inequalities
limit
real analysis
real analysis unsolved

Let $ \ell_{1}(\mathbb{N}) \equal{} \left\{(x_j) \in \mathbb{R}: \sum_{j\equal{}1}^{\plus{}\infty}|x_j| < \plus{}\infty\right\}$ and $ \ell_{2}(\mathbb{N}) \equal{} \left\{(x_j) \in \mathbb{R}: \sum_{j\equal{}1}^{\plus{}\infty}|x_j|^{2} < \plus{}\infty\right\}$. Prove that the unitary closed ball in $ \ell_1(\mathbb{N})$, $ B[0,1] \equal{} \left\{(x_j): \sum_{j\equal{}1}^{\plus{}\infty}|x_j| \leq 1\right\}$ is closed in $ \ell_{2}(\mathbb{N})$ with empty interior.

I really need some help!!!!! Someone can help me, please!!!!

pick $ (x_j) \in B[0,1] \to \sum_{j \equal{} 1}^\infty |x_j| \equal{} \sigma \le 1$

$ \forall \epsilon > 0, \exists N_\epsilon$ s.t. $ \sum_{j \equal{} N_\epsilon}^\infty |x_j| < \epsilon$

$ \sum_{j \equal{} 1}^\infty |x_j|^2 \equal{} \left( \sum_{j \equal{} 1}^\infty |x_j| \right)^2 \minus{} \sum_{j \equal{} 1}^\infty \sum_{k\ne j} |x_j\parallel{}x_k| \equal{} \sigma^2 \minus{} \sum_{j \equal{} 1}^\infty \left[ |x_j| \sum_{k\ne j} |x_k|\right]$

$ \beta_j \equal{} \sum_{k\ne j} |x_k| < \sigma, \forall x_j \ne 0 \to\sum_{j \equal{} 1}^\infty \beta_j |x_j| \equal{} \sigma ' < \sigma^2$
$ \sum_{j \equal{} 1}^\infty |x_j|^2 \equal{} \sigma^2 \minus{} \sigma ' \le 1$ so $ (x_j) \in l_2(N)$

so $ B[0,1] \subset l_2 (N)$

how do you measure in $ l (N)$?

$ \|(x_j)\|_1 \equal{} \sum_{j\equal{}1}^{\plus{}\infty}|x_j|$ is a metric in $ \ell_1(\mathbb{N})$ and $ \|(x_j)\|_2 \equal{} \sqrt{\sum_{j\equal{}1}^{\plus{}\infty}|x_j|^{2}}$ is a metric in $ \ell_2(\mathbb{N})$.

It is easy to see that $ \|(x_j)\|_{2} \leq \|(x_j)\|_{1}$.

Pick a point $ x \equal{} (x_j)$ in $ \overline{B_{\ell_1}[0, 1]}$ where the closure is taken in $ \ell_{2}(\mathbb{N})$. Then, there is a sequence $ (x^{n}_{j})$ in $ B_{\ell_1}[0,1]$ such that

$ \|x^n \minus{} x\|_2 \equal{} \|(x^n_j) \minus{} (x_j)\|_2 \rightarrow 0$  when $ n \rightarrow \plus{}\infty$.

But I cant see how to prove that $ x \equal{} (x_j) \in B_{\ell_1}[0, \, 1]$!!!!!!!!!!!!!!

[hide="Hint"]Assuming $ \|x\|_1>1$ it holds that $ \sum_{j\equal{}1}^n|x_j|>1\plus{}\varepsilon$ for some $ \varepsilon>0$ and $ n>0$. Now look at the sequence $ (x^n_j)$ restricted to the first n entries, i.e. for $ j\equal{}1,\ldots,n$ to get a contradiction.

What happens to the inequality of norms if you assume that the ball contains a non-void open subset of $ l^2$?[/hide]

Sorry, I tried your suggestion...but I cant see yet...

Did you use that $ \lim_{n\to \infty}x^n_j\equal{}x_j$ for every $ j$?

Use fatou's lemma and convergence in each component to conclude $ \sum |x_n| \leq \liminf_{k} \sum |x^k_n| \leq 1$


Let  x and y  be natural numbers.Find the number of solutions  of equation
         5.x+4.y=3643

$ y\equal{}{3643\minus{}5x\over 4}\equal{}910\minus{}x\plus{}{3\minus{}x\over 4}$

Therefore $ 3\minus{}x\equal{}4t\implies x\equal{}\minus{}4t\plus{}3\implies y\equal{}5t\plus{}907$ where $ t$ is an integer.

$ x\geqslant 1\implies\minus{}4t\geqslant\minus{}2\implies t\leqslant 0$

$ y\geqslant 1\implies 5t\geqslant\minus{}906\implies t\geqslant\minus{}181$

So $ t$ can take $ 182$ values for which we get natural pairs $ (x,y)$, hence that's the number of solutions.


In Geometry, what I consider as a nice theoretically explained book is Geometry Revisited, and as a very good problem book Prasolov.
Can you tell me some inequality e-books which are like those 2 in geometry? Thanks.

e-books is a hard one because most english books are protected heavily by copyright laws.


How many positive integers less than 575 are divisible by $ 2$,$ 3$, and $ 5$?

$ 2\cdot 3\cdot 5\equal{}30$
every multiple of $ 30$ is divisible by $ 2,3,5$
There are 19 multiples of 30 less than 575 :)


real analysis
trigonometry
integration
limit
function
abstract algebra
logarithms

Define $P : (r, \theta) \mapsto \frac{1}{2\pi}\frac{2(-r+\cos \theta)}{1+r^{2}-2r\cos\theta}$. If $\int_{-\pi}^{\pi}g(\varphi) \, d\varphi = 0$, why is $\lim_{r \to 1^{-}}\int_{-\pi}^{\pi}P(r, \theta-\varphi) g(\varphi) \, d\varphi = g(\theta)$?

Note that this is directly relevant to solving the Neumann problem with respect to the Laplacian. Normally you can use the simple approximate identity argument, but the family of functions $P(r, \theta)$ doesn't work like it does for the Poisson kernel -- for example $\int_{-\pi}^{\pi}P(r, \theta) \, d\theta = 0$ for $0 \le r < 1$.

It may or may not be useful to observe that $\frac{\partial}{\partial r}\left[-\frac{1}{2\pi}\log (1+r^{2}-2r\cos\theta)\right] = \frac{1}{2\pi}\frac{2(-r+\cos \theta)}{1+r^{2}-2r\cos\theta}$.

Sorry guys, this is actually trivial -- I'm so stupid. Rewrite $P(r, \theta) = \frac{1}{r}\left(Q(r,\theta)-\frac{1}{2\pi}\right)$, where Q is the Poisson kernel given by $Q(r,\theta) = \frac{1}{2\pi}\frac{1-r^{2}}{1+r^{2}-2r\cos\theta}$. Then $\lim_{r\to 1^{-}}\left(\frac{1}{r}\int_{-\pi}^{\pi}Q(r,\theta-\varphi)g(\varphi) \, d\varphi-\frac{1}{2\pi}\int_{-\pi}^{\pi}g(\varphi) \, d\varphi\right)$ is clearly $g(\theta)$.


Hey can someone please help me out with this problem!!!

Prove that the intersection points of the opposite sides of an inscribed hexagon are collnear(lie on one line)

This is a standard result known as [url=http://mathworld.wolfram.com/PascalsTheorem.html]Pascal's Theorem[/url].


group theory
abstract algebra
superior algebra
superior algebra solved

Is there a group of infinite order with only finitely many subgroups?

Of course not :). If it has an element of infinite order, just take infinitely many subgroups of the group generated by that element. On the other hand, if it only has elements of finite order, take infinitely many subgroups generated by such elements.


abstract algebra
group theory
superior algebra
superior algebra unsolved

1. Viewing Z and Q as additive groups, show that Q/Z is a torsion group, which has one and only one subgroup of order n for each integer $ n \ge 1$ and that this group is cyclic.

2. Let H be a subgroup of a finite abelian group G. Show that G has a subgroup that is isomorphic to G/H.

3. Show that the action of the alternating group $ A_n$ on {1,...,n} is (n-2)-transitive.

1. which multiple of $ a/b \plus{} \mathbb{Z}$ is zero in $ \mathbb{Q}/\mathbb{Z}$? show that $ 1/n \plus{} \mathbb{Z}$ has order $ n$, hence generates a subgroup of order $ n$, which equals the $ n$-torsion subgroup $ T_n$. now if $ U$ is an arbitrary subgroup of order $ n$ in $ \mathbb{Q}/\mathbb{Z}$, then $ U \subseteq T_n$ by lagrange and even equality by comparings orders.

2. use the structure theorem.

3. let $ \{1,..,n\} \equal{} \{x_1,...,x_n\} \equal{} \{y_1,...,y_n\}$, we look for a $ \sigma \in A_n$ which maps $ x_i$ to $ y_i$ for $ i\equal{}1,...,n\minus{}2$. if we consider all $ \sigma \in S_n$ with this property, we have only two possibilities, namely $ x_{n\minus{}1} \to y_{n\minus{}1}$ and $ x_{n\minus{}1} \to y_n$. note that these permutations have a different sign, so one must be in $ A_n$.

Can anyone explain 1 and 2 in more detail?

it's your exercise, isn't it? I've given you enough hints to getting started.


Given real $a_1,\cdots, a_n>0$ and $p>1$ prove that
$\sum_{k=1}^n (\frac {a_1+\cdots +a_k}k)^p \leq (\frac p{p-1})^p \sum_{k=1}^n a_k^p$

it looks strange. What is the equality case?
I think it is a strict inequality
\[ (\frac{p}{p-1})^p \sum a_k^p > \sum a_k^p \ge k(\frac{a_1+...+a_k}{k})^p > \sum \frac{a_1+...+a_k}{k})^p \]
?

Hardy's inequality is proved somewheer on site. Actually, it was first topic on ML I saw when I visited it first time :) .
It will not a big surprize for me, if I wrote the proof here, since I "invented" ( :P ) and proved Hardy's inequality some years ago.

[quote="Megus"]siuhochung: your inequalities doesn't always hold - look that it is said that $a_1,...,a_n$ are reals[/quote]

[quote="iura"]Given real $a_1,\cdots, a_n>0$ [/quote]

apart from that, i really don't understand your proof:
first i think that it is $\sum_{k=1}^n {a_k^p}\geq n(\frac{a_1+a_2+...+a_n}{n})^p$ instead of
$\sum_{k=1}^n {a_k^p}\geq k(\frac{a_1+a_2+...+a_k}{k})^p$ because k is variable, and wouldn't have any sense...
 finally, i don't see why it is so clear that $n(\frac{a_1+a_2+...+a_n}{n})^p > \sum_{k=1}^n {(\frac{a_1+a_2+...+a_k}{k})^p}$
am i missing something?

ahhhhhhhh..
sorry i misread the problem.........


geometry
geometric transformation
rotation
3D geometry

[quote="unimpossible"]1. Bill buys wigits for 3 dollars, gigts for 8 dollars, and pigits for 11 dollars. If Bill bought exactly 100 aliens for 400 dollars, and he purchased as few as possible, how many of each type of alien did he buy?
[/quote]
What did he purshase as few as possible?

[quote="unimpossible"]1. Bill buys wigits for 3 dollars, gigts for 8 dollars, and pigits for 11 dollars. If Bill bought exactly 100 aliens for 400 dollars, and he purchased as few (EDIT: gigits) as possible, how many of each type of alien did he buy?

2. On a 5 by 5 grid of unit squares, one unit square is colored blue, one unit square is coloered red, and the rest of the unit squares are white. Grids are considered different only if no rotation could turn one into the other.  How many different grids are there?

3. The cities of Smallville and Largeville are 300  miles apart.  Jim left from Smallville to go to Largeville at 10 a.m.  Mickey left Largeville to go to Smallville at 10:30 a.m. on the same day.  Jim traveled at a constant speed that was twice mickey's constant speed, and they both arrived at a point 90 miles from Largeville at the same time.  what was MIckey's constant speed, in miles per hour?

4.  How many sets of 7 integers are there such that the mean is 20, the median is 21, the mode is 22 and unique, and the range is 10?

5.  What is the least positive integer that cannot be expressed as the sum of 8 or less cubes of positive integers?[/quote]

[hide="1"]

w, g, and p for 3, 8 and 11 dollars.  He purchased as little g as possible.
Set up equations:

w+g+p=100
3w+8g+11p=400

Multiply the top eq. by 3, subtract from the second equation to get 5g+8p=100.  From here, you can just use some guess and check to find ordered pairs of (g,p) where g is as small as possible.

You get g=4, p=10.  Therefore, w=86.

86 wigits, 4 gigits and 10 pigits.
[/hide]
[hide="3"]

Simply S*T=D.

Whenever I do these problems i find it [i]very useful [/i] to make an organized table of the speeds, distances and times.

[code]
  |  D  |  T   |  S  |
J | 210 |   t   |  s   |
M| 90  |  t-1/2 |  s/2  |
[/code]

(eww, ugly table XP)

Then,

$ \frac{s}{2}(t\minus{}\frac{1}{2})\equal{}90$
$ ts\equal{}210$

$ \frac{st}{2}\minus{}\frac{s}{4}\equal{}90$
$ \frac{210}{2}\minus{}\frac{s}{4}\equal{}90$
$ 105\minus{}\frac{s}{4}\equal{}90$
$ \frac{s}{4}\equal{}15$
$ \boxed{\frac{s}{2}\equal{}30}$

Which is Mickey's constant speed, which we were asked to find.

[/hide]

ok thanks.  I think i got caught up in some algebra errors for number 3.

[hide="2"]
There are 25 places you can put the blue square and 24 places you can put the red square after placing the blue square. However, there are 4 different rotations for each possible grid, so we have to divide by 4.
The answer is $ \frac{25\times24}{4}\equal{}\boxed{150}$
[/hide]

[hide="5"]
We can make all the numbers from 1-8 using ones only.
Using an 8 and up to 7 ones, we can make the numbers from 9-15
Using two 8's and up to 6 ones, we can make the numbers from 16-22.
However, we cannot make 23, since three 8's is too big, and the next cube, 27, is also too big, so the answer is 23.
[/hide]

[quote="unimpossible"]1. Bill buys wigits for 3 dollars, gigts for 8 dollars, and pigits for 11 dollars. If Bill bought exactly 100 aliens for 400 dollars, and he purchased as few as possible, how many of each type of alien did he buy?

2. On a 5 by 5 grid of unit squares, one unit square is colored blue, one unit square is coloered red, and the rest of the unit squares are white. Grids are considered different only if no rotation could turn one into the other.  How many different grids are there?

3. The cities of Smallville and Largeville are 300  miles apart.  Jim left from Smallville to go to Largeville at 10 a.m.  Mickey left Largeville to go to Smallville at 10:30 a.m. on the same day.  Jim traveled at a constant speed that was twice mickey's constant speed, and they both arrived at a point 90 miles from Largeville at the same time.  what was MIckey's constant speed, in miles per hour?

4.  How many sets of 7 integers are there such that the mean is 20, the median is 21, the mode is 22 and unique, and the range is 10?

5.  What is the least positive integer that cannot be expressed as the sum of 8 or less cubes of positive integers?[/quote]

[hide]

4. Just list

so the middle number has to be 21, they have to add up to 140, the mean is 22, and the range has to be 10. So we'll go like this

_,_,_,21,_,_,_

So, let's say that all of the 3 numbers above 21 are 22.

_,_,_,21,22,22,22

That means that the bottom number is 12
12,_,_,21,22,22,22

So those numbers have to be 20, 21.

12,20,21,21,22,22,22

so that is one way.

It can't be _,_,_,21,_,22,22 because then that blank in between 21 and 22 would have to be 21, which would make the mode not unique.

So _,_,_,21,22,22,_ is the only remaining option.

So, the highest that the highest number can be is 28, or the mode wouldn't be unique.

So 18,19,20,21,22,22,28 = 150
Except, it is too high, so 28 is to high of a highest number.

15,17,18,21,22,22,25 so 25 is the highest high number.

15,16,19,21,22,22,25
14,17,20,21,22,22,24
14,18,19,21,22,22,24
13,19,20,21,22,22,23

so

12,20,21,21,22,22,22
15,17,18,21,22,22,25
15,16,19,21,22,22,25
14,17,20,21,22,22,24
14,18,19,21,22,22,24
13,19,20,21,22,22,23

the answer is 6.[/hide]


Let $ \mathcal{P}(\mathbb{R})$ be the set of all measures on $ (\mathbb{R}, \mathcal{B}(\mathbb{R}))$.
Is there any measure $ \mu$ on $ (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ such that $ P\ll \mu$ for every $ P\in \mathcal{P}(\mathbb{R})$?

The counting measure will certainly do the job. I am just wondering whether there is any other measure with the above property?


let x,y,z be positive reals. prove that

 xyz(x+2)(y+2)(z+2)<=(1+2(xy+xz+zy)/3)^3  :huh:

[quote="Leonhard Euler"]It is kmo problem. see http://www.mathlinks.ro/viewtopic.php?t=111065[/quote]

thanks


Actually, we only need to consider the case $ x_{1}=x_{2}=...=x_{n-1}$. I solved by induction. This problem was once posted by me and the detailed solution will be found in my future book this IMO.


Looking at each solution to all the AIME's from 1983, most geometry problems seem to require only a very fundamental understanding on the precalculus methods applied in a very clever way, rather than the use of advanced and esoteric formulas. Can anyone elaborate or comment?

That's basically what the AIME tests. Cleverness.

The most complicated things you need to know are probably: Stewart's Thm (rarely), Ptolemy's Thm, Ceva's Thm, Menelaus's Thm, and a few more which other people can kindly state at their will.

All the lovely inequalities may come in handy like AM-GM, rarely Cauchy, and others.

Oh, yeah, there is very little in the AIME that is advanced in terms of the standard school curriculum. Rather, it is deeper, more based on multistep problems that require you to keep track of what you are doing and not make false assumptions. Much of real-world math is not advanced, in the curricular sense, but it is misunderstood because too many school-trained individuals have only seen easy textbook exercises and not challenging real-world problems.

I'd agree, and I rather like it this way...It makes high scores on the test accessable, even if you are not taking high-level mathematics.  On the AIME, all the problems are doable with basic methods, and a lot of thought and problem-solving techniques.

JB

Even though many problems can be done woth an advance thm,or higher level principal,but they are always not required,it might juse be more complicated and hard to think of......

that is because we dont see these problems often in our daily lives.

actually, ive seen these types of problems in easier tests, ie mu alpha theta, but the problems on AIME ask for different or more complicated solutions.


Consider a semicircle(C) with diameter AB. Let w be a circle tangent to AB at L and to C at K. Let l be a line perpendicular to AB (at T) and tangent to w. l intersects C at Q. Prove that the bisectors of <AQT and AKB meet on AB.

Let $\ell$ touch $\omega$ at $S$. Now consider the inversion of pole $B$ and power $BQ^2$. It turns the circle of diameter $AB$ into the line $TQ$ and vice-versa, and it invariates $\omega$. This means that it turns $K$ into $S$ and vice-versa, so $B,K,S$ are collinear. We also have $BL^2=BQ^2$, so $BL=BQ$ because $L$ must be invariant.

We have $\angle BKL=\angle SLT=45^{\circ}=\frac{\angle AKB}2$, so the bisector of $\angle AKB$ cuts $AB$ in $L$.

From the fact that $KL$ is the bisector of $\angle AKB$ we derive $\frac{LA}{LB}=\frac{KA}{KB}=\frac{TS}{TB}$, and, since $LT=TS$, we have $\frac{LA}{LT}=\frac{BL}{BT}=\frac{BQ}{BT}=\frac{QA}{QT}$, so $QL$ is the bisector of $\angle AQT$, and the conclusion follows (the two bisectors cut each other in $L$, which is on $AB$).

I used an inverson with pole $L$. Denote by $X'$ the point(circle,semi-circle) $X$ under this inversion. Then $w'$ will be a line parallel to $AB$ and tangent to $C'$,and $(TQ)'$ will be a circle passing through $L$ also tangent to $w'$ and with center on $AB$. Hence the semi-circle of $(TQ)'$ at the same side of $AB$ as $C'$, will be equal to $C'$. Then we easily get that $\angle{AQL}=\angle{Q'A'A}=\angle{Q'T'L}=\angle{TQL}$ and we are done.

I read an article in Gazeta Matematica No. 5-6/2004 today (written by Neculai Roman), in which the author establishes a generalization of this problem (actually, a generalization of the fact that $AL$ bisects $\angle AQT$):

Let $PQ$ be a chord of the circle $(O)$, and let $\Omega$ be a circle internally tangent to $(O)$ and tangent to $PQ$ at $C$. Take $B\in PQ$ s.t. $AB\perp OQ$ and $AB$ is tangent to $\Omega$. Then $AC$ bisects $\angle PAB$.

Grobber solution is excellent as always.
A remark to the Sailor's solution:
I think in the last sentence he meant <Q'A'A = Q'B'B or <Q'A'L = Q'B'L
which gives <AQL = BQL by inverse inversion and problem is solved.

There was a typo but I have corrected it.


limit
trigonometry
calculus
calculus computations

$lim_{x \to 0} \frac{x^2}{\sqrt{1+sinx}-\sqrt{cosx}}=?$

$\lim_{x\rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{\cos x}}{x}=\lim_{x\rightarrow 0} \frac{1+\sin x-\cos x}{x(\sqrt{1+\sin x}+\sqrt{\cos x})}=\lim_{x\rightarrow 0} \frac{2\sin ^2\frac{x}{2}+\sin x}{x(\sqrt{1+\sin x}+\sqrt{\cos x})}$

${=\lim_{x\rightarrow 0} \frac{2\sin \frac{x}{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}{x(\sqrt{1+\sin x}+\sqrt{\cos x})}=\lim_{x\rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\cdot \frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\sqrt{1+\sin x}+\sqrt{\cos x}}}$

$=1\cdot \frac{0+1}{1+1}=\frac{1}{2},$ thus $\lim_{x\rightarrow 0} \frac{x^2}{\sqrt{1+\sin x}-\sqrt{\cos x}}=\lim_{x\rightarrow 0} \frac{x}{\frac{\sqrt{1+\sin x}-\sqrt{\cos x}}{x}}=\frac{0}{\frac{1}{2}}=\boxed{0}.$


blogs
function
integration
calculus
calculus computations

This was a problem of the week from the blog Topological Musings. I was reminded of it when I saw the post about the beta function (hint, hint  :wink: ). Of course you'll ruin the fun if you look it up on that blog since the answer has been posted. (There are at least three fundamentally different ways to do it).

What is $ \sum_{k\equal{}0}^\infty \frac{1}{{n\plus{}k \choose k}}$ for $ n>1$?

i am not sure why you gave the hint, you  [i]spoiled[/i] it for me. :( i typically like to solve problems without anyone giving me any hints... unless, of-course, i ask for it !

i'll think of another way to solve it later, but here is the one that i used now.

$ \sum_{k\equal{}0}^\infty \frac{1}{\dbinom{n\plus{}k}{k}}\equal{}\sum_{k\equal{}0}^\infty \frac{n!\;k!}{(n\plus{}k)!}\equal{}\sum_{k\equal{}0}^\infty \frac{n\Gamma(n) \; \Gamma(k\plus{}1) }{\Gamma(n\plus{}k\plus{}1)}\equal{}n\;\sum_{k\equal{}0}^\infty\textbf B(k\plus{}1,n)$

$ \equal{}n\;\sum_{k\equal{}0}^\infty\int_0^1x^k\;(1\minus{}x)^{n\minus{}1}\;dx\equal{}n\int_0^1(1\minus{}x)^{n\minus{}1}\left(\sum_{k\equal{}0}^\infty \;x^k\right)\;dx$

$ \equal{}n\int_0^1(1\minus{}x)^{n\minus{}2}\;dx \equal{}\minus{}n\int_0^1(1\minus{}x)^{n\minus{}2}\;d(1\minus{}x)\equal{}\minus{}n\;\frac{(1\minus{}x)^{n\minus{}1}}{n\minus{}1}\bigg|_0^1\equal{}\boxed{\frac{n}{n\minus{}1}}$

OK, OK... :D let me ask a similar kind of q

solve, for $ |k|<1$,

$ \sum_{i\equal{}2}^\infty\; \frac{1}{i!}\; \sum_{j\equal{}0}^\infty\; \frac{k^j}{\dbinom{i\plus{}j}{i}}$


geometry
ratio
probability
real analysis
analytic geometry
invariant
parameterization

I alluded to a long analysis in previous posts.  I'm trying to start a new thread in the Round Table to allow me to present this, have a discussion on what I present and perhaps learn something from this community.

I present things in a roundabout order, but I believe that this order will help some follow the discussion and build things that are used later.  I would appreciate your patience as I do this the best way I know.

Galileo Galilei is reported to have written, "Philosophy is written in this grand book - I mean the universe - which stands continually open to our gaze, but it cannot be understood unless one first learns th comprehend the language and interpret the characters in which it is written.  It is written in the language of mathematics, without which it is humanly impossible to understand."

While math is illuminating, it also is limiting, as is any language or tool.  For example, the Greeks, whom I believe got their math from the Egyptians, and whom provided the math to the Romans and the Europeans through the dark ages had several constraints that came out of their use of numbers, geometry (calculating land areas) and accounting for business: they didn't have the concept of zero or negative numbers, there preference was for integers and decimals were disfavored, even for at least some who using Babylonian decimals (Base 60) for calculations (there were many conversions).

Let's examine Xeno's paradox that seemed to have puzzled the world for 2,500 years.  While everyone knew it wasn't true, the error in Xeno's analysis was not pointed out until long after Xeno was dead.  In this version, let's assume that our Greek hero, Achilles is in a race with a tortoise.  Achilles starts out one unit behind the tortoise, they both start at the same time, running at a constant speed towards the finish line and Achilles runs twice as fast as does the tortoise.  Xeno suggested that logic would require that Achilles could never pass the tortoise, as when Achilles reached the point where the tortoise started, the tortoise would have moved half a unit towards the finish line, remaining ahead of Achilles.  The next cycle is just the same as the first, but with half the distance.  Xeno suggested that no matter how many cycles the analysis took, Achilles would always be behind the tortoise.

Many of us have run a similar experience and know that a leader in a race is not always destined to finish first.  This the Greeks knew, but the flaw in Xeno's analysis escaped detection for quite a while.  Perhaps we find this not so puzzling as did the Greeks because either we have already heard the answer or perhaps because we are more comfortable in dealing with fractions and infinite series.

If you solve for the time that Achilles and the tortoise are at the same place in the race at the same time, you will see that this occurs when Achilles and the tortoise are both two units from where Achilles started.  The flaw is that no matter how many of Xeno's cycles were calculated, the race never progressed past the point where both were at the same location.

This shows us that the sum of the infinite series of 1/i is two.  That is if we add infinite number of the decreasing numbers in this series (1, 1/2, 1/4, 1/8, . . .).  I'll get back to this known sum of an infinite series that solving Xeno's paradox generated.

A mathematician named Georg Kantor was born near the Russian/German border in 1845.  He worked with infinite numbers and his work is the inspiration for this next step.

If one wanted to determine which of two sets was larger, one could count both sets and compare the sums.  However, this is at least inconvenient in some situations,  For example, if one wanted to know if there were more fans in a soccer (football for our non-NA readers) stadium or seats, one could also ask everyone to sit down and count the standing fans or empty seats.  If there are standing fans, then the set of fans is larger.  If there are empty seats, then the set of seats is larger.

Following this logic, if we imagined a horizontal line, with the integers marked at some equal units from a zero, in both a positive and negative direction, we would have a line with all the numbers on it.  This line would have an infinite number of points.  If we imagined a vertical line, with the integers marked at the same equal units from the zero in both a positive and negative direction, we would have the equivalent of a two dimensional axis system.

Now, imagine another horizontal line, identical and parallel to the first horizontal line, displaced so that it crossed the vertical line at the point marked with a "1."  Imagine a system of such horizontal lines, each crossing the vertical line at each point where an integer is marked.

Next, imagine a similar set of vertical lines, each crossing the first horizontal line at each of the marked integers.  We now have a system of boxes, much larger, but similar to the outline of the squares on a chess or checker board.  Label each of the points where the number lines cross, other than the crossings involving the first horizontal and first vertical lines, with a ratio.  Use the integer where the vertical line crosses the first horizontal line as the numerator of the ratio.  Use the integer where the horizontal line crosses the first vertical line as the denominator of the ratio.

We will now be imagining a system of ratios, if we start at the intersection of the line through both the vertical and horizontal points for the integer 1 and follow a line in the direction of both positive integers, we find the ratios start with 1/1 and continue 1/2, 1/3, 1/4, 1/5, . . . .  If we return to the same point and follow the other line forming the 1/1 ratio in the direction of both positive integers, we find the ratios start again with 1/1 and continue 2/1, 3/1, 4/1, 5/1 . . . .

If we consider all the ratios, we have a set that includes all rational numbers.  (Note that this set of ratios is much larger than the set of rational numbers, because some rational numbers are represented more than once.  5/2 and 10/4 are the same rational number.  -1/3 and 1/-3 are the same rational number.  217/6973  and -217/-6973 are the same rational number.  However, the important thing for me is that all rational numbers are included in the set of these ratios.

Now, let's return to infinite series we used in examining Xeno's paradox.  Let's start with 1/1 and imagine a section of space centered on the intersection that is labeled 1/1.  This is sort of like covering a spot with a carpet square.  This space has an area equal to 1, the first number in the series.  We now move to 1/2 and similarly cover it with a piece of space with an area equal to 1/2, the second number in the series.  There are a sufficient number of spaces to cover all the ratios on this line.

Now, let's slide spaces over one and place the space with area 1 over the point marked 1/3.  Since infinity minus one is still infinity, all of the points that have ratios on the first line are still "covered."  We can repeat this an infinite number of times to "cover" all of the points on the second line that have ratios.  We can repeat this for each of the other lines, still "covering" all of the points on each line that have ratios.

We now have an upper bound on the area consumed by rational numbers on a number line: 2.

We can now decide that we want a closer upper bound on the area consumed by rational numbers on a number line.  Let's remove the first area and "pull" the other areas over one.  Since infinity minus one is still infinity, all of the ratio points are still covered.  We now have a second upper bound on the area consumed by rational numbers on a number line: 1.

We can now decide that we want an even closer upper bound on the area consumed by rational numbers on a number line.  We can remove the largest numbers from this series without limit, just as we removed the first number from the series.

What is area consumed by the rational numbers on a number line if we have an upper bound that we can reduce without limit?  Isn't this zero.  That means that the area consumed by the rational numbers on a number line is zero.

This means that if we select a number at random from the set of real numbers (the number line), the probability that it is rational is zero, not low, but zero.  In most instances, this is not relevant, since we can't measure selected numbers very accurately.  It also is not relevant if the selection is not from the set of real numbers, such as how many quarter do I have in my pocket or what numbers will come up on a six, or ten sided die?

The equation for the force exerted by electric charges between two objects that are sufficiently separated so they act as if they were point source is F = k q1 *q2 /(r) ^2.  In this equation, k is determined by the units used in the measurements and the force, so it is definitional and not a fundamental part of the laws of physics. (if we used different units, we would get a different k).  F, q1, q2 and r are variables, inputs to the equation and are not a fundamental part of the laws of physics.

However, the fact that the two masses between which the force is calculated, m1 and m2 are raised to the first power (exponents) and the fact that the distance between the masses, r is raised to the -2 power (exponent) are fundamental parts of the law of physics.  Anyone, anywhere with the same laws of physics, would determine the same three exponents.

Taking these two points and seeing how precisely these exponents are measured as rational numbers, gives us an upper bound on the probability that they were randomly selected, at the beginning of the big bang, rather than selected (perhaps by God).

The measurements reported in the Phy. Rev. A 33, 79-761 (1986) reports that the exponent on r is measured as - 2, with an experimental error of 1.2 x 10 - 16.  The assumption is that the exponents on q1 and q2 are precisely 1.

If the probability of the big bang randomly assigning three exponents as integers is zero, how unlikely is it that the actual real number, just happens to appear to be integers to twelve places?  I think that that probability is less than one-half, the only number I received from you gentle readers as what the probability needs to be to get your attention.

One may find similar results in the ongoing measurements from the Apache Observatory as part of the Apollo program where the earth lunar distance is being measured to confirm that the exponents for m1 and m2 are 1 with r being -2 to something like 12 places.  This is NASA and NSF supported research.  The project's webpage is http://www.physics.ucsd.edu/~tmurphy/apollo/apollo.html.

I thought this is interesting and I'd wanted to share this with you all.  Again, as I've said before, I don't think this (or anything) is a proof of God.  However, it's a thought I had that cause pause to the belief that all is just random arrangements, without any guidance.

Please point out any errors, concerns or questions.  Hopefully politely, but I'll take what I can get.

(This is my first attempt at submitting a new topic, so please let me know if I've not done it as I should have.)

[quote="wizard2378"]This is my first attempt at submitting a new topic, so please let me know if I've not done it as I should have.[/quote]

I actually haven't read most of your post yet, which is why I am addressing this first. Microsoft Word counts 2024 words, and I count about 29 paragraphs of content. Usually when you begin a discussion, posters should be able to find the purpose and topic of discussion [b]very[/b] quickly.

You'll find that [b]hide tags[/b] are your friends. You can write points, or macroscopic steps in your arguments, or major pieces of evidence, and then hide the description. This way, people can get an overview of what you're going to talk about, and then go in and read your discussion.

It's not that people aren't willing to read 2024 words, but they need to be indexed and in some sort of visual organization, so that people can get a scope of what you're saying.

My 2¢

Now, on to reading the thing  :wink:

So, to make sure I understand, the argument consists of the following points:

With respect to the Lebesgue measure on the real line, the rationals have measure zero.

In Coulomb's Law, the (coordinate invariant) exponents are widely believed to be integers and have been proved as such to a very high degree of accuracy.

What are the chances??

Assuming my comprehension has not run faulty, I will reply, again, that such probabilistic arguments almost always (not in a measure theoretic sense; in an idiomatic sense) are either bluntly invalid or largely misleading. Perhaps the most elementary mistake when one attempts to cite probabilistic arguments is a lack of a basic understanding of the foundation of any such argument, namely the probabilistic distribution in question. Any argument lacking the probability distribution is missing at least half the content. In the first statement above, I believe the probability measure was simply standard Lebesgue measure on the real line, but there is no mention of anything analogous when the eventual question is posed. Are we to assume that if the universe were indeed randomly selected, the parameters in question would have been uniformly selected from the real line with respect to Lebesgue measure? It's very difficult how such a question could even be sensibly posed, let alone give us some insight as to religion and philosophy.

Furthermore, just because something's an integer does not necessarily mean it's a tremendous coincidence. There are a plethora of examples coming, of course, from various aspects of quantisation: for example, if there is some gauge freedom in some theory, the physically meaningful quantities must be well-defined independent of this gauge freedom, and it commonly happens that this is equivalent to some quantisation of some physical property. (a good argument that runs along these lines is dirac's explanation of quantisation of electric charge assuming the existence of magnetic monopoles. unfortunately, this doesn't hold as magnetic monopoles are widely believed to not in fact exist, which sucks since it's an awesome argument. there are other similar examples that are physically valid but more mathematically involved.) Now, off the top of my head, I can't think of such a pure reason of this sort for why the exponents in the Coulomb law are integers, but I'd be willing to bet that the nice form of the law is responsible for some nice cancellations which is in turn responsible for nice conditions for life and so we could probably recourse to the anthropic principle.

You seem to be saying that the real world abides by mathematical principles.

How do we know the real world matches the mathematical world? So you [i]think[/i] that it does, but how do you know you weren't indoctrinated as a child to think that? And what does experimental accuracy mean? Does that mean *wave hands* *take average* *do other things that are purported to "reduce error" but all we really know is that it makes "nicer" numbers in the end*?

For instance, one taste plus one taste is one taste. One male persimmon tree plus one female persimmon tree equals many many persimmon trees. That doesn't seem to obey the laws of arithmetic.

(I personally do believe in the laws of physics, but I don't believe that science is infallible.)

I have no idea what point you were attempting to make there.

personally i don't think the OP is going far enough.

think of the uncountable number of possible functions of three variables.  what are the odds that the force between two objects is of the form $ q_1^aq_2^br^c$?  very small, i would think!

think of the uncountable number of properties of two objects.  what are the odds that there are EXACTLY three properties of the two objects such that the force between them is determined by those three properties?

there certainly are some philosophical considerations behind what you mention, but i don't see the connection to religion -- only if you believe that any phenomenon unexplainable by science is de facto proof of (the judeo-christian) god could you really argue such a thing.  the anthropic principle, as mystic terminator mentions, is one (not completely satisfying) explanation which does not resort to invoking god.

p.s: there's no translation invariant probability measure on the entire real line so it's not really well defined to say 'what are the odds of randomly picking the real number -2 from the set of all real numbers?'

I would refute any philosophical implications at this point. The question has yet to be phrased in a way that even makes sense and until then it would be patently unsound to attempt to derive any further conclusions.

Everyone brace yourselves for what I'm about to say. I agree with MysticTerminator here. The only sense I've pulled from this leads me to say, "And..." because I fail to see a well founded argument.

haha

and why would this require such collective bracing

Hmm...maybe [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211179&start=120]posts #123-153[/url]? Although, as I always say, a nice healthy dose of arguing can only make you a better person.  :)  We need people to challenge our views, or else we would be in a horrible place right now. You've definitely presented the first argument for how God can avoid an omnipotence/free will paradox that seems reasonable. I'm still not sure it actually avoids the paradox, though  :P

Please don't comment on this people. I'll probably get yelled at for changing the subject if this gets brought up again...

Sorry for the delay in responding.  Other things limit my ability to participate.

TZF - Thanks for the tip on the hide tags.  How would I go about using them?  (I'm asking about the mechanics of putting them into a post, not when.)  I thought that this would be long, but I didn't want to shorten things so much my thoughts wouldn't be able to be followed.  Also sorry for the lack of outline.  A valid point.

Mystic - The whole probability issue comes from the concept that the exponent on the separation are randomly selected in the Big Bang theory.  I know that we have only the physical properties of this universe to measure, so the method of selection of the laws of physics can't be addressed by measurement, only theory.  When I was taught the theory I was told that they were randomly selected.  What distribution is another question about which I am clueless.

It seems that you grasp the point I'm making.  Note what I am not saying.  I'm not saying that this is a proof of anything, but when I saw Georg Kantor's proof, I remembered what I had been thought about the Big Bang and random laws and thought that this was interesting.  Like you, I don't know of any reason that it should be an integer.

In the discussion of whether everything happens by chance or not, this seemed to have the chance of being something of interest to the discussion.

The anthropic principle is similar to the reality principle.  Something had to happen and in hindsight, the probability of what happened, no matter how improbable is 1.  If it were necessary for us to exist so we could contemplate it, we always end up with a probability of 1 and no insight to the larger system or what could have, but didn't happen.

Solas- I guess at one level my point is that the real world doesn't seem to abide by mathematical principles.  If the probability of the selection of an integer was 0.85 I wouldn't have seen this as anything special.

As Mystic pointed out when math deviated from the real world we tend to change the math, to keep it useful - so if the math generated and validated against part of the real world is extrapolated to another part of the real world, one can usually start with the idea that the math is still valid.  The other assumption leaves us with so many fewer tools to use in examining the extrapolated area.

Your point that we may be looking at things "wrong" and biased was part of my reason to start with Zeno's paradox.  This may well be what is happening here, I just don't see it.  I think that the experimenter's reductions of errors just means that we can rely that their measurements are accurate, at least to the degree of reduced error.

I agree that real science is changing to include new measurements, which happened to be outside the range used to derive the existing theory and happen not to match the theory that used to explain all observations or is awaiting for the new measurement that causes an enhancement of the science.

blah - That is a broader question, but just like the joke of looking for the keys where the light is, rather than where one lost the keys, I don't have anything to address that question.

Not knowing what a "translation invariant probability measure" is, I missed the last point.  It may that I don't know the probability function of selecting numbers for the formula, so the probability is beyond my ability to calculate or estimate.

Hibert- I'm glad you don't see a well founded argument, as I didn't try to make an argument, just share something I thought was interesting and see what the rest of you thought about it, if anything.

I agree that discussing things with reasonable people has often increased my understanding of what I thought I knew.

I'm really not at all sure where the idea that the initial parameters were randomly selected is coming from, and I don't think it can be stated in a way that makes any sense. I know you're not doing a formal proof, but in general, I tend to be wary of thinking that cannot be rigourised as it can be quite misleading, even in the "informal" implications that one attempts to draw. Another statement I'd like to see a bit more substance to - what exactly do you mean by "the real world doesn't seem to be abide by mathematical principles" ? This seems to be another phrase that is easy to state but (at least for me) very difficult to give actual substance to.

Eh, I don't really know what to say about the anthropic principle. Like most physicists, I'm not too fond of it, but it does seem like we're stuck with it for the time being (in a way that does make sense but that I admittedly cannot make precise -- I don't know the exact physical phenomenon here). So ... yeah.

I think blah^3 was just underscoring a technical point that I overlooked, namely that there's no uniform (or more properly, "translation invariant") probability measure on the real line. I had implicitly assumed Lebesgue measure, but I assumed we were working on a closed interval; on the real line, of course, the total measure is clearly not one.

I don't know the source of my instructor's statement, but if there is a range or distribution of potential formulas, it seems that there are only two possible ways for our reality to be determined, some random selection or something selecting.  If we are assuming that there is no God, therefore no selection and seeing if this assumption leads to a logical contradiction, a random selection would seem to me to be the only alternative left.

Examples of math models of the real world being contradicted by the real world leading to more complex math models would include Newton's physics and the orbit of Mars - leading to Einstein's physics, ect.;  genetic inherence   and regression towards the mean; medical testing (TB) and the consideration of Type II errors; and blood transfusions and blood types/RH factors.

I agree that the number line would not seem to be a closed interval.

These are examples of specific models being proven wrong. This is hardly surprising and, as far as I can tell, has nothing to do with the statement that "the real world does not adhere to mathematical principles". I'm also not, at this point, asking for evidence for why you think the statement may or may not be true -- I'm just asking for what exactly you mean when you throw it out there. If you do in fact mean the statement as "simplified models of the world turn out to not be perfect", then I can only respond by noting that the original statement uses some very misleading language.

I'm really confused with the attempt to parse how the universe was chosen. It's like some people were sitting around some table and drawing up some plans for the universe, being all like "okay, we've got this far; we need to determine these physical parameters now. How should we do it? Set it ourselves or choose them randomly?" If you're using this as a model of how the universe was created, then sure, one might postulate that there are only two alternatives here. But seriously? Why would we assume that even the tiniest aspect of our intuition in this regard extends to [i]before the birth of the universe[/i]?

Furthermore, it certainly does not need to be either -- perhaps the very early universe just started out with some silly constants and over the first moments the parameters flowed to come to a stable point, thence settling on their values henceforth.

I'll try to answer your questions.

My statement, "the real world does not adhere to mathematical principles," was a statement reflecting my belief that math is not just a neat logical system.  It also must be useful.  Where math and the real world conflict - you expressly asked me not to talk about where that occurs - math is changed to conform to the real world, because math not only is a neat logical system it also has been construed to be useful.  So math;s congruence with the "real world" is, at least in part, because math was/is constructed/modified by people to be useful in representing the real world, while still being a neat logical system.

Please provide the answer to which you refer.  If case 1 is a random selection of the physical laws of the universe and case 2 is God (or some other intelligence) selecting the physical laws of the universe, what is case 3, independent of the first two cases?  (For this point it doesn't matter what made the selection.)  An assumption underlying my initial thought was showing case 1 to be more unlikely has to increase the likelihood of case 2.  If there are more than two case (possible ways the physical laws were determined) than any discussion such as that I proposed needs to address these additional cases.  I just don't know what they might be.

Your suggestion that the "very early universe just started out with some silly constants and over the first moments the parameters flowed to come to a stable point, thence setting on their values henceforth." passes over the question I have.

First, I don't care, for this discussion, and don't know how anyone could measure or know what the parameter were in the very early universe.  We might see "residual radiation" or something that well could have come from something else to imply former physical laws.  I have enough trouble dealing with the current physical laws.  Regardless of intermediate values, or future values - how were the current physical laws determined?  Again, it seems to me that it was either random (Case 1 - similar to the question what part of a sidewalk gets wet when a rainstorm only covers 10% of the surface)  or a decision of an intelligence (Case 2 - similar to the question what part of a sidewalk gets when I wet 10% of the sidewalk's surface from a hose or bucket).

I think the original poster is making one of the most basic errors in trying to work out what is going on.

He wants a certain set of things to be true, and thus looks for supporting arguments.

For anyone else who is leaning towards a Creationist view of the world, let us look at a bit of probability, using the sort of sporting analogy that is the upper bound of the scope of understanding by Intelligent Design advocates.

Consider golf, more importantly consider golf played by someone as useless at golf as me.
The ball can go pretty much anywhere, anf the chances of it landing on any particular blade of grass is about 10 million to 1.
I guess it would take me at least 100 strikes to make it round a course, so the set of grass blades is 10 million to the power of 100. That's a really big number.

The chances of any particular set of blades being hit, is very very small, and from the perspective of someone after I had played, they might well say that it is so improbable as to be impossible, since of all the games that might played we don't even have a name or physical expression for such a big number. The number of atoms in the universe is pathetically small compared to this, and since Creationists need to believe the universe has only a 6,000 light year radius a [i]lot[/i] bigger than their universe.

You see this issue.
The odds of humans evolving, from the perspective of the start of the Earth's formation is very close to zero.
This is the same as someone trying to say "Dominic will hit this set of grass blades, in this order".

After the event, this really is not the same thing.

Under various interpretations of quantum theory, you don't even need evolution. There is a finite, non zero probability that we just appeared this morning. OK, it is not a [i]big[/i] probability, but it is non zero.

At one level the anthropic idea is almost a tautology, and a form of existentialism, since to exist it must be possible that we exist.

But the Creationist fallacy is that the possibility of our coming into existence by a given mechanism can be deduced by the probability of a given path.

[/i]

Dominic - If one assumes for the moment that all of us are acting rationally, looking at arguments that support our previous positions more favorably than those that require a change in position to accept, I don't understand how the validity of the argument changes.

Ignoring your attempt to characterize everyone who advocates Intelligent Design, you can not believe for a moment that your statements are anywhere consistent.  If the odds of you hitting every blade of grass was even close to equal, it would take a lot more than 100 strokes to finish a round of golf.  (It would be grand to see you hit a blade of grass that happened to be more than 1,000 meters from where you hit it, even once, let alone over half the time.)

Please explain your statement that Creationists need to believe the universe has only a 6,000 light year radius.

Are you suggesting because the laws of physics have already been set, nothing can be concluded by examining them?

Where are you getting the statement about the probability of our coming into existence?  Is this from something else you read?


show that $ 1 \plus{} \frac {1}{2} \plus{} \cdots \plus{} \frac {1}{n}$ is not an integer


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Hi all,

We have enabled post ratings on the site. You need to have javascript enabled (and for best viewing we recommend you to install one of the new browsers: Internet Explorer 7, Firefox 2 or Opera 9.2).

I will answer all your questions in this thread, but make sure you've read all the previous answers (and this post in particular) before asking a question.

 :arrow: post ratings are enabled to increase the quality of general posting on the site; they determine a users' posting quality, and very soon you will be able to select from your profile a minimal rating for a post to be displayed (for example you will be able to hide/ignore all the posts with rating less than 2 in a thread)

 :arrow: there are some measures in place to prevent abuses (there are limits to how many posts you can rate per hour (it's set to 5 right now for users with average post rating below 4), and how many you can rate during that hour from the same poster (!)

 :arrow: any attempt to rate posts based on anything but the post itself (in the topic context) is forbidden (for example boosting your friends ratings or something); users that will do that will be discovered and their rating privileges will be removed (that means you will not be able to rate anything if we think you are not rating the post, but the person!)

I was going to ask why bees took over ML.  :ninja:
Anyway, does the amount of ratings decrease for those with less than 4 posts a day? It already tells me I've rated  too many and I thought I just rated 1!

There might be some small bugs that I could have not foreseen in the tests I made before I put this live.

[quote="funcia"][quote="AstroPhys"]Had the same problem when I tried to rate the initial post in this thread.  :maybe:[/quote]
Yes, that was what happened to me, however when I tried to rate another post later it worked out fine. Is there a limit to how many people can rate posts?[/quote]No, there is no limit on that. However there's a limit on how many posts you can rate an hour, if your rating is not big enough.

So for example if the average of your posts is $\geq 4$ (cubes or hexagons :P) then you won't have that limit. If it's less then you will have a limit of posts per hour you can rate (now it's 5).

This is to prevent abuses.

I haven't rated any posts, and it tells me already that I've rated too many in this hour. I tried it on IE, but it doesn't work on that either.

Yeah some people are getting that. I do not know why yet, I'm working on it.

So basically, it's like Digg?

I haven't tried it out yet, but it seems interesting.

The one problem I can see with the system is this: When you hide all the posts below a certain viewing threshold, don't you lose some of the continuity of the thread?

(Of course, there's probably already a system to remedy this.)

[quote="mathnerd314"]The one problem I can see with the system is this: When you hide all the posts below a certain viewing threshold, don't you lose some of the continuity of the thread?

(Of course, there's probably already a system to remedy this.)[/quote]I'll probably make a user option to remove the post completely from the view or just put it all under a hide tag by default (so if you really want to see it, you can click on the tag).

This is probably getting a rating of one but...since some of us can read extremely fast like Data will the amount of ratings we are able to make ever increase(maybe if we are found to not abuse rating)?

If your average post rating is $\geq 4$ you will have much more possible ratings per hour.

Erm, sorry for my stupidity, but how exactly do you submit your rating? :blush: Is it just by clicking on the appropriate cube at the top? Seems like everybody else knows how to do it but I'm still in the dark :?. Could you post some detailed description of what is what in this misterious sequence of 6 cubes and what one is and is not supposed to do with them? Also, I hope we can try to "test rate" the posts on the test subforum and those ratings will not count as well as "abuses" of the rating system made there :).

You can't test rate just yet :) You just select your rating, click on it and voila you've rated (a confirmation message will appear). Hover your mouse on the cubes (hexagons for ML skin) and you will see the rating.

The darker shaded poly-s mean that you've voted for that particular post in the past, while the ligher shadow means that other users have rated that post (but not you).

i tried to rate a post for the first time, and it said that i've rated too many posts within the current hour.

Interestingly, I can see it now...
(And I surely did not rate my own post)

So yes it's possible that it really is the number of ratings, but then I somehow have a feeling that near to no post in the olympiad section gets enough ratings.

[quote="ZetaX"]So yes it's possible that it really is the number of ratings, but then I somehow have a feeling that near to no post in the olympiad section gets enough ratings.[/quote]
In fact, I can't remember the last time I saw a rating withing 4 days. And I browse many forums.

[quote="ZetaX"]No I see nothing.[/quote]But I do, I see that post as rated 6.

What is actually the problem you're having with the ratings? Here everything seems to work?

You see it now, but as I said, it was not visible some time before and is visible now.

How many rated posts have you seen in the Olympiad section recently¿

Every posts to be rated a couple of times before it becomes visible.

Why would you exactly rate a post if you can see with your own eyes right in front of you?

To tell the author what you think about that post, for example.

I like the new design (your rating and overall rating in separate sets of boxes) -- thanks!

[quote="AIME15"]Me too, but when I view a new thread, it lags a bit and displays some text there before showing the cubes, and my computer is pretty fast.[/quote]

Yeah, I get the same thing, too.

[quote="stevenmeow"]an acquaintance of mine only has 16 posts but (s)he can now rate posts. does the new rating system allow for this?[/quote]She can't rate posts. The backend forbids it (even if it looks like she can -- I will look into that tomorrow)

[quote="AIME15"]Me too, but when I view a new thread, it lags a bit and displays some text there before showing the cubes, and my computer is pretty fast.[/quote]

By text do you mean those little drop-down menu things that say stuff like "Not Rated Yet"? If so, it happens to me too, but it's not a big problem.

[quote="TZF"]I like how you can remove your rating in the new system.

However, I like the 1/3 and 2/3 cubes in the old system, as opposed to the strange 1/2 cube in the new system.[/quote]There was no 1/3  or 2/3 cube in the old system. It was 1/2 as well. This is not yet well tweaked (that's why you see the select boxes before the page loads -- I will fix that at some point -- right now it doesn't affect the functionality.)


algebra
polynomial
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algebra unsolved

I'm trying to find the irreducible polynomials in $ \mathbb{F}_5$ of deg. 2. I know there are 10 of them. But so far I can only come up with 9. A polynomial is not irreducible if it has a zero $ {0,1,2,3,4}$.

What is the $ 10^{th}$? Is there also a fast way to determine exactly what the irreducible polynomials are?

Thanks in advance,

Los

Correction: x^2+2, x^2+3, x^2+3x+3, x^2+x+1, x^2+4x+1, x^2+2x+3, x^2+3x+4, x^2+x+2, x^2+4x+2, x^2+2x+4

By analogy with the Sieve of Eratosthenes, multiply out all the linear polynomials to get all of the reducible polynomials, and what's left is the irreducible polynomials.  Of course, this is extremely tedious to do by hand.

Alternately, find all values of $ b, c$ such that the discriminant of $ x^2 \plus{} bx \plus{} c$ is not a square root $ \bmod 5$.

Edit:  Your list looks wrong.  There should be two irreducible polynomials for every choice of $ b$ by my logic.

hey, is factoring $ x{}^p{}^n-x$ over a finite field used in any way in BCH codes?

I'm reading a book independently, where these topics appear in the same chapter, but I don't see any connection between the two.

For any odd prime $ p$ take some non-square $ d\in\mathbb{F}_p$ and $ s\in\mathbb{F}_{p^2}$ with $ s^2\equal{}d$.
Then the elements of $ \mathbb{F}_{p^2}\setminus \mathbb{F}_p$ can be written uniquely as $ a\plus{}bs$ with $ a,b\in  \mathbb{F}_p,b\not\equal{}0$,
which has minimal polynomial $ (X\minus{}a\minus{}bs)(X\minus{}a\plus{}bs)\equal{}X^2\minus{}2aX\plus{}a^2\minus{}b^2d\in \mathbb{F}_p[X]$.
Also only $ a\pm bs$ have the same minimal polynomial.
So we find all $ {p(p\minus{}1)\over 2}$ irreducible monic quadratic polynomials over $ \mathbb{F}_p$ by
$ X^2\minus{}2aX\plus{}a^2\minus{}b^2d\equal{}(X\minus{}a)^2\minus{}b^2d$
for $ a\equal{}0,\pm 1,\ldots ,\pm {p\minus{}1\over 2}$ and $ b\equal{}1,\ldots ,{p\minus{}1\over 2}$.

In case $ p\equal{}5$ we can take $ d\equal{}2$, which gives
$ X^2\pm 2,(X\pm 1)^2\pm 2,(X\pm 2)^2\pm 2$.

1) Hi Olorin; it's a pretty approach but $ in$ $ fine$ it's the same idea because you find the polynomials in the form $ (x\minus{}a)^2\minus{}\alpha$ where $ \alpha$ isn't a square.
2) Another idea:
$ x^{5^2}\minus{}x$ is the product of the irreducible polynomials of degree $ p$ s.t. $ p$ divides $ 2$ ($ ie$. $ p\equal{}1$ or $ p\equal{}2$). Thus the decomposition in irreducible of $ \dfrac{(x^{25}\minus{}x)}{x(x^2\minus{}1)(x^2\minus{}4)}$ gives the $ 10$ irreducible of degree $ 2$.
Remark: practically it isn't better.


Trying to help son on Alegebra 2 HW.

Problem: solve for u

Y=(u+1)/(u+2)

There must be a trick Im just not seeing - HELP!!!

\begin{eqnarray*}y & = & \frac{u+1}{u+2}\\ (u+2)y & = & u+1 \\ uy+2y & = & u+1 \\ uy-u & = & 1-2y \\ u(y-1) & = & 1-2y \\ u & = & \frac{1-2y}{y-1}\end{eqnarray*}

See, the trick is to remove all fractions from the equation, isolate all the terms with u on side of the equation, and then solve for u.


P(x) is a polynomial with integer coefficients.

P(21) =17,  P(32)= -247  and P(37) =33

Prove that if P(N) =N+51 for some integer N then N=26

Hint:
[hide]
use $(m-n)|(P(m)-P(n))$ for all integers $m,n$
[/hide]

Let N be such integer that $P(N)=N+51$. And let for another integer $x$ be $P(x)=y$. Then we know that $(N-x)|(P(N)-P(x))$. We can modify it to: $(N-x)|(N+51-y) => (N-x)|(51-y+x)$.

Now set $x=21,37$. From this flows $(N-37)|55, (N-21)|55$. Divisors of 55 are -55,-11,-5,-1,1,5,11,55.
$N-37$ and $N-21$ are divisors of 55 with distance 16. We can see that there are only 2 possibilities: $N-37=-11,N-21=5 => N=26$ or $N-37=-5,N-21=11 => N=32$. But second possibility can't be true, because we know that $P(32)=-247$.

It's all :)


Let a,b, c be the sides of a triangle.
    Prove that:

                     a^3 + b^3 + 3abc > c^3

Let's push this a little harder. As a new problem:

Suppose $a, b, c$ are the three sides of a triangle. Prove that

$a^3+b^3+2(a+b)c^2 \ge c^3 + 2(a^2+b^2)c+abc$.

(Edited: I meant to have a 2 on the left hand side, not a 3.)

This one is even harder (from Kiran Kedlaya's inequality notes)

(MOP, 1999) Let $a,b,c$ be lengths of the the sides of a triangle. Prove that

\[
a^3 + b^3 + c^3 - 3abc \geq 2 \max \{|a-b|^3, |b-c|^3, |c-a|^3\}.
\]


How many perfect squares are there between 200 and 300?

sqrt(200)= 14.14.....
sqrt(300)=17.32....

In between the squares of 15, 16, and 17 exist. That's $ \boxed{3}$ numbers


All right.
This is a new game I'm trying.
[hide="Objective"]
To make TPAM (the person above me) ashamed of doing something. [/hide]
[hide="Rules"]
The first person (me) makes a declaration of doing something. They do not have to actually do it, but it must be reasonable (which makes this game slightly unbalanced and vulnerable to spam).[/hide]

My first declaration is:
[hide="click this"]
[hide="click again!"]
[hide="Am I annoying you?"]

I will pass the SAT when I take it.[/hide][/hide][/hide]

This is basically a Corrupt a Wish Game.

It is quite easy to pass the SAT but much harder to make a perfect. :wink: THBAPSOTPAM

I will find all zeros of the Riemann zeta function over the summer. :|  :rotfl:

I will ignore the Riemann zeta function and not make myself confused over it.

I will be first in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213378]FTW Butterscotch tourney[/url] by [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=40870]isabella2296[/url].

btw, your computer being bigger than mine doesnt mean anything. It could be one of those REALLY old computers that are soooooooo big they could fill a room.

HINT: Improve your Chinese.

Computer is "计算机"

Calculator is "计算器" :wink: :wink: :wink:

I used my native skills. you probably used a translator. AND TRANSLATORS ARE WRONG!!!!!!!!!! HAHAHAHA!

I learn to do everything well since you didnt post anything.

(sorry if you didnt use one, but you did)

Um, no, my Chinese "native skills" are just better than yours, no offense.  And you don't need a translator to say something that basic. :wink:

oh. i just saw the chinese flag at the bottom of your profile :blush:

yeah, math geek has the chinese right, ji suan qi is calc, ji suan ji is comp, don't feel like writing the chinese

ok

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


Scores on the PSAT range from 20 to 80. For a certain population of students, the mean is 52 and the standard deviation is 9.
Which of the following scores can be expected to occur [i]less than[/i] 3% of the time? (1) 39, (2) 47, (3) 65, (4) 71
[color=blue]I think the answer could be (1) and (4) because there could be 97% grades are higher than the 3%'s which is about 39, Or there could be 3% grades are higher than 97% grades which is 71, so I don't know how to choose? [/color]
Thanks for help!

well my guess would be to use the 68-95-99.7 rule.
this basically says that
68% is with in 1 standard deviation of the mean. 95% is within 2 and 99.7 is withing 3.

so use that and some guess work i suppose.


tan(x) = (1 + sin(40) - cos(40)) / (1 + sin(40) + cos(40)). Find all values of x if 0 <= x < 360 and all angles are measured in degrees.

Well practically everyone was still in Math Jam when you posted this, so no wonder no one replied...I didn't realize you posted this question until now either.

As for the question, [hide] should I use tanx=sinx/cosx or should I start with simplifying the numerator and the denominator?[/hide]

Actually, i haven't done this problem myself, either.
THis was one of Joel's posts...

Wait, isn't that kind of stealing? and where's the original post by Joel?

joel originally posted 15 questions into one post titled "stuff"
i broke up that one post into 15 individual posts

Eeeeeeeh, so you split it into 14 posts and kept a problem for yourself as your cut, smart one doc!
And I apologize if the above wasn't the case.

ok this is as far as i got:

[hide]tanx=(root[2]-sin5)/(root[2]+cos5)[/hide]

Does that get me anywhere?

Don't look at me! :)
Can anyone give us a hint or lead us in the right direction?

Alright guys, here is the solution i devised while in the shower:

[hide]1+sin40-cos40=cos0-cos40+sin40

1+sin40+cos40=cos0+cos40+sin40

NOw, by the sum to product formulas, the two expressions become

2(sin20)^2+sin40

2(cos20)^2+sin40.

Breaking up sin40 into 2sin20cos20

We can factor out a 2(sin20+cos20) from both numerator and denominator, which leaves us with

tanx=sin20/cos20 -->x=20

So since 0<=x<360, and noting that tanx is positive,

x=20, 200.[/hide]

um can you post the product-to-sum/sum-to-product formulas please? I have a hard time remembering them for some reason. I know they're in a book in my house, but I don't know which book, so since you guys know everything, I'll ask you.

LoL!! I kno hardly everything....

Anyway, here are the sum-to-product formulas:

sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]
sinA-sinB=2cos[(A+B)/2]sin[(A-B)/2]
cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2]
cosA-cosB=(-2)sin[(A+B)/2]sin[(A-B)/2]

The product to sum formulas are:
sinacosb=(1/2)*(sin(a+b)[i]+[/i]sin(a-b))
cosasinb=The same except it's - in the italic positive sign.
cosacosb=The same except the sins are coss
sinasinb= The same except the sins are coss and it's - instead of the italic positive sign.

I know, it would've been a lot faster to just copy them and tweak them a bit...

Tell me if you need to know any formula because I got a Japanese list of formulas. (fyi, those are [b]really [/b]helpful since they list [b]all [/b]the formulas:))

thanks, you guys. um, no thanks, I don't need any more formulas, since I either remember those, or can derive those fairly easily with complex numbers. No one ever told me how to derive these, so I have a hard time remembering.

darkquantum wrote:I thought u were korean.....

No, (would all Korean people disreagard the following sentence, please)[hide] I don't eat dogs.[/hide]

Im dumb. If its not too much trouble, would someone mind deriving the first Prosthaphaeresis formula? I am refering to sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]. The rest follow easily from this one.

Um, fiery, why don't u try to prove that the identity holds, and then the derivation should be clear to you (working backwards)

simplify the right hand side first.

Yeah, I tried that. I used the sum of sin/cos formulas, and then the half-angle formulas, but got something messy after that. It ended up as needing to prove sin(A)^2 + sin(B)^2 = 1 - cos(A+B)cos(A-B)

Write A as (A+B)/2+(A-B)/2 and B as (B+A)/2+(B-A)/2. Then simplify the LHS and voila! the RHS appears.

[color=cyan]This problem is also doable without said formulae.  I certainly didn't know them when I made up the question.  Any takers for non-sum-to-product-formula method?[/color]

hmm, i think i better not answr this one again. LEave it to simon or mathfanatic, or maybe TripleM

Interesting. I was planning on leaving it to you.

Well, no one should depend on me anyways, so at least I don't have that sorta burden of someone else leaving a problem for me.  :D

[color=cyan]Well, I'm leaving the problem for all of you, so there.

Tare suggested in spoiler an excellent start to the non-sum-to-product solution, and one of darkquantum's posts utilized a second method that will come in useful later.  Although, as usual, he didn't bother to explain where he got what he got from . . . [/color]

What is that supposed to mean?

And what did i say that i didn't bother to explain?

ANd as for Tare's inadvertent hint, u mean simplifying thhe numerator and denominator, or using tan=sin/cos?

[color=cyan]You often post things without explanation, that's all.

Well, you only posted so many things on this thread -- go find the one without explanation.

Yes, maybe.[/color]

[quote="darkquantum"]ANd as for Tare's inadvertent hint, u mean simplifying thhe numerator and denominator, or using tan=sin/cos?[/quote]
The "inadvertent" part sticks in my ears (or on my eyes), but oh well.
Besides, those weren't even hints. It was a question that no one answered. I just kept on playing around with the original equation, and realized that those two things could lead to the solution, but I wasn't sure about it.


geometry
3D geometry
octahedron
ratio
algebra
polynomial
trigonometry

1. An octagon is inscribed in circle O. The sides of the octagon have lengths a,a,a,a,b,b,b,b respectively. What is the area of circle O in terms of a and b.

2. Two circles intersect in A and B, the measure of the common chord AB is 10. The line joining the centers cuts the circle in P and Q. If PQ=3, and the measure of the radius of one circle is 13, find the radius of the other circle.

3. Find all integer x such that x^3-3x+10 is a perfect square.

4. Given cube X, octahedron Y is formed by joining the center of the faces of cube X where Y is a regular octahedron. The cube Z is formed by joining the center of the faces of octahedron Y. FInd the ratio of the volume of Y to the volume of Z.

5. What is the largest integer that is a divisor of (n+1)(n+3)(n+5)(n+7)(n+9) for all positive even integers n?

6, A subset of B of the set of integers from 1 to 2008 , inclusive, has the property that no two elements of B sum to 3000. What is the maximum possible number of elements in B?

7. Let P(x) and Q(x) be polynomials of degree 4 such that P(n)=Q(n)=n^3-n for n=1,2,3,4,5. If P(6)=105, find all values of Q(6).

8. To be different, John uses a 24-hour analog watch, very similar to a 12-hour one, only numbered up to 24. Each full revolution of the minute hand still corresponds to one hour., and the hour hand moves in proportion to the movement of the minute hand. What is the measure of the smaller angle between the hands of Iking's clock at 7:40PM?

9. Let a, b, and c be digits such that there are exactly three 3-digit numbers that can be formed using these digits. If the sum of the three 3-digit numbers os 1,998, determine all values of abc.

10.How many ordered triples (j,a,m) of real numbers satisfy: ja=m , mj=a, am=j?

[hide="1"]
order the sides a,b,a,b... and we get an octagon. make 8 isosceles triangles by drawing lines from the vertecies to the center. Therefore in each quadrant of the circle there is two triangles a-r-r and b-r-r.

Call the two angles which are the same, in triangle a-r-r; x and b-r-r ; y
$ \Rightarrow 2(x\plus{}y) \equal{} 360\minus{}90 \longrightarrow (x\plus{}y) \equal{} 135$

$ \Rightarrow 2r^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos(135)$
$ \Rightarrow$ Area of circle = $ \frac{1}{2}\pi (a^2 \plus{} b^2 \minus{} \sqrt{2}ab)$ [/hide]

[hide="10"]

Multiply the three equations to get
$ (jam)^2 \equal{} jam$
$ jam(jam \minus{} 1) \equal{} 0$

So $ jam \equal{} 0$ or $ jam \equal{} 1$
for $ jam \equal{} 0$, we get $ j \equal{} a \equal{} m \equal{} 0$
for $ jam \equal{} 1$, we get $ j \equal{} a \equal{} m \equal{} \pm1$

So the ordered triples are $ (0,0,0),(1,1,1),(1, \minus{} 1, \minus{} 1),( \minus{} 1, \minus{} 1,1),( \minus{} 1,1, \minus{} 1)$

[/hide]

EDIT: oops, sorry, I've made a terrible mistake....thanks for reminding me

@Attila the Hun : if I'm not wrong, there isn't any other triplets......

[hide="2"]
Let $ O_1$ be the center of the first circle and $ O_2$ be the center of the other circle.
And $ R$ be the midpoint of chord $ AB$
Assume the radius of the first circle is $ 13$ and the other circle is $ r$
From $ \triangle AO_1R$, we have
$ AO_1^2\equal{}AR^2\plus{}O_1R^2$
$ 169\equal{}25O_IR^2$
$ O_1R\equal{}12$

So $ PR\equal{}2$

Then, from $ \triangle AO_2R$, we get
$ AO_2^2\equal{}AR^2\plus{}O_2R^2$
$ r^2\equal{}25\plus{}(r\minus{}2)^2$
$ r\equal{}\frac{29}{4}$

The radius of the other circle is $ \frac{29}{4}$

[/hide]

[quote="IW@IT"][hide="10"]

Multiply the three equations to get
$ (jam)^2 \equal{} jam$
$ jam(jam \minus{} 1) \equal{} 0$

So $ jam \equal{} 0$ or $ jam \equal{} 1$
for $ jam \equal{} 0$, we get $ j \equal{} a \equal{} m \equal{} 0$
for $ jam \equal{} 1$, we get $ j \equal{} a \equal{} m \equal{} \pm1$

So the ordered triples are $ (0,0,0) and (\pm1,\pm1,\pm1)$

[/hide][/quote]
Your answer would be wrong because it would include the triples of...
[hide](-1,-1,-1), (-1,1,1), (1,-1,1), and (1,1,-1), but those values
would "get you into a jam."  You must have an even number
of negatives.

With your plus or minus sign at each location, that tells me
that each 1 would take that on for a total of 8 triples, which is not
the case.[/hide]

[quote="IW@IT"][hide="10"]

Multiply the three equations to get
$ (jam)^2 \equal{} jam$
$ jam(jam \minus{} 1) \equal{} 0$

So $ jam \equal{} 0$ or $ jam \equal{} 1$
for $ jam \equal{} 0$, we get $ j \equal{} a \equal{} m \equal{} 0$
for $ jam \equal{} 1$, we get $ j \equal{} a \equal{} m \equal{} \pm1$

So the ordered triples are $ (0,0,0) and (\pm1,\pm1,\pm1)$

[/hide][/quote]

aren't there any other triples such that jam = 1?

[hide]1)

let θ be the angle at the centre subtended by the side of length a and α be the angle subtended by the sides of length b.

4α + 4θ = 360

θ + α = 90.

radius (r) of the circle is given by

2r^2 (1 - cosθ) = a        [Law of cosines]

r = sqrt [a / 2(1 - cosθ)].

also,

2r^2 (1 - cosα) = b

r = sqrt [b/2(1 - cosα)].

the area of the circle is

A = pi*a/2(1 - cosθ)...........(1)

A = pi*b/2(1 - cosα)

   = pi*b/2(1 - sinθ)..................(2)

eliminating θ between (1) and (2) gives A in terms of a and b.

[/hide]

Thank you to those who responded.

1. The answer of ocha is correct. Thank you.

2. 29/4 -> correct

10. There are 5 triples.

The other numbers do you have any idea?  :idea:

here's number 4

[hide="four"]

let cube $ x$ have sidelength $ 2$
therefore the sidelength of $ Y$ is $ \sqrt {2}$

Now the center of a face on $ Y$ occurs at an altitude of $ \frac {1}{\sqrt {6}}$ from the bottom
And therefore from similar triangles we can see that the side length of $ Z$; $ z'$ is equal to

$ \frac {z'}{2} \equal{} \frac {1}{3}$

Volume of $ Y \equal{} 2\left(\frac {1}{3} (\sqrt {2})^2 (1)\right) \equal{} \frac {4}{3}$

Volume if $ Z \equal{} \left(\frac {2}{3}\right)^3 \equal{} \frac {8}{27}$

$ Y: Z \equal{} 9 : 2$

[/hide]

EDIT: fixed

[hide="six"]
consider numbers $ 1\rightarrow 1500$, no two 'distinct' numbers add to $ 3000$

The set can be changes by exchanging an element $ \in [1501,2008]$ with its 'partner' element in $ [1,1500]$ that adds to $ 3000$. However the subset can never have anymore than $ 1500$ elements

Answer: $ 1500$[/hide]

[hide="5)"]First of all, since n is even, (n+1)(n+3)(n+5)(n+7)(n+9) is odd. Odd multiples of 5 differ by 10, and the least and greatest factors of (n+1)(n+3)(n+5)(n+7)(n+9) differ by 8. Since we are dealing with numbers that differ by 2, we can use the pigeonhole principle to find that (n+1)(n+3)(n+5)(n+7)(n+9) is divisible by $ \boxed{5}$ for any even number n.[/hide]
[hide="6)"]Since n+(3000-n)=3000, there can be at most 3000/2 elements in B. otherwise, by the pigeonhole principle, there must be a pair (n, 3000-n). So there can be at most $ \boxed{1500}$ elements in set B.[/hide]
[hide="7)"]There is an important equality principle that says:
If P(x) and Q(x) are two polynomials of degree at most n and P(x) = Q(x) for n + 1 distinct
values of x, then P(x) and Q(x) are the same polynomial.
That is exactly the case for this problem. So $ \boxed{Q(6) \equal{} 105}$[/hide]
[hide="8)"]7:40 PM --> 19:40 --> hour hand at 59/3 out of 24 hours minute hand at 40 out of 60 minutes
295-240=$ \boxed{55^\circ}$[/hide]
[hide="3)"]i dont think there any integer x because the polynomial isnt a perfect square polynomial.[/hide]
[hide="9)"]if a, b , and c are different, there are 6 three digit numbers. therefore, two of them must be the same. now, note that 1998=18*100+18*10+18. so, we must have a+b+c=18. Assume WLOG c=a. Then we have 2a+b=18. So we have a=5,b=8; a=7, b=4; a=8, b=2.
The three digit numbers are $ \boxed{558,585,855,774,747,477,882,828,288}$.[/hide]

[quote="funtwo"][hide="3)"]i dont think there any integer x because the polynomial isnt a perfect square polynomial.[/hide][/quote]
Just because a polynomial is not a perfect square does not mean it can't equal a perfect square. For example, when $ p(x) \equal{} x^3 \plus{} 1$, $ p(2) \equal{} 9 \equal{} 3^2$. However, for this problem, I too think that there is no solution.

[hide="Progress?"]$ x^3 \minus{} 3x \plus{} 10 \equal{} k^2$
$ x^3 \minus{} 3x \equiv k^2 \mod 5$.

The residues of $ (x^3 \minus{} 3x) \mod 5$ are $ 0$, $ 2$ and $ 3$. The residues of $ k^2\mod 5$ are $ 0$, $ 1$, and $ 4$. So, if a solution for $ x$ exists, it has to be a multiple of $ 5$. Doing some more mod stuff reveals that $ x$ has to be a multiple of $ 10$. Let $ x \equal{} 10p$.
The problem now becomes proving that there are no integer solutions to $ k^2 \equal{} 1000p^3 \minus{} 30p \plus{} 10$...which is more complex than the original question  :( [/hide]

EDIT: This is called a Weierstrass elliptic curve and unless there is a clever solution, I don't think that this can be solved by elementary methods.

It should have been

2. Find all integer x such that x^2-3x+10 is a perfect square.

Answer to
9. 128, 196, 200

I think the rest of the problems have been solved except for #4.

Thank you very much!

I've fixed my solution to number 4 and it's now correct (althought not very detailed)
If you just make a small sketch you should see that it is all pythagoras and similar triangles

oh i thought abc in #9 meant the three digit number...fail
a weird solution for #3:
[hide]complete the square to get (x-1.5)^2+7.75. now, note that the squared term is of the form n.5 (because x is an integer). when n.5 is squared, it yields n(n+1).25 or n(n-1).25 if n is negative (CHeck this by expanding (n+.5)^2 and (-n+.5)^2). Adding 7.75 yields n(n+1)+8 and n(n-1)+8. n=0,1,2 do not yield perfect squares (so n=-1,-2 do not either). Now the squares are far enough apart so that the equations must equal the next highest square after n(n+1) or n(n-1). Solving for the lower of the factors of n(n+1) or n(n-1):
n(n+1)+8=(n+1)^2
n(n+1)+8=n(n+1)+n+1
n=7
n(n-1)+8=(n-1)^2
n(n-1)+8=n(n-1)-(n-1)
n-1=-8
adding 2 to these values, we get $ \boxed{x\equal{}\minus{}6,9}$[/hide]

[hide="#3 again, a little more formally"]
If $ x > 9$, then
$ (x\minus{}2)^2 \equal{} x^2 \minus{} 4x \plus{} 4 < x^2 \minus{} 3x \plus{} 10 < x^2 \minus{} 2x \plus{} 1 \equal{} (x\minus{}1)^2$.
But since $ x$ is an integer, $ x^2 \minus{} 3x \plus{} 10$ is squeezed between two consecutive perfect squares and cannot be a perfect square itself.

If $ x < \minus{}6$, then
$ (x\minus{}2)^2 \equal{} x^2 \minus{} 4x \plus{} 4 > x^2 \minus{} 3x \plus{} 10 > x^2 \minus{} 2x \plus{} 1 \equal{} (x\minus{}1)^2$.
For the same reason here, $ x^2 \minus{} 3x \plus{} 10$ cannot be a perfect square.

It remains to simply check the integers from $ \minus{}6$ to $ 9$, from which it is found that the only solutions are $ \boxed{x \equal{} \minus{}6, 9}$.[/hide]

Just one more and everything is ok! Thanks to those who helped me out.

Anyone else who can help me out with the other problems?

Mathematica says that $ f(n)\equal{}n^3\minus{}3n\plus{}10$ is a squarefree expression. Furthermore, we know that if it were possible, then $ m$ and $ n$ would be found with $ \{m\to10m,n\to \{5n,10n\}\}$. Making those substitutions, we encounter a contradiction to both cases anyways. I think its safe to say the solution set is empty.


How can I solve this or this kind of problems

[url=http://img216.imageshack.us/my.php?image=76931634ac1.jpg][img]http://img216.imageshack.us/img216/1040/76931634ac1.th.jpg[/img][/url]

m1=m2=m    and    k1=k2=k

Find the frequency of system.

http://en.wikipedia.org/wiki/Hooke%27s_law

learn hooke

I already know Hooke's law but this problem is bit complicated

see wikipedia the derivation of Multiple springs in series

with a small generalization you can put a mass between the 2 springs to know the force.

afterwards is just rotate the problem and put a additional force gravitational.

this is a little bit troublesome and have many calculations, but can be donne, dont give up


Hi guys! I'm new to the forum and i don't know how to properly use LaTeX in here, can someone post a link to a page where I can learn how to use it in message writing?
So, here it goes, not yet in good LaTeX:

To which values for $c \in N : 40 \leq c \leq 70$ are there integer solutions $x, y \in Z$in the equation:
$754x+962y = c$

PLEASE, do not just post the plain answer; *PLEASE* explain how to solve this kind of problems.

[color=darkblue][b]Theorem.[/b] Let $a,b,c$ be three integer numbers so that $abc\ne 0\ .$
Then the equation $ax+by=c$ has at least an integer solution if and only if
the best common divizor of the numbers $a$ , $b$ divides the number $c\ .$

[b]Proof.[/b] The implication $''\Longrightarrow ''$ is evidently. Suppose that the best common divisor $d$ of the numbers $a,b$ divides the number $c\ .$
Then exist the integer numbers $u,v$ sothat $ua+vb=d\ .$ If $c=df\ ,$ then $(uf,vf)$ is a solution of the given equation. [/color]


The roots of $ f(x) \equal{} 3x^{3} \minus{} 14x^{2} \plus{} x \plus{} 62 \equal{} 0$ are $ a,b$ and $ c$.  Find the value of
                           $ \frac {1}{a \plus{} 3} \plus{} \frac {1}{b \plus{} 3} \plus{} \frac {1}{c \plus{} 3}$.

I did this by doing a polynomial transformation, and NOT EXPANDING.  I just found the $ x^{2}$ and $ x^{3}$ terms, and then used Vieta's.  How would you do it (aside from expanding, again)?

($ MA\theta$ $ 1991$)

[b]Lemma 1:[/b]  The polynomial with roots $ a \plus{} 3, b \plus{} 3, c \plus{} 3$ is $ f(x \minus{} 3)$.

[b]Lemma 2:[/b]  The polynomial with roots $ \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ is $ x^3 f \left( \frac{1}{x} \right).$

I just did the whole thing at once; namely, I just plugged in the inverse of the expression $ j(x) \equal{} \frac {1}{3 \plus{} x}$, or $ j^{ \minus{} 1}(x) \equal{} \frac {1}{x} \minus{} 3$ into the original function $ f(x)$.

Yes, that's what I was getting at.  It's a good way to do these types of problems.  You can also add the three fractions and then write the numerator and denominator as polynomials in $ a \plus{} b \plus{} c, ab \plus{} bc \plus{} ca, abc$.


From a hot air balloon 2 km high, the angles of depression of 2 towns east fo the balloon and in line with the ballon are $81.2^{\circ}$ and $13.5^{\circ}$. What is the distance between the two towns?

[hide="partial solution"]Let $x$ be the distance between town  closer to the balloon and the point on the ground below the hot air balloon.  Let $y$ be the distance between the two towns.  Then we have
$\tan81.2=\frac{2}{x}$
$\tan13.5=\frac{2}{x+y}$
which you can solve for $y$.
[/hide]


How many faxes do you guys estimate USAMTS received today? The line was busy for like the whole night.  :)


inequalities
function
integration
Support
real analysis
real analysis unsolved

Let $f\in C_{0}^{1}(\mathbb{R}^{n})$ i. e $f$ is compactly supported once differentiable function and $p>1$. Prove that there exists a constant $c\in\mathbb{R}$ such that
\[\int_{\mathbb{R}^{n}}(Mf(x))^{p}\text{ d}x\leq c\int_{\mathbb{R}^{n}}(M(|\nabla f(x)|))^{p}\text{ d}x.\]
Here $Mf(x)$ is the Hardy-Littlewood maximal function of $f$.

[quote="alekk"]The contant does not depend on $f$ isn't it ?[/quote]
Actually I'm not sure. I think it depends on some property of $f$, but still I don't know how to prove it.

if it does not depend on $f$, it is trivial because both side are finite :lol:

It looks like that $c$ depends on the $n$ and $p$, since I found the following inequality: Let $Q=(0,r)^{n}$ and let $u\in C^{1}(Q)$ and $u=0$ in every set $A\subset Q$ where $m(A)\geq \mu m(Q), \mu\in (0,1].$ Then for every $p\geq 1$ we have
\[\parallel u\parallel_{L^{p}(Q)}\leq c(n,p,\mu)r\parallel \nabla u\parallel_{L^{p}(Q)}. \]

You can drop $M$ on both sides since for $p>1$ the $L^{p}$ norms of $g$ and $Mg$ are comparable. Thus, the required inequality is equivalent to
\[\int_{\mathbb{R}^{n}}|f(x)|^{p}\, dx\le c\int_{\mathbb{R}^{n}}|\nabla f(x)|^{p}\,dx\quad (*) \]
Let's see how (*) behaves under scaling: let $f_{\epsilon}(x)=f(\epsilon x)$ and note that $|\nabla f_{\epsilon}(x)|=\epsilon |\nabla f(\epsilon x)|$. So, if (*) is true, then
\[\int_{\mathbb{R}^{n}}|f(x)|^{p}\, dx\le c\epsilon^{p}\int_{\mathbb{R}^{n}}|\nabla f(x)|^{p}\,dx\quad \]
for any $\epsilon>0$. This does not look good, does it?

My speculation: whoever gave you this assignment wanted you to combine the maximal function inequality with the Poincare inequality; for this to work, you need a restriction on the support of $f$.

[quote="mlok"]for $p>1$ the $L^{p}$ norms of $g$ and $Mg$ are comparable.[/quote]
Does that mean $\parallel g\parallel _{L^{p}}=\parallel (Mg)\parallel _{L^{p}}+O(1)$ or something similar? Comparable is a new term for me in this situation.

[quote="mlok"]My speculation: whoever gave you this assignment wanted you to combine the maximal function inequality with the Poincare inequality; for this to work, you need a restriction on the support of $f$.[/quote]

I was thinking also Poincaré's inequality. However, I'm still a very beginner  on such inequalities.

I meant "comparable" in the multiplicative sense: $f_{1}$ and $f_{2}$ are comparable (somewhere) if $C^{-1}\le \frac{f_{1}}{f_{2}}\le C$ for some $C>0$. In this case $\parallel g\parallel _{L^{p}}\le \parallel Mg\parallel _{L^{p}}\le C_{p}\parallel g\parallel _{L^{p}}$. The best possible valued of $C_{p}$ is still unknown.

[quote="mlok"]let $f_{\epsilon}(x)=f(\epsilon x)$ and note that $|\nabla f_{\epsilon}(x)|=\epsilon |\nabla f(\epsilon x)|$.[/quote]
This implies $\epsilon=0$. I think it should be $|\nabla f_{\epsilon}(x)|=\epsilon |\nabla f(x)|$.


3 MEN GO INTO A MOTEL. THE MAN BEHIND THE DESK SAID THE ROOM IS 30 dollars,
SO EACH MAN PAID 10 dollars AND WENT TO THE ROOM.

A WHILE LATER THE MAN BEHIND THE DESK REALIZED
THE ROOM WAS ONLY 25 dollars, SO HE
SENT THE BELLBOY TO THE 3 GUYS' ROOM WITH 5 dollars. ON THE WAY, THE BELLBOY
COULDN'T FIGURE OUT
HOW TO SPLIT 5 dollars EVENLY BETWEEN 3 MEN,
SO HE GAVE EACH MAN A 1 dollars AND
KEPT THE OTHER 2 dollars FOR HIMSELF.

THIS MEANT THAT THE 3 MEN EACH PAID 9 dollars FOR THE ROOM,
WHICH IS A TOTAL OF 27 dollars,
ADD THE 2 dollars THAT THE BELLBOY KEPT = 29 dollars.

WHERE IS THE OTHER DOLLAR?

sorry about the caps.

in this case you wouldn't add the two dollars the bellhop has, but rather subtract it from the 27 dollars the three men have to get 25 dollars, the price of the room

There's been a bunch of problems identical to that all over the place.


Body A, of mass 2.0kg, has a light spring attached to it. Body A moves with a velocity of 3.5ms^-1 over a frictionless plane directly towards a body B of mass 5.0kg which is at rest. At a certain time during the collision, the spring undergoes maximum compression and the two bodies A and B have a common velocity V. Find the value of V.

My question: How is it possible to find V if the length of spring compression and spring constant are not given?

 Isn't it suppose to be Lost in Kinetic energy of A=Gain in K.E of A and B+Elastic potential energy as a result of compression?

[hide]see u have  to assume  the spring constant  and also max elongation. they get eliminated while solving the problem.[/hide]

can someone show me the solution to this problem

you look at the center of mass reference frame. In the cm frame, if both A & B are stationary with respect to each other in the lab frame, they must be stationary in the center of mass frame. Then it is a simple matter to find the velocity of the center of mass frame, which is V, the speed of A and B when they are both moving with the same velocity.

is it 1m/s?  :maybe:
i am not sure though


[quote="arqady"]Prove that for all real $a,$ $b$ and $c$ the following inequality holds:
$a^{4}b^{2}+b^{4}c^{2}+c^{4}a^{2}\geq a^{3}c^{2}b+b^{3}a^{2}c+c^{3}b^{2}a.$[/quote]
by $AM\geq GM$:
$a^{4}b^{2}+c^{4}a^{2}\geq 2a^{3}c^{2}b$
$a^{4}b^{2}+b^{4}c^{2}\geq 2b^{3}a^{2}c$
$b^{4}c^{2}+c^{4}a^{2}\geq 2c^{3}b^{2}a$
and by the sum of these inequalities we're done.
I think the similar to this one has been discussed before :)

OK, sinajackson.
$a^{4}b^{2}+b^{4}c^{2}+c^{4}a^{2}\geq a^{3}c^{2}b+b^{3}a^{2}c+c^{3}b^{2}a\Leftrightarrow\sum_{cyc}\left(\alpha a^{2}b+\beta b^{2}c+\gamma c^{2}a\right)^{2}\geq0,$
where $\alpha+\beta+\gamma=0.$ :wink:

It's trival by Cauchy.

Let $\vec u = <a^{2}b, b^{2}c, c^{2}a>$.

Then form $\vec v$ by shifting over each element of $\vec u$.

Apply Cauchy and you're done.


Let $a,b,c$ be postive numbers such that $a^4+b^4+c^4=a+b+c$.Prove the following Inequality.

\[\frac{a}{c^2(b^2+ca)}+\frac{b}{a^2(c^2+ab)}+\frac{c}{b^2(a^2+bc)}\geq \frac{3}{2}\]

[quote="mecrazywong"]By Cauchy Schwarz, $\frac{a}{c^2(b^2+ca)}+\frac{b}{a^2(c^2+ab)}+\frac{c}{b^2(a^2+bc)}\ge\frac{(\sum\frac{a}{c})^2}{\sum a^2(b+c)}\ge\frac{9}{\sum a^2(b+c)}$. It remains to prove\[\frac{9}{\sum a^2(b+c)}\ge\frac{3\sum a}{2\sum a^4}\]which is trivial.[/quote]
It should be $\frac{a}{c^2(b^2+ca)}+\frac{b}{a^2(c^2+ab)}+\frac{c}{b^2(a^2+bc)}\ge\frac{(\sum\frac{a}{c})^2}{2\sum a^2c}\ge\frac{9}{2\sum a^2c}$.
[quote="Soarer"][quote="mecrazywong"]By Cauchy Schwarz, $\frac{a}{c^2(b^2+ca)}+\frac{b}{a^2(c^2+ab)}+\frac{c}{b^2(a^2+bc)}\ge\frac{(\sum \sqrt{\frac{a}{c}})^2}{\sum a^2(b+c)}$[/quote]
 :P[/quote]
In this case, it should be $\frac{a}{c^2(b^2+ca)}+\frac{b}{a^2(c^2+ab)}+\frac{c}{b^2(a^2+bc)}\ge\frac{(\sum \sqrt{\frac{a}{c}})^2}{2\sum a^2b}$.
Really funny for it not becoming symmetric... :D
Anyway, the proof still works :)


http://en.wikipedia.org/wiki/Muirhead's_inequality
In this site the author gives this example $ x^2\plus{}y^2 \ge 2xy$. He uses bunching to show it. But why does the sequence (2,0) majorizes the sequence (1,1)?

[quote="danilorj"]http://en.wikipedia.org/wiki/Muirhead's_inequality
In this site the author gives this example $ x^2 \plus{} y^2 \ge 2xy$. He uses bunching to show it. But why does the sequence (2,0) majorizes the sequence (1,1)?[/quote] because $ 2\ge 1$ and $ 2\plus{}0\equal{}1\plus{}1$

I was trying to solve a problem and then I got to this:
$ a^2b^2\plus{}b^2c^2\plus{}a^2c^2\minus{}ab^2c\minus{}abc^2\minus{}a^2bc \ge 0$.
$ \sum_{sym} a^2b^2c^0 \ge \sum_{sym} a^2b^1c^1$
Then the sequence (2,2,0) majorizes (2,1,1). Is that correct?
I still didn't understand why a sequence majorizes the other?

[quote="danilorj"]I was trying to solve a problem and then I got to this:
$ a^2b^2 \plus{} b^2c^2 \plus{} a^2c^2 \minus{} ab^2c \minus{} abc^2 \minus{} a^2bc \ge 0$.
$ \sum_{sym} a^2b^2c^0 \ge \sum_{sym} a^2b^1c^1$
Then the sequence (2,2,0) majorizes (2,1,1). Is that correct?
I still didn't understand why a sequence majorizes the other?[/quote]
$ a_1\ge a_2\ge...\ge a_n$ and $ b_1\ge b_2\ge...\ge b_n$ than
$ (a_1,a_2,...,a_n) \succ (b_1,b_2,...,b_n)$ if
for any $ i<n$
$ a_1\plus{}a_2\plus{}...\plus{}a_i\ge b_1\plus{}b_2\plus{}...\plus{}b_i$
and $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}b_1\plus{}b_2\plus{}...\plus{}b_n$
 for your example $ 2\ge 2$
$ 2\plus{}2\ge 2\plus{}1$
$ 2\plus{}2\plus{}0\equal{}2\plus{}1\plus{}1$

I hope everything is clear here:

http://www.mathlinks.ro/viewtopic.php?t=14975