having been suggested - facetiously - as names for the 5th and 6th derivatives"
[/quote]
Wikipedia


inequalities
logarithms
inequalities proposed

Let $ a,b,c\in (0,1)$ and $ x,y,z\in (0, \plus{} \infty)$ be six real numbers such that
\[ a^x \equal{} bc , \quad b^y \equal{} ca , \quad c^z \equal{} ab .\]
Prove that
\[ \frac 1{2 \plus{} x} \plus{} \frac 1{2 \plus{} y} \plus{} \frac 1{2 \plus{} z} \leq \frac 34 .\]

[i]Cezar Lupu[/i]

so nice Cezar....
You are really a master on ineqs....

put $x=\frac{k+l}{m}$ and so on.....

Is this the way to proceed?? :)

[quote="Valentin Vornicu"]$m,n,p\in (0,+\infty)$[/quote]
Where do we need this assumption?

$x=\frac{\log{bc}}{\log{a}}$ and so on. Let $p=-\log{a}, q=-\log{b}, r=-\log{c}$, then $x=\frac{q+r}{p}$ and so on. ($p,q,r$ are positive)
$\sum \frac{1}{2+x}=\sum \frac{p}{2p+q+r}=3-\sum \frac{p+q+r}{2p+q+r}=3-(p+q+r) \sum \frac{1}{2p+q+r} \leq 3-(p+q+r) \frac{9}{4p+4q+4r}$ by AM-HM inequality or cauchy. RHS is equal to $\frac{3}{4}$.

Suppose $a,b,c\in (0,1)$ or $a,b,c\in (1,\infty)$ and define $u=|\lg a|$, $v=|\lg b|$, $w=|\lg c|$. Thus $x=\frac{v+w}{u}$ a.s.o. and $\frac{1}{2+x}=$ $1-\frac{u+v+w}{2u+v+w}$ a.s.o. Therefore, $\sum \frac{1}{2+x}=$ $3-\sum u\cdot \sum \frac{1}{2u+v+w}=$ $3-\frac 14\sum (2u+v+w)\sum \frac{1}{2u+v+w}\le$ $3-\frac 94\le \frac 34$ $\Longrightarrow$ $\boxed {\sum \frac{1}{2+x}\le \frac 34}\ .$

$a^x=bc$ , so $a^{x+1}=abc$ , so $x+1=\log_a abc$. put $x+1=p , y+1=q , z+1=r$ etc.
so , note that , $\sum {\frac{1}{p}}=1$
now , we consider $f=\frac{3}{4}-[\sum{\frac{1}{p+1}}]$ , $g=[\sum{\frac{1}{p}}]-1=0$
now ,$L=f-\lambda g$.calculating $\frac{\partial L}{\partial p}$ etc ,and using LM method,
we arrive at our desired conclusion! :D


A chemist has 10 mL of a 20% salt solution. How many milliliters of water should she add to produce a 5% salt solution?
(yes I know this is a question from last year's Feb regional)

At first there are $ 10\cdot0.2\equal{}2$ milliliters of salt. Thus, the amount of water we need to add satisfies the equation

$ 0.05\equal{}\frac1{20}\equal{}\frac2{10\plus{}x}\implies20\equal{}\frac{x\plus{}10}2\equal{}\frac x2\plus{}5\implies15\equal{}\frac x2\implies x\equal{}\boxed{30}$.

Have you ever heard of the "bucket" method? It may be a good place to start.


What do you expec your score to be in the State Mathcounts this year?

Uhh... I think this *might* warrant a lock because A) it's sorta spammy and B) state competitions will start March 3rd, so some people would vote AFTER  the competition... you know?

Estimating your score for the State competition reveals information about the difficulty of Chapter competition.  The only time a poll would be permitted is during the (very) short window between the close of the Chapter window and the start of the State window.  Poll deleted.  Score estimates deleted.  Thread locked.


function
calculus
derivative
complex analysis
complex analysis unsolved

Let $f$ be an anlytic function in the open unit disk of the complex plane such that $|f(z)|\leq \frac{C}{1-|z|}$ for all $z$ in the unit disk, where $C$ is a positive constant.Prove that $|f(z)'|\leq \frac{C}{1-|z|}$

This is false, as is shown by the simple example of $f(z)=\frac1{1-z}.$ I assume that this whole topic is simply a typographical error and you posted the correct problem as another topic.

Yes,at first it was as an error, and after i decided not to ask moderators to remove the topic; if you gave an example i think it was a bit useful error


[quote="ashrafmod"]$\frac{1}{log_{\frac{1}{2}}{x}}+\frac{1}{log_{\frac{2}{3}}{x}}+\frac{1}{log_{\frac{3}{4}}{x}}....+\frac{1}{log_{\frac{9}{10}}{x}}=1$[/quote]
[hide]With the inverse property we are allowed to swithch the log and base since the reciprocal is being taken.
So, this equation is equal to $\log_{x}\frac{1}{2}+\log_{x}\frac{2}{3}+...+\log_{x}\frac{9}{10}=1$
The product property tells us that $\log_{x}\frac{1}{10}=1$, so $x=\frac{1}{10}$[/hide]


1.  Show that there are no positive integers $m, n$ such that $4m(m+1) = n(n+1)$

Consider the equation as a quadratic in $n$. Solve for $n.$ Since $m > 0,$ we get $n=m$ as the only possible solution. But substituting $n$ for $m$ in the original equation yields a contradiction.

I remember this being posted somewhere else before.

[hide]$(2m+1)^{2}=n^{2}+n+1$
$n^{2}<n^{2}+n+1<(n+1)^{2}$

$n^{2}+n+1$ is a square between two consecutive squares, a contradiction.[/hide]

[quote="SplashD"]I remember this being posted somewhere else before.[/quote]Yeah... http://www.mathlinks.ro/Forum/viewtopic.php?p=581542#581542

Follow-up:

Describe the possible integer values of $\frac{n(n+1)}{m(m+1)}, n, m \in \mathbb{N}$.


function
number theory open
number theory

Find all positive numbers $n$ such that $\phi (n)$ divides $n-1$, where $\phi (n)$ is Euler's totient function.

I am not sure that this problem isn't very easy or hasn't been discussed before, but i would like to know.
Thanks

It is an open problem. D.H.Lehmer conjectured that there is no composite value of $n$ such that $\phi(n) | n-1$. Such an $n$ must be a Carmichael number.

See, for example, http://mathworld.wolfram.com/LehmersTotientProblem.html or Guy UPINT3 B37.


calculus
function
real analysis
real analysis unsolved

i was doing my calculus exam last week, and i got to the final step of :
0 = e^2x^2 which in English is "e" to the power of 2x squared, find x.  I couldn't find x, i tried to "ln" both sides, but ln0 is undefined, so how do i find x?

Use parentheses; you wrote $ e^2\cdot x^2$.

An exponential is never zero.

it does not exist, $ e$ is a strictly positive function


Given 4 points $ O(0,\ 0),\ A(3,\ 0),\ B(2,\ 2),\ C(4,\ 1)$ in the $ xy$ plane.

(1) The point $ P(x,\ y)$ satisfies the relation $ |3\overrightarrow {OP}\minus{}\overrightarrow{OA}\minus{}\overrightarrow{OB}\minus{}\overrightarrow{OC}|\equal{}3$.  Find the equation for which $ x,\ y$ should satisfy.

(2) When the point $ P$ moves satisfying the condition of (1), find the coordinate of $ P$ for which the area of $ \triangle{PAC}$ is maximized.

Here's what I have for (1), I'll post (2) when I get around to it

[hide="(1)"]
Squaring the given relation yields
\[ 9OP^2 \plus{} 2(OA^2\plus{}OB^2\plus{}OC^2)\minus{}3\vec{OP}(\vec{OA}\plus{}\vec{OB}\plus{}\vec{OC})\minus{}\frac{1}{2}(AB^2\plus{}BC^2\plus{}AC^2)\equal{}9\]
Clearly, $ OP^2 \equal{} x^2 \plus{} y^2$. Also, we can find that \[ \vec{OP}\cdot\vec{OA}\equal{}\displaystyle\frac{OP^2\plus{}OA^2\minus{}AP^2}{2}\equal{}\displaystyle\frac{(x^2\plus{}y^2)\plus{}(3)^2\minus{}[(x\minus{}3)^2\plus{}(y\minus{}0)^2]}{2}\equal{}3x\]
Similarly, $ \vec{OP}\cdot\vec{OB}\equal{}2x\plus{}2y$ and $ \vec{OP}\cdot\vec{OC}\equal{}4x\plus{}y$. Therefore,
\[ 9(x^2\plus{}y^2)\plus{}2(9\plus{}8\plus{}17)\minus{}3(3x\plus{}2x\plus{}2y\plus{}4x\plus{}y)\minus{}\frac{1}{2}(5\plus{}5\plus{}2)\equal{}9\]
\[ \boxed{9x^2\plus{}9y^2\minus{}27x\minus{}9y\plus{}53\equal{}0}\]

[/hide]

[quote="Tenoreoz"]Here's what I have for (1), I'll post (2) when I get around to it

[hide="(1)"]
Squaring the given relation yields
\[ 9OP^2 \plus{} 2(OA^2 \plus{} OB^2 \plus{} OC^2) \minus{} 3\vec{OP}(\vec{OA} \plus{} \vec{OB} \plus{} \vec{OC}) \minus{} \frac {1}{2}(AB^2 \plus{} BC^2 \plus{} AC^2) \equal{} 9
\]
Clearly, $ OP^2 \equal{} x^2 \plus{} y^2$. Also, we can find that
\[ \vec{OP}\cdot\vec{OA} \equal{} \displaystyle\frac {OP^2 \plus{} OA^2 \minus{} AP^2}{2} \equal{} \displaystyle\frac {(x^2 \plus{} y^2) \plus{} (3)^2 \minus{} [(x \minus{} 3)^2 \plus{} (y \minus{} 0)^2]}{2} \equal{} 3x
\]
Similarly, $ \vec{OP}\cdot\vec{OB} \equal{} 2x \plus{} 2y$ and $ \vec{OP}\cdot\vec{OC} \equal{} 4x \plus{} y$. Therefore,
\[ 9(x^2 \plus{} y^2) \plus{} 2(9 \plus{} 8 \plus{} 17) \minus{} 3(3x \plus{} 2x \plus{} 2y \plus{} 4x \plus{} y) \minus{} \frac {1}{2}(5 \plus{} 5 \plus{} 2) \equal{} 9
\]

\[ \boxed{9x^2 \plus{} 9y^2 \minus{} 27x \minus{} 9y \plus{} 53 \equal{} 0}
\]
[/hide][/quote]

Why did you go about such a long method when u could have used rectangular components $ \hat{i}$ and $ \hat{j}$??

Does anyone have a solution for the second part? I couldn't think of a less calculation-intense way...


complex analysis
complex analysis unsolved

this problem is also of two parts:

1) show that the complementary components of a closed curve in the complex plane are simply connected.

2) show that the single valued analytic branches of logz,z^a,and z^z can be
defined in any simply connected region that does not
contain the origin.

thx

Please: change the headings of your posts!
All the same 'Complex analysis', it's hard to distinguish them!

Sorry u are right i did not pay attention that all the headings of my posts have the same name

sorry again but i do not know how to change the headings.
Second time i will pay attention.
thx for ur advice

I'm not totally sure, but clicking on 'Edit' in the first post and changing the text in 'Subject' could work.


f deined on[0,1] be twice differentiable function such that [size=150]mod(f''(x))<=1 [/size]for all x in [0,1].

if f(0)=f(1) then show that mod(f'(x))<1 for all x in[0,1]

First note by applying the Mean Value Theorem to f, there exists

$ \theta \in (x,y)$ such that $ \frac {f(y) \minus{} f(x)}{y \minus{} x} \equal{} f'(\theta)$

this is also true for x = 0 and y = 1 so that we have $ 0 \equal{} f(1) \minus{} f(0) \equal{} f'(\theta)$.

Note that f' is continuous on [0,1] and f''(t) exists for every $ t \in (0,1)$ imply that for distinct values x, y in [0,1] there exists $ \theta$ between x and y such that one has:

$ f(y) \equal{} f(x) \plus{} f'(x) (y \minus{} x) \plus{} f''(\theta)\frac {(y \minus{} x)^2}{2}$

$ \implies |f'(x)| \equal{} |\frac {f(y) \minus{} f(x)}{y \minus{} x} \minus{} f''(\theta)\frac {(y \minus{} x)}{2} |$
$ \equal{} |f'(\theta) \minus{} f''(\theta)\frac {(y \minus{} x)}{2}| \leq |f''(\theta)| \leq 1$

I don't think this is right.

The $ \theta$ depends on x and y.

I'll follow akech's idea:

$ 0\equal{}\frac{f(1)\minus{}f(0)}{1\minus{}0}\equal{}f'(\xi),\xi\in(0,1)$

$ \Rightarrow \left|\frac{f'(x)\minus{}f'(\xi)}{x\minus{}\xi}\right|\equal{}\left|f''(\theta)\right|<1$

Now if $ f'(x)>1,x\in[0,1]$ you get a contradiction.

Since $ f(0)\equal{}f(1)$,

By $ Rolle's$ we must have some $ x\equal{}c$ where $ c\in(0,1)$ for which $ f'(c)\equal{}0$

Consider the interval $ [x,c]$,for any arbitrary $ x\in[0,c]$.By $ LMVT$ for $ y\equal{}f'(x)$ on $ [x,c]$ we have

$ f''(\theta)\equal{}\frac{f'(c)\minus{}f'(x)}{c\minus{}x}$,where $ \theta \in(x,c)$

$ f''(\theta)\equal{}\frac{f'(x)}{x\minus{}c}$

$ f'(x)\equal{}(x\minus{}c)f''(\theta)$

$ |f'(x)|\equal{}|x\minus{}c\parallel{}f''(\theta)|\leq 1$

Similarly,

Consider the interval $ [c,x]$,for any arbitrary $ x\in[c,1]$.By $ LMVT$ for $ y\equal{}f'(x)$ on $ [c,x]$ we have

$ f''(\alpha)\equal{}\frac{f'(x)\minus{}f'(c)}{x\minus{}c}$,where $ \alpha \in(c,x)$

$ f''(\alpha)\equal{}\frac{f'(x)}{x\minus{}c}$

$ f'(x)\equal{}(x\minus{}c)f''(\alpha)$

$ |f'(x)|\equal{}|x\minus{}c\parallel{}f''(\alpha)|\leq 1$

Thus, $ |f'(x)|\leq 1$ for all $ x\in (0,1)$


geometry
incenter
cyclic quadrilateral
geometry proposed

Let $ ABC$ be a triangle and incenter touch $ BC,CA,AB$ at $ A',B',C'$ let $ D$ is projection of $ A'$ on $ B'C'$ let $ Ge$ is Gergonne point of $ ABC$ assume that, the circles $ (DGeB)\cap (DCA) \equal{} \{D,E\},EC\cap (DGeB) \equal{} \{E,M\},EA\cap (DGeB) \equal{} \{E,N\}$ prove that $ MN \equal{} BGe$.

It is complicated  :(
could you give me some clue?

[quote="encyclopedia"]Hi dear kaka__2004, they are in a same circle so you should prove measure of arc $ MN$ equal measure or ar $ BGe$.[/quote]
I don't know how to use these conditions :blush:
could you please post your solution for this problem and [url]http://www.mathlinks.ro/viewtopic.php?t=263806[/url]
hanks,dear [b]encyclopedia[/b]
ps:how did you find these nice problems?by your own or books? :)

I created this problem own, here is my solution

Actually this problem is true for all $ P$ in plane, Gergone point only the concurrent point of $ AA',BB',CC'$, I can write it as

Problem. Let $ ABC$ be a triangle and point $ P$, $ A'B'C'$ is Cevian triangle of $ P$, $ D$ is projection of $ P$ on $ B'C'$ assume that, the circles $ (DPB)\cap (DCA) = \{D,E\},EC\cap (DPB) = \{E,M\},EA\cap (DPB) = \{E,N\}$ prove that $ MN = BP$.

Lemma. $ \angle BDP+\angle ADC=\pi$

Solution of lemma. Let $ A''=PA\cap B'C'$ let $ DA\cap BC=\{K\}$, we can prove $ (APA''A')$ is harmonic, $ DA'\perp DB'$ this mean $ DA'$ is bisector of $ \angle KDC$, more if $ B'C'\cap BC=\{A'''\}$ then $ (BCA'A''')$ is harmonic and $ DA'\perp DB'$, this mean $ DA'$ is bisector of $ \angle BDC$, from two claim we can show $ \angle BDP=\angle KDC$ but $ \angle KDC+\angle ADC=\pi$, we are done.

Solution my problem. $ ADEC$ is cyclic quadrilateral we have $ \angle ADC=\angle AEC$ but $ \angle MEA+\angle AEC=\pi$ from lemma we get $ \angle MEN=\angle BDP$ this is succifient to show $ MN=BP$ because they are the arc of same circle.


Round 7 Matchups

Lucky707  vs.  mlee06

yifman12  vs.  hashbrownie

mat_88 is the only one left in the tournament who hasn't been beaten.

I will get the problems out today or tomorrow.

Bah, I got number three wrong. Noticed it when I sent it.

could you post an updated bracket please?

If I have time I ll do it tonight.  If not I will try to do it tomorrow morning.

Results for this round thus far:

yif man12 4    vs.   Hashbrownie 4

I will get tiebreakers out hopefully soon.

mlee06  3    vs.  Lucky707

Lucky, please get your answers in ASAP.

I will post the updated bracket at the conclusion of this round.

It s still going.  I haven't gotten the tiebreakers back.  But don't worry it is still in progress.  Hopefully I can get it done soon.

Most recent news:

Lucky707  def.  mlee06 4-3

There is, tomorrow. I may have time tonight after I finish my research and homework, but I'll definitely have time tomorrow night. Sorry about the delays on the tie-breakers. School's a killer.

Hashbrownie def.  yif man12

It was a narrow margin but Hashbrownie pulled it out.

Next matchup is

Hashbrownie vs. Lucky707

I ll get the problems out hopefully soon.

Btw,  here are the problems from last round:

1.	Is the number $\binom{1000}{500}$  divisible by 7?

2.	Which of the following triples could not be the lengths of the three altitudes of a triangle?

(a)	1, $\sqrt{3}$, 2

(b)	 3, 4, 5

(c)	5, 12, 13

(d)	7, 8, $\sqrt{113}$

(e)	8, 15, 17

3.	Show that $n^2+n-1$ and $n^2+2n$ have no common factor.

4.	Let $f(x)=|x-2|+|x-4|-|2x-6|$.  Compute the sum of the largest and smallest values of $f(x)$.

5.	If n is a positive integer such that 2n has 28 divisors and 3n has 30 divisors, then how many divisors does 6n have?

Tiebreakers

1.	If $h_a$ is the altitude to side a in triangle ABC and likewise with $h_b$ and $h_c$, then prove that \[\frac{1}{h_a}+\frac{1}{h_b}>\frac{1}{h_c}.\]

2.	Find the maximum value of the function, $f(x)=\frac{x}{x^2+1}$, which is defined over the reals.

3.	P(7,12,10), Q(8,8,1) and R(11,3,9) are three vertices of a cube.  What is the surface area of the cube?

This tournament is almost over.  I doubt that I will be able to host any tournaments for a while due to time constraints (i.e. school).  So unless someone else would like to host one you may have to wait a while.

I don't know whether joml88 will host anymore tournaments soon but if you want to do some math competition, you can always join to any competition in this site. (no prize for most cases)

Like mine :)


Is there a characterization of those subsets $A$ of the unit circle with the following property: there exists a power series $\sum_{n\geq 0}{a_n z^n}$ with radius of convergence 1 and such that the series defining it converges only on point of $A$ on the unit circle?

I checked with a few experts and they all said that full characterization of such sets is still unknown. So, I'm moving this topic to the Open Questions subforum. Of course, everybody is welcome to prove me wrong. :P


Minimize $\frac{a^{2}}{b}+\frac{(1-a)^{2}}{1-b}$ for positive reals $a,b<1$.

I'd have thought AM-GM would kill it straight away, but I'm having problems. Is there something simple that I'm missing?

By Cauchy in Engel form:
$\frac{a^{2}}{b}+\frac{(1-a)^{2}}{1-b}\geqslant \frac{(a+1-a)^{2}}{b+1-b}= 1$.
The equality holds for $a=b=\frac12$.  :)


$ 99099$ is the largest $ 5$-digit palindrome that is a multiple of $ 99$.  Compute the smallest $ 5$-digit palindrome that is a multiple of $ 99$.

[hide="solution"]
Let the palindrome be $ \overline{abcba}$.
note that $ 99\equal{}9\times 11$,so if $ 99|n$ then $ 9|2a\plus{}2b\plus{}c$ and $ 11|\minus{}2a\plus{}2b\minus{}c$.

Let us consider in $ 1\le a\le 4$ first,because we are now seaching the least,then
$ 2\le 2a\plus{}2b\plus{}c\le 35$ and $ \minus{}17\le \minus{}2a\plus{}2b\minus{}c\le 16$,so we have
$ 2a\plus{}2b\plus{}c\in \{9,18,27\}$ and $ \minus{}2a\plus{}2b\minus{}c\in \{\minus{}11,0,11\}$,
and easily see that there are no solution of $ a,b,c$ such that $ 1\le a\le 4$.(note that sum of $ 2a\plus{}2b\plus{}c$ and $ \minus{}2a\plus{}2b\minus{}c: 4b$ is divisible by $ 4$).

if $ a\equal{}5$,we have $ 10\le 10\plus{}2b\plus{}c\le 37$ and $ \minus{}19\le \minus{}10\plus{}2b\minus{}c\le 8$,hence
$ 2a\plus{}2b\plus{}c\in \{18,27,36\}$ and $ \minus{}2a\plus{}2b\minus{}c\in \{\minus{}11,0\}$,and easily give that only $ (a,b,c)\equal{}(5,4,9)(5,9,8)$ work.

we conclude the smallest palindrome is $ 54945$.[/hide]

[quote="Octahedr0n"][quote="Anavel_Gato"]if $ 99|n$ then $ 9|2a \plus{} 2b \plus{} c$ and $ 11| \minus{} 2a \plus{} 2b \minus{} c$.[/quote]

I don't get how you got this.  Could you please explain a bit more?  Thanks![/quote]

$ n\equal{}\overline{abcba}$.

I used the [b]divisibility test[/b] for 9 and 11.(proofs of which are both easy.)

A positive integer is divisible by $ 9$ if and only if the sum of its digits is also divisible by $ 9$,
thus if $ 9|\overline{abcba}$ then $ 9|a\plus{}b\plus{}c\plus{}b\plus{}a\equal{}2a\plus{}2b\plus{}c$.

similarly,A positive integer is divisible by $ 11$ if and only if the difference between the sum of the odd numbered digits and the sum of the even numbered digits is divisible by $ 11$,
thus if $ 11|\overline{abcba}$ then $ 11|(b\plus{}b)\minus{}(a\plus{}c\plus{}a)\equal{}\minus{}2a\plus{}2b\minus{}c$.

now you can see that?

[quote="Octahedr0n"][quote="Anavel_Gato"]$ 2a \plus{} 2b \plus{} c\in \{9,18,27\}$ and $ \minus{} 2a \plus{} 2b \minus{} c\in \{ \minus{} 11,0,11\}$,[/quote]

Yes, I get the divisibility stuff now.  However, I don't get the above.  Thanks![/quote]

We showed $ 2a\plus{}2b\plus{}c$ is a multiple of $ 9$,and now $ 2\le 2a\plus{}2b\plus{}c\le 35$,hence the possible cases are
$ 2a\plus{}2b\plus{}c\equal{}9,18,27$.

we can do the same way for $ \minus{}2a\plus{}2b\minus{}c$.


When the numerator of a common fraction is simplest form is decreased by 6 and the denominator is decreased by 9, the resulting fraction is equivalent to the original fraction. Express the original number as a common fraction.

We see that if the original is $ \frac{x}{y}$, the new one is $ \frac{x\minus{}6}{y\minus{}9}$, so $ \frac{x}{y}\equal{}\frac{x\minus{}6}{y\minus{}9}$. Cross multiplying gives that $ xy\minus{}9x\equal{}xy\minus{}6y$, so $ 6x\equal{}9y$. Dividing by 6y gives $ \frac{x}{y}\equal{}\boxed{\frac{2}{3}}$.

I am still  a bit  :huh:, please explain?


what does $\bar{w+z}=\bar{w}+\bar{z}$ means? I need help with what the notation meant.

Use \overline instead of \bar for those big ones.

If $z=a+bi=re^{i\theta}$, then $\bar{z}=a-bi=re^{-i\theta}$. $\bar{z}$ is called the complex conjugate of $z$.

$\overline{(w+z)}=\bar{w}+\bar{z}$ is one of the properties of complex conjugates. It means that the complex conjugate of a sum of two complex numbers is the sum of the complex conjugates of those numbers.

Proof.
If $w=a+bi$ and $z=c+di$, then $w+z=(a+c)+(b+d)i$.
$\overline{(w+z)}=(a+c)-(b+d)i=(a-bi)+(c-di)=\bar{w}+\bar{z}$

Basically, the conjugate of $a+b\sqrt{c}$ is $a-b\sqrt{c}$. In this case, $c=-1$. So $\overline{z}=\overline{a+bi}=a-bi$.

Right, ok I also can't follow through this:

If the complex number $z=a+bi$ is a root of $f(x)$, we have
$f(z)=a_{n}z^n + a_{n-1}z^{n-1} + ... + a_{1}z + a_{0} = 0$

Now we use some of our useful propertyies of complex numbers, such as $\overline{w+z}=\bar{w}+\bar{z}, \bar{z^k}=(\bar{z})^k,$ and w=z implies $\bar{w}=\bar{z}$. Applying these principles to $f(z)$ we have

here we go:

got to go, I will post more later


function
real analysis
real analysis solved

P0(x)=0
Pn+1(x) = Pn(x) + (x^2-Pn(x)^2)/2, n=0,1,2,...
Show that the sequence of functions {Pn(x)} uniformly converges to |x| on [-1,1]

Yes, he is sure.

Since Pn(x) is even function for every n it is sufficient to prove statement for x \in [0,1].

We have if 0 \leq Pn(x) \leq x then Pn(x) \leq Pn+1(x) \leq x. So Pn(x) is monotonic in each point x. It follows that Pn(x) -> P(x) in each point x, and P(x)+(x2-P2(x))=P(x) ==> P(x)=x.

Thus sequence Pn(x) is monotonic and converges to x on [0,1]. Due to Dienes's theorem Pn(x) uniformly converges to x on [0,1].


hey here is a fun thing to do whenever you feel bored
http://www.artofproblemsolving.com/Forum/weblog.php?w=1040 come here and spam by pressing f5 and get more views! also, there is a f5 spam-fest on friday the thirteenth at 7:30 et so if you can come please do so!

7:30 evening or morning?

[hide]
Just wanted to post something, don't really care  :D
[/hide]

If you come and get at least 5K views for him, tell we'll tell stevenmeow to make you a contributor :!:  :!:  :!:

Texas is in central time zone, not eastern. why's he saying it in eastern?

um obviously because texas people aren't the only people who go to stevenmeow's blog and because et is universal
and clearly you didn't even read my post because it says 7:30 PM

[quote="ra5249"]hey here is a fun thing to do whenever you feel bored
http://www.artofproblemsolving.com/Forum/weblog.php?w=1040 come here and spam by pressing f5 and get more views! also, there is a f5 spam-fest on friday the thirteenth at [b]7:30 et[/b] so if you can come please do so![/quote]

 :wink:  :roll:  :P

DON'T GIVE VIEWS TO STEVENMEOW'S BLOG!!!!!!!!!!!!!!!!!!!!!

It is a waste of time. You could be learning just a little more math. So on the day that you die, you will have just the knowledge to prove the Riemann Hypothesis, or some other famous conjecture, and the world will have everlasting peace. But if you waste your time pressing F5, you will just be behind a little on your math, and then you will not make it on your dying day, and the world will be doomed. This is the generation of past, present, and future people. The world will depend on us. Yes, YOU. and YOU.

 [hide]
 :rotfl:
[/hide]

erm by post i meant post on his blog
which clearly you were too lazy to click on :O

it was a big hit, stevenmeow is now first in views!!!

btw stupidityismygam should be a contributor

I'll pretend that I'm not confused.

 :huh:


The sequences R_n are defined as follows. R_1 = (1). If R_n = (a_1, a_2, ... , a_m), then R_(n+1) = (1, 2, ... , a_1, 1, 2, ... , a_2, 1, 2, ... , 1, 2, ... , a_m, n+1).

For example, R_2 = (1, 2), R_3 = (1, 1, 2, 3),
R_4 = (1, 1, 1, 2, 1, 2, 3, 4).

Show that for n > 1, the kth term from the left in R_n is 1 iff the kth term from the right is not 1.

This was on IMO short or long-list in the 1990's, so probably already solved on ML.


How many two-digit positive integers have at least one $ 7$ as a digit?

$ \textbf{(A)}\ 10\qquad
\textbf{(B)}\ 18\qquad
\textbf{(C)}\ 19\qquad
\textbf{(D)}\ 20\qquad
\textbf{(E)}\ 30$

[hide]
There are 9 numbers with a 7 in the units digit, 9 numbers with a 7 in the tens digit and some other number in the units, and 1 number with a 7 in the tens digit and a 7 in the units. This makes a total of 18, so B is our answer.
[/hide]

[hide="Did you make some typoes?"]
There are $ 10$ two-digit numbers with a $ 7$ as their $ 10$'s digit and $ 9$ two-digit numbers with $ 7$ as their units digit.  Because $ 77$ satisfies both of these, we subtract it out once (for double-counting) so the answer is $ 10\plus{}9\minus{}1\equal{}18$, or $ \boxed{\textbf{(B)}}$.
[/hide]


Two fair coins are flipped. What is the probability that at least one coin is heads? Express your answer as a common fraction.

Use complementary probability.

Find the probability that none of the flips were heads.  This means that both the flips were tails.

$ \frac12 \times \frac12 \equal{} \boxed{\frac14}$

You then subtract this from 1 to get 3/4  :P


Prove that polynomial $ f\equal{}X^n\plus{}a_{1}X^{n\minus{}1}\plus{}...\plus{}a_{n\minus{}3}X^3\plus{}a_n \in R[X]$ can't have all real roots, for  $ a_n \neq 0$.

[quote="soruz"]Prove that polynomial $ f \equal{} X^n \plus{} a_{1}X^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 3}X^3 \plus{} a_n \in R[X]$ can't have all real roots, for  $ a_n \neq 0$.[/quote]
but $ f\equal{}(X\plus{}1)^n$ have all real roots

[quote="soruz"]Prove that polynomial $ f \equal{} X^n \plus{} a_{1}X^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 3}X^3 \plus{} a_n \in R[X]$ can't have all real roots, for  $ a_n \neq 0$.[/quote]

Let $ r_i$ the roots (complex or real) of $ f$. $ a_n\neq0$ implies $ r_i\neq 0$

Coefficient of $ x$ in $ f$ is zero so $ (\prod_i r_i)(\sum_i\frac{1}{r_i})\equal{}0$ and so $ A\equal{}\sum_i\frac{1}{r_i}\equal{}0$
Coefficient of $ x^2$ in $ f$ is zero so $ (\prod_i r_i)(\sum_{i\neq j}\frac{1}{r_ir_j})\equal{}0$ and so $ B\equal{}\sum_{i\neq j}\frac{1}{r_ir_j}\equal{}0$

Then we have $ A^2\equal{}\sum_i\frac{1}{r_i^2}\plus{}2B$ and, since $ A\equal{}B\equal{}0$ : $ \sum_i\frac{1}{r_i^2}\equal{}0$

And so, since all $ \frac{1}{r_i}\neq 0$, Some $ r_k$ are complex numbers.


Solve the equation $ x^4\plus{}1\equal{}2\cdot\sqrt [4]{2x\minus{}1}$ .

$ x\equal{}1$ is a solution. Write the equation as $ x^4\minus{}1\equal{}2(\sqrt[4]{2x\minus{}1}\minus{}1)$, so that
\[ (x\minus{}1)(x\plus{}1)(x^2\plus{}1)\equal{}\frac{4(x\minus{}1)}{(\sqrt[4]{2x\minus{}1}\plus{}1)(\sqrt{2x\minus{}1}\plus{}1)}\]
Any solution other than 1 will satisfy
\[ (x\plus{}1)(x^2\plus{}1)(\sqrt[4]{2x\minus{}1}\plus{}1)(\sqrt{2x\minus{}1}\plus{}1)\equal{}4\]
The left side is increasing for $ x\geq \frac{1}{2}$. So there's a unique solution other than 1 that works.

To find the other solution, note that $ f(f(x))\equal{}x$ where $ f(x)\equal{}\frac{x^4\plus{}1}{2}$. So if $ a$ satisfies $ f(a)\equal{}a$, then $ a$ is a solution. This is true for
\[ 0\equal{}a^4\plus{}1\minus{}2a\equal{}(a\minus{}1)(a^3\plus{}a^2\plus{}a\minus{}1)\]
So the other solution is the unique real root of $ x^3\plus{}x^2\plus{}x\minus{}1$.

[hide="Another hint"]Denote $ \sqrt [4]{2x \minus{} 1} \equal{} y$ then we have $ \Bigg\{\begin{aligned} x^4 \plus{} 1 \equal{} 2y \\
y^4 \plus{} 1 \equal{} 2x\end{aligned}$[/hide]


This is just a topic for jokes....

I'll start

Three scientists were stranded on a native island. Suddenly, they were captured by a tribe of natives. The chieftain said, "If you can find the 10 of the holy fruit of the island, I will let you go." So the three scientists went out to search for the fruit.

Scientist 1 came back with plums. The chieftain said, "Now, shove the 10 plums up your bottom, without flinching or making any sound." Scientist 1 tried, but squeaked when he shoved the 6th plum up.

Scientist 2 was given the same task. But he had brought back small berries. On the 9th one, he laughed so hard that the chieftan executed him.

In heaven, Scientist 1 said to Scientist 2, "What were you doing? 1 more berry and you would've escaped!"

"Well, I would've, but I looked over the chieftain's shoulder and saw Scientist 3 carrying an armful of pineapples!"  :D

[quote="Treething"][hide]man that's nasty

here's a cleaner one

so a man's at a bar for the second consecutive night and he tells the bartender "man last night i was so drunk i blew chunks"

and the bartender goes "that's nothing, this man a week or so ago was so drunk that he crashed his car into his house, knocked over a candle, and burned the whole thing down"

then the man goes "no you don't understand, chunks is my dog"[/hide][/quote]

...that's not exactly clean either...

NO MORE DISGUSTING JOKES PLEASE

oh man did i say "blew chunks"?

errrgh i meant to say "blew up chunks"

freudian slip

okay here's a good one that's not sick

bushie walks into a bar and buys everybody a drink.  somebody says, "hey bushie, why are you so happy?"
then bushie said, "well, for the last six months, I've been working on a puzzle.  I didn't really spend any time doing presidential stuff and the country went bozonkers, but thats ok.  so, I finally finished the puzzle after working on it 24/7
and on the side of the box, it said, "for two to four years"

Three goons were taking a man to their boss.

man- But I'm not Dick Cheney! I'm Osama Bid Laden!
1st goon- The same Osama Bid Laden that crashed planes into the 1st world trade center?
man- Actually, my name is Bart Gambe.
2nd goon- The same Bart Gambe that's been taking pictures of my sister?
man- Well, ACTUALLY, my name is Klisov Gronae.
3rd goon- The same Klisov Gronae that banned cigarettes from the town until his wife persuaded to put cigarrettes back on buisness?
man- NOW I remember my name! Benedict Arnold!
all goons- The same Benedict Arnold that surrendered West Point to the hated British?

At the hideout:

boss- Nice to meet you. I'm Frolov Garona.
man- Dick Cheney.

ok. Here's a better one.

You know Pythagoras? He wasn't greek. He was a native american. He has 3 wives, or squaws, if you will. He put the first squaw on a deer hide, and she produced a bouncing baby boy. He put the second squaw on a bear hide, and she produced another bouncing baby boy. He put the third squaw on the hippopatomas, and she bore twins.

Conclusion: The squaw of the hippopatomas equals the squaws of the two hides.

So a man was good in his life, so he went up to heaven after he died. There was a long line at the gate though, so as he was waiting he noticed a bunch of clocks on the wall.

He asks, "What are those clocks for?"

Response: "Notice under each clock there is a name. Everytime that person tells a lie, his clock goes up by one minute. There, you can see Mother Theresa's. Hers hasn't moved one minute. There you can see Ghandi's. His has only moved up about three minutes."

The man looks around, and asks, "Where's George Bush's?"

Response: "Oh, that's in Moses's room. He's using it as a ceiling fan"

A similar one:

A wife, dying some time after her husband's death and reaching heaven, naturally starts looking for him.

Wife: Have you seen my husband?
Inquiry: His name please?
Wife: John.
Inquiry: Sorry too many John's here.
Wife: John Waters.
Inquiry: Even that is too common a name. Anything else you can specify?
Wife: Umm ... he told me before he died that for each time he has been unfaithful to me he'll turn around in his grave.
Inquiry: (brightening) Oh you mean John 'flywheel' Waters?

Bill Clinton always kept a box under his bed, and asked his wife never to look at what's inside. But one day, she did, and she found $\$123$ and two beer cans inside.

She asks what this means, and Bill explains, "Well,  I got a confession to make. Every time I cheated on you, I put a beer can inside the box."

Then she asked what the money was doing inside, and he said,  "Oh, I recycled some of the cans"


ARML
AMC
USA(J)MO
USAMO
geometry
3D geometry
Duke

As I already noted in the other topic, DE will definitely win.

Uhh.. Florida. Duh. We rock.  Well. If there was a category for style points.  we'd so win.

[quote="Karth"]Who will win ARML 2007?  :D[/quote]

We need to do well this year. I think last year we got big heads right before the competition, which hurt us slightly (even though we weren't cocky).

[quote="MysticTerminator"]there is no question

as long we return in the pink shirts, none can stop us[/quote]
yes pink rocks.

I hope SFBA wins

Do you mean Sam Keller as the secret weapon?

[quote="MysticTerminator"][quote="t0rajir0u"]WA has a secret freshman weapon, and Adam and Jeffrey are at their peaks.  This will be an [b]interesting[/b] year.  :wink:[/quote]

freshman + adam + jeffrey < berman + jeremy + noah + yasha[/quote]

Hey. I decline to be part of the less than side of an inequality.

[quote="t0rajir0u"]There is another :D[/quote]

Kerry Xing?

[quote="paladin8"]
Hey. I decline to be part of the less than side of an inequality.[/quote]

1/Jeffrey > 1/Arnav

(Jeffrey,Arnav > 0, of course)

I'm pretty sure a flying blue whale with eyes that can shoot laser beams could pwn you Arnav. Um... To answer your question...

I don't have an answer. Maybe because you pwn the IMO... although you might argue that pwn = 42, in which case I think otherwise, but whatever. I mean, Lehigh still has like... 8 USAMO qualifiers. That's decent. No IMOers, but then again, there's only six of them. We have a strong base. I believe.

That's why I say we should definitely kidnap Zeb and drag him to ARML with us....

Then we'll pawn EVEN MORE :D

[quote="Karth"]
dude

but java majorizes c++

because you can't deploy c++ on the internet without installing a whole bunch of stuff[/quote]

What? JAVA was written in C++, so it's sort of like gg.

And besides, C++ is about 3 times faster than JAVA.

And C++ is easier to write in.

And C++ has next_permutation().

And C++ has draw_teacup().

And C++ is allowed at the IOI.

And C++ is not used by CollegeBoard.

And C++ has the Standard Template Library.

And C++ does not have functions like Integer.parseInt().

And C++ has #define statements.

And C++ has explicit pointers.

And C++ has inline functions.

And C++ evaluates bools as ints and ints as bools.

And C++ has <?= and >?=.

And C++ calls bools bools, as opposed to booleans.

And C++ allows you to abuse switch statements by messing with the scope.

So you see there really is no reason to use JAVA unless your only goal is to do stuff to strings, and you don't care how fast your program takes to run, and you don't happen to know / be able to use Python. Or if you want to use the BigInteger class that JAVA has, though that itself is pretty awkward, and if you just write a BigInteger class for C++ once then you can use templates and stuff and be happy.

On another note, Haitao beat Arnav at Duke. Also, Haitao goes to TJ and so is automatically cooler than Arnav, who does not go to TJ.

Haitao may have beaten Arnav at Duke, but East Chapel Hill High beat TJ  :wink:

[quote="MysticTerminator"]ok guys guys what the heck why have I become the standard of what various people can or can not pwn[/quote]
An IMO Gold Medal automatically earns you a Bull's Eye on your back :).

[quote="pkerichang"]That's why I say we should definitely kidnap Zeb and drag him to ARML with us ... [/quote]
To Be Or Not To Be?

Well, it's Not TO BE, but S Cal will be alright. If all fail, S Cal can still Cube all others to their knees.

[quote="jb05"]What is it with you Californians and cubing? You don't see anybody else cubing......[/quote]

It's just that most people who did cube gave it up after a while (I'm assuming).  At least I haven't cubed for almost a year...

[quote=JSteinhardt]On another note, Haitao beat Arnav at Duke. Also, Haitao goes to TJ and so is automatically cooler than Arnav, who does not go to TJ.
[/quote]

erm no... TJ<NC by definition

and 2 NC people got perfect scores at Duke

so HA

Um yes and then Haitao owned them both in the Tiebreaker round. So your point is what?

And I think you need to fix your equation. It's more like

$o(TJ) < o(NC)$

!?!?!?!?!?!

ooh one person can own 2 USAMO winners+like 10 other USAMO qualifiers! Obviously

Well apparently he can, since he got 1st place at Duke.

Also TJ had 13 USAMO qualifiers + 2 MOPpers last year. Also, I don't think that there is much difference between a MOPper and a USAMO winner in terms of ARML stuff.

... sure, but look at the page before this; one person can't make THAT much of a difference.

[quote="not_trig"]Well, NC [i]did[/i] win last year, so we obviously have a bunch of good people... (yasha, n^4+4, jb05, teki-teki... plus others, and of course Arnav)[/quote]
Agh! You guys stole Yasha from us. Not cool.

[quote="JSteinhardt"]Also TJ had 13 USAMO qualifiers + 2 MOPpers last year. Also, I don't think that there is much difference between a MOPper and a USAMO winner in terms of ARML stuff.[/quote]

Err... so basically you agree that 6 MOSPers pwns 2 MOSPers, since we're ignoring USAMO winners.


Let $ F_0 \equal{} 0,F_1 \equal{} 1, F_k \equal{} F_{k \minus{} 1} \plus{} F_{k \minus{} 2},\, k\ge2$ be a Fibonacci sequence. Find all positive integers $ n$ for which the polynomial $ F_n x^{n \plus{} 1} \plus{} F_{n \plus{} 1} x^n \minus{} 1$ is irreducible in the ring of polynomials with rational coefficients $ \mathbb{Q}[x]$. (R. Ushakov)

[hide]
Let $ P_n(x) \equal{} F_n x^{n\plus{}1} \plus{} F_{n\plus{}1} x^n \minus{} 1$. Note that $ P_n(x) \minus{} P_{n\minus{}1}(x) \equal{} F_n x^{n\minus{}1} (x^2 \plus{} x \minus{}1)$. It follows that $ P_n(x) \equal{} (x^2 \plus{} x \minus{} 1)(F_1 \plus{} F_2 x \plus{} ... \plus{} F_n x^{n\minus{}1})$. So $ P_n(x)$ is irreducible over $ \mathbb{Q}$ only for $ n \equal{} 1$.
[/hide]


inequalities
function
linear algebra
matrix
inequalities proposed

Simply apply the 3-dimensional Jensen inequality (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31173]here[/url]) for the convex function $f(x,y,z)= x\left(1+2x+yz\right)^{\frac{3}{2}}$ and you are done.

I also used that method, but I' m wondering whether there is more basic method :?:

[quote="N.T.TUAN"]Why  $f(x,y,z)= x\left(1+2x+yz\right)^{\frac{3}{2}}$ is convex function?[/quote]
Rzeszut means that $f(x)=x^{\frac{3}{2}}$ is convex function. :wink:

[quote="arqady"]
Rzeszut means that $f(x)=x^{\frac{3}{2}}$ is convex function. :wink:[/quote]
No, I think $f(x,y,z)=(1+2x+yz)^{\frac{3}{2}}$ is convex. But why? Again  :maybe:
You used theorem '' twice differentiable function of several variables is convex on a convex set if and only if its Hessian matrix is positive semidefinite on the interior of the convex set''. ? (from http://en.wikipedia.org/wiki/Convex_function )

Is there a better solution here? This one seems to be so advance.

I understand. But in normal weighted Jensen, we must prove
\[\sum (1+2a+bc) \ge 16/3\]
which is false.

[quote="GacKiem"]I understand. But in normal weighted Jensen, we must prove
\[\sum (1+2a+bc) \ge 16/3 \]
which is false.[/quote]
Maybe $\sum (1+2a+bc) \ge 16/9$ $?$ :wink:


i have a question for you >Can you help me?
Let 1+1/2+1/3+.....+1/n=a(n)/b(n)   where (a(n),b(n))=1
a, Show that for infinitely many n Such that a(n) is not a power prime
b,p is aprime ,m is a positive interger
Prove that there exist N / with n>=N we have b(n) divisible by p^m

       the a(n) is not a power prime

Part (a) is Fedor Petrov's question for ARO 2002.  His solution can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1778]here[/url].

And b) is an open question (Eswarathasan-Levine, 1991).  :)  I can show it for $p \in \{2,3,5,7,13\}$, in case you're interested.

--Vesselin

I'll be thankful to see it so if you can Vesselin ...  :)


function
inequalities
inequalities proposed

ahh!!!!! :(  What a lot of nonesense I wrote
I edited...

WLOG $a\geq b\geq c$
Case 1: $a\geq 2$.
For $x\geq 2$ the function $\frac{x}{2x^2+1}$ is decreasing. It follows that $\frac{a}{2a^2+1}\leq \frac{2}{9}$. So it is enough to prove
\[ \frac{b}{2b^2+1}+\frac{c}{2c^2+1}\leq \frac{7}{9}  \]
But by AM-GM we have
\[ \frac{b}{2b^2+1}+\frac{c}{2c^2+1}\leq\frac{b}{2\sqrt{2}b}+\frac{c}{2\sqrt{2}c}=\frac{1}{\sqrt{2}}<\frac{7}{9}  \]

Case 2: $a,b,c\leq 2$
We have
\[ \frac{a}{2a^2+1}\leq -\frac{a}{9}+\frac{4}{9}\iff (a-1)^2(2-a)\geq 0  \]
So
\[ \sum \frac{a}{2a^2+1}\leq -\frac{3}{9}+\frac{12}{9}=1  \]

I have a different idea but .....
We have to prove that  $\frac{a}{2a^2+1}+\frac{b}{2b^2+1}+\frac{c}{2c^2+1}\leq1.$

this is equivalent to $\sum (\frac{a}{2a^2+1}-\frac{1}{3}+\frac{1}{9}(a-1))\leq 0$

We consider the function  $f(x)=\frac{x}{2x^2+1}-\frac{1}{3}+\frac{1}{9}(x-1)$

$f'(x)=0$  is equivalent to $x=1$  and $x=\sqrt{\frac{5}{2}}$ so we cannot aply the method that Arqady uses to prove the first barra's problem ?

Very nice proof, barra!:)

[quote="silouan"]
$f'(x)=0$  is equivalent to $x=1$  and $x=\sqrt{\frac{5}{2}}$ so we cannot aply the method that Arqady uses to prove the first barra's problem ?[/quote]
We cannot apply the barra's method of the first  problem,
 since expression $\frac{a}{2a^2+1}-\frac{1}{3}+\frac{1}{9}(a-1)=\frac{2(a-1)^2(a-2)}{9(2a^2+1)}$ can be positive.

I think that following inequality is  harder:
Let $a,$ $b,$ $c$ are non-negative real numbers and $a+b+c=3.$
Prove that $\frac{a}{11a^2+4}+\frac{b}{11b^2+4}+\frac{c}{11c^2+4}\leq\frac{1}{5}.$  ;)

Generalization:
If $a,$ $b,$ $c$ are non-negative real numbers, such that $a+b+c=3$
and $0\leq k\leq\frac{13+4\sqrt{10}}{9}$  then $\sum_{cyc}\frac{a}{ka^2+1}\leq\frac{3}{k+1}.$

[quote=arqady]$a,$ $b,$ $c$ are non - negative real numbers and $a+b+c=3.$
Prove that $\frac{a}{2a^2+1}+\frac{b}{2b^2+1}+\frac{c}{2c^2+1}\leq1.$[/quote]
It's true for all reals $a$, $b$ and $c$ such that $a+b+c=3.$


vector
symmetry
analytic geometry
abstract algebra
linear algebra
linear algebra unsolved

Let V be n-dimensional vector space with inner product $ (,)$ and let $ \Lambda^k V$ be its external product. I want to show that $ \Lambda^k V$ can be equipped with an inner product given by the following condition: if $ \alpha \equal{} \alpha_{1} \wedge \dots \wedge \alpha_{k}$ and $ \beta \equal{} \beta_{1} \wedge \dots \wedge \beta_{k}$ for some $ \alpha_{1}. \dots, \beta_{k} \in V$, then $ (\alpha,\beta) \equal{} det( (\alpha_{i}, \beta_{j}) )$ (it suffices to define the product for vectors which admit such decomposition). However, I don't know how to prove that this is well-defined, i.e. every decomposition of $ \alpha$ gives the same value, without ugly calculations involving determinants etc. Any help?

I would, on your place, start with a different definition of the scalar product on $ \wedge^k V$: Let $ \left\{e_1,e_2,...,e_n\right\}$ be a basis of $ V$. For any $ k$-element subset $ S$ of $ \left\{1,2,...,n\right\}$, denote by $ e_S$ the wedge product $ e_{s_1}\wedge e_{s_2}\wedge ...\wedge e_{s_k}$, where $ s_1$, $ s_2$, ..., $ s_k$ are the $ k$ elements of $ S$ in increasing order. Then, $ \left\{e_S\mid S\text{ is a }k\text{ \minus{} element subset of }\left\{1,2,...,n\right\}\right\}$ is a basis of $ \wedge^k V$.

Now, define a scalar product on $ \wedge^k V$ by setting $ \left(e_I,e_J\right) \equal{} \det\left(\left(\left(e_{i_u},e_{j_v}\right)\right)_{1\leq u\leq k,\ 1\leq v\leq k}\right)$ for any two $ k$-element subsets $ I$ and $ J$ of $ \left\{1,2,...,n\right\}$, where $ i_1$, $ i_2$, ..., $ i_k$ are the $ k$ elements of $ I$ in increasing order, and $ j_1$, $ j_2$, ..., $ j_k$ are the $ k$ elements of $ J$ in increasing order. Thus you have defined the scalar product for any two elements of the basis $ \left\{e_S\mid S\text{ is a }k\text{ \minus{} element subset of }\left\{1,2,...,n\right\}\right\}$ of $ \wedge^k V$. Extend this by linearity (bilinearity actually) to a scalar product for any two elements of $ \wedge^k V$. You get a scalar product on $ \wedge^k V$ (the axioms of a scalar product are easily checked - the only one not completely obvious is the symmetry, but it is enough to check it for basis elements, where it follows from the invariance of the determinant under transposition of the matrix). It coincides with the scalar product that you wanted to define, because you can easily see that $ \left(\alpha_1\wedge\alpha_2\wedge ...\wedge\alpha_k,\beta_1\wedge\beta_2\wedge ...\wedge\beta_k\right) \equal{} \det\left(\left(\left(\alpha_u,\beta_v\right)\right)_{1\leq u\leq k,\ 1\leq v\leq k}\right)$ for any $ \alpha_1$, $ \alpha_2$, ..., $ \alpha_k$, $ \beta_1$, $ \beta_2$, ..., $ \beta_k$ from $ V$ (in fact, because of the multilinearity of both the scalar product and the determinant, it is enough to prove this for $ \alpha_1$, $ \alpha_2$, ..., $ \alpha_k$, $ \beta_1$, $ \beta_2$, ..., $ \beta_k$ being elements of the basis $ \left\{e_1,e_2,...,e_n\right\}$ of $ V$, but then this coincides with our definition of the scalar product, modulo a possible permutation of the basis elements where you have to use the fact that determinants change their sign under transposition of rows or of columns).

Note that the only time that I used the symmetry of the scalar product on $ V$ was when I proved the symmetry of the induced scalar product on $ \wedge^k V$. So you can start with any bilinear form on $ V$ and construct an induced bilinear form on $ \wedge^k V$, which will then satisfy $ \left(\alpha_1\wedge\alpha_2\wedge ...\wedge\alpha_k,\beta_1\wedge\beta_2\wedge ...\wedge\beta_k\right) \equal{} \det\left(\left(\left(\alpha_u,\beta_v\right)\right)_{1\leq u\leq k,\ 1\leq v\leq k}\right)$ for any $ \alpha_1$, $ \alpha_2$, ..., $ \alpha_k$, $ \beta_1$, $ \beta_2$, ..., $ \beta_k$ from $ V$.

Sorry for brevity. I guess this all would follow much easier from universal definitions, but I am too tired to work that out now.

  Darij

[quote="neworder"]Ok, but this doesn't prove that this inner product on $ \Lambda^k V$ is defined coordinate-free, i.e. that it doesn't depend on an arbitrary choice of basis.[/quote]

You obtain this immediately when you finally show that $ \left(\alpha_1\wedge\alpha_2\wedge ...\wedge\alpha_k,\beta_1\wedge\beta_2\wedge ...\wedge\beta_k\right) \equal{} \det\left(\left(\left(\alpha_u,\beta_v\right)\right)_{1\leq u\leq k,\ 1\leq v\leq k}\right)$ for any $ \alpha_1$, $ \alpha_2$, ..., $ \alpha_k$, $ \beta_1$, $ \beta_2$, ..., $ \beta_k$ from $ V$. It is just easier to start with the non-coordinate-free definition and then show that it allows for a coordinate-free equivalent definition than doing this the other way round.

[b]EDIT:[/b] On a second thought, your objection has a point - for finite-dimensional vector spaces $ V$ we can use bases as we wish, but what happens for arbitrary modules over rings? We still can try to define a scalar product by $ \left(\alpha_1\wedge\alpha_2\wedge ...\wedge\alpha_k,\beta_1\wedge\beta_2\wedge ...\wedge\beta_k\right) \equal{} \det\left(\left(\left(\alpha_u,\beta_v\right)\right)_{1\leq u\leq k,\ 1\leq v\leq k}\right)$, and now we have to show that it is indeed bilinear. Any ideas? Probably we will finally need diagram chasing here.

[b]EDIT 2:[/b] I have written the above posting 3 years ago. Now I realize that, while everything I wrote here is correct mathematically, it is not quite correct pedagogically. In fact, the coordinate-free definition is better than the non-coordinate-free definition in every possible sense; in particular, only the coordinate-free definition works for arbitrary modules over (commutative) rings.

  darij


modular arithmetic
algebra
polynomial
number theory unsolved
number theory

Prove that if $ p$ is an odd prime and $ k$ is an integer satisfying $ 1 \leq k \leq p\minus{}1$, then
\[ {{p\minus{}1} \choose k} \equiv (\minus{}1)^k \pmod{p}.
\]

[i]First solution.[/i]  $ (p \minus{} 1)(p \minus{} 2)...(p \minus{} k) \equiv ( \minus{} 1)( \minus{} 2)...( \minus{} k) \equiv ( \minus{} 1)^k k! \bmod p$.  Divide by $ k! \not \equiv 0 \bmod p$ and the result follows.

[i]Second solution.[/i]  $ (x \plus{} 1)^p \equiv x^p \plus{} 1 \bmod p$ (this is well-known).  Polynomial division by $ x \plus{} 1$ yields

$ (x \plus{} 1)^{p \minus{} 1} \equiv \frac {x^p \plus{} 1}{x \plus{} 1} \equiv x^{p \minus{} 1} \minus{} x^{p \minus{} 2} \pm ... \plus{} 1$.

The result follows by comparing coefficients.

If you aren't familiar with the proof that $ (x \plus{} 1)^p \equiv x^p \plus{} 1$, all it comes down to is that if $ k \neq 0, p$ then $ p | {p \choose k}$.


I don't know if I  can post the solution? I think I can't post the solution in the link .But this topic isn't locked.

You can post your solution after 6.11 ;)

[quote="arqady"][quote="village"]$a,b,c> 0$ $prove$ $that$ $\frac{1}{3\sqrt[3]{abc}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq$ $\frac{1}{a+ab+abc}+$ $\frac{1}{b+bc+abc}+$ $\frac{1}{c+ca+abc}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116648[/quote]

not hard: $\frac{1}{a+ab+abc}+$ $\frac{1}{b+bc+abc}+$ $\frac{1}{c+ca+abc}\geq \sum \frac{1}{3\sqrt[3]{a^{3}b^{2}c}}$
so we only need to prove:$\sum \frac{1}{\sqrt[3]{a^{2}b}}\leq \sum \frac{1}{a}$
$\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\geq \frac{3}{\sqrt[3]{a^{2}b}}, \frac{1}{b}+\frac{1}{b}+\frac{1}{c}\geq \frac{3}{\sqrt[3]{b^{2}c}}, \frac{1}{c}+\frac{1}{c}+\frac{1}{a}\geq \frac{3}{\sqrt[3]{c^{2}a}},$


If 74 hens lay 74 dozen eggs in 74 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?

[hide]this simplifies to 1 hen lays a dozen eggs in 74 days.  it eats 1 kilogram of wheat in 37 days, so it eats 2 kilograms in 74 days.so, the answer is 2 kilograms.[/hide]

[quote="Klebian"][hide]OK. Simplify each.

1 hen lays 1 dozen eggs in 74 days, 1 hen eats 1 kg of wheat in 37 days
so 1 hen eats 2 kg of wheat in 74 egs.

So our answer is 2.[/hide][/quote]

Sorry that I am so lazy, I did it the same way, and my answer is

$2$

Me too.

$2$

Oh, so that's how you do BB Code. I never knew what that meant.

[hide]I just multiplied by the two 37s by 2 to get 74 hens eat 148 kilos in 74 days and lay 74 eggs

2[/hide]

[hide]if 37 hens eat 37 kilograms of wheat in 37 days, then 74 hens eat 2*74 kg of wheat in 74 days. We know that 74 hens lay 74 dozen eggs in 74 day, so a dozen eggs takes 2 kg of wheat.[/hide]

I can't believe this is in Nationals! I have never done Mathcounts, but I hope to get my school going next year...

This was so easy... :rotfl:

I think that a lot of us can solve every question in a National Written test barring mistakes given ample time.  But can you do them under time pressure (80 secs might seem like a lot, but a hard problem would burn your time), fatigue (you head has seen 30 questions, some that have frustrated you and you get like 5 min before the target), nervousness (some people like me only get one trip to nationals)...?


group theory
superior algebra
superior algebra unsolved

alternating group means even permutation...

my doubt is ...what even permutation means ? give example foe symmetric group as well as for alternating group...

give the conjugacy classes of A4?
 help me ...i'm in confusion...

Wiki is your friend.You can find almost everything there.

See [url=http://en.wikipedia.org/wiki/Alternating_group]here[/url] and [url=http://en.wikipedia.org/wiki/Even_permutation]here[/url]


It is very easy for you all but I am not getting it: :blush:

I have 6 friends and during a certain vacation, i met them during several dinners. I found that I dined with all the 6 on 1 day, with every 5 of them on 2 days, with every 4 of them on 3 days, with every 3 of them on 4 days, and with every 2 of them on 5 days. Further every friend was absent at 7 dinners. How many dinners did I have alone?

I'm not sure if this is correct, but...

Since each of your 6 friends was absent 7 times, there were a total of 6 x 7 = 42 absenses.

When you dined with 6 of them on 1 day, nobody could have been absent.
When you dined with 5 of them on 2 days, there was a total of (6-5) x 2 = 2 absenses.
When you dined with 4 of them on 3 days, there was a total of (6-4) x 3 = 6 absenses.
When you dined with 3 of them on 4 days, there was a total of (6-3) x 4 = 12 absenses.
When you dined with 2 of them on 5 days, there was a total of (6-2) x 5 = 20 absenses.

Adding these up, your friends were absent a total of 2 + 6 + 12 + 20 = 40 times. However, since we already decided that there was a total of 42 absenses, you must have dinner by yourself 42 - 40 = 2 times.

i don't understand how you are counting the absences :huh:

[quote]with every 5 of them on 2 days, with every 4 of them on 3 days, with every 3 of them on 4 days, and with every 2 of them on 5 days[/quote]
here it is mentioned 'with every' against each part. so, for case 2,
no. of combinatuons of absentees =6*5/2=15   and the combination for one of them is 5  and so if we add rest of them it would exceed 7. so, i think the question needs some correction

if dirtlamb interpreted it properly then[quote]Adding these up, your friends were absent a total of 2 + 6 + 12 + 20 = 40 times. However, since we already decided that there was a total of 42 absenses, you must have dinner by yourself 42 - 40 = 2 times.[/quote]till 40 times solution is correct but if he has dinner by himself then absentees would increase by 6 for every dinner extra
hence if 2 dinners were taken alone then absentees would be 40+12=52, which is incorrect

[quote]Further every friend was absent at 7 dinners.[/quote]
This my just be me being nitpicky, but it never said that every friend was absent at [i]exactly[/i] 7 dinners. So for the last two friends to be absent 7 times I'd think he'd need to eat dinner alone twice. But yes, this question was worded rather badly.

i think[quote]Further every friend was absent at 7 dinners.[/quote]
means exactly 7 dinners. but we need euclid's intervention for correction here.

If you did not figure the question yet, here's a hint[hide]One day $ (1$ time$ )$, all $ 6$ friends $ (A, B, C, D, E,$ and $ F)$ were present (all together).
Each subset of $ 5$ of them $ (\{A, B, C, D, E\}$ for example$ )$, was present $ 2$ times, but that includes the $ 1$ time all $ 6$ of them were present.
Each subset of $ 4$ of the friends $ (\{A, B, C, D\}$ for example$ )$, was present $ 3$ times, but that includes the times a total of $ 5$ or $ 6$ of them were present.
And so on, and so forth, ...[/hide]
I calculated my own solution as[hide]He had $ 1$ dinner with all $ 6$, $ 6$ dinners with groups of $ 5$ (one friend missing each time) and $ 6$ dinners all alone.[/hide]