Hey, My name is Jim. I found this board today whilst searching for some career advice.
Background:

I have just completed the second year of an undergraduate degree in mathematics. I haven't really specialized in anything concrete, so it's not too late to change paths.

I wanted to major in mathematics since high school because I always found it to be very easy and straight forward. So when I applied to university, I chose courses centered around mathematics. I decided that I wanted to become an actuary because I thought that making a large amount of money would be a good thing.

However, my views on the world have dramatically changed. Over the course of the past year I have dedicated my free time to becoming informed about the horrors of the world. I hate feeling powerless to do anything when there is so many problems we could fix if only people would open their eyes.

The problem lies in that now I think I will not be able to find a job when I graduate because my strict morals will not allow me to justify working in any corporation that either directly or indirectly supports environmental or social violations and alas most do. Also I don't want to work for a university because it is my personal view that teaching is a waste of my talents.

So my dream job would be to use my logic, problem solving, etc. skills that mathematicians are very good with and use them to aid social and environmental causes dedicated to making the world a better place or maybe become a spy and take down the system from the inside. I don't know which is more plausible.

I don't know if anything out there exists for me. Maybe some of you know of some fellow mathematicians that have gone on to help change the world.

If you're looking to change the world through problem solving skills, there are many options: engineering, inventing, venture capital (with your philosophy of responsibility), law (with your philosophy of responsibility), software, etc.

It's not hard to imagine jobs that are part of building a better world, even if the rhetoric you hear from many college peers echoes the doomsayers of a lifeless corporate future.  It's all a matter of finding a problem that you will enjoy throwing your mind and energy at and then doing it.

Though, just to play devil's advocate, you could always work as an actuary (or anything else) for a few years to build up capital before using that capital to help fulfill your personal initiatives.  Richard Rusczyk might have something to say about that course.

Thanks for the advice. I've never heard of venture capital though. Would you mind explaining it.

[quote="MCrawford"]work as an actuary (or anything else) for a few years to build up capital before using that capital to help fulfill your personal initiatives.[/quote]

I resemble that remark, not that it was my conscious plan.

More seriously, look into careers in areas you enjoy.  But I suggest you look closely before crossing things out.  For example, if you want to work in alternate energies, you may have to work for a major oil company.  But that may be your best bet of ever successfully bringing about a revolution in energy usage.  (This is just a simple example.)

Mainly, I'd advise against adapting strict rules like those you suggest - nearly every opportunity has some good and some ill.

And spend some time studying the good things in the world, too.

Venture capitalists have money (theirs or somebody else's).  They look for people with good ideas to invest in.  Often they will seed startups, some of which fail, but when they succeed, the return on the initial investment is high.  If a venture capitalists invests in 10 such projects and 5 succeed (and 1 or 2 in a big way), the venture capitalist might be well rewarded.

Venture capital is an "economic technology" that matches monetary and intellectual capital resources.  It would be hard to accomplish without diversifying investments because startups tend to be boom or bust [relative to initial investment] during the first few years -- at least, a fairly high percentage of the time.

[quote="annabelle's temptation"]
So my dream job would be to use my logic, problem solving, etc. skills that mathematicians are very good with and use them to aid social and environmental causes dedicated to making the world a better place or maybe become a spy and take down the system from the inside. I don't know which is more plausible.
[/quote]

My first thought is that you should take a look into the world of socially responsible investing.
Links:
http://www.socialinvest.org/areas/sriguide/default.htm
http://www.sustainablebusiness.com/
http://www.domini.com/
http://www.cleanedge.com/
http://www.kld.com/

My second thought is that you might look at the world of social entrepreneurship.
http://www.ashoka.org/fellows/social_entrepreneur.cfm
google-ing social entrepreneurship will give you lots of interesting hits.

Thanks a lot. Those socially responsible investments links look really interesting, I'll check them out when I have more time.

[quote="annabelle's temptation"]
However, my views on the world have dramatically changed. Over the course of the past year I have dedicated my free time to becoming informed about the horrors of the world. I hate feeling powerless to do anything when there is so many problems we could fix if only people would open their eyes.

The problem lies in that now I think I will not be able to find a job when I graduate because my strict morals will not allow me to justify working in any corporation that either directly or indirectly supports environmental or social violations and alas most do. Also I don't want to work for a university because it is my personal view that teaching is a waste of my talents.

So my dream job would be to use my logic, problem solving, etc. skills that mathematicians are very good with and use them to aid social and environmental causes dedicated to making the world a better place .[/quote]

When there are natural or man-made disasters, when there is natural, accidental or intentional injury to people or property, there is one industry that helps people resume their lives:  insurance.  There would be no doctors, medicine, cars or computers, just to name a few things, without insurance.  There would be no newspapers, freedom of the press, and buildings to work in without insurance.  And actuaries help make the insurance industry financially viable to pay for accidents when they occur.

Yes, the insurance point might be right but I'd rather stop problems from continuing to occur rather than lessening the impact of the ones that already happen. The more I think about it, the less I want tobe an actuary. I just need more excitment in my life rather than always being risk averse, which is the viewpoint of actuaries, I think.

I think there will be some major scientific break throughs in the next decade. Hopefully more beneficial to society, but that's probably unlikely as long as we have militant states.  Anyway, I'm digressing.

Now, I'm contemplating becoming involved in technological research, like space or the more science fictionesque type research.

I'm pretty sure mathematicians would be useful in those type of jobs because I'm sure the science would involve very complex calculations.

Does anyone know what my chances are for getting into something like that? or What what kind of educational background it would require?

Thank you so much for being so helpful thus far.

[quote="annabelle's temptation"]The more I think about it, the less I want tobe an actuary. I just need more excitment in my life rather than always being risk averse, which is the viewpoint of actuaries, I think.
[/quote]

I'll take a moment to challenge this point-of-view.

Avoiding unnecessary risks allows us to take more risks in our life.  Many risks come with payouts that substantially outweight the volatility incurred.

The Sharpe Ratio gives us an idea as to how volatility "eats value."  Take two hypothetical companies that each average 30% return on capital each year.  One of the companies is very stable and profits alternate between around 20% and 40% from year to year.  The other company is more volatile and alternates between 0% and 60%.

The first company earns a 68% return on capital every two years while the second earns 60%.

However, if we could aggregate the risks of several volatile companies, the year to year return on capital would begin to approach that of the more stable companies.

This is the exact same value-added process insurance companies are involved in.  Insurance companies regulate cash flows such that there is more value for the insureds.  Insurance is not simply a matter of avoiding risk.  In fact, it's a matter of funding more risks from which people can create more value.

In fact, if you look at investment behavior on both the level of labor and liquid capital, it's hard to dispute that insurance has created an environment in which people regularly take more and greater risks and are rewarded for it.

[quote="MCrawford"]
The Sharpe Ratio gives us an idea as to how volatility "eats value."  Take two hypothetical companies that each average 30% return on capital each year.  One of the companies is very stable and profits alternate between around 20% and 40% from year to year.  The other company is more volatile and alternates between 0% and 60%.

The first company earns a 68% return on capital every two years while the second earns 60%.
[/quote]

This is basically AM-GM inequality, right?

[quote="themonster"]
This is basically AM-GM inequality, right?[/quote]

No.  Though I didn't point it out here, the Sharpe Ratio gives us a precise measure of the value lost (given some usually reasonable assumptions of the distribution of the outcomes).

[quote="annabelle's temptation"]
Now, I'm contemplating becoming involved in technological research, like space or the more science fictionesque type research.

I'm pretty sure mathematicians would be useful in those type of jobs because I'm sure the science would involve very complex calculations.

Does anyone know what my chances are for getting into something like that? or What what kind of educational background it would require?

[/quote]

Hi,

It may be fun to do proofs of theorems, but that doesn't usually pay the bills.  What pays the bills for a mathematician is being able to APPLY the mathematics to real problems.  And to understand those real problems, it is the subject matter of those real problems that you need to learn.

So perhaps you need to study astrophysics, or biology, or environmental science, if you want to apply mathematics to those areas.  Choose whichever application interests you, and study it.  Without this additional study, you are basically a theoretical mathematician in the eyes of these other fields.

As an actuary, I have chosen to study insurance.  For many years I could not talk intelligently to insurance people about insurance until I studied that subject.  At this point in my career I have studied as much about insurance, if not more, than the typical non-mathematical insurance professional.

[quote="ComplexZeta"]Proving theorems can definitely pay the bills![/quote]

Yes, but for a small minority of mathematically-trained minds.  Most people need to plan for the event that they will not be professional mathematicians which are rapidly becoming a smaller and smaller proportion of that group.

[quote="mahras"]Off topic: However speaking of sharpe ratio's the problem is that it punishes positive deviation of returns as well as negative ones. For example, say on a daily basis a fund returns 2%, 6%, 9%, 3% vs say a fund that returns 2.5%, 3%, 1.5%, 1%. As the deviation from the mean is greater on the first fund it is punished with a harsher shape ratio. Sometimes risk may be necessary.

Calculation of the risk to reward ratio should be key in all situation in finance as well as in life. One must calculate and evaluate the risk involved for a certain reward in all situations. It varies by person to person so one person's perspective of risk taking may be dramatically different than anothers.

Just a viewpoint. (BTW this was my first post  :D )

-Bobby[/quote]

Welcome to the forums.

Yes, risk is necessary.  The point of the Sharpe ratio is to evaluate the relative reward of those risks so that we have a more firm way of evaluating the value of risk.


I recently upgraded from Fedora 8 to Fedora 9, and from Firefox 2.0 to Firefox 3.0. Whereas before I could view other people's Geogebra but couldn't make anything, now I can't do either. Please help!

I have a similar bug. I have XP, I think latest Java, and Opera 9.0 but all I can see on geogebra posts are Applet Loaded and there is nothing there...


Let $a,b,c$ be distinct real numbers  such that the equations $x^2+ax+1=0$ and $x^2+bx+c=0$ have  a  common real root, and the equations $x^2+x+a=0$ and $x^2+cx+b=0$ also have a  common real  root. Compute the sum $a+b+c.$
(All Russian MO, 2000)

We are to obtain the product of two common roots is 1.

I don't know if there is a shorter solution to this problem
(I think that there is, because the title says [color=blue][b]Easy algebra[/b][/color]  ) :D

[hide="Here is my solution"]
$x^2+ax+1=0 \ \ (1)$
$x^2+bx+c=0 \ \ (2)$
$x^2+\ x+a=0 \ \ (3)$
$x^2+cx+b=0 \ \ (4)$

The equation $(3)$ has real roots, so $1^2-4a\geq 0\Rightarrow a\leq  \frac{1}{4}$

So $a \neq 1$

Also, if $c=1$ then the equations $(3),(4)$ can't have a common root (because $a \neq b$). So $c\neq 1$

Let $x_1$ be the common root of $(1)$ and $(2)$
and $x_2$ the common root of $(3)$ and $(4)$

$\begin{array}{cc} x^2+ax+1=0 & (1) \\ x^2+bx+c=0 & (2) \end{array} \}\Rightarrow  (b-a)x_1+(c-1)=0 \Rightarrow  \boxed{x_1 = \frac{c-1}{a-b}}$

$\begin{array}{cc} x^2+x+a=0 & (3) \\ x^2+cx+b=0 & (4) \end{array} \}\Rightarrow  (c-1)x_2+(b-a)=0 \Rightarrow  \boxed{x_2 = \frac{a-b}{c-1}}$

Hence, $x_1x_2=1$, as [b]kunny[/b] pointed out

Since $x_1x_2=1$, it follows that $x_2$ is also the second root of $(1)$

So $x_2$ is a common root of $(1),(2),(3)$

$\begin{array}{cc} x^2+ax+1=0 & (1) \\ x^2+1x+a=0 & (3) \end{array} \}\Rightarrow (a-1)x_2=a-1\Rightarrow \boxed{x_2=1}$

$x_2 = \frac{a-b}{c-1} \Rightarrow  1 = \frac{a-b}{c-1} \Rightarrow  a-b=c-1 \Rightarrow b+c=a+1\Rightarrow  \boxed{\boxed{a+b+c=2a+1}}$

So we need to find $a$

Apply $x=1$ to $(3)\Rightarrow$

$1+1+a=0\Rightarrow \boxed{a=-2}$

So finally $a+b+c=2a+1 =-4+1 \Rightarrow  \boxed{\boxed{a+b+c=-3}}$

Notice that $(1)$ became $x^2-2x+1=0$, so $x_1=x_2=1$.

In order to find the sum $a+b+c$, I first found $a$.

[b]Nickolas[/b], is there a simpler way to solve this problem?
[/hide]

one thing that you could do is multiply the respective equations, and then they will have a double root, and to satisfy that, f(r)=0, and f'(r)=0, but this seems that it would get messy

[quote="Altheman"]f'(r)=0[/quote]Refrain from using calculus

No  pontios, i dont think there is  an easier solution. Mine was  almost  the same with yours.

If you don't know, then can use this lemma (what prove easily):

[color=red]Two equations $ax^2+bx+c=0$ and $a'x^2+b'x+c'=0$ have at least
a common root if and only if $\boxed {\ \left|\begin{array}{cc} a & c\\ a' & c'\end {array}\right| ^{\ 2}=\left|\begin{array}{cc} a & b\\ a' & b'\end{array}\right|\cdot \left|\begin{array}{cc} b & c\\ b' & c'\end{array}\right|\ }$
where $\left|\begin{array}{cc} X & Y \\ Z & T\end{array}\right| =XT-YZ.$[/color]

[quote="shobber"]May I know which grade the problem was in? 8th? 9th?[/quote]
Αt $9^{th}$ grade.


hmm how do you solve this problem??

Triangle ABC has vertices A(0,0), B(0,3) and C(5,0).
A point P inside the triangle is $ sqrt10$ units from point A and $ sqrt13$ units from point B. How many units is P from point C? Express your answer in simplest radical form.

Help me.

ok

we first draw a diagram (i cant post one)

P is $ /sqrt{10}$ units away from A

the only possible integer way you could do that is legs of 3&1

P is $ \sqrt{13}$ units from B

the only possible way that w/ integers is legs of 3&2
since P is in the triangle, it must be 2 away from (0,3) and 3 up, so the coordinates of P are (1,3) (check it, it works)
distance from P to C

draw perpindiculars as legs

the legs are of length 1&2

so PC=$ \sqrt{4\plus{}1}\equal{}\sqrt{5}$


limit
logarithms
algebra unsolved
algebra

Let $ x_{1},x_{2},...$ is sequence of positive integer number such that $ x_{m+1}>x_{m}\forall m\geq 1$ and $ x_{m}|x_{m+1}$ and

$ \lim_{n\to\infty}\frac{\ln x_{m+1}}{\ln x_{m}}=1$. Let $ P\in\mathbb{Z}[x], \deg{P}=1982$ such that $ P(x_{m})$ is perfert square for all $ m$. Prove that exists $ Q\in\mathbb{Z}[x]$ such that $ P\equiv Q^{2}$.


:?:
Given the curved arc length and central rise (distance from centre of chord to centre of arch), how is the radius caculated?

Well, the equation I got seemed to be really difficult to solve for with general constants, but a numerical solution is not bad with a calculator.

Let $r$ be the radius and $2\theta$ be the angle that subtends the arc. Draw a radius that bisects the arc and draw a segment from one endpt. of the arc down to intersect the new radius perpendicularly. This creates a right triangle with angle $\theta$. Given a length for the arc, $s$, we know that $\theta=\frac{s}{2r}$. Also, given the length of the "central rise", $x$, we can write the equation $rsin(\theta)+x=rsin(\frac{s}{2r})+x=r$. Now you can solve for $r$.

[quote="gsandy"]Thanks for the formula. I am now trying to get that into a spreadsheet and having trouble inserting your solution. My maths is a little rusty.

I tackled the problem by trying to find the arc bisect angle "B" and got:

B/(2 - cosB) = (s x 90)/(Pi x X):  B in radians, s = arc length and X = rise

Once B is known the radius can be cacluated.

But once again my maths let me down.

For example I do know that for X (rise) of 291 and s (arc length) of 5360
The radius is 12300 and B (bisect angle) = 24.97 degrees[/quote]

Sorry that should read:

B/(1-cosB) = (s x 90)/(pi x X): B in degrees

Any help to obtain B = ??

If your given arc length $s$ and radius $r$, then subtended angle is $s/r$. Then you can divide by 2 for bisect angle, if I'm interpreting the term correctly.

I havnt got the radius (or should I say I could not get it from your formula)

I only have s (arc length) and X (rise)

Thats why I submitted my formula to solve for angle B (which is half bisect angle) and from that I can obtain the radius

Thanks for you help so far please persevere

[quote="gsandy"]Thanks for the formula. I am now trying to get that into a spreadsheet and having trouble inserting your solution. My maths is a little rusty.

I tackled the problem by trying to find the arc bisect angle "B" and got:

B/(2 - cosB) = (s x 90)/(Pi x X):  B in radians, s = arc length and X = rise

Once B is known the radius can be cacluated.

But once again my maths let me down.

For example I do know that for X (rise) of 291 and s (arc length) of 5360
The radius is 12300 and B (bisect angle) = 24.97 degrees[/quote]

When you have an equation like that for $B$, mixing $B$ with a trigonometric function of $B$, you usually [b]can't[/b] get an exact solution. Unless the answer could be figured out by guess-and-check (such as $\pi /3$), you would have to resort to a calculator.

Thanks Jason,

If that is the case, I will have to try and find another way to solve the problem

Sandy :?


calculus
integration
limit
function
derivative
calculus computations

Hi! How calculate $\int_{x_0}^{x_0+h} (2t^4)/(\sqrt{t^4+9}) dt$?:blush:
Thank you for your help... :D

[quote="bénabar"][quote="Kent Merryfield"]Just a thought... bénebar, was this by any chance not the whole problem? Were you perhaps asked the following question?

$\lim_{h\to0}\frac1h\int_{x_0}^{x_0+h}\frac{2t^4\,dt}{\sqrt{t^4+9}}$[/quote] Here is the complete statement of the exercise: Let $F$, a primitiv of the function $x \mapsto f(x) = \sqrt{x^4+9}$.
I want prove that: $F(x_0+h)-F(x_0)-hf(x_0) < 7h^2/4$.

I apply Taylor's formula to $ [x_0; x_0+h] $ in the order $2$, thus $F(x_0+h)-F(x_0)-hf(x_0) = \int_{x_0}^{x_0+h} (x_0+h-t) F''(t)dt$. Thus I want prove $\int_{x_0}^{x_0+h} (x_0+h-t) F''(t)dt < 7h^2/4$. ($F''(t) = 2x^3/ \sqrt{x^4+9}$).

Maybe I have badly acts? :blush:  :?  :([/quote] Am I clear? (I have difficulties with the English language).  :blush:

[quote="fedja"][quote="bénabar"]

I apply Taylor's formula to $ [x_0; x_0+h] $ in the order $2$, thus $F(x_0+h)-F(x_0)-hf(x_0) = \int_{x_0}^{x_0+h} (x_0+h-t) F''(t)dt$. Thus I want prove $\int_{x_0}^{x_0+h} (x_0+h-t) F''(t)dt < 7h^2/4$. ($F''(t) = 2x^3/ \sqrt{x^4+9}$).

Maybe I have badly acts? :blush:  :?  :([/quote]

No, what you are doing is perfectly fine. Also your English is quite clear. But something is wrong with your exercise as you present it: if $x_0$ can be very large, the derivative $F''(t)$ may also be very large on $[x_0,x_0+h]$ and no estimate independent of $x_0$ is possible. Is there any bound on $x_0$ in your book?[/quote] Sorry, Give for $x_0 = 2$, $|h|<1/10$.


arithmetic sequence
combinatorics unsolved
combinatorics

Consider the following array of (5*5):
[b]1    2    3    4    5
6    7    8    9   10
11  12  13  14  15
16  17  18  19  20
21  22  23  24  25 [/b]
show that no matter which 5 numbers are removed from this array, there exists a 5-term Arithmetic progression among the remaining 20 numbers

if some row or column doesn't contain any of the 5 removed numbers, the conclusion follows by virtue of the fact that every column and every row forms an arithmetic progression.

so assume that every column and every row contains exactly one of the 5 numbers. if $ a_{i1}$ is removed for any $ i$ from 1 to 4, then the next five numbers following $ a_{i1}$ are all not removed and they form an arithmetic progression. so assume that $ a_{51}$ is removed. if $ a_{i2}$ is removed for any $ i$ from 1 to 3, then the 5 numbers following $ a_{i2}$ are all not removed and they form an arithmetic progression. so assume that $ a_{51}$ and $ a_{42}$ are removed. continuing this reasoning, we may assume that the diagonal from the lower left to the upper right is removed - i.e., the numbers 5, 9, 13, 17, 21. but then the numbers 2, 4, 6, 8, 10 have not been removed and form an arithmetic progression. so in every case a 5-term arithmetic sequence remains.


Three people, Alan, Beth, and Cindy, pooled their money to purchase lottery tickets. Alan invested $ \$25.00$. Beth invested $ \$20.00$ and Cindy invested $ \$35.00$. Their winnings totaled $ 6.4$ million dollars. How many dollars was Beth's share of the winnings if the amount received was proportional to the amount invested?

The total amount the three of them invested is $ 25\plus{}20\plus{}35\equal{}80$.  This means Beth's proportion is $ \frac{20}{80}\equal{}\frac{1}{4}$.  To find out how much she wins, we do $ \frac{1}{4}6400000\equal{}\boxed{1600000}$


calculus
integration
number theory unsolved
number theory

Let $ x,y$ be real numbers such that the numbers $ x\plus{}y, x^2\plus{}y^2, x^3\plus{}y^3, x^4\plus{}y^4$ are all integers. Prove that for all positive numbers $ n$, the number $ x^n \plus{}y^n$ is an integer.

[hide="solution"]Since $ x \plus{} y\in \mathbb{Z}$, we can rewrite $ x \equal{} \frac {k}{2} \plus{} \frac {\sqrt {a}}{2}$, $ y \equal{} \frac {k}{2} \minus{} \frac {\sqrt {a}}{2}$ for some $ k\in \mathbb{Z}$, $ a\in \mathbb{R}^ \plus{}$.  From $ x^2 \plus{} y^2 \equal{} \frac {1}{2}(k^2 \plus{} a)\in \mathbb{Z}$ we conclude that $ a\in \mathbb{N}$.  Let $ p(n) \equal{} x^n \plus{} y^n$, then it is easy to check that $ p(n)$ satisfies the recursion equation
\[ p(n \plus{} 2) \minus{} kp(n \plus{} 1) \plus{} \frac {k^2 \minus{} a}{4}p(n) \equal{} 0\]
This implies that $ \frac {k^2 \minus{} a}{4}p(1) \equal{} \frac {k(k^2 \minus{} a)}{4}\in \mathbb{Z}$ and $ \frac {k^2 \minus{} a}{4}p(2) \equal{} \frac {(k^2 \minus{} a)(k^2 \plus{} a)}{8}\in \mathbb{Z}$.  Let $ k^2 \minus{} a \equal{} c\mod 8$, then we have the system:
\[ c(c \plus{} 2a) \equal{} 0 \mod 8 \\
k^2 \minus{} a \equal{} c \mod 8 \\
kc \equal{} 0 \mod 4\]
It's easy to verify that all solutions satisfy $ c \equal{} 0\mod 4$, so $ \frac {k^2 \minus{} a}{4}\in \mathbb{Z}$, and therefore $ p(n)\in \mathbb{Z}$ for all $ n\in \mathbb{N}$.[/hide]

You need not complicate it so much.
Using $ x \plus{} y, x^{2} \plus{} y^{2}, x^{3} \plus{} y^{3}, x^{4} \plus{} y^{4} \in \mathbb{Z}$, we have $ 2xy \in \mathbb{Z}$(from $ x \plus{} y, x^{2} \plus{} y^{2} \in \mathbb{Z}$) and $ 6x^2y^2 \in \mathbb{Z}$(from $ x \plus{} y, x^{4} \plus{} y^{4} \in \mathbb{Z}$ and the previous statement.
These together imply that $ xy \in \mathbb{Z}$.)
Now since $ x^n \plus{} y^n$ is symmetric, it can be written in terms of $ x \plus{} y$ and $ xy$.(which $ \in \mathbb{Z}$)
Therefore, $ x^n \plus{} y^n$ is itself an integer.

The given data implies that $xy$ is integral.

In the binomial expansion of $(x+y)^n$ [Integral], we can group the terms other than $x^n+y^n$ in terms of $xy$ and $x+y$ [Both are integral].

Thus, $x^n+y^n$ is also integral.


In the year $ 2001$, the United States will host the International Mathematical Olympiad. Let $ I$, $ M$, and $ O$ be distinct positive integers such that the product $ I\cdot M \cdot O \equal{} 2001$. What's the largest possible value of the sum $ I \plus{} M \plus{} O$?

$ \textbf{(A)}\  23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\  99 \qquad \textbf{(D)}\  111 \qquad \textbf{(E)}\ 671$

[hide="Solution"]I got E.

This occurs when the numbers I, M< and O are 667,3, and 1.[/hide]

[hide]2001=3*23*29
all prime, so the only possible answer
3+23+29=55

[b]B[/b]
/edit: nevermind, I completly forgot about 1 as a factor >_<

[/hide]

That's wrong.
No offense.
It is 1*667*3=2001
1+667+3=671
E is the correct answer.

[quote="Pakman2012"]Oh man, you're so hardcore.  Solved an IMO Problem!!  :rotfl:[/quote] :first:

this is an AMC 10 question. AMC 10 2000 #1, actually, lol


i don think the yeild will be much...

Think of better methods....
I'll post the solution later...

actually it's essential to know where u wanna start with to talk about yield

Ok start with Diethyl maonate and diethyl fumarate....and those who answer please propose the mechanism....

@chemrock-- no answering this question please..

then it's just a malonic ester synthesis :)

[quote="Euclidean Geometer"]start with propene----> allylic Bromination using NBS, followed by Sn2 using KCN and hydrolyse it. <At this stage i dunno if its necessary to protect dis COOH frm oxidn>

Then, Use dat OsO4 thing/ PBA followed by hydrolysis to get vicinal diol.
Will Vigorous oxidation lead us to da product????  :maybe:  :?:[/quote]
nope, it wont. as it has two carbons less. :D
Pardesi is rite, it’s a malonic ester synthesis, treat di ethyl malonate with a base and react with diethyl fumarate, foll by hydrolysis and heat
Here’s another method: (kinda longish but uses only diethyl malonate)
Diethyl malonate+ NBS:
Followed by hydrolysis and double arndt-eistert synthesis
Then treat with KCN and hydrolyse


Does AoPSWiki employ such tags? It'll make everything a lot easier for the common viewer.

Example of such tags: ("http://en.wikipedia.org/wiki/Palmersville%2C_Tennessee")

We can have:
Deletion
Neutrality dispute
Cleanup/Polish
Inaccuracy dispute
In use (for someone who is currently majorly editing the page)

Then, we can have portal page for each article in each category for  review (say AoPSWiki:Deletion, AoPSWiki:Neutrality, etc.).

If AoPSWiki already uses this please ignore this post.  :lol:

See:[url=http://en.wikipedia.org/wiki/Wikipedia:Maintenance]Wikipedia:Maintenance[/url]

We can actually use there tag images/text for our Wiki (many other wiki's already do).

The AoPS wiki is heavily stubbed.  Many of the others are not relevant to a wiki composed almost entirely of mathematics and math problems.  Look at SpecialPages:Categories to see a list of the categories in use, many of which have associated templates (such as {{stub}}).

surge, as the AoPSWiki develops, we'll expand the tags and procedures for working through disputes.  However, it may yet be weeks or months before some of these are useful.  The reason is that AoPSWiki has very little "bait" for politically slanted article writers.  It will be used mostly by math students, and then by high school science and computer science students as it grows.  Nearly all such material is uncontroversial (except perhaps debate centering around the Flying Spaghetti Monster).

Just curious:

Was "flying spaghetti monster" something you made up, something you heard about, or something that actually happened on AoPS.  The [url=http://en.wikipedia.org/wiki/Flying_Spaghetti_Monster]reason[/url]....


A sequence $\{ a_n\}$ is defined by $a_{n+1}=2(a_n) ^2$, where $n$ is a natural number and $a_1=1$.

Find the maximum possible value of $n$ such that $a_n<10^{60}$.

Let $a_n = 2^{b_n}$. Then we have the recurrence $b_{n+1}=2b_n+1$, or even better, $b_n = 2^{n-1}-1$, so we want to find the smallest $n$ such that

\[ 2^{2^{n-1}-1}<10^{60}  \]

Some simple algebraic manipulations with logs (remember, $\log 2 = .3010$ !!) yields that $n=8$ is our maximum.


[quote="chien than"]Let $a;b;c;d>0$ such that $a+b+c+d=4$.Prove that:
$ab+bc+ca+ad \geq ab^{2}c+bc^{2}d+cd^{2}a+da^{2}b$[/quote]
The $LHS$ is $ab+bc+cd+da$, isn't it?
[hide="if the $LHS=ab+bc+cd+da$"]$RHS=(ab+cd)(bc+da) \le \frac{(ab+bc+cd+da)^{2}}{4}$
But $ab+bc+cd+da=(a+c)(b+d) \le \frac{(a+b+c+d)^{2}}{4}=4$
So we get the desired result.[/hide]


A tank contains 10,000 gallons of water at the beginning of the day on
June 1. Each day, 1% of the water in the tank at the beginning of the day is lost to
evaporation. How much water is left at the end of the last day of June? Express
your answer to the nearest whole number.

Ok do you have to keep multiplying it by 0.01   30 times? I don't understand any other way to do it.

Yes.  This is $ (.99)^{30} \times 10000 \approx \boxed{7397}$.  Use a calculator (this is either a target, team, or workout problem).


analytic geometry
function
geometry
trigonometry
rotation

the positive x&y quadrent is number I
The negative x&positive y quadrent is number II
The negative x&y quadrent is number III
The positive x&negative y quadrent is number IV

I think what you mean is why they don't number the quadrents from left to right.  I think that this is because the 1st quadrent is the one we usually work in, and the 2nd is the next common, because we use that when we mirror the 1st quadrent in functions and parabollas and stuff like that.  3rd quadrent is the next common, because it is the opposite of the 1st quadrent.  4th quadrent is the 4th because we use the least important functions that include quadrent I in this quadrent.

This is probably wrong, it is why i think they have numbered the quadrents as they have.  I would love to learn that this is correct, and i would also like to learn the real reason if this is incorrect :) good question

[quote="supersonicnapkin"]Hey does any one know why the quadrants in coordinate graphing are backwards??[/quote]

I guess the mathmaticians just don't want to make things easy  :rotfl:

But seriously, I don't know either.  :?

maybe the guy who invented the coordinate graph was from a country where they count backwards.

wow I never knew it was this complicated!! the other day we were talking about it in class, and we came up with these..haha:

1) the guy was left-handed ( I think my teacher said that.... I don't know what that's got to do with it but...)
2) the guy was weird and insane. haha :lol:

I think all that trig is a little advanced for this forum  ;)  :roll:

[quote="nvidura"]This is just because q geometry was found to define Trigonometry for non right angles. When you are defining trigonometric for none right angular angles

You have to assume that a point moving in a counter clockwise circular path, with the radius of 1 unit and center is the origin, starting from (1, 0). The angle between the point, origin and the positive side of the x-axis is the variable. This can be 0 to 2П or sometimes more than that (1st round max 2П, 2nd round max 4П …). Since the point is moving with radius 1 it certainly, passes the following co-ordinates. (1, 0), (0, 1), (-1, 0), (0,-1). In the very beginning the point is in (1, 0) and the angle is (1, 0)(0, 0)(1, 0) which is 0. When the point is in (0, 1) the angle is (0, 1)(0, 0)(1, 0) which is П/2

When it is passing quadrant I
x value: 1→ 0
y value: 0 → 1

When it is passing quadrant II
x value: 0→ -1
y value: 1 → 0

When it is passing quadrant III
x value: -1→ 0
y value:  0→ -1

When it is passing quadrant IV
x value: 0 → 1
y value:  -1 → 0

Sin, cos, tan, cossec, sec & cot graphs will be drawn with this assumption. Because of this reason the y = sin x graph cuts the origin and y = cos x graph do not.

also ( x , y ) its quadrant I then,

Sin x (0 < x < П/2) = Positive
Cos x (0 < x < П/2) = Positive
Tan x (0 < x < П/2) = Positive
Cossec x (0 < x < П/2) = Positive
Sec x (0 < x < П/2) = Positive
Cot x (0 < x < П/2) = Positive

If ( -x , y ) its quadrant II then,

Sin x (П/2 < x < П) = Positive
Cos x (П/2 < x < П = Negative
Tan x (П/2 < x < П = Negative
Cossec x (П/2 < x < П) = Positive
Sec x (П/2 < x < П = Negative
Cot x (П/2 < x < П = Negative

If ( -x , -y ) its quadrant III then,

Sin x (П < x < 2П/3) = Negative
Cos x (П < x < 2П/3) = Negative
Tan x (П < x < 2П/3) = Positive
Cossec x (П < x < 2П/3) = Negative
Sec x (П < x < 2П/3) = Negative
Cot x (П < x < 2П/3) = Positive

If ( -x , y ) its quadrant IV then,

Sin x (2П/3 < x < 2П) = Negative
Cos x (2П/3 < x < 2П) = Positive
Tan x (2П/3 < x < 2П) = Negative
Cossec x (2П/3 < x < 2П) = Negative
Sec x (2П/3 < x < 2П) = Positive
Cot x (2П/3 < x < 2П) = Negative

In addition, that can be easily understood if you try to draw the Sin, Cos and Tan curves. [size=150][b]If you notate that in any other methods the explanations of mathematics would be different than now we are using. [/b][/size][/quote]

Wow...I understand like 2% of that... :?  I think the theory about like common/not common makes sense. that's what I always thought.


A two-inch by six-inch board costs 24 cents per linear foot.
Jennifer needs to purchase 18 boards that are each
10 feet long. What is the total cost of the lumber?

[asy]unitsize(0.5inch);
draw((0,0)--(12,0)--(12,4)--(0,4)--cycle);
draw((0,4)--(18,20)--(30,20)--(12,4));
draw((30,20)--(30,16)--(12,0));
label("6''",(6,0),S);
label("2''",(0,2),W);
label("24 cents per foot",(21,8),E);[/asy]

Unless I'm interpreting this incorrectly, this is just $ 18\cdot 10\cdot 24\equal{} \boxed{\$ 43.20}$ since I'm pretty sure they would ask for dollars not cents.


What does it mean for a variable $ {x}$ to be uniformly distributed between 0 and 1.

It would help to know the context of the statement, but what that would seem to imply is that $ x$ is chosen at random between 0 and 1.


Find the domain and range of:
$ f(x) \equal{} \sqrt{x^2 \plus{} 2x \plus{} 2}$
:nhl:  :pilot:  :bruce:

hello, we have $ x^2\plus{}2x\plus{}2\equal{}(x\plus{}1)^2\plus{}1 \geq1$ so it is
$ \sqrt{x^2\plus{}2x\plus{}2}\geq1$ for all real $ x$. For $ x\equal{}\minus{}1$ we have $ f(\minus{}1)\equal{}1$.
Sonnhard.


CAN ANYONE HELP ME WITH MISCELLANEOUS PROBLEMS. I DONT REALLY UNDERSTAND THEM. =(

1. WHAT'S THE RATIO OF A TO B?
2A-3B=5
A+B=10

2. ON A TEST WHICH 200 STUDENTS TOOK, 48% OF THE STUDENTS ANSWERED QUESTION A CORRECTLY, AND 55% OF THE STUDENTS ANSWERED QUESTION B CORRECTLY. IF 23% OF THE STUDENTS ANSWERED BOTH QUESTIONS WRONG, FIND THE NUMBER OF STUDENTS WHO ANSWERED BOTH QUESTIONS A AND B CORRECTLY.

3. IF FIVE WRESTLERS ARE COMPETING WITH ALL OTHER WRESTLERS ONCE, HOW MANY MATCHES ARE THERE IN THE GAME?

4. WHEN A DIE IS THROWN TWO TIMES, TEH FIRST NUMBER IS X AND THE SECOND NUMBER IS Y ; FIND THE PROBABILITY OF HAVING 2X+Y=5

5. IN A SEQUENCE OF NUMBERS, THE FIRST NUMBER IS 2 AND EACH NUMBER AFTER THE FIRST IS 4 LESS THAN 3 TIMES THE PRECEDING NUMBER. WHAT IS THE FOURTH TERM IN THE SEQUENCE.

[hide="1"]multiply the second equation by 2 and subtract to get -5B=-15.
B=3, and from the second original equation, A=7
A:B=7:3[/hide]

[hide="2"]100-23=77% of the students got something right. 48+55=103 (added percentage). There must be something repeated because 103>77, so we subtract 77 from 103 to see that 26% of people (52 people) got both right. [/hide]

[hide="3-wrong read chess64s solution"]Within 5 wrestlers, let's have 4 play matches (2 games), to leave three people left. Two of them play a match (1 game) to leave two left. Then those two play a final match for a total of 4 matches? [/hide]

[hide="4"]There are 36 ways to throw a dice. When X=1, Y has to be 3. when X=2, Y has to be 1. There are two ways. 2/36=1/18 [/hide]

[hide="5"]The first term is two, so the second term is 2(3)-4=2. That's nice. All the numbers are 2. The fourth term is 2. [/hide]

PLEASE AVOID THE CAPS! IT SEEMS LIKE YOU'RE SHOUTING!

1)
$2a-3b=5$
$2a+2b=20$
$-5b=-15 \rightarrow b=3 \rightarrow a = 7$
$a/b=7/3$

2)
$.48 \cdot 200 = 96$ A
$.55 \cdot 200 = 110$ B
$.23 \cdot 200 = 46$ NONE
$200 = 96+110+46-x \rightarrow x=52$

3) Let's pretend we're picking the matches. We can choose from 5 people for the first contestant and 4 for the second. That means $20$. However, A vs. B is the same as B vs. A, so we have to divide $20/2=10$.

4)
$2x+y=5 \rightarrow (x,y)=(2,1),(1,3)$ are the only possibilities.
$2 \cdot \frac{1}{36}=\frac{1}{18}$

5)

$\{ 2, 3 \cdot 2 - 4=2, 3 \cdot 2 - 4 = 2 \ldots \}$

So $2$ because $3 \cdot 2 -4 =2$ so it's going to stay 2 for the whole sequence.

Sorry about number 3, I got confused by what "all other wrestlers" meant. It should be "if each of the five wrestlers competed with each other once, ..."


Crystal has 3 nickels, 2 dimes and 1 quarter. How many different sums of money can Crystal obtain by using exactly four of these coins?

w/ q.

1q 2d 1n
1q 1d 2n
1q 3n

w/o q

2d 2n
1d 3n

5.

[quote=Possible]Don't Bump 12 year old threads[/quote]

It's fine as long as it contributes to the discussion.

[quote=coolmath2017]is there way w/o listing?[/quote]

I think listing is the fastest way because there are only $5$ different sums.

If you were to do it without listing, there are $2^6 - 1 \implies 63$ possible combinations, but then you would have to correct for a lot of overcounting. I think listing is easier.


Sa se determine relatia dintre $a>1$ si $\alpha \in R^{*}$  astfel incat ecuatia:
               $a^{x^{2}+\alpha y}+a^{y^{2}+\alpha x}= 1$
sa admita solutie unica in $R$x$R$.

Evident $x=y$
Daca $x \neg y$ atunci avem solutiile $(x,y)$ si $(y,x)$ ceea ce contrazice enuntul problemei
Din $x=y\Rightarrow 2a^{x^{2}+\alpha x}=1 \Leftrightarrow a^{x^{2}+\alpha x}=\frac{1}{2}\Leftrightarrow log_{a}\frac{1}{2}=x^{2}+\alpha x$ unde obtinem ecuatia de gradul doi $x^{2}+\alpha x-log_{a}\frac{1}{2}=0$ unde avem solutie unica daca si numai daca $\Delta=0\Leftrightarrow \alpha^{2}+4log_{a}\frac{1}{2}=0$ unde obtinem relatia dorita: \[log_{a}\frac{1}{2^{4}}=-\alpha^{2}\]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Q. If
\[ K \triangleleft H \triangleleft G,
\]
then is $ K \triangleleft G ?$

Take a look at the following thread:

[url]http://www.mathlinks.ro/viewtopic.php?t=256519[/url]

Regards,  :wink:
José.

but in this case it is not given that $ K$ is characteristic subgroup in G.

Forget all about the word [i]characteristic[/i], dear Rahul... Just take a look at the fourth post in the thread.

If still in doubt, check this out:

[url]http://www.mathlinks.ro/viewtopic.php?t=250127[/url]

Regards,
José.


X and Y are positive real numbers such that

X^2 + Y^3 >= X^3 +Y^4

Prove X^3 + Y^3  <= 2

Solution appears on page 161 Mathematical Olympiads Problems and Solutions from around the world by T Andreescu and Z. Feng

:cool:  Hey man ! What's an open question??  :cool:
.........

Thank you Huhuhu
Congratulations on your nice solution.!

Michael Niland

[quote="huhuhu"]x^3+x^3+1>=3X^2,
y^4+y^4+y^4+1>=4y^3
so:2x^3+3y^4+2>=3x^2+4y^3=3(x^2+y^3)+y^3>=3(x^3+y^4)+y^3,
so:2>=x^3+y^3.
OK
Very nice,isn't it?[/quote]

What a problem!! What a solution!! Are those type of inequalities solving majorly consisting of experimenting?

[quote=huhuhu]$$x^3+x^3+1 \geq 3x^2$$,
$$y^4+y^4+y^4+1\geq 4y^3$$
so:$$2x^3+3y^4+2 \geq 3x^2+4y^3=3(x^2+y^3)+y^3\geq 3(x^3+y^4)+y^3$$,
so:$$2\geq x^3+y^3$$.
OK
Very nice,isn't it?[/quote]

Very nice. I don't know how I missed it.  I was very close . I proved it exactly as in #2 post.

[quote=Michael Niland]$x$ and $y$ are positive real numbers such that $x^2 +y^3\geq x^3 +y^4.$ Prove $$x^3 + y^3\leq  2$$[/quote]
My solution:
$$x^{3}+y^{3}\leq  x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3}
\implies x^{3}+y^{3}\leq 2.$$
[url=https://artofproblemsolving.com/community/c6h142627p5810520]Hungary-Israel Binational  2002[/url]
[url=https://artofproblemsolving.com/community/c6h2540883p21655496]Moldova 2000:[/url]
Let $x<1<y$ and $2\log x+\log(1-x)\ge3\log y+\log(y-1).$ Prove that$$x^3+y^3<2$$


algebra
function
domain
analytic geometry
graphing lines
slope
calculus

2 questions that came up in my pre-calc class today that got me really curious:

[b]1.[/b] Given the theorem that $ \sqrt{a}*\sqrt{b}\equal{}\sqrt{ab}$ and considering that $ \sqrt{2} * \sqrt{2} \equal{} \sqrt{2*2} \equal{} \sqrt{4} \equal{} 2$

it can be assumed that $ \sqrt{\minus{}2} * \sqrt{\minus{}2} \equal{} \sqrt{\minus{}2*\minus{}2} \equal{} \sqrt{4} \equal{} 2$

but $ \sqrt{\minus{}2} * \sqrt{\minus{}2} \equal{} i\sqrt{2} * i\sqrt{2} \equal{} \minus{}1*2 \equal{} \minus{}2$.

Which one is correct, and why is it that the other is incorrect?

[b]2.[/b] Consider the graph $ y\equal{}x^3\plus{}2x^2\minus{}8x\plus{}2$ with a domain of the interval $ (\minus{}4,\infty)$. Is there a local minimum at $ (\minus{}4,2)$? Does the point on the graph immediately after $ (\minus{}4,2)$ count as a local minimum?

[hide]1.) I believe that that theorem only works if both a and b are positive, so -2 would be the correct product.

2.)The line of the graph has a positive slope when x=-4, so it seems to me that the next point would have a greater value for y, which would make (-4, 2) the local minimum.

I haven't done work with local minimums, so it may be that the graph actually has to turn for it to count.
[/hide]

There's no point on the graph immediately after $ (\minus{}4, 2)$. And you can't have a local minimum at a point not in the domain of the function.

I'm not sure whether by "after" Captain_p meant the point with an infintesimally greater value for x, or an infintesimally lesser one. One of them isn't in the domain, and the other has a greater value for y.

For [b]#1[/b]:

You should get $ \pm 2$, and therefore, both ways work. Remember that $ \sqrt {x^2} \equal{} \pm x$ or $ |x|$.

[b]#2[/b]:

$ \frac {dy}{dx} \equal{} 3x^2 \plus{} 4x \minus{} 8$. So the local minimum/maximum is when $ \frac {dy}{dx} \equal{} 0$. So, $ 3x^2 \plus{} 4x \minus{} 8 \equal{} 0$, so $ x \equal{} \frac { \minus{} 2 \pm 2 \sqrt {7}}{3}$ but neither of these values are in the domain, so there is [b]no local minimum[/b].

*note: I guess derivatives aren't in Pre-cal so forget about my solution, just the answer.

[quote="modularmarc101"]Remember that $ \sqrt {x^2} \equal{} \pm x$ or $ |x|$.[/quote]

Actually, standard usage details that $ \sqrt{x^2}$ can [b]only[/b] express the same thing as $ |x|$, never $ \pm x$. To express plus or minus square root, use $ \pm \sqrt x$.

The "function" $ \sqrt {\cdot}$ is a real function (no quotation marks needed) when its domain and range are both $ {\bf R}_{\geq 0}$.  In this case, the "theorem" stated in the O.P. is a theorem (no quotation marks needed).  On $ \bf R$ (as a "function" mapping into $ \bf C$), $ \sqrt {\cdot}$ is not actually a function: it assigns to every number except 0 exactly two values, and there is no useful concept like the ordering of the reals that lets us choose one of these values in a natural way.  In this context, the "theorem" stated in the O.P. can still be true, but you have to think harder about what it actually means -- in particular, it [i]should[/i] give precisely two values.  (If you try to choose $ \sqrt {\cdot}$ as one branch of the square root function, the "theorem" will just be false.)

Can we please kill off quickly the notion that the real numbers contain some object that can be formally refered to as an "infinitesimal"?  There are no gaps in the real numbers, and there's no way to move to "the next" real number without passing by a continuum of other reals.

By the way, \cdot is a much better multiplication symbol than * in LaTeX.  Also, can we get more informative titles on your posts in the future?  ("Precalc questions" contains almost 0 useful information about the content of the post and would not be easy to find again in the future.)


For $ a>0,\ b>0,\ c>0$, prove that $ \frac{a^3\plus{}b^3\plus{}c^3\plus{}6}{ab\plus{}bc\plus{}ca}\left(\frac{b^2}{a}\plus{}\frac{c^2}{b}\plus{}\frac{a^2}{c}\right)\geq 9$.

[quote="kunny"]For $ a > 0,\ b > 0,\ c > 0$, prove that $ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 6}{ab \plus{} bc \plus{} ca}\left(\frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c}\right)\geq 9$.[/quote]
Nice
1) By Angel $ \frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c}\ge a\plus{}b\plus{}c$
2)$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6\equal{}(a^3\plus{}1\plus{}1)\plus{}(b^3\plus{}1\plus{}1)\plus{}(c^3\plus{}1\plus{}1)\ge 3(a\plus{}b\plus{}c)$ by AM-GM
3) $ \frac{3(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ac}\ge 9$

My Solution:

$ a^3\geq 3a \minus{} 2\Longleftrightarrow (a \minus{} 1)^2(a \plus{} 2)\geq 0$ e.t.c. we have $ \frac {b^2}{3a} \plus{} \frac {c^2}{3b} \plus{} \frac {a^2}{3c}\geq \frac {b^2}{a^3 \plus{} 2} \plus{} \frac {c^2}{b^3 \plus{} 2} \plus{} \frac {a^2}{c^3 \plus{} 2}\geq \frac {(a \plus{} b \plus{} c)^2}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6}\geq \frac {3(ab \plus{} bc \plus{} ca)}{a^3 \plus{} b^3 \plus{} c^3 \plus{} 6}$.

$ \therefore (a^3 \plus{} b^3 \plus{} c^3 \plus{} 6)\left(\frac {b^2}{a} \plus{} \frac {c^2}{b} \plus{} \frac {a^2}{c}\right)\geq 9(ab \plus{} bc \plus{} ca)$.


integration
real analysis
function
real analysis unsolved

For each n, define $ h_n \equal{} \sum_{j \equal{} 1}^{n}( \minus{} 1)^{j}X_{[(j \minus{} 1)/n,j/n]})$. Show that $ \int_{0}^{1}fh_ndm\rightarrow 0$  as n goes to $ \infty$ for every  lebesgue integrable function f on [0, 1] . Here $ X_{E}$ is characteristic function of the set E

You probably wanted the symbol $ \chi$ (\chi) there.

Let's rewrite that $ h_n$. Define $ h(x)\equal{}\begin{cases}\minus{}1&0<x<1\\1&1<x<2\\0&x\in\mathbb{Z}\end{cases}$ on $ [0,2]$, and then extend it to all of $ \mathbb{R}$ by making it periodic with period $ 2$. Then $ h_n(x)\equal{}h(nx).$
The one relevant fact about $ h$: it has average value zero. Then this is the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]Riemann-Lebesgue lemma[/url].

Hello jmerry,

Thanks. But I do not get the point why did you consider the interval [0, 2], because the function defined on [0, 1]. Also How do I show the limit of $ h_{n}$ is zero

Actually, I defined $ h$ on $ \mathbb{R}$. It has to be, too, because I'm taking its values arbitrarily far out.

Also, $ h_n$ does not tend to zero.

Hi Jmerry,
 I Just want to know can we apply dominated convergence theorem here

We do not have pointwise convergence, convergence in $ L^1$, convergence in measure, or anything except convergence of the overall integral.

So, no.


inequalities
limit
logarithms
real analysis
real analysis unsolved

Prove that

Lim   ( Sigma(ai^s,1,n) / n ) ^ (1/s) = (a1*a2*...*an) ^ (1/n)
s->0

where a is a positive series.

It's from the extent form of Average Inequality, which is

Let s>r

(Sigma(ai^s,1,n) / n) ^ (1/s) >= (Sigma(ai^r,1,n) / n) ^ (1/r)

When s=0, the left formula equals (a1*a2*...*an) ^ (1/n)

do you mean prove $\lim_ {s \to 0} (\frac{a^s_1+\cdots+a_n^s}{n})^{1/s}=\sqrt[n]{a_1a_2 \cdots a_n}$?
and here is a proof:
we should only to  calculate $\lim_ {s \to 0} \frac{\ln \frac {a^s_1+\cdots+a_n^s}{n}}{s}$
it is not hard

Err..
I'm a newer so would you mind giving me some details?

Is it deducted from Robida theorem? (err.. I forget his English name.. that is
lim   f(x)/g(x) = lim  f'(x) / g'(x)  )
x->0                x->0

[quote="Agnimon"]Err..
I'm a newer so would you mind giving me some details?

Is it deducted from [b]Robida[/b] theorem? (err.. I forget his English name.. that is
lim   f(x)/g(x) = lim  f'(x) / g'(x)  )
x->0                x->0[/quote]

 You should have said L'Hospital  :roll:


vector
calculus
integration
calculus computations

I am having a problem and it may be that I am just wiped out from being sick....

This problem asks me to get the Arc Lenght of a vector curve
r(t)=<sqrt(2)*t, e^t, e^(-t)> for -10<t<10
Since Arc Lenghth = the integral of the magnitude of r'(t)
I got:
r'(t)=<sqrt(2), e^t, -e^(-t)>
magnitude of r'(t)=sqrt[sqrt(2)^2+e^(2t)+e^(-2t)]=sqrt[2+(e^t+e^(-t))^2]
At this point both my book and calculator agree that:
sqrt[2+(e^t+e^(-t))^2]=e^t+e^(-t)  ??????????

What happened to the 2?
I understand how:
sqrt[(e^t+e^(-t))^2]=e^t+e^(-t)
That's basic Algebra. Is this a property of e that i missed?

One of the problems with trying to write mathematics in straight text is that it can be very hard to see parentheses and even harder to see lack of parentheses. That's one of the reasons the regulars on this site have all learned $\text{\TeX}$ - it just makes things easier to read.

$r'(t)=\langle\sqrt{2},e^t,e^{-t}\rangle.$ Now take the magnitude:

$\dfrac{ds}{dt}=|r'(t)|=\sqrt{2+(e^t)^2+(-e^{-t})^2}$

$=\sqrt{e^{2t}+2+e^{-2t}}=\sqrt{\left(e^t+e^{-t}\right)^2}=e^t+e^{-t}.$

The key is the expansion of the perfect square:

$\left(e^t+e^{-t}\right)^2=(e^t)^2+2e^te^{-t}+(e^{-t})^2=e^{2t}+2+e^{-2t}.$

Sweet thanks, I guess in my sickness I just missed that....

btw, how do i put TEX into a message?

Put the code inside dollar signs- the forum software does the rest.


geometry
rectangle
incenter
ratio
geometry proposed

Let $ ABC$ be an acute-angled triangle such that there exist points $ D,E,F$ on side $ BC,CA,AB$, respectively such that the inradii of triangle $ AEF,BDF,CDE$ are all equal to $ r_0$. If the inradii of triangle $ DEF$ and $ ABC$ are $ r$ and $ R$, respectively, prove that \[ r\plus{}r_0\equal{}R.\]
[i]Soewono, Bandung[/i]

[quote="Raja Oktovin"]Let $ ABC$ be an acute-angled triangle such that there exist points $ D,E,F$ on side $ BC,CA,AB$, respectively such that the inradii of triangle $ AEF,BDF,CDE$ are all equal to $ r_0$. If the inradii of triangle $ DEF$ and $ ABC$ are $ r$ and $ R$, respectively, prove that
\[ r \plus{} r_0 \equal{} R.\]
[i]Soewono, Bandung[/i][/quote]
Indonesia IMO 2010 TST  :?:  :?:
it is too early to choose team IMO 2010  :lol: .

cool enough uh?  :(

fyi,
regional selection test was in april 2009,
national olympiad was in august 2009,
about 30 contestants go to 1st stage
the 1st stage of tst is in october-november 2009,
about 12 contestants go to 2nd stage
the 2nd stage of tst will be in about march 2010 (together with APMO 2010),
the 2nd stage is to choose 6 IMO 2010 contestants
the 3rd stage is the last preparation to IMO.

:)

Let $ O_1,O_2,O_3$ denote the centers of the incircles of $ \triangle AFE,$ $ \triangle BDF,$ $ \triangle CED.$ $ (O_1),(O_2),(O_3)$ touch $ EF,FD,DE,$ at $M,N,L.$ $ (O_2),(O_3)$ touch $ BC,$  at $P,Q.$ $ (O_3),(O_1)$ touch  $ CA$ at $R,S$ and $ (O_1),(O_2)$ touch  $ AB$ at $T,U.$ Since $ O_1O_2UT,$ $ O_2O_3QP$ and $O_3O_1SR$ are rectangles and $ O_1,O_2,O_3$ lie on the internal bisectors of $ A,B,C,$  it follows that $ \triangle ABC$ and $ \triangle O_1O_2O_3$ are centrally similar through their common incenter $ I.$ The similarity coefficient equals the ratio between their perimeters/inradii

$ \frac {r \minus{} r_0}{r} \equal{} \frac {O_1O_2 \plus{} O_2O_3 \plus{} O_3O_1}{a \plus{} b \plus{} c} \equal{} \frac {PQ \plus{} RS \plus{} TU}{a \plus{} b \plus{} c}$

$\frac {r \minus{} r_0}{r} \equal{} \frac {DN \plus{} DL \plus{} EL \plus{} EM \plus{} FM \plus{} FN}{a \plus{} b \plus{} c} \equal{} \frac {DE \plus{} EF \plus{} FD}{a \plus{} b \plus{} c} \ (1)$

On the other hand, we have:

$r(a \plus{} b \plus{} c) \equal{} r_0(a \plus{} b \plus{} c \plus{} DE \plus{} EF \plus{} FD) \plus{} \varrho(DE \plus{} EF \plus{} FD)$

$\Longrightarrow \ r \minus{} r_0 \equal{} \frac {(DE \plus{} EF \plus{} FD)( r_0 \plus{} \varrho)}{a \plus{} b \plus{} c} \ (2)$

Substituting $ (r \minus{} r_0)$ from $ (1)$ into $ (2)$ yields $ r \equal{} r_0 \plus{} \varrho.$


function
trigonometry
geometry solved
geometry

Let A be a fixed point on the circle has diameter BC . M,N move on this
circle such that:

tg(BC,AM)*tg(BC,AN)=constant .

Prove MN passing through a fixed point

Note that : use inversion

It's mean tang : is a fucntion trigonometry

Oh, I was confused because I didn't know you were referring to the angle between the two lines.  :? Thought you were taking the tangent of two lines. !!!

[quote="A1lqdSchool"]Note that : use inversion[/quote]

Thanks for the kind hint, but before I succeed to explain the inversive solution (which uses an involution as well, so it is not that elementary), I probably have explained the elementary one three times ;) . But first I slightly rewrite your problem (you know, rewriting others' problems is my hobby):

[color=blue][b]Problem.[/b] Let A be a fixed point on a fixed circle k, and let g be a fixed line. Also, let t be a fixed real number. Two points M and N move on the circle k and satisfy the condition $\tan\measuredangle\left(g;\;AM\right)\cdot\tan\measuredangle\left(g;\;AN\right)=t$, where the angles are directed angles modulo 180°. Prove that the line MN passes through a fixed point.[/color]

[i]Solution.[/i] Let the parallel to the line g through the point A meet the circle k at a point P (apart from A), and let the perpendicular to the line g through the point A meet the circle k at a point Q (apart from A). Since AP || g and $AQ\perp g$, we have $AP\perp AQ$, so that < PAQ = 90°. Since the points P, Q, A and M lie on one circle (namely, on the circle k), we have < PMQ = < PAQ; thus, < PMQ = 90°.

Since AP || g, we have < (AP; AM) = < (g; AM) and < (AP; AN) = < (g; AN). In other words, < PAM = < (g; AM) and < PAN = < (g; AN). Thus, the equation $\tan\measuredangle\left(g;\;AM\right)\cdot\tan\measuredangle\left(g;\;AN\right)=t$ becomes $\tan\measuredangle PAM\cdot\tan\measuredangle PAN=t$.

Let the line PQ meet the line MN at a point R. We will now use non-directed angles, since we will work with sines.

Applying the sine law to triangles PRM and QRM, we get $\frac{PR}{RM}=\frac{\sin\measuredangle PMR}{\sin\measuredangle RPM}$ and $\frac{RM}{RQ}=\frac{\sin\measuredangle RQM}{\sin\measuredangle QMR}$. Thus,

$\frac{PR}{RQ}=\frac{PR}{RM}\cdot\frac{RM}{RQ}=\frac{\sin\measuredangle PMR}{\sin\measuredangle RPM}\cdot\frac{\sin\measuredangle RQM}{\sin\measuredangle QMR}$.

Now, since the points A, P, Q, M, N all lie on the circle k, the chordal angle theorem yields < PMN = < PAN and < PQM = < PAM. First, this implies < PMR = < PMN = < PAN. Further, since < PMQ = 90°, the triangle PQM is right-angled at M, so that < QPM = 90° - < PQM; thus, < RPM = < QPM = 90° - < PQM = 90° - < PAM. Also, clearly < RQM = < PQM = < PAM. Finally, < QMR = < PMQ - < PMN = 90° - < PAN. Thus, the above equation becomes

$\frac{PR}{RQ}=\frac{\sin\measuredangle PAN}{\sin\left(90^{\circ}-\measuredangle PAM\right)}\cdot\frac{\sin\measuredangle PAM}{\sin\left(90^{\circ}-\measuredangle PAN\right)}=\frac{\sin\measuredangle PAN}{\cos\measuredangle PAM}\cdot\frac{\sin\measuredangle PAM}{\cos\measuredangle PAN}$
$=\frac{\sin\measuredangle PAM}{\cos\measuredangle PAM}\cdot\frac{\sin\measuredangle PAN}{\cos\measuredangle PAN}=\tan\measuredangle PAM\cdot\tan\measuredangle PAN=t$.

Thus, the point R divides the segment PQ in a fixed ratio. Since the segment PQ itself is fixed, this yields that the point R is a fixed point, and thus the line MN passes through a fixed point, namely the point R.

Problem solved.

  Darij


A gardener wishes to enclose a rectangular area forming 2 sperate gardens. The gargage will be along one side so no fence is needed there. If she has 900 meters of fencing material, please determine the following.

[img]http://home.comcast.net/~krunner03/jhjhjhjhjhjh000000000000000000000000.JPG[/img]

1.) Write a function for the area of this region in terms of x.

I worte 900m = 3x + y and solved for y. Since xy = area, I wrote the function as x(300 - 3x). (300-3x) is what I got for y.

Part two of this problem is giving me trouble. It says :

2.) What dimensions for the total enclosed area would give an area of 64,800 square meters?

y hasn't been labled on the diagram but I assume it to be the bottom length.

Solving:
$900 = 3x+y$
for y would give you:
$y=900-3x$
So the equation for the area is:
$x(900-3x)$
So you want to know when this is equal to 64,800.
$x(900-3x)=64800$
$3x^{2}-900x+64800=0$
$x^{2}-300x+21600=0$
$(x-120)(x-180)=0$
And now it should be obvious.


We have a square grid of 4 times 4 points. How many triangles are there with vertices on the points? (The three vertices may not lie on a line.)

A. 252
B. 256
C. 360
D. 516
E. 560

[hide]There are $ \binom{4^2}{3} \equal{} 560$ ways to select $ 3$ points. We need to subtract the ones that are collinear, and there are $ 2\cdot 4\cdot \binom{4}{3} \plus{} 2\cdot\left(2 \plus{} \binom{4}{3}\right) \equal{} 44$ of them.

The answer is $ 560 \minus{} 44 \equal{} 516 \Rightarrow \boxed{D}$. [/hide]

[size=59]Solution hid by Silverfalcon[/size]


function
real analysis
real analysis unsolved

Assume f is continuous at at least one point x'.  Also let f(x+y)=f(x)+f(y).

Show there exists a constant m such that f(x)=mx for all x.

I'm not sure why I need continuity at one point??

You need it to extend your expression from $\mathbb{Q}$ to $\mathbb{R}$ by continuity.

There are nowhere continuous functions $f\colon\mathbb R\to\mathbb R$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb R$. Unfortunately, it seems impossible to present an explicit example of such a function - one must use a [url=http://mathworld.wolfram.com/HamelBasis.html]Hamel basis for $\mathbb R$ over $\mathbb Q$[/url].


[quote="rrusczyk"]Mathew Crawford has left the AoPS full-time staff.  He will still be working on some projects with AoPS, including teaching WOOT classes and writing more books.
We wish him success in his future projects and thank him for all his contributions to the site.[/quote]
I wish him all the best too. He was an excellent administrator.

I was wondering why his name wasn't in [color=orange]orange[/color] anymore...

Just thanks for everything, and good luck from now on

[quote="solafidefarms"]Eh, I wondered why you depriveledged him on the wiki some time ago. Now I know...[/quote]

me?? I don't remember using the wiki

I wondered why [i]Mr. Rusczyk[/i] depriveledged Mr. Crawford on the wiki recently.

Why is he doing this? It's not a position I would be leaving.

Oh, that's sad. Good luck, Mr. Crawford, and thank you. You've been a great teacher.

[quote="SplashD"][quote="nebula42"]I'll still be seeing him around in WOOT classes  :) , but I just want to say this: good luck for the future, and thank you for all of your hard work here.[/quote]

Wait, will he still be teaching WOOT :?:[/quote]
[quote="rrusczyk"]Mathew Crawford has left the AoPS full-time staff.  He will still be working on some projects with AoPS, including [b]teaching WOOT classes [/b]and writing more books.

We wish him success in his future projects and thank him for all his contributions to the site.[/quote]

...

[quote="kyyuanmathcount"]Mr. Crawford was a great teacher, I learned a lot from him and I'm really sad he's leaving... :([/quote]

Well as long as he's teaching WOOT, I'll still be able to take his classes...  Good luck Mr. Crawford, and I'll see you next year in WOOT!  :wink:

I don't think so, because he is not listed in the Administrators usergroup or the Moderators usergroup.  Of course, he could be hidden...

Thanks everyone.

Just to be clear, I'm not leaving the education community.  In addition to writing books, I'm trying to put together a local program to strengthen math education in the San Diego community.

I wish you luck as you pursue a larger project.

aww that's so sad. You should come back.

Eh I'm sure he has other things he wants to focus his time on. But I'd say nothing to fear, it seems like he's still going to stop by AoPS here and there  :)


1. good textbooks
2. good teachers
3. a group of 'like minded' peers
4. persistence for understanding material
5. a methodical regime of practice
6. other?

How would you rate the importance of each of these factors in learning mathematics?

Shyeah! Lockhart's Lament! Definitely read that!

On another note, math education has become more than your list, and emcompasses a wide variety of topics.  You might want to put down other environmental factors, like parental units. A motivation to find these 'good textbooks' or 'good teachers'. Just some thoughts.

I think that student motivation is the most important thing. If you don't have motivation, with all the money in the world, the best materials and best schooling, you'll still FAIL.

Then again, I'm biased. But who isn't? :lol:

Good luck defining this.

I think a key factor to good education is to keep students of one level together. This doesn't drag back higher level students and doesn't go too fast for lower level students. Also, problem solving is not taught in school. It should be.

I'd rate them as

1) Good textbooks
2) Persistence
3) Method
4) Like-minded peers (though this can increase pressure)
5) Good teachers.

there is actually no need of a teacher for a motivated student. a good textbook will do.

[quote="Nebraska boy"]I think that one of the common mistakes is this process of learning math is that the profesors teach math like any other subject , and do not promote thinking, students are becoming more lazy , and im talking about  highschool level because im a highschool student and this is what i see.[/quote]
I think this has to do with that fact that most subjects are more complicated than math but do not require as complicated a level of thinking as math. Teachers are used to teaching "stuff" and that "stuff" is usually stuff to memorize, not to think about. If the stuff is complicated, then there is a lot to memorize. If it is not complicated, then there isn't a lot to memorize.

Nebraska boy, I think the problem is rather that none of the high school teachers are anywhere near being professors. It is hard to teach a subject without extensive knowledge.

[quote="Attila the Hun"]there is actually no need of a teacher for a motivated student. a good textbook will do.[/quote]

A textbook is a static collection of material.  You learn what's written in the book in the order that the book is written.  If you get stuck, you have to veer away from the textbook.

A teacher on the other hand is dynamic.  She can see when you're struggling and find another way to present the information.  She can get at the root of your misunderstanding more quickly than you're able to by going through random books or searching online.

A motivated student deserves to have a good teacher to guide him through the course and to bring more life into it than a book can ever do.  That said, textbooks are great.  They preserve the outline of topics and mass of problems.  A teacher can't retain a database like memory of problems and topics.  The combination of a good textbook and a good teacher is ideal, yet rarely seen in K12 education.

[quote]Case in point: your math teacher is Mr./Mrs./Ms. and not Prof./Doc.[/quote]

You don't need an extensive graduate level degree to be able to teach mathematics to K12 kids effectively.  The marginal utility at that level does nothing to replace genuine interest in teaching and a true love for mathematics that you need to inspire others.

I think that a good teacher is more valuable than any textbook can be.  As joml88 said, a textbook is just a collection of the material.  Going through a textbook can often be confusing or headache-inducing (at least for me :wink: ) whereas an exciting teacher can present the material in an interesting manner.  They can also create motivation in students where none otherwise existed.

[quote="Attila the Hun"]4) Like-minded peers (though this can increase pressure) [/quote]

What's wrong with increased pressure?  I would think students would have more motivation to work under pressure.

[quote="Attila the Hun"]4) Like-minded peers (though this can increase pressure) [/quote]

What's wrong with increased pressure?  I would think students would have more motivation to work under pressure.[/quote]

People work harder (and in my opinion better) under pressure. Also, if there is one person that is better than you in your school, you would get extremely motivated and want to do even better (at least in my case :P )

Sadly, it's like that in every topic. My social studies teacher said that Copernicus created the heliocentric world view...  :maybe:

Edit: I meant geocentric.

[quote="blahblahblah"][quote="joml88"]
You don't need an extensive graduate level degree to be able to teach mathematics to K12 kids effectively.[/quote]

That being said, a huge proportion of K12 teachers have such poor mathematics skills that they really can't teach the subject effectively.[/quote]

Also, a lot of math teachers arent into what they teach, and as a result neither are the kids. Thats probably a huge reason that math is the least favorite subject for many kids.

[quote="Ihatepie"]My social studies teacher said that Copernicus created the geocentric world view.[/quote]
I took an 8th grade Social Studies test. I expected to have some of US and World History on it, but noooooooo.

All you do is look at tables, determine stuff about them, come to logical (in my opinion trivial) conclusions, and use common sense for most of the problems. Also know what income tax is (they asked me that).
But you might want a few history facts memorized, like what the Emancipation Proclamation was for, and the expedition of Lewis and Clark.

I've seen some history books saying Christopher Columbus (English Explorer who discovered Central America) was the first person to sail around the world (Ferdinand Magellan, Portuguese Explorer).

Do you think that mathematics isn't the only subject that isn't being taught well?

What are the formal requirements placed on high school teachers in the US? Here HS teachers need about the equivalent of a Bachelor's degree for each subject they teach in HS. Earlier they needed a Masters, but it was changed - as more and more people go on to study in high school and the status and salary of being a HS teacher drops, why would one get a Master's degree? Still, teachers need at least 2 subjects so they spend a lot of time studying at university, probably one of the worst educations you can get at university career-wise.

I've taken classes with many aiming to be teachers, and while I dare to generalize and say that they're not the top tier, many are certainly quite able. After all, they have to pass. So maybe a large part of it is the lack of stimulus that makes them not so knowledgeable as time passes? After all, it is easy to forget things as the years go on, and skills probably deteriorate if you don't actively use them. So maybe the overall culture can be blamed here too - exactly how do you participate in exciting math-related things outside of academia?

In the United States, everything about education, including teacher credentialing standards, is regulated by the states, and states may differ from each other. Because of that, I can only speak about the California rules.

In California, any teacher who teaches only a single subject - math, science, English, history, etc. - is supposed to have a "single subject credential". This applies to both high school teachers and middle school teachers. (Elementary school teachers would typically have a "multiple subject credential.") The specified requirements for a single subject credential are very close to a bachelors degree in that subject plus an additional year of coursework and training in education.

That's the theory. In practice, the number of math and science teachers who got their credentials in that manner is far smaller than the number of active math and science teachers. In some cases, teachers have assignments that they're not credentialed to cover. But in other cases, the state has provided alternative paths to the credential. Probably the most important such path is an examination (perhaps in several parts) called the CSET. In our program, of the people entering the education courses for the math credential, more are planning to take the CSET than have the mathematics course program (the program that's nearly the same as the major), and many of those have very, very little in the way of collegiate mathematics background. And many of them will pass the exam, which largely tests mathematics as a secondary school level - you don't need to know much beyond that level (maybe a little calculus).

Of course, given the imminent meltdown of government finances at all levels in California, we're going to see a lot of teachers losing their jobs in the next year or two. I doubt that the order in which they lose their jobs will bear much relationship, if any, to the level of their qualifications.


What's PIE?
I see lots of people using it to solve problems, is it a trick like SFFT or a method like balls and urns?
Can someone please explain? :huh:
No, I don't mean $\pi$.

I know it stands for Principal Theory of Inclusion and Exclusion...but I'm not sure how you actually use it

Ok, thanks, I'll do a search on the AoPS Wiki and edit if I find anything.
 :)

EDIT: I found it, yay! :D
[url=http://www.artofproblemsolving.com/Wiki/index.php/Inclusion_Exclusion_Principle]PIE in AoPS wiki[/url]

[url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]AoPS Wiki page[/url]

EDIT: shucks. ya beat me to it.

[quote="Aryth"]a type of pastry that consists of an outer crust and a filling; extended to other, non-pastry dishes that maintain the general concept of a shell with a filling[/quote]

not funny.

Basically you would use it for something like how many numbers less than 100 are multiples of 2 or 3?

No it's not funny.  :spam:

LeetSpeak can you elaborate on that please?

I believe what leet meant was that you would use PIE on a problem that asked "How many number less than 100 are multiples of 2 or 3?" Because, to solve it, you would find the number of integers less than 100 that are multiples of 2, then find the number of integers less than 100 which are multiples of 3, add those two numbers, and then subtract the number of integers less then 100 that are multiples of both 2 and 3.

[quote="13375P34K43V312"][quote="Aryth"]a type of pastry that consists of an outer crust and a filling; extended to other, non-pastry dishes that maintain the general concept of a shell with a filling[/quote]

not funny.
[/quote]
Actually I'm going to have to disagree and say that that was pretty funny.

[quote="CircleSquared"]It was SPAM.   :spam:

Last time I checked, SPAM is against the rules.

Again, I'm not talking about food.

Thank you again to everyone who gave a SERIOUS post.[/quote]

wow look at that anger..want some pie to calm you down?

No thanks, I don't need the pie. :|

Besides, isn't being angry at spammers okay :?:
...This is going off topic.  I have a feeling a mod is going to lock this soon.


inequalities
geometry
inequalities proposed

[quote="Virgil Nicula"]Let $ABC$ be a equilateral triangle $(AB=a)$ and a inner point $P$.
Denote the area $\sigma\equiv \sigma \left( PA,PB,PC\right)$ w.r.t. the point $P.$ Prove that :

$\blacktriangleright\ 1^{\circ}.\ 4\sigma \sqrt 3\le a^2\le PA^2+PB^2+PC^2\ .$

$\blacktriangleright\ 2^{\circ}.\ PA^4+PB^4+PC^4\ge \frac 13\cdot a^4\ .$[/quote]

I added the first inequality (and in the first message). The triangle for which the lengths of the sides are equal with $PA$, $PB$, $PC$ is named the [u]Pompeiu's triangle[/u].

[quote="Virgil Nicula"]Let $ABC$ be a equilateral triangle $(AB=a)$ and a inner point $P$.
Denote the area $\sigma\equiv \sigma \left( PA,PB,PC\right)$ w.r.t. the point $P.$ Prove that :

$\blacktriangleright\ 1^{\circ}.\ 4\sigma \sqrt 3\le a^2\le PA^2+PB^2+PC^2\ .$

$\blacktriangleright\ 2^{\circ}.\ PA^4+PB^4+PC^4\ge \frac 13\cdot a^4\ .$[/quote]
These problems will be not hard if we use  the barycentrical coordinates.
Let $P(x,y,z)$,$x,y, z \ge 0,x+y+z=1$
then you will find:
$PA^2=(y^2+yz+z^2)a^2$
$PB^2=(z^2+zx+x^2)a^2$
$PC^2=(x^2+xy+y^2)a^2$
then
$4\sigma \sqrt 3\le a^2$
$\iff 9\sum_{cyc}x^2y^2+18\sum_{cyc}xyz^2 \le (x+y+z)^4$
obvious by expand.
$a^2\le PA^2+PB^2+PC^2$
$\iff (x+y+z)^2 \le y^2+yz+z^2+z^2+zx+x^2+x^2+xy+y^2$
obvious .
$PA^4+PB^4+PC^4\ge \frac 13\cdot a^4$
$\iff 6\sum_{cyc}x^4+9\sum_{cyc}x^2y^2+6\sum_{sym}x^3y \ge (x+y+z)^4$
obvious by expand.

[quote="Virgil Nicula"]Zhaobin, see my last reply from the topic
http://www.mathlinks.ro/Forum/viewtopic.php?t=67254[/quote]
Thanks.And I think the indentity also can be solved by the barycentrical coordinates.
 :)


What product(s) might be expected when benzene reacts with 2,2-dimethyloxirane and some HCl as catalyst?

First the $ \ce{H\plus{}}$ attacks the epoxide and cause ring opening to form $ \ce{OH\minus{}CH_2\minus{}C(CH3)_2^\plus{}}$ and then electrophilic substitution in the benzene ring.
The product:
$ \ce{C_6H_5\minus{}C(CH3)_2\minus{}CH_2\minus{}OH}$

another product is a ketone due to migration of H- (pinacol rearrangement)

Migration of $ \ce{H^\minus{}}$?? :huh:  :huh:
Can u explain it please varun?

there can also be this possibility...........

the $ \ce{H\plus{}}$ attacksthe side containg the 2 methyl groups .....resulting in the formation .
$ \ce{\plus{}CH2\minus{}C(CH3)2\minus{}OH}$ ........
after that a methyl shift occurs..............

But tertiary carbocation is more stable due to Hyperconjugative resonance :huh:

[quote="VARUNARASU"]since OH-CH2-C(CH3)2+ is formed, the hydrgoen attached to the carbon bearing the oh group migrates with its lone pair as H- to the neighbouring electron deficient carbon and you get a ketone. the migration takes place bcoz, the carbocation alpha to OH group is more stable, due to electron releasing ability of oxygen(+R effect).
 the final product is O=CH-CH(CH3)2[/quote]
Yes that is surely correct man :clap: But ur product is wrong
Great thinking that is. But carcul surely resonance is more than hyperconjugation so varun's product is stable.

the migratory aptitude is H>aryl>alkyl.
whats the reason, carcul? the question says all possible products doesnt it

The protonated epoxide ring (or the tertiary carbocation) is already stabilized.

that doesnt mean rearrangement cannot take place, if that was so, pinacol rearrangement would never take place


In my thesis I must have figure and equation number dependant of section. I tried something like this:
[code]\usepackage{chngcntr}
\counterwithin{figure}{section}
\counterwithin{equation}{section}
\renewcommand{\theequation}{\thesection.\arabic{equation}}[/code]

When I refer to some equation it prints out "2.1" (for first equation in second section), as wanted. I tried to make same effect for figures in the same way:[code]\renewcommand{\thefigure}{\thesection.\arabic{figure}}[/code]
but when I refer to some figure it still prints out only "1" (for first figure in second section). Can you help me figure reference have the same style as equation reference? I tried with \numberwithin also, but it gives the same result.

How can I make figure caption like "Figure xxxx. caption" instead of "Figure xxxx: caption" using caption package?

EDIT: Interesting, that try with [code]\renewcommand{\thefigure}{\thesection.\arabic{figure}}[/code]
succeded :). Figure environment must be typed like this:[code]\usepackage{float}
\usepackage[font=normalsize,format=plain,labelfont=bf,up,textfont=it,up]{caption}

\begin{figure}[H]
	\centering
	\includegraphics[scale=0.55]{bitstring.png}
	\caption{Caption.}
	\label{fig:bitstring}
\end{figure}
[/code]

The second issue remains.

[quote="kokan"]How can I make figure caption like "Figure xxxx. caption" instead of "Figure xxxx: caption" using caption package?[/quote]See [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/caption/caption-eng.pdf]caption package documentation[/url] page 26 \DeclareCaptionLabelSeparator


Can we possibly use a simple language to explain the difference between logitudinal and transverse waves?

Is there any difference in the mathematical equation between the two?
f they are two different kinds of phenomenon, I wonder why they have the same partial differential equation?

A transverse wave osscilates perpendicular to the direction of travel, whereas a longitudal wave osscillates along the direction of travel.

They are described by the same PDE, because PDEs usually have very general solutions, if no initial or boundary conditions are given.

In fact, this PDE: $a^{2}\frac{\partial^{2}u}{\partial x^{2}}= \frac{\partial^{2}u}{\partial t^{2}}$,
can be factored as: $(a \frac{\partial}{\partial x}+\frac{\partial}{\partial t})(a \frac{\partial}{\partial x}-\frac{\partial}{\partial t})u = 0$,
where $a$ is the velocity.

The only difference between the two operators on $u$ is a matter of orientation along the x-axis.  Furthermore, since one of them gives $0$ as a result when operating on $u$, thus, the function $u$ is invariant if $\frac{x}{a}-t$ is constant.

Thus, if $x = at$, the wave at position $x$ and time $t$ remains constant.

In other words, a "wave" in general is just a traveling disturbance through some medium, which retains its "shape" as it travels.

Yes, so should we fairly say that logitudinal and transverse waves are actually the same wave motion but viewed at different angles?
For example when we see water waves from the above, it looks moving up and down which is a logitudinal wave in the up and down direction. When we take a look form it side, it has trough and crest and the pattern moving to one direction which is a transverse wave?  :?:

No no, when we look at a water wave from the side or above, we see the wave moving horizontally and the crests rising vertically, perpendicular to the direction of travel.

Longitudal waves involve the medium making small oscillations along the direction of travel.  However, you can think of the resulting pressure as having peaks and troughs due to this, which appears similar graphically to transverse waves, should u choose height to represent the magnitude of the pressure.

So, if we are simply given the solution of the wave equation

$\psi({\tilde x},t)=A\exp(i({\tilde k}\cdot{\tilde x}-\omega t))$

where ${\tilde x}\in{\mathbb R}^{3},t\in{\mathbb R}$ are the space and time parameters.Could we tell whether $\psi$ is a transverse or longitudinal wave from the given coefficients $A,{\tilde k}$ and $\omega$ only?

No. Consider the case of sound waves.  The air oscillates longitudally, and the pressure varies with the same frequency.

So the pressure would have "peaks" and "troughs" just as there are peaks and troughs on a water wave or an electromagnetic wave.

In fact, to further generalize the wave equation, any function $f(x-at)$ (or $f(x+at)$ if considering the other direction as distinct) satisfies the wave equation.

So in essence, $\psi (x, t) = \sqrt(kx-\omega t)$ is a wave, as is $\phi (x, t) = \ln(kx-\omega t)$.

The most general "definition" of a "wave motion" is any vector in the Hilbert space. For example, any componentf of a Fourier expansion could be taken as a "wave motion" so that the superposition principle is satisifed.
Not all solutions of PDEs are taken as wave motion, those PDEs with solution set not satsfies the superposition principle should not be taken as wave motion.
However, Every Physical phenomenon must has its mathematical formulation. As we can tell "Transverse wave has Crest and troughs", "Logitudinal waves have Compressions and rarefractions" ... so and so and "water waves are transverse", "sound waves are logitudinal" as physical examples, there must be a correct mathematical expression which tells us "what kind of motion described by $({\tilde x},t)\in{\mathbb R}^{4}$ are "transverses waves" and what other kinds of motions described by $({\tilde x},t)$ are "logitudinal waves". We should be able to tell the others transverse and logitudinal waves are two disjoint subsets of the set of all wave motions and there is no other kind of wave motions.
So, a mathematical formulation to distinguish the two must exists :(

According to wikipedia, transverse waves are described by a vector wave equation and are associated with the $%Error. "curl" is a bad command.
$ operator (ie. Maxwell's equations), whereas longitudal waves are described by a scalar wave equation and associated with the $\div$ operator (for example, in fluid mechanics).

So it actually has nothing to do with the constants.


inequalities
rearrangement inequality
inequalities proposed

$ a$ , $ b$ , $ c$ are the  lenghts of a triangle , prove that :

$ \frac{abc}{(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a)} \geq \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{8abc}$

[quote="y-a-s-s-i-n-e"]$ a$ , $ b$ , $ c$ are the  lenghts of a triangle , prove that :

$ \frac {abc}{(a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)} \geq \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159482

thank's arqady

is there a nice solution for the problem

This is just about the same thing as Jankri's.
Use the same substitution $ a: = x + y,b: = y + z,c: = z + x$,
but once we write the required inequality as

$ (x + y)^{2}(y + z)^{2}(z + x)^{2}%Error. "geqxyz" is a bad command.
(x + 2y + z)(x + y + 2z)(2x + y + z)$
we instead write it as
$ (x^{2} + 2xy + y^{2})(y^{2} + 2yz + z{2})(z^{2} + 2zx + x^{2}) = (x^{2} + 2xy + zx)(y^{2} + 2yz + xy)(z^{2} + 2zx + yz)$
which makes it much easier because we have a "substitution"
$ A: = x^{2} + 2xy$
$ B: = y^{2} + 2yz$
$ C: = z^{2} + 2zx$
and then it's just Rearrangement Inequality:
$ (A + y^{2})(B + z^{2})(C + x^{2}) \geq (A + zx)(B + xy)(C + yz)$.

thank's    the.sceth

but what if we put $ a\equal{}x\plus{}y\minus{}z$ , $ b\equal{}z\plus{}x\minus{}y$ , $ c\equal{}y\plus{}z\minus{}x$ ?

i think this condition is good


Given $ x\neq0$, find the positive value of $ b$ such that the equation $ \frac 3x\plus{}\frac x3\equal{}b$ will have exactly one solution.

AM-GM:

$ \frac 3x \plus{} \frac x3 \geq 2$

And it is obvious that we can replace 3/x with x/3 for the same result, unless they are equal.

Therefore we must have 3/x=x/3 or x^2=3^2, or x=3.


trigonometry
inequalities
inequalities unsolved

Determine the smallest value of the expression
$ M\equal{}\frac{ac\plus{}ab}{bc \cos A}\plus{}\frac{ba\plus{}bc}{ca \cos B}\plus{}\frac{cb\plus{}ca}{ab \cos C}$
where a, b, c are three sides of a triagle $ ABC$ and $ abc\equal{}1$

Use $ bc \cos A \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2}$ and Cauchy-Schwarz.

The condition for $ abc\equal{}1$ is not really necessary in this problem because we obtain the homogeneous inequality already. Therefore, all we need to do is to prove

$ (abc)^2 \ge (a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)(c^2\plus{}a^2\minus{}b^2)$

[quote="great math"]The condition for $ abc \equal{} 1$ is not really necessary in this problem because we obtain the homogeneous inequality already. Therefore, all we need to do is to prove

$ (abc)^2 \ge (a^2 \plus{} b^2 \minus{} c^2)(b^2 \plus{} c^2 \minus{} a^2)(c^2 \plus{} a^2 \minus{} b^2)$ [color=red][size=125](1)[/size][/color][/quote]

I don't understand  :( . Can you help me again ?
Why do you have  [color=red][size=125](1)[/size][/color]  :?:

ok, here is my answer
$ (a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2) \le \left(\frac{a^2\plus{}b^2\minus{}c^2\plus{}b^2\plus{}c^2\minus{}a^2}{2} \right)^2\equal{}b^4$

[quote="great math"]ok, here is my answer
$ (a^2 \plus{} b^2 \minus{} c^2)(b^2 \plus{} c^2 \minus{} a^2) \le \left(\frac {a^2 \plus{} b^2 \minus{} c^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}{2} \right)^2 \equal{} b^4$[/quote]

Oh no  :huh:  :huh:
I don't ask how to prove that, and I ask you that how to do have (1)


I found that typing $k$ instead of typing your greek $\lambda$ is a lot fast, so,  :D

WLOG $x_i=y_i$ where $i \in Z^{+}, i \le n$
I got this idea:
$\sqrt{\frac{1+k}{1+kx_i}} +\sqrt{\frac{1+k}{1+ky_i}} = \frac{\sqrt{1+kx_i}+\sqrt{1+ky_i}}{\sqrt{1+k^2+(x_i+y_i)k}} \le \frac{2}{\sqrt{1+k}}$
By squaring both sides, we get : ${\frac{k(x_i+y_i)+2 + 2\sqrt{k^2+(x_i+y_i)k+1}}{\sqrt{k^2+(x_i+y_i)k+1}} \le \frac{4}{1+k}}$
Through $LHS/RHS$ we can easily prove that's greater than one.
This can be fairly useful when $2|n$

I see that It's mathlinks constest. More general, but the solution's the same.

? I think It's Harazi problem?

The mathlinks contest's $RHS$ is different and the problem in ML is different.
But these problems look like


This has probably been discussed before, but after a search I didn't see anything recent. I'm a math major really interested in the math going on behind string theory. I've been told that physicists really need mathematicians to be doing developmental research alongside the string theorists. Does anyone know what area of math you should specialize in to go into this type of research? I'm guessing something like algebraic geometry or something like that due to the heavily geometric nature of the subject, but there seem to be topological or PDE sides to it as well. I know Brian Greene is known for his physics, but he is a very good mathematician as well, what is his area of math?

Take a look at this website:

http://superstringtheory.com/math/math2.html


How many subsets of $ \{1, 2, 3, 4, 5\}$ with two or more elements have the property that the sum of their elements is a positive, even number?

[hide]The number of subsets of {1,2,3,4} is $ 2^4\equal{}16$. 5 can be included or excluded depending on the current parity of the subset. {},{2},and {4} break the at least 2 elements rule.
16-3=13[/hide]


[quote="duy_nguyen"]Này bạn. Tui bj't bạn là người Việt Nam, Đúng ko.Đây là bài trên THTT số 359, lại đem lên đây nhờ giúp.là 1 HS chuyên toán bạn có thấy xấu hổ ko.hix....Po tay.... :oops:[/quote]
to duy_nguyen: you should type English in this forum, try to do it as soon as possible :)
to thao_39: you mustn't post that kind of problems. It is on a mathematical magazine in Vietnam and haven't expired yet. :D
Can any moderators lock this topic?

have to say that , i've seen many problems in this 4rum which are from the newest magazine in VN , but all of them still be unlocked ... i guess the rule in ML dont forbid it  :roll:

In fact, this is not fair. You should notice the moderators in Vietnam national forum to help. This [b]must[/b] be locked

uhm.Yes, but I think you're come from Viet Nam,yes.So, We should use VietNamese,hehe


[quote="chien than"]Prove that:$3(a+b+c) \geq 2(\sqrt{a^{2}+bc}+\sqrt{b^{2}+ca}+\sqrt{c^{2}+ab})$
where $a;b;c$ are non-negative[/quote]
$\sqrt{a^{2}+bc}+\sqrt{b^{2}+ca}\leq \sqrt{2(b^{2}+c^{2})+2a(b+c)}$
and then prove
$\sqrt{2(b^{2}+c^{2})+2a(b+c)}+\sqrt{c^{2}+ab}\leq LHS$
(not hard)
Lol, it's not easy if u change $a^{2}+bc$ by $ka^{2}+bc$ and find max.

Vasc's proof see:
http://www.mathlinks.ro/Forum/viewtopic.php?t=63645
Another proof see:
http://www.mathlinks.ro/Forum/viewtopic.php?t=124812
[quote="kcbu"]
Lol, it's not easy if u change $a^{2}+bc$ by $ka^{2}+bc$ and find max.[/quote]
$k=\frac{578}{1143}$ see:
http://www.mathlinks.ro/Forum/viewtopic.php?t=124116


inequalities
AMC
AIME
AoPS Books
triangle inequality
\/closed

[quote="mel"]Hey.

If I use the AoPS books, Zeitz, and Kiran Kedlaya's inequalities manuscript, will I be able to go through enough of the material that is covered in the Olympiad Inequalities course to do well on the AIME?[/quote]

For the AIME? The AIME isn't an Olympiad level test...so you do not have to have knowledge of stuff like Holder, Jensen, power mean, Bernoulli and so forth.  The only ones that might show up are one's like AM GM, the triangle inequality, or perhaps even Cauchy might be useful.  So, yes.  In fact, the AoPS books and Zeitz should be plenty for the AIME (Kedlaya's notes are olympiad level).

Yeah, I ask because it says that the Olympiad course takes problems from the AMC to Olympiad level.

Thanks, joml.

Yeah, you don't have to worry about the AIME if you understand this material.
I believe that Kedlaya's notes alone are even more than sufficient for the IMO!

Actually, while I wouldn't recommend Olympiad Inequalities as a method for studying for the AIME, there are some AIME problems included in the classes and inequality/optimization problems do show up on the AIME.  Students who take Olympiad Inequalities should have an easier time with those problems, which tend to be among the more challenging on the AIME.


geometry probability symmetry Support percent calculus integration

Well,INChO was too difficult
INAO was also the same

INPhO was wayyyy too easy---and surprisingly objective   :)  :( ...i expect the cut off over 70/80..and many must get 80/80

INBO--well,great paper ..but i dont think i stand any chance :(

How hav u guys performed????

:(
ChO was ok...
Physics I Dont Expect To Cross 60..
so all in all
i dont stand a chance :(
nyways
rituraj and co..have fun at the OCSC...all the best...i wont even open the post that comes to me :(

comparing to last years cutoff (that was 62).

this year it must be around 74/80.

i might get above 70.
when are answers/results gettin out!!

Sanchit Mittal

wasnt the restitution thing wrong?
noone of the options match...??
and evn tht potential..protons and positrons
O_O
noone of the options match :((
i spend @#$&*) lotta time on those..and i left 6marks in the last subjective one and i left some 6-8questions in first part also..:(

Hi guys I am new to this forum.
I'm Sagar Vaze from Pune.
I am scoring around 70 to 75 in INChO. What are my chances of getting through ?

well i m hardly gonna get past 50 in INPhO
so i m not thinking of OCSC :P

well resrtitution ans is c (actually i was getting 11/34 and did approximation :P )

and positron one is d...it came in VMC TS or sth

@sanchit--superb!!!

tuhin says A1 guys of kota expect 50-60...i m surprised.Seems i'll hav to talk!!

well let the solutions come,  b'coz i have not counted silly mistakes yet. that might tighten some of my marks.  where do i check my marks. are answers available on net somewhere. or do someone can do this. i m really feeling hurry for that. :roll:

Why dont you guys post some of the interesting questions?

i dont kno abt questions ..but here are the answers---

[quote="tuhin_fire"]1-c, 2- b, 3 d,4 b, 5 c, 6 a , 7 d(?), 8 d, 9 d, 10 d (?),11 c, 12 d, 13,b, 14 c, 15 a, 16 a , 17 a,18 a , 19 a 20 a, 21 c ,22 a,23c, 24 c, 25 c, 26 a, 27 1 28 d, 29 c, 30 c, 31 d, 32 c,33 b, 34 a , 35 a , 36 a, 37 c ,38 b

B

1c,2a,3d,4b,5e,6e,7e,8a,9a,10b,11e,12a,13a,14b,15b,16c,17d,18a,19a,20c
[/quote]
My doubts
part A
3--i've done d but ppl say ans is b (mass-energy)
7--b :)
9--c
14--a?--dc circuits..will we consider flux
18 ans looks to be b--but i've marked c :(
20--(a) r u sure??? it had come in VMC TS and ans was (d)  [i also hav marked a]
33--d

PartB--18a?????

Post ur variations too!!!

LOL you dont get prize for giving answers without questions. :P

its interesting only one out of 300+ tried to post the answers. Do the real champs do not come to web??

Anyone having any ans. ,, please post them.

Well acc. to my talk with a prof.  cutoff must go beyond 74 marks. that means ticking just 3-4 wrong questions inspite of all absolute correct! You are not selected.

All the Best

i remember leaving 6questions in part a
:(
and i dont have the question paper with me...one of my friends wanted it
nyways...ritu i am sure u ll get thru ...
all the best with ocsc atleast in bio and physics

Sanchit,y dont u post ur answers?? :D  :)

And i think the real champs rather dont hav time to waste :rotfl:  (not true really)

And if the cutoff is 74,i'll hav no regrets..as i m scoring quite lower

hey vinit,u r having a go at IChO as well???

best of luck :D

@ rituraj..
no...once you're selected in a team you cant attend other camps...

dis is wrong  :mad:

in other countries(most performing better than india,students are allowed go 4 more than 1 olymp)

and is sahal in other camps??

i m supporting him wholeheartedly(i want one 11thie in indian team,it wud have been a history)

bad 4 him

Had he made in the team,it wud have been a history

no 11thie till now has been selected in indian team(PCB)

when is bio camp??/
and chem???

chem camp is on

and there is a 11thie in bio as well (frm 3rd jun...)

and i think sahal must b the 5th (unknown one) in phy team...coz Delhi TOI said "13 yr old bio going to mexico.. :P "

WOW that's amazing..
Congrats einstein

[quote]and i think sahal must b the 5th (unknown one) in phy team...coz Delhi TOI said "13 yr old [b][u]bio[/u][/b] going to mexico..  "[/quote]

Is he going to bio camp as well???

congrats einstien!!!!!
 :winner_first:  :thumbup:  :first:
make sure u get gold!!!

no he isnt going to bio

this time there's a girl

and mexico is hosting IPhO :D

well getting the gold wont b difficult 4 einstein :) usually india rocks in physics

A girl...that's interesting! where is she from?
Btw--How many girls in Bio camp this yr?

14 girls

and that 11thie is again from Delhi :D

Goodluck to all Artists of problem solving :D

i hope i dont find ne problem in the thing we were discussing just a post earlier... :D