Does .9 repeating = 1

i personally have to say this:

in "theoretical math" .9 repeating does = 1

because

1/3 = .3 rep

2/3 = .6 rep

3/3 = .9 rep ... but also 3/3 is known to be one...

but in "real" math...

3/3 is the same as saying 3 divided by 3... known to be one...

and if .9 equals 1... then shouldnt 1 - .9 rep = 0 ??? (to make it easier 1 = 1.0 = 1.0 rep...)

thus  shouldnt 1.0 rep - .9 rep = 0???

wouldnt it be something more like

[img]http://www.teaminsurgency.com/images/infinity_new.jpg[/img]

im open to ideas and reasons... so please let me knw wat ure thinking...

Yes it IS equal to zero, by the simple logic of 1 - 1 = 0. And also, I personally find that your proof isn't the one with most depth and gives you most understanding; so here's another proof.

x = 0.9999....
10x = 9.99999.....
10x = 9 + x (make sure you see why; note that we originally assigned x to equal 0.99999... and 9.99999... = 9 + 0.99999....)
9x = 9
x = 9/9
x = 1

Also, what do you mean by "real" and "theoretical?" Could you elaborate more on that? Also, I don't see what you mean by 1 - 0.999... would be 1/infinity. Could elaborate more on that, too?

By theoretical math, i mean to point out the type of mathematical thinking which allows for 1 = .9 rep

and "real" math is the customary math we know and love, simple, built on rules... and everything has proof... the dispute between 1 and .9 rep is something a theoretical mathematician would try to answer...

when i said 1/infinity'th, i meant this...

.9 rep is a rational number somewhere on a number line... 1 is a separate rational number somewhere on a number line... they happen to be close... but they are different numbers...

in todays calculations, it seems the difference wouldnt matter at all beween the two, so we consider them the same... but a small difference still exists... tehy are two different rational numbers...

thus... me and my imagination could only show this difference as 1 / infinity'th

kinda hard to grasp... ive never been a good explainer... but try to think about it from a few different perspectives...

This question has been discussed several times on this forum (one example is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=476762#476762]here[/url]).

Easy solution:

$ \frac{1}{3}\equal{}.\overline3$
$ \frac{1}{3}\equal{}.\overline3$
$ \frac{1}{3}\equal{}.\overline3$
$ \frac{1}{3}\plus{}\frac{1}{3}\plus{}\frac{1}{3}\equal{}.\overline3\plus{}.\overline3\plus{}\overline3$
$ 1\equal{}.\overline9$.

Another proof:

$ x\equal{}.\overline9$
$ 10x\equal{}9.\overline9$
$ 10x\equal{}9\plus{}x$
$ x\equal{}1$.

There you are.

btw

$ \frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}\frac{1}{16}\plus{}\cdots\equal{}1\minus{}\frac{1}{2^{\infty}}$,

but does $ \frac{1}{2^{\infty}}$ really matter? depends on your point of view.

[quote="BOGTRO"]$ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} \cdots \equal{} 1 \minus{} \frac {1}{2^{\infty}}$,

but does $ \frac {1}{2^{\infty}}$ really matter? depends on your point of view.[/quote]

NO!  This is meaningless, and it certainly does not depend on your point of view.  There is no "$ 2^{\infty}$ term at the "end" of this series!  It's an [i]infinite series[/i] -- it goes on forever and doesn't end, so it [i]doesn't have a last term[/i]!  The symbol $ \infty$ is not an integer, not a real number, and isn't at the "end" of the integers.  Its literal meaning in the summation $ \sum_{n \equal{} 1}^\infty \frac{1}{2^n}$ is that you sum starting from $ n \equal{} 1$ and you keep on going forever.  You don't ever reach infinity because there is no infinity there to reach.

The real numbers [i]do not contain[/i] either infinite or infinitesimal quantities.  There are mathematical systems that do (e.g. the surreal numbers), but the real numbers are not among them.  Treating $ \infty$ as if it were a real number is just an abuse of notation (although a sometimes useful one, e.g. when applying L'Hopital's rule in calculus).

But don't you think that .9 rep is a rational number... and so is 1

and on a number line each rational number has 1 and only 1 numerical value...

thus doesnt this mean that .9 and 1 are both different numbers...

and algebra has glitches... which is why i dont believe in the x = .9 rep equation above... its not that its not mathematically incorrect... but just like ext. roots... its a glitch...

mostly because math is a human creation, and its a system... in some cases a flawed system...

and no one has answered me... if wat i said about the number line is true... then what is the difference between 1 and .9 rep...

0.9 (repeating) and 1 have the same value.  They are the same number.

What do you mean that "algebra has glitches"?

[quote="avolkov"]and on a number line each rational number has 1 and only 1 numerical value...

thus doesnt this mean that .9 and 1 are both different numbers... [/quote]

No: decimal representations are unique, [i]except[/i] that every number with a terminating decimal expansion actually can be expressed in two different ways.  The two different ways to express $ 1$ are as $ 1.000\ldots$ and as $ 0.999\ldots$.  Similarly, the two different ways to express $ \frac{5}{8}$ are $ 0.625000\ldots$ and $ 0.624999\ldots$.  Decimal expansions are only unique if you forbid an infinite repeating tail of $ 9$s.  The only difference between $ 1$ and $ 0.999\ldots$ is that they are two different ways of writing the same number.

and, btw, algebra has NO glitches lolz.

it's all based on a rigid and accurate system.
it's all rigorous also....exept for the basic axioms(assumptions) of math. everything else is built off of that just like a building....

[quote="avolkov"]and algebra has glitches... which is why i dont believe in the x = .9 rep equation above... its not that its not mathematically incorrect... but just like ext. roots... its a glitch... [/quote]

Extraneous roots are a result of the fact that the function $ f(x) \equal{} x^2$ is not one-to-one.  Once you understand this concept, they make perfect sense and are not "glitches" at all.

[quote="avolkov"]mostly because math is a human creation, and its a system... in some cases a flawed system... [/quote]

Mathematics is based on [b]proof.[/b]  Correct mathematics is correct because we have defined our terms and axioms to make them correct and because proof guarantees certainty.  Don't claim this until you understand [url=http://en.wikipedia.org/wiki/Real_number#Axiomatic_approach]the axiomatic foundation of the real numbers[/url].

The issue is not any distinction between "real" and "theoretical" math.  Math is math.

[quote="avolkov"]and no one has answered me... if wat i said about the number line is true... then what is the difference between 1 and .9 rep...[/quote]

The real number line is a much less tame beast than you imagine it to be (and not only for the reason JBL has already stated; your assumption that decimal representations are unique is incorrect).  Here is an example:

Consider a function $ c(x)$ defined as follows:  $ c(x) \equal{} 1$ if $ x$ is rational and $ c(x) \equal{} 0$ if $ x$ is irrational.

The graph of $ c(x)$ appears to consist of two horizontal lines, one at $ y \equal{} 0$ and one at $ y \equal{} 1$.  But you know that the top line consists only of points with rational $ x$-coordinate and the bottom line consists only of points with irrational $ x$-coordinate, and you know the following two facts:

- Between two rational numbers there are an infinite number of irrational numbers.
- Between two irrational numbers there are an infinite number of rational numbers.

So the graph of $ c(x)$ consists of two lines with an infinite number of holes, [b]no matter how closely you look at it[/b], that are also infinitely small.  And all we did was strain out the rational numbers from the real number line.

Wow, JBL and t0rajir0u pretty much said all there is to be said.

avalkov, just make sure to know that just because something is counter-intuitive (or at least, to your intuition) doesn't mean it's wrong or "algebra has glitches" (which I don't think is proper use of the word glitch, anyway).

Also, about you're thing about the number line, on a less technical note, your logic is flawed, since you... well, let's look at it again.

[quote].9 rep is a rational number somewhere on a number line... 1 is a separate rational number somewhere on a number line... they happen to be close... but they are different numbers...[/quote]
I think you've made this flaw several times in this thread already: you back up your points by assuming 1 and 0.9999.... are different. You state that we "consider them the same number because the difference is so little." Actually, this is not true. We do not consider them the same because they are ALMOST the same. We consider them the same because we can prove it and use logic to flawlessly confirm it. I think your main conceptual error is that there is a last 9 and that it eventually ends. However, this is not true; the 9's go on forever.

Also:
[quote]
By theoretical math, i mean to point out the type of mathematical thinking which allows for 1 = .9 rep

and "real" math is the customary math we know and love, simple, built on rules... and everything has proof... the dispute between 1 and .9 rep is something a theoretical mathematician would try to answer...
[/quote]
A few comments. First of all, I still do not see the distinction between "theoretical" and "real." t0raj0u, agreed. Math is math. Second of all, there is no dispute. Another point, you say you use "real" math and "real" math is based upon rules and proof, then you go on to assume that .999... and 1 are distinct. Am I missing something?

To be honest none of these proofs seem to make much sense to me.

What is 0.9999...?  It seems like it should be the limit of the sequence 0.9,0.99,0.999, whose k-th term is 0.99...9 (with k nines)

This is a monotone increasing bounded sequence whose supremum is 1, so 0.999 ... = 1.

$ \frac{1}{infty}\equal{}0$

Guys... keep up with your math

Here is a good book to read about zero and infinity:

Zero: A biography of a dangerous idea.

[quote="JBL"]The symbol $ \infty$ is not an integer, not a real number, and isn't at the "end" of the integers.  Its literal meaning in the summation $ \sum_{n \equal{} 1}^\infty \frac {1}{2^n}$ is that you sum starting from $ n \equal{} 1$ and you keep on going forever.  You don't ever reach infinity because there is no infinity there to reach.
[/quote]

QFAT (Quoted for Absolute Truth)

[quote="PI-Dimension"]$ \frac {1}{infty} \equal{} 0$

Guys... keep up with your math

Here is a good book to read about zero and infinity:

Zero: A biography of a dangerous idea.[/quote]

ehhh.. As JBL said we do NOT use infinity like this. The closest you can say is that as $ x$ [i]approaches[/i] infinity, $ 1/x$ approaches $ 0$.

My god, this reminds me of the non-rigorous tripe we had to go through with some of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184288]binomial_4eva's threads.[/url]

[quote="t0rajir0u"][quote="avolkov"]mostly because math is a human creation, and its a system... in some cases a flawed system... [/quote]

Mathematics is based on [b]proof.[/b]  Correct mathematics is correct because we have defined our terms and axioms to make them correct and because proof guarantees certainty.  Don't claim this until you understand [url=http://en.wikipedia.org/wiki/Real_number#Axiomatic_approach]the axiomatic foundation of the real numbers[/url].

The issue is not any distinction between "real" and "theoretical" math.  Math is math.
[/quote]

Well, he isn't totally off here. We cannot prove that the axioms of real numbers are consistent.

That's not quite accurate -- we can't prove the axioms of arithmetic consistent relying only on the axioms of arithmetic, but there might be some other consistent set of axioms from which we could prove the real numbers consistent.  (Of course, we couldn't prove [i]that[/i] system consistent from within itself either, and so on.)

But the comment you've quoted is still bogus: if the real numbers are inconsistent, math based on the real numbers isn't flawed "in some cases," it's a completely meaningless exercise, while if the real numbers are consistent then they don't admit any "glitches."

In addition, although I admire your attempt charitably interpret the nonsense of the OP to render it less embarassing, I think it's pretty clear that the OP couldn't possibly have been making the argument you suggest.

Besides, mathematics isn't created by humans, it's discovered by humans with humans simply using mathematical tools to represent things (i.e. decimal system). It the most perfect science of all.

[quote="JBL"]That's not quite accurate -- we can't prove the axioms of arithmetic consistent relying only on the axioms of arithmetic, but there might be some other consistent set of axioms from which we could prove the real numbers consistent.  (Of course, we couldn't prove [i]that[/i] system consistent from within itself either, and so on.)[/quote]

Okay. How is that a proof? It relies on something that is not provable...

And okay. It is a stupid argument.

AND does anybody want to check what I wrote in the wiki...the second part of what I added might be pointless...I was happy that I found a way to use the least upper bound axiom in an interesting way though...

[quote="Altheman"][quote="JBL"]That's not quite accurate -- we can't prove the axioms of arithmetic consistent relying only on the axioms of arithmetic, but there might be some other consistent set of axioms from which we could prove the real numbers consistent.  (Of course, we couldn't prove [i]that[/i] system consistent from within itself either, and so on.)[/quote]

Okay. How is that a proof? It relies on something that is not provable...[/quote]

How is what a proof?  You probably want to look at [url=http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems]Godel's incompleteness theorems[/url] in more detail.

Yes...I am familiar with that. It is what motivated my discussion. I'm just saying that as long as he is arguing my point that you cannot fully prove something, it is nonsense to say well if we assume this, we can prove it.

A proof is a valid chain of reasoning beginning with axioms.  The validity of the proof doesn't rest upon the consistency of the axioms.  (In particular, systems with inconsistent axioms admit valid proofs of everything.)  Godel tells us that we cannot prove the consistency of a set of sufficiently interesting axioms from those axioms, but there might be other axioms from which we can prove consistency.  If you insist that we choose axioms that imply their own consistency, you are locking yourself in to an extremely limited corner of mathematics.  (I think you probably know this already.)


I was thinking that it might be cool to have a kind of a riddle challenge. I'll post a riddle, and then people can pm me the answer, and I'll send them another challenge. There will be five challenges in all. I would probably give hints daily, and people could race through the riddles. This would be kinda totally based on the webriddler website that I showed you, except more competitive against eachother. Post if you're interested or have questions.

In.

The message is too small. Please make the message longer before submitting.

I'm just going to post the first one, and people can join if they are interested, so that you can see what it's going to be like. I'll make a new topic for it.


integration
calculus
calculus computations

Let $ C$ be the curve of intersection of $ y\plus{}z\equal{}0$ and $ x^2\plus{}y^2\equal{}a^2$ oriented in the clockwise direction when viewed from a apoint high on the z-axis.  Compute $ \int_{C}(xz\plus{}1)dx\plus{}(yz\plus{}2x)dy$ by stokes' thm.


calculus
integration
trigonometry
limit
inequalities
function
calculus computations

Another proof :

By integral mean value theorem

$\\ I_{n}= \cos ( c_{n}-2006 ) \int_{2006}^{2006+\frac{1}{n}} x \, dx = \cos ( c_{n}-2006 ) \frac{4012 n+1}{2n^{2}}$

for some $c_{n} \in [2006,2006+\frac{1}{n} ]$ .

Now $\lim_{n \to \infty } n I_{n} = 2006$ since $\lim_{n \to \infty }\cos ( c_{n}-2006 ) =1$

 by continuity of cosine function .

 :)

[quote="kunny"]Your answer is corect, Ramanujan. :)
We have another basic sulution.[/quote]

Yes we can calculate the integral .Set $y =x-2006$ .

It's  $n I_{n} = \frac{2}{n}+1003 +\frac{\cos \frac{2}{n}-1 }{4 \frac{2}{n}} + \frac{\sin \frac{2}{n}}{4} +1003 \frac{\sin \frac{2}{n}}{\frac{2}{n}}$

and the limit is clearly $2006$ .

 :)

This solution is O.K. You can use $f'(a)=\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}.$


Um, I think you have to use logarithms - unless there's a typo somewhere.

[quote="Quevvy"]How would you solve this type of problem?
$6+6=7^{x}$[/quote]

This is, in my opionion, above MC level, as mathcounts does [i]not[/i] require the use of logarithms.

[hide="Above MC Solution so please move thread to HSB"]
As mentioned above, you will have to use $\text{logs}$ to solve these.

$6+6=7^{x}$
$12=7^{x}$
$\log_{7}12=x$

By Change of Base Formula, $\log_{7}12=\frac{\log 12}{\log 7}$.

Now, use a log table or a calculator.

$\log 12\approx 1.07918125$
$\log 7\approx 0.84509804$

So, the original equation becomes:

$\frac{\log 12}{\log 7}\approx\frac{1.07918125}{0.84509804}\approx 1.27698941$

Therefore, $x=\boxed{1.27698941}$[/hide]

I wasn't sure how difficult this problem would be but someone posted it in the LaTeX Testing forum and I didn't know how to solve it.
Does $\log_{7}12$ mean the log of 12 in base 7?

Yes.

$\log_{a}b$ means the Log $b$ base $a$.

if this has served it's purpose (i think it has :D )...
this should probably be locked
-jorian


I did a search, nothing turned up.

When checking out, clicking "submit" on the page with my credit card number and my address makes verisign give me "Credit Card Type Not Supported" or something similar. Going back to the page with credit card numbers, I see no place to specify the credit card type.

I'm using firefox 1.0.6 on linux.

You shouldn't need to specify the type.

A couple of obvious things:

- did you try to pay with American Express?  (We only accept Mastercard, Visa, and Discover)

- is the expiration date valid?

If you still can't get it to work, you may phone in your order at (619) 659-1612.

[quote="DPatrick"]
- did you try to pay with American Express?  (We only accept Mastercard, Visa, and Discover)
[/quote]

I remember reading that a couple times, but I guess it didn't go through my head.

Thanks! :D


two nafar typist mitavanand 2 safhe ra dar modate 5 daghighe type konand.
chand nafar typist ehtiaj ast ke 20 safhe ra dar modete 10 daghighe type konand?

[hide]10 nafar! :) har jofte 2 nafari ha 2 bar kar mikonand! badihie ke kamtaram nemishe!
fekr nemikonam nokteye khassi dashte bashe ke man nafahmidam! age hast begin:D[/hide]

ابتدا ببینیم 2 نفر ، 20 صفحه را در چند دقیقه تایپ میکنند.
با توجه به فرض و یک تناسب ساده بدست می آید 50 دقیقه.
حالا وقتی 2 نفر،20 صفحه را در 50 دقیقه تایپ می کنند، با یک تناسب [b]معکوس[/b]
مشاهده می شود که به 10 نفر نیاز است.
ب[/b]


Let $p(x)$ be a $1999$ degree polynomial with integer coefficients that is equal to $\pm 1$ for $1999$ different integer values of $x$. Show that $p(x)$ cannot be factored into the product of two polynomials with integer coefficients.

Has been posted several times before.

[hide="Solution"] Let $p(x) = f(x) g(x)$, and WLOG let $g(x)$ be the polynomial of lesser degree, so it has degree at most $999$.

But $g(x) = \pm 1$ for at least $1999$ integers $x$, so it must take one value or the other $1000$ times, so it is a constant polynomial. [/hide]

Sorry, didn't realize it had been posted before  :blush: Thanks for the solution.


algebra
polynomial
number theory proposed
number theory

Let $P(x)$ be a polynomial with integer coeffiences satisfying : $100 \leq P(x) \leq 999$ , $\forall x \in (1,2,...,1999)$ . Prove that : $P(x)$ has no integer root .

Assume that $P(a)=0$ for some integer $a$. Let $r \in \{1,2, \cdots, 1999 \}$ such that $a = 1999q+r$ where $q$ is an integer.
Then $P(r) = P(a) = 0 \mod [1999].$
This contradicts that $100 \leq P(r) \leq 999$.

Pierre.


I keep on getting a "file not found error" in latex when I try inserting an image even though the image is in the specified directory...what happened?

Here is some of the abbreviated latex:

\documentclass{report}
\usepackage{setspace}
\usepackage{graphicx}
\doublespacing
\author{.....}
\title{.....}
\date{......}
\begin{document}
\maketitle
\tableofcontents
\begin{abstract}
........
\begin{figure}
\centering
\includegraphics{C:/Program Files/Report Graphics/Fig1.eps}
\label{fig:Fig1}
\end{figure}

\end{abstract}
\end{document}

I don't think I'm doing anything wrong in my coding, but texnic center keeps on saying that the Fig1.eps is not found when I double-checked that the file is still in the specified drive.  Any ideas on why this is happening?

Thanks in advance for any help or suggestions!!!!

That's because C:/Program Files/Report Graphics/Fig1.eps contains spaces. It is important when using TeX related programs not to use spaces in file names or paths (some programs work with spaces but many don't).

The easiest thing is to put the tex file and the graphics in the same place so you can then use \includegraphics{Fig1.eps} but if you need to put it in another directory then use something like
\includegraphics{C:/graphics/fig1.eps}.

If you use lower case for all file names then you can still use the tex file on non-Windows systems where the file names are case-sensitive.


linear algebra
matrix
group theory
abstract algebra
function
superior algebra
superior algebra unsolved

Let $ A=\left\{1,2,\cdots,n\right\}$. Let $ S_{n}$ be the set of all permutaions of $ A$. For example for $ n=3$, $ S_{n}$ is:

$ \begin{pmatrix}1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix},\;\;\begin{pmatrix}1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix},\;\;\begin{pmatrix}1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix},\;\;\begin{pmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix},\;\;\begin{pmatrix}1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix},\;\; \begin{pmatrix}1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}$

It'll be easily proved that $ S_{n}$ with the operation of Production of permutaions is a Group!

Now lets define a $ \boxed{\text{cycle}}$. A $ \boxed{\text{cycle}}$ is a permuatation. In a cycle each member is mapped into the next member (except the last one that is mapped to the first member) and the other numbers not in the cycle are keeped constant in the permutation.
for example if $ A=\left\{1,2,3,4,5\right\}$, $ (1,3,5,4)$ is a cycle with lenght $ 4$ and:

$ (1,3,5,4)=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 3 & 2 & 5 & 1 & 4 \end{pmatrix}$

Each cycle withy lenght $ 2$ is called a $ \boxed{\text{transposition}}$. We can write each permutation with atleast $ 2$ members (from a finite set) as the product of a number of transpositions.
For example:

$ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 2 & 4 & 3 & 1 \end{pmatrix}=(1,6)(2,5)(2,3)$

These were just to remind... :)
Now Some Problems:

$ 1.$ Prove that for each subGroup of $ S_{n}(n\geq2)$ like $ H$, we have one of the following conditions:

$ a)$all the permutations of $ H$ are $ \boxed{\text{even}}$.
$ \text{ or }b)$exactly half of the permutations of $ H$ are $ \boxed{\text{even}}$.

$ \text{Note: }$ an $ \boxed{\text{even}}$ permuatation is a permutation that can be written as the production of an even number of transpositions.

$ 2.$Let $ \sigma$ be a permutation of set $ A$. We'll say "$ \sigma\text{ }\boxed{\text{moves}}\text{ the member }a\in A$" if $ a\sigma\neq a$. If $ A$ is a finite set, how many members of $ S_{A}$ will be moved by a cycle $ \sigma$ with lenght $ n$?

$ 3.$ Let $ A$ be an infinte set. Let $ H$ be the set of all permutations $ \sigma$ of $ S_{A}$ that $ \boxed{\text{move}}$ only a finite number of members of $ A$. Prove that $ H$ is a subGroup of $ S_{A}$.

 :)

1.
First of all for every subgroup the subset of the even elements is always a subgroup of that subgroup.
Let $ J$ be the subset of $ H$ that contains it's even permutations.
We have $ J<H$ and more, it's easy to prove that $ J$ is a normal subgroup of $ H$.
If $ J$ is the whole $ H$ than $ H$ only has even permutations, if not, consider $ H/J$, let's us prove: $ |H/J|=2$
Let $ a$ be a odd element of $ H$, we want to prove that every odd element $ b$ belongs to $ aJ$, it's easy just consider $ b=a(a^{-1}b)$ because $ a^{-1}b$ is in $ J$, so $ |H/J|=2$.
2.
I might not have understood this question, it look's too trivial. (but I think the answer is $ n$)
3.
Well, it's easy, $ \forall_{a,b \in H}$ we have $ ab^{-1}\in H$

Another problem about permutations:
$ A_{n}$ is the subgroup of the even permutations of $ S_{n}$.
An isomorphism is a one to one function that preserves the operation (in this case the composition of permutations) $ f(ab)=f(a)f(b)$

4. Prove that: $ S_{n}$ is isomorphic to a subgroup of $ A_{n+2}$

5. Find all naturals $ n$ such that $ S_{n}$ is isomorphic to a subgroup of $ A_{n+1}$


limit
real analysis
real analysis unsolved

Assume $ D \equal{} \{(x,y)|a\leq x\leq b,0\leq y < y_0\}$

$ f(x,y)$ is continuous on $ D$ of variable $ x$,[color=red](edit)[/color]

when $ y \rightarrow y_0$ monotone increasing.

$ f(x,y) \rightarrow \varphi(x)$

show that:

$ \lim_{y \rightarrow y_{0}}f(x,y) \equal{} \varphi(x)$  for $ x \in [a,b]$ uniformly

[quote="Arrigo-Sacchi"]How is $ f(x,y)$ connected to $ f(x)$?[/quote]

Sorry for my carelessness. It's $ f(x,y)$ :)


Are we allowed to scan our handwritten solutions and email them?

Also, the deadline to email or fax the solutions is 11:59 P.M. right?

thanks in advance

That should be fine. Remember to write dark enough. Note that USAMTS only accepts the following file types: DVI, PDF, RTF, TEX, TXT, TIF, GIF, BMP, JPG, EPS.

I think the deadline is somewhat loose, but you really shouldn't test its extent.

Good luck!

[quote="Max300"]Also, the deadline to email or fax the solutions is 11:59 P.M. right?[/quote]Yes.  But we do not recommending waiting until the last possible moment.  In particular, we cannot guarantee that you will not get a busy signal if you try to call our fax machine at 11:55 P.M.

[quote="Max300"]Are we allowed to scan our handwritten solutions and email them?[/quote]Yes, but to clarify this -- scan all of your solutions into [b]one[/b] file.  Do not send every page as a separate file.

[quote="towersfreak2006"]Pacific time. (It was mentioned in one of DPatrick's posts.)[/quote]
So is that 9:00 EST???

No, that's 3AM EST.  You've got a few hours left.


synthesize $ \ce{NBS}$ from $ \ce{H2C \dbond CH2}$  :D

[hide="Answer"]1) $ 2\,CH_2 \equal{} CH_2 \plus{} transition\,\,metal\,\,catalyst \longrightarrow CH_2 \equal{} CH \minus{} CH \equal{} CH_2$

2) $ CH_2 \equal{} CH \minus{} CH \equal{} CH_2 \plus{} light \longrightarrow cyclobutene$

3) $ cyclobutene \plus{} KMnO_4/HO^ \minus{} /heat \longrightarrow HO_2C \minus{} CH_2CH_2 \minus{} CO_2H$

4) $ HO_2C \minus{} CH_2CH_2 \minus{} CO_2H \plus{} NH_3 \longrightarrow succinimide$

5) $ succinimide \plus{} Br_2 \longrightarrow NBS$.

In step 2 by "light" I mean light of appropriate wavelenght in order to trigger the photochemical isomerization of butadiene to cyclobutene.[/hide]


I have started to make a mock test with both sprint and target rounds because that's how bored I am. Any questions can be posted here, you do not I have to sign up. I will post the test like a week after mathcrazed posts his. You will be able to do the test at anytime but please be honest and don't cheat. After I think most of people have sent me their answers, I will post the scores here.

Okay, I've finished making the sprint round. The first person to pm me get to proofread it(note: this makes you ineligible to take the test. So preferably someone thats going to 9th grade next year.)

Yongyi is currently proofreading my test. Thanks! But because I have no idea how many people will actually take the time to do my test, I ask that you sign up if you are interested. Thank!

1)math154
2)Urc
3)ernie
4)ihatepie
5)ShortGeek
6)Yongyi781
7)ajain
8)neurosurgeon1118
9)BOGTRO
10)Quickster94?
11)mewto55555
12)heneieneh
13)ernie
14) 12markkram34
15) RisingFlame

Okay, I've decided to post the test on Tuesday(2 days from now) for those of you who are too lazy to go to my blog. You will have until next Sunday to submit your answers which is a VERY long time so I hope everyone can do it. Any questions about the test can be posted here.

Okay, here's the mock test. Remember you have until Sunday night to pm me your answers. All Mathcounts rules apply. You are on the honor system. Have fun!

Downloaded 9 times yet no submissions? :huh:

Well, you see... it takes at least 40 minutes to do the problems. :D

Wow, this is hard, even harder than a normal MC Nats round.

Really? Well, I tried to make it easier than some of the other mock tests that have been posted here where like half of them are way too hard.

Yes, this is quite hard. When I compare it to an actual MC Nats Round, the questions are much more complex and time-consuming. Even look at #1 on your sprint round. Although there is a trick to solving it, it does take a good minute or more of thinking to see that trick. On the MC round, the first question is usually a simple problem that can be solved by mere computation.

Really? I got a score of $ \ge 46$ on that test. I thought it was easier than Nationals, in my opinion. :?:

*cough* geometry??? *cough* anyway I didn't do it.

I put way more geometry problems in there than a normal MC test.

Thanks to everyone who participated!!!(cough cough cough).

I will now announce the top 2:

In second place..... with a score of 45... SHORTGEEK!!!! :winner_second:

and in FIRST PLACE.... with a perfect score...YONGYI781!!!!! :winner_first:

(mods feel free to delete this topic now)

Um I randomly scrolled through your test and like there was this problem with like the harmonic/geometric mean or something of the previous 17 answers there was a 0 in those...right??? :P

nope, there was no 0 any of the previous answers.

Of course, when I proofread his test, one of the answers was "ambiguous" because the question had infinite answers. And you cant find the harmonic OR geometric mean of any set with a non-number in it. :P

Did you get these problems from a book?  I have seen the exact same problems before (other than the fact that you used diiferent nouns (andrew instead of bionic women, apples instead of eggs...)

These are MUCH harder than the actual MathCounts questions at Nationals (or maybe it's just because there were so many geometry problems!)

Anyway, I liked number 4 on the target round.

P.S. You always have to include the year in one of the problems!

Apparently people either get perfect scores or low 30's. Like it all depends how good you are at geo.


Today is a Saturday in March. What day of the week will it be one year from today since next year is not a leap year?

Next year is not a leap year so one year from today is $ 365$ days from now. Since $ 365 \div 7\equal{}52$ remainder $ 1$, one year from today will be exactly $ 52$ weeks $ 1$ day from today so it is a [b]Sunday[/b].


Two pages that might be useful:
http://home.earthlink.net/~w6rmk/math/matinv.htm
http://mathworld.wolfram.com/MatrixInverse.html


Let $A = \left( \begin{array}{cccc}1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right)$

Find all complex matrices B such that $e^{B}= A$. In particular, is the set of such B empty? finite, but not empty? countably infinite? uncountably infinite?

We can diagonally block-structure $A$ as $\begin{bmatrix}U^{T}&0\\0&U\end{bmatrix}$ where $U=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$ It follows that if $e^{X}=U$ then if $B=\begin{bmatrix}X^{T}&0\\0&X\end{bmatrix}$ , $e^{B}=A.$

By direct calculation, if $X=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ then $e^{X}=U.$

It can be shown that $X$ is the only real $2\times 2$ matrix with $e^{X}=U.$ But with complex entries we can use that $e^{X+2\pi inI}=U$ as well. We will have a countably infinite set of solutions.

In fact, we should have $B=\begin{bmatrix}X^{T}+2\pi inI&0\\0&X+2\pi imI\end{bmatrix}.$


induction
USAMTS
number theory
greatest common divisor
relatively prime

Hey guys, so for this round I feel I should've gotten:

1: 4: because I got all 11 cases, just happened to add another.
2: 5 because I did do induction, and prove the period, just not actually state I inducted.
3: at least 4  because I got the right answer, gave an example, and proved (albeit confusingly) through casework that it was minimal. I didn't explicitly give the a,b,c, but I gave k=a'b'c', a',b',c'=(2,3,5), a,b,c=k(a',b',c').
4: Oh I'm not protesting this random guess that earned me 3 points.
5: 2 or 3: I managed to say it was kb instead of k(b-1), which sort of derails everything.

The problem is, I got 3/4/2/3/1, which is at least 2 less than I expected as my lower bound, and I'm somewhat perturbed. Before going off on general unfairness, I want to get ya'll's opinion of whether a 13 is justified. [I'll readily say that I barely had time to do it and wrote it up with approximately no attempt at clarity or coherence, just a solution for each problem save 4, which I guessed correctly on.]

Attached are my solutions. [In my 5 years of USAMTS, I've gotten 2 12s and another 13 for 4 problems submitted; nothing this low for a submission on all 5 problems ever.]

My comments are at [url=http://usamts.org/MyUSAMTS/Comments/Year20/Round3/3b1aaa65f98c05d14242395db97c4bfe.jpg]here[/url].

Many thanks for your input.

[Fixed attach]

That's the wrong year you attached.  :wink:

Edit: #1,2,3 seem to be arguable for maybe a point.

For #3, the error lies in the statement that $ (jkl)^2 | abc\text{gcd}\,(j,k,l)$, because as $ j$ and $ l$ might not be relatively prime, there is no reason to belive $ jl | a$ etc (try $ (a,b,c) \equal{} (2^43^2, 2^33^3, 2^23^4)$, for example).

#5 is wrong since "obtaining all multiples of $ b$" is much stronger than the original problem and is unfortunately not true (the false statement is that adding $ (b\minus{}1)a$ or $ ba$ guarantees another multiple of $ b$), so the given solution in effect becomes a re-statement of the problem.

hmm i was only one point away from you
12...
3/3/2/1/3

I guess I'm kinda similar in the way that I write up my solutions too late cuz I'm a procrastinator...But yeah. It's best not to procrastinate especially on USAMTS, where you need to put in good care in your solutions.

Agreed. Last year I did each round early and checked my work over several times and ended getting a perfect. This year...I did each round a couple day before due and wrote up at the due day, yeah, and I missed points on two #1's.


Simplify $ (22 \plus{} 2.22)\div 0.2$. Express your answer as a decimal.


While AMC A day is approaching, I'm going to shamelessly take advantage of the high volume of AMC forum activity...

Basically, I'm planning to make this whole Mock USAMO thingy for the purpose of giving people a chance to have a sort of "warm-up" for USAMO as well as proof-writing practice. While this might be more timely some time in late March, I'm too impatient to wait that long.

Thus, I'm planning to hold it this weekend, i.e. Feb. 4-5 (assuming I can count, which is probably a bad assumption, especially when jokers and cards are involved). Anyway, the test is going to be similar to a real USAMO, except I won't hold you to any strict time limits. Instead, I'm just going to give Day 1 on Friday night and collect it roughly 24 hours later, and I'll do a similar thing for Day 2.

If you're going to send me proof(s) (which I'll be happy to grade as long as there aren't way too many), then you'll have to LaTeX it, or scan it if you're not as lazy as I am. I'll be somewhat lenient on the rigor for LaTeX'd proofs when it's clear that a diagram would have helped but would be too hard to make.

It's already almost done: 5 of the problems can be used for sure, and the last problem has some backups too. There's a chance that I'll have to delay it, but that chance is very small. However, the test, although non-trivial, may be appreciably easier than an actual USAMO. So if you're interested, please post, as I have no idea how many people are interested (I might have to move this to the WOOT forum).

I would be willing to take. My only concern is that I will be out of town on friday and saturday so I will probably need to take it on Sunday.

[quote="amirhtlusa"][quote="paladin8"]Sounds good to me, but I have to say it'll be a stretch to manage 9 hours of testing in this coming weekend; I already have a math competition to go to...[/quote]

or you can just not go the competition! :D  and have your entire team not go to it, :D
then i might win something.[/quote]

Right...  :P. Unfortunately, I'd let down all nine teams that we're sending. Which is a school record I think :!:. Not that the school really cares about that, but oh well.

I'd be happy to do some USAMO practice...I really need it...

Btw, what math competition r u guys talking about?

If #1 looks approachable, I might do it :D

Well it will be fine if you don't do the problems until Sunday. You can just do one day if you don't have time for both, and of course I will always be willing to accept stuff after Sunday, but you might not be guaranteed eternal (i.e. about a week) fame in the results that I plan to post.

lol, I doubt I would be able to receive a high enough score to have "fame." But thanks alot.

[quote="Boy Soprano II"]Will this be happening next weekend, then, as it is clearly nearly impossible to give it today and tomorrow?[/quote]

Actually, I just gave it to the WOOT people instead. But you can still see it here: http://www.artofproblemsolving.com/Forum/files/day1_425.pdf

Edit--and here (Day 2):
http://www.artofproblemsolving.com/Forum/files/day2_107.pdf

I just thought I'd mention that the mock USAMO has not died... the links to the problem are above, and you may also see the dropped problems [url=http://www.artofproblemsolving.com/Forum/files/dropped_173.pdf]here[/url].

[quote=2015WOOTer]I realize this is a 9.5 year bump, but can someone give a non-broken link to the dropped problems mentioned in above post?

(To clarify: Day 1 and Day 2 in Post 13 are still attached properly, but the link in post 14 is broken and I want to see the problems.)[/quote]

I used the wayback machine. The attachment and the link is the same

https://web.archive.org/web/20060209034111/http://www.artofproblemsolving.com/Forum/files/day1_425.pdf

https://web.archive.org/web/20060326223932/http://www.artofproblemsolving.com/Forum/files/day2_107.pdf

problems have been posted in separate threads inside aops and are collected [url=https://artofproblemsolving.com/community/c3625013_2006_mock_usamo_1]here[/url] (Mock USAMO #1)
(among others [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url], among USA Mocks/ USAMO Mocks)

Related post [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url]

The proposer's second Mock USAMO (#2) for 2006 has also been collected [url=https://artofproblemsolving.com/community/c3625014_2006_mock_usamo_2]here[/url].

I also attach the pdf with the 6 dropped problems , mentioned in #14


Please help me solve two problems below:
1/Let a,b,c are real numbers and a+b+c=2(a,b,c >0)
Find the smallest value of S=a^2+b^2+c^2+2abc
2/Let a,b,c are reals numbers and a+b+c=1.Prove that
(a-bc)(b-ac)(c-ab)<8(abc)^2

[quote="SherlockBond"]Please help me solve two problems below:
2/Let $ a,b,c$ are reals numbers and $ a\plus{}b\plus{}c\equal{}1$.Prove that
$ (a\minus{}bc)(b\minus{}ac)(c\minus{}ab) \le 8(abc)^2$[/quote]

That inequality equivolent to:
$ (a^2\plus{}ab\plus{}ac\minus{}bc)(b^2\plus{}ab\plus{}bc\minus{}ac)(c^2\plus{}ac\plus{}bc\minus{}ab) \le 8a^2b^2c^2$
or
$ \sum_{sym}{a^4b^2}\plus{}\sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^2b^2c^2} \ge \sum_{sym}{a^4bc}\plus{}2\sum_{sym}{a^3b^2c}$
which is right becouse:
$ \sum_{sym}{a^4b^2} \ge \sum_{sym}{a^4bc}$ by Muirhead inequality.
$ \sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^2b^2c^2} \ge 2\sum_{sym}{a^3b^2c}$ by Schur inequality.

[quote="SherlockBond"]
1/Let $ a,b,c$ are real numbers and $ a\plus{}b\plus{}c\equal{}2$ (a,b,c >0)
Find the smallest value of $ S\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2abc$
[/quote]
The answer is $ \frac{52}{27}.$


Euler
function
calculus
integration
analytic geometry
derivative

[quote="meinstein"]can anyone help me with the so called langrangian method???? i cant find out its all concept...
what is the euler langrange equation and what it is used for??

thank you in andvance[/quote]

I'll give it a shot.

One can derive a function for any dynamical system called the Lagrangian, $ \mathcal{L}$, that is a function of $ q$, $ \dot{q}$, and $ t$ -- position, velocity, and time respectively.

The Lagrangian for a given system is $ \mathcal{L}\left(q,\dot{q},t\right) \equal{} T \minus{} V$.

$ T$ denotes the kinetic energy of the system.

$ V$ denotes the potential energy.

Why those particular variables? I don't know why, other than these are the variables that have historically been used to designate these functions.

In the calculus of variations there is an integral called the action-integral:

$ J \equal{} \int_{t_1}^{t_2} \mathcal{L}\left(q,\dot{q},t\right)\, dt$

The goal is to find an equation involving $ q$, $ \dot{q}$, and $ t$ for which $ J$ is fixed.

That means $ \delta J \equal{} 0$.

$ \delta J \equal{} \int_{t_1}^{t_2} \delta\mathcal{L}\left(q,\dot{q},t\right)\, dt \equal{} 0$

$ \Rightarrow \delta J \equal{} \int_{t_1}^{t_2} \left\{\frac{\partial \mathcal{L}}{\partial q}\delta q \plus{}\frac{\partial \mathcal{L}}{\partial \dot{q}}\delta \dot{q} \plus{} \frac{\partial\mathcal{L}}{\partial t}\delta t\right\}\,dt \equal{} 0$

$ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}} \delta q\right) \equal{} \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}}\delta q \plus{} \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta \dot{q}$

In the differential equation above the term in parentheses on the left hand side has units of action, the very thing are attempting to hold fixed. Hence, this term is equal to zero. This fact implies $ \minus{}\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}}\delta q \equal{} \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta \dot{q}$.

Substituting this equation into the integral equation above:

$ \delta J \equal{} \int_{t_1}^{t_2} \left\{\frac{\partial \mathcal{L}}{\partial q}\delta q \minus{}\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}}\delta q \plus{} \frac{\partial\mathcal{L}}{\partial t}\delta t\right\}\,dt \equal{} 0$

The end-points (limits of integration) are fixed and thus $ \delta t \equal{} 0$ therefore the integal becomes:

$ \delta J \equal{} \int_{t_1}^{t_2} \left\{\frac{\partial \mathcal{L}}{\partial q} \minus{}\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}}\right\}\,\delta q\,dt \equal{} 0$

Since neither $ \delta q$ nor $ dt$ equals to zero then this implies

$ \frac{\partial \mathcal{L}}{\partial q} \minus{}\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}}\equal{}0$.

This is the Euler-Lagrange equation. (It is usually written with the terms reversed, which is okay since the equation is homogeneous.)

$ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}} \minus{} \frac{\partial \mathcal{L}}{\partial q}\equal{}0$.

Now let's try applying it to a classic system: A spring with a mass attached.

First, $ T\left(\dot{q}\right) \equal{} \frac{1}{2} m\dot{q}^2$ and $ V\left(q\right) \equal{} \frac{1}{2}kq^2$.

Next, $ \mathcal{L}\left(q,\dot{q}\right) \equal{} T\left(\dot{q}\right) \minus{} V\left(q\right) \equal{} \frac{1}{2} m\dot{q}^2 \minus{}\frac{1}{2}kq^2$.

Now, $ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}} \equal{} m\ddot{q}$

and, $ \minus{}\frac{\partial \mathcal{L}}{\partial q} \equal{} kq$

Therefore, $ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}} \minus{} \frac{\partial \mathcal{L}}{\partial q}\equal{} m\ddot{q}\plus{} kq \equal{} 0$

This final equation, $ m\ddot{q}\plus{}kq \equal{} 0$, is an equation of motion, as $ \ddot{q}$ represents acceleration. The coordinate $ q$ and its derivatives $ \dot{q}$ and $ \ddot{q}$ are generalized coordinates. Applying the Euler-Lagrange equation to whatever coordinates one is using to depict one's system, it is possible to write similar equations of motion in those coordinates.


I was viewing Xevarion's blog and the times were a bit off. The most recent entry said it was posted yesterday but the second recent entry said it was posted today.

I am using Firefox, 2.0.0.11 and Windows XP.

Please always give links when posting a bug report.

[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=363[/url]

Might as well include a screenshot too...

I do not see the error.

Windows Vista IE7.

Filed means he looked at a certain entry during that time. Posted implies that he when he made that entry.

At least that is how I understand it.

I've been having the same issue, and for quite some time.  [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=634]Most of my recent entries[/url] claim to have been filed today.  (Firefox 2.0.0.11, Windows XP, although I don't think it's a client-side issue.)


Find all the extrema in the interval [0,2(pi)] for y=cosx-x

hello, do you mean the following problem
1. Find all extrema of $ y\equal{}\cos(x)\minus{}x$ in the interval $ [0,2\pi]$?
Sonnhard.


Can I know my usamo grade details (means which question I got what score, such as #1, 2 points, #2, 5 points....) after they mail the result to my teacher? Or can I know that any way? I think this is very important to improve my proof and performance for next year. Also I can know which part of proof is my inferiority.
Thanks.

The AMC director has already addressed this. You will not get your scoring details.

If you can replicate your proofs on the board some way I'm sure you'll get plenty of feedback as to where you lost your points.


We have a ball on rectangular table with borders. Table is of size  $V$x$H$. We put the ball in the center of the table.

We know that ball bounces $B_V$ times from vertical borders and $B_H$ times from horizontal borders and then back to the center of the table. How to find angle $\alpha$ on which the ball has to roll so that the number of bounces can be reached?
And the second question: How to find distance the ball will roll?

We know that $\alpha$ is in range $<0^o;90^o>$. We can assume that there is no power loss when bouncing and that the velocity of the ball is equal all the time.

EDIT: We also assume that the radius of the ball is equal to zero ($r=0$)

I think this one should be in the physics forum.  :)

Here's a very similar problem: Imagine you have the same table, ball is somewhere on it, and there is a red dot on the table. Your task is to hit the ball so that it bounces (once) from a given border and then stops at the red dot. If you can solve this problem, solving the original problem requires no more than using the same approach only on bigger scale.

[hide="Hint"]Draw a mirror image of the table so that it has one common side with the original table...[/hide]
[hide="Solution"]... and connect the position of the white ball on the original table with the red dot on the mirror image table.[/hide]
[hide="Solution to the multi-bounce problem"]You do the same thing. Draw a $B_V$ x $B_H$ grid of tables and connect the centres of tables that are in opposite corners of your grid. What you get is the trajectory of the ball.

$\tan \alpha=\frac{VB_V}{HB_H}$

$l=\sqrt{(VB_V)^2+(HB_H)^2}$[/hide]


inequalities
calculus
calculus computations

If $a, m, n$ are real numbers and $a>1$, show that $a^{m+n}=a^ma^n$ if we define $a^x =$sup $A(x)$ for all real $x$, where $A(x)$ is the set of numbers $a^t$, $t$ rational and $t \leq x$.

Right now, I am trying to show that $a^{m+n}\leq a^ma^n$ and $a^{m+n}\geq a^ma^n$. The second inequality is the one I am having trouble with.

Thanks.

the first one is tricky, but the second one is easy. Its just hard to find the right words.

SInce you have problem with the secind one. Let $r,s$ be any rational numbers such that $r\leq m$ and $s\leq n$. Then clearly $r+s\leq m+n$ and by definition of $a^{m+n}$ we have $a^{r+s}\leq a^{m+n}$. We assume that the multiplication rule of exponential holds for all rational numbers (if not prove that one first :D ). So
\[
a^{r+s}=a^ra^s\leq a^{m+n}\Leftrightarrow a^r\leq\frac{a^{m+n}}{a^s}
\]
By definition of $a^m$ and since $r\leq m$ arbitrary then we have
\[
a^m\leq \frac{a^{m+n}}{a^s}\Leftrightarrow a^s\leq \frac{a^{m+n}}{a^m}
\]
and now by definition of $a^n$ we conclude that
\[
a^n\leq \frac{a^{m+n}}{a^m}\Leftrightarrow a^ma^n\leq a^{m+n}
\]
and we are done.

[quote="Fierytycoon"]barra,

Thanks for your post. However, I do not understand how you got $\displaystyle a^m\leq \frac{a^{m+n}}{a^s}$. By the definition of $a^m$, we know that $\displaystyle a^r \leq a^m$. You also already showed that $\displaystyle a^r\leq\frac{a^{m+n}}{a^s}$, but...that doesn't help us prove that $\displaystyle a^m\leq \frac{a^{m+n}}{a^s}$, right?[/quote]

If we have
\[
a^r\leq \frac{a^{m+n}}{a^s}
\]
for every $r\leq m$ rational, then it means that $\frac{a^{m+n}}{a^s}$ is an upper bound of $A(m)$. Since $a^m$ is the smallest upperbound, then we must have
\[
a^r\leq a^m\leq \frac{a^{m+n}}{a^s}
\]

Ah, I understand now. Thanks for all your help.


Here's what I know:
Some problems require multiple skills to solve, and thus are listed in multiple categories. Categories disappear if the system deems you "not ready" for the subject because you didn't pass them. How they reappear, I don't know, but my guess is that subjects reappear if you get a correct answer on a problem that requires a different skill and the skill which you "failed".
Yes, you can watch a video more than once-I've done it.
The problem discussion button is at the lower right-hand corner of the solution page of the problem. If you can't see it then you should report it to the admins.

The "Post problem on Forum" button is located either in the bottom left or the bottom right corner of the solution for the problem.


algebra
polynomial
superior algebra
superior algebra theorems

Let $ \mathbb{F}_q$ be a finite field, $ \overline{\mathbb{F}}_q$ its algebraic closure.

Is it true that for $ x \in \overline{\mathbb{F}}_q$ we have $ x^q\equal{}x$, as we have in $ \mathbb{F}_q$ ?

If not then, is it true that $ (x^q)^q\equal{}x^q$ ?

[quote="Matnomi"]Let $ \mathbb{F}_q$ be a finite field, $ \overline{\mathbb{F}}_q$ its algebraic closure.

Is it true that for $ x \in \overline{\mathbb{F}}_q$ we have $ x^q \equal{} x$, as we have in $ \mathbb{F}_q$ ?

If not then, is it true that $ (x^q)^q \equal{} x^q$ ?[/quote]

No, and no.  Take q=2, then F_8 is contained in the algebraic closure of F_2, but (x^2)^2 = x^4 ≠ x for any x in F_8 - F_2, since those elements have order 7 in the multiplicative group.

The set of x in the algebraic closure of F_q such that x^(q^n) = x forms a field of order q^n.

Clearly no claim of this form can be true since polynomials over a field have finitely many roots and $ \overline{ \mathbb{F}_q }$ is infinite.


f(0,r)=r , f(n+1,r)=r + f(n,r)/f(n,r+1)  (n∈N)
lim f(n,1) = ?  (n goes to +∞)

 :help: please~   If I get solution,I will blow my kiss  :)

Suppose that for every $ r$ there exists a function $ L$ such that

$ \lim_{n \rightarrow \infty}f(n,r) \equal{} L(r)$

Then, we have

$ L(r)\equal{}r\plus{}\frac{L(r)}{L(r\plus{}1)}$

$ L(r\plus{}1)\equal{}\frac{L(r)}{L(r)\minus{}r}$

What you're searching for is $ L(1)$ so we have:

$ L(1)\equal{}\frac{L(0)}{L(0)\minus{}0}\equal{}1$

I appreciate your help.
I'm so happy to get the solution for this problem !  :)

For big $ r$: $ |L(r)|<r^{\minus{}\alpha}\;\textbf{for} \;\alpha>0$. This gives $ \lim_{r\to\infty}{L(r)}\equal{}0$.


integration calculus function trigonometry logarithms Putnam derivative

$\int_{0}^{\infty}\frac{sin(x)}{x}dx = \frac{\pi}{2}$
why?
Also: I've seen the integral of ln(sin(x)) often lately, and I was wondering if the  technique used to evaluate it applied to other integrals as well...and in general what other(non antiderivative) methods are there for evaluating definite/improper integrals and may I have some examples of some standout methods...sorry for being vague or a pest, just always been curious

You can find some answers at:
[url]http://press.princeton.edu/books/maor/chapter_10.pdf[/url]

RicOh

Your integral is a particular case of a not very well known result which is called the Lobachevschi`s formula, namely:

Assume that $f$ is a function defined on $[0,\infty)$ which satisfies the following two conditions:

$f(x+\pi)=f(x)$ and $f(\pi-x)=f(x)$.

Assuming that the left hand side integral exists, the following formula holds:

$\int_{0}^{\infty}f(x)\frac{\sin x}{x}dx=\int_{0}^{\frac{\pi}{2}}f(x)dx$.

In your case $f=1$ hence $\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$.

It is an easy result if you know residue calculus.

Another idea is to write $\frac{1}{x}= \int_{0}^\infty e^{-xy}\,dy$ and change the order of integration, but the integrand will not be absolutely convergent so I personally don't know how to justify the change of order of integration (I suppose I could if I really wanted to, but I haven't seen any type of discussion of conditional convergence in higher dimensions, so I can't refer you to a theorem).

It is easier to do a partial integration to get a similar integral which is very absolutely convergent and then use a similar technique to what is described above.

Here are a couple of links:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73681]Link 1.[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41288]Link 2.[/url]

My favorite method is spread between these two links without quite being entirely in either; it's essentially what Kalle said in his last post.

wow fubini's theorem is awesome, so many possibilities

For $\int_{0}^{\frac{\pi}{2}}\ln \sin x \ dx$ check out this link

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125460

yeah, I know that one already, I was asking if any other integrals can be solved similarly; the general ? was what methods could be used to evaluate definite integrals without antidifferentiation, and I got a good answer so far with Fubini's and somewhat with solving for the unknown integral, but are there any other methods?

Sure, there is a technique called parameterization.  Let $I = \int_{a}^{b}f(x) \ dx$.  Then introduce another variable $t$ such that $I(t) = \int_{a}^{b}f(x,t) \ dx$.  Then, if done correctly, we can evaluate $I'(t)$, then find $I(t)$ from the resulting differential equation.

This comes from Leibniz's Rule:
\[\frac{\partial}{\partial z}\int_{a(z)}^{b(z)}f(x,z)\ dx = \int_{a(z)}^{b(z)}\frac{\partial f}{\partial z}\ dx+f(b(z),z)\frac{\partial b}{\partial z}-f(a(z),z)\frac{\partial a}{\partial z}\]
(Sorry if this sounds confusing, an example should make it clear):

Try $\int_{0}^{1}\frac{x-1}{\ln x}\ dx$, let $I(t) = \int_{0}^{1}\frac{x^{t}-1}{\ln x}\ dx$

Also, try this on the 2005 Putnam A5.

Do you already know how to evaluate $\int_{-\infty}^\infty e^{-x^{2}}\ dx$?  If not, let $I$ be the desired integral, then find $I^{2}$, then switch to polar coordinates.

Finally, you might want to learn a little residue calculus, because many integration formulas can be found  very quickly that way.

yeah, a friend showed me e^-x^2, thanks.  I don't fully understand how to set up a differential equation for that integral, can you please explain? would it be something like this:
$I'(t) = \int_{0}^{1}\frac{\ln(x)(tx^{t-1})-(x^{t}-1)(\frac{1}{x})}{\ln(x)^{2}}dx$

No, differentiate with respect to $t$ not $x$, and note that $I(0)=0$.

[quote="Dvagner"]$I'(t) = t\int_{0}^{1}\frac{x^{t-1}}{\ln(x)}dx$
now what?[/quote]

Actually, the derivative of $\frac{x^{t}-1}{\ln{x}}$ with respect to $t$ would be

$\frac{\ln{x}\cdot x^{t}}{\ln{x}}= x^{t}$,

since the $t$ is in the exponent. The rest should be fairly clear.

again, sorry for the careless error, I don't know what I was thinking; thanks a lot!


logarithms
probability
expected value
combinatorics unsolved
combinatorics

Hello,

I have to solve a problem quite similar to the well known "coupon collector" problem but with a little variant.

The variant compared to the coupon collector problem is that now more than one coupon is draw for each realization.

More precisely and if it can help, for my practical example, exactly N/2 coupons are drawed each time. So in average, what is the expected number of draw to get the whole N coupons ?

Best Regards.

Have you tried the problem when drawing exactly 2 coupons each time?

For any such problem, it is not to hard to set up a recursion (since expectation is linear), but I don't see a nice way to solve the recursions in general.

The following Mathematica code should produce the values you seek for a given number of coupons and coupons-drawn, but I haven't been able to recognize a pattern even for drawing just 2 at a time.  (If I had to guess, I'd say it's about $\frac{1}{2}n \log n$, or about half what you get if you only draw one coupon.)

P[n_, t_, k_, i_] = If[t == i == 0, 0, Binomial[n - t, i]*Sum[(t + i - j)^k * Binomial[i, j] * (-1)^j, {j, 0, i}] / n^k];

(this is the probability that, holding t ticket types out of n total and drawing k, you end up with i new types)

e[t_,n_,k_] := e[t,n,k] = If[t≥n, 0, (1 + Sum[e[t+i,n,k]*P[n,t,k,i], {i,1,k}]) / (1-P[n,t,k,0]) ]

(this is the expected number of turns to finish if you hold t of n types and draw k each turn).

For your particular example, this gives the values

$e(0, 2, 1) = 3$
$e(0, 4, 2) = \frac{463}{105}\approx 4.41$
$e(0, 6, 3) = \frac{3891393}{743470}\approx 5.23$
$e(0, 8, 4) = \frac{16334427070903}{2810973990825}\approx 5.81$.

Incidentally, I was assuming here that the $k$ tickets drawn could have repetitions; I suppose you could ask the same question where the $k$ tickets drawn are guaranteed to be of $k$ different types.  Also, of course, I assumed that each type of ticket was equally likely.


I've always [b]dreamt of MIT[/b] and want to go there more than anything in the worls (even if  I get Caltech or Princeton or Stanford etc.)I live in India and somebody told me just 2 students are going to be taken from india this year.Is this true?????????

I've come among the [b]top 40 Young Scientists in India[/b] under the KVPY Scholarship conducted by the [b]IIT's[/b](the best and respected worldwide)and the IISc , BAngalore in India(it';s conducted by the govt.).I'm also a National Talent search Scholar and have also been selected for the Singapore International Airlines Scholarship.I've always been top in class and scored 94% in ALL India Board CBSE exams.(regarding very high)
Ive also got 2280 in SAT(new)(800-math CR-700 :(  and weiting - 780 (10/12))and am quite confident about SAT II's i'll give next year.I'm also expecting near 100% in TOEFl
I was also adjudged the best All round student in my school in Grade 10.(i had to leave that scholl coz it did'nt have classes beyond grade 10 I go to another school now)

Yet that 2 students taking thing is really a blow to me .Can u guys help me out.This is  a long post but plz rate my chances and what else you think need to be done to reach my goaL.I'd like to know the achievements etc. of other international students who have got in.

[b]THANK YOU[/b]

On the one hand, you need to have the courage to be yourself. Not everyone who desires to get into MIT gets in, and especially for international students MIT admission is very competitive. So "to thine own self be true," as I recall Shakespeare wrote. Don't try to set your whole course for the next year or two of your life just to get into MIT--there isn't any sure formula for admission anyway.

On the other hand, you need to be at peace with yourself and accept that not getting into MIT is not the end of the world. Have some desires and goals that aren't strictly connected with what school you are applying to, and pursue those. That maximizes your chances of acceptance at any school. If you do what you do only to look good to a college admissions officer, you will look like a poseur. If you do what you do to pursue a personal passion for excellence, you will look like an interesting person, and at least have a maximum (small) chance of standing out in the highly competitive applicant pool of the top colleges in the United States.

Good luck in your application.

After edit: I'll lock this thread and let the other thread (from the double post) keep going.


A rectangle has a diagonal of 14 feet and one side of length $ 7\sqrt{3}$ feet. What is the area of the rectangle, in square feet? Express your answer in simplest radical form.

The triangles created by the diagonal of the rectangle are 30-60-90 triangles so the unknown leg length is equal to half the hypotenuse.

$(14\div2)\times7\sqrt3= 49\sqrt3$

[quote=PersisII]I'm pretty sure I typed 49sqrt3 but it didn't work. Does anyone know why? :([/quote]

Are their answers the right answers?


A 3x3x3 cube is formed by gluing together 27  standard cubical dice. (on a standard die, opposite sides add to 7). The smallest possible sum of all the numbers on the surface of the 3x3x3 cube is____?

[hide="My answer"]90[/hide]

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. the largest integer that can be an element of this collection is____?

[hide="My answer"]14[/hide]

Last one...

Tina randomly selects two distinct numbers from the set {1,2,3,4,5}, and Sergio randomly selects a number from the set{1,2,3,....,10}. What's the probability that Sergio's number is greater than the sum of the 2 numbers chosen my Tina?

[hide="My answer"]2/5[/hide]

Note that these problems are all in the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2002]resources section[/url]. You can find answers by clicking on the problem number, which will bring you to the original thread, which often has solutions. You can also check answers [url=http://www.unl.edu/amc/e-exams/e5-amc10/archive10.shtml]on the AMC website[/url].


[quote="GameBot"]Find $ x \plus{} y \plus{} z$ in the arithmetic sequence $ \{2, x, y, z, 10, \ldots\}$[/quote]

Since it's an arithmetic sequence $ x \minus{} 2 \equal{} z \minus{} y \equal{} 10 \minus{} z$, i.e. they have a common difference. If you picture this numbers as evenly spaced coordinates on a number line, we see that we moved up 8 points from 2 to 10. There are 4 "spaces" between the coordinates, so the common difference if 2. Thus $ x \equal{} 4, y \equal{} 6$, and $ z \equal{} 8$, and their sum is $ \boxed{18}$.

[quote="GameBot"]Find $ x \plus{} y \plus{} z$ in the arithmetic sequence $ \{2, x, y, z, 10, \ldots\}$[/quote]

Alternatively, we note that the mean of 2,10 is 6. $ 6 \cdot 3\equal{}18$.


Dear friends,

I don't know if this inequality holds, but I'm sure it does (I couldn't find a counter-example). Here it is:
$ (a_1^k \plus{} a_2 ^k \plus{} ... \plus{} a_n^k)(a_{n \plus{} 1}^k \plus{} a_{n \plus{} 2}^k \plus{} ... \plus{} a_2n^k)....(a_{(k \minus{} 1)n \plus{} 1}^k \plus{} ... \plus{} a_kn^k) \ge (a_1*a_{n \plus{} 1}*....*a_{(k \minus{} 1)n \plus{} 1} \plus{} ... \plus{} a_n*a_{2n}*....*a_{kn})^k.$
If you guys do find a counter-example, I'm sorry for wasting your time.

This is Holder's inequality.

Cheers,

Hunter

Oh, ok (sorry!) I should've researched before posting.


I'm told (in a physics problem) that:

$ {\sqrt{2s / a}  / {\sqrt{2s / a'} = {\sqrt{a'/a}}}}$

Can anyone tell me the steps to get that result please?

[b]Division by a fraction:[/b]  $ \frac{ \frac{a}{b} }{ \frac{c}{d} } \equal{} \frac{ad}{bc}$.

Proof:  $ \frac{ \frac{a}{b} }{ \frac{c}{d} } \equal{} \frac{ \frac{a}{b} \cdot bd}{ \frac{c}{d} \cdot bd} \equal{} \frac{ad}{bc}$.

Thanks very much!  I didn't realize it worked the same for variables
with radicals :)

Before you do any fraction simplification, you must first do this:

$ \frac {\sqrt {2s\div a}}{\sqrt {2s\div a'}} \equal{} \sqrt {\frac {2s\div a}{2s\div a'}}$

Then you can proceed to simplifying the inside of the square root. :)


in past years, the NSA has posted solutions for the questions. Are you planning to post solutions to this past year's contest?

If NSA provides them to us, then yes (I think they are still working on them, but I am not sure).  We will, of course, post full solutions for next year as the contest proceeds.


inequalities
calculus
integration
inequalities proposed

[quote="Vasc"]If $a,b,c$ are positive numbers such that $abc=1$, then

$\frac 1{\sqrt[3]{a^{2}+25a+1}}+\frac 1{\sqrt[3]{b^{2}+25b+1}}+\frac 1{\sqrt[3]{c^{2}+25c+1}}\ge 1$.[/quote]
Ok.Nice inequalities,Vasc! I will try it.

We perform the substitution $a=\frac{x^{2}}{yz}$, $b = \frac{y^{2}}{zx}$, $c ={z^{2}}{xy}$, then the initial inequality is equivalent to
$\sum \frac{1}{\sqrt[3]{\left(\frac{x^{2}}{yz}\right)^{2}+25 \frac{x^{2}}{yz}+1}}\ge 1$
$\Leftrightarrow \sum \frac{1}{\sqrt[3]{\frac{x^{4}}{y^{2}z^{2}}+25 \frac{x^{2}}{yz}+1}}\ge 1$
$\Leftrightarrow \sum \frac{yz}{\sqrt[3]{x^{4}yz+25x^{2}y^{2}z^{2}+y^{3}z^{3}}}\ge 1$

Applying Holder's inequality, left hand side is
$\ge \frac{(xy+yz+zx)^{4}}{(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})+(x^{4}y^{2}z^{2}+x^{2}y^{4}z^{2}+x^{2}y^{2}z^{4})+25(x^{2}y^{3}z^{3}+x^{3}y^{2}z^{3}+x^{3}y^{3}z^{2})}$
This is larger than 1, by expanding $(xy+yz+zx)^{4}$ and Muirhead's inequality. This is true because $(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})+(x^{4}y^{2}z^{2}+x^{2}y^{4}z^{2}+x^{2}y^{2}z^{4})$ appears in the expansion of $(xy+yz+zx)^{4}$ and thus will be cancelled, and $(3,3,2)$ is the "weakest" integral order 8 sums. Therefore we can prove it by Muirhead's inequality.

[quote="Vasc"][quote="Zamfirmihai"]Very nice...

$a^{2}+25a+1\leq (a+2)^{3}$ because it's equivalent to $(a-1)^{2}(a+7)\geq0$.
So our inequality reduces to $\sum_{cyc}\frac{1}{a+2}\geq1$ which is something more familiar :D[/quote]
Actually, we have
 $\sum_{cyc}\frac{1}{a+2}\leq 1$. :!:[/quote]

Indeed, by clearing the denominator, it suffices to show
$(ab+bc+ca)+4(a+b+c)+12 \leq abc+2(ab+bc+ca)+4(a+b+c)+8$
$\Leftrightarrow ab+bc+ca \ge 3$
In general, what is the maximum value of $\sum_{cyc}\frac{1}{a+k}$ given that $abc=1$? For $k \ge 2$, the value is $\frac{3}{k+1}$.

I think the minimum and maximum value of 1/a+k+1/b+k+1/c+k with abc=1is
(2k^2/(1+k^3)+1/(k^4+k)),3/(k+1),4/k)if I am not mistaken.

(inequality holds at (1/k^2,1/k^2,k^4)or(1,1,1)or(zero,zero,infinity)
(the last one is a limit)
I am not sure if 1/k^2,1/k^2,k^4 can be the maximum or the minimum.
Too lazy to check... :blush:

[quote="arqady"][quote="Soarer"]
Applying Holder's inequality... [/quote]
Or Jensen's inequality. :D[/quote]
Plaese can u show me how ? I cant solve it by Jensen...

[quote="Sunjee"]
Please can u show me how ? I cant solve it by Jensen...[/quote]
Let $xy+xz+yz=1.$
$f(x)=\frac{1}{\sqrt[3]x}$ is  convex function.
Hence, $\sum_{cyc}\frac{yz}{\sqrt[3]{x^{4}yz+25x^{2}y^{2}z^{2}+y^{3}z^{3}}}\geq\frac{1}{\sqrt[3]{\sum(x^{4}y^{2}z^{2}+25x^{2}y^{3}z^{3}+y^{4}z^{4})}}.$ :wink:

[quote="Vasc"]The following inequality also holds for $abc=1$:

$\frac 1{\sqrt[4]{a^{2}+79a+1}}+\frac 1{\sqrt[4]{b^{2}+79b+1}}+\frac 1{\sqrt[4]{c^{2}+79c+1}}\ge 1$.[/quote]
Hum.... we have:
\[\frac{1}{\sqrt[4]{a^{2}+79a+1}}\ge \frac{5-a}{12}\]
In fact, this inequality is equivalent to:
\[\frac{(a-1)^{2}(20111+1308a^{2}-a^{4}-61a^{3}-8653a)}{12^{4}(a^{2}+79a+1}\ge 0\]
I think my solution have make mistake  :P I will try again


[quote="car202"]I'm considering the major between these two. Very similar, but different.
Can anyone explain the similarity and difference, strength and weakness, etc for these two?[/quote]

What are your long-term goals?

You'll likely do more modeling in physics, which will therefore make you more employable outside academia (assuming you're going to go all the way to a PhD in said subjects).

Why not do both.  I know alot of people who are doing mathematical physics. I don't know if the place your studing offers that degree. You could look for places that do.

Well, a career in physics is not linked to as many careers or course of study as mathematics. Thus, you cannot change your major so easily (inferring to academical stress) if you take physics. However, that is just my personal opinion, so...

As the saying goes, "go with your heart's desire"... and I know that sound's corny, but it is the truth. Good luck having a hard time choosing then ^_^

p.s. -- you [b]know[/b] that CHEMISTRY, of course, is very good for everyone *ahem*... just kidding


calculus
derivative
integration
calculus computations

If integral f(x)dx=g(ax)+C, where a is a nonzero constant, what is g'(x)?

Differentiate both sides, using the FTC and the chain rule.


Has anyone taken these? What did you think?

I'm going to be a senior this fall and wanted to learn some programming over the summer since I'll have no time when school starts. Are the courses more for younger students? Are they an adequate intro if I want to participate in USACO?

I've had a friend take them. Are you learning algorithms or programming? If the target is USACO, you want to learn algorithms. EPGY is mostly programming - too much stuff only tangentially related to USACO. Thus, a cheaper and more efficient method is:

Get The C Programming Language, by Kernighan and Ritchie. Study, do the problems. That'll teach you C and basic algorithms.
Work on working through the training pages. and simultaneously,
Get Introduction to Algorithms, Cormen, Leiserson, Rivest, and somebody. It'll help you up through when you're Gold.

You have now spent 1800 dollars less (and a little more time) and gotten a much better start on USACO.

I'm taking the EPGY course right now - it's horrible. The program and examples used in the lectures are different from the program you are supposed to use, making the instructions very confusing. Don't take that class.


I think I'll open a thread for the "nerdy sport"...well, at least that's the way quite a few people here think about it. Anyway, this thread is for any DDR discussion.

So, let's start. I'm wondering if anyone out there knows how to unlock the six Groove Radar specials on SuperNOVA2? I've completed all missions up to Area J without Custom Modules, but there are quite a few missions in Area K and Area L (especially the bosses) that are way to hard to beat without Custom Modules, and I believe there's a certain number of those that need to be beat without Custom Modules. Is anyone really pro out there that has unlocked the Groove Radar specials?

Oh man.
DDR, is a full out habit of mine. (not hobby... [b]habit[/b])
I use to do it a lot when I was growing up, then I stopped for a couple years.... then I saw one of those machines and found out that I'm rusty, so... I like do random moves in my spare time when no one's looking :lol:

Ooh I love DDR! I used to play like every day when they had it at the nearby mall (then it got moved :mad: ). I wasn't awesome but I was good enough to impress people.

And btw for whatever reason I really can't stand SuperNova. It must be the soundtrack and the weird background thing it's got going on, but I can only ever play Extreme. Any tips on how to survive Heavy Mode? I just beat Max on Standard the other day.

I usually play the game "In the Groove 2" at home. It's basically a clone of DDR that is currently owned by Konami because RoXoR Games got the [whatever it is] sued out of them. And they don't have Couple mode either. T_T

I play on keyboard, too... ^_^

I can usually do 7-foot, 8-foot, some 9-foot songs. But because I don't have that bar, I sometimes lose balance when I attempt 10-footers.

I used to have the old rubber mat at home  :rotfl:
Wow it's been long...

Anyone wish to answer my original question?

Groove Radar.. hrm... that rings a bell.... OH! Right, I spent most of my childhood trying to unlock that.  :rotfl:

sdkudrgn88, do you use stepmania to play ITG2? Because I have ITG 1 for the PS2, and I really wish that Roxor would publish ITG 2 for playstation, but they're obviously not going to anymore. :(
ITG is a LOT better than DDR, because of tons of more modifiers, a greater variety of difficulty (like, the difficulty goes up to 12, but a 10 in ITG is still slightly harder than a 10 in DDR), less distracting backgrounds, better scoring (uses percentage, has more than just "perfect, great, good, miss"), and some cooler features, like some songs having mines (can't step on them) and hands (where 3 or 4 arrows go up at the same time).
[quote="sdkudrgn88"]But because I don't have that bar, I sometimes lose balance when I attempt 10-footers.[/quote]
Heh, I just put a chair behind me to use as a bar for the harder songs.

And anyone that went to AMP texas probably knows that I like playing DDR a lot, because of the DDR machine in the SU. :D (yay I got an A on Max 300 heavy there)

Edit: I only have Max, Max 2, and ITG, so I can't really answer your question about Supernova, although you could probably google it and find out how
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Cache-Control: max-age=0

-----------823378840143542612896544303
Content-Disposition: form-data; name="fileupload"; filename=""
Content-Type: application/octet-stream

Sorry, I don't have Supernova either, so I can't answer that question. XD

[quote="andersonw"]sdkudrgn88, do you use stepmania to play ITG2?[/quote]
As a matter of fact, I do. I also modified the source code a little bit (No cheating, I promise!) to change Survival mode around, but I can't test my changes on a pad because my computer's upstairs... T_T

[quote]Because I have ITG 1 for the PS2, and I really wish that Roxor would publish ITG 2 for playstation, but they're obviously not going to anymore. :([/quote]
Because they got sued, of course.

[quote]ITG is a LOT better than DDR, because of tons of more modifiers, a greater variety of difficulty (like, the difficulty goes up to 12, but a 10 in ITG is still slightly harder than a 10 in DDR), less distracting backgrounds, better scoring (uses percentage, has more than just "perfect, great, good, miss"), and some cooler features, like some songs having mines (can't step on them) and hands (where 3 or 4 arrows go up at the same time). [/quote]
Actually, ITG pretty much ripped off the DDR concept except that they changed Oni (Challenge) mode to Survival mode, and introduced the new modifiers that the people who made Stepmania coded. Basically, ITG 2 [i]is[/i] Stepmania, only with a custom theme. ^_^ And they took away Couple mode... ¬_¬

Actually, the "Fantastic, Excellent, Great, Decent, Way Off, Miss" metric is a direct equivalent to "Marvelous, Perfect, Great, Good, Boo, Miss". It's just that the times are spread out a bit more.

Also, their songs are longer, which is a plus.

[quote][quote="sdkudrgn88"]But because I don't have that bar, I sometimes lose balance when I attempt 10-footers.[/quote]
Heh, I just put a chair behind me to use as a bar for the harder songs.[/quote]

But I'm afraid that the chair will slip back... ._.

Here's a reference list of the judgment times:

In DDR:
Marvelous = $ \pm$0.015 s, 3 dance points in Oni mode, 2 dance points otherwise
Perfect = $ \pm$0.030 s, 2 dance points
Great = $ \pm$0.09 s, 1 dance point
Good = $ \pm$0.12 s, 0 dance points
Boo = $ \pm$0.15 s, -4 dance points
(Miss = -8 dance points)
O.K.! = $ \pm$0.20 s, if I'm not mistaken, and it gives you 6 dance points.
(N.G. = 0 dance points)

In ITG:

Fantastic = $ \pm$0.0215 s, 2.5 dance points
Excellent = $ \pm$0.043 s, 2 dance points
Great = $ \pm$0.102 s, 1 dance point
Decent = $ \pm$0.135 s, 0 dance points
Way Off = $ \pm$0.180 s, -3 dance points
(Miss = -6 dance points)
Yeah! = $ \pm$0.320 s, 2.5 dance points, but I made it 4 in my customization, to keep it more in line with the DDR scoring system
(Bad = 0 dance points)
Hit mine = -3 dance points

I'm not going to get into the grade rankings, there are way too many of those already.

[quote="r15s11z55y89w21"]Yeah, I've gotten A on Max 300, too, but I somehow can only get a B on The Legend of Max. The Highest AAA I've gotten was on an 8-footer.[/quote]

The highest AAA I've ever gotten (all Perfect, I'm assuming) was on a 5- or 6-footer. I'm not a pro, but I guess I'm okay.

There's no survival mode in ITG1 :(. And I don't have stepmania either, so yeah.

In Max, there isn't any marvelous, so that's probably why it always seemed easier to get perfects in DDR than fantastics in ITG.
I can get AAs on most 8 and 9-step songs, but I haven't ever gotten a triple A yet, although I haven't really played that many 6-step or less songs for a while.

I tried Legend of Max at AMP for the first and only time, and failed when it got to the slow part and sped up again. :P

[quote="andersonw"]There's no survival mode in ITG1 :(.[/quote]
Ah, the joys of ITG 2. Get Stepmania, it's free (and open-source) software. (actually, you should already have it because you have ITG, but if I know my stuff, ITG 1 is based on Stepmania 3.9, so you need Stepmania 3.95 or OpenITG for ITG 2 to work).

Compared to all of you, I'm a noob :( . I don't have DDR, so I only get to play it at school on free play days in P.E.

I've never gotten an AAA (but I think I could if I dropped down to beginner), my only AA was on a 3-footer (under the sky difficult should be basic), and I still can't pass any of the SuperNova 2 boss songs on difficult (except NGO). At least I can pass 7 and 8 footers now...

Isn't Groove Radar just the diagram with the levels of stream, voltage, air, freeze, and chaos?

Once, as a joke, I played Fascination -eternal love mix- on challenge.  :stretcher:

Well.. if you're bored, look up darkdieuguerre on YouTube.  :oops:

Or WinDeu, for that matter. He makes stepfiles as well, with stuff like scripted modifiers (effects that change in the middle of the song.)

[shameless plug]Heh, I have some videos of myself on youtube as well. Search "anderson wang ddr" or something. Although, they're like two years old now.[/plug]

[quote="sdkudrgn88"]actually, you should already have it because you have ITG[/quote]
I have the home version of ITG for the PS2, not for the PC. I looked up stepmania, and it looks like a really nice program, although I can't seem to figure out where to find any ITG2 songs for stepmania.

[quote="andersonw"]I have the home version of ITG for the PS2, not for the PC. I looked up stepmania, and it looks like a really nice program, although I can't seem to figure out where to find any ITG2 songs for stepmania.[/quote]

Whoops, forgot it was a PS2 game.

There used to be some at jeffrey1790.com, but that website's down now for some reason. Oh wait, it's back up again.

man, i dont even HAVE any of those games :(

i got a 360 from christmas, but the best game is probably halo 3 or madden 08.

Madden08 Failed. i have 09
DDR well i fail.

i rox no body's sox at DDR.

i rox no body's sox at Madden08

Again;
DDR and I are long lost rivals;
Madden and I are complete enemies, I tried playing all the plays and stuff on my cousins console, yet I fail. Hard.  :rotfl:

Anyone really really pro at DDR Extreme 2 to help me here?

I've beaten the entire Dance Master Mode Game, so now just for the heck of it I am attempting to get all hidden connections. Currently, I have all except for two, as far as I know--I know there's the connection from the upper handicap mission to Mission D-30, which shouldn't be all that hard (hard, yes, but not as much as the other one), and I know there's the connection from Mission D-48 to Mission D-47, which seems to be the hardest one to get in the entire game. Anyone know of how to get it? (By the way, for those who are wondering how in the world to even get to D-48, it's a really hard and hidden route, but you have to get there via the hidden connection from Mission D-18 to Mission X-32, and when you pass X-32 (which is really hard), you get to D-48 (which is even harder).)

Again (maybe i didn't tell you this),

My PS2 broke because my sister STEPPED on it while i was playing.

So, i can't play DDR any more. :(


Game: Post a sentence but only show the vowels; replace all consonants with a hyphen ("-"). Other people must try to guess what the sentence is. Once you get a sentence, post one of your own.

Additional Rules:
-To avoid confusion, Y is considered a vowel in all cases.
-No proper nouns.
-No foreign words. (EDIT: speak only English)
-Sentences should be long enough to actually figure them out
-Nothing completely random- sentences should be at least coherent. (This is a step up from just "grammatically correct")

For example, the sentence "I eat, drink, and breathe mathematics." would become "I ea-, --i--, a-- --ea--e -a--e-a-i--."

Okay, here goes.

--y -a- --e a-o- -o -u-e i- -a- -o-- a- e-e---o-? I- -a- -o-i-i-e!

"No foreign words" is written in English, and it would be slightly metaphysical to think of it in any other way.

but in other words just stay reasonable,  no impossible sentences

(If the first one is too hard -- it's a rather common chemistry joke)

I -i-- -o ea- -o-- o- -oo-

[hide="answer"]

I like to eat lots of food

Does this obey the rules?

[/hide]

@Prime factorization: You should have more long words in your sentence to make it unambiguous. For example, your sentence also could have been

I like to eat lots of poop

[quote="1=2"][quote="randomdragoon"]-No foreign words.![/quote]

Some Yugoslavian could read that and speak Yugoslavian. Please tell me that you meant : "Only speak English".[/quote]

Just a remark: as a "Yugoslavian" (as there is no more Yugoslavia, it is a bit weird to say that I'm Yugoslavian, but... I was born in Yu  :lol: ) I can with certainity say that there is no such thing as Yugoslav language... I had an urge to say this, for the record...  :lol:

[quote="Prime factorization"]-e --o --o--a--i-a-e- --a-- -ai- -a-y e-a--

[hide="hint"]

This is a preachy message about doing work late

[/hide][/quote]
I'm almost positive that the 3rd word is procrastinate, and am pretty sure that the first word is Be or We (could be he or me also). Otherwise, i have no clue..

The answer is "He who procrastinates shall fail many exams"  :D

[quote="1=2"]-o -e o- -o- -o -e, --a- i- --e -ue--io-.[/quote]

To be or not to be, that is the question.


Let $ ABC$ be an acute triangle,and let $ D$ be the point on the side $ BC$ such that $ \angle BAD\equal{}\angle CAD$.Points $ E$ and $ F$ are the foot of perpendiculars from $ D$ to $ AC$ and $ AB$,respectively.Let $ H$ be the intersection of segments $ BE$ and $ CF$.The circumcircle of triangle $ AFH$ meets the line $ BE$ again at $ G$.Prove that segments $ BG,GE,BF$ can be the sides of a right triangle.


Hello everybody,

what is your favourite CAS (= Computer Algebra System). Famous ones are Maple, Mathematica, Derive and MuPAD.

Maybe we could make a new poll on this topic.

I prefer Maple and MuPAD, both great programs...

You may say, real mathematicians don't use CAS's  :D but it saves much time in calculating trivial things like expanding (a+b+c)^3  :D

I have Mathematica, but I can't exactly say I know how to use it :D:D.

[quote="Daniel Gutekunst"]Hello everybody,

what is your favourite CAS (= Computer Algebra System).[/quote]

My brain. Not much calculative power, but an ability to provide some output known as "proofs" usually not expected from computer calculation systems. Has proven itself in a number of mathematics olympiads.

Well, seriously, of course, algebra systems are often helpful. I use a neolithic version of Maple imbedded in Scientific WorkPlace; I also have Matlab installed on my PC but I am too lazy to learn how to use it :-(

  dg

Laziness seems to be my gratest problem :D. I'd love to know lots of tricks in Mathematica, but I can never get past the basics because I'm too lazy.

Who else scrolled all the way down here?


AMC
USA(J)MO
USAMO
quadratics
algebra unsolved
algebra

Find all positive integers $n$ such that there are $k \geq 2$ positive rational numbers $a_1, a_2, \ldots, a_k$ satisfying $a_1 + a_2 + \ldots + a_k = a_1 \cdot a_2 \cdots a_k = n.$

Eh...

[hide="my bad partial solution"]
First, for n composite, let $n=ab$, then the terms can be a,b,1,1,.....,1
Then for n prime greater than 7.  have the terms be $\frac n2, \frac 12, 4, 1,1,1,.....,1$
And for 7 there is a special case.
And for 1,2,3,5 prove by contradiction using am-gm.

[/hide]

[hide="Solution"]
[u]Case 1[/u]: $n$ is composite.

Let a factor of $n$ be $p\ne 1$. Set $a_1=p$, $a_2=n/p$, and $a_3=a_4=\cdots=a_{n-p-(n/p)+2}=1$. Clearly the condition is satisfied, so all composite integers $n$ can be written in this form.

[u]Case 2[/u]: $2\le n\le 5$ is prime.

By AM-GM, \[ a_1+a_2+\cdots+a_k\ge k\sqrt[k]{a_1a_2\cdots a_k}, \] or \[ n\ge k\sqrt[k]{n} \iff n \ge k^{k/(k-1)}. \] But when $k=2,3,4$, or $k\ge 5$, we see that $n>5$, so this doesn't work.

[u]Case 3[/u] $n\ge 11$ is prime.

Let $a_1=n/2$, $a_2=1/2$, $a_3=4$, and $a_4=a_5=\cdots=a_{(n-9)/2}=1$, so that $a_1+a_2+\cdots=(n+9)/2+(n-9)/2=n$, and $a_1a_2a_3\cdots=(n/2)(1/2)(4)=n$, as desired.

[u]Case 4[/u] $n=7$.

In this case we can have $a_1=\frac{9}{2}$, $a_2=\frac{7}{6}$, and $a_3=\frac{4}{3}$, which works.

Thus the complete solution set is $\{4,6,7,8,9,\ldots\}$.
[/hide]

Sorry, but can somebody answer Differ's question? I have the same question.

[quote="Differ"]How did you people figure out the solution to 7?[/quote]

I guess the motivation is as follows. Proving that 2, 3, 5 don't work isn't too bad, and you'll (obviously) notice that the proof doesn't work for 7. It would seem that you need something pretty contrived or powerful in order to prove that stuff doesn't work in a different way, but this is #4. Always remember that #1's and #4's for the most part have solutions that don't involve anything particularly tricky if you think in the right general direction for long enough. Also, they won't use big theorems or ideas. So assume that there's some easy thing to do. Maybe it's one of those annoying problems where the answer is "yes, it exists." This should at least cross your mind.

Anyway, you first try k=2. It's easy to see by solving the quadratic with that you get irrational numbers. If you play around with k=3, you'll eventually be led to the solution...I guess the key word is play, keep experimenting. As said before, being #4, the answer probably won't be something ridiculous like a_1 = 3912/37, a_2 = 3010395/1337, whatever.

I did n=1,2,3,5 similar to chess64 with AM-GM, then found the same case for n=7. But instead of splitting the rest by composite and prime, I did even and odd:
For $ n\equal{}2m$, consider the numbers $ 2,m,1,1,1,...$.
For $ n$ odd, consider $ \frac{n}{2},4,\frac{1}{2},1,1,...$

The first case works when $ 2\plus{}m\leq 2m \iff m\geq 2 \iff n\geq 4$, the second when $ \frac{n}{2}\plus{}4\plus{}\frac{1}{2} \leq n \iff n\geq 9$.

Finding the case for $ n\equal{}7$ was particularly unhappy for me. I easily limited it down to $ k\equal{}3,4$ by AM-GM arguments and excluding $ k\equal{}2$. Then I tried $ k\equal{}3$, attempting to solve $ a\plus{}b\plus{}c\equal{}abc\equal{}7$. I ended up with an argument like this: for fixed $ a$, we have the equations $ b\plus{}c\equal{}7\minus{}a, bc\equal{}\frac{7}{a}$. Then, $ b,c,$ are roots of the quadratic $ x^2\minus{}(7\minus{}a)x\plus{}\frac{7}{a}$, with discriminant $ a^2\minus{}14a\plus{}49\minus{}\frac{28}{a}$ necessarily a perfect square of a rational. Then, knowing there had to be at least one fraction with 7 (or a multiple of 7) in the numerator, I tried $ a\equal{}\frac{7}{i}$ for $ i\equal{}2,3,4,5,6$ until I hit the jackpot with 7/6.

Another construction for the case $n=7$ is $\frac 76 + \frac 43 + \frac 32 + 3 = 7$. Indeed it's a matter of trying things until you get something that works, since it seems pretty hard for a #4 to try and prove that $7$ is not constructible.

All $n$ except for $1$, $2$, $3$, $5$ work.

[list]
	[*]For composite $n=xy$, since $x+y\le xy$, we can take $x+y+1+\cdots+1=xy\cdot 1\cdots 1=n$.
	[*] For prime $n\ge 11$, we can take
	\[\frac n2+\frac12+4+1+\cdots+1=n\] since $\frac{n+1}{2}+4\le n$.
	[*] For $n=7$, we have
	\[\frac76+\frac92+\frac43=\frac76\cdot\frac92\cdot\frac43=7.\]
        [*] For $n=1,2,3,5$, we have that
	\[\frac nk\ge\sqrt[k]n.\] This is clearly not true for $k\ge n$, so we only have to check $k$ such that $2\le k<n$.
	[list]
		[*] $n=1,2$. No values of $k$ exist to check.

		[*] $n=3$. We have $3\ge k\sqrt[k]{3}$, which does not hold for $k=2$.

		[*] $n=5$. We have $5\ge k\sqrt[k]{5}$, which does not hold for $k=3,4$. For $k=2$, we have
		\[a_1+a_2=a_1a_2=5,\] but solving gives \[\{a_1,a_2\}=\left\{\frac{5-\sqrt5}{2},\frac{5+\sqrt5}{2}\right\},\] which are irrational.
	[/list]
[/list]

The answer is all positive integers except $1,2,3,5.$\\\\ $\textbf{Claim: }$ All odd $n\ge 9$ work.\\\\ $\emph{Proof: }$ Let $m=\frac{n-9}{2}.$ We have $$\underbrace{1+1+\dots+1}_{m\text{ ones}}+\frac{n}{2}+\frac{1}{2}+4=n,$$ $$\underbrace{(1)(1)\dots (1)}_{m\text{ ones}}\left(\frac{n}{2}\right)\left(\frac{1}{2}\right)(4)=n.$$ Therefore, $(a_1,a_2,\dots,a_{m+3})=\left(1,1,\dots,1,\frac{n}{2},\frac{1}{2},4\right)$ works, so we are done. $\blacksquare$\\\\ $\textbf{Claim: }$ All even $n\ge 4$ work.\\\\ $\emph{Proof: }$ Let $m=\frac{n-4}{2}.$ We have $$\underbrace{1+1+\dots+1}_{m\text{ ones}}+\frac{n}{2}+2=n,$$$$\underbrace{(1)(1)\dots (1)}_{m\text{ ones}}\left(\frac{n}{2}\right)(2)=n.$$ Hence, $(a_1,a_2,\dots,a_{m+2})=(1,1,\dots,1,\frac{n}{2},2)$ works. $\blacksquare$\\\\$\textbf{Claim: }$ $n=7$ works.\\\\ $\emph{Proof:}$ The triple $(a_1,a_2,a_3)=(\frac{9}{2},\frac{7}{6},\frac{4}{3})$ works. $\blacksquare$ \\\\ $\textbf{Claim: }$ $n=1,2,3,5$ don't work.\\\\ $\emph{Proof: }$ Suppose $a_1+a_2+\dots+a_k=a_{1}a_{2}\dots a_{k}=n.$ Then, by AM-GM, $$\frac{n}{k}\ge\sqrt[k]{n}\implies n\ge k^{k/(k-1)}.$$ Note that $x^{x/(x-1)}$ increases as $x$ increases. Therefore, since $3^{3/2}>5,$ we must have $k=2.$\\\\ But it is easy to check that there are no $a_1,a_2$ such that $a_1+a_2=n$ and $a_{1}a_{2}=n$ for $n=1,2,3,5,$ as desired. $\blacksquare$

For even numbers $n\geq 4$, observe that $\frac{n}{2}\cdot 2\cdot \underbrace{1\cdot1\cdots \cdot 1}_{\frac{n}{2}-2}=n=\frac{n}{2}+\underbrace{1+1+\cdots +1}_{\frac{n}{2}-2}+2.$\\For odd $n\geq 9, \frac{n}{2}\cdot\frac{1}{2}\cdot 4\cdot\underbrace{1\cdot1\cdots \cdot 1}_{\frac{n-9}{2}}=n=\frac{n}{2}+\frac{1}{2}+4+\underbrace{1+1+\cdots +1}_{\frac{n-9}{2}}.$\\For $n=7,$ one can see that $\frac{7}{6}\cdot\frac{4}{3}\cdot\frac{9}{2}=7=\frac{7}{6}+\frac{4}{3}+\frac{9}{2}.$\\For $n\in\{1,2,3,5\},$ assume there are $\ell$ positive rationals $\geq 1$ named $G_1,G_2,...,G_{\ell}$ (If there are none then the product is less than $1$) then $\ell < n$ and $G_1+G_2+...+G_{\ell}\geq \ell \sqrt[\ell]{G_1G_2...G_{\ell}}\geq \ell \sqrt[\ell]{n}> n,$ contradiction.

s t o r a g e

The answer is all $n$ except $1,2,3,5$.

First, we note that all composite $n$ work: factor $n$ into $p\cdot q$ for $p,q\geq 2$. Taking $p$, $q$, and $n-p-q$ ones works, since \[n-p-q=pq-p-q=(p-1)(q-1)-1\geq 0\] as $p,q\geq 2$.

We also note that all primes $\geq 7$ work. For $7$, simply take the set $\frac{7}{6},\frac{4}{3},\frac{9}{2}$. For all primes $p$ greater than $7$, we can take the construction $\frac{p}{2}$,$\frac{1}{2}$, $4$, and $\frac{p-9}{2}$ ones.

Now I'll show that $1,2,3,5$ don't work. Clearly $k=2$ does not yield a valid solution for all four of these numbers. Suppose $k\geq 3$. By AM-GM we have that \[n\geq k\cdot n^{\frac{1}{k}}\] One can show that the right hand side is an increasing function in $k$, and that minimum occurs at $k=\ln n$. Since $\ln n<3$ for $n=1,2,3,5$, it suffices to check that $n\geq 3\cdot n^{\frac{1}{3}}$. However, this is only true when $n\geq 3\sqrt 3$, so there does not exist a valid $k\geq 3$ for $n=1,2,3,5$, and thus they don't work.

The answer is all positive integers except for $1,2,3,5$.

[b]Case 1:[/b] $n<4$
First note that by AM-GM, $\frac nk\ge\sqrt[k]n$, so $n\ge k^{\frac k{k-1}}$. Since $k\ge2$, this implies $n\ge4$.

[b]Case 2:[/b] $n=5$
Then $k\le2$ from the aforementioned bound, so $k=2$. If we let $a_1+a_2=a_1a_2=5$, then solving we get:
$$a_1,a_2=\frac{5\pm\sqrt5}2$$
which are both irrational.

[b]Case 3:[/b] $n$ is composite
Let $n=xy$ for positive integers $x,y\ge2$. Let $k=n+2-x-y$, $k$ is positive since:
$$x+y\le x\cdot\frac y2+y\cdot\frac x2=n.$$
Then, let $a_1=x$, $a_2=y$, and $a_i=1$ otherwise.

[b]Case 4:[/b] $n$ is prime and $n\ge11$
Let $k=\frac{n-1}2-1$ and $a_1=\frac n2$, $a_2=\frac12$, $a_3=4$, and $a_i=1$ otherwise.

[b]Case 5:[/b] $n=7$
Let $k=3$ and $a_1=\frac92$, $a_2=\frac43$, $a_3=\frac76$.

Using $AM-GM$ gives that $\sum a_i \ge k \sqrt[k]{\prod a_i} \Rightarrow n^{k-1} \ge k^k \ge 4$, which gives that $n \neq 1,2,3,5$

Observe that we can use the number $1$  to full the difference between the sum and $n$ while not affecting $n$, Hence it motivates us to select some $a_1,a_2,..,a_i$ such that their sum and product is integer, hence it motivates us to use $n$'s factors

      Let $n=ab, a,b \in \mathbb{N}$, then let $a_1,a_2=a,b$, then take enough $1$'s so that $a+b+m=n$. But observe that $a+b>ab$ if $a=1$, hence it is only true for composite numbers.
For prime numbers, select $a_1,a_2,a_3= \frac{n}{2},\frac{1}{2},4$ and fill the gaps with $1$s. (note that $2|n+1$).  But the case $n=7$ is a exception since sum gives $8$. Hence it is enough to show a specific configuration for $7$.

   It is easy to prove $k=2$ is impossible. Hence assume that it has configuration for $k=3$. Then $a_1,a_2,a_3=\frac{7}{x} , \frac{xz}{y} ,\frac{y}{z}$
Now, it is matter of calculations ( starting from $\sum a_i =7$, to $2$). Starting from $x=2$ to $x=6$. Solving for $z$ for $x=2,3,4,5$ makes $z$ irrational, but if $x=6$, we have $z=\frac{3y}{4}, \frac{2y}{9}$ , since $y,z$ are integers, choose some $y$ that is divisible by $3$ or $2$,  which gives $(y,z)=(4m,3m),(9m,2m)$, therefore we have $(a_1,a_2,a_3)=(\frac{7}{6},\frac{9}{2},\frac{4}{3})$ which clearly works, hence we are done . $\square$

Solved with [b]megarnie[/b].
Used the 30% hint on ARCH.

The answer is $n=4$ or $n\ge 6$.

If $n\le 3$, we have
\[\sqrt[k]{n}=\sqrt[k]{a_1a_2\dots a_k}\le\frac{a_1+a_2+\dots+a_k}{k}=\frac{n}{k}.\]
Thus if $k>n$, this is impossible. We can then manually check that this inequality is never satisfied when $2\le k\le n\le 3$.

If $n=5$, with the inequality, we can check that only $k=2$ works out of $k\in\{2,3,4,5\}$. But solving for $a_1$ and $a_2$ result in irrational numbers, so this doesn't work either.

For composite numbers $n$, let $p$ be its smallest prime factor and let $n=mp$. Then set $k=n-p-m+2$, $a_1=p$, $a_2=m$, $a_3=a_4=\dots=a_k=1$. Clearly this works(we only need to check that $n\ge p+m$ which is obvious).

For prime $n\ge 11$, set $k=\frac{n-3}{2}$, $a_1=\frac n2$, $a_2=\frac12$, $a_3=4$, $a_4=a_5=\dots=a_k=1$. Clearly this works(we only need to check that $k\ge 3$ which is obvious).

For $n=7$, use $a_1=\frac76$, $a_2=\frac43$, and $a_3=\frac92$. This completes our proof. $\blacksquare$

[hide=Remark]I really did think that $n=7$ was impossible for a while until I read the ARCH hint. Then I proved that $k=3$ or $k=4$, but couldn't finish until I decided to mess around with $\nu_p$ stuff. It turned out that when $k=3$, we need the numbers to look like $\frac{7a^2}{bc},\frac{b^2}{ac},\frac{c^2}{ab}$. Then we get $7a^3+b^3+c^3=7abc$, which I couldn't find a solution to until [b]megarnie[/b] pointed out that $(1,2,3)$ works.[/hide]

All $n$ work except for $1$, $2$, $3$ and $5$.
Clearly $1$ does not work.
By AM-GM we have $\sqrt[k]{n} \geq \frac{n}{k}$. Let $p$ be some prime dividing $n$. Then if $n$ is composite then $a_1 = \frac{n}{p}$, $a_2 = p$, and then letting the rest of $a_i$ being $1$(until the sum of $a_i$ is $n$) works.
If $n$ is prime then AM-GM gives $\sqrt[k]{n} \geq \frac{n}{k} \implies n \geq k^{k/(k-1)}$. From this we see that $n = 2, 3$ doesn't work. We also can find that $n = 5$ doesn't work by checking cases.
If $n = 7$ then $a_1 = \frac{7}{6}$, $a_2 = \frac{3}{2}$, $a_3 = \frac{4}{3}$ and $a_4 = 3$ works.
If $n$ is prime and is $\geq 11$, then $a_1 = \frac n2$, $a_2 = \frac 12$, $a_3 = 4$, and letting the rest of $a_i$ being $1$ works(until the total sum is $n$).
Really boring problem


function
Asymptote
calculus
derivative
logarithms
calculus computations

Find the global maximum and minimum of the function

f(x) = 1/2 (e ^ 2x) - 5(e^x) + 4x

on the interval [ln1/4 , ln 5].

Find the vertical and horizontal asymptotes of the function

y=(e^-1/x)

and sketch the graph.

Let y = ln| (x^2) - 3x + 1 | , find all critical points of y, determine its local extme values and intercaptes with the x axis , find the equation of its asymptotes.

[quote="GEEK007"]Find the global maximum and minimum of the function

$f(x) = \frac1{2}e^{2x}-5e^{x}+4x$

on the interval $\left[ \ln\frac1{2}, \ln 5\right]$. [/quote]

To determine the global maximum and minimum of the function, we first look for extrema (maximums and minimums) inside the given interval, and check the end points of the interval for possible global extrema. So look for the roots of the derivative, that is solve:

\[f^{\prime}(x) = e^{2x}-5e^{x}+4=0\Rightarrow \left( e^{x}-1\right)\left( e^{x}+4\right) = 0\Rightarrow e^{x}-1=0\mbox{ or }e^{x}+4=0\]

now of these, namely $e^{x}-1=0$, produces $x=\ln 1 = 0$, and the second of these, namely $e^{x}+4=0$, has no solution (since $e^{x}>0$ for all real x). So the only critical point is $x=0$, and this is inside our interval since $\left[ \ln\frac1{2}, \ln 5\right] \approx \left[-.69 , 1.61\right]$. Note that $f(0)=\frac1{2}-5=-\frac9{2}=-4.5$. Remember to check the end points:

\[f\left(\ln\frac1{2}\right) = \frac1{2}e^{2\ln\frac1{2}}-5e^{\ln\frac1{2}}+4\ln\frac1{2}= \frac1{2}e^{\ln\left(\frac1{2}\right)^{2}}-5\left(\frac1{2}\right)+4\ln\frac1{2}\]
\[=\frac1{2}\left(\frac1{4}\right)-\frac5{2}+4\ln\frac1{2}=-\frac{19}{8}+4\ln\frac1{2}\approx-5.15\]

\[f\left(\ln 5\right) = \frac1{2}e^{2\ln 5}-5e^{\ln 5}+4\ln 5 = \frac1{2}e^{\ln 5^{2}}-5\cdot 5+4\ln 5\]
\[=-\frac{25}{2}+4\ln 5 \approx-6.06\]

So we see that $f(0)=-4.5$ is the global maximum and that $f\left(\ln 5\right) \approx-6.06$ is then the global minimum. Here is a plot of the function over the given interval:

[url=http://ezimagecenter.com/out.php/i187_Calcpraticeexamproblems.JPG][img]http://ezimagecenter.com/out.php/i187_Calcpraticeexamproblems.JPG[/img][/url]

[quote="GEEK007"]Find the vertical and horizontal asymptotes of the function

$y=e^{-\frac1{x}}$

and sketch the graph.[/quote]

Here is a plot of the function

[quote="GEEK007"]Let y = ln| (x^2) - 3x + 1 | , find all critical points of y, determine its local extme values and intercaptes with the x axis , find the equation of its asymptotes.[/quote]

[color=red]Note[/color]: may finish later.