[hide]1, because 42 is a multiple of 7.[/hide]

[quote="rd5493"]yeah, since if u multiply anything by 42 it becomes a multiple of 7, so obviously the answer is 0.  come on, that was too easy.[/quote]

I'm sorry that my problems that can evidently be solved by not-so-smart people are bouncing off your large head.

If you think all these problems are too easy go to a different forum

i dont think that these are too easy, but if i can get them (you know me being the not einstein that i am) then its probably not that hard.  im not what u would call einstein

[quote="13375P34K43V312"]So basically in a nutshell here's what you're doing

"that was too easy"
"i don't think this is too easy"

If someone else doesn't get it they're going to feel discouraged that they can't get a problem that's "too easy"[/quote]

sorry, as i review what i said, i sound like a [cannot say this on forum]
what i meant was that i was surprised at how quick i could get it and that it made me feel smart.  sorry for sounding like a jerk

[quote="13375P34K43V312"]A number is multiplied by 42 and converted to base 7. What is the probability that it ends in a zero base 7?[/quote]

Since there's already a solution, and this topic has gone way off-topic, complaining the question is too easy or too hard, let me offer some suggestions on how to improve the question.

First of all, most really good MATHCOUNTS questions have many strategies to a solution.  The most advanced students might find a strategy (say an obscure theorem) that yields the answer instantly.  A somewhat less advanced student might find a strategy (say algebra) that yields the answer, but takes a little time.  An even less advanced student might find a strategy (say guess-and-check) that yields the answer, but takes a long time.

Students who know more mathematics and more advanced strategies are rewarded with extra time to work on other problems.

Let's consider LeetSpeak's problem.  When you refer to "a number", you are inviting an immediate generalization.  [They may assume you mean "a randomly selected positive integer".]  Those who see the generalization will say "this is easy".  Those who don't see the generalization probably can't solve it.  There's not an inviting middle-ground for those who don't see the problem.  (I.e., they might try a bunch of numbers, but it's not clear how many numbers they will need to try, and it's not very inviting.)  Others, who are good at "test-taking strategies", might note the selection of "a number" is not well-defined, so the "probability" is bound to be zero or one.

So, here's a suggested rewording:
[quote]1.  The number $15_{10}$ is multiplied by $14_{10}.$  If the product is converted to base $6,$  What is the units "digit" of the base-6 product?[/quote]

A slightly harder variant:
[quote]2.  The number $30_{5}$ is multiplied by $24_{5}.$  If the product is converted to base $6,$  What is the units "digit" of the base-6 product?[/quote]

For these problems, if you don't know what "base-6" means, you probably can't solve them.  So, this question would be intended to test if you understand base-6.

For number 1, the most astute readers might factor the multiplicand and the multiplier.  Seeing the product contains a multiple of 6, and knowing that base-6 numbers that end with zero are multiples of 6, the answer is clear.

A clever student might multiply 15 by 14 to get 210.  They notice this is 6 less than 216, which is a power of 6.  Aha!  $216_{10}=1000_{6}.$  Subtracting six in base-6, we get $1000_{6}-10_{6}=550_{6}.$  Now, they can answer the question.

A more typical student might multiply 15 by 14 to get 210.  Then they will convert to base-6.  $210_{10}=5\times 6^{2}+5\times 6^{1}+0\times 6^{0}=550_{6}.$  Now, they can answer the question.

For number 2, a typical student must understand how to convert both from a base, other than 10; and to convert to a base, other than 10.  It's a harder variant.

The most astute reader might recognize that $30_{5}$ is a multiple of 3, and $24_{5}$ is a multiple of 2.  So the product is a multiple of 6.  But, as with Number 1, a typical student will convert from base-5 to base-10; then multiply; then convert from base-10 to base 6.  There's nothing wrong with this solution.  It's just that it will take a minute or two, and give you less time on the other questions.

It's very hard to write a high-quality MATHCOUNTS question.  It's hard to provide a question with multiple solutions.  It's hard to provide a question at the "right" level.  It's hard to be completely unambiguous.

So, let's all chill-out a little bit.  This problem is hard for some.  Easy for others.  Most people still on this Forum are among the best in the nation.  What's easy for a Forum member in late April is very, very hard for a typical Forum member in October.

Thanks to LeetSpeak for providing a question to challenge the community.  And thanks to those who provided a solution.

Assuming that n is a natural number...
[hide="solution"]Since n is a multiple of 7, the probability is 1[/hide]


Determine the minimum of $a^{2}+b^{2}$ where $a,b$ are reels, for wich the equation
$x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least one reel root.


$ a,b,c \geq 0$ , $ a\plus{}b\plus{}c\equal{}1$ . Prove that :
$ \sqrt{a^2\plus{}b}\plus{}\sqrt{b^2\plus{}c}\plus{}\sqrt{c^2\plus{}a} \geq 2$

Maybe Minkowski inequality will help.

$ \sqrt {a^2 \plus{} b} \plus{} \sqrt {b^2 \plus{} c} \plus{} \sqrt {c^2 \plus{} a} \geq \sqrt {1 \plus{} ({\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2}}$ $ \equal{} \sqrt {2(1 \plus{} {\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})}}$

Then you should prove that $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\geq 1$  :maybe:

You're wrong .It holds when $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ , $ a\equal{}b\equal{}0;c\equal{}1$ ...

[quote="conan_naruto236"]$ a,b,c \geq 0$ , $ a \plus{} b \plus{} c \equal{} 1$ . Prove that :
$ \sqrt {a^2 \plus{} b} \plus{} \sqrt {b^2 \plus{} c} \plus{} \sqrt {c^2 \plus{} a} \geq 2$[/quote]
My solution use only AM-GM.
Can you wirte your solution ? My solution will post later.

[quote="mitdac123"]I have seen this ineq in cacbaitoanhay.pdf ^^ ( of can_hang2007) .Of course ,his solution was Am-Gm ^^.
I want to see your solution  .(If it's nice :D)[/quote]
realy mitdac123! i have the file ''cacbaitoanhay.pdf of can_hang2007" but i don't see that.
the inequality which conan_naruto post before is the problem of Phan Thanh Nam and his solution used to lemma :if $ a,b,c,d\geq 0$ such that $ a\plus{}b\equal{}c\plus{}d$ and $ |a\minus{}b|\leq |c\minus{}d|$ then an inequality is always true $ \sqrt {a}\plus{}\sqrt{b}\geq\sqrt{c}\plus{}\sqrt{d}$

[quote="conan_naruto236"]$ a,b,c \geq 0$ , $ a \plus{} b \plus{} c \equal{} 1$ . Prove that :
$ \sqrt {a^2 \plus{} b} \plus{} \sqrt {b^2 \plus{} c} \plus{} \sqrt {c^2 \plus{} a} \geq 2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106782


When $ x \equal{} 1, y \equal{} \frac{\sqrt {10 } \plus{} \sqrt {5 } \plus{} \sqrt {2 } \plus{} 1}{2}, z \equal{} x \plus{} y$,

simplify $ f(x,y,z) \equal{} \bigg|{\frac {(x \minus{} y)(y \minus{} z)(z \minus{} x)}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}}\bigg|$ .

The answer is $ \frac {8 \sqrt {2 } \minus{} 5 \sqrt {5 }}{3}$.

But I don't know how to get it without using calculator.

Any good idea?

It seems to be nothing more that long, but straightforward computations.  Nothing particularly tricky, just tedious.

Maybe it helps that $ y\equal{}\frac{1}{2}\cdot (\sqrt{2}\plus{}1)(\sqrt{5}\plus{}1)$...

[quote="Altheman"]Maybe it helps that $ y \equal{} \frac {1}{2}\cdot (\sqrt {2} \plus{} 1)(\sqrt {5} \plus{} 1)$...[/quote]I noticed that too, but it doesn't help that much... I mean, the expression just simplifies to $ \left|\frac{(y\minus{}1)(y)}{(y\plus{}1)(y\plus{}2)(2y\plus{}1)}\right|$ which is relatively inscrutable... I don't like this problem unless there is a very clever solution.

[quote="mathland"]When $ x \equal{} 1, y \equal{} \frac {\sqrt {10 } \plus{} \sqrt {5 } \plus{} \sqrt {2 } \plus{} 1}{2}, z \equal{} x \plus{} y$,

simplify $ f(x,y,z) \equal{} \bigg|{\frac {(x \minus{} y)(y \minus{} z)(z \minus{} x)}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}}\bigg|$ .

The answer is $ \frac {8 \sqrt {2 } \minus{} 5 \sqrt {5 }}{3}$.

But I don't know how to get it without using calculator.

Any good idea?[/quote]

This problem is actually from the book "Problem Solving Strategies, 7.Inequalities Problem 58, (c)"
The book gives only answer.
I checked it's true by using Maple.
But I don't know by hands.

If there is no good idea, this problem is just brutal-force calculation.
I wondered whether there is a fantastic idea for it.
Still I expect *^^*~


$ x,y,z>0$ and $ x\plus{}y\plus{}z\plus{}2\equal{}xyz$. Prove that:

$ \frac{\left ( x\minus{}1 \right )^2}{x^2\plus{}2}\plus{}\frac{\left ( y\minus{}1 \right )^2}{y^2\plus{}2}\plus{}\frac{\left ( z\minus{}1 \right )^2}{z^2\plus{}2}\geq \frac{1}{2}$

[quote="Itachi.vn"]$ x,y,z > 0$ and $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$. Prove that:

$ \frac {\left ( x \minus{} 1 \right )^2}{x^2 \plus{} 2} \plus{} \frac {\left ( y \minus{} 1 \right )^2}{y^2 \plus{} 2} \plus{} \frac {\left ( z \minus{} 1 \right )^2}{z^2 \plus{} 2}\geq \frac {1}{2}$[/quote]
It's true for all reals $ x,$ $ y$ and $ z$ such that $ x\plus{}y\plus{}z\plus{}2\equal{}xyz.$ :wink:

@arcady, could you please explain in greater detail why the inequality holds for all real numbers such that $ x\plus{}y\plus{}z\plus{}2\equal{}xyz$?

[quote="Itachi.vn"]$ x,y,z > 0$ and $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$. Prove that:

$ \frac {\left ( x \minus{} 1 \right )^2}{x^2 \plus{} 2} \plus{} \frac {\left ( y \minus{} 1 \right )^2}{y^2 \plus{} 2} \plus{} \frac {\left ( z \minus{} 1 \right )^2}{z^2 \plus{} 2}\geq \frac {1}{2}$[/quote]

with $ x,y,z > 0$ and $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$, we can set that $ x \equal{} \frac {a \plus{} b}{c},y \equal{} \frac {b \plus{} c}{a},z \equal{} \frac {c \plus{} a}{b},a,b,c > 0$.
ineq <=>\[ \sum \frac {(a \plus{} b \minus{} c)^2}{(a \plus{} b)^2 \plus{} 2c^2} \ge \frac {1}{2}\]<=>\[ \sum f(a,b,c)(a \minus{} b)^2 \ge 0\]$ f(a,b,c) \equal{} \frac {4a^2 \plus{} 4b^2 \plus{} 3ac \plus{} 3bc \plus{} ab \minus{} c^2}{((b \plus{} c)^2 \plus{} 2a^2)((a \plus{} c)^2 \plus{} 2b^2)}$.
assume $ a\ge b\ge c$, easy to prove that $ f(c,a,b)\ge 0,f(c,a,b) \plus{} f(a,b,c)\ge 0$, use "[url=http://www.mathlinks.ro/viewtopic.php?p=458553#458553]S.O.S Theorem[/url]", just need to prove
$ f(c,a,b) \plus{} f(b,c,a)\ge 0$,
actually,
$ f(c,a,b) \plus{} f(b,c,a)$
$ \equal{}\frac {(a \plus{} b)^2 (2 a \plus{} b) (a \plus{} 2 b) \plus{} 3 (a \plus{} b) \left(5 a^2 \minus{} 2 a b \plus{} 5 b^2\right) c \plus{} \left(17 a^2 \plus{} 18 a b \plus{} 17 b^2\right) c^2 \plus{} 12 (a \plus{} b) c^3 \plus{} 8 c^4}{\left((a \plus{} b)^2 \plus{} 2 c^2\right) \left(2 b^2 \plus{} (a \plus{} c)^2\right) \left(2 a^2 \plus{} (b \plus{} c)^2\right)}$
$ \ge 0$,
so the inequality holds.

[quote="Spribo"]@arcady, could you please explain in greater detail why the inequality holds for all real numbers such that $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$?[/quote]
Certainly!
For $ x\equal{}\minus{}1$  an original inequality is true.
Let $ x\neq\minus{}1,$ $ x\equal{}\frac{b\plus{}c}{a}$ and $ y\equal{}\frac{a\plus{}c}{b}.$
Hence, $ a\plus{}b\plus{}c\neq0$ and $ x\plus{}y\plus{}z\plus{}2\equal{}xyz$ gives $ z\equal{}\frac{a\plus{}b}{c}.$
Thus, we need to prove that $ \sum_{cyc}\frac{(a\plus{}b\minus{}c)^2}{(a\plus{}b)^2\plus{}2c^2}\geq\frac{1}{2}.$
But $ \sum_{cyc}\frac{(a\plus{}b\minus{}c)^2}{(a\plus{}b)^2\plus{}2c^2}\equal{}\sum_{cyc}\frac{(a\plus{}b)^2(a\plus{}b\minus{}c)^2}{(a\plus{}b)^4\plus{}2c^2(a\plus{}b)^2}\geq$
$ \geq\frac{\left(\sum((a\plus{}b)^2\minus{}c(a\plus{}b))\right)^2}{\sum((a\plus{}b)^4\plus{}2c^2(a\plus{}b)^2)}\equal{}\frac{4(a^2\plus{}b^2\plus{}c^2)^2}{\sum(2a^4\plus{}4a^3b\plus{}4a^3c\plus{}10a^2b^2\plus{}4a^2bc)}.$
Id est, it remains to prove that
$ 4(a^2\plus{}b^2\plus{}c^2)^2\geq\sum_{cyc}(a^4\plus{}2a^3b\plus{}2a^3c\plus{}5a^2b^2\plus{}2a^2bc),$
which is obviously true for all reals $ a,$ $ b$ and $ c.$


Let $P,Q,R$  be polynomials of degree at least $1$ with integer coefficients such that for any real number $x$ holds:  $P(Q(x))\equal{}Q(R(x))\equal{}R(P(x))$. Show that the polynomials $P,Q,R$ are equal.

TRong Cuong, dao nay van tot chu!
I think the condition the coeficient of P,Q,R are integer is not  indispensable

It is easy to see that P,Q,R have a same degree, and the coefficient of $ x^n$  (deg P=deg Q=deg R=n)
let K(x)=P(x)-Q(x), and we have deg K =d$ \le n \minus{} 1$;
since P(Q(x))=Q(R(x)), we infer that Q(Q(x)) +K(Q(x))=Q(R(x));
it follows that K(Q(x))=Q(R(x)) - Q(Q(x));
the degree of LHS = nd ;
degree of RHS equals the degree of $ R(x)^n \minus{} Q(x)^n \equal{} (R(x) \minus{} Q(x))(R(x)^{n \minus{} 1} \plus{} ... \plus{} Q(x)^{n \minus{} 1})$(1)
+)case 1
P(x)=Q(x) we infer that K(x)=0 and we have Q=R=P;
+)case 2
P(x) and Q(x) are different ,we infer that the degree of RHS of (1) is at least n.(n-1) but degree of LHS = nd and d$ \le n \minus{} 1$
so we attain d=n-1 and degree of R(x)-Q(x) is 0.Actually, R(x)=Q(x)+c;
and similarly if we let H(x)=Q(x)-R(x) ... we attain that P=Q=R or P(x)=R(x)+e (e=const) so we only consider in the case P(x)=R(x)+e
we infer that P(x)-Q(x)=const=K(x), but it is contrary to the degree of K(x)=d=n-1;
so, P(x)=Q(x)=R(x)

[quote="kihe_freety5"]TRong Cuong, dao nay van tot chu! [/quote]
Chu thay anh post bai the nay thi phai doan anh rat tot chu. Tinh hinh thi doi tuyen the nao roi(hom nao thi vong 2).Nick anh : ltcuongdhqg


let $ a,b,c$ be sides of a triangle.Prove:

$ \sum \frac {b\plus{}c}{\sqrt {bc(b \plus{} c \minus{} a)(a \plus{} b \plus{} c)}} > \sum \frac {1}{a}$

[quote="Amir.S"]let $ a,b,c$ be sides of a triangle.Prove:

$ \sum \frac {1}{\sqrt {bc(b \plus{} c \minus{} a)(a \plus{} b \plus{} c)}}\ge \sum \frac {1}{a}$[/quote]

Try $ a \equal{} b \equal{} c > 1$.  :wink:


geometry
geometric transformation
reflection
AMC
USA(J)MO
USAMO

Today at an awards ceremony I did not receive many awards, which is something i'm not accustomed to. Usually, I am among the top students in my class. This time however, I only received awards in math and running. For the most part, this reflects my decision to focus more on my passions, and worry less about these types of awards.

Still, it is kind of discouraging when you know that you could do better than the people winning some awards. So basically, I used to consider myself an all-around overachiever (which is not necessarilly something I'm trying to put down), but now I have focused more on my passions.

I think that, in the end, what will matter is your individual drive. Receiving accolades or awards is not the most important thing (yet still a plus) when you know deep down what you want to do with your life.

But what about you guys? I'm sure a lot of you are really big overachievers. What do you guys think about overachieving? How much importance do you place on overachieving and in the end, how does it fit in the grand scheme of things?

I think it's hard to be an all-around achiever, let alone an all-around overachiever.  You'd probably be more satisfied and do better only focusing on your passions and getting very proficient at them.

Well, I have no clue of what happens after one dies and what effect, if any, the actions one takes in their life will have on them after they die.  So whenever I have to choose between two things I try to take the option that will most likely give me the most personal satisfaction in the long run.  To clarify a little, if I had the opportunity to earn money quickly but illegally or go to college instead, I would probably choose going to college since in the former scenario guilt would outweigh worldly possessions.

So I suggest doing what you believe will ultimately make you as happy as possible.

I hate the term 'overachiever'.  It is most oftenly used in a derogatory manner.  It also implies that the person is doing wrong by surpassing what others deem as the normal expectations.  "Overachievement" should be encouraged.  But instead it seems that everyone is being pushed toward the middle.  Some people discourage competition in school.  And then there is No Child Left Behind...  It's intended to raise the overall level of American education but I see it as focusing resources on making bad students mediocre instead of conceding that not everyone needs to (or should) go through the same exact high school program.  Vocational schools anyone?  Voc schools aren't perfect, but they are in many ways better than ramming everyone through the same high school curriculum.

In high school I was well rounded academically but focused in on mathematics.  Around 11th grade, I gave up trying to be the best in my english class and just aimed for an A while allowing myself more room to study math.  I gave up trying to be the best in other areas earlier.  Trying to be an all-around "overachiever" seems to be too stressful to me.  Being well rounded yet standing out in a few areas suited my personality perfectly.

[quote="JBL"]Oh, good, more threads to fill up the Self-Pity forum.[/quote]

Hehe, I remember when I was like this. I'd stress over minor things (math competitions for example). Now, I've just pulled an Office Space. I don't care. I do what's fun, instead of just what I'm good at. That means Muay Thai [i]and[/i] math, games [i]and[/i] programming.

[quote="yif man12"][quote="JBL"]Oh, good, more threads to fill up the Self-Pity forum.[/quote]

Just like it's brother, the look-at-me-I-should've-made-USAMO-but-didn't forum.[/quote]

Well, I figure we can toss all of those threads in there, too.  The people who actually go to the AMC forum for content will be very pleased.

[quote="JBL"][quote="yif man12"][quote="JBL"]Oh, good, more threads to fill up the Self-Pity forum.[/quote]

Just like its brother, the look-at-me-I-should've-made-USAMO-but-didn't forum.[/quote]

Well, I figure we can toss all of those threads in there, too.  The people who actually go to the AMC forum for content will be very pleased.[/quote]

people do that? whoa

[quote="yif man12"]
Just like it's brother, the look-at-me-I-should've-made-USAMO-but-didn't forum.[/quote]

That's me :D

Methinks overachieving isn't possible...


calculus
derivative
function
calculus computations

A particle moves on x-axis with position function
x = x(t) = t^3 - t^2 - 5t + 10; t=>0.
Determine when speed increases/decreases.

I can solve for the derivative (velocity) and the second derivative (acceleration). But how do I know when the speed is going up or down?? What numbers should I plug in the x??

Thanks!!!! GBU.  :)

Right, so your question (not the math problem you're trying to solve, but your question about it) indicates a lack of understanding of the relationship between the formulas you're manipulating and their [i]meaning[/i].  We have a formula for $ x(t)$, and this formula gives (for example) $ x(0) \equal{} 10$ and $ x(1) \equal{} 5$.  What does this mean about the position of the particle, and its behavior over the first second?

You've computed the velocity -- if we obtain that for a certain value of $ t$, the velocity is $ 2$, what does this mean about how (e.g., how fast) the particle is moving?  What if instead we find that its velocity is $ \minus{} 2$?

Now, suppose that its velocity at some value of $ t$ is positive, and its acceleration is also positive; what does this mean about how it's behaving?  What if the velocity is positive and the acceleration is negative?  If the velocity is negative and the acceleration is negative?  If the velocity is negative and accleration is positive?  Finally, how can we figure out when the particle has positive (or negative) velocity or acceleration?

Once you understand the answers to these questions, knowing how to answer the math problem you've given (and others like it) will be immediate.  If you post your thoughts here, either I or others will help you along.  (You may also find it helpful to look at a graph of the behavior of the function in the $ t,x$-plane.)


my recursive sequence is a1=-7  and a2=(1/5)(3(a1)+4)

      I need to change this recursive sequence into a sequence in terms of "n"

     the first 6 terms are as follows"

       a1 = -7

      a2 = -17/5

     a3 = -31/25

     a4 = 7/125

    a5 = 521/625

    a6 = 4063/3125

could use help in solving this one

thanks Lorie

It is an trivial problems .
Let $ x_n\equal{}a_n\minus{}2$
Then $ x_{n\plus{}1}\equal{}\frac{3x_n}{5}$

3x(n)/5 does not work for all n if it does please explain

The recursion is $ a_{n\plus{}1}\equal{}\frac{3a_n\plus{}4}5.$

That's something sort of like $ a_{n\plus{}1}\equal{}\frac35\,a_n,$ only there are some extra constants around. We assume that we have something like geometric growth, $ \left(\frac35\right)^n.$

So I make the following structured guess: $ a_n\equal{}\alpha\left(\frac35\right)^n\plus{}\beta.$

Let's put this into the recursion:

$ \alpha\left(\frac35\right)^{n\plus{}1}\plus{}\beta\equal{}
\frac35\left[\alpha\left(\frac35\right)^n\plus{}\beta\right]\plus{}\frac45$

$ \alpha\left(\frac35\right)^{n\plus{}1}\plus{}\beta\equal{}
\alpha\left(\frac35\right)^{n\plus{}1}\plus{}\frac35\beta\plus{}\frac45$

$ \frac25\beta\equal{}\frac45$ or $ \beta\equal{}2.$

So now we have that $ a_n\equal{}\alpha\left(\frac35\right)^n\plus{}2.$

(By the way, this is what TTsphn said. He is correct, he just didn't show how to finish the problem.)

We do have an initial condition, namely $ a_1\equal{}\minus{}7.$ Then

$ \minus{}7\equal{}\frac35\alpha\plus{}2$

$ \frac35\alpha\equal{}\minus{}9$

$ \alpha \equal{}\minus{}15.$

So now our complete formula is $ a_n\equal{}\minus{}15\left(\frac35\right)^n\plus{}2.$

Check this with $ n\equal{}2,3,4,5,6$ to see that it agrees with what you have.

Note also that $ \lim_{n\to\infty}a_n\equal{}2,$ which we could have found even without the explicit formula for $ a_n.$


$ {x^3y^{'''}}\plus{}x^2y''\minus{}2xy'\plus{}2y\equal{}x^{\minus{}2}$, $ x>0$

How would I solve this differential equation

Just a rough guide will do! No need for the full soln.

hello, note that $ y_1(x)\equal{}\frac{1}{x},y_2(x)\equal{}x,y_3(x)\equal{}x^3$ are solutions of the homogenius part of your equation.
Sonnhard.

$ L[y] \equal{} x^3y''' \plus{} x^2y'' \minus{} 2xy' \plus{} 2y \equal{} x^{ \minus{} 2}$

$ y \equal{} x^r \to y' \equal{} rx^{r \minus{} 1} \to y'' \equal{} r(r \minus{} 1)x^{r \minus{} 2} \to y''' \equal{} r(r \minus{} 1)(r \minus{} 2)x^{r \minus{} 3}$

$ L[x^r] \equal{} [r(r \minus{} 1)(r \minus{} 2) \plus{} r(r \minus{} 1) \minus{} 2r \plus{} 2]x^r \equal{} 0$

$ \to 0 \equal{} (r \minus{} 1)(r \plus{} 1)(r \minus{} 3) \to r \equal{} \pm 1, 3$

are the homogenous solutions...

trying on a hunch, $ r \equal{} \minus{} 2$,

$ L[x^{ \minus{} 2}] \equal{} ( \minus{} 2 \minus{} 1)( \minus{} 2 \plus{} 1)( \minus{} 2 \minus{} 3)x^{ \minus{} 2} \equal{} \minus{} 15x^{ \minus{} 2}$

using linearity and all of this together we get:

$ y(x) \equal{} \frac {A}{x} \plus{} Bx \plus{} Cx^3 \minus{} \frac {1}{15x^2}$


in scientific notation for all...
please show work. thx =)

edit:sorry attachments didn't work...

edit:ok.. sorry about that  :blush:


[hide] From the equation we have $x^{2}-x+1=0$ which solves for $x=e^{\pm\frac{i\pi}{3}}$.  Putting this in the expression yields $e^{\pm 667i\pi}+e^{\mp 667i\pi}=2\cos667\pi=-2$
[/hide]

[quote="ashrafmod"]for the real number x
if
$x+x^{-1}=1$
calculate
$x^{2001}+x^{-2001}$[/quote]

No real $x$ is a root of the given equation, since $x+\frac1x \geq 2$ by AM-GM.   :P  (edit: for positive $x$. Generally, it is $\left| x+\frac1x \right| \ge 2$.)

this has been posted a millino times...search

[quote="paladin8"][quote="cincodemayo5590"][quote="ashrafmod"]for the real number x
if
$x+x^{-1}=1$
calculate
$x^{2001}+x^{-2001}$[/quote]

No real $x$ is a root of the given equation, since $x+\frac1x \geq 2$ by AM-GM.   :P[/quote]

That's only true for positive numbers, though the no real roots statement is valid. You could also add that

$x+\frac{1}{x}\le-2$

for negatives $x$.[/quote]

Oh oops, well there's an easy way to fix that. Just add absolute value bars.  :)


inequalities
LaTeX
inequalities unsolved

Prove for all positive reals a,b,c,d:

$ \frac{a\minus{}b}{b\plus{}c}\plus{}\frac{b\minus{}c}{c\plus{}d}\plus{}\frac{c\minus{}d}{d\plus{}a}\plus{}\frac{d\minus{}a}{a\plus{}b} \geq 0$

[quote="Matematika"]Prove for all positive reals a,b,c,d:

$ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} a} \plus{} \frac {d \minus{} a}{a \plus{} b} \geq 0$[/quote]

Let $ a\geq b \geq c \geq d$ then $ LHS \geq \frac{a\minus{}b}{a\plus{}b} \plus{}\frac{b\minus{}c}{a\plus{}b} \plus{}\frac{c\minus{}d}{a\plus{}b} \plus{}\frac{d\minus{}a}{a\plus{}b}\equal{}\frac{0}{a\plus{}b}\equal{}0$

You aren't allowed to assume $ a \geq b\geq c\geq d$ cause inequality isn't symmetric.
As much as I know you can't assume that if inequality is cyclic.

Why not , do you have any mathematical explanation for it ?

[quote="enndb0x"][quote="Matematika"]Prove for all positive reals a,b,c,d:

$ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} a} \plus{} \frac {d \minus{} a}{a \plus{} b} \geq 0$[/quote]

Let $ a\geq b \geq c \geq d$ then $ LHS \geq \frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{a \plus{} b} \plus{} \frac {c \minus{} d}{a \plus{} b} \plus{} \frac {d \minus{} a}{a \plus{} b} \equal{} \frac {0}{a \plus{} b} \equal{} 0$[/quote]
Can you repeat your argument when I assume $ a\ge c\ge b\ge d$?.My solution :
The inequality is equivalent to
\[ \frac{a\plus{}c}{b\plus{}c}\plus{}\frac{b\plus{}d}{c\plus{}d}\plus{}\frac{c\plus{}a}{d\plus{}a}\plus{}\frac{d\plus{}b}{a\plus{}b}\ge 4
\]
Apply Cauchy Schwarz inequality in Engel form, we have
\[ LHS\ge \frac{(2(a\plus{}b\plus{}c\plus{}d))^2}{(a\plus{}c)(b\plus{}c)\plus{}(b\plus{}d)(c\plus{}d)\plus{}(c\plus{}a)(d\plus{}a)\plus{}(d\plus{}b)(a\plus{}b)}
\]
It remains to prove
\[ \frac{(2(a\plus{}b\plus{}c\plus{}d))^2}{(a\plus{}c)(b\plus{}c)\plus{}(b\plus{}d)(c\plus{}d)\plus{}(c\plus{}a)(d\plus{}a)\plus{}(d\plus{}b)(a\plus{}b)} \ge 4
\]
But in the fact: we have
\[ (a\plus{}b\plus{}c\plus{}d)^2\equal{}(a\plus{}c)(b\plus{}c)\plus{}(b\plus{}d)(c\plus{}d)\plus{}(c\plus{}a)(d\plus{}a)\plus{}(d\plus{}b)(a\plus{}b)
\]
It ends our proof.

Can you elaborate more how you won that equivalence ?

that was mine solution on competition as well.
mathematical proof for that assuming is simply counter example you have expression 2(a-b) (it is cyclic for 2 arguments)
and if you could assume $ a \geq b$ you would get that it is bigger then 0 and you can take b less than a and that's not true.
and that assumption comes from the fact that in symmetric inequality if you change two variables you don't change a thing but in cyclic ones you do change
inequality.

just one comment you don't need to add 4 cauchy in engel form is correct for negative values of nominators (denominator must be positive)

There is faster way to prove $ \frac {a \plus{} c}{b \plus{} c} \plus{} \frac {b \plus{} d}{c \plus{} d} \plus{} \frac {c \plus{} a}{d \plus{} a} \plus{} \frac {d \plus{} b}{a \plus{} b}\ge 4$.
We have \[ \frac {a \plus{} c}{b \plus{} c} \plus{} \frac {b \plus{} d}{c \plus{} d} \plus{} \frac {c \plus{} a}{d \plus{} a} \plus{} \frac {d \plus{} b}{a \plus{} b} \equal{} (a \plus{} c)(\frac {1}{b \plus{} c} \plus{} \frac {1}{d \plus{} a}) \plus{} (b \plus{} d)(\frac {1}{c \plus{} d} \plus{} \frac {1}{a \plus{} b})\geq \frac {4(a \plus{} b \plus{} c \plus{} d)}{a \plus{} b \plus{} c \plus{} d} \equal{} 4\] which is just $ \frac {1}{a} \plus{} \frac {1}{b}\geq \frac {4}{a \plus{} b}$ for $ a,b > 0$.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102924

i use cauchy ineequality at first
let $ \frac{a\minus{}b}{b\plus{}c}\plus{}\frac{b\minus{}c}{c\plus{}d}\plus{}\frac{c\minus{}d}{d\plus{}a}\plus{}\frac{d\minus{}a}{a\plus{}b} \equal{}k$
$ k.((a\minus{}b)(b\plus{}c)\plus{}(b\minus{}c)(c\plus{}d)\plus{}(c\minus{}d)(d\plus{}a)\plus{}(d\minus{}a)(a\plus{}b)) \geq 0$
if a=1 b=2 c=3 d=4 then $ ((a\minus{}b)(b\plus{}c)\plus{}(b\minus{}c)(c\plus{}d)\plus{}(c\minus{}d)(d\plus{}a)\plus{}(d\minus{}a)(a\plus{}b))<0$then k<0 !!!!![/quote]

[quote="Mathias_DK"][quote="table1388"]i use cauchy ineequality at first
let $ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} a} \plus{} \frac {d \minus{} a}{a \plus{} b} \equal{} k$
$ k.((a \minus{} b)(b \plus{} c) \plus{} (b \minus{} c)(c \plus{} d) \plus{} (c \minus{} d)(d \plus{} a) \plus{} (d \minus{} a)(a \plus{} b)) \geq 0$
if a=1 b=2 c=3 d=4 then $ ((a \minus{} b)(b \plus{} c) \plus{} (b \minus{} c)(c \plus{} d) \plus{} (c \minus{} d)(d \plus{} a) \plus{} (d \minus{} a)(a \plus{} b)) < 0$then k<0 !!!!![/quote][/quote]
Cauchy requires positive, or at least non-negative, numbers.[/quote]

i dont think so,i studied my book again arent you wrong ?....

[quote="table1388"][quote="Mathias_DK"][quote="table1388"]i use cauchy ineequality at first
let $ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} a} \plus{} \frac {d \minus{} a}{a \plus{} b} \equal{} k$
$ k.((a \minus{} b)(b \plus{} c) \plus{} (b \minus{} c)(c \plus{} d) \plus{} (c \minus{} d)(d \plus{} a) \plus{} (d \minus{} a)(a \plus{} b)) \geq 0$
if a=1 b=2 c=3 d=4 then $ ((a \minus{} b)(b \plus{} c) \plus{} (b \minus{} c)(c \plus{} d) \plus{} (c \minus{} d)(d \plus{} a) \plus{} (d \minus{} a)(a \plus{} b)) < 0$then k<0 !!!!![/quote]
Cauchy requires positive, or at least non-negative, numbers.[/quote]

i dont think so,i studied my book again arent you wrong ?....[/quote]
I'm sure ;) You just found a counter example, when they were negative, didn't you?

[quote="Mathias_DK"][/quote][quote="table1388"][quote="Mathias_DK"][quote="table1388"]i use cauchy ineequality at first
let $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} = k$
$ k.((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) \geq 0$
if a=1 b=2 c=3 d=4 then $ ((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) < 0$then k<0 !!!!![/quote]
Cauchy requires positive, or at least non-negative, numbers.[/quote]

i dont think so,i studied my book again arent you wrong ?....[/quote][quote="Mathias_DK"]
I'm sure ;) You just found a counter example, when they were negative, didn't you?[/quote]
yes... :idea: .. you are right ...thanks.

I think this problem is from Hung's book and limes123's solution is beautiful enough as given in the book. :)

[quote="Zarif"]Very nice and easy inequality. This is problem equivalently to Baltic Way 1995.[/quote]

You just reposted a solution...

I`m sorry Pain Rinnegan. I don`t know english and Latex!

I think AG is enough
the ineq is equivalent to
$\sum\frac{a+c}{b+c}\ge 4$or
$(a+b+c+d)(\frac{a+c}{(a+d)(b+c)}+\frac{b+d}{(a+b)(c+d)})\ge 4$
by AG,LHS$\ge 4(a+b+c+d)(\frac{a+c}{(a+d+b+c)^2}+\frac{b+d}{(a+b+c+d)^2})=4$.


Prove that the center of the sphere circumscribed around a
tetrahedron ABCD coincides with the center of a sphere inscribed in
that tetrahedron if and only if AB = CD, AC = BD, and AD = BC.
by the way, what is an exscribed circle?


[quote="Michael"]Let $a,b,c$ be three positive real numbers. Prove the following:
\[ {{1}\over{a+b+c+1}}-{{1}\over{(a+1)(b+1)(c+1)}}\leq {1\over 8}.  \][/quote]
 This is a very old and easy inequality. ;)

This is a very old and easy inequality. ;)[/quote]

So, post your "nice" solution :P

[quote="Michael"]
So, post your "nice" solution :P[/quote]
Look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67113  :)


I'm unable to proceed , please help . :blush:

In the first compound, there is no chiral carbon, so there must be some typo.

In the second compound, ask yourself: which is more reactive, a ketone or an esther? [hide="Hint"]Ressonance[/hide]

In the third reaction, because of the high steric hindrance, we will have an SN2 mechanism, giving ethyl methyl ether and 2,4,6-tri-tert-butylbenzoate anion.

I just had a thought. In the second one it has 1 eq of Grignard reagent so wouldn't that imply reaction with the ketone? Seeing that Grignard reaction of ester requires 2 eq.

[quote="BanishedTraitor"]Seeing that Grignard reaction of ester requires 2 eq[/quote]

That's not correct. A Grignard reagent almost always add to the carbonyl group of an esther. However, if the purpose is to obtain a tertiary alcohol from an esther then obviously we must use 2 eqs of organometalic reagent (and this what you had in mind), but of course there is also reaction with just one equivalent.


Here is the solution :
 $S=\sum\frac{a^3(a+c)(a+b)}{(b+c)^3}\geq \frac{(a+b+c)^2}{6}$

$Then$ $S\geq \frac{\left( \sum a^3(a+c)(a+b)\right)*\left(\sum\frac{1}{(b+c)^3}\right)}{3}\geq \sum \frac{a^3}{b+c}$

 $\sum \frac{a^3}{b+c}\geq \frac{a^3}{a+c}+\frac{b^3}{a+b}+\frac{c^3}{b+c}$

 $\sum \frac{a^3}{b+c}\geq \frac{c^3}{a+c}+\frac{a^3}{a+b}+\frac{b^3}{b+c}$

 $\sum \frac{a^3}{b+c}\geq \frac{\frac{a^3+c^3}{a+c}+\frac{a^3+b^3}{a+b}+\frac{b^3+c^3}{b+c}}{2}\geq \frac{a^2+b^2+c^2}{2}\geq \frac{(a+b+c)^2}{6}$
 I hope my solution is true  :)

[quote="CTLHP"]For any three positive reals $ a$, $ b$, $ c$, prove the inequality

$ \sum\frac {a^3\left(a \plus{} b\right)\left(a \plus{} c\right)}{\left(b \plus{} c\right)^3}\geq\frac {\left(a \plus{} b \plus{} c\right)^2}{6}$,

where the $ \sum$ sign stands for cyclic summation.[/quote]

[i]Proof.[/i] We will prove this inequality with the help of a real MathLinks specialty, namely the [b]Vornicu-Schur inequality[/b]:

[color=blue][b]Lemma 1 (generalized Schur inequality / Vornicu-Schur inequality).[/b] Let a, b, c, x, y, z be six nonnegative reals such that $ a\geq b\geq c$, and such that either $ x\geq y\geq z$ or $ x\leq y\leq z$. Then,

$ x\left(a \minus{} b\right)\left(a \minus{} c\right) \plus{} y\left(b \minus{} c\right)\left(b \minus{} a\right) \plus{} z\left(c \minus{} a\right)\left(c \minus{} b\right)\geq 0$.[/color]

For the proof of this inequality, look at http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9.

Now, in order to prove your inequality, we WLOG assume that $ a\geq b\geq c$. Consider the positive reals $ x \equal{} \frac {a^3}{\left(b \plus{} c\right)^3}$, $ y \equal{} \frac {b^3}{\left(c \plus{} a\right)^3}$, $ z \equal{} \frac {c^3}{\left(a \plus{} b\right)^3}$. Since $ a\geq b$, we have $ b\leq a$ and thus $ b \plus{} c\leq c \plus{} a$, so that $ \left(b \plus{} c\right)^3\leq\left(c \plus{} a\right)^3$ and $ \frac {a^3}{\left(b \plus{} c\right)^3}\geq\frac {a^3}{\left(c \plus{} a\right)^3}$. On the other hand, $ a\geq b$ yields $ \frac {a^3}{\left(c \plus{} a\right)^3}\geq\frac {b^3}{\left(c \plus{} a\right)^3}$. Thus, $ \frac {a^3}{\left(b \plus{} c\right)^3}\geq\frac {b^3}{\left(c \plus{} a\right)^3}$. In other words, $ x\geq y$. Similarly, using $ b\geq c$, we can prove $ y\geq z$. Thus, $ x\geq y\geq z$. Combining this with $ a\geq b\geq c$, we see that we can apply Lemma 1 to the nonnegative reals a, b, c, x, y, z and get

$ x\left(a \minus{} b\right)\left(a \minus{} c\right) \plus{} y\left(b \minus{} c\right)\left(b \minus{} a\right) \plus{} z\left(c \minus{} a\right)\left(c \minus{} b\right)\geq 0$.

In other words,

$ \frac {a^3}{\left(b \plus{} c\right)^3}\left(a \minus{} b\right)\left(a \minus{} c\right) \plus{} \frac {b^3}{\left(c \plus{} a\right)^3}\left(b \minus{} c\right)\left(b \minus{} a\right) \plus{} \frac {c^3}{\left(a \plus{} b\right)^3}\left(c \minus{} a\right)\left(c \minus{} b\right)\geq 0$.

We rewrite this using the $ \sum$ notation:

$ \sum\left(\frac {a^3}{\left(b \plus{} c\right)^3}\left(a \minus{} b\right)\left(a \minus{} c\right)\right)\geq 0$.

Thus,

$ \sum\frac {a^3\left(a \plus{} b\right)\left(a \plus{} c\right)}{\left(b \plus{} c\right)^3} \minus{} \sum\frac {2a^4}{\left(b \plus{} c\right)^2} \equal{} \sum\left(\frac {a^3\left(a \plus{} b\right)\left(a \plus{} c\right)}{\left(b \plus{} c\right)^3} \minus{} \frac {2a^4}{\left(b \plus{} c\right)^2}\right)$
$ \equal{} \sum\left(\frac {a^3}{\left(b \plus{} c\right)^3}\left(a \minus{} b\right)\left(a \minus{} c\right)\right)\geq 0$,

and hence,

$ \sum\frac {a^3\left(a \plus{} b\right)\left(a \plus{} c\right)}{\left(b \plus{} c\right)^3}\geq\sum\frac {2a^4}{\left(b \plus{} c\right)^2}$.

Hence, instead of proving the initial inequality

$ \sum\frac {a^3\left(a \plus{} b\right)\left(a \plus{} c\right)}{\left(b \plus{} c\right)^3}\geq\frac {\left(a \plus{} b \plus{} c\right)^2}{6}$,

it will be enough to verify the stronger inequality

$ \sum\frac {2a^4}{\left(b \plus{} c\right)^2}\geq\frac {\left(a \plus{} b \plus{} c\right)^2}{6}$.

Upon division by 2, this inequality becomes

$ \sum\frac {a^4}{\left(b \plus{} c\right)^2}\geq\frac {\left(a \plus{} b \plus{} c\right)^2}{12}$.

This rewrites as

$ \sum\frac {\left(a^2\right)^2}{\left(b \plus{} c\right)^2}\geq\frac {\left(a \plus{} b \plus{} c\right)^2}{12}$.

But by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url], we have

$ \sum\frac {\left(a^2\right)^2}{\left(b \plus{} c\right)^2}\geq\frac {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2}{\left(b \plus{} c\right)^2 \plus{} \left(c \plus{} a\right)^2 \plus{} \left(a \plus{} b\right)^2}$.

Thus, it remains to prove

$ \frac {\left(a^2 \plus{} b^2 \plus{} c^2\right)^2}{\left(b \plus{} c\right)^2 \plus{} \left(c \plus{} a\right)^2 \plus{} \left(a \plus{} b\right)^2}\geq\frac {\left(a \plus{} b \plus{} c\right)^2}{12}$.

This inequality is clearly equivalent to

$ 12\left(a^2 \plus{} b^2 \plus{} c^2\right)^2\geq\left(a \plus{} b \plus{} c\right)^2\left(\left(b \plus{} c\right)^2 \plus{} \left(c \plus{} a\right)^2 \plus{} \left(a \plus{} b\right)^2\right)$.

But this inequality is true, since

$ 12\left(a^2 \plus{} b^2 \plus{} c^2\right)^2 \minus{} \left(a \plus{} b \plus{} c\right)^2\left(\left(b \plus{} c\right)^2 \plus{} \left(c \plus{} a\right)^2 \plus{} \left(a \plus{} b\right)^2\right)$
$ \equal{} \left(\left(b \minus{} c\right)^2 \plus{} \left(c \minus{} a\right)^2 \plus{} \left(a \minus{} b\right)^2\right)\left(5a^2 \plus{} 5b^2 \plus{} 5c^2 \plus{} 2bc \plus{} 2ca \plus{} 2ab\right)\geq 0$.

Hence, your inequality is proven.

  Darij

have never heard about Shur-Vornicu before so i need time to understand your proof.
But my solution just uses cauchy and nesbit and i will post it tomorrow


[size=150][b][color=red]DEL[/color][color=white]EGA[/color][color=blue]CIÓN[/color] [color=red]PAR[/color][color=white]AGU[/color][color=blue]AYA[/color][/b][/size]

[b][i]Participantes:[/i][/b]
1. Caló Caligaris, Maurizio.
2. Arias Michel, Rafael Guillermo.
3. Torres Herebia, Darío de Jesús.
4. Jacquet Montiel, Jorge Daniel.

[b][i]Suplentes:[/i][/b]
1. Benitez Maidana, Rodrigo Daniel.
2. Zena Turró, Juan Guido.

El equipo de El Salvador es:

1. Alfonso Abraham Alvarenga Gamero
2. Luis Damián Reyes Rodríguez
3. Mario José Castro Hernández
4. Johnatan Josué Gámez Rodríguez

Para Colombia, como siempre, dos de IMO y dos nuevos:

1.Santiago Cuéllar
2.Federico Castillo^2
3.Gustavo Nicolás Paez
4.Hayden liu

Y aquí el equipo chileno

1. Jaime Soza
2. Sebastián Henríquez
3. Estefanía Vidal
4. Sebastián Illanes (yo!  :) )

Saludos

Jorge Velásquez

Vandria Álvarez

Hugo García

Esteban Arreaga

 :)

[color=blue]Hace tiempo que terminó la Olimpiada…

En Paraguay y en Portugal hubo mucho de que festejar. Ambos equipos tuvieron resultados históricos el año pasado.[/color]


Could you explain me like 'talking-to-a-complete-idiot-style' what are polynomials? :oops: Please give a few examples of them

Why did you bump this topic to ask this question?

A polynomial is just any expression that you can get by repeatedly adding and multiplying numbers together with $x$.

For example, if I took an $x$, added $3$, multiplied by $x$ again, and added $-4$, I would get

$(x+3) \cdot x-4 = x^{2}+3x-4$

Which is an example of a polynomial.

Another example:  If I took a $10$, added $x$, multiplied by $7$, added $-2x$, and multiplied by $\frac{1}{3}$, I would get

$\left( (10+x) \cdot 7-2x \right) \cdot \frac{1}{3}= \left( 70+7x-2x \right) \cdot \frac{1}{3}= \frac{70}{3}-\frac{5}{3}x$

More generally, then, a polynomial is any expression of the form

$P(x) = a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}$

Where the $a_{0}, a_{1}, ... a_{n}$ are some numbers.  (Depending on the context, they are usually real numbers, but sometimes a problem will stipulate that they must be rational or integer; but really the concept of polynomial is very powerful, and actually they can be anything depending on the context.)

And note that for example $\text{cos }t\text{ }xy+y^{2}+2$ is a polynomial with respect to $x$ and $y$ but not with respect to $t$. In mathematics, it is possible to deal with several variable polynomials too.


function
number theory unsolved
number theory

Find all positive integer solutions (a,b,c) to the function a^2+b^2+c^2=2007, where a<=b<=c

There are no solutions, because $ 2007\equal{}7\mod 8$.


algebra
polynomial
linear algebra
linear algebra unsolved

$P=A+iB \in GL_{n}(C)$ with $A,B \in M_{n}(R)$
suppose $A,B,A+B,A-B,A+2B,A-2B$ are singular

Prove that $n\geq 6$

Consider $p(x,y) = \det(yA+xB)$ as a homogeneous binary form in $x,y$ of degree $n$.  We're told that $A+iB$ is invertible so $p(1,i) \not= 0$ and in particular $p \not= 0$.  Further, $0,\pm1,\pm2$ are roots of the univariate polynomial $p(x,1)$ and $0$ is a root of $p(1,y)$: so $p$ has degree at least 6 and hence $n \ge 6$.


function
quadratics
algebra proposed
algebra

Let $ a,b,c,d $ be distinct real numbers . Prove that :
 $ \sum \frac{a^4+1}{(a-b)(a-c)(a-d)} = a+b+c+d $

bring everything to LHS, make everything a bigass fraction, and we prove the identity if we can show that the numerator always=0.

when writing the numerator out, you can write it in terms of a function of $a^2$, all the terms will have $a^4$ or $a^2$ terms, so it's a quadratic in terms of $a^2$, and you can show it has 3 distinct roots, which means it'll always be zero so the identity holds.


Find all natural $ x,y$ such that $ xy\plus{}x$ and $ xy\plus{}y$ are both squares .

there is a similar problem as follow:
[b]Suppose $ x,y$ are positive integers such that both $ xy\plus{}x$ and $ xy\plus{}y$ are squares. Prove that exactly one of the $ x,y$ is a perfect square.[/b]
which was pretty difficult itself,now about finding all pairs of $ x,y$ which the propperty holds must be very difficult to solve,anyway at least we know that exactly one of $ x,y$ is a perfect square... :D

babak ghalebi, this similar problem you mention... do you have a solution or a link to a solution?

http://www.mathlinks.ro/viewtopic.php?t=6505
THere the problem babak ghalebi was talking about, is posted.


Να λυθεί  στους ρητούς η εξίσωση
$30(tanx)^3-113(tanx)^2+107tanx-30=0$

[quote="kostas"]σαν εξισωση δε μου λεει κατι.σαν ταυτοτητα μπορει, σαν πολυωνυμο ομως...  ;)[/quote]

Την έχεις λύσει ??Αν ναι βάλε τη λύση σου  ;)

Έστω η συνάρτηση $f(y) = 30y^{3}-113y^{2}+107y-30$

Παρατηρούμε ότι το $y_{0}=2.5$ είναι ρίζα της $f\ \ ($ μη με ρωτήσετε πώς το παρατηρούμε  :lol: $)$

Με σχήμα Horner βρίσκουμε ότι η $f$ γράφεται $f(y) = 2(y-2.5)(15y^{2}-19y+6)$

Το τριώνυμο έχει ρίζες $y_{1}=0.6$ και $y_{2}=0.\overline{6}$

Τέλος, βρίσκουμε το $x$ από τη σχέση $\tan x = y$

[quote="pontios"]( μη με ρωτήσετε πώς το παρατηρούμε  :lol: )[/quote]

Απλά κοιτάμε ποιες μπορεί να είναι οι ρητές ρίζες της εξίσωσης.  :) . Άρα όπως είπες
$f(y)=30y^{3}-113y+107y-30=(-5+2 y) (-2+3 y) (-3+5 y)$

Μετά όμως έκανες λάθος πέφτοντας στην παγίδα της άσκησης  :wink: .

Loipon tha grapsw ena thewrima to opoio einai poli poli palio kai arketa endiaferon gia na to psaksete (osoi den to exete ksanadei).

Leei oti : " Ta monadika rita pollaplasia $x$ tou diastimatos $\big(0,\frac{\pi}{2}\big)$ gia ta opoia to $cosx$ einai riza ritou einai ta $\big\{\frac{\pi}{6}, \frac{\pi}{4},\frac{\pi}{3}\big\}$".

Afoy loipon i parapanw eksiswsi mas dinei $tanx=\frac{5}{2},\frac{2}{3},\frac{3}{5}$ ara $tan^{2}x=\frac{25}{4},\frac{4}{9},\frac{9}{25}$ ara $cosx=\pm\sqrt{\frac{4}{29}},\pm\sqrt{\frac{9}{13}},\pm\sqrt{\frac{25}{34}}$ (an oles oi prakseis einai swstes).

Sinepws anagastika to $x$ (ws ritos arithmos), prepei na einai kapoio apo tis gnwstes gwnies $\frac{\pi}{6}, \frac{\pi}{4},\frac{\pi}{3}$, pragma pou den isxiei kai etsi den exei rites liseis i parapanw eksiswsi.

Alexandros

Πολύ σωστά .(αν και είχα στο μυαλό μου ένα αντίστοιχο θεώρημα για τις εφαπτομένες .  :wink:

to prosarmozeis antistoixa opws theleis kai gia imitono kai gia o,ti thes! einai poli flexible oi trigonometrikes sinartiseis :)

Alexandros

[quote="silouan"]
Μετά όμως έκανες λάθος πέφτοντας στην παγίδα της άσκησης  :wink: .[/quote]

Α! ήταν στους ρητούς. ok!
Όντως έπεσα στην παγίδα   :D


linear algebra
matrix
algebra
polynomial
logarithms
floor function
number theory

From jmerry's post you can derive a solution of the general term as:

$ a_n \equal{} \exp\left(\ln 2 \left\lfloor \frac{\sqrt{8\,n\plus{}1}\plus{}1}{2}\right\rfloor\right) \plus{} 3 \exp\left(\ln 2 \left( n \minus{} \frac{1}{2} \left\lfloor \frac{\sqrt{8\,n\plus{}1}\plus{}1}{2}\right\rfloor \left\lfloor \frac{\sqrt{8\,n\plus{}1}\minus{}1}{2}\right\rfloor\right)\right)$

for $ n \geq 0$. The $ \exp(\ln 2 (\cdot))$ part is just for nicer representation - it's just the power of 2.

How did I see it? I looked at the big numbers, saw the doubling in the differences, and traced it back.

Now, the second set:
The first one: Start with a base of $ 19, 23, 29, 31, 37, 47, 49, 53, 65, 79, 85, 89,\cdots$. Then, take the powers of $ 2$ times those. I have no idea what's generating those numbers; why $ 65$ and not $ 41$?

The other three follow the same doubling principle. Understand the odd ones, and you understand everything.

Again, I see no discernible pattern in the odd ones. Don't expect anything without some idea of where these are coming from.

The actual Sequence is the following:

4.
\[ \left( \begin{array}{cccccccccccccccc} 0 & 1 & 2 & 4 & 8 & 5 & 7 & 10 & 11 & 14 & 20 & 19 & 23 & 29 & 38 & 65 \end{array} \right)\]
5.
\[ \left( \begin{array}{cccccccccc} 0 & 1 & 2 & 4 & 8 & 16 & 5 & 7 & 10 & 11 \\
14 & 19 & 20 & 22 & 28 & 40 & 19 & 23 & 29 & 31 \\
37 & 38 & 46 & 49 & 58 & 76 & 65 & 73 & 85 & 103 \end{array} \right)\]
6.
\[ \left( \begin{array}{cccccccccc} 0 & 1 & 2 & 4 & 8 & 16 & 32 & 5 & 7 & 10 \\
11 & 14 & 19 & 20 & 22 & 28 & 35 & 38 & 40 & 44 \\
56 & 80 & 19 & 23 & 29 & 31 & 37 & 38 & 46 & 47 \\
49 & 53 & 58 & 62 & 65 & 74 & 76 & 89 & 92 & 98 \\
116 & 152 & 65 & 73 & 85 & 89 & 101 & 103 & 119 & 125 \\
130 & 143 & 146 & 170 & 179 & 206 & 260 & 211 & 227 & 251 \end{array} \right)\]
etc.

I wanted a general term for each one of this sequences. there are $ 2^n$ numbers in each one of them.
If we make the plot of the sequence, any of them, we have Jumps.
I noted that the jumps are at determined points, so I made this sequence in subsequences.

the zero one is just 0
the first one is simple 1,2,4..
the second one is the 5,7, 10

etc.

Now, thanks to you, I noted one thing.
The sequence of the jump $ p$ is determined by:
$ \sum _{i \equal{} 1}^p \alpha _i 2^{\kappa _i \minus{} 1}$

If we have $ 1 \leq \kappa_{i \plus{} 1} \leq \kappa_i$ we have also $ \alpha_{i} > \alpha_{i \plus{} 1}$

[quote="EulerIntegral"]
milin, your formula is not in order, but I'm now aware that order is not important in this matter[/quote]

How is it not? Am I missing something?

$ 2^{\left\lfloor \frac{1}{2} \left(\sqrt{8 n\plus{}1}\plus{}1\right)\right\rfloor }\plus{}3 2^{n\minus{}\frac{\left\lfloor \frac{1}{2} \left(\sqrt{8   n\plus{}1}\minus{}1\right)\right\rfloor  \left\lfloor \frac{1}{2} \left(\sqrt{8 n\plus{}1}\plus{}1\right)\right\rfloor }{2}}$

Using it:

{5, 7, 10, 11, 14, 20, 19, 22, 28, 40, 35}

see? 20,19 and others.


Let $N = 111...1222...2$, where there are 1999 digits of 1 followed by 1999 digits of 2. Express $N$ as the product of four integers, each of them greater than 1.

[hide="My solution"]

The sum of digits of $N$ is 1999(1+2) or 5997. Since 5997 is divisible by 3, $N$ is divisible by 3 as well. Also, since last digit of $N$ is 2, it is divisible by 2 as well. This can be shown as this:

$N = (11...11) \cdot 10^{2000} + (22...22) \cdot 10^0$

Taking 2 out, we get:

$N = 2[(11..11) \cdot 10^{1999} \cdot 5 + (11...11) \cdot 10^0]$

Inside bracket, we can take $(11...11)$ out as well.

$N = 2(11...11)[10^{1999} \cdot 5 + 1]$

Now, let's consider the number inside the bracket in base 3.

$10^{1999} \equiv 1^{1999} \equiv 1 \mod 3$
$5 \equiv 2 \mod 3$

Thus, $10^{1999} \cdot 5 \equiv 2 \mod 3$. Adding this with $1 \mod 3$ gives $3 \mod 3$ or 0. So, $3|N$. This is:

$N = (2)(3)(11...11)[\frac{10^{1999} \cdot 5 + 1}{3}]$

To calculate the number inside bracket, we test for smaller powers of 10.

$10^1 \cdot 5 +1 = 51$ and $\frac{51}{3} = 17$
$10^2 \cdot 5 + 1 = 501$ and $\frac{501}{3} = 167$
$10^3 \cdot 5 + 1 = 5001$ and $\frac{5001}{3} = 1667$

So, 0's are replaced by 6's in quotient. Hence, $10^{1999} \cdot 5 + 1 = 50...01$ and $\frac{50...01}{3} = 16...67$ where there are  1998 6's.

Therefore, $N = (2)(3)(11...11)(16...67)$ [/hide]

Is this right?  :lol: (Took me sometime to type this..)

[quote="xxreddevilzxx"]Well one comment to Silverfalcon:

We are done once we can find 3, because then we can find the fourth one by division, etc. we know that right away it is divisible by $2$ and $1111.....1111$ (1999 1's). Then by your method, it is also divisible by $3$. Then we know that the number can be written as $2\cdot 3 \cdot 111......111 \cdot x$, where $x$ could be easily found.[/quote]

True. I knew I did more work than I needed but it helped me to see the right way.

Just wondering, how would anyone (this is not necessary on the problem) prove that $11...11$ (1999 1's) is pirme?

I think it's prime but I just can't prove that it is..  :? Or prove that it's not prime.

[quote="Silverfalcon"]
Just wondering, how would anyone (this is not necessary on the problem) prove that $11...11$ (1999 1's) is pirme?

I think it's prime but I just can't prove that it is..  :? Or prove that it's not prime.[/quote]

I don't know what to tell you about that, because 1999 is prime.

[quote="Silverfalcon"]
Just wondering, how would anyone (this is not necessary on the problem) prove that $11...11$ (1999 1's) is pirme?

I think it's prime but I just can't prove that it is..  :? Or prove that it's not prime.[/quote]

No it  ($11...11$ (1999 1's)) is [b]NOT[/b], it is divisible by [b]366008861997013 [/b]
 And there are stlill more  factors.

Also BTW the original  USAMTS problem has another factor $23$
And by the way $(10^{1999}+2) /(3*2*23).$ is not prime either!

I remember this question as,  I corrosponded with  Dr. Kelly ( Then the Director)  about not-trivial other factors of the question. !

If you want to  know some trivia  :
only known primes of the type $1111 \dots $    (or n 1's) are when you repeat it either  2 or 19 or 23 or 317  or 1031 or 49081 or 86453 times .. for all other cases  (less than 100,000 1's) they are  not prime.

[quote="Silverfalcon"]You said "23 is ANOTHER factor."

So does that mean my answer is incorrect? I don't see why because no matter how long they are, they ARE still integers..[/quote]

NO, , your answer is [b]not[/b] [b]wrong[/b], as it did not ask you to find PRIME  factors.  .. In the original contest  a few students did find that 23 was a factor (to get a full score, it was not necessary to find <small> factors other than 2 and 3)

 Of couse, the factor (A 15 digit number) I put above,  most likely,  very few, if [b]any[/b] ((except a few with whom I shared the answer before, or who have seen this post),  in the world knows.

For really large numbers:
( With some advance math tricks - It is easy to prove that the number is not prime, it is much harder to find its factors)

Wait.... look back at the problem. It says to express the number in product of four integers, not four PRIME integers. I am not sure why you are looking for prime factors, Silverfalcon.

EDIT: u said it had notthing to do with the problem. sorry.


AMC
AIME
algebra
difference of squares
special factorizations

In square $ ABCD$, BC is the radius of a circle centered at C. a point on circle C is 1 unit away from both AD and AB. What is the radius of this circle?

[hide]
Let the point on circle $ C$ that is one unit away from both $ AD$ and $ AB$ be $ M$, the perpendicular from $ M$ to $ AD$ be $ E$, and the perpendicular from $ M$ to $ AB$ be $ I$. Then we have two squares, $ ABCD$ and $ AIME$. We know that the side length of $ AIME$ is 1, so the diagonal $ AM$ is $ \sqrt {2}$. The diagonal $ AC$ is $ r\sqrt {2}$, and $ MC$ is $ r$. So we set up an equation:

$ r \plus{} \sqrt {2} \equal{} r\sqrt {2}$

from which we can deduce that $ r$ is $ \boxed{2 \plus{} \sqrt {2}}$.

[/hide]

Edit: Of course, there is always the possibility that $ M$ could be halfway around the circle... in which case,

$ 1 \minus{} \frac{r\sqrt{2}}{2} \equal{} r$

[Magic algebra]

$ r \equal{} 2 \minus{} \sqrt{2}$

So our final answer is $ r \equal{} \boxed{2 \pm \sqrt{2}}$.

if we set the x axes and y axes as AB and AD, then the circle C contains point (1,1) and is tangent to and x and y axes. The center of the circle is point (x,x) and radius is x so if we plug in the variables to the equation of a circle, we get $ (x \minus{} 1)^2 \plus{} (x \minus{} 1)^2 \equal{} x^2$ ==> $ x^2 \minus{} 2x \plus{} 1 \plus{} x^2 \minus{} 2x \plus{} 1 \equal{} x^2$==> $ x^2 \minus{} 4x \plus{} 2 \equal{} 0$= = >$ x \equal{} \frac {4 \pm \sqrt {16 \minus{} 4(1)(2)}}{2} \equal{} \frac {4 \pm \sqrt {8}}{2} \equal{} \frac {4 \pm 2\sqrt {2}}{2} \equal{} [2 \pm \sqrt {2}].$

edit: i think there are two answers. In this case, $ 2 \plus{} \sqrt{2}$ and $ 2 \minus{} \sqrt{2}$.

Look at it this way:

Case 1: $ M$ is [i]inside[/i] square $ ABCD$. Then, the answer is $ 2 \plus{} \sqrt{2}$.
Case 2: $ M$ is [i]outside[/i] (to be exact, $ 225^\circ$ from the polar axis) square $ ABCD$. Then, the answer is $ 2 \minus{} \sqrt{2}$.

Case 2 assumes that your statement "... is 1 unit away from both AD and AB" means line $ AD$ as opposed to line segment $ AD$, same for $ AB$.

my bad. i forgot to say it had to be inside the square
EDIT: how does it make a difference. i cant picture it.

[quote="uldivad9"]Well, let the radius be x. We then have sqrt2 + x = x sqrt 2, by drawing the square of side 1 and using it's diagonal. Do all the iffy algebra and you get x = 2 + sqrt2.[/quote]

wait, when I did the problem, I got sqrt2 + x = xsqrt2, but how would you solve for x? (I haven't really learned how to solve this yet.) I did a really complicated way, but I believe that there is a simpler way.

[hide="what I did"]
I changed sqrt2+x=xsqrt2 into xsqrt2-x-sqrt2=0

I added 1 to each side and got xsqrt2-x-sqrt2+1=1. Then I factored the left side to get (x-1)(sqrt2-1)=1 Then I divided sqrt2-1 from both sides to get
x-1=1/sqrt2-1 Multiplying the top and bottom of the fraction by sqrt2+1, I got  x-1=sqrt2+1/1
so x=sqrt2+2.
[/hide]

Basically my question is, is there a simpler way?

Yes:

You can manipulate the equation to say:
$ x\sqrt {2} - x = \sqrt {2}$
Then factor $ x$ out of the left side:
$ x(\sqrt {2} - 1) = \sqrt {2}$.
Then divide both sides by $ \sqrt {2} - 1$:
${ x = \frac {\sqrt {2}}{\sqrt {2} - 1}}$
Rationalize denominator:
${ x = \frac {\sqrt {2}(\sqrt {2} + 1)}{(\sqrt {2} - 1)(\sqrt {2} + 1)}} = \frac {2 + \sqrt {2}}{2 - 1} = \boxed{2 + \sqrt {2}}$

I'm not sure if that's simpler, but it's a more direct approach.

[quote="Ihatepie"]
wait, when I did the problem, I got sqrt2 + x = xsqrt2, but how would you solve for x? (I haven't really learned how to solve this yet.) I did a really complicated way, but I believe that there is a simpler way.
[/quote]

Right you are, there's an easier way.
sqrt2 = x(sqrt2 - 1)
x = sqrt2/(sqrt2-1)
For Simplest Radical Form, we multiply top and bottom of the fraction by (sqrt2 + 1) [we do this because difference of squares conveniently removes the sqrt]
and we get (2 + sqrt2)(2-1) = x = 2+sqrt2.

Aww man, Yongyi beat me... and his is prettier because it has LaTeX... :(


$p$ is a prime and $a, b, c$, are distinct positive integers less than $p$ such that $a^{3}= b^{3}= c^{3}(\mod p)$. Show that $a^{2}+b^{2}+c^{2}$ is divisible by $a+b+c$.

[quote="Rust"]Therefore $a+b+c=p \ or \ 2p$ [/quote]

because $a,b,c$ are roots of $x^{3}-k=0$ in $\mathbb{Z}_{p}$ therefore $p|a+b+c$  :wink:


blogs
quadratics
algebra
polynomial
modular arithmetic
quadratic formula
special factorizations

[quote="maokid7"]How would you solve a problem like
$4x^{2}-4x-1\equiv 0\bmod 7$.
[hide="here is what i have"]
after reading a post on t0rajir0u's blog i got to
$(x-r_{1})(x-r_{2})\equiv 0\bmod 7$ where $r_{1},r_{2}$ are the roots of $4x^{2}-4x-1$ (i leave it to u to find them if you wish)
so we have either $x-r_{1}\equiv 0\bmod 7$ or $x-r_{2}\equiv 0\bmod 7$
one of these leads to solving
$2x+\sqrt{2}\equiv-1\bmod 7$
since $\sqrt{2}\equiv 3\bmod 7$ (for more information go to this link http://www.artofproblemsolving.com/Forum/weblog.php?w=634 )
this leads to $2x\equiv-4\bmod 7 \Longrightarrow x\equiv 5\bmod 7$
[/hide]
but that doesn't work. can any one tell me where i went wrong and how to fix the problem.[/quote]

multiply by 2, to get

$7|(x^{2}-x-2)$, so $7|(x-2)(x+1)$, since $7$ is prime, we have either $x\equiv 6\mod 7$, or $x\equiv 2\mod 7$

[quote="maokid7"]thanks. but how would you attack the general form $ax^{2}+bx+c\equiv 0\bmod p$ where $p$ is an odd prime number?[/quote]

a has an inverse mod p, multply by that to get a new quadratic: x^2+b'x+c'\mod p, test x=0,1,2,...,p-1 mod p for solutions, then you can add multiples of p to the coefficients so the polynomial factors that way

The quadratic formula also works modulo primes.  In this case

$\ x \equiv \frac{4 \pm \sqrt{32}}{8}\equiv \frac{4 \pm 12}{8}\equiv-1, 2 \mod 7$

[hide]
It's very straightforward by completing the square:
\[(2x-1)^{2}\equiv 2\equiv 9\pmod{7}\]
So $2x-1\equiv \pm 3$, which gives $x\equiv 2$ or $x\equiv 6$.
[/hide]

[quote="mna851"]The quadratic formula also works modulo primes.  [/quote]

This is the point I was trying to get at in my blog post.   :)


The point $ (0,8)$ lies on the graph of the curve defined by $ y\equal{}c\cdot2^x$. What is the value of $ y$ when $ x\equal{}2$?

hello, because $ (0,8)$ lies on the graph we have $ y\equal{}c\cdot2^0\equal{}c$, hence we get the function $ f(x)\equal{}8\cdot2^x$, so our value is $ f(2)\equal{}8\cdot4\equal{}32$, the answer is $ \boxed{32}$
Sonnhard.


geometry calculus derivative search integration probability CMU

[quote="Fierytycoon"]Do most aspiring quants study a lot of financial mathematics before sending their resumes to financial firms?

If so, what are some good (introductory) books for financial mathematics?

Thanks for any help you can provide.[/quote]

More than they used to.  I didn't know a thing before I started.

The place to start for really diving in to the mathematics behind asset pricing is John Hull's Options, Futures, and Other Derivatives.  Some of the math may be a bit daunting for high school, but it will suggest areas of math for you to learn.

[quote="ffdbzathf"]there appears to be a 4th and 5th edition available on amazon of that book.  does it make a difference (sometimes earlier editions are better than newer ones) in this case?

there is also another book by the same author which is supposedly for undergraduates instead of graduates, Fundamentals of Futures and Options Markets.  know anything about this?[/quote]

I don't know anything about the fundamentals book, but I'll bet it's the same as the other, but with training wheels on.

I would get the cheapest edition I could find.

[quote="rrusczyk"]I would get the cheapest edition I could find.[/quote]

I looked up the book at my university's library earlier today, and there is indeed a difference between the different editions. The author has been adding a few new chapters every time he publishes a new edition.

If there's not much price difference, go ahead and get the most recent.  But if you can pick up a cheap used one, I'd do that.  It's not a cheap book!

Hull's OFOD is quite heavy on the mathematical theory, and would be very intimidating to someone trying to learn about derivatives; it's more of a reference to someone who is familiar with the subject.

FFOM is much more user friendly.  I would recommend it to a beginner, particularly the chapters on binomial trees.

A fun book:

"My life as a Quant"
I really enjoyed that book. Deals a lot about finance, physics, relationships with both, firm/university politics etc. Highly recommend.

Here is possibly the most comprehensive quant reading list I found on the web:

http://finmath.com/

Enjoy! Happy reading (for quite a while to come!). Problem is the dent it has on your wallet. ouch!

Also consider Paul Wilmott on Quantitative Finance" I think its pretty expensive though.  I just came a cross a nice set of notes by Lane Hughston and Christopher Hunter of Kings College, (search for Lane Hughston Kings College - they are on his site under publications i think) They get quite technical but seem to go about it in a really nice way.

No matter which book you read the continuous-time/stochastic calculus stuff is going to be a problem at the begining , I suggest really focussing on the binomial model at first as it allows you to get a feel for alot of important ideas behind the Black-Scholes result without much more than school maths and some stats.  hope that helps.

[quote="Fierytycoon"]Do most aspiring quants study a lot of financial mathematics before sending their resumes to financial firms?

If so, what are some good (introductory) books for financial mathematics?

Thanks for any help you can provide.[/quote]

Here are two books I have read:

1.  Salih N. Neftci ---Introduction to the Mathematics of Financial Derivatives

   This book introduces stochastic calculus through financial examples how the stochastic is used in the option pricing theory. A Ph.D. will have no problem follow the book chapter by chapter and do the exercises after each chapter to get falimilar with the conecepts and principles. Among them, you should know the definitions of Markov, Martingale, Brownian motion, Ito's Integral, Ito's lemma, Girsanov theorem, Feyman-Kac.

2. Steven E. Shreve --- Stochastic Calculus for Finance II : Continuous-Time Models (Springer Finance)
    I personally believe this is one of the best books on stochastic calculus for finance applications. It is accessible for anyone who has a background in calculus, probability, algebra and finance. If you want to go beyond the Neftci level, you should use this textbook. This book is the volume II of Shreve's two volume series of stochastic calculus for finance. Volume I is Stochastic Calculus for Finance I: The Binomial Asset Pricing Model. The difference betwen the two volumes is that first discusses discrete-time models while the second discusses continuous-time models. Shreve has been writing his book for many years dated back to 1980s when option pricing theory was getting popular. He cofounded Carnegie Mellon University's Master’s program in Computational Finance and this book evolves from the class he taught there over ten years. Over the time, he distributed the draft freely to many quantitative finance students who have benefited a lot from his notes. The volume II is good enough for a quant to learn suffice stochastic calculus. Chapter 1, 2 3 present basic background in probability theory such as probability measure, conditional expectation and sigma-algebra, abstract concept for information. Chapter 4 discusses stochastic calculus, including Brownian motion, quadractic variation, Ito Integral, Ito's lemma, stochastic differential equation (SDE) and derivation of famous Black-Scholes-Merton equation etc. Chapter 5 discusses risk neutral measure, Girsanov theorem and chapter 6 discusses Feyman-Kac theorem which relates risk-neutral measure to partial differential equations. Chapter 7-11 discusses exotic options, American options, change of numeriaire, term structure of interest rate and jump diffusion process. A quant should finish the first six chapters to have a solid background in stochastic calculus. A quant would better finish the exercises in the book as well.

There are some other book reviews in the website:

 [url=http://www.quantfinancejob.com]QuantFinanceJob.com[/url]

[quote="River Phoenix"]Hull Hull Hull, wonderful Hull Hull Hull :)[/quote]

I am not quite excited about Hull's book. Well, it is the bible on derivatives for practitioners who wants to yield a proper understanding of those instruments but for somebody like me who is really interested in those issues, the book is not sufficiently theoretically founded. They don't offer all proofs etc. But in order to gain an initial understanding the book is probably appropriate.

I second the recommendation for Sheve's Volume II.  It is not an easy read by any means, but its treatment of key topics (i.e. all of stochastic calculus) is much more complete than Hull's.  This is to be expected, since Shreve's book is really a math book, where Hull's is not.

Hi Rich,

Hull's 6th edition is the most recent, as far as I know. It's quite expensive. I've got a hardcover version. But one can get significantly cheaper versions online, probably softcovers printed outside of US.

From what I've heard, it's a great investment if the return you're looking for is to land a quant job.

By the way, how can one land a quant job without going through a recruiter?

Also, how difficult is it for one to land a quant job with just average math/programming skills?

I'm not nearly at the USMO level skills as most on this board possess. But I do have a physics and an apllied math background at the ABD level from a mid-tier Ivy. I also have a genuine interest in physics, math and programming and I have a strong commitment to learn finance.

As you may have guessed, my goal is to position myself to make a living doing something that I'm passionate about. Now that would be a wonderful thing.

Also, does 1-2 year of credit risk managment in a top tier firm help or hurt my chances of landing a quant job?

I'd appreciate any input one can offer. Thanks ver much.

Cheers,
Mike.

[quote="mikeymike"]

By the way, how can one land a quant job without going through a recruiter?

Also, how difficult is it for one to land a quant job with just average math/programming skills?

Also, does 1-2 year of credit risk managment in a top tier firm help or hurt my chances of landing a quant job?

[/quote]

For the first question: networking.

Second: Hard.  You're probably best off improving your coding skills.

Third: Not sure.  Depends on what you call a quant job.  Middle-office risk evaluation of internal positions might be possible, but it's hard to say.

[quote="saturnlife"]Is coding really considered more important than math skills for quants?[/quote]

Hard to say which is more 'important', except that coding is essential in the job.  Certain areas of math are necessary for certain jobs, but you'll have a hard time even getting an interview unless you can show some experience with coding.  (There are lots of good non-quant trading jobs that don't require coding, of course.)

To be honest, if Hull's book seems too mathematical then you are a considerable distance from being ready to start on a quant career, since it is seen by some as their first book. Some of my the books written by my colleague Paul Wilmott are also very popular, but the maths content is approximately the same.

In our Guide, there is a reading list, and to be honest it's neither short nor easy. You don't have to read them all, but to stand a realistic chance you need to have a good go.

A version is at http://www.amazon.com/Reading-nbsp-List-nbsp-for-nbsp-a-nbsp-Job-nbsp-in-nbsp-Quantitative-nbsp-Finance/lm/R2JIGJY4H5FIO9/ref=cm_srch_res_rpli_alt_2

The amount of coding varies a lot, the generally accepted average is 60% but 95% is not unknown, and certainly there will be phases where that is true.
That's not quite the same as important. Certainly you ought to start on C++ ASAP, [i]not[/i] Java, but you need very good maths more.


Solve for x:

log_2(log_4(x))+log_4(log_2(x))=2.

Yup, fiery

In case u guys wonder how fiery simplified the logs, here's a nice, um , thing:

log_(a^b) of (c^d) is equal to (d/b)*log_a of c

Ok here it goes.

[hide]log2(log4x) + log4(log2x)=2

((log4(log4x))^2) + log4(log2x) = 2

log4((log4x)^2)*(log2x)=2

((log4x)^2)*(log2x)=16

2((log4x)^2)(log4x)=16

(log4x)^3=8

log4x=2

x=16[/hide]

I know I posted late.  Sorry, I'm a day ahead!

Your right! Good job. (but don't notice that this is, like, the only thing positive I wrote and said all day :) )

Another way to do it.

[hide]

Note that log_2 x = log_4 x:^2:

Reducing the original equation, we get

log_4[(log_4 x:^2:)]:^2: = 2

16 = (log_4 x:^2:):^2:

log_4 x:^2: = 4

4:^4: = x:^2:

x = 16

[/hide]

wow, southpaw
u actually posted a full solution
im impressed!! :D


geometry
circumcircle
incenter
trapezoid
trigonometry
geometry proposed

[color=darkblue] Let $ ABCD$ be a quadrilateral with $ \angle{DAB} \equal{} 60^\circ$, $ \angle{ABC} \equal{} 90^\circ$, $ \angle{BCD} \equal{} 120^\circ$. Diagonals of the quadrilateral intersect at $ M$, such that $ MB \equal{} 1$, $ MD \equal{} 2$. Calculate tha area of $ ABCD$.
[/color]

I got a horrible solution, using Pythagoras on ABC, power of point and Stewart's Theorem applied on ABC with cevian AM.
Anyone have a nice solution, or is this problem so ugly?

Obviously BD = 3 and, with <BAD = 60 degs we get the radius of (ABCD) R = sqrt(3). Applying Stewart in triangle OBD with cevian OM (O being the circumcenter) we get OM = 1, then we easily get cos (<OMB) = -1/2, that is, <(AC,BD) = 120 degs. But AC is a diameter of the circle, hence
AC = 2 sqrt(3), further the area ABCD , S = (AC.BD.sin(AMB))/2 = 4.5.
Junior level...

Best regards,
sunken rock

I know that it's junior level,  but our solutions are ugly :P

I'll try and make it more beautiful then ;D

Let $ G$ be the midpoint of $ \overline{BM}$ , and $ P$ a point such that $ MP \equal{} 1$ . We easily find that $ \widehat{BPG} \equal{} 90^\circ$ .

Since opposite angles sum $ 180^\circ$ , quadrilateral $ ABCD$ is cyclic.

Through simple angle chasing, we find that $ \widehat{DBA} \equal{} 45^\circ$ , which will imply that $ \triangle MPB \cong \triangle BPG$ , $ \triangle APD$ and $ \triangle PDC$ are isosceles, and point $ P$ is the center of the circumference.

Through Pythagoras, $ AD \equal{} DC \equal{} \sqrt {6}$ and $ [ADC] \equal{} \frac {1}{2}\cdot \sqrt {6}\cdot \sqrt {6} \equal{} 3$

Since $ \triangle ADB \sim \triangle BMC$ , we have the following proportion:

$ \frac {AC}{3} \equal{} \frac {1}{BC}\therefore AC\cdot BC \equal{} 3\therefore [ABC] \equal{} 1.5$

And finally $ [ABCD] \equal{} [ABC] \plus{} [ADC] \equal{} 3 \plus{} 1.5 \equal{} 4.5$

$ U \equiv AB \cap CD,$ $ V \equiv BC \cap DA,$ $ W \equiv AC \cap UV.$ $ C$ is orthocenter of $ \triangle AUV$ and incenter of its orthic triangle $ \triangle WDB$ $ \Longrightarrow$ $ \angle BWD \equal{} 180^\circ \minus{} 2 \angle VAU \equal{} 60^\circ$ and $ \frac {WD}{WB} \equal{} \frac {MD}{MB} \equal{} 2$ $ \Longrightarrow$ $ \angle DBW \equal{} 90^\circ$ $ \Longrightarrow$ $ \angle ABD \equal{} 45^\circ.$ Perpendicular to $ AB$ through $ D$ cuts $ AB$ at $ N$ and the circumcircle $ \odot (ABCD)$ again at $ E.$ $ EBCD$ is isosceles trapezoid $ \Longrightarrow$ $ CE \equal{} BD$ and $ \angle ECB \equal{} \angle DBC \equal{} 45^\circ$ $ \Longrightarrow$ $ CE \perp BD.$ Right $ \triangle DNB$ is isosceles with $ ND \equal{} NB$ $ \Longrightarrow$ $ \triangle AND \cong \triangle ENB$ $ \Longrightarrow$ $ S(ABCD) \equal{} S(EBCD) \equal{} \frac {_1}{^2} BD^2 \equal{} \frac {_9}{^2}.$

Due to fact that BD = 3 is opposite to angle <BAD = 60 degs it becomes obvious that OB = sqrt(3)  ( 1 ) and <OBD = 30 degs. See that OB^2 = BM.BD ( 2 ), hence triangles OBM and BDO are similar and angle <BOM = 30 degs, consequently <BAO = 15 degs and
<DAO = 45 degs, the right angled triangle ADC has the area S = (AC^2)/4 = 3. Further see that S[ACD]/S[ABC] = DM/BM = 2, hence [S[ACD] + S[ABC]]/S[ACD] = (MB + MD)/DM ( 3 ), but S[ACD] + S[ABC] = S[ABCD] = 4.5

Best regards,
sunken rock

i've got he feeling that i saw this problem in the highschool section... :?:

[quote="Ahiles"][color=darkred] Let $ ABCD$ be a quadrilateral with $ \angle{DAB} \equal{} 60^\circ$, $ \angle{ABC} \equal{} 90^\circ$, $ \angle{BCD} \equal{} 120^\circ$. Diagonals

of the quadrilateral intersect at $ M$, such that $ MB \equal{} 1$, $ MD \equal{} 2$. Calculate tha area $ S$ of $ ABCD$.
[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] Observe that the quadrilateral $ ABCD$ is inscribed in a circle $ w\equal{}C(O,R)$ with the diameter $ [AC]$ , where $ R\equal{}\sqrt 3$ .

From the power of $ M$ w.r.t. the circle $ w$ obtain $ MB\cdot MD\equal{}R^2\minus{}MO^2\implies MO\equal{}1$ . Thus, $ \triangle ODM$ is $ O$ - rightangle

because $ OM^2\plus{}OD^2\equal{}MD^2$ and $ DM\equal{}2\cdot OM\implies m(\angle OMD)\equal{}60^{\circ}$ . In conclusion, $ S\equal{}\frac {AC\cdot BD\cdot\sin\widehat {OMD}}{2}\equal{}\frac 92$ .[/color]


algebra
polynomial
Ross Mathematics Program
graph theory

I just came home from the district science fair. I'm going on to regionals. I hope I can continued to be succesfful at regionals and go onto state and be succesful there and there go to nationals (follow the footsteps of a fellow classmate last year). Do you guys have any experience and tips with science fairs?

I have a couple that I learned since doing them in 3rd grade.
Wear a Suit, Shirt and Tie
Have a very complex question and project!
Have a neat and concise board.
Know what you are going to say!

It just makes me wonder what kind of math belongs in science fairs. The nature of the subject doesn't seem to belong.

That's because math isn't necessarily a "science". The mathematical study of problems can be approached from a lot of different angles - philosophical, artistic, historical, or scientific. But generally speaking, the mathematical process is quite different than the scientific process.

I have seen some very good math projects at science fairs though - ranging from the elementary school level up to the high school level. Most math projects at science fairs tend to be studies of computational or statistical aspects of something. Last year I saw a very good project on some of numerical/statistical features related to Ulam's Conjecture.

Every now and then you'll get someone who actually proves a theorem at a science fair. But that's very rare.

Of the 6 math papers at Intel last year, we had 2 game theory proofs, one graph theory, one on cryptography, one on (I hope this is right) continued fractions for polynomials, and one on knot theory.  They were not all that computationally based, actually, if I remember correctly.  There may have been some computer-based stuff, but for the most part it wasn't.

New York City has a Math Fair in addition to the science fair, which is nice.

yea, my counselor from the ROSS program was Lester Mackey who placed sixth in intel with the graphy theory project last year.

OH YEAH! I made it to the Intel International Science Fair in Oregon with a project in Biochemistry! Nobody does math over here. So if you did you would of automatically got 1st in your division.


What is the remainder when $2^{87} + 3$ is divided by 7?

[hide="Solution"]2^n in mod 7 looks like this: 2,4,1,2,4,1,...
2^(3m) in mod 7 will always equal 1.
So 2^(87) in mod 7 equals 1.
3 in mod 7 equals 3.
1+3 in mod 7 is 4.[/hide]

[hide="My Answer."]Make a table of the powers of two mod 7.

$\begin{array}{cc}2^n&2^n(mod\ 7)\\\\
2^0&1\\
2^1&2\\
2^2&4\\
2^3&1\\
2^4&2\\
2^5&4\end{array}$
This repeats every three.  Therefore, $2^{87}(mod\ 7)=1$.  The problem asks for $2^{87}+3(mod\ 7)$.  This is $1+3=4$.  4 is the answer.[/hide]

[hide="My answer"]

:phi: (7)=6. So 2^87  :equiv: 2^3  :equiv: 1 (mod 7). Then the answer is 4.[/hide][/hide]


Suppose triangle ABC has a side of a=12 and angle B=36 degrees.  Find the range of values for side b so that:
a.  There are two possibilities for triangle ABC.
b.  There is only one possibility for triangle ABC.
c.  Triangle ABC cannot actually exist.

any ideas on how to solve this.
I think b is equal to or less than 0 for part c.

I assume that side b is opposite angle B, etc.

Draw segment CB (side a) with a fixed length 12.

Now draw a ray that is 36 degrees offset from side CB.

Finally, draw a circle centered at point C with radius b. The intersections of this circle with the ray in the previous step give the possible locations of A.

- If the circle is large enough to intersect the ray twice, there are two possible triangles.
- If the circle is tangent to the ray, there is one possible triangle, and it is a right triangle (why?)
- If the circle is too small to reach the ray, then there are no possible triangles.

If we find the value of b that results in tangency, we know that any larger value will result in two possible triangles, and any smaller value will result in no possible triangles.

If there is tangency, then note that $ b \equal{} 12 \sin 36^{\circ} \equal{} 3 \sqrt{10 \minus{} 2 \sqrt{5}}$

[quote="The Zuton Force"]- If the circle is large enough to intersect the ray twice, there are two possible triangles.
- If the circle is tangent to the ray, there is one possible triangle, and it is a right triangle (why?)
- If the circle is too small to reach the ray, then there are no possible triangles.[/quote]
But also note if the circle extends beyond point A (i.e., b>a), then there can only be one possible triangle, since one direction will be overreaching it.


Find the smallest positive integer x such that x is divisible by 8, (x-1) is divisible by 9, and (x-2) is divisible by 10.
   The solution is 352, but how do we approach this kind of problem?

Note that x+8 is divisible by 8,9 and 10. The least positive value for x+8 is hence gcd(8,9,10) = 360. Hence x = 360-8 = 352

Super thanks, hsbhatt. That was elegant!!!


see explosion involves heat light....potential energy due to the internal attraction of carts and son on so u can definitely say their momen is conserved but some energy may go 'wasted' as sound heat and light enenrgy

What potential energy are we talking about?

In a basic physics course, we generally restrict ourselves to gravitational, spring, and possibly electrical potential.  The energy required for an explosion, on the other hand, is likely to be [i]chemical[/i] potential energy.

If we are not talking about chemical potential, and you accept that potential energy is conserved, then mechanical energy cannot be conserved because kinetic energy is not conserved:  there is none before the explosion and there is some after.

Moreover during the explosion, energy goes away as heat....

On the other hand, on an explosion (for example, a granade) there is conservation of the position and velocity of the center of mass of the system of particles that makeup the device, and in an explosion there is conservation of linear momentum.

[quote="climskywell"]is mechanical energy conserved in an explosion?

say two carts are attached together and someone come and breaks them in to two.
the two carts would gain kinetic energy and travel into opposite direction in certain velocities.  In this case would the mechanical energy be conserved.  (the potential energy is conserved because this is a linear momentum, and the carts are traveling on a horizontal surface)

plz explain!![/quote]

Momentum is conserved in all circumstances. Kinetic energy is not.

Consider the senario you posited. Suppose further that the two carts have equal mass, for the sake of simplicity without loss of generality.

The momentum of the two carts while coupled is $ \vec{P}$.

When uncoupled each cart has its own momentum, ergo,

$ \vec{P} \equal{} \vec{p}_1 \plus{} \vec{p}_2$.

The kinetic energy of the system before separation is

$ KE_{\text{before}} \equal{} \frac {1}{2}\frac {P^2}{2m} \equal{} \frac {1}{4m}\left(\vec{p}_1 \plus{} \vec{p}_2\right)\cdot\left(\vec{p}_1 \plus{} \vec{p}_2\right) \equal{} \frac {1}{4m}\left(p_{1}^2 \plus{} 2\vec{p}_{1}\cdot \vec{p}_{2} \plus{} p_{2}^2\right)$.

The kinetic energy of the system after separation is

$ KE_{\text{after}} \equal{} \frac {1}{2m}\left(p_{1}^2 \plus{} p_{2}^2\right)$

The difference in kinetic energy is

$ \Delta KE \equal{} KE_{\text{after}} \minus{} KE_{\text{before}} \equal{} \frac {1}{4m}\left(p_{1}^2 \minus{} 2\vec{p}_1\cdot\vec{p}_2 \plus{} p_{2}^2\right)$.

This difference is absorbed by the environment as heat.


vector
analytic geometry
floor function
linear algebra

Let S contained in R^n be the set of vectors whose coordinates are palindromes.  Show that S is a subspace and find a basis for S. Find a second subspace T such that R^n=S+T (+ denotes the direct sum of S and T).

I don't understand where to start.

This should probably be moved to linear algebra.
[Moderator: done]

The zero vector is in $ S$, the sum of two palindrome vectors is a palindrome, and scalar multiples of palindromes are palindromes, so $ S$ is a subspace.  It's easy to verify that $ S$ is spanned by $ \left \{{e_k + e_{n + 1 - k}: 1\leq k\leq \left \lfloor \frac {n + 1}{2}\right \rfloor \right\}}$, where $ \{e_k: 1\leq k\leq n\}$ is the standard basis on $ \mathbb{R}^n$.  This spanning set is linearly independent, so it forms a basis of $ S$.  Verify that $ T = \text{span}\left(\left\{e_k: 1\leq k\leq n - \left \lfloor \frac {n + 1}{2}\right \rfloor\right\}\right)$ satisfies $ T\oplus S = \mathbb{R}^n$.


Um... they deleted the website or something? :huh:

If you want to view it use wayback machine or google cache.

or you can just try to find the website's new address and edit ur sig
*yes i know its on ur sig*

[quote="PI-Dimension"]What happened to this site: http://courses.ncssm.edu/goebel/statecon/state.htm ???  :o

What does this mean?!?!?![/quote]
http://courses.ncssm.edu/goebel/STATECON/state.htm
I'm surprised people at NCSSM took down the old website and put up a new one just so that STATECON could be in capitals.


function
search
algebra
polynomial
combinatorics unsolved
combinatorics

For $r \in N_0$, let $a_r$ denote the number of ways of selecting $4$ distinct integers from $(1,2,...,r)$ such that no two are consecutive. Find the generating function for $(a_r)$ and deduce that $a_r=C_{r-3}^4$

Well $C_{r-3}^4$ means 4 distinct numbers $a_0,...,a_3$ from 1 to r-3
then $a_i+i(i=0,1,2,3)$ makes 4 non-consecutive numbers from 1 to r. And it's easy to prove that it's a bijective.

Yes, I had the same idea, but the question is asking to find a generating function.

a generating function? I really misunderstood you. What is it? Could explain it more particular?

Use the search button. Using generating functions is an important method to solve some combinatorics problems, although I agree that it's easier to solve the problem without generating functions in this particular case.

Edit: Sorry, I used a wrong definition of generating functions :)

Well, in that case the generating function would be \[\sum{{r - 3 \choose 4}x^r},\] isn't it?

http://mathworld.wolfram.com/GeneratingFunction.html

Mathworld shares my opinion :)

[quote="Arne"]Well, in the lectures about generating functions I got in the IMO training, I was taught that the generating function of some sequence $\{a_i\}_{i \geq 0}$ is the polynomial \[\sum_{i = 0}^{+ \infty}a_ix^i,\] i.e. to get $a_i$, you need to find the coefficient of $x^i$ in the power series.[/quote]
That is _not_ a polynomial :? but you are right. Apparently it is something else than I thought. :) (I thought it was f(x), where a_i = f(i), with f(x) in sum notation)

So, in that case, I appear to come to the answer $\frac{-x^7}{(x-1)^5}$, with $r$ starting from 3 here instead of 0, as it is not defined for $r<3$. :) Is this correct?

[quote="Peter VDD"]That is _not_ a polynomial :?[/quote]

Look, I know that, but some people call this a polynomial, just to make things not too complicated. If you want to be very precise, no, it's not a polynomial. Let's just call it a power series or a generating function then :D

Yes, Arne. You can easily find the generating function thanks to $a_r=C_{r-3}^4$.
But, I am interesting in "how do you find this generating function without the idea $a_r=C_{r-3}^4$?".


$i,j \in \math{Z}, \ \ 0\leq i < j \leq 3332$ Find the probability that,
$\cos{\frac{2\pi(j-i)}{3333}} \geq \frac{\sqrt{3}}{2}$

Working on this for a bit, we get :
$j-i=<277$
$j-i\geq 1389$
The number of solutions to the thing above is 2776389, while total number of solutions for $0=<i=<j=<3332$ is 5552778. The probability is 1/2 suggesting a pretty solution. :?


floor function
combinatorics unsolved
combinatorics

There are $1999$ people participating in an exhibition. Among any $50$ people there are two who don't know each other. Prove that there are $41$ people, each of whom knows at most $1958$ people.

Call $X$ the set containing the $1999$ people and label its elements as $a_{1}, a_{2}, \cdots, a_{1999}$. Now choose one of them, wlog $a_{1}$. By the hypotesis we know that for any $50$ elements there are two people that don't know each other. So there are at least $\lfloor\frac{1998}{49}\rfloor=40$ people that $a_{1}$ doesn't know, or, equivalently, $1958$ people at most that he knows.
So let's call $Y$ the subset of $X$ containing $a_{1}$ all the people that $a_{1}$ doesn't know. We immediately have $|Y|=41$.
Let's now put in the worst case in which every element $\in Y$ knows any element $\not \in Y$ and there are not three elements of $Y$ in every $50$ people chosen. Thus we should have $\binom{41}{2}$ different couples of elements chosen in $Y$. But it is clear that it is possible to choose a different group of $50$ people for every couple since $\binom{41}{2}\leq \binom{1958}{48}$.
To conclude the proof let's dispose the $1999$ people on a circumference and label every $a_{1+k*49}^{th}$ person, with $0\leq k\leq 40$. Every group of $50$ must be chosen in this way. Let's choose $a_{1+k*49}$ and $a_{1+j*49}$ and the $48$ elements preceeding $a_{1+j*49}$. So every group is different from the others and the proof is complete.

[quote="venatrix"]By the hypotesis we know that for any $50$ elements there are two people that don't know each other. So there are at least $\lfloor\frac{1998}{49}\rfloor=40$ people that $a_{1}$ doesn't know[/quote]

I do not think this follows.

Here is less constructive proof.   Observe graph $G$ with $1999$ vertices.

Say $X = \{x\in G;d(x)\leq 1958\}$ and $Y = \{x\in G;d(x)\geq 1959\}$.

We have to prove $|X| \geq 41$.  Assume $|X|\leq 40$, then $|Y|\geq 1959$.

Since there is at most $|X|\cdot 1958 \leq 40\cdot 1958$ edges between $X$ and $Y$
there is, by Handshake lemma, at least
\[ {1\over 2}(|Y|\cdot 1959 - |X|\cdot 1958) \geq {1\over 2}(1959^2- 40\cdot 1958)\] edges between vertices in $Y$.

But since there is no $K_{50}$ in $G$ and there for no $K_{50}$ in $Y$ we
have, by Turan theorem, at most:
\[ {48\over 2\cdot 49}\cdot 1959^2\]
edges in $Y$. So we have (after some transformations) $1959^2\leq 1960\cdot 1958
= 1959^2-1$ which is impossible. So $|X| \geq 41$.

Suppose otherwise. Then we have a minimum of $1959$ people who do not know at most $39$ people.

Let $S$ be a set of $1959$ people all who do not know at most $39$ people.

Let $Q_i$ be the set which contains all the people that a person $P_i$ does not know.$|Q_i|\le 39$

Let $G_n$ be a group of people such that if $P_i\in G_n$ and $P_j\in G_n$, then $P_i$ knows $P_j$.

Clearly $G_2$ exists.
If $G_n$ exists with people $P_1,P_2,\ldots,P_n$, we consider the set $T=S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n)$.
If $T$ has at least one person $P_{n+1}$, we include it in $G$ to give us $G_{n+1}$.
But $|T|=|S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n)|\ge |S|-|Q_1|-|Q_2|-\cdots-|Q_n|\ge 1959-39n\ge 1$ for $2\ge n\ge 49$.

So we have a valid $G_{50}$. Contradiction.

EDIT: thanks to some external guidance, here is the more accurate version:

If $G_n\subset S$ exists with people $P_1,P_2,\ldots,P_n$, we consider the set $T=S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n\cup G_n)$.
If $T$ has at least one person $P_{n+1}$, we include it in $G$ to give us $G_{n+1}\subset S$.
But $|T|=|S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n)|\ge |S|-|Q_1|-|Q_2|-\cdots-|Q_n|-|G_n|\ge 1959-39n-n\ge 1$ for $2\le n\le 49$
So we have a valid $G_{49}\subset S$. As for $G_{50}$ we just consider some guy not in $Q_1\cup Q_2\cup Q_3\ldots\cup Q_{49}\cup G_{49}$ and add
him in, and since $1999-49 \times 39 - 49 \ge 1$ that guy must exist. We have a valid $G_{50}$, contradiction!


what is it?

the wiki page looks a bit confusing

What is it exactly that you don't understand? We can't really help you if we don't know. You ask what it is, well, it states it right there.

However, here's a method for a combinatorial proof not shown in the Wiki.

We wish to find the number of solutions to a_1+a_2+...+a_n <= k, for nonnegative a_i and fixed k. We can do this in two ways.

Way 1: Casework on what the LHS equals: 0, 1, 2, ..., k. Balls and urns.
Way 2: Add a dummy variable b, such that a_1+a_2+...+a_n+b=k. Balls and urns.

These ways count the same thing, so their results, which are in different forms, will have to be equal.

it's a combinatorics identity that says the following:
$ \binom{n}{n}\plus{}\binom{n\plus{}1}{n}\plus{}\binom{n\plus{}2}{n}\plus{}...\plus{}\binom{k}{n}\equal{}\binom{k\plus{}1}{n\plus{}1}$
the reason it's called the hockey stick is because if you connect the various binomial values in the above identity on Pascal's triangle, they look like a hockey stick.
hoped that helped!


calculus
real analysis
advanced fields
advanced fields unsolved

First, let me say that I am only entering college and have done no coursework beyond advanced calculus :sleep2:. I hope this will explain the terminology I use as well as my inability to solve this problem.

Suppose we have a set $W$ of compact regions in $\mathbf{R}^{3}$, each with nonzero volume, such that the intersection $R\cap{R}'$ of any $R,R'\in{W}$ has zero volume. Is $W$ necessarily countable?

I'm pretty sure the answer is yes in $\mathbf{R}^{n}$ for all $n$.

The notion of "volume" you want is Lebesgue measure, and we don't really need compactness.

There are two versions: we need either that volume is additive and the regions in $W$ are bounded or that volume is countably additive. We also need some "obvious" properties of volume; volume is nonnegative, boxes have finite volume, and intersections/unions of regions with volume have volume (In the second case, make that countable intersections). I'll do the first, generalized to $\mathbb{R}^{n}$.

Let $v(R)$ be the volume of $R$. Let $B_{N}=\{(x_{1},x_{2},\dots,x_{n}): |x_{i}|\le N\}$.
Choose positive integers $N$ and $M$. Let $W_{N,M}=\{R\in W: v(R\cap B_{N})>\frac1M\}$, and let $R_{1},R_{2},\dots,R_{k}$ be the elements of $W_{N,M}$.
By inclusion/exclusion, $v(\cup W_{N,M}\cap B_{N})=v(R_{1}\cap B_{N})+v(R_{2}\cap B_{N})+\dots+v(R_{k}\cap B_{N})$
$-v(R_{1}\cap R_{2}\cap B_{N})-\dots-v(R_{k-1}\cap R_{k}\cap B_{N})+v(R_{1}\cap R_{2}\cap R_{3}\cap B_{N})+\dots$
$=v(R_{1}\cap B_{N})+v(R_{2}\cap B_{N})+\dots+v(R_{k}\cap B_{N})$
$>\frac{k}{M}$.
This volume must also be no greater than $(2N)^{n}$, so $(2N)^{n}>\frac{k}{M}$ and $k<M\cdot(2N)^{n}$. That's finite. Since $W=\bigcup_{N,M}V_{N,M}$, a countable union of finite sets, $W$ is countable.


number theory
prime numbers
number theory open

Sorry for the length of this question...

The product $N_0$ of consecutive primes minus one (i.e. $N_0 = P_1P_2P_3...P_n - 1$, {where $P_1<P_2<P_3<...<P_n$}) cannot have any of those consecutive primes as a factor. Hence $N_0$ is either a prime larger than $P_n$ or, the product of two or more primes larger than $P_n$.

The above statement also holds true for the difference $N_1 = P_1P_3 - P_2...P_n$ or any other such arrangement of the consecutive primes (including any positive integer exponents of each prime).

We can thus see that the use of a prime on [b]one[/b] side (only) of the minus sign thereby excludes it from being a factor of $N_1$ (i.e. it effectively 'filters out' that prime).

Now, if $N$ is not prime and is a product of two or more primes larger than $P_n$ , then the smallest possible such product is $(P_{n+1})^2$ , hence if $N$ is less than $(P_{n+1})^2$ , $N$ can only be prime.

i.e. $N = P_1P_3 - P_2...P_n$ 	 any $N < (P_{n+1})^2$ is definitely prime.

(Note that if the next larger prime $(P_{n+1})$ is not known it is impossible to test whether $N$ is $< (P_{n+1})^2$, however, even when unknown, $P_{n+1}$ is always greater than or equal to $P_n + 2$, hence if $N$ is ${< (P_n+2})^2$ it is most definitely prime).

As can be seen, the formula above can be used to generate proven prime numbers that are greater than $P_n$ and less than $(P_{n+1})^2$.

Simple examples:
$3 = 2^2 - 1$
$5 = 2.3 - 1 = 3^2 - 2^2$
$7 = 3^2 - 2 = 2^3 - 1$
$11 = 3^3 - 2^4 = 2^2.3 - 1$
$13 = 2^4 - 3$
$17 = 2.3^2 - 1 = 3^4 - 2^6$
$19 = 3^3 - 2^3$
$23 = 3^3 - 2^2 = 2^5 - 3^2$
$29 = 2.3.5 - 1 = 2^5 - 3$

The problem lies in being able to 'minimize' the difference in order to prove that $P_{n+1}$ can always be generated in such a manner. Any suggestions?

This related idea might help.

If N is the product of the first n primes and r is such that 1 < r < (Pn + 2)^2 and gcd(N, r) = 1 then r is prime. The smallest such r is the next prime after Pn. If there are k of these r's they are the k primes which follow Pn.

The question is, is N + r prime for all/any r?

I don't see how you can 'minimize' your N values as they are a highly erratic sequence of integers.


hi! hungkhtn I hope you will like and give a nice solution to this inequality:

Show that if $x_i$ are positive reals $i=1,2,...,n$ then :
$n(n-1)+x_1 x_2 \cdots x_n \left( \frac{1}{x_1 ^n}+\frac{1}{x_2 ^n}+\cdots +\frac{1}{x_n ^n}\right)\geq (x_1+x_2+\cdots +x_n)\left( \frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}\right)$

Yes, thank you because of your nice inequality. I will think about it.

Replacing each $x_i$ by $1/x_i$, we find a known inequality.
See
http://www.mathlinks.ro/Forum/viewtopic.php?p=105910#p105910

wow of course! :blush:
how could I not see that !

Yes, last night I also realize that this one is equivalent to VASC's problem.


Evaluate $ (\cos\pi/3)(\cos\pi/6) \minus{} (\sin \pi/6)(\sin \pi/3)$.

[hide="jeez you should start a marathon"]
Product to sum identities
$ \frac{1}{2}(\cos\frac{\pi}{2}\plus{}\cos\frac{\pi}{6})\minus{}\frac{1}{2}(\cos\minus{}\frac{\pi}{6}\minus{}\cos\frac{\pi}{2})$
Don't forget $ \cos x\equal{}\cos \minus{}x$
$ \frac{1}{2}(2\cos\frac{\pi}{2})$
$ \cos\frac{\pi}{2}$
$ \equal{}0$

[/hide]

[hide="Easier"]The given equation is equal to $ \cos(\pi/3 \plus{} \pi/6)$ by the addition formula.[/hide]

What's the addition formula?
EDIT: never mind, I found it.

[quote="CircleSquared"]jeez you should start a marathon
[/quote]

Well, I got a contest on all of these next Saturday, and I am stuck on a lot of problems. Thankfully, I doesn't have to wait for every weekly math team practice to get help.  :D


What is the area of the projection of x^2+y^2-4x+8y-16=0 onto the xy plane after it has been rotated 60 degree out of the plane?

Before you solve the problem, could you explain me how is the image look like?

We have $ (x \minus{} 2)^2 \plus{} (y \plus{} 4)^2 \equal{} 6^2$.

If I understand correctly, your area should just be the area of the circle times $ \pi 6^2 \cos \frac{\pi}{3} \equal{} 18 \pi$

First you find the projection of the center, and that is the center of the projection. Then, the circle will now be compressed by a factor of $ \cos \frac{\pi}{3}$ in the direction perpendicular to the axis of rotation. In other words, it's an ellipse.

Of course, I don't know what your axis is -- I'm assuming the axis is some line in the xy plane.

It doesn't matter what axis you rotate about as long as it is in the xy plane.


What is the Tutte theorem ? plz help me  :huh:


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ P\equal{}\sum_{k\equal{}1}^n a_kX^{i_k}$, with $ a_k \in \mathbb{R}^*$, and $ i_k \in \mathbb{N}$ pairwise different.
It seems that $ P$ has at most $ 2n\minus{}1$ different real roots.
Prove or disprove.

Descartes' Rule of Signs tells you that $ P$ can have at most $ n \minus{} 1$ positive real roots and at most $ n \minus{} 1$ negative real roots, and it can also have the root $ 0$.  (The rule of signs counts multiplicities, so if you disregard the $ 0$ you can claim something stronger.)

Can you claim something stronger without disregarding 0?

Not in the obvious way; when I say "stronger" I mean "with multiplicity" and there we have the counterexample $ P(x) \equal{} x^{i}$ for arbitrary $ i$.  I'd be interested if there was a less obvious way to claim something stronger.

I doubt that there is a stronger general statement, for instance $ X^3\minus{}X$ has 3 different real roots, and $ X^5\minus{}3X^3\plus{}X$ has 5 different real roots.
I suspect we can find explicitly $ P \equal{} \sum_{k \equal{} 1}^n a_kX^{i_k}$, with $ a_k \in \mathbb{R}^*$, and $ i_k \in \mathbb{N}$ pairwise different such that $ P$ has exactly $ 2n \minus{} 1$ different real roots.


The number $85^9 - 21^9 + 6^9$ is divisible by an integer between 2000 and 3000. compute that integer.

[hide]
If $n$ is a positive odd integer and $a$ and $b$ are integers, then $a^n+b^n$ is divisible by $a+b$ and $a^n-b^n$ is divisible by $a-b$, which can be shown easily through factoring. Then $85^9-21^9$ is divisible by 64, and since $6^9$ is also divisible by 64, so must $85^9-21^9+6^9$. Similarly, by factoring $85^9+6^9$ and $6^9-21^9$, we can show that $85^9-21^9+6^9$ is divisible by 7 and 5.

Since $85^9-21^9+6^9$ is divisible by 5, 7, and 64, it must be divisible by their least common multiple, $\boxed{2240}$.
[/hide]

[hide="Leonardoty"]
No, he's saying that since $85^{9}-21^{9}$ is divisible by $85-21=64$ AND $6^{9}= 2^{9}\cdot 3^{9}$ is divisible by $64$, the expression $85^{9}-21^{9}+6^{9}$ must also be divisible by $64$...[/hide]

if im not wrong , 2240 is the only one factor of n that between 2000 and 13000
we can prove that all prime from 11 to 41 isnt divisible the number shown ....

it is from 2000 to 3000...

if anybody cares, =2^6*5*7*79699*1297382249

i don't really like this problem because it is totally unrigirous