logarithms
calculus
calculus computations

$ x(n)\equal{}\frac{(n!)^{1/n}}{n}$

Find the limit of x(n) denoted by L

Find an equivalent of $ x(n)\minus{}L$

There are many different ways to get that $ L\equal{}\frac1e.$

To do the second part, I think we need Stirling's approximation, or at least most of it.

By Stirling, $ \frac{(n!)^{1/n}}n\approx e^{\minus{}1}(2\pi n)^{\frac1{2n}}.$ If we subtract $ e^{\minus{}1}$ from that, we get $ \frac{\ln(2\pi n)}{2ne}$ as the equivalent Moubinoul was asking for. Or, since, $ \ln(2\pi)$ is constant, we could give that equivalent as $ \frac{\ln n}{2ne}.$


There are fewer than 6 dozen rocks in a collection.

-Grouping the rocks by 2s leave one extra
-Grouping by threes leave 2 extra.
-Grouping by fours leave 3 extra.
-Grouping by fives leaves four extra.

How many rocks are in the collection?

I just did a problem just like this one....

[hide]The LCM of the numbers (2-6) is 60. Subtract one, since if you add one more rock to each collection, there will be 0 left over, and you get 59. [/hide]

[hide]
I did this problem by tring to find four consecutive numbers that were multiples of 5, then 4, 3 and 2 by guess and check. I could make it much easier by noticing that the first number had to end with a 5, it couldn't end with a 0 or the next number wouldn't be divisible by 4, it would be odd. So I got 55, 56, 57 and 58.

The answer is 59[/hide]

p.s. yes, math92 posted a similar problem... I think it is chicken wings or something?

[quote="pascal_1623"]p.s. yes, math92 posted a similar problem... I think it is chicken wings or something?[/quote]

Yes, except it had the difference that it had to be divisible by 7.


Find the answer to the problem.
sin-1(1)=

Um. Isn't it 90? You really don't have to tell us this. I don't think this goes in the MOEMS forum too. If you have something productive, post. But if you don't, then don't post.

@Coptie - He meant $ \sin ^{\minus{}1} (1)$.

@whoever started the topic

why is trignometry on he moems forum.

this should be on the highschool topics.

this isn't even trigonometry, this is some useless problem that can:
1) be solved easily with a calculator
2) something you probably automatically know if you've ever looked at trigonometry in your life

uh I know it's 90 but my $ n$spire says otherwise ;)
apparently it says it's $ 1.57079632679$ for $ \sin^{\minus{}1}(1)$ and $ 0.017454178738$ for $ \sin^{\minus{}1}\left(\frac{\pi}{180}\right)$

Are you sure you have it said to degrees and not gradians or whatever?

angle is set to radians.
So apparently $ 1^{\circ}\neq\frac{\pi}{180}?$

1 degree is defined to be $ frac{\pi}{180}$ radians.

Some calculators define the cosecant to be sin^-1, and they make a different function for the arcsin.

In most problems, $ \sin^{\minus{}1}$ is defined to be the arcsin.

-- Someone who has learned trigonometry.


I have no idea how to use computer program to graph this function , Would you ?

$f(x)=\sum_{n=1}^{\infty}\frac{1}{n^{x}}\ \ (x>1)$

You could probably truncate the expression somewhere to get a fairly good approximation. The rest is simple graphing.

If you want to follow the above suggestion:

- Taking a length 10 partial sum will be 99% accurate after $x \approx 2.63854$
- Taking a length 100 partial sum will be 99% accurate after $x \approx 1.90038$
- Taking a length 1000 partial sum will be 99% accurate after $x \approx 1.61954$

- You'll need at least 61 iterations if you want it to be 99% accurate by $x=2$

Don't try length 1000 on your calculator, and probably not length 100 either, it will not like you.

It's much better to grab hold of a program, like Mathematica, that will just graph the zeta function for you.


find all labelled simple connected graphs s.t.
 label of any vertex=k(sum of labels of its adjacent vertices) and all labels are positive real nos.,k a fixed real no.

There's no solution for $k = 2$.  Summing up both sides for all vertices, we get

$\sum_{v \in V}f(v) = k \sum_{v \in V}\deg (v) f(v)$

so clearly $k < 1$.


function
integration
logarithms
Putnam
calculus
calculus computations

Here is an interesting little problem I read on another forum.

"Find all differentiable functions such that $f'(\frac{a}{x})=\frac{x}{f(x)}$

a positive".

The chain rule was applied:

$f'(\frac{a}{x})=f'(x)(\frac{-a}{x^{2}})=\frac{x}{f(x)}$

$f(x)f'(x)=\frac{-x^{3}}{a}$

Let $u=f(x)$ and $du=f'(x)$

$-\int{u}du=-\int{\frac{x^{3}}{a}}$

$-\frac{u^{2}}{2}=\frac{x^{4}}{4a}$

$-u^{2}=\frac{x^{4}}{2a}$

Since $\sqrt{(-u)^{2}}=|u|$

$|u|=\frac{x^{2}}{\sqrt{2a}}$

$|f(x)|=\frac{x^{2}}{\sqrt{2a}}$

I checked and this works out. My question, is my use of the absolute viable?. If I merely eliminate the negative without the absolute it works out, but that isn't a good way to go about it, is it?.

Are there other functions not accounted for?. Any ideas as to how to work this fun little problem?. That negative tripped me up.

Others seem to believe there is more to it. This is not homework, just a curious problem I thought was interesting.

what do you mean $f'(\frac{a}{x})$?
Is it $\frac{df(\frac{a}{x})}{dx}$, or $\frac{df(t)}{dt} |_{t=a/x}$?

I suppose it's the former. That is how the problem was posed. $f'(\frac{a}{x})=\frac{x}{f(x)}$

Differentiable with respect to x.

Sorry, i don't  understand your words [quote] this works out[/quote]  :blush:
If you mean that this answer is correct , than \[ f(\frac{a}{x})=\frac{a^{3/2}}{x^2\sqrt{2}} \], and
$\frac{df(\frac{a}{x})}{dx} = -\frac{2a^2}{x^3\sqrt{2a}}\ne \frac{x\sqrt{2a}}{x^2}.$
But i think you mean that solution isn't correct. Maybe because $f'(a/x)$ is $\frac{df(t)}{dt} |_{t=a/x}.$ not $\frac{df(\frac{a}{x})}{dx}$
In this case i think i have full solutin:
Answer:
\[ f(x)=\frac{x^k}{\sqrt{ka^{k-1}}} \]
Solution:
[hide]
First of all let write $a/x$ instead of $x$. We get $f'(x)f(\frac{a}{x})=\frac{a}{x}$.
And we knew that $f'(\frac{a}{x})f(x)=x$.
Now \[ \frac{d}{dx}(f(\frac{a}{x})f(x))=f'(\frac{a}{x})(\frac{-a}{x^2})f(x)+f'(x)f(\frac{a}{x})= \]
\[ =-\frac{a}{x}+\frac{a}{x}=0. \]
So $f(\frac{a}{x})f(x)=\text box{const} = c.$ Note that $f(x)\ne 0,x\ne 0$ because $f'(\frac{a}{x})f(x)=x\ne 0,x\ne 0$
Thus $\frac{a}{x}=f'(x)f(\frac{a}{x})=f'(x)\frac{c}{f(x)}$.
From this diff. equation we get
$\int{\frac{df}{f}}=\int{\frac{a}{cx}}dx+b. b\in R$
or $\ln f = k\ln x + b$ (where $k=a/c \in R.$)
Then \[ f(x)=e^b x^{k}=Ax^k. \]
Put this solution in the first equation we get
$Ak(\frac{a}{x} )^{k-1}=\frac{x}{Ax^k}$
and $A^2=\frac{1}{ka^{k-1}}$. It follows that $A=\frac{1}{\sqrt{ka^{k-1}}}.$
And functions
\[ f(x)=\frac{x^k}{\sqrt{ka^{k-1}}} \] satisfy the first equation.
[/hide]

Thank you very much. That's clever.

Here's what I meant when I checked:

$f(x)=\frac{x^{2}}{\sqrt{2a}}$

$f'(x)=\frac{\sqrt{2}x}{\sqrt{a}}$

$f'(\frac{a}{x})=\frac{\sqrt{2a}}{x}$

$\frac{x}{f(x)}=\frac{\sqrt{2a}}{x}$

$f'(\frac{a}{x})=\frac{x}{f(x)}$

I thought I was onto something but fell kind of short.

This was a problem on this year's Putnam, only not quite completely stated: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64444]Putnam 2005 B3.[/url]

Thank you so very much for the link, your time, and expertise.


geometry
algebra
system of equations
geometry proposed

Let $ D,E$ and $ F$ be points on the sides $ BC,CA$ and $ AB$ respectively of a triangle $ ABC$, distinct from the vertices, such that $ AD,BE$ and $ CF$ meet at a point $ G$. Prove that if the angles $ AGE,CGD,BGF$ have equal area, then $ G$ is the centroid of $ \triangle ABC$.

I think the problem must be:
The triangles $ AGE,CGD,BGF$ have the equal area.

Here is my solution:
Denote $ x \equal{} S_{AGB},y \equal{} S_{BGC},z \equal{} S_{CGA}$. We need to prove that $ x \equal{} y \equal{} z$

Since $ S_{AGE} \equal{} S_{AGC}.\frac{AE}{AC}$
$ S_{CGD} \equal{} S_{CGA}.\frac{GD}{GA}$ and $ S_{AGE} \equal{} S_{CGD}$, which implies $ \frac{AE}{AC} \equal{} \frac{GD}{GA}$

By the other hand:
$ \frac{AE}{AC} \equal{} \frac{S_{AGE}}{S_{AGC}} \equal{} \frac{S_{ABE}}{S_{ABC}} \equal{} \frac{S_{ABE} \minus{} S_{AGE}}{S_{ABC} \minus{} S_{AGC}} \equal{} \frac{S_{AGB}}{S_{AGB} \plus{} S_{BGC}} \equal{} \frac{x}{x \plus{} y}$

$ \frac{GD}{GA} \equal{} \frac{S_{BGC}}{S_{AGB} \plus{} S_{CGA}} \equal{} \frac{y}{z \plus{} x}$,

which implies
$ \frac{x}{x \plus{} y} \equal{} \frac{y}{z \plus{} x}$,
$ (x \minus{} y)(x \plus{} y) \equal{} x(y \minus{} z)$

Similarity we obtain the system of equations:
$ (x \minus{} y)(x \plus{} y) \equal{} x(y \minus{} z)$
$ (y \minus{} z)(y \plus{} z) \equal{} y(z \minus{} x)$
$ (z \minus{} x)(z \plus{} x) \equal{} z(x \minus{} y)$

If $ x \equal{} y$, we can easily solve the problem. Assume that $ x,y,z$ are different, multiply three equations yields
$ (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} xyz$, which is obviously wrong.

So $ x \equal{} y \equal{} z$, which implies $ G$ is the centroid of triangle $ ABC$


My philosophy: If people fail to appreciate classical music, then they probably do not deserve to hear it in the first place.

Well usually a lot of people I know are inherently against classical music but most of them think its pretty cool if I play the right things.

I play classical piano and have little interest in any modern music except country. I am homeschooled, but my public school friends think I am wierd because I play classical piano. One of them stared at me and said: Wow. I don't know what to say. When I told her I played classical piano.

i enjoy the piano....kinda...but i LOVE modern music...i dont blast it though...i dont hav a single classical song i listen to out side of piano though...i just dont enjoy classical that much...i have nothing against it though

well...get them to listen to the allegro con fuco part in Dvorak's From the new world
ive made countless converts that way :-)

classical music is NICE, i use it to calm myself down or to fall asleep

modern music (as in pop music) is to listen at parties or dances

but modern music is good as long as you don't full blast it on your iPod

Most people I know at my school don't like classical music. They say things like "it's too calm" or "it's boring" and stuff like that. Personally, I think that they just don't understand classical music. To them, classical music is Mozart and Bach and all the old school stuff... and that's totally cliched. I doubt they've heard the "cooler" side, like 19th and 20th century music. But when I try to explain, they're like "yeah, who cares?" And that's pretty annoying to me....

Well...yeah, me too. A bunch of people at my school hate Pachelbel's Canon in D and Reverie by Debussy, my absolute favorites. And the closest I've gotten to pop music is alt rock.
Don't worry about what they think. I went to a recital at my piano teacher's church and I met a LOT of people who appreciated classical music. Or if you REALLY want them to like your taste, just pick out a few songs that I call "classical music pop." For example, Alejandro by Lady Gaga. You hear Czardas throughout the song in between pop, which i think is okay.

i think you should try listening to more modern music.  you could play some classical music when some of your freinds are around, and slowly get them to appreciate it

Why all this hating?  Modern music is awesome.  Just listen to Schoenberg's Piano Concerto. :P

@Alan Li: New World is good, but Liszt's Totentanz works, too.  :P

[quote="tennis123"]Is anyone else in this situation?
At my school, probably only one other person(the orchestra teacher) has heard of Heifetz, Milstein, or Kreisler. All the other kids (for some reason) just enjoy listening to modern music at mind-blowingly loud levels. Don't get me wrong, I don't have anything against modern music, I just don't want to lose my hearing at age 20. Also, whenever somethings announced about the music programs, people usually say ...band and chorus...... How can I get people to realize that orchestra and classical music have value?[/quote]
You don't always have to love classical music that is fancy or deep. Lots of kids in my orchestra don't listen to classical music either, but they love the pieces we're playing in orchestra. There was one kid who was humming the Barber of Seville in math class and told me she couldn't get it out of her head (in a good way :)).
I guess that the reason why I like classical music is that I play the violin and it sounds so exhilarating. But maybe kids find it easier to connect with modern music because it has a catchy tune and words they can relate to.
For me, I really like both. I find it better to wind down by listening to modern music and just relax instead of classical.
I really don't know what to say about the music program anouncements. At my school they announce all three programs...

I like today's music, but during youth symphony, when I can play viola surrounded by musical people for 3 hours, it's so fun!

Same at our school. Everyone's heard of band and choir, but I'm pretty sure half the school doesn't even know we have an orchestra.

Also, The Medic Droid ftw. :P

@3333, I'm not saying that jazz and/or classical music is bad, simple, or boring in any way.  In fact, I really enjoy listening to jazz (specifically Miles Davis and Pat Metheny.)  And, I do prefer it over rap or dubstep.    Really, I'm just saying that I think that if you look hard enough, you can find something to like in almost all kinds of music.

@3333: I, too, prefer the likes of Mingus to modern musicians, but I don't think you can say that jazz is "better" in an objective sense then modern music.  You think that it is, and I do to, but these are [i]opinions[/i].  That said, there are many modern musicians whom I love and who draw extensively on their knowledge of older music (like Radiohead).

One thing about jazz [and this is just my opinion], sometimes, the sounds made in most jazz is kind of, well, "off". It's a little hard to explain, but it's something around the fact that the beat, as in counts, is usually really loose.

[quote="jag"]@EpicSkills32: That's called swing, and it's kinda the essence of most jazz.  It has to do with putting the emphasis on the upbeats, rather than the downbeats.[/quote]

I kind of knew that....
And I'm not saying it's bad or anything.

I think that jazz is pretty awesome.
Rap is catchy, but I don't think that it is really good music.
Most raps have bad words in them  and I don't enjoy that.
I get teased a lot at school for enjoying classical music, but I really don't care.

That's awesome
 I think way more people should appreciate classical music. a lot of music besides rap have bad words too though lol.

Idk, being esoteric is a good thing usually. Think exclusivity. :D

Except when the industry starts going bankrupt (NY opera house; and I'm sure many others).

But seriously guys, "bad words" are cool. Grow up.

All music is awesome.

[quote="gapoc459"]
But seriously guys, "bad words" are cool. Grow up. [/quote]

No not cool. The people who think bad words are cool , are the ones who need to [quote="gapoc459"]
 Grow up. [/quote]

No, I'm not defending pop or anything; I'm just saying there's nothing wrong with "bad" words. Their intrinsic taboo nature is what causes them to be "bad"; it's a tautology of sorts. Hence, when people, viz. rappers use them, it is generally considered cool that they are, in a sense, breaking the de facto morals of our (already very flawed) society. In a way, it's (rightly) mocking the ridiculousness of the taboo with which we associate these "bad" words. This brave mockery is what I consider cool in popular music, rather than the "bad words" per se.

[quote="vonneumann13"]So to clarify, gapoc, you think that the social defiance, not the words themselves, is cool?[/quote]

yes, that is also very "bad".

[quote="gapoc459"]But seriously guys, "bad words" are cool. Grow up. [/quote]
Bad words are signs of poor intelligence. You use them when you can't explain what you want to say with normal language while expressing your feelings. These words were originally intended to offend people. People use them to get attention and make their statements sound stronger.

OK, that's kind of off topic.

Yes, precisely, vonneumann.

I'm going to make a new thread to reply; because this debate is kind of off topic.

/bump

Does it look like for everybody that the human value of 'eshay cool stuff' has increased, while the human value of real ART is regressing rapidly?


It's not very hard.
We have $(\sum_{k=1}^n a_{k}^{2n})(\sum_{k=1}^n a_{k}^{-2n})-n^2= \sum_{1 \leq i<j \leq n}(\frac{a_{i}^{2n}}{a_{j}^{2n}}+\frac{a_{j}^{2n}}{a_{i}^{2n}})+n-n^2= \sum_{1 \leq i<j \leq n} (\frac{a_{i}^{n}}{a_{j}^{n}}-\frac{a_{j}^{n}}{a_{i}^{n}})^2$.

Now it is easy to see that for $x\ge1$, we have $f(x): =x^n-\frac1{x^n}- n(x-\frac1x)\ge0\ge f(\frac1x)$. Indeed, $f^\prime(x)\ge0 \Leftrightarrow (x^n-x)(x^{n+1}-1)\ge0$ for all $x>0$ and $f(1)=0.$ This means for all $x>0$

$(x^n-\frac1{x^n})^2\ge n^2(x-\frac1x)^2$. Put $x=\frac{a_i}{a_j}$ and sum up. Done.

[quote="nhat"]Let $a_{1},a_{2},...,a_{n}$ be positive real number $(n \geq 2)$,not all equal,such that $\sum_{k=1}^n a_{k}^{-2n}=1$,prove that:
                   $\sum_{k=1}^n a_{k}^{2n}-n^2.\sum_{1 \leq i<j \leq n}(\frac{a_{i}}{a_{j}}-\frac{a_{j}}{a_{i}})^2 >n^2$[/quote]

For Lagrange's identity,

$\sum_i{a_i^{2n}}\sum_j{a_j^{-2n}} - n^2 = \sum_{1 \le i < j \le n}{{\left(\left(\frac{a_i}{a_j}\right)^n - \left(\frac{a_j}{a_i}\right)^n\right)}^2}$

So we only need to prove

$\forall x \in \mathbb{R^+}$

$\left(x^n - x^{-n}\right)^2 \ge n^2\left(x - x^{-1}\right)^2$

It's equivalent to

$x^{n-1} + x^{n-3} + \cdots + x^{-(n-1)} \ge n$

By AM-GM, we know that

$x^r + x^{-r} \ge 2$.


A fast food restaurant specializes in ham sandwiches. A customer may choose to add any of the following: mayonnaise, mustard, lettuce, tomato or cheese. How many different ham sandwich combinations are possible?

Each of the five toppings are added or not added, giving $ 2^5\equal{}32$ total combinations.

I'm sorry to revive this, but I did this problem a different way.  I just got this problem in a game, and here is what I did:

[hide="My way"]$\dbinom{5}{0}+\dbinom{5}{1}+\dbinom{5}{2}+\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5}$

$\dbinom{5}{0}=\dbinom{5}{5}=1$

$\dbinom{5}{1}=\dbinom{5}{4}=5$

$\dbinom{5}{2}=\dbinom{5}{3}=10$

$1+5+10+10+5+1=\boxed{32}$

The way I solved that was casework.
For just 0  toppings, there is $\dbinom{5}{0}$ ways.
For just 1 topping, there is $\dbinom{5}{1}$ ways.
For 2 toppings, there is $\dbinom{5}{2}$ ways.
For 3 toppings, there is $\dbinom{5}{3}$ ways.
For 4 toppings, there is $\dbinom{5}{4}$ ways.
For 5 toppings, there is $\dbinom{5}{5}$ ways.[/hide]

That is a combinatoric identity.

For any n, $nC0 +nC1 + nC2 + nC3... nCn= 2^n$
sorry, latex fail.


search
algebra
function
domain
inequalities
inequalities proposed

For $x_1,x_2,...,x_n$ positive numbers which add up to 1, find the minimum value of
   $(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n})(1-x_1)(1-x_2)...(1-x_n)$. Hope it doesn't follow immediately from I don't know what VGI or PSG or...

And when you think it is a direct application of a "wierd" problem we discussed a lot... ;)

0. Minimum value of the expression when one is variables is near to zero is $\approx (n-1)^2$. It is greater than value for $x_1=...=x_n=1/n$. So we are to search for minimum inside of domain $x_1+...+x_n=1$.
1. Using Lagrange multipliers we immediately obtain that optimal set $(x_1,...,x_n)$ takes at most two different values for its variables.
2. For two different values we write down the corresponding function of one variable and determine its extremums using derivative. Fortunately, there is only one required extremum $\frac{1}{n}$.

Therefore answer is $\frac{(n-1)^n}{n^{n-2}}$.

P.S. I really don't want to give detailed explanations, since they are far from beauty (I mean step 2).

A great french mathematician showed me the same solution and indeed step 2 wasn't simple at all. In fact, it was very technical. Here is my solution:
  Remember the intriguing inequality from mathlinks contest, which can be written in the homogenized form as:
  $\frac{a_2^2+...+a_n^2}{n-1}...\frac{a_1^2+...+a_{n-1}^2}{n-1}\cdot(1/a_1+...+1/a_n)^2\geq n^2\cdot (\frac{a_1^2+...+a_n^2}{n})^{n-1}$?
     Then using Cauchy-Schwarz we have $(1/a_1+...+1/a_n)^2\leq n(1/a_1^2+...+1/a_n^2)$ and now if we take $a_i=\sqrt{x_i}$ we can conclude.


I've asked several people about this and I haven't been able to get an answer I really understand.

How do you find the number of even factors?
How do you find the number of odd factors?
How do you find the sum of the even factors?
How do you find the sum of the odd factors?

These questions may seem basic..but I need help heh.

Say we have a number $75,600$
$2^4*3^3*5^2*7^1$

How would you find it?

Hmmm.... you know that you can multiply the primes in the factorization to come up with factors. Odd numbers can only result from the product of odd numbers.
so in the number 75600, we want to see how many ways you can multiply the odd primes in the factorization.

The odd primes are 3, 5, and 7. We use the same method as the one we use for counting factors (exponents). We have 4*3*2 = 24 odd factors (add 1 to exponents and multiply them). The number of even factors is just the total number of factors minus the odd factors.

I'm not very sure about this, because it's also the first time I've encountered such a problem. I have no clue about the sum.

I don't know if this helps, but to find the sum of all the factors of, for example, 24, whose prime factorization is $2^3 * 3$ we would do $(2^0 + 2^1 + 2^2 + 2^3)(3^0 + 3^1) = 60$
the sum of [b]all[/b] the factors of 75,600 is therefore 307520...i think
not very helpfu l - I bet you already knew that  :roll:

Well lets see... wouldn't the sum of the even factors be  $(sum of all the factors)-(sum of odd factors)$

Using the same method 236factorial sorta used.. just say its $(3^0+3^1+3^2+3^3)(5^0+5^1+5^2)(7^0+7^1)$?

Edit: I was just checking to see if the words would be all bunched together...guess so and yeah it is odd.

Try using bold text rather than Latex. It looks similar, you can use spaces, and it won't look italicized.

xenon, your technique looks correct... however, it would get pretty crazy for some numbers. Try finding the sum of the factors of $7,938,000,000$. :lol:

[quote="236factorial"]The odd primes are 3, 5, and 7. We use the same method as the one we use for counting factors (exponents). We have 4*3*2 = 24 odd factors (add 1 to exponents and multiply them). The number of even factors is just the total number of factors minus the odd factors.

I'm not very sure about this, because it's also the first time I've encountered such a problem. I have no clue about the sum.[/quote]
A small correction here -- you must subtract one from that total in order to account for the null set. Basically, by multiplying together the exponents of each of the prime divisors, you are making a permutation of all the possible factors (aka, I have 3 3's. I can take 0, 1, 2, or 3 of them. That's 4 choices, so multiply by 4, etc). However, if you take 0 for all of them, that's 0 which is not a factor of 75,600 (or anything, for that matter) . So, actually, it should be 23 odd factors, not 24.

No, that's not right. If you take 0 3's, that's actually a 1, since you have $3^0$.

[quote="Hikaru79"] However, if you take 0 for all of them, that's 0 which is not a factor of 75,600 (or anything, for that matter) [/quote]

If you take 0 for all of them, you'll get 1. ($n^0=1$)


Which would be worth more, a pound of
10 dollar pure gold coins or half a pound of 20 dollar pure gold coins or would they be the same? Explain.

[quote="ngade"]Obviously, the same.[/quote]
    wrong

[quote="OHO"][hide]a pound of 10[/hide][/quote]

explain  :lol:

[hide]Is the gold worth more or less than the denomination? If the coins were worth more than their denomination, the amount of gold would be the determining factor.
If, however, the coins were worth more than the gold in them, we need more information about how much each denomination weighs.
If the weights are equal, then the two options are equal. It's more likely that ten dollar coins are lighter, in which case they are worth more. In the really unlikely case that the twenty dollar coins are lighter, they are worth more. [/hide]

A pound of pure gold is usually more valuable than half a pound of pure gold. Given no other information, I'd take the sack of 10 dollar coins.

[quote="7h3.D3m0n.117"][quote="OHO"][hide]a pound $>$half a pound[/hide][/quote]

that doesn't matter because the half a pound bag contains 20 dollar coins...which makes up the difference for half a pound[/quote] sorry but he's right

[quote="randomdragoon"]Also: What if a single 20 coin weighed twice as much as a 10 coin? Since they're made out of a commodized, precious metal, it's very possible.[/quote]

u would still choose the 10 coin because its 2 times more


geometry
geometric transformation
real analysis

I received a question form my lecturer and I thought it's a tough one.

Let a, b, c are elements of R (real munber set)

Prove that

1) a+B = b+a - Commutative law
2) (a+b)+c = a+(b+c) - Associative law
3) a(b+c) = ab+ac - Distributive law

I've learn only to memorize but not to prove this. So, please help me.

So, I see that our teacher tricks us. Oh my god.

[quote="DPatrick"]You can't "prove" these statements -- these statements are axioms.  We assume them in order to make the real numbers work.[/quote]

Sure you can prove those statements--I've seen it done. Landau proves all of those statements from the Peano axioms in his [url=http://www.amazon.com/exec/obidos/tg/detail/-/0828401411/learninfreed]Grundlagen der Analysis[/url], and anyone else who constructs the real number system from the Peano axioms does too.

The typical pedagogical approach in high school is to take the field properties of the real number system as axioms, but the typical approach in a foundations course, or in some real analysis courses, is to take the simpler Peano axioms as axioms and derive the field properties from those.

I'm not sure that this is a Getting Started question, other than that we tend to allow informational questions in whatever forum they land in. As a challenge problem, I would classify the question as no harder than pre-Olympiad.

I've moved this to Other Topics because this is not a "problem" in the usual sense (and in any event it's not Getting Started).

It is true, as tokenadult said, that you can define the natural numbers via the Peano axioms, define addition on the natural numbers via an inductive-type argument, and then prove that the addition you've just defined is commutative, associative, and distributive.  Then you define the rationals as the field closure of the naturals, and then the reals as the topological closure of the rationals (so that all Cauchy sequences have a limit), and then prove that what you get is a field.

I'm pretty sure that this is not what the original poster had in mind!  (Maybe he did, in which case, it definitely should not have been posted in the Getting Started forum!)

As a practical (or pedagogical) matter, we generally assume the field properties of the real numbers as axioms, but definitely, if you want to construct mathematics from the ground up, you would need to prove that the real numbers form a field.

Alternatively, you can define the reals as the (unique) totally-ordered field which is Dedekind-complete (that is, every non-empty bounded subset has a unique least-upper bound.)  In this case, you do indeed assume commutativity, etc., as axioms.

If you don't understand what tokenadult and I are talking about, don't worry!  Unless you are willing to construct the set of real numbers from scratch, you can assume that a+b=b+a, etc.  This stuff is (typically) undergraduate-level real analysis and abstract algebra.

Sorry for the confusion that my original (overly simplistic) reply caused.

[quote="tokenadult"]Sure you can prove those statements--I've seen it done. Landau proves all of those statements from the Peano axioms in his [url=http://www.amazon.com/exec/obidos/tg/detail/-/0828401411/learninfreed]Grundlagen der Analysis[/url], and anyone else who constructs the real number system from the Peano axioms does too.
[/quote]

What language is the mentioned book in?

[quote="RC-7th"][quote="tokenadult"]Landau proves all of those statements from the Peano axioms in his [url=http://www.amazon.com/exec/obidos/tg/detail/-/0828401411/learninfreed]Grundlagen der Analysis[/url].
[/quote]

What language is the mentioned book in?[/quote]

The original language is German. The same book is available in English translation with the title [url=http://www.amazon.com/exec/obidos/tg/detail/-/082182693X/learninfreed]Foundations of Analysis[/url]. There are various other books that show the steps for construction of the complex number system from the Peano axioms.


A, B, C are distinct non zero digits. Find the smallest three digit number ABC such that the average of the numbers, ABC, ACB, BAC, BCA, CAB, and CBA has 8 as its units digit.

You could write the average as 37(A+B+C). If the units digit is 8, then A+B+C must = 4, 14, or 24. It cannot = 4 because A,B,C are distinct and it. It could equal 14. To minimise ABC, A=1, C=9, therefore B=4.
So 149 is the least such number.

how do you mean we write the average as that?

I checked it out and it is true, but how did you know that?

edit:: figured it out for anyone who wants to know..

[hide="how 37(a + b + c) = average"]

there are 3! = 6 ways we can write a number with 3 digits using the same 3 numbers.

if you line up all the possibilities

abc
acb
bac
bca
cab
cba

you notice that in each column there is  2a, 2b and 2c, thus the sum of these numbers is

$(2a+2b+2c)+10(2a+2b+2c)+100(2a+2b+2c)$

$2(a+b+c)+20(a+b+c)+200(a+b+c)$
$222(a+b+c)$

divide the sum by 6 for the average,

$\frac{222(a+b+c)}{6}= 37(a+b+c)$

very cool [/hide]

my question is, did you know this already, or did you figure this out on the fly?

It is remarkably similar to a question that appeared in the UK Intermediate mentoring scheme recently, which asked mentees to prove that the sum of 6 such numbers was 222x(the sum of the digits). Therefore the average is 37x(sum of digits).

[hide="proof of answers"]Well, yes the sum is 222(a+b+c) and yes, the average is 37(a+b+c). Here's the proof.

Let $\overline{abc}$ equal a three digit number with distinct digits. The number can also be written as $100a+10b+c$. So we write the sum as

$\overline{abc}+\overline{acb}+\overline{bac}+\overline{bca}+\overline{cab}+\overline{cba}=100a+10b+c+100a+10c+b+100b+10a+c+100b+10c+a+100c+10a+b+100c+10b+a=222a+222b+222c=222(a+b+c)$

Ending proof with Q.E.D.

Now we want the average, which is 37(a+b+c). For it to have a units digit of 8, (a+b+c) has to have a units digit of 4. It can't be 4, since the digits must be distinct. It can't be 24, since we are looking for the smallest.

So we are looking for the least three digit number with diistinct digiits with the diigits summing up to 14. The least one is 149(by inspection).[/hide]


(a) Solve this equation: 2x2 – x = 1999000.
(b)  Factor the expression 12x2 - x - 1, write this polynomial as a product of polynomials of the first degree.
(c) Using the previous items, verify that the number is 1999000 + 9999999 compound.

[quote="luiseduardo"](c) Using the previous items, verify that the number is 1999000 + 9999999 compound.[/quote]
Whats compund :wink:

Composite probably.

a) $ x(2x\minus{}1)\equal{}1999000$, factor RHS

b) $ 12x^2\minus{}x\minus{}1\equal{}(ax\plus{}b)(cx\plus{}d)$

Then $ ac\equal{}12,bd\equal{}\minus{}1,ad\plus{}bc\equal{}\minus{}1$

Let $ b\equal{}1,d\equal{}\minus{}1$. $ ac\equal{}12,a\minus{}\equal{}1$, so $ a\equal{}4,c\equal{}3$.

$ 12x^2\minus{}x\minus{}1\equal{}(3x\minus{}1)(4x\plus{}1)$

c) ??? I can prove that it's composite by other means, but by using a)... How?

For part a) just use quadratic formula. Gives $ x \equal{} \frac { \minus{} 1999}{2}$ or $ x \equal{} 1000$

Or just solve like normal quadratic: $ 2x^2\minus{}x\minus{}1999000 \equal{} (2x\plus{}1999)(x\minus{}1000) \equal{} 0$

Roots are $ x \equal{} \frac { \minus{} 1999}{2}$ or $ x \equal{} 1000$


Who is going to state in southern California?

umm wouldn't it be better to put this in the california forum?

me!!!! ill see you there. im mihail eric!


Mods, please do not move this, the CS forum is a wasteland.

How do I install the gcc compiler? The instructions on the website are really really complicated for me and I need help.

Thanks

On Windows install mingw
On OS X, install Xcode
On 'nux, apt-get

so,
what's the difference between mingw and gcc?

mingw = Minimal Gnu for Windows -> GCC for Windows. See http://www.mingw.org/wiki/HOWTO_Install_the_MinGW_GCC_Compiler_Suite for installing.

I'm more a fan of Cygwin, though; it's a fairly complete Linux userland on windows [X! Everything else if you want it!], and works well. (According to [url=http://gcc.gnu.org/install/binaries.html]gcc.gnu.org[/url], Cygwin and  mingw are your two binary choices.)

Also the CS forum isn't that unread.


This is a most unusual paragraph. How quickly can you find out what is so unusual about it? It looks so ordinary that you would think that nothing is wrong with it at all, and, in fact, nothing is. But it is unusual. Why? If you study it and think about it, you may find out, but I am not going to assist you in any way. You must do it without any hints or coaching. No doubt, if you work at it for a bit, it will dawn on you. Who knows? Go to work and try your skill. Good luck!

What is unusual about the above paragraph?

[hide="it took me 10 seconds"]it has no "e"[/hide]


Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this cryptarithmetic equation, where [b]M[/b] is nonzero.

([b]MMETH[/b])^2 = [b]MODERATES[/b]

The answer to the above "cryptarithmetic problem" is as follows:

[b] (1 1 8 6 7) ^ 2 = 1 4 0 8 2 5 6 8 9 [/b]

where

[b] M= 1 ; E=8 ; T=6 ; H=7 ; O=4 ; D=0 ; R=2 ; A=5 ; S=9 [/b]

This conforms to all the rules as stated in the problem, which thereby solves the problem.


trigonometry
function
algebra unsolved
algebra

Solve the equality:
$ \frac{5^{sinx}}{sinx}\equal{}\frac{5^{cosx}}{cosx}$

Since $ \sin{x} , \cos{x} \in \left[\minus{}1,1\right]$ and the function $ f\left(x\right)\equal{}\frac{5^{x}}{x}$ is injective on $ \left[\minus{}1,1\right]$.
Hence, $ \sin{x}\equal{}\cos{x}$. This implies $ x\equal{}\frac{\pi}{4}\plus{}k\cdot \frac{\pi}{2}$ for all $ k \in \mathbb{Z}$ and it can be easily checked that the values are fine.
 :D

NO!I don't think so.NOTE:5>$ e^\sqrt{2}$
This equation have 3 solution but I don't know what it is?
Let help me!

[quote="epsilonist"] and the function $ f\left(x\right) \equal{} \frac {5^{x}}{x}$ is injective on $ \left[ \minus{} 1,1\right]$.[/quote]
could you give a proof?


Given $\triangle ABC$ incircle $(I)$ with any line through centroid $G$ cuts $(I)$ at $M,N$ prove that
$MA^{2}+NA^{2}+MB^{2}+NB^{2}+MC^{2}+NC^{2}=3MN^{2}+(s-a)^{2}+(s-b)^{2}+(s-c)^{2}$
with $s$ is semiperimeter of $\triangle ABC$

Yes you are right, we can change incenter by nine-point circle we will have the similar result:
$MA^{2}+NA^{2}+MB^{2}+NB^{2}+MC^{2}+NC^{2}=3MN^{2}+\frac{a^{2}+b^{2}+c^{2}}{4}$

 :)  :)


Can you make an option, so we can edit the CSS of the pages that are not like the index. What I mean is that the CSS of the index page also apply to the other pages? Is that possible?

[quote="N.T.TUAN"]I can't delette a topic in my Blog. WHY?  :o[/quote]

Can you help me?  :(

I don't think you can delete topics in blogs.

 :(

HI there but can an admin please try to answer my question


For positive integers $ n$, the sequence $ a_{1}, a_{2}, a_{3}, ..., a_{n}, ...$ is defined by

$ a_{1}\equal{}1, a_{n}\equal{}(\frac{n\plus{}1}{n\minus{}1})(a_{1}\plus{}a_{2}\plus{}a_{3}\plus{}...\plus{}a_{n\minus{}1}), n>1$.

Determine the value of $ a_{1997}$.

[quote="AndrewTom"]For positive integers $ n$, the sequence $ a_{1}, a_{2}, a_{3}, ..., a_{n}, ...$ is defined by

$ a_{1} \equal{} 1, a_{n} \equal{} (\frac {n \plus{} 1}{n \minus{} 1})(a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n \minus{} 1}), n > 1$.

Determine the value of $ a_{1997}$.[/quote]

So $ a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n \minus{} 1}\equal{}\frac{n\minus{}1}{n\plus{}1}a_n$
And $ a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n \minus{} 1} \plus{} a_n\equal{}\frac{n}{n\plus{}2}a_{n\plus{}1}$

Subtracting the first from the second gives $ \frac{n\minus{}1}{n\plus{}1}a_n\plus{}a_n\equal{}\frac{n}{n\plus{}2}a_{n\plus{}1}$ and so $ a_{n\plus{}1}\equal{}2\frac{n\plus{}2}{n\plus{}1}a_n$ and so $ a_n\equal{}2^{n\minus{}2}(n\plus{}1)a_1$

And so $ a_{1997}\equal{}1998\cdot 2^{1996}$

Hence the result : $ \boxed{a_{1997}\equal{}999\cdot 2^{1997}}$


$\text{The Phantom Mark Spaceship approached the planet yDome where the temperature } \dagger \text{(in tLoomas) is expressed in terms of the temperature m (in mallots) by the formula } \dagger = \frac{4}{11} (3m + 8).\text{ For what value of }\dagger \text{ will the temperature } \dagger \text{ equal the temperature m?}$

$\text{BTW: The dagger symbol in LATEX is used to represent t, or tLoomas.}$

[hide]Write the formula in terms of m.
we have $m=\frac{11}{12}t-\frac{8}{3}$
since we want m=t, whether we use m or t in the formula makes no difference.

So $\frac{4}{11}(3m+8)=\frac{11}{12}t-\frac{8}{3}$

$\frac{-23}{132}m=\frac{184}{33}$

$m=-32$[/hide]


AMC AIME geometry rectangle probability algebra polynomial

Ok I am going to start pm ing the second round questions out to those involved.  Remember respect the time limit!  There is no way for me to know if you cheat or something so we are going to do this on a trust system.  I am not worried though.  All of you have enough dignity not to cheat.  No calculators.  All that you can use is your brain, paper, and pencil i.e. no internet resources, help from others etc.  The time limit for this round will be 25 minutes.  All answers must be in on or before Monday.

Second Round Matchups

LonelyTraveler8 vs. Dune

RC-7th vs. EFuzzy

nr1337 vs. yifman12

tandjsnell vs. Peter VDD

Singthesorrow vs. Lucky707

white_horse_king88 vs. LynelleYe

HashBrownie vs. Genius1601

Mat_88 vs. Soccerplayer

I also have the updated bracket attached.

What were the answers to the tiebreakers?

If anyone wants to try the tiebreakers they are in the thead for Round 1.  I will post solutions if no one does them.

That was a fun round.  I think I missed a couple though :? .  Good luck to my opponent, Dune :cool:

Omg.  I'm up against a AIME qualifier!  How scary.  Well, um, GOING TO BE ELIMINATED.  Haha, this tournament was fun anyway.

Again, my thanks to joml88. Number 5 was especially nice, and the tricks for the first few were quite interesting.

joml88, since i'm eliminated from winner's bracket, could you PM the questions for the round as they go by, so I can see what they are, thanks!

Questions answered: two
Questions answered [i]correctly[/i]: none
The first one is a known fact (pathetic, I know). The second is my educated guess. Hoping the loser's bracket will have easier questions.

Made such a stupid mistake on the second one... maybe that was why it was so hard :P

[quote="tandjsnell"]Where did you get all of the questions?[/quote]

I'm pretty sure joml88 send them to you by pm at whatever time you ask him to.

I've already done them. I just wanted to know where joml88 got them from.

I made them myself.  They are mostly based on problems I have seen before.

There's a trick for #2, which joml will probably post on the answer key.

Is there a shortcut for number 3?  cuz I DID list all the factors, but I didn't want to give myself credit for it.

Since the round has finished, can some one post the problems?

If the round is over - I don't know whether this is true or not. Here is the trick to number 3. It is in spoiler, so if someone hasn't taken the test, don't look until joml88 has officially modified this response!

3) 300 prime factorized is 2^2*3*5^2. There are then, 3 * 3 * 2 factors, because you can choose either 1, 2, or 4 for the first power of two, and so on for the rest of the primes. 3*3*2 = 18. To answer the next question: What is the sum of the factors? Use the above mentioned pricinciple. You have (1+2+4)(1+3)(1+5+25) = 868. If you expand that, you get a list of all the factors

So is the round officially over because I want to start posting some answers :)

soccerplayer hasn't send me his answers yet... so the round isn't over yet.

might want to edit white_horse_king88's for the time being then.  Just so that that answer isn't given away.  Up to you though.

[quote="RC-7th"]Since there isn't a time limit, besides that we have to send in the answers by a certain date, can't you post the problems?[/quote]

I thought that there was a time limit.  About 5 minutes per question.

Maybe there is a little bit of a misunderstanding.  What I am going to do and have started doing for this round is I ll send you the questions and when you get them you do them for the time limit which I allot.  If you don't have enough time when you first get them DONT look at them.  Wait until you have enough time to do them all at once, do them, and send in the answers/solutions promptly.  The time limit for this round is 25 minutes.  I m sorry for any misunderstanding that might have arisen during this round.

Round 2 is over!

Final match up

mat_88    def.    soccerplayer  8-2

I will start sending out round 3 questions tomorrow morning.  For now you all can discuss the problems.  I have the solutions in a pdf but I ll let you all talk about them first.

For those who weren't part of the round here are the questions:

1.	(1 point)What is the sum of the reciprocals of the roots of \[x^2+12x+22\]

2.	(1 point)What is the volume of a rectangular box if one face has area $4\sqrt{2}$, another has area of 45, and a third has area of $20\sqrt{2}$?  ( the 6 faces of a rectangular box are all rectangles and opposite faces are congruent).

3.	(2 points) * How many factors does 300 have?  What is the sum of the factors of 300?

4.	(2 point)A hat has 7 fair coins and 3 biased coins.  A fair coin comes up with heads  the time.  A biased coin comes up with heads  the time.  What is the probability that if I choose one coin from the hat and flip it that it comes up heads?

5.	(2 point)* How many elements are in the union of the sets A, B, and C where A is the set of all whole numbers $\leq200$ that are divisible by 2, B is the set of all whole numbers $\leq200$ that are divisible by 3, and C is the set of all whole numbers $\leq200$ that are divisible by 5?

     (In other words how many whole numbers less than or equal to 200 are divisible by 2,3, and/or 5.)

1.Let y=1/x. Then the equation becomes $\frac{1}{y^2}+\frac{12}{y}+22$
Multiplying by $y^2$ we get $22y^2+12y+1$
The sum of the roots is $-12/22=-6/11$

2.If a,b,and c are the lengths of the sides of the rectangle, one of the faces must be a*b, another must be b*c and the last must be a*c. Multiplying the numbers given we get 7200, which equals a*a*b*b*c*c. Taking the squareroot gives 60sqrt(2)=abc

4. The probablity that you get a fair coin is 7/10. Multiplying that by 1/2 gives 7/20, the probablity that you flip a fair coin and get heads. Similarily, 3/10*3/4=9/40 is the probablity that you flip a biased coin and get heads. Adding the two fractions gives 23/40

5.146

Since RC-7 th has posted solutions I will let everyone see mine.  I did numbers 1 and 5 differently than he did.  I like the way he did number one.  Transforming polynomials is pretty cool.  Basically he did number 5 the same as me but he has a nice geometric representation(venn diagram) to go along with it.

For number you use what is called the principle of inclusion-exclusion.  It s a cool principle.  I think that white_horse_king88 had some trouble with this one.  Look at RC-7th s or my solution and if you don't understand it I 'll try explaining PIE (principle of inclusion-exclusion) better.

I have a little trick that someone showed me for number 1.  This person is full of tricks so I often go to him when I'm stuck with a problem.  I got this one right but this method would have cut the time down to seconds.

[quote][quote]Hello, lonelytraveler8!

These are great problems!
I'll do a few of them now . . . (the rest tomorrow)

[quote]1. (1 point)What is the sum of the reciprocals of the roots of [tex]x^2 + 12x + 22[/tex][/quote]
There's an incredible trick for this one!

Did you know this about the roots of a quadratic?
Let [tex]a[/tex] and [tex]b[/tex] be the two roots.
Then [tex]a + b = -12,\;\;ab = 22[/tex]

Now [i][b]divide the two equations[/i][/b]: [tex]\frac{a + b}{ab} = \frac{-12}{22}[/tex]

The left side is: [tex]\frac{1}{a} + \frac{1}{b}[/tex]
The right side is: [tex]-\frac{6}{11}[/tex] . . . [i]There![/i]
[/quote]

This person also showed me a little forumla for problems like number 5.  I of course don't take credit for the solution because I did not do it, it was showed to me afterwards and I thought it would be worth showing on here.

[quote][quote]5. How many elements are in the union of the sets A, B, and C
where A is the set of all whole numbers [tex]\leq 200[/tex] that are divisible by 2,
B is the set of all whole numbers [tex]\leq 200[/tex] that are divisible by 3,
and C is the set of all whole numbers [tex]\leq 200[/tex] that are divisible by 5?
(In other words how many whole numbers less than or equal to 200 are divisible by 2,3, and/or 5.)[/quote]
This one is tricker than it looks . . .

There is a formula for the union of [i][b]three[/i][/b] sets: [tex]n(A \cup B \cup C)[/tex]
[tex].\;\;\;= \;n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A\cap B\cap C)[/tex]
(Add the "singles", subtract the "doubles", add the "triple".)

Now we have to do some serious Counting.

Let [tex]n(2)[/tex] = number of elements divisible by 2.
Since [tex]200\div2 = 100[/tex] there are 100 of them: [tex]n(200) = 100[/tex]

Find [tex]n(3)[/tex]
Since [tex]200\div3 = 66.66...[/tex] there 66 of them: [tex]n(3) = 66[/tex]

Find [tex]n(5)[/tex]
Since [tex]200\div5 = 40[/tex] there are 40 of them: [tex]n(5) = 40[/tex]

Find [tex]n(2 \cap 3)[/tex] . . . These are number divisible by [b]6[/b].
Since [tex]200\div6 = 33.33...[/tex] there are 33 of them: [tex]n(2 \cap 3) = 33[/tex]

Find [tex]n(3 \cap 5)[/tex] . . . These are number divisible by [b]15[/b].
Since [tex]200\div15 = 13.33...[/tex] there are 13 of them: [tex]n(3 \cap 5) = 13[/tex]

Find [tex]n(2 \cap 5)[/tex] . . . These are number divisible by [b]10[/b].
Since [tex]200\div10 = 20[/tex] there are 20 of them: [tex]n(2 \cap 5) = 20[/tex]

Find [tex]n(2 \cap 3 \cap 5)[/tex] . . . These are numbers divisible by [b]30[/b].
Since [tex]200\div30 = 6.66... [/tex] there are 6 of them: [tex]n(2 \cap 3 \cap 5) = 6[/tex]

Now, we can drop these numbers into that formula:
[tex]n(2 \cup 3 \cup 5) \;= \;100 + 66 + 40 - 33 - 13 - 20 + 6 \;= \;{\bf146}[/tex][/quote][/quote]

Also I should mention that I got this one wrong when I did it :maybe:

Anyways, I thought I should mention these techniques since they would have made the time for my solutions a lot shorter(and I would have gotten number 5 correct) if I had known them.

Yeah, what you were shown for number 5 is the principle of inclusion-exclusion like I was taling about.  In the problem I gave it was for three sets.  I would show you the general formula but it would take a while to type and I am sure that you wouldn't understand it (it has a bunch of sigmas and stuff).  So I'll just show you how it works for the first couple cases and I think you will see the pattern(which is a lot easier than trying to understand it by looking at the formula!).

Let $|A|$ mean the number of elements in set A.  $|A\cup B|$ means the number of elements in the union of A and B.  $|A\cap B|$ means the number of elements in the intersection of A and B and so forth...  If someone doesn't understand intersection or union ask and I'll explain it.

So for 1 set we have
$|A|=|A|$ obviously

For 2 sets we have

\[|A\cup B|=|A|+|B|-|A\cap B|\]

For 3 sets we have

\[|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|\]

I ll stop at 4 sets

\[|A\cup B\cup C\cup D|=|A|+|B|+|C|+|D|-|A\cap B|-|A\cap C|-|A\cap D|-|B\cap C|-|B\cap D|-|C\cap D|+|A\cap B\cap C|+|A\cap B\cap D|+|B\cap C\cap D|-|A\cap B\cap C\cap D|\]

Ok, I hope you get the idea... that was a lot of typing!!!  Oh yeah,  the person you got this from said $n(A)$ was the number of elements I believe that $|A|$ denotes the number of elements and $n(A)$ is the number of subsets... just thought I would point that out.

I couldn't look at the pdf file because I had to reinstall acrobat reader 2 days ago and it told me I had to restart my computer for it to take affect but I've been too lazy :D


real analysis
combinatorics unsolved
combinatorics

Hello everyone,

Prof Merryfield proposed a problem in the Spanish Communities Forum (very easy, but I made a mistake the first time  :blush: ), nevertheless, trying to generalize the resoult, I got to a sum that I can't find a "nicer" expression (an expresion more "simplyfied"), then, I thought that it would be a nice problem to propose here.

Simplify $\sum_{k=0}^{n}\sum_{j=0}^{k}\binom{n}{k-j}\binom{k}{j}$.

Best,

This is very well known.  It's Vandermonde's convolution formula.  It sums to (n+k)Ck

This is non-sense since you are summing over $k$, so that it can't appear in the final answer.

Pierre.

Well, at least it may help something, because, acconding to what I found in the web, a special case of the Vandermonde's formula can be written as $\binom{n+m}{k}=\sum_{j=1}^{\max(m,k)}\binom{n}{k-j}\binom{m}{j}$, and this is the inner sum in the expression, for $m=k$, then, we can simplify to

\[\sum_{k=0}^{n}\sum_{j=0}^{k}\binom{n}{k-j}\binom{k}{j}=\sum_{k=0}^{n}\binom{n+k}{k}\]

Can someone simplify this even further?

Best,

the last sum,namely $\displaystyle \sum_{k=0}^{n} \binom{n+k}{k}$ is simply $\binom{2n+1}{n+1}$.this is easily checked but it is easy to see this combinatorially also;starting with the set $\{1,2,\ldots,2n+1\}$ we wish to draw a set of size $n+1$,which of course is done in $\binom{2n+1}{n+1}$ ways.alternately if we fix the last element of the set to be $n+k+1$(where $0 \leq k \leq n$) then we choose the other elements as an $n$ set in $\{1,2,\ldots,n+k\}$ and the total number of choices is just $\displaystyle \sum_{k=0}^{n} \binom{n+k}{n}$ and this is just the first sum.


I have to import a figure into my latex file. I have the figure both in .png and .eps formats.
My output is .pdf. so i tried to import the figure with .png. I did as follows but the figure did not come out.

\documentclass{article}
\usepackage{graphicx}
\begin{document}
\begin{figure}{h}
\caption{energy chart}
\centering
\includegraphics{energy charter.png}
\end{figure}
\end{document}

what is wrong with the above? Thanks.

It isn't working because you have spaces in your filename. LaTeX is an old system so [i]always[/i] avoid spaces in all filenames and directories. Rename the file and it wil be fine, or if you must have spaces, you could try adding \usepackage{grffile} and that may well solve the problem.

Thanks first.
 I tried to fix the problems as you said.
Rename my figures with no spaces. And add the usepackage.
But the figure still does not show up in the output.
Any more ideas?


Place the numbers $ 1, 2, 3, 4, 5, 6, 7, 8$ at the vertices of a cuboid such that the sum of any $ 3$ numbers on a side is not less than $ 10$.  Find the smallest possible sum of the 4 numbers on a side.

Let $ a_1$, $ a_2$, $ a_3$, $ a_4$ the numbers on a side (face).
Let $ b_1\equal{}a_1\plus{}a_2\plus{}a_3$, $ b_2\equal{}a_1\plus{}a_2\plus{}a_4$, $ b_3\equal{}a_1\plus{}a_3\plus{}a_4$, $ b_4\equal{}a_2\plus{}a_3\plus{}a_4$
Then we have that all $ a_i$ are distinct, so all $ b_i$ are distinct and $ C\equal{}b_1\plus{}b_2\plus{}b_3\plus{}b_4\equal{}3(a_1\plus{}a_2\plus{}a_3\plus{}a_4)$
We have to find the minimum of $ C$.
W.L.O.G. $ b_4>b_3>b_2>b_1 \ge 10$ Then $ C\ge 46$
But $ C$ is a multiple of $ 3$, then $ C\ge 48$, and I have an example for $ C\equal{}48$
$ (0;0;0)\equal{}1$; $ (0;0;1)\equal{}7$; $ (0;1;0)\equal{}6$; $ (1;0;0)\equal{}4$; $ (0;1;1)\equal{}3$; $ (1;1;0)\equal{}5$; $ (1;0;1)\equal{}8$; $ (1;1;1)\equal{}2$
The minimum possible is $ 16$


fidn the circumference of a circle inscribed in an isoscles triangle with legs 1.

The inradius of the given triangle is $ \frac{2\minus{}\sqrt{2}}{2}$, yielding the answer is $ (2\minus{}\sqrt{2})\pi$.

I used the fact : the inradius of the right-angled triangle with legs length of $ a,\ b$ and hypotenuse of $ c$ is $ r\equal{}\frac{a\plus{}b\minus{}c}{2}$.

kunny, how do you know it's a right triangle?

The problem said legs, so kunny probably assumed it was a right triangle. Otherwise, with only two side lengths of a triangle given, there are multiple triangles which would satisfy the problem.

Didn't the problem specify it as an isosceles triangle?  :huh:

EDIT: Unless you consider a 45-45-90 as an isosceles.

[quote="7h3.D3m0n.117"]Didn't the problem specify it as an isosceles triangle?  :huh:[/quote]
Yes, but not necessarily a right triangle. I don't think the original problem has enough information to be solved, unless you take the leap of faith that Rising Star did in saying that "legs" implies a right triangle. Though this may make sense, I guess...

[quote="7h3.D3m0n.117"]EDIT: Unless you consider a 45-45-90 as an isosceles.[/quote]

A $ 45^o\minus{}45^o\minus{}90^o$ triangle is an isosceles triangle.


Let $ a,b,c,d$ be rational numbers with $ a>0$. If for every integer $ n\ge 0$, the number $ an^{3} \plus{}bn^{2} \plus{}cn\plus{}d$ is also integer, then the minimal value of $ a$ will be

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\  \frac{1}{2} \qquad\textbf{(C)}\  \frac{1}{6} \qquad\textbf{(D)}\ \text{Cannot be found} \qquad\textbf{(E)}\ \text{None}$

[hide="Giveaway"] [b]Claim:[/b]  A polynomial takes on integer values at every integer if and only if it is a $ \mathbb{Z}$-linear combination of the polynomials $ {x \choose k} \equal{} \frac {x(x \minus{} 1)...(x \minus{} (k \minus{} 1))}{k!}$.  See:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941 [/hide]

Let $n=0$, $d$ will be an integer.
Let $n=1$, $a+b+c$ will be an integer because $d$ is an integer.
Let $n=2$, $8a+4b+2c$ will be an integer.
$(8a+4b+2c) - 2(a+b+c) = 6a+2b$ will be an integer.
Let $n=3$, $27a+9b+3c$ will be an integer.
$(27a+9b+3c)-3(a+b+c)=24a+6b$ will be an integer.
$4(6a+2b)-(24a+6b)=2b$ will be an integer.
$(6a+2b)-2b=6a$ will be an integer.
Since $a>0 \Rightarrow 6a>0 \Rightarrow 6a \geq 1 \Rightarrow a \geq \frac 16$. The answer is $\boxed{C}$.


Here is CGMO 2004 problems, in Chinese :blush:

1) If there exists a permutation $a_1, a_2, \ldots, a_n$ of $1, 2, \ldots, n$ such that $k + a_k$ ($k = 1, 2, \ldots, n$) are all perfect squares, we call $n$ a [i]good[/i] number.
In the set $\{11, 13, 15, 17, 19\}$, which are good numbers, and which are not?

2) Let $a, b, c$ be positive reals.  Find the smallest possible value of \[\frac{a+3c}{a+2b+c} + \frac{4b}{a+b+2c} - \frac{8c}{a+b+3c}.\]

3) It is known that the circumradius of the obtuse angled triangle $ABC$ is $1$.  Prove that $\triangle ABC$ can be covered by an isosceles right triangle with hypothenuse of length $\sqrt{2} + 1$.

4) There are $32$ cards in a pack of cards with suits of $3$ colours.  Of these, there are $10$ red, yellow and blue cards, each set being numbered $1, 2, \ldots, 10$. There is also one "big" card and one "small" card, both with the numbers $0$.  From this pack of cards, some are drawn, and their value is calculated as follows: each card numbered $k$ gets $2^k$ points.  If the sum of points is $2004$, the cards drawn are called a [i]good[/i] set.
Find the number of good sets.

5) Let $u, v, w$ be positive reals such that $u\sqrt{vw} + v\sqrt{ww} + w\sqrt{uv} \geq 1$.
Find the smallest possible value of $u + v + w$.

6) $O$ is the circumcentre of an acute triangle $ABC$.  $AO$ intersects $BC$ at $D$.  The variable points $E, F$ are on $AB, AC$ respectively such that $A, E, D, F$ are concyclic.    Prove that the length of the projection of the segment $EF$ on the line $BC$ is constant.

7) It is known that $p$ and $q$ are positive integers with the same parity, and $n$ is a non-negative integer.  How many integers can be represented in the form $ip + jq$, where $i$ and $j$ are non-negative integers, and $i + j \leq n$.

8) We call a [i]cross[/i] the shape obtained by removing the four corners from a $3 \times 3$ grid.  How many non-overlapping crosses can be placed on a $10 \times 11$ chessboard, if each cross must cover exactly $5$ squares?

I'm not a girl, but still let me reply...

[quote="Valiowk"]6) $O$ is the circumcentre of an acute triangle $ABC$.  $AO$ intersects $BC$ at $D$.  The variable points $E, F$ are on $AB, AC$ respectively such that $A, E, D, F$ are concyclic.    Prove that the length of the projection of the segment $EF$ on the line $BC$ is constant.[/quote]

[Solution moved to http://www.mathlinks.ro/Forum/viewtopic.php?t=66417 .]

  Darij

[quote="darij grinberg"]I'm not a girl, but still let me reply...[/quote]

Darij, I don't want to have to translate [b]and[/b] solve all of them alone ;) .

Thank you very much, Valiowk!
You have done a great work!

5th problem is clear from
$ (u+v+w)^2 \geq 3(u\sqrt{vw}+v\sqrt{uw}+w\sqrt{uv}) \geq 3 $
which you can get by adding this AM-GM
$\frac{1}{2}(u^2+u^2+uv+uw+vw+vw) \geq \frac{6}{2}\sqrt[6]{u^6v^3w^3} = 3u\sqrt{vw} $
and other two cyclic with this one.

What is the meaning of Girls' olympiad?
Is there a boys' olympiad as well in China, or is
it that the national competition has mostly male
competitors?

btw, I know of at least two Chinese women
who were winners of the national olympiad, and
one or both got medals at the IMO.

To my knowledge, top math olympians are predominantly male (this is true in most, if not all countries, just look at the IMO figures). So China apparantly wants to encourage female participation in math competitions, hence the olympiad.

Similarly, there're the China West MO, to encourage participation from the west.

When I talked to the chinese IMO team in Athens, they said that there hadn't been a girl on the team for years.

[quote="fleeting_guest"]Did the Chinese IMO'ers say how many years it
was since a woman was on the team?[/quote]

I'm not sure, can't remember. Though they said that only one girl qualified to the selection camp this year through the main Chinese olympiad.

Is is OK with the problem 7? If p and q have different parity, we may always replace them by 2p and 2q respectively. So this condition does not seem to be essential. Maybe, p and q are of different parity?


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integration

As anyone can clearly see from this Careers in Mathematics Forum, there are plenty of job opportunities available to smart, math-loving people outside of academia.  However, what about jobs in academia?  Does anyone have any information about life as a mathematics professor?  How does it compare to other mathematical careers?

[quote="bubala"] Does anyone have any information about life as a mathematics professor?  How does it compare to other mathematical careers?[/quote]

The answer to your first question is Yes.  (I am a math professor, so I can certainly provide my own perspective, and I keep tabs on the job market from year to year as well.)  I can't say much about how it compares to other math careers, because I haven't had any; it'd all be guesswork on my part.

Could you ask some more specific questions?  Then academics and nonacademics can post responses from which you can make the comparisons.

Sorry I wasn't specific enough, but I don't really have any specific questions.

Can you just generally describe what being a professor is like?  How hard do professors work, for example?

I was on the academic career path for a while (Ph.D, postdoc; I left academia in 2001).  There are basically two main aspects of your job: teaching and research.  For me personally, I enjoyed the teaching (which is why I'm here at AoPS!) but never really blossomed on the research side.

Some of the plusses of academia:
- for the most part you have the freedom to work on whatever you want to work on
- except for when you have to teach, you can keep your own hours
- you often get summers off (this is by far the part I miss the most!)
- lots of opportunity to travel if you want to
- once you get tenure, you have job security

Some of the minuses:
- pay is pretty low to start, and never really that great (compared to non-academic careers)
- the entry-level job market is brutal
- there's a fair amount of early-career pressure: "publish or perish"
- academic politics are horrible
- many people don't thrive very well in a directionless environment: since there's nobody telling you what to do, you have to be highly self-motivated

Most of these points (plus and minus) are pretty minor though when compared to the big question: Is doing mathematics research and/or teaching high-level mathematics something that you'd enjoy doing?

If you'd like to get some idea of the life of a professional academic mathematician, I would recommend the book [i]I Want to Be a Mathematician[/i] by Paul Halmos.  It is basically Halmos's autobiography (or, as he calls it, "automathography"), from his childhood in Hungary, through college, graduate school, and his multi-decade career as a professor.  Although parts are a bit dated (he went to college in the 1930s if I recall correctly, and a lot has changed since then), it does give an excellent flavor of what it's like to pursue a career in academia.

Something else I'll add is that I always thought I wanted to be a mathematics professor until I got to college.  It fits some personalities and not others.

Also, your interests might change over time.  By the time you are done with your undergraduate work, there will probably be a dozen possible career paths that don't even exist yet.  Not to mention that many math people find computer science more appealing at that level.  It's all matter of interests that do indeed change.

Halmos' career largely coincided with the old days of relatively plentiful jobs and Sputnik funding.
His memoir accordingly omits the Darwinian pressures that are at the forefront these days.

[quote="bubala"]
Can you just generally describe what being a professor is like?  How hard do professors work, for example?[/quote]

Probably the best answer here is "it depends."

There are three major components to a professor's job: teaching, research, and service.  There are academic jobs out there with whatever mix of those three components you'd like.  However, the prestige, pay, etc. for these jobs varies with that mix.  Other things that vary are the number of students per class and the type of support you receive for each of the activities you do.  By the way, "service" refers to everything from committee work with in your department to refereeing papers for a journal.  The definitions of teaching, research, and service vary from institution to institution.

It's a partial myth that faculty have summers off.  What's true is that few faculty are required to teach during the summer.  However, most of us are expected to do research during the summer because we don't have enough time to do it during the academic year.

The bulk of the faculty I know who like their jobs work REALLY hard.  Some people don't, either because they don't need to or because they don't want to.

Some specifics, just from my job as an example: I have complete control over my schedule, and near complete control over the courses I teach (subject to the condition that at most 1/3 of the courses can be for majors).  So on MWF, I'll teach from 8:30 - 11:30, three classes back-to-back.  One will be an upper-level course and the other two will either be watered-down calculus or discovery-based courses of my own design in super-elementary graph theory, knot theory, 4-d geometry, etc. Each class has 10 - 20 students. The remainder of the morning and all of lunchtime is spent preparing for the next round of classes.  I'll spend the afternoon grading and in meetings and maybe I'll get to see students a little bit.  A bunch of time will be spent in answering student emails and such.  Often the evenings are spent on various service activities, like preparing for meetings or doing assignmnents I've been given from meetings.  On Tuesday mornings I try to think about the long-range activities of my classes, and whether I should be doing things differently, or whether there are certain students who are struggling, etc.  Often this is pre-empted by students coming by to talk about research projects or to work on independent studies... colleagues come by to discuss departmental or college matters (one of the joys of my job is when a colleague from biology stops by and tells me stuff about the chemical structure of membranes, or metabolic processes or whatever, and I get to learn something!) and in the afternoon I generally hang out with students who need help with homework and such.
Thursday mornings I try to do research, but often there's yet more grading to do or assignments to prepare or, despite the fact that everyone knows it's my research day and my door is closed, colleagues come in to talk to me... Thursday afternoon is sometimes meetings, and often I actually get to do research then.

But that's just me, and that's just the weekdays!  I've had two permanent jobs, and they were very different.    Each of my departmental colleagues has a different schedule, and across the profession it all varies widely.  Again, the above is just ONE example.  Do not generalize.

[quote="smbelcas"]The bulk of the faculty I know who like their jobs work REALLY hard. Some people don't, either because they don't need to or because they don't want to. [/quote]
That's a good quote. Many of the details of my life differ from the details that smbelcas provided, but she still captures the flavor of it pretty well. I found my way to an institution - an "urban comprehensive university" - which does value all three legs of the teaching, research, service triad with more than lip service, and perhaps teaching most of all. The latter part of my career has been devoted mostly to teaching and service - among other things, a lot of academic advising.

[quote="DPatrick"]- the entry-level job market is brutal [/quote]
In my case, I got my first - and last - "tenure track" (meaning potentially permanent) job a full 8 years after I had earned my doctorate, when I was already in my mid-30's. And at that, I felt lucky. The last time we advertised for a tenure track job opening in "pure" mathematics, we had 200 applicants. I can assure you that there were candidates with very attractive-looking resumes who never even got a phone interview.

Don't delude yourself into thinking that your communication skills, oral and written, don't matter, that all you need to do is solve mathematical problems. Everyone in that pool of candidates could solve mathematical problems. Communication skills played a very large part in choosing whom we hired.

I should also mention that it's an inflexible career path. Changing jobs is as hard as getting one in the first place, and you may have little choice about what part of the country you'll live in. Every career decision requires very long lead times and can only happen at certain times of the year.

[b]smbelcas [/b]is she, a female math professor.

Isn't pure theoretical research in mathematics a very hard thing to do, since it is already a very advanced field?  Or maybe I'm completely flawed in my conception of "research".

Yes, it is hard, but it can be done. You can spend a long time working on a problem you find interesting until you (hopefully) solve it. One thing a professor told me is that if you're doing research, you're much better off coming up with your own problems and solving them than trying to work on someone else's problems. In the latter case, people will already have worked on them, so they will tend to be much more difficult.

I'm a belgian student in mathematics, now soon beginning his third year (out of four)
so this year I have to start thinking about a thesis subject

so uhm, how does 'research' like that go
my father told me : you have to find something and you're the first to do so
is that correct?
doesn't sound too easy !

so how does that go? do you have a lot of freedom in that? do you get help/hints from professors/assistants?
how new is your thesis?

by the way I study pure mathematics, computer science and stuff is not really my thing so it certainly won't be
as a humble student now, I find number theory and geometry very interesting

[quote="bubala"]However, what about jobs in academia?  Does anyone have any information about life as a mathematics professor?  [/quote]

In France you can do after your baccalaureat

1/ Two years preparatories classes MPSI and MP* in
Louis Le Grand high school or Henri IV high school in Paris

2/ Pass Concours Grandes Ecoles ENS Ecole Normale Supérieure
study fours years,  pass Agrégation Externe Mathématiques with a rank between 1 and 20, complete with a phD in pures mathematics

3/ Apply to the office of Inspection Générale Mathématiques to get a job
such as professeur de mathématiques en Classes Préparatoires MPSI

[quote="fredbel6"]...so this year I have to start thinking about a thesis subject
...so how does that go? do you have a lot of freedom in that? do you get help/hints from professors/assistants?
how new is your thesis?
[/quote]

This all depends heavily on your institution.  At most places, an undergraduate thesis is not required to be original.  (At many places, it's rare for undergraduates to write original theses.)  Degree of freedom in choosing a topic, amount of help one gets in choosing a topic and/or in working on it, etc. depend not only on the institution but on the faculty member.

My advice: pick your favorite professor and ask him/her these questions, as s/he will probably know the lay of the land at your institution and within the department.

What do you all think about teaching in a community college?  It seems like it would be a fairly easy and enjoyable job that pays pretty well (I've heard you can get 80 or 90 thousand dollars a year in California with a Ph.D), and has good vacations, and also leaves a lot of time and energy for any creative pursuits you might have.  But then again, maybe I don't know what it's really like.

You don't have to have a Ph.D. to teach at a community college. Such colleges hire many faculty members with masters degrees, and it's not clear that the Ph.D. gives that much of an advantage over the M.S. in the hiring process.

The teaching load for a full-time faculty member at a community college tends to be pretty heavy, since teaching is the bulk of the job.

What mathematics classes do community colleges teach? It's a pyramid, with the most sections at the lowest levels. At many urban community colleges, a majority of the sections taught are in classes at a level too low to transfer for college credit: Intermediate Algebra, Geometry, Elementary Algebra, even Pre-Algebra. And even among the classes that do transfer, such classes as College Algebra, Finite Mathematics (for prospective business majors), "math for poets," Business Calculus, and so on, are more common than the Calculus sequence that science and engineering majors would take.

Most community colleges hire a large number of part-time faculty members. It's fairly common for those part-timers to work simultaneously at more than one institution; they become known as "freeway fliers." The permanent, full-time faculty members probably get more choice about what they teach than do the part-timers. How often do permanent jobs open up? In that, the community colleges go through boom-and-bust, feast-or-famine cycles closely tied to the budgetary status of the governments that support them.

[quote]... and also leaves a lot of time and energy for any creative pursuits you might have.[/quote]
The ones I know don't seem to have all that much excess time and energy. But then, the ones I know also tend to belong to the "freeway flier" category.

[quote="The Mathematician"]What do you all think about teaching in a community college?  It seems like it would be a fairly easy and enjoyable job... leaves a lot of time and energy for any creative pursuits you might have.[/quote]

My mom teaches at a community college (in the Midwest), and has been there as a full-time masters-level employee for a couple of decades.  Her teaching load is more comparable to that of a high-school teacher than a college professor---as Kent says, it's quite high---and she never feels like she has enough time to do a good job.  I think she feels that it's an easy job compared to that of a college professor, and she enjoys it; however, she is paid for exactly 40 hours per week and always has to put in more time than that in order to get her job done.  She has time and energy for relaxing (non-intensive) creative pursuits, but it's certainly not a lot of time.

I'm a college professor and business school administrator at a small, public, regional university.  Many of the professors at our school derive the greatest job satisfaction from the accomplishments of our graduates rather than from research or other activities.  Applied research is typically required at smaller regional schools, but mainly to inform teaching.

Generally, our professors tend to get more respect and enjoy their work when they teach as an integral part of a major program such as math, business, computer science etc.  If you become a college professor, try to find an employer that has a successful major program in your field of study.

Even though we have a small undergraduate math program at my school, many of the math professors teach primarily service courses to non-majors.  Like many regional state schools, we're not too selective in admissions.   Therefore, the students in the service classes tend to be relatively unmotivated with serious secondary school mathematics deficiencies.  Successful, happy math professors at my institution view this as a challenge, find ways to motivate these students and feel good about improving the numerical literacy of average college student.  Of course, not everyone is successful and happy.

If teaching unmotivated students isn't for you, then you'll need to find employment at a research institution or at a highly selective teaching school - jobs that can be very difficult to obtain.

Sembelcas and Kent Merryfield have some job perks that aren't available to everyone.  Our professors have some input into their teaching schedules but ultimately, administrators determine when classes are scheduled.  We have night programs at our school and most professors must teach at least one evening course per year.  This can be difficult for those with children and family responsibilities.

Overall, most faculties control of a substantial amount of their time and are given leeway (not much "direction") in what they do with this time.  Many are drawn to the profession by this freedom.  However, as others have attested, it's not for everyone.

For the most part, college professors are in a service profession - the satisfaction derived from the job comes from helping others learn and developing academic programs that meet the needs of society.  A very select few will truly push back the frontiers of knowledge.  As Ph.D. students, we were trained to believe that job satisfaction would come from research accomplishments - ultimately, this probably leads to some dissatisfaction among professors with the realities of most college teaching environments.

For those who can be happy teaching and influencing the lives of many aspiring young people, being a college professor at a regional institution might be a good match.  If you're not sure, build some other skill sets (programming, ...) that you use to develop alternate career paths.

I am a math professor at a community college and what Kent says about the job is correct.

The pay can be competitive with 4-year colleges and reach 6 figures (the highest salaries are in California).  Full time faculty typically teach 15 hours a week, plus office hours, preparation and grading.  My college is on a semester system with 16 weeks of classes followed by a week of finals.  There is no obligation to do research and scholarship, and neither does tenure or promotion depend on these activities.  There is a significant collegial governance component, which takes you outside your department.  The minimum qualification for the job is a master's in math, though 25% of the full-time faculty in my department have doctorates (all in math).

We teach the lower division curriculum, plus lots of remedial students.  Student success rates are low - 57% in our department, which is actually several percentage points above the statewide average for math students at community college, but much lower than any other department on campus.  Many of our math students don't like math and are only taking it because it is a requirement, which is probably true at any level above 6th grade.

The full-time math professors in my department love their jobs are are dedicated to their students.  The department has always been well run, and therefore has little interference from administrators, and the faculty have an excellent relationship with the administrators.  This is not universally true at other colleges.

Being a part-time teacher, and some are freeway flyers as Kent mentioned, is a far less attractive and well-paid prospect, but departments rely heavily on these teachers.  Also, nearly all the full-timers had to start out as part-timers for a few years in order to get the experience needed to become competitive for the full-time jobs.

So if you like teaching and like math, community college is a good career choice.  You can still get the freedom that 4-year-college math faculty enjoy.

I can post more specific details about community colleges if anyone wants.

[quote="xfactor"]I can post more specific details about community colleges if anyone wants.[/quote]

Thanks for your interesting account. Please go ahead! We are always interested to get to know the various career options one may pursue.


complex analysis
complex analysis solved

Determine explicitly the largest disc about the origin whose image under the mapping w=z^2 + z is one to one.

Appreciate ur help

Let $f(z)=z^2+z$. Then $f(w)=f(z)\implies w^2+w=z^2+z\implies (w-z)(w+z)=z-w\implies w+z=-1$ (implications are if and only if) so that for such $w,z$, $|w|+|z|\le |w+z|=1$. It follows that, $R\ge 1/2$. If you have a circle of radius $>1/2$ it's not hard to find $w\neq z$ such that $f(w)=f(z)$.


Sorry because my language isn't English but I think you can understand it easily!

[img]http://i227.photobucket.com/albums/dd72/TPFC/vd11nl8-1.png[/img]

I must have seen this somewhere else on the forum before:

[hide="Brute Force"]

It's equivalent to proving that

$ \sum_{cyc.} a^{5}bc^{2} \plus{} \sum_{cyc.}a^{5}b^{0}c^{3} \plus{}2( \sum_{cyc.}a^{4}b^{3}c^{2} \plus{} \sum_{cyc.}a^{3}b^{2}c^{2} \plus{} \sum_{cyc.}a^{3}b^{3}c^{3}) \geq 9\sum_{cyc.}a^{4}b^{2}c^{2}$

Which is true by AM-GM Inequality.[/hide]

Yes,actually it can be done with am-gm only,just square both side. :wink:
[hide="To kim77"]
Where are you from?[/hide]

[quote="BanishedTraitor"]Also you can find stronger inequalities on the forum I think.[/quote]
Yeah, and here is it
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\left( \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\right)^{2/3}\]
:)


calculus
derivative
inequalities proposed
inequalities

Let $a;b;c;d \in R$ such that:$ad-bc=1$.Find min
$A=a^{2}+b^{2}+c^{2}+d^{2}+ac+bd$

$A\geq 2\sqrt{x^{2}+1}+x$, where $x=ac+bd.$

The answer is min $A=\sqrt{3}$ with equality $x=\frac{1}{\sqrt{3}}.$
I don't use differentiation or discrminant. :D

[hide]
Since $1+(ac+bd)^{2}=(ad-bc)^{2}+(ac+bd)^{2}=(a^{2}+b^{2})(c^{2}+d^{2})$, we have $a^{2}+b^{2}+c^{2}+d^{2}\geq 2\sqrt{1+t^{2}}$ where $t=ac+bd$.

Now we must find minimum of $2\sqrt{1+t^{2}}+t$. Letting $t=-u$, it becomes $2\sqrt{1+u^{2}}-u$. But Cauchy gives
\[\sqrt{(1+u^{2})(3+1)}\geq \sqrt{3}+u \]
Therefore, the lower bound is $\boxed{\sqrt{3}}$, and one instance where it's attained is
\[a=\frac{\sqrt{2}}{\sqrt[4]{3}},\ b=0, \]

\[c=-\frac{1}{\sqrt{2}\cdot \sqrt[4]{3}},\ d=\frac{\sqrt[4]{3}}{\sqrt{2}}\]
[/hide]

To be honest, first I imagined the use of Cachy-Shwarz because the sum of the squared of real numbers implies the squared of a distance, but I couldn't solve using this. Anyway nice solution! scorpious119. :lol:


Suppose we want to divide the 10 dogs into three groups, one with 3 dogs, one with 5 dogs, and one with 2 dogs.  How many ways can we form the groups such that Fluffy is in the 3-dog group and Nipper is in the 5-dog group?

What does (8) x (6) = 480 mean?
                4      4

We place Fluffy and Nipper in their respective groups. Thus, we now have $ \dbinom{8}{2}\cdot\dbinom{6}{4}\cdot\dbinom{2}{2} \equal{} (28)(15) \equal{} 420$.

chubbyg--what the heck?

[quote=econaxis]Why can't it be $\binom82\cdot\binom84\cdot\binom82$[/quote]

Note that after the first choice we have took away $2$ dogs.

[quote=V_V]why doesn't this work:

put fluffy in 3 dog group

(9 choose 2) = 36

put nipper in 5 dog group
7 dogs remaining but subtract 1 for nipper

(6choose 4) = 15

36* 15 =540
(i left the implied 2 choose 2 out)[/quote]

The issue is that when you have $\binom{9}{2}$, you could have Nipper be chosen for the 3-dog group too. You have to place both dogs first, then choose the others.

[hide=sol]place fluffy and nipper in their respective groups by default
then we can count the ways to place the rest to be 8c2*6c2*4c4=420[/hide]

omg the answer is 420


I am going to ISEF in Arizona . I will be in mathematics, who else is going and in which catergory  ;) ??

That's funny.  We don't even have our district fair until tomorrow,  but I have a great project this year.  I'll probably be going in computer science.

i have a friend who is going, his name is Donald Eng he's gonna be in Engineering... Manuel, could you explain your project to me? I was going to do a mathematics project this year, but or science fair coordinator said they didnt want to put a mathematics category if there was only gonna be one for regionals... So, I got stuck in physics and i barely made it to state fair....  ;)

I'm interested in attending - but this year's too late. Is a mentor virtually required?

my project is about combinatorial geometry. It is about structures determined by inifinite configurations in n-dimensional euclidian space.

[quote="PenguinMath"]Explain your project if you don't mind.  I'm interested in doing Comp Sci next year.[/quote]
Sorry I haven't replied for a while.  I've been to busy working on my project.  I researched a relatively new form of programming called genetic algorithms that model biological evolution to solve a problem.  These algorithms, along with a genetic program are used to make things like autopilot systems and speech recognition software.  I made one of these algorithms and tested thousands of different coding arrangements (programmatically of course) to determine the most efficient arrangement.  I was able to make a program to solve the Traveling Salesman problem for 30 cities in less than a minute! Do some research into it yourself.  It's all very interesting.
[quote="manuel"]Hope to see you there diophantus![/quote]
Hope to see you there too, but I think it is likely we wouldn't recognize each other.

It's interesting that you were able to obtain quick results with your algorithm; I thought that one of the disadvantages of using genetic algorithms was that they were relatively slow.

edit: I looked into ga a bit and saw that the tsp is one of its applications; I'll have to look into ga a little.

did anybody go to isef in the  math category?


conics
parabola
geometry unsolved
geometry

Consider the tangency point as allways poin E.Choose a fifth point on the parabola.Then the points are marked A,B,C,D and E counterclockwise.So the metod is:
P=ABxDE
Q=AExCD
X=BCxPQ

The line XE is he answer

x is intersection

the parabola line is not a active line so i can't say E point on this parabola line without construct the point E with the other 4 points A, B, C, D

 :(

OK.I THOUGHT THAT THE LINE EXIST.I'LL THINK ANOTHER SOLUTION


limit
trigonometry
real analysis
real analysis solved

a) prove that the ecuation $x=\tan(x)$ has only one solution on every interval of the form: $((2k-1)\frac{\pi}{2},(2k+1)\frac{\pi}{2})$, with k $\in$ Z.
b) we denote with $x_{n}$ the unique solution on the interval corespunding to k=n. Show that:
$\displaystyle \lim_{x\to \infty} n((2n+1)\frac{\pi}{2}-x_{n})=\frac{1}{\pi}$

hi moubi... your solution is pretty nice but this problem was in one of the tests for entrance in an university (Cluj-computer science) and it should have a more elementary solution (this one is too advanced for the level of the contest)
thx

Hi Kuba,

This problem is classical one in preparatories classes
" Classes Prparatoires Mathmatiques Suprieures MPSI "
It is also a classical problem for Mathematic Oral Examination, Grandes Ecoles
ENS (Ecoles Normales Suprieures), X Ecole Polytechnique, Mines-Ponts,
Ecole Centrale.

I wanted to post an elementary solution for this one (ohne Landau) but instead of pressing submit I closed the IE window.. can anyone be more brilliant than that? :D

If anyone really wants to see the solution which I found, I will post it next week.

 :lol:

Now I can post:

Let $a_{n}=x_{n}-n\pi$, thus we have $\tan a_{n} = \tan x_{n}= x_{n} = a_{n} + n\pi$ and $\displaystyle a_{n} \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Using all these we compute out limit: \[\displaystyle lim_{n \to \infty} n[(2n+1) \frac{\pi}{2}-x_{n}] =lim_{n \to \infty} n( \frac{\pi}{2}-a_{n})= lim_{n \to \infty} \frac{\frac{\pi}{2}-a_{n}}{\tan( \frac{\pi}{2}-a_{n})} \cdot lim_{n \to \infty}  [n \tan(\frac{\pi}{2}-a_{n})]=lim_{n \to \infty}  n \tan (\frac{\pi}{2}-a_{n})= lim_{n \to \infty} \frac{n}{ \tan a_{n}}= lim_{n \to \infty} \frac{n}{a_{n}+n\pi}=\frac{1}{\pi} \]

I hope it is okay!

cheers! ;)


I'm enrolled in Olympiad Inequalities now and was wondering when my class page would show up for that forum.

Edit: Also, will you be offering Intermediate counting and Algebra  simultaneously from April to June?

The Message Board doesn't show up until the instructor cycles the usergroup.  Since you enrolled today, I had not yet cycled the usergroup since you enrolled.  You should now see the Message Board.

We are uncertain as of yet when we will offer the next Intermediate classes, but currently we have no plans to begin any classes between January and around June except the AIME Problem Seminar.

We would consider more classes only if demand seemed to warrant them.


inequalities
calculus
derivative
inequalities unsolved

Is the inequality flipped? Also, what is IMAR? (I think it is a real inequality since equality holds where expected, it is just a typo on your part :lol: )

If a+b+c=1, where  a,b,c>0,then inequality

   3*(a*b+b*c+c*a)*((a+b+c)^3+(48-t)abc]>=(75-t)abc

holds ,if and only if  t>=0.

If a+b+c=1, where a,b,c>0,then inequality

3(ab+bc+ca)[1+(48-t)abc]>=(75-t)abc

holds ,if and only if t>=0.

We prove that $(ab+bc+ca)(1+48abc) \geq 25abc$ if $a,b,c \geq 0$ and $a+b+c=1.$

Assume WLOG that $a\leq b\leq c,$ and let $a=u-v,$ $b=u+v,$ $c=1-2u,$ with $0 \leq u,v \leq 1/3.$  Then

$(ab+bc+ca)(1+48abc)-25abc= \newline 48v^{4}(1-2u)-2v^{2}(96u^{3}-144u^{2}+73u-12)+2u(144u^{4}-168u^{3}+73u^{3}-14u+1).$

We show that this quantity decreases as $v$ decreases to $0,$ implying that its minimum occurs at $v=0,$ when $x=y.$ Its derivative with respect to $v$ is $-4v\left(48v^{2}(2u-1)+(96u^{3}-144u^{2}+73u-12)\right).$  Clearly $2u-1 \leq 0.$  We write

$96u^{3}-144u^{2}+73u-12= \newline-1/9-32(1-3u)\left( (u-7/12)^{2}+1/32 \right),$

which is negative since $u \leq 1/3.$  Thus the derivative is non-negative, and the minimum of the expression for $0 \leq v \leq 1/3$ occurs when $v=0.$

Now let $a=u,$ $b=u,$ $c=1-2u,$ so that $(ab+bc+ca)(1+48abc)-25abc=2u(144u^{4}-168u^{3}+73u^{2}-14u+1)=2u(1-3u)^{2}(1-4u)^{2}\geq 0,$

and we are done.

[quote="arqady"][quote="pohoatza"]Let $a,b,c$ pozitive real numbers with :  $a+b+c=1$ . Prove the following inequality :

$3(ab+bc+ca)(1+52abc)\geq 79abc$[/quote]
It's wrong.
But the following inequality is true and nice:
Let $a,$ $b$ and $c$ are positive  numbers such that  $a+b+c=1.$
Prove that $(ab+bc+ca)(1+48abc)\geq 25abc.$ :wink:[/quote]

yes, arqady , you were right, the best case is for $48$.

Let a,b,c  are positive numbers such that a+b+c=1,
Prove that  3(ab+bc+ca)(1+48abc)>=75abc.


[hide]$x-\sqrt{x}-2=0$
$\sqrt{x}=x-2$
$x=(x-2)^{2}$
$x=x^{2}-4x+4$
$x^{2}-5x+4=0$
$(x-4)(x-1)=0$
$x$ cannot be $1$ as this is rejected when it is put into the original equations.
Therefore $x=4$[/hide]

[hide]Let $\sqrt{x}=a$ so you have $a^{2}-a-2=0$ thus you get that $a=2,-1$ but square roots can't be negative so $a=2$. Now $\sqrt{x}=2$ and therefore $\boxed{x=4}$.[/hide]


is this possible to partetion $ \mathbb{R}^2$ into 3 or more  connected dense sets?

we can  partition the plane into $ c$  arcwise connected dense sets,
and it is not possible to  partition the plane into countable number $ n\geq 3$  arcwise connected dense sets. :wink:


algorithm
number theory
greatest common divisor
Euclidean algorithm
number theory unsolved

Is there any definite algorithm for determining the inverse , please illustrate by finding the inverse of $ 23(mod100)$.

Yes, there is!

Basically you do the Euclidean algorithm to compute the gcd of 23 and 100, and keep track of some extra metadata,
writing 23 = 1*23 + 0*100 and 100 = 0*23 + 1*100 to start -- you keep track of these extra coefficients as you go
to solve a*23 + b*100 = 1, then a*23 == 1 (mod 100).

A cute way of doing this by hand:

Use the Euclidean algorithm to find the gcd of 23 and 100:

100 = 23*4 + 8
23 = 8*2 + 7
8 = 7*1 + 1
7 = 1*7 + 0

Now look at partial quotients (7,1,2,4).

Make a table

$ \begin{array}{rrrrrrrrr} & & 7 & 1 & 2 & 4 \\
0 & 1 & 7 & 8 & 23 & 100 \\
1 & 0 & 1 & 1 & 3 & 13 \end{array}$

Where you start with
0 1
1 0

and obtain the next column by adding a multiple of the previous column to the second to last column gotten so far.

Then 3*100 - 13*23 = 1. So (-13)*23 == 1 (mod 100).

Basically, I need the algorithm for solving system of congruences using the Chinese Remainder Theorem.

Thanks a lot.....I see the algorithm uses Bezout's theorem as well.

I used to find the inverse using the $ \phi$ function...this would come in handy....

A question: can we similarly use the algorithm to express the GCD of two numbers as a linear combination of the numbers..?What modifications should be made then?

Well lets look at 6 and 10, for example; do Euclid on that:

10 = 6*1 + 4
6 = 4*1 + 2
4 = 2*2 + 0  DONE!

Then
$ \begin{array}{rrrrrrrrrr} & & 2& 1& 1 \\
0 & 1 & 2 & 3 & 5 \\
1 & 0 & 1 & 1 & 2. \end{array}$

Notice we end in (3,5) now instead of (6,10), because this is what is left when dividing out by the GCD (2). Then we get 3*2 - 5*1 = 1; bringing the 2 back into the picture gives 6*2 - 10*1 = 2. The last two coefficients are again on that bottom line.


[b]A.)[/b]If Today is Monday, January 1 2009.  Then what day of the week will be $ 2{}^9{}^8{}^3{}^3$ days from now??

#3 What's a through?
[hide="#1"]
$ 2^{9833}\equal{}(2^3)^{3277}\cdot 4\equiv 4 (mod 7)$
So, it will be Friday[/hide]

[quote="Bugi"]#3 What's a through?
[hide="#1"]
$ 2^{9833} \equal{} (2^3)^{3277}\cdot 4\equiv 4 (mod 7)$
So, it will be Friday[/hide][/quote]
A trough I think you mean...
A trough is a basin.

A trough is a long, narrow, generally shallow receptacle for holding water.

[quote="maximos"][b]B.)[/b] A tank contains 100 gallons of water. Brine having a concentration of $ A$ pounds of salt per gallon of solution enters the tank at the rate of 3 gal/min. The mixture leaves the tank at a rate of 2 gal/min. If the the concentration of the solution in the tank is to be 2 lbs/gal at the end of 20 minutes, find $ A$. "Assume instantaneous mixing".

[b]C.)[/b] A triangular trough is 5 meters long, 2 meters wide at the top and 2 meters deep. If industrial strength acid is flowing at a rate of 3 cubic meters per minute. How fast is the area of the surface of the liquid increasing when the acid is 90 centimeters deep??[/quote]
C is a calculus problem if I'm not mistaken.
As for B:
There will be $ (3\minus{}2)(20) \plus{}100 \equal{} 120$ gallons at the end of mixing. We thus need $ (2)(120) \equal{} 240$ lbs of salt in the tank at the end. Since $ \frac {2}{3}$ of the salt will exit the tank, there needs to be $ \frac {(240)(3)}{2} \equal{} 360$ lbs of salt total entering the tank. If 2 gallons enter the tank each minute and there are 20 minutes then 40 total gallons are going to enter the tank. $ \frac {360}{40} \equal{} 9$ lbs per gallon is the concentration of the salt.

Ooppsss... I placed problems [b]B[/b] and [b]C[/b] on a wrong section....
They've been deleted already though...

Anyway Bugi ....That's probably the shortest way...
I'll Show another way which might be a longer way in finding for the answer. But, without using Modulus arithmetic...

[hide="Click to show content."]
First: Find a pattern/expression for $ 2^n$ that can be a multiple of $ 7$(so that it will be easier to know what day will it be if a number of days exceeds a week).

Notice,

$ 2^3 = 8 ; 8 - 1 = 7$
$ 2^6 = 64 ; 64 - 1 = 63$
$ 2^9 = 512 ; 512 - 1 = 511$

Implies that $ 2{}^3{}^n - 1$ is a multiple of $ 7$.

Therefore, $ 2{}^3{}^n - 1 + 1 = 2{}^3{}^n$ is one day after $ 7$ days. Since, $ 9833$ is not divisible by $ 3$, look for the highest number less than $ 9833$ that is divisible by $ 3$. That number is $ 9831$.

$ 2{}^3{}^n = 2{}^9{}^8{}^3{}^3 = 2^3^({}^3{}^2{}^7{}^7^)^ + ^ 2 = 2{}^9{}^8{}^3{}^1 \cdot 2^2 = 2{}^9{}^8{}^3{}^1 \cdot 4$

And since $ 2{}^3{}^n$ is one more than a multiple of $ 7$, If you multiply $ 4$ to it, we have:

$ 4(7k + 1) = 28k + 4 = 4(7k) + 4$

So, $ 2{}^9{}^8{}^3{}^3$ days after a Monday is $ 4$ days after another Monday.

And that day is $ \boxed{\text{Friday}}$

[/hide]


let us choose arbitarily n vertices of regular-2n-gon and color them red. the remainig vertices are colored blue . we arrange all red-red distances into a non- decreasing sequence and do the same for the blue-blue distances. prove that the sequences are equal

we could solve this problem by using graph theory and ramsey's numbers R(p,q).

? I will be interested to see how you put Ramsey numbers here....

Pierre.


inequalities
algebra
polynomial
inequalities unsolved

prove that forall $a,b,c\in(0,1]$ such that $a^{2}+b^{2}+c^{2}=2$ there is
$\frac{1-a^{2}}{c}+\frac{1-b^{2}}{a}+\frac{1-c^{2}}{b}\leqslant\frac{5}{4}$
actually take $x=\sqrt{1-a^{2}},y=\sqrt{1-b^{2}},z=\sqrt{1-c^{2}}$ we have
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leqslant\frac{5}{4}$
forall $x,y,z\geqslant0$ such that $x+y+z=1$. The equality holds when $x=1/4,y=0,z=3/4$
This is an old problem (but I don't know the answer and failed in the exam today). Could anybody tell me how to do it?
And there is another question: forall $x,y,z,w\geqslant0$ such that $x+y+z+w=1$ there is
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+w}}+\frac{w}{\sqrt{w+x}}<\frac{3}{2}$
In addition, is there
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+w}}+\frac{w}{\sqrt{w+x}}\leqslant\sqrt{2}$
If this is right, the equality holds when $x+y=y+z=z+w=w+x=1/2$

[quote="shalex"]
And there is another question: forall $x,y,z,w\geqslant0$ such that $x+y+z+w=1$ there is
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+w}}+\frac{w}{\sqrt{w+x}}<\frac{3}{2}$
[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98684

[quote="arkhammedos"][quote]actually take $x=\sqrt{1-a^{2}},y=\sqrt{1-b^{2}},z=\sqrt{1-c^{2}}$ [/quote] i think that $x=1-a^{2},y=1-b^{2},z=1-c^{2}$.in order to have [quote]$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leqslant\frac{5}{4}$[/quote][/quote]
Yeah you're right.

in
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=98684[/url]
there is
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}+\frac{w}{\sqrt{w+x}}<\frac{3}{2}$
and take
$x=0,y=1/2,z=1/3,w=1/6$
the inequality
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}+\frac{w}{\sqrt{w+x}}\leqslant\sqrt{2}$
is wrong. So what's the value of
$\sup\left\{\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}+\frac{w}{\sqrt{w+x}}\right\}$

[quote="shalex"]$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leqslant\frac{5}{4}$
forall $x,y,z\geqslant0$ such that $x+y+z=1$[/quote]

Could someone find a solution to this one using Iura's method ?(the one at the link given by Arqady )

Thank you very much. We use Pascal and get similar stuff.

But oh what a large polynomial!

[quote="shalex"]prove that forall $ a,b,c\in(0,1]$ such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 2$ there is
$ \frac {1 \minus{} a^{2}}{c} \plus{} \frac {1 \minus{} b^{2}}{a} \plus{} \frac {1 \minus{} c^{2}}{b}\leqslant\frac {5}{4}$
actually take $ x \equal{} \sqrt {1 \minus{} a^{2}},y \equal{} \sqrt {1 \minus{} b^{2}},z \equal{} \sqrt {1 \minus{} c^{2}}$ we have
$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leqslant\frac {5}{4}$
forall $ x,y,z\geqslant0$ such that $ x \plus{} y \plus{} z \equal{} 1$. The equality holds when $ x \equal{} 1/4,y \equal{} 0,z \equal{} 3/4$
This is an old problem (but I don't know the answer and failed in the exam today). Could anybody tell me how to do it?
[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152848]See my solution here[/url] :)

I saw your proof from pvthuan's book. It's really a nice proof (though by some calculations, I think it's simply enough). Cheers! :D