function
algebra
partial fractions
complex analysis

Find the Taylor series expansions $ f(z)\equal{}\frac{1}{z(z\minus{}1)}$ around the point $ Z_0\equal{}1$

I obtained $ \minus{}\frac{1}{z}\sum_{k\equal{}0}^\infty (z\minus{}1)^k (\frac{1}{2})^{k\plus{}1}$ which I believe is wrong.

Please offer suggestions.

Most definitely since it's not a Taylor expansion... Partial fractions would be a good first step.. Try that and see if you can make it from there..

$ \ f(z)$ is analytic in $ |z \minus{} 1| < 1$ and hence we obtain a Laurent Expansion in powers of |z-1| in this region :) .

$ \frac {1}{z(z \minus{} 1)}$= $ \frac {1}{z \minus{} 1}$  x  $ \frac {1}{1 \plus{} (z \minus{} 1)}$

Expanding we get..

$ \frac {1}{z \minus{} 1} \minus{} 1 \plus{} (z \minus{} 1) \minus{} (z \minus{} 1)^2 \plus{} ....$.

Dear DineshManohar,

This is a Taylor expansion exercise. I cannnot solve this using Laurent Series.  :wink:

In Taylor series, the power should be the same. I indeed obtain your solution which I expressed as $ \sum_{k\equal{}0}^\infty (\minus{}1)^{k} (z\minus{}1)^{k\minus{}1}$ but the power is different. It should be K for both if I am not mistaken.

Taylor Series expansion of a function exists about a point $ z \equal{} a$  iff the function is analytic about that point. In our case $ z\equal{}1$ is a pole.  :)


Is there anywhere I can ask for some verification on the AoPS books?  I think one of the answers in the solution book is wrong, but I might have an older edition so it may have been corrected since then.  Thanks

There are a few that have persisted - shoot me an email about it and I'll let you know (rusczyk@artofproblemsolving.com).

Or just post the problem and your solution here. People will be quickly able to confirm/point out a mistake.

[quote="mirebeka"]Thanks-it's  volume 1, chapter 6, exercise 2, part iii (Middle of p.73):

Find the solutions to 2x :^2: +1/3x-2/3=0

I got (-2/3, 1/2)
The solutions book says (-3/2, 1/2) but since it's just the reciprocal I think it's a typo.
The solution is on p. 38[/quote]
You are correct, since -3/2 gives a value of 10/3, which is clearly not 0. Unless you made a typo in the problem itself... it'd have to be a pretty large typo, though.


let 's P be a polynomial such that:P is monic , all the coefficient of p are the integer numbers.There are exist two rational number a,b such that
 a+b* \sqrt 2 is a root of P(x).
Prove that: a,b are integer.
 :)  :)  :)  :)  :)  :)

One very well-known thing is that a-b*\sqrt2 is also a root of the polynomial. Maybe this helps?

P(x) obviously have a factor (x-a+b \sqrt 2)(x-a-b \sqrt 2)=x 2 -2ax+(a 2 -2b 2 ).
suppose a=p/r,b=q/r such that (p,q,r)=1, we have a integer-coefficient factor of f: (r 2 x 2 -2prx+p 2 -2q 2 ).
But since P is monic, all prilimitive integer-coefficient factor of P are also monic.
Thus we have r=1, and a,b are integers.


calculus
integration
geometry
trigonometry
limit
calculus computations

Let $a_{n}$ be the area surrounded by the curves $y=e^{-x}$ and the part of $y=e^{-x}|\cos x|,\ (n-1)\pi \leq x\leq n\pi \ (n=1,\ 2,\ 3,\ \cdots).$
Evaluate $\lim_{n\to\infty}(a_{1}+a_{2}+\cdots+a_{n}).$

Yes, I know, a revival of the dead, but no one had answered anyways and there aren't that many good problems nowadays.

\[a_{n} = \int_{\left(n - 1\right)\pi }^{n \pi } \! \left(1 - \left|\cos{x}\right|\right){e }^{-x} \, \mathrm{d}x = \underbrace{{e }^{\left(1 - n\right)\pi } \int_{0}^{\pi } \! \left(1 - \left|\cos{y}\right|\right){e }^{-y} \, \mathrm{d}y}_{x = \left(n - 1\right)\pi  + y}\]
$\int_{0}^{\pi } \! \left|\cos{y}\right| {e }^{-y} \, \mathrm{d}y = -\int_{0}^{\frac{\pi }{2}} \! \cos{y} {e }^{-y} \, \mathrm{d}y + \int_{\frac{\pi }{2}}^{\pi } \! \cos{y} {e }^{-y} \, \mathrm{d}y$
By Parts
$\int \! {e }^{-y} \cos{y} \, \mathrm{d}y = {e }^{-y} \sin{y} + \int \! {e }^{-y} \sin{y} \, \mathrm{d}y$
$\int \! {e }^{-y} \sin{y} \, \mathrm{d}y = -{e }^{-y} \cos{y} - \int \! {e }^{-y} \cos{y} \, \mathrm{d}y$
$\int \! {e }^{-y} \cos{y} \, \mathrm{d}y = \frac{1}{2} {e }^{-y}\left(\sin{y} - \cos{y}\right)$

\[\sum_{n = 1}^{\infty } \! a_{n} = \left(\sum_{n = 1}^{\infty } \! {e }^{\left(1 - n\right)\pi }\right)\left(\int_{0}^{\pi } \! \left(1 - \left|\cos{y}\right|\right){e }^{-y} \, \mathrm{d}y\right) = \frac{{e }^{\pi }}{{e }^{\pi } - 1}\left(\frac{1}{2} - \frac{{e }^{-\pi }}{2} - {e }^{\frac{-\pi }{2}}\right)\]


Let $ a,b,x\in \mathbb N^*$, such that $ x^{a\plus{}b}\equal{}a^b\cdot b$. Prove that $ b\equal{}x^x$ and that $ x\equal{}a$.

[quote="Inequalities Master"]Let $ a,b,x\in \mathbb N^*$, such that $ x^{a \plus{} b} \equal{} a^b\cdot b$. Prove that $ b \equal{} x^x$ and that $ x \equal{} a$.[/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150685


First, we know the perimeter is 40.
Second, we know that there are 8 unit lengths on the outside to the whole shape
Third, we know that all the sides are the same
fourth: we know that each of the quadrilaterals are squares.

First step: We have to find out how many centimeters each side length is.
We do: 40/8=5 so each side length is 5.
Second step: Now that we know the side lengths, we also know the area because it is just 5^2=25 since it is a square.
Third step. Since we know on square's area we just multiply it by 3
 25 X 3= 75

Consider point I, intersection of (undrawn) lines AB and GH.

The perimeter of hexagon ABFGHD is equal to square BFGI. Thus the sidelength of the square is 10, so the square's area is 100.

But the hexagon is 3/4 of that area, so 75.


inequalities
induction
quadratics
floor function

For $ n\geq2$ let $ a_1, a_2, \ldots a_n$ be positive real numbers such that
\[ (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)\left(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \cdots \plus{} \frac {1}{a_n}\right) \leq \left(n \plus{} \frac {1}{2}\right)^2.
\]
Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$.

outline:

[hide]
Order them so that $ a_1 \le \ldots \le a_n$. Assume for contradiction that $ a_n > 4a_n$. Then

LHS > $ n \plus{} 2\binom{n\minus{}2}{2} \plus{} 5(n\minus{}2) \plus{} (4 \plus{} \frac{1}{4}) \equal{} \left(n\plus{}\frac{1}{2}\right)^2$, contradiction.

(lemma: a+1/a is increasing for a>1 => a/b+b/a is increasing in a if a>b. first term is from summing 1's, 2nd from summing a_i/a_j + a_j/a_i for 1<i<j<n, 3rd is from summing a_1/a_k+a_k/a_n + a_n/a_k+a_k/a_1, am-gm, and lemma, 4th is from a_n/a_1+a_1/a_n.)
[/hide]

[hide]Lemma: $ u(x)\equal{}x\plus{}\frac{1}{x}$ is increasing for $ x\geq 1$.
Proof: (calc)

WLOG let $ a_1$ be the max and $ a_2$ be the min. Let the left hand side be $ F_n$. We prove the contrapositive by induction: if $ a_1 > 4a_2$, then $ F_n > (n\plus{}\frac{1}{2})^2$.

Base: $ n\equal{}2$. Note $ \frac{a_1}{a_2} > 4$. Simplifying, we get $ F_2\equal{}2\plus{}u(\frac{a_1}{a_2}) > 2\plus{}u(4) \equal{} (2\plus{}\frac{1}{2})^2$ by Lemma.

Inductive Step: suppose for $ n\minus{}1$ the claim is proven. Simplifying,
$ F_n \equal{} F_{n\minus{}1}\plus{}1\plus{}\sum_{i\equal{}1}^{n\minus{}1}\frac{a_i}{a_n} \plus{} \frac{a_n}{a_i}$
By AM-GM, $ \sum_{i\equal{}3}^{n\minus{}1}\frac{a_i}{a_n} \plus{} \frac{a_n}{a_i} \geq \sum_{i\equal{}3}^{n\minus{}1} 2\sqrt{\frac{a_i}{a_n} \cdot \frac{a_n}{a_i}} \equal{} 2(n\minus{}3) \equal{}2n\minus{}6$.

We now prove that $ \frac{a_1}{a_n} \plus{} \frac{a_n}{a_1} \plus{} \frac{a_2}{a_n} \plus{} \frac{a_n}{a_2} \geq 5$.
If $ \frac{a_2}{a_n} > \frac14$, then $ 4\cdot \frac{a_2}{a_n} > 1$. Also, $ \frac{a_1}{a_n} \equal{} \frac{a_1}{a_2}\cdot \frac{a_2}{a_n} > 4\cdot \frac{a_2}{a_n}$. Thus, by Lemma, $ \frac{a_1}{a_n} \plus{} \frac{a_n}{a_1} > \frac{4a_2}{a_n} \plus{} \frac{a_n}{4a_2}$. Therefore,
$ \frac{a_1}{a_n} \plus{} \frac{a_n}{a_1} \plus{} \frac{a_2}{a_n} \plus{} \frac{a_n}{a_2} > \frac{5a_2}{a_n} \plus{} \frac{5a_n}{4a_2} \geq 2\sqrt{\frac{5a_2}{a_n} \cdot \frac{5a_n}{4a_2}} \equal{} 5$.
If $ \frac{a_2}{a_n} \leq \frac14$, then $ \frac{a_n}{a_2} \geq 4$. Since $ \frac{a_1}{a_n} \plus{} \frac{a_n}{a_1} \geq 2$ by AM-GM, $ \frac{a_1}{a_n} \plus{} \frac{a_n}{a_1} \plus{} \frac{a_2}{a_n} \plus{} \frac{a_n}{a_2} > 6$.

Thus,
$ F_n \equal{} F_{n\minus{}1}\plus{}1\plus{}\sum_{i\equal{}1}^{n\minus{}1}\frac{a_i}{a_n} \plus{} \frac{a_n}{a_i}$
$ > (n\minus{}\frac{1}{2})^2\plus{}1\plus{}5\plus{}2(n\minus{}3) \equal{} (n\plus{}\frac12)^2$.

Induction complete.

[/hide]

A friend of mine somehow used Jensen's. I'm not sure if it works; any idea on how it would run?

Induction: base case for $ n\equal{}2$ is simple. Let $ b$ be the minimal element. Now suppose $ a_1 > 4b$.
Let $ (a_1\plus{}a_2...b)$ excluding $ a_n$ be $ r$, and let the sum of the reciprocals of those numbers be $ r'$.

So we need to show that $ (r\plus{}a_n)(r'\plus{}1/a_n) > (n\plus{}\frac{3}{2})^2$

By induction, $ rr' > (n\plus{}\frac{1}{2})^2$. So we have to prove that $ r/a_n \plus{} a_nr' \geq 2n\plus{}1$.
Now using the fact that $ x \plus{} 1/x \geq 2$, we may simplify this equation to:
$ a_1/a_n \plus{} a_n/a_1 \plus{} b/a_n \plus{} a_n/b \geq 5$. We notice that since $ b$ is the smallest element, the sum is minimized by taking $ b$ as large as possible. But we assumed that $ b<a_1/4$. So take $ b\equal{}a_1/4$ and we need to show that
$ a_1/a_n \plus{} a_n/a_1 \plus{} a_1/4a_n \plus{} 4a_n/a_1 \geq 5$. Now let $ a_1/a_n \equal{} q \geq 1$.

Then $ q\plus{}1/q\plus{}q/4\plus{}4/q \geq 5 \equal{}> 5q^2 \minus{} 20q \plus{} 20 \geq 0$ or $ q^2\minus{}4q\plus{}4 \geq 0$ which is true.

[hide="three lines"]
WLOG $ a_1$ is smallest, $ a_2$ is largest.
$ (a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)(\frac {1}{a_1} \plus{} \ldots \plus{} \frac {a}{1_n}) \geq (\sqrt {(a_1 \plus{} a_2)(\frac {1}{a_1} \plus{} \frac {1}{a_2})} \plus{} (n \minus{} 2))^2$ by Cauchy

Suppose for the sake of contradiction that $ a_2 > 4a_1$, then $ (a_1 \plus{} a_2)(\frac {1}{a_1} \plus{} \frac {1}{a_2}) > \frac {25}{4}$, so the whole thing is greater than $ (n \plus{} \frac {1}{2})^2$, which contradicts the problem statement.
[/hide]

[quote="JSteinhardt"]
$ \frac {a}{1_n}$[/quote]
One of the more humorous typos I've seen in awhile. And it's from Jacob Steinhardt too! First you let $ 1 \equal{} d$, and now you're putting subscripts on your $ 1$s. Next thing I know you'll be defining a function $ 1(x)$.

Anyways, that's pretty nice, concise version of mine and some others I've seen, but you did skip over the proof of the $ n \equal{} 2$ case which you used at the end. Pretty big part of the problem, as that solution shows.

Well, that part was pretty trivial if you scaled both variables down, so that it was (x + 1)(1/x + 1) which is obviously increasing.

Also, that was a pretty awesome proof :) .

[hide="More than 3 lines"]
WLOG $ a_1\geq a_2 \geq \dots \geq a_n$. Assume $ a_1 = ka_n$ for some $ k > 4$. Then substitute $ a_1 = ka_n$ and $ \frac 1{a_n} = \frac k{a_1}$. Then we have:
\[ (a_1 + \dots + a_n)\left(\frac 1{a_1} + \dots + \frac 1{a_n}\right) \& = (a_2 + \dots + (k + 1)a_n)\left(\frac {k + 1}{a_1} + \dots + \frac 1{a_{n - 1}}\right) \\
\& = (a_2 + \dots + a_{n - 1})\left(\frac 1{a_2} + \dots + \frac 1{a_{n - 1}}\right) + \\
\& + (k + 1)a_n\left(\frac 1{a_2} + \dots + \frac 1{a_{n - 1}}\right) + \\
\&+ \frac {k + 1}{a_1}(a_2 + \dots + a_{n - 1}) + (k + 1)^2\frac {a_n}{a_1}
\]

The first term $ \geq (x-2)^2$ by Chebyshev's, combining the second and third terms, and applying AM-GM, yields second + third $ \geq (n-2)\frac{k+1}{\sqrt k}$, and the last term equals $ \frac{(k+1)^2}{k}$. This equals $ \left((x-2)+\frac{k+1}{\sqrt k}\right)^2$, and for $ k>4$, this $ > \left(x+\frac 12\right)^2$, contradiction, QED.
[/hide]

my first reaction was to write out Lagrange's Identity, except I cited it as 'Legendre's identity'  :roll:  .. of course it was completely unnecessary, but I didn't feel like writing it over.

Also looks very much like [[2004 IMO Problems/Problem 4]].

[quote="azjps"]
Also looks very much like [[2004 IMO Problems/Problem 4]].[/quote]

lol yes that is the first thing I saw when I saw the problem

[hide="solution"]We will prove the contrapositive: let $ a_1, a_2, \ldots a_n$ be positive real numbers such that
\[ \max(a_1, a_2, \ldots a_n)> 4\min(a_1, a_2, \ldots a_n)\]
Prove that $ (a_1\plus{}a_2\plus{}a_3\ldots a_n)\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\frac{1}{a_3}\ldots \frac{1}{a_n}\right)> \left(n\plus{}\frac{1}{2}\right).$

Notice first that we can scale all $ a_k$ by the same factor and keep $ (a_1\plus{}a_2\plus{}a_3\ldots a_n)\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots \frac{1}{a_n}\right)$ the same.  So we scale them so $ \min(a_1, a_2, \ldots a_n)\equal{}1$.  We can also order them in any order we wish, so let $ a_1\leq a_2\leq a_3\ldots a_n$.  Then $ a_1\equal{}1$ and $ a_n> 4$ by assumption.

We proceed by induction.

[u]Base case[/u]: $ a_1\equal{}1$ and $ a_2> 4$, so let $ a_2\equal{}4k$ with $ k> 1$.  We try to prove that $ (1\plus{}4k)\left(\frac{1}{1}\plus{}\frac{1}{4k}\right)> \left(2\plus{}\frac{1}{2}\right)\equal{}\frac{25}{4}$.  If we combine the second multiplicand over one fraction we try to show: $ \frac{(1\plus{}4k)^2}{4k}> \frac{25}{4}$.

Multiply by $ 4k$ (this doesn't change the inequality direction because $ 4k> 0$) and get
\[ (1\plus{}4k)^2\equal{}16k^2\plus{}8k\plus{}1> 25k\]

Subtract $ 25k$ from both sides and complete the square to get $ 16k^2\minus{}17k\plus{}1\equal{}(4k\minus{}\frac{17}{8})^2\minus{}\frac{225}{64}> 0$

Now we plug in $ k> 1$ to get $ (4k\minus{}\frac{17}{8})^2\minus{}\frac{225}{64}> (4\minus{}\frac{17}{8})^2\minus{}\frac{225}{64}\equal{}\left(\frac{15}{8}\right)^2\minus{}\frac{225}{64}\equal{}0$

Thus the base case is proven.

[u]Inductive step[/u]: Suppose that for some $ n$,
\begin{equation}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\ldots \frac{1}{a_n}\right)(a_1+a_2+a_3\ldots a_n)> (n+\frac{1}{2})^2\end{equation}

when $ a_1\equal{}1\leq a_2\leq a_3\leq\ldots a_n$ and $ a_n> 4$.

If we multiply $ (1)$ by 4, we have
\[ (2n\plus{}1)^2< 4(a_1\plus{}a_2\plus{}a_3\ldots a_n)\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\frac{1}{a_3}\ldots \frac{1}{a_n}\right)\]
or $ (2n\plus{}1)^2\minus{}4(a_1\plus{}a_2\plus{}a_3\ldots a_n)\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots \frac{1}{a_n}\right)< 0$.

We interpret this as a discriminant:
\[ \left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots \frac{1}{a_n}\right)x^2\minus{}(2n\plus{}1)x\plus{}(a_1\plus{}a_2\plus{}a_3\ldots a_n)\] has no solutions in real numbers.  Since the coefficient of $ x^2$ is positive (because it is the sum of reciprocals of positive reals), we have \begin{equation}\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots \frac{1}{a_n}\right)x^2-(2n+1)x+(a_1+a_2+a_3\ldots a_n)> 0\end{equation} for all $ x\in\mathbb{R}$, and specifically $ x\in\mathbb{R}^{\plus{}}$.  We will work only in positive reals, so when dividing/multiplying, we don't flip the inequality.

Let $ a_1< x< a_n$.  We divide $ (2)$ by $ x$:
$ \left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots \frac{1}{a_n}\right)x\plus{}\frac{(a_1\plus{}a_2\ldots a_n)}{x}> 2n\plus{}1$
Now add $ (a_1\plus{}a_2\plus{}\ldots a_n)\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\ldots \frac{1}{a_n}\right)\plus{}1$:

\begin{align*}(a_1+a_2+a_3\ldots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots \frac{1}{a_n}\right)+\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots \frac{1}{a_n}\right)x+\frac{(a_1+a_2+a_3\ldots a_n)}{x}+1&> 2n+1+(a_1+a_2+\ldots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots \frac{1}{a_n}\right)+1\\
&>(n+\frac{1}{2})^2+2n+2\\
&=n^2+3n+2.25\\
&=((n+1)+\frac{1}{2})^2\end{align*}

We factor the left side:
\[ ((a_1\plus{}a_2\plus{}a_3\ldots a_n)\plus{}x)(\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\ldots \frac{1}{a_n}\right)\plus{}\frac{1}{x})> ((n\plus{}1)\plus{}\frac{1}{2})^2\]

We know that $ x< a_n$ so we let $ x\equal{}a_{n\plus{}1}$ then arrange the $ a_k$'s so that they are in order again.  Thus the max is still greater than 4, and the min is still 1.  Thus we have shown that if $ (1)$ is true for $ n$, it is true for $ n\plus{}1$ so, by induction, it is true for all integers $ n\geq 2$.  Since this is equivalent to the given problem, the given problem is also true.  $ \boxed{\text{QED}}$.[/hide]

This is my solution verbatim.  What do you think it will get?

By the way, I left out the dollar signs around the equation and array environments.  How do you make this work on AoPS?

[hide="Outline."]
=====
Label the elements of sets so that $ i<j \implies a_i\leq a_j$. We prove the contrapositive: given a set $ \{a_1,\dots ,a_n\}$, if $ a_n>4a_1$, then $ \left(\sum_{i\equal{}1}^n a_i\right)\left(\sum_{i\equal{}1}^n \frac{1}{a_i}\right)>(n\plus{}0.5)^2$. We induct.

For $ n\equal{}2$, we get $ (a_1\plus{}a_2)\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\right)\equal{}2\plus{}\frac{a_2}{a_1}\plus{}\frac{a_1}{a_2}$. If we consider $ f(x)\equal{}x\plus{}\frac{1}{x}$, we can show that it is increasing on $ (4,\infty)$. Since $ \frac{a_2}{a_1}>4$, the quantity is greater than $ 2\plus{}4\plus{}\frac{1}{4}\equal{}\frac{25}{4}\equal{}(2.5)^2$.

Now suppose our claim is true for any set of size $ k$. Now for any set of size $ k\plus{}1$, write the expression out but pull out $ a_2$. The remaining set $ \{a_1,a_3,\dots a_{k\plus{}1}\}$ allows us to use the inductive hypothesis. We end up getting:

$ (a_1\plus{}a_2\plus{}\dots \plus{}a_{k\plus{}1})(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\dots \plus{} \frac{1}{a_{k\plus{}1}})$
$ \equal{}(a_1\plus{}a_3\plus{}\dots \plus{}a_{k\plus{}1})(\frac{1}{a_1}\plus{}\frac{1}{a_3}\plus{}\dots \plus{} \frac{1}{a_{k\plus{}1}}) \plus{} a_2(\frac{1}{a_1}\plus{}\frac{1}{a_3}\plus{}\dots \plus{} \frac{1}{a_{k\plus{}1}})\plus{}\frac{1}{a_2}(a_1\plus{}a_3\plus{}\dots \plus{} a_{k\plus{}1})\plus{}1$
$ \equal{}(a_1\plus{}a_3\plus{}\dots \plus{}a_{k\plus{}1})(\frac{1}{a_1}\plus{}\frac{1}{a_3}\plus{}\dots \plus{} \frac{1}{a_{k\plus{}1}}) \plus{} (\frac{a_2}{a_3}\plus{}\frac{a_3}{a_2})\plus{}\dots \plus{} (\frac{a_2}{a_k} \plus{} \frac{a_k}{a_2})\plus{}(\frac{a_1}{a_2}\plus{}\frac{a_2}{a_{k\plus{}1}})\plus{}(\frac{a_{k\plus{}1}}{a_2}\plus{}\frac{a_2}{a_1})\plus{}1$

We use the inductive hypothesis for the first product, $ x\plus{}\frac{1}{x}\geq 2$ for the next $ k\minus{}2$ sums, and for the next two terms, use:
$ (\frac{a_1}{a_2}\plus{}\frac{a_2}{a_{k\plus{}1}})\plus{}(\frac{a_{k\plus{}1}}{a_2}\plus{}\frac{a_2}{a_1}) > 2\sqrt{\frac{a_1}{a_{k\plus{}1}}}\plus{}2\sqrt{\frac{a_{k\plus{}1}}{a_1}}\geq 2(2\plus{}0.5)$,
where we use AM-GM and the same $ f(x)\equal{}x\plus{}\frac{1}{x}$ trick with $ \sqrt{\frac{a_{k\plus{}1}}{a_1}}>\sqrt{4}\equal{}2$.

$ >(k\plus{}0.5)^2\plus{}2(k\minus{}2)\plus{}2(2\plus{}0.5)\plus{}1$
$ \equal{}(k^2\plus{}k\plus{}0.25)\plus{}2k\plus{}2$
$ \equal{}(k^2\plus{}2k\plus{}1)\plus{}(k\plus{}1)\plus{}0.25$
$ \equal{}((k\plus{}1)\plus{}0.5)^2$,

as desired.
=====
[/hide]

[hide="Generalization"]
Scale the whole equation and rename the variables so that the given becomes $ (x + a_1 + a_2 + a_{n-2} + 1) (\frac{1}{x} + \frac{1}{a_1} + ... + \frac{1}{a_{n-2}} + 1  ) \leq  ( n + \frac{1}{2}  )^2$, and such that $ x \geq a_1 \geq ... \geq a_{n-2} \geq 1$. It suffices to show that that $ x \leq 4$.

Then $  ( n + \frac{1}{2}  )^2 \geq  ( x + \sum a_i + 1  )  ( \frac{1}{x} + \sum \frac{1}{a_i} + 1  )$
$ = 1 + \sum \frac{x}{a_i} + x + \sum \frac{a_i}{x} + \sum_{i,j} \frac{a_i}{a_j} + \sum a_i + \frac{1}{x} + \sum \frac{1}{a_i} + 1$
$ = 2 + x + \frac{1}{x} + \sum \frac{x}{a_i} + \sum a_i + \sum \frac{a_i}{x} + \sum \frac{1}{a_i} + \sum_{i,j} \frac{a_i}{a_j}$
$ = 2 + x + \frac{1}{x} + \sum  ( \frac{x}{a_i} + a_i  ) + \sum  ( \frac{a_i}{x} + \frac{1}{a_i}  ) + (n-2) + \sum_{i \neq j} \frac{a_i}{a_j}$
$ \geq 2 + x + \frac{1}{x} + (n-2)2\sqrt{x} + (n-2)\frac{2}{\sqrt{x}} + (n-2) + 2\binom{n-2}{2}$
$ =  ( \sqrt{x} + \frac{1}{\sqrt{x}}  )^2 + 2(n-2) ( \sqrt{x} + \frac{1}{\sqrt{x}}  ) + (n-2)^2$
$ =  ( n - 2 + \sqrt{x} + \frac{1}{\sqrt{x}}   )^2$

So $  ( n - 2 + \sqrt{x} + \frac{1}{\sqrt{x}}   )^2 \leq  ( n + \frac{1}{2}  )^2 \Rightarrow \sqrt{x} + \frac{1}{\sqrt{x}} \leq 2 + \frac{1}{2} \Rightarrow \sqrt{x} \leq 2 \Rightarrow x \leq 4$, as desired.
[/hide]

This solution also uses nothing but AM-GM, which is interesting.

[quote=PCChess]
For sake of contradiction assume that $\frac{a_n}{a_1}>4.$ [b]Then, $\frac{a_1}{a_n}>\frac{1}{4}$[/b][/quote]

The bolded statement is completely wrong (try $a_n/a_1=5$), but it's not hard to fix using the fact that x+1/x is strictly increasing in $(4,\infty)$.

Pretty cool problem. Similar to many above solutions - posting for storage.
[hide=Solution]
Without loss of generality assume $a_n \geq ... \geq a_1$ and let $$k = \frac{a_n}{a_1} $$ we shall assume for the sake of contradiction that $k > 4$. \\
Then we have:
$$(a_1+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n}) = (a_2+...a_{n-1})(\frac{1}{a_2}+..+\frac{1}{a_{n-1}}) + (\sum_{j=2}^{n-1}\frac{a_1}{a_j}+\frac{a_j}{a_i}+\frac{a_n}{a_j}+\frac{a_j}{a_n}) + \frac{a_1}{a_1} + \frac{a_n}{a_n}+\frac{a_1}{a_n}+\frac{a_n}{a_1}$$
$$\geq (n-2)^2 +2 + k + \frac{1}{k}+ (\sum_{j=2}^{n-1}\frac{a_1}{a_j}+\frac{a_j}{a_i}+\frac{ka_1}{a_j}+\frac{a_j}{ka_1}) > n^2-4n+10+\frac{1}{4}+ (\sum_{j=2}^{n-1}\frac{(k+1)a_1}{a_j}+\frac{(k+1)a_j}{ka_i})$$
$$> n^2-4n+10 +\frac{1}{4} + 5(n-2) = n^2+n+\frac{1}{4} = (n+\frac{1}{2})^2$$
where we use that $\frac{1}{k}+k > \frac{17}{4}$ as $k > 4$ and $\frac{(k+1)a_1}{a_j}+\frac{(k+1)a_j}{ka_i} \geq \frac{2(k+1)}{\sqrt{k}} > 5$ by AM-GM and again using the fact that $k > 4$. \\

This ultimately gives a contradiction meaning that $a_n \leq 4a_1$. $\blacksquare$ \\[/hide]
[hide=Remark]
I think that this problem is slightly easier than IMO 2004, P4. This is simply because here, one is required to prove that $4a_1 \geq a_n$ (in the previous ordering of the reals), meanwhile; in the previously mentioned IMO problem you are not able to prove something of the form $a_1+a_2 \geq 2a_1 \geq a_n$ and are required to instead prove that $a_1 + a_2 \geq a_n$, nevertheless, both problems are relatively straight forward so maybe it is not the greatest of ideas to compare their difficulty. [/hide]

Wlog, $\max(a_1, a_2, \ldots, a_n)=a_n$ and $\min(a_1, a_2, \ldots, a_n)=a_1$, hence we need $4 a_1\geq a_n$.

Using Cauchy-Schwarz,
$$ \left(n + \frac {1}{2}\right)^2 \geq \left(\sum_{i=1}^{n}a_i \right) \cdot  \left(\sum_{i=1}^{n}\frac{1}{a_i} \right) \geq \left(\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}}+n-2 \right)^2. $$
Thus, we have $$\frac {5}{2}\geq \sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}}\Longleftrightarrow \frac{17}{4}a_1a_n\geq a_1^2+a_n^2\Longleftrightarrow \left(\frac{a_n}{4}-a_1\right)(4a_n-a_1)\leq 0$$
Note that as $4a_n\geq a_n\geq a_1$, we must have $\dfrac{a_n}{4}-a_1 \leq 0\Longleftrightarrow 4a_1\geq a_n$. Done.

[hide=Solution]
Assume without loss of generality $a_1$ is the smallest number and $a_2$ is the largest number. Suppose that $4a_1>a_2$. We will show that $(a_1+a_2+\ldots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\ldots+\frac1{a_n}\right)>\left(n+\frac12\right)^2$. Let $x=a_3+a_4+\ldots+a_n$, and let $y=\frac1{a_3}+\frac1{a_4}+\ldots+\frac1{a_n}$. Then, we have
\begin{align*}
(a_1+a_2+x)\left(\frac1{a_1}+\frac1{a_2}+y\right)&=1+1+xy+\frac x{a_1}+\frac x{a_2}+ya_1+ya_2+\frac{a_2}{a_1}+\frac{a_1}{a_2}\\
&\geq2+xy+2\sqrt{xy}\left(\sqrt{\frac{a_2}{a_1}}+\sqrt{\frac{a_1}{a_2}}\right)+\frac{a_2}{a_1}+\frac{a_1}{a_2}\\
&>2+xy+5\sqrt{xy}+\frac{17}4.
\end{align*}
Since $xy\geq(n-2)^2$, this means that we have
\begin{align*}
2+xy+5\sqrt{xy}+\frac{17}4&\geq2+(n-2)^2+5(n-2)+\frac{17}4\\
&n^2-4n+6+5n-10+\frac{17}4\\
&=n^2+n+\frac14\\
&=\left(n+\frac12\right)^2.
\end{align*}
Therefore, this is a contradiction, so we must have $\max(a_1,a_2,\ldots,a_n)\leq4\min(a_1,a_2,\ldots,a_n)$.

WLOG let $a_1\leq a_2\leq\dots \leq a_n$ and $a_1=1$. Now let $a_n>4$. We have $$(a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right)\geq \left(n-2+\sqrt{(1+a_n)\left(1+\frac{1}{a_n}\right)}\right)^2,$$ to show a contradiction we only need to show that $\sqrt{(1+a_n)\left(1+\frac{1}{a_n}\right)}>\frac{5}{2}$ or $a_n+\frac{1}{a_n}>\frac{17}{4}=4+\frac{1}{4},$ which is trivial. $\blacksquare$

WLOG $a_1\le a_2\le\ldots\le a_n$. Assume FTSOC that $a_n>4a_1$. Denote $x=a_2+\ldots+a_{n-1}$, $y=\frac1{a_2}+\ldots+\frac1{a_{n-1}}$, and $z=\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}}$. We have $z^2-2=\frac{a_1}{a_n}+\frac{a_n}{a_1}$ and $xy\ge(n-2)^2$ by AM-HM. After:
\begin{align*}(a_1+a_n+x)\left(\frac1{a_1}+\frac1{a_n}+y\right)&=2+\frac{a_1}{a_n}+\frac{a_n}{a_1}+ya_1+ya_n+\frac x{a_1}+\frac x{a_n}+xy\\
&\ge2+z^2-2+2z\sqrt{xy}+xy\\
&=(z+n-2)^2\end{align*}
it suffices to show that $z\ge\frac52$. But this is true since:
$$z=\sqrt{\frac{a_1}{a_n}+\frac{a_n}{a_1}+2}=\sqrt{\frac{(a_n-4a_1)\left(a_n-4a_1+\frac{15}4a_1\right)}{a_1a_n}+\frac{25}4}\ge\sqrt{\frac{25}4}.~\square$$

Assume FTSOC that there exist indices $k$ and $\ell$ such that $\frac{a_k}{a_{\ell}} = d > 4$ and WLOG let $k=1$, $\ell=2$. Write $\left(\sum a_i\right)\left(\sum \frac{1}{a_i}\right)$ as

$\frac{a_1}{a_2} + \frac{a_2}{a_1} + \sum_{i=3}^n\left( \frac{a_1}{a_i} + \frac{a_i}{a_1} + \frac{a_2}{a_i} + \frac{a_i}{a_2}\right) + \text{other stuff}$.

We will deal with each term separately. First, $\text{other stuff}$ can be partitioned into pairs $\frac{a_i}{a_j} + \frac{a_j}{a_i}$, so the third term is at least $n^2 - 4n + 6$ by AM-GM.

Next, by AM-GM $\frac{a_1}{a_i} + \frac{a_i}{a_2} > 2\sqrt{d}$ and $\frac{a_2}{a_i} + \frac{a_i}{a_1} > \frac{2}{\sqrt{d}}$. Since the function $x + \frac{1}{x}$ is increasing on the interval $(2,\infty )$ (just differentiate), the middle term is at least $5n - 10$.

Finally, using the fact that $x + \frac{1}{x}$ is increasing on $(4,\infty )$, $\frac{a_1}{a_2} + \frac{a_2}{a_1} > \frac{17}{4}$.

Putting it all together,

$\left(\sum a_i\right)\left(\sum \frac{1}{a_i}\right) > \frac{17}{4} + 5n - 10 + n^2 - 4n + 6 = \left(n + \frac{1}{2}\right)^2$

and we have a contradiction, as desired.

WLOG $a_1\le a_2\le \cdots \le a_n$. Suppose FTSOC that $a_n>4a_1$. We have \[\left(\sum a_i\right)\left(\sum \frac{1}{a_i}\right)\ge \left(\sqrt{\left(a_1+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)}+(n-2)\right)^2=\left( \sqrt{2+\frac{a_n}{a_1}+\frac{a_1}{a_n}}+(n-2)\right)^2 \]by Cauchy.

Claim: $x+\frac{1}{x}\ge 4+\frac{1}{4}$ if $x\ge 4$.
Proof: We need to show that in the interval $[4,\infty)$, $x+\frac{1}{x}$ is increasing. We have the derivative of $x+\frac{1}{x}$ is $1-\frac{1}{x^2}$. Since that is always greater than $0$ for $x\ge 4$, $x+\frac{1}{x}$ is increasing over $[4,\infty)$. $\blacksquare$

If we let $k=\frac{a_n}{a_1}$, we have $k>4$. Also, \begin{align*}
\left(\sqrt{2+k+\frac{1}{k}}+(n-2)\right)^2 \\
>\left(\sqrt{2+4+\frac{1}{4}}+(n-2)\right)^2 \\
=\left(\sqrt{\frac{25}{4}}+(n-2)\right)^2 \\
=\left(\frac{5}{2}+(n-2)\right)^2 \\
=\left(n+\frac{1}{2}\right)^2 \\
\end{align*}a contradiction.

Assume WLOG $a_1 = \min(a_i)$ and $a_n = \max(a_i)$. From Cauchy, note that
\[ \left(a_n + a_2 + \dots + a_{n-1} + a_1\right)\left(\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}\right)\ge \left(n-2 + \sqrt{\frac{a_n}{a_1}} + \sqrt{\frac{a_1}{a_n}}\right)^2.\]
Thus,
\[ \left(n +\frac{1}{2}\right)^2\ge \left(n-2 + \sqrt{\frac{a_n}{a_1}} + \sqrt{\frac{a_1}{a_n}}\right)^2\implies \frac{5}{2}\ge \sqrt{\frac{a_n}{a_1}} + \sqrt{\frac{a_1}{a_n}}.\]
Substitute $t = \sqrt{a_n/a_1}\ge 1$. Now, define the function $f(t) = t + 1/t$ for $t > 0$. Note that $f$ is increasing in $[2,\infty)$, so
\[ t + \frac{1}{t}\le \frac{5}{2}\implies f(t)\le f(2)\implies t\le 2,\]
as desired.

Note $x+\frac{1}{x}$ is increasing over $(1,\infty)$.\\

Define $Gain(a,b)$ for positive real numbers $a,b$ as $$Gain(a,b)=\frac{a}{b}+\frac{b}{a}-2.$$ Clearly, $Gain(a,b)$ is always nonnegative. The condition is equivalent to $$\sum_{1\leq i<j\leq n}Gain(a_i,a_j)\leq n+\frac{1}{4}.$$ For the rest of the solution, WLOG that $a_1=1$ is the smallest and that $a_n$ is the largest, since this is both symmetric and homogenous. Assume for the sake of contradiction that $a_n>4$. Then, clearly, $$Gain(a_1,a_n)> 4+\frac{1}{4}-2=\frac{9}{4}.$$

We then claim that for any $2\leq k\leq n-1$, we have $$Gain(a_1,a_k)+Gain(a_k,a_n)>1.$$ This is just showing that $$\frac{1}{a_k}+a_k+\frac{a_k}{a_n}+\frac{a_n}{a_k}>5.$$ Regroup this as $$a_k(1+\frac{1}{a_n})+\frac{1}{a_k}(1+a_n)>5.$$ By AM-GM, $$a_k(1+\frac{1}{a_n})+\frac{1}{a_k}(1+a_n)\geq 2\sqrt{(1+\frac{1}{a_n})(1+a_n)}$$ $$=2\sqrt{2+\frac{1}{a_n}+a_n}>2\sqrt{2+4+1/4}=5,$$ which shows the claim.\\

This means that $$Gain(a_1,a_n)+\sum_{k=2}^{n-1}Gain(a_1,a_k)+Gain(a_k,a_n)>\frac{9}{4}+(n-2)=n+\frac{1}{4}.$$ This is only a subset of the total gain over all pairs, so the total gain is obviously larger than $n+\frac{1}{4}$ as well, contradiction, so we are done.

WLOG the $a_i$ are increasing. Observe the estimate $$(a_1+a_2+\cdots+a_n)\left(\frac 1{a_1} + \frac 1{a_2}+\dots + \frac 1{a_n}\right) \geq \left(n-2 + \sqrt{\frac{a_1}{a_n} + \frac{a_n}{a_1}}\right)^2.$$ Set $k = \frac{a_n}{a_1}$ and analyzing the RHS in terms of $k$ yields $\frac 14 \leq \leq 4$, hence the result.

WLOG suppose $a_1$ is the maximum and $a_2$ is the minimum. Denote $x = \sqrt{\frac{a_2}{a_1}} + \sqrt{\frac{a_1}{a_2}}$. The problem statement reduces to proving $x \leq \frac 52$.

The trick is to switch $a_1$ and $a_2$, then apply Cauchy to get
\[\left(a_2 + a_1 + a_3 + \ldots + a_n\right) \left(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \ldots + \frac{1}{a_n}\right) \ge (n+x-2)^2.\]

Hence, using our given bound, we get $x \leq \frac 52$.

FTSOC + WLOG, assume that $a_1=1$, and $a_2=4+\epsilon$ for some real number $\epsilon>0$. This ensures that $4\min(a_1,a_2,\dots,a_n)<\max(a_1,a_2,\dots,a_n)$. Then, we have the equation
\[\left(1+[4+\epsilon]+a_3+a_4+\dots+a_n\right)\left(1+\frac{1}{4+\epsilon}+\frac{1}{a_3}+\frac{1}{a_4}+\dots+\frac{1}{a_n}\right)\]
\[=\left([5+\epsilon]+a_3+a_4+\dots+a_n\right)\left(\left[\frac{5+\epsilon}{4+\epsilon}\right]+\frac{1}{a_3}+\frac{1}{a_4}+\dots+\frac{1}{a_n}\right),\]
and by Cauchy-Schwarz;
\[\left([5+\epsilon]+a_3+a_4+\dots+a_n\right)\left(\left[\frac{5+\epsilon}{4+\epsilon}\right]+\frac{1}{a_3}+\frac{1}{a_4}+\dots+\frac{1}{a_n}\right)\]
\[\geq \left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+1+1+\dots+1\right)^2 = \left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+n-2\right)^2.\]
However, since $\epsilon>0$, we have that
\[4\epsilon^2+15\epsilon>0,\]
\[\iff 100+40\epsilon+4\epsilon^2>25\epsilon+100,\]
\[\iff 4(5+\epsilon)^2>25(4+\epsilon),\]
\[\iff \frac{(5+\epsilon)^2}{4+\epsilon}>\frac{25}{4},\]
\[\iff \sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}>\frac{5}{2},\]
which implies that
\[\left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+n-2\right)^2>\left(n+\frac{1}{2}\right)^2.\]
Using this, we have that
\[(a_1+a_2+\dots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\geq \left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+n-2\right)^2>\left(n+\frac{1}{2}\right)^2,\]
a contradiction. Therefore, we must have $\max(a_1,a_2,\dots,a_n)\leq 4\min(a_1,a_2,\dots,a_n)$, finishing the problem.

We will prove the contrapositive; assuming that $\max > 4 \min$, we will show $\left( \sum a_i \right) \left( \sum \tfrac{1}{a_i} \right) > \left( n + \tfrac{1}{2} \right)^2$.

WLOG $a_1 \leq a_2 \leq \dots \leq a_n$. Let $i$ be an index such that $a_1 \leq a_i \leq a_n$. Then, we have
\begin{align*}  & \left( \frac{a_1}{a_i} + \frac{a_i}{a_1} \right) + \left( \frac{a_n}{a_i} + \frac{a_i}{a_n} \right)\\
= ~ & \left( \frac{a_1}{a_i} + \frac{a_i}{a_n} \right) + \left( \frac{a_n}{a_i} + \frac{a_i}{a_1} \right) \\
\geq ~ & 2 \sqrt{ \tfrac{a_1}{a_n}} + 2 \sqrt{\tfrac{a_n}{a_1}} \\
> ~ & 2 \sqrt{ \tfrac{1}{4}} + 2 \sqrt{\tfrac{4}{1}} \\
= ~ & 5.
\end{align*}

Also, note that $\tfrac{a_1}{a_n} + \tfrac{a_n}{a_1}$ is strictly greater than $4 + \tfrac{1}{4} = \tfrac{17}{4}$.

So,
\begin{align*}
& \left( \sum a_i \right) \left( \sum \tfrac{1}{a_i} \right) \\
= ~ & \sum_{1 \leq i \leq n} ( a_i \cdot \tfrac{1}{a_i} ) + \sum_{1 < i < j < n} \left( \tfrac{a_i}{a_j} + \tfrac{a_j}{a_i} \right) + \sum_{i \neq 1, n}  \left( \tfrac{a_1}{a_i} + \tfrac{a_i}{a_1} + \tfrac{a_n}{a_i} + \tfrac{a_i}{a_n} \right) + \left( \tfrac{a_1}{a_n} + \tfrac{a_n}{a_1} \right) \\
> ~ & n + \binom{n-2}{2} \cdot 2 + 5(n-2) + \tfrac{17}{4} \\
> ~ & \left( n + \tfrac{1}{2} \right)^2.
\end{align*}


Find all funktions, that satisfy  f(a)-f(b)=a^3-b^3       if this is a real funktion.

Do you want to find a function $ f(x)$ such that .
IF this it is quite easy.
Let $ g(x)\equal{}f(x)\minus{}x^3$
Then $ g(a)\equal{}g(b),\forall a,b\in R$
So $ g(x)\equal{}c$
Imply that
$ f(x)\equal{}x^3\plus{}c$


A "rising number" is a positive integer with multiple digits such that each digit is larger than the one to its left.
(ex. 578, 145 are, but 242, 144, and 321 aren't)

What is the total number of "rising numbers"?

EDIT: I meant to post this into HSB; i posted it into there just now, can someone delete this?

This probably belongs better here, anyway. HSB is meant for VERY basic questions.

[hide] I think the answer is 512-1-9=502. (note a bijection between rising numbers and subsets of $\{1,2,...9\}$. [/hide]

[hide]
Case 1: First digit 1. If the second digit is 2, then there are 7 possibilities for the last digit, if the second digit is 3, then there are 6 possibilities, etc.

Case 2: First digit 2. If the second digit is 3, there are 6 possibilities, if the second digit is 4, there are 5 possibilities, etc.

...

Case 7: First digit 7, one possibility.

So now we're evaluating $\binom{8}{2}+\binom{7}{2}$ etc. By the hockey stick identity, this is $\binom{9}{3}$, so the answer is $84$.[/hide]

[hide]Given a subset of {1,2,3,4,5,6,7,8,9}, there is one and only one way to make a rising number. There are $2^{9}=512$ subsets, subtract $9$ for the numbers 1 through 9, since they don't have multiple digits, and 1 for the nullset, gives $502$. [/hide]
EDIT:[quote="Max300"]pianoforte, you forgot to subtract the null set so I think the answer is one less than what you got.[/quote]

Oops, I thought it meant only 3 digit numbers.

Great solution anyway!... I think there was a very similar problem (maybe the same) in the final of the italian national olympiads (I mean the team competition, of course!)...  :)

Yes, the corollary was falling numbers in a Math League #6, with which the answer was 1013.


If six coins are tossed, what is the probability that the outcome will show 3 heads and 3 tails? Express your answer as a common fraction.

We simply have $ \left ( \frac {1}{2} \right )^6 \times \binom{6}{3}$ because there are $ \binom{6}{3}$ ways to make this, and a $ \left ( \frac {1}{2} \right )^6$ that it will happen.  Simplifying, we get $ \frac {1}{64} \times 20 \equal{} \boxed{\frac {5}{16}}$.

EDIT: Thanks.  I calculated $ \binom{6}{3}$ incorrectly.

I believe this is given by
$ \binom{6}{3}\cdot\left(\frac {1}{2}\right)^{3}\left(\frac {1}{2}\right)^{3} \equal{} \binom{6}{3}\frac {1}{2^{6}} \equal{} \frac {20}{2^{6}} \equal{} \frac {20}{64} \equal{} \frac {5}{16}$

AIME15, I think $ \binom{6}{3}\equal{}20$, not 15.


probability
conditional probability
probability and stats

f(x,y,z)=2 for 0<x<y<1 and 0<z<1  and 0 otherwise

Find P(3X>Y|1<4Z<2)

first integrate to show that its a density.
then find the marginal distribution of Z.
(think Z is uniform (0,1)).
find P(1/4<Z<1/2).
next find the joint probability
P(3X>Y,1/4<Z<1/2).
hence find the conditional probability by division. :)

think the answer will be 2/3.


function
linear algebra
matrix
integration
trigonometry
algebra
functional equation

Supposed that $f(x),g(x)$ are real-valued functions at $-\infty <x< \infty$ and $f(0)^2+g(0)^2>0$.
Find the $f(x)^2+g(x)^2$ which satisfies the following conditions for all real numbers $x,y$.
\[f(x+y)=f(x)f(y)-g(x)g(y)\]
\[g(x+y)=f(x)g(y)+f(y)g(x)\]
\[f'(0)=2,g'(0)=1\]

Short answer:

[hide]$f(x)^2+g(x)^2=e^{4x}.$[/hide]

A little more elaboration on that answer. I'll keep it hidden, becuase it still makes a good exercise for someone.

[hide]Define $F:\mathbb{R}\to\mathbb{C}$ by $F(x)=f(x)+ig(x).$ The two real functional equations can now be written as one complex functional equation: $F(x+y)=F(x)F(y).$

If we set $x=y=0$ and note the extra hypothesis $F(0)\ne0$, we get that $F(0)=1.$

Now take difference quotients:

$\frac{F(x+h)-F(x)}h=\frac{F(h)-1}h\cdot F(x).$

Taking limits, we see that if $F'(0)$ exists, then $F$ is differentiable everywhere and $F'(x)=F'(0)F(x).$ Finally, we are told that $F'(0)=2+i,$ so $F'(x)=(2+i)F(x).$ This linear differential equation (with initial condition $F(0)=1$) has the unique solution $F(x)=e^{(2+i)x}.$

Then $f(x)^2+g(x)^2=|F(x)|^2=|e^{2x+ix}|^2=|e^{2x}|^2|e^{ix}|^2=e^{4x}.$

We could have accomplished the same thing with the function from $\mathbb{R}$ to the $2\times 2$ matrices whose values are

$\begin{bmatrix} f(x) & -g(x) \\ g(x) & f(x) \end{bmatrix}.$

Of course, that's the same thing as the complex numbers.

If we took away the hypothesis of differentiablity at 0, would there be any solutions to this functional equation other than $F(x)=e^{rx}?$ Yes, there are, but the others are all non-measurable. They're obtained by using a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ to construct an additive function on $\mathbb{R},$ then taking the exponential of that.[/hide]

The statement implies $f,g$ are continuous

$x=y=0$ in the second equation $g(0)=2f(0)g(0)$

If $f(0)=1/2$ with the first equation $f(0)=f(0)^2-g(0)^2$ so $g(0)^2=1/4 - 1/2$ impossible. We have $g(0)=0$

$f(0)=f(0)^2$ if $f(0)=0$ we have a contradiction with $f(0)^2+g(0)^2>0$

So $f(0)=1 , \; g(0)=0$

Since $f'(0)=2$
$f$ is derivable around 0, it is continous around 0, there exist $a>0$

such that $\int_{-a}^{a}f(t)\geq 2a>0$

We integrate the first equation according $y$

$\int_{-a}^{a}f(x+y)dy=\int_{-a-x}^{a+x}f(t)dt=Af(x)-Bg(x)$ (1)
where $A=\int_{[-a,a]}f , \; B=\int_{[-a,a]}g$

derivate (1) $Af'(x)-Bg'(x)=f(a+x)+f(-a-x)$

Integrate the second equation leads to (2) $Bf'(x)+Ag'(x)=g(a+x)+g(-a-x)$

Consider (1) and (2) like a system of equation unknown are $f'(x) , \; g'(x)$
the system is Cramer since $A^2+B^2 \neq 0$, $f',g'$ are $C^{0}$ functions
$f,g$ are $C^{1}$ and again with (1),(2)
$f,g$ are $C^{\infty}$

Derivate the first and the second equation in $y$ variable and fixed $y=0$
with $f'(0)=2 ,\; g'(0)=1$ we have

$f'=2f-g$
$g'=f+2g$

$f''=2f'-g'$
$g'=f+2g$

we get

$f''=4f'-5f$ similarly $g''=4g'-5g$

$f,g$  verify the differentail equation

$y''-4y'+5y=0$
$y=me^{2x}\cos x +ne^{2x}\sin x$
with initial conditions

$f(0)=1, \; f'(0)=2 ; \;g(0)=0 ,\; g'(0)=1$

$f(x)=e^{2x}\cos x$
$g(x)=e^{2x}\sin x$


I know that the following sum can be simplified like thus:

$ \sum^c_{b\equal{}0}\sum^b_{a\equal{}0}a \equal{} \sum^c_{b\equal{}0}\frac{b(b\plus{}1)}{2} \equal{} \sum^c_{b\equal{}0}\frac{b^2 \plus{} b}{2} \equal{} \frac{c(c\plus{}1)(2c\plus{}1)}{12} \plus{} \frac{c(c\plus{}1)}{4} \equal{} \frac{c^3\plus{}3c^2\plus{}2c}{6}$

But what if we have something like:

$ \sum^g_{f\equal{}0}\sum^f_{e\equal{}0}\sum^e_{d\equal{}0}\sum^d_{c\equal{}0}\sum^c_{b\equal{}0}\sum^b_{a\equal{}0}a$

Is there an easy way to get the closed-form formula, or am I limited to working this out by hand? Please do respond if you have any idea as to how to approach this. Thank you.

In your case the answer is $ \frac{g(g\plus{}1)(g\plus{}2)(g\plus{}3)(g\plus{}4)(g\plus{}5)(g\plus{}6)}{7!}$ I hope you can see a pattern. Once you determine the pattern prove it by mathematical induction!


probability
LaTeX
number theory
prime numbers

#2 what is the probability that the sum of the digits of a randomly chosen 3 digit # is a prime

#4 How many integers between 0 and 100 exclusive, can be expressed as the sum of 2 squares

lol abacadaea, wow, how do you forget a question you did 2 hours ago!!!!
the #4...it should read:

#4--What is the probablity that a number selected from the set $ 0 \angle x \angle 100$ can be represented as the sum of two perfect squares?

(sorry for bad latex)

and i dont remember #2 what test was it from!

oh and if your wondering why you cant find these problems online--its because theyre from the 1990's(holy cow that was a long time ago!)

2) Just try to find the prime numbers obtainable from the sum of three positive integers, the rest should be simple
4) Basicaly, you have 10 squares, and you need to choose 2 of them, then you need to subtract the ones that would give you a sum above 100.

4. I'm too lazy to do it, but don't forget that some things, like 25+0, equals some other sum, like 16+9.

[quote="Math Geek"]I'm doing number 2 off the top of my head, but I think you should list out all possible prime numbers (which would be from 2 to 23).  Then, find the number of ways for three digits to equal 2, 3, 5, ...23, where the digits are all nonzero, using balls and urns, and then count the number of ones with at least one zero in them, using casework.  I guess you can't just go only with balls and urns b/c there might be some cases where the first digit is 0, which is incorrect.  idk, but that's how I would do it. :)[/quote]

If you understand the idea of balls and urns, you can modify it to get what you want :)


superior algebra
superior algebra unsolved

Let $ A$ be a ring, and let $ e$ be an element in A such that $ e^2 \equal{} e$
$ eb \in eA, eb' \in A$ why if $ eb \equal{} eb'$ then $ b \equal{} b'$?

I know it is untruth. Is any way to remove this post?

As a general rule, no. Only moderators can delete threads.

Just for completeness, here's a counterexample:
$ A\equal{}\mathbb{Z}^2$ with pointwise multiplication, $ e\equal{}(1,0),$ $ b\equal{}(0,0),$ $ b'\equal{}(0,1).$ Both $ eb$ and $ eb'$ are zero.
In fact, nontrivial idempotents are always zero-divisors.


Find the area of the region determined by the system:
$ y \ge |x|\\
y \le \minus{}|x\plus{}1|\plus{}4$
Express your answer as a decimal.
[hide="Answer"]7.5[/hide]
Can you do this without graphing?

[hide="Solution"]Find the intersections of the two graphs $ y\equal{}|x|$ and $ y\equal{}\minus{}|x\plus{}1|\plus{}4$.
$ |x|\equal{}\minus{}|x\plus{}1|\plus{}4$

Consider $ x\ge 0$

Then we have that $ x\equal{}\minus{}x\plus{}3$
$ 2x\equal{}3$
$ x\equal{}\frac{3}{2}$

Thus our first intersection point is $ (\frac{3}{2},\frac{3}{2})$.

Now Consider $ \minus{}1\le x\le 0$. This doesn't work.

If $ x\le \minus{}1$

Then we have that $ \minus{}x\equal{}x\plus{}5$
$ 2x\equal{}\minus{}5$
$ x\equal{}\minus{}\frac{5}{2}$

Which gives us the Intersection point $ (\minus{}\frac{5}{2},\frac{5}{2})$

The other two boundary points are $ (0,0)$ from the first equation $ y\ge x$ and $ (\minus{}1,4)$ from the second equation.  These are the other two points because they are the turning points of both equations.

Thus, we have that the points are $ (0,0), (\frac{3}{2},\frac{3}{2}), (\minus{}1,4), (\minus{}\frac{5}{2},\frac{5}{2})$

You can the use shoelace: $ \frac{1}{2}*|(0\plus{}6\minus{}\frac{5}{2}\plus{}0)\minus{}(0\minus{}\frac{3}{2}\minus{}10\plus{}0)|\equal{}\frac{1}{2}*|15|\equal{}\boxed{7.5}$ As desired.[/hide]


let $ 4^n\plus{}2^n\plus{}1$ is a prime number.prove : there is a natural number like k such that: \[ n\equal{}3^k\]

Put $ n\equal{}3^k \cdot t, (3,t)\equal{}1$. $ 0, t, 2t$ is a complete residue system mod $ 3$ and we get $ 4^{3^k}\plus{}2^{2^k}\plus{}1|4^n\plus{}2^n\plus{}1$ (more nicely: $ x^2\plus{}x\plus{}1 | x^{2t}\plus{}x^{t}\plus{}1$ when $ (t,3)\equal{}1$). When $ t>1$ it contradicts primality, thus $ t\equal{}1$ and then $ n$ is indeed of the form $ 3^k$.

- Stupid title.
- Triple-post.
- Posted before.

See http://www.mathlinks.ro/viewtopic.php?t=135914 .


trigonometry
function
geometry
3D geometry
geometry unsolved

Hello
I'm new here and a try to find a solution.
My problem is that y cant find a formula for calculating the angle Alpha like in the attachment.
I' now that the angle is aprox. 32 degreese, but I could't find the mathematik way.
It would be wery nice if anybody could help me.
In the second attachment I startet to get to the formula, but i'm not sure if it's the right way.
Thanks

Hello
I saw that many people entered at my topic, but it looks like that nobody could find a solution.
Does anybody has a Idea how to solve it.
Please help me!

The points in the diagram appear to nonconstructible by ruler and compass (for given generic values of $a,b,c$), but are easily constructed if $\alpha$ is also given. If that is true, i.e. the equations that appear are irreducible, then $\alpha$ is nonconstructible and trigonometric functions of $\alpha$ will involve cube roots.


Please convert the following:

4.2 ft.lb(force)/lb(mass).F to m^2/s^2.K

Please reply to me ASAP.

[quote="t0rajir0u"][quote="Inspired By Nature"]lb(force)/lb(mass)[/quote]

Pounds aren't a valid unit of mass.  The corresponding unit of mass is the slug.[/quote]
My chemistry teacher told me that in chemistry (but not in physics), you can treat mass and weight as the same thing. Is that true?  :huh:
Also, this seems to me like it might actually be more like a physics problem. (It might be a chemistry problem, though, I'm not really sure.  :P )

well, I guess you can, although why you would want to express ANYTHING in english units is beyond me:).  as long as you are on earth and resonably close to sea level, to convert between mass (kg) and force (N) you just need to multiply by 9.8 m/s^2.  After that its easy -- just convert between N and lb, a force to force conversion (although I don't know the exact factor without calculating.  use google if you need it...).

Hence, mass and weight are "the same" according to someone at sea level on earth in the sense that given one, it is possible to determine the othere.  however, keep in mind that they have different units and measure different things.

There are several english engineering units.  One is pounds (force).  Another is pounds (mass).

Independent of a discussion about the merits of the problem, if one knows that 2.2 kg (mass) equals 1 pound (mass) and produces 1 pound (force) at the earth's surface when, g is 9.8 m/s/s and 32 f/s/s and spends a bit of time, this problem should be solved.

[quote="wizard2378"]if one knows that 2.2 kg (mass) equals 1 pound (mass)[/quote]

If one knows that, one is wrong :)

However, if one knows that 2.2 pounds equals 1 kg, one is much closer to the real thing ;)


i hate gambits, b/c u have to give up pawn(s) just for some center control

no, if u gambit for center, thats a good thing. i play lots of gambits

Well a gambit usually involves the sacrifice of something, so a lot of experianced players will not take.

I use gambits if I have too, or it will make a great play. I usually don't sacrifice my pawns.

[quote="hunter34"]I usually don't sacrifice my pawns.[/quote]

I like playing a closed position because I am better at placing my knights in good squares than my bishops. The sacrifice of pawns (for example, in Queen's or King's Gambit) will throw me off a little bit while I adjust to an "open" game (of course it still benefits me as I retake the pawn and develop a bishop).

[quote="Temperal"][quote="shaggy75"][quote="7h3.D3m0n.117"]I play Queen's gambit and if they take the "free" pawn you get the pawn back anyway while you develop a bishop, which gives you a tempo.[/quote]

that does not happen very often[/quote]

Uh, "does not happen very often"? It's impossible to stop the capture of the pawn. But you really shouldn't take it, there are more important things that pawns.

I don't think you really understand gambits.[/quote]

oops, sorry. i phrased that wrongly. what  i meant to say is that when others play the queen's gambit against me, i like to play c5 after they play c4 instead of taking the pawn

[quote="mz94"]not if u play 3:...e7-e6. thats the tarrasch(if he doesnt take pawn)[/quote]

Yeah, that variation's kind of annoying. I do not like Tarrasch's obsession with pawns.

[quote="pianoforte"]I like to move out all of my pawns out, starting from the sides, and then use my queen for utter PWNAGE.[/quote]

Uh.... right. Is that supposed to be a pun?

Someone should actually try that against a good player, there'll be PWNAGe, all right, just the other way around.

wow, that pun was not intended. cool, accidental punning  :yinyang: .

For me, it depends. I'm not too good and when I play people that are better than me, I screw up and lose with gambits

[quote="7h3.D3m0n.117"]Yeah, that's called the Benoni. Well actually it's a variation of the Benoni.

I hate playing symmetrical games, it is way too annoying. That's why I stopped playing the English in tournaments because a lot of people would counter with Sicilian.[/quote]

I once lost a game to a kid who was coping my move.

 :(

i did that once and lost, except that time, i was the copier :)

I play the Sergei Urusov gambit as white frequently, and occasionally the Evans, but in general I avoid playing gambits. I just happen to like those two.

Vienna Gambit = Epic Pwn

(in my opinion)

[quote="pythag011"]I play Benko gambit[/quote]

Darn, you play chess? *gasp*

I play gambits occasionally like Smith-Morra stuff. fun.

[quote="bobbym"]I like sacrificing the other guys pawns and pieces much better.[/quote]

Try playing Black vs. me when I play Smith-Morra. It won't be fun for you :P

I have given the Kings Gambit a whirl now and then. With mixed results.

[quote="bryan_hooi"][url]http://en.wikipedia.org/wiki/Blackmar-Diemer_Gambit[/url] (1. d4 d5 2. e4 dxe4 3. Nc3 Nf6 4. f3) is also quite fun. Very aggressive for white and makes for open positions and lots of tactics. At lower skill levels one lost pawn hardly matters anyway :P[/quote]
you rock
i play the bdg in tournaments (competitive USCF 1800+ games) and it's very effective; i recommend it to people

Noooooooooooo...

I think gambits are good as long as they aren't BS.  By BS I mean "they can be refuted by knowing a little bit of theory".

Also...the queen's gambit isn't really a gambit :P

Yes, I love to use the queens gambit to gain strength on the left side of the board and continue to force the king to castle kingsides.

[quote="n0vad3m0n"]I play Queen's gambit and if they take the "free" pawn you get the pawn back anyway while you develop a bishop, which gives you a tempo.[/quote]

The queen's gambit is a misnomer. It got it's name way before much of chess strategy was developed, in the romantic era.

Sure, most gambits are technically "unsound". However, below master level gambits can be deadly, if the Gambiteer's style is suited to the gambit he plays. The fact is that below master level players don't know openings all that much. When an unprepared club-level player is faced with a gambit, he will almost certainly lose.

This having been said, however, many grandmasters use gambits as surprise weapons. Witness Kasparov's use of the Evans Gambit and Fischer's use of the Kings Gambit!

Gambits work against confused people and people who don't know the opening.  I think that the Najdorf variation works better(for black) against e4 rather than waiting for a move like c4


trigonometry
function
inequalities proposed
inequalities

$ x,y,z$ are nonegative real numbers with $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$

Then prove:
$ x^2 \plus{} y^2 \plus{} z^2 \ge \frac {3}{4}$

$ 1\equal{}2xyz\plus{}\sum xy\ge 2xyz\plus{}3(xyz)^{\frac{2}{3}}\Rightarrow xyz\le \frac{1}{8}$

Assume the contrary.
$ \frac{3}{4}>\sum x^2\ge \sum xy\equal{}1\minus{}2xyz\ge \frac{3}{4}$. Contradiction.

[quote="borislav_mirchev"]I have a question: What is the source of this problem? Is it a problem composed by you?[/quote]

I created it.

$ x \plus{} y \plus{} z \ge \frac {3}{2}$ is also true :P .

and $ x^2\plus{}y^2\plus{}z^2 \ge 4xyz \plus{} \frac{1}{4}$ too.

Let $ x \plus{} y \plus{} z \equal{} k$. Then $ k^2\ge 3(xy \plus{} yz \plus{} zx)$ and $ k^3\ge 27xyz$. Then we have $ \frac {k^2}{3} \plus{} \frac {2k^3}{27}\ge xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$ and thus $ 2k^3 \plus{} 9k^2\ge 27$. Assume the contrary, that $ k < \frac 32$. Then $ 2k^3 < \frac {27}{4}$ and $ 9k^2 < \frac {81}{4}$ which imply $ 2k^3 \plus{} 9k^2 < \frac {108}{4} \equal{} 27$ contradiction.

Obvious problems,just use that
if $ xy\plus{}yz\plus{}zx\plus{}2xyz\equal{}1,x,y,z\geq 0$,then there exist $ \alpha,\beta,\gamma\in [0,\frac{\pi}{2}]$,such that
$ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$,and $ \sqrt{xy}\equal{}\cos{\alpha},\sqrt{yz}\equal{}\cos{\beta},\sqrt{zx}\equal{}\cos{\gamma}$.

[quote="Erken"]Obvious problems,just use that
if $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1,x,y,z\geq 0$,then there exist $ \alpha,\beta,\gamma\in [0,\frac {\pi}{2}]$,such that
$ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$,and $ \sqrt {xy} \equal{} \cos{\alpha},\sqrt {yz} \equal{} \cos{\beta},\sqrt {zx} \equal{} \cos{\gamma}$.[/quote]

Or use the fact that if $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1,x,y,z\geq 0$ then there exists $ a,b,c > 0$ such that: $ x\equal{}\frac{a}{b\plus{}c}$.

[quote="FOURRIER"][quote="Erken"]Obvious problems,just use that
if $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1,x,y,z\geq 0$,then there exist $ \alpha,\beta,\gamma\in [0,\frac {\pi}{2}]$,such that
$ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$,and $ \sqrt {xy} \equal{} \cos{\alpha},\sqrt {yz} \equal{} \cos{\beta},\sqrt {zx} \equal{} \cos{\gamma}$.[/quote]

No, if you put $ \sqrt {xy} \equal{} \cos{\alpha}$ and similars, you can't compute $ x$ in function of $ \alpha,\beta , \gamma$.[/quote]
Why?
$ x\equal{}\frac{\cos{\alpha}\cos{\gamma}}{\cos{\beta}}$.

[quote="Virgil Nicula"][quote="FOURRIER"]$ \left\|\begin{array}{c} \{x,y,z\}\subset [0,\infty ) \\
\ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1\end{array}\right\|$ $ \implies$ $ x^2 \plus{} y^2 \plus{} z^2 \ge \frac {3}{4}$[/quote]
[color=darkblue][b][u]Proof.[/u][/b] Let $ t^3 \minus{} mt^2 \plus{} (1 \minus{} 2n)t \minus{} n \equal{} 0$ the equation with the roots $ x,y,z$ , where $ x \plus{} y \plus{} z \equal{} m$ , $ xy \plus{} yz \plus{} zx \equal{} 1 \minus{} 2n$ and $ xyz \equal{} n$ . Observe (similarly to [b]SplasD[/b]'s idea) that $ 1 \minus{} 2n\ge 3\sqrt [3]{n^2}$ Denote $ p \equal{} \sqrt [3]n$ . Thus, $ 2p^3 \plus{} 3p^2 \minus{} 1\le 0$ $ \implies$ $ (p \plus{} 1)^2(2p \minus{} 1)\le 0$ $ \implies$ $ p\le\frac 12$ $ \implies$ $ \boxed {\ n\le\frac 18\ }$ . Therefore, $ \sum x^2 \equal{} m^2 \minus{} 2 \plus{} 4n$ and $ f'(x) \equal{} 3t^2 \minus{} 2mt \plus{} (1 \minus{} 2n)$ with $ \Delta' \equal{} m^2 \minus{} 3(1 \minus{} 2n)\ge 0$ , i.e. $ \boxed {m^2\ge 3(1 \minus{} 2n)}$ . Therefore, $ \sum x^2 \equal{} m^2 \minus{} 2 \plus{} 4n\ge 3(1 \minus{} 2n) \minus{} 2 \plus{} 4n \equal{} 1 \minus{} 2n\ge \frac 34$ because $ n\le \frac 18$ .[/color][/quote]

Nice,
I think that : $ \Delta' \ge 0$ because f(t) have 3 roots so it can't be strictly increasing, is that it?

[quote="Erken"]
Why?
$ x \equal{} \frac {\cos{\alpha}\cos{\gamma}}{\cos{\beta}}$.[/quote]

Ok , but in this way it becomes:
$ 4 \sum cos^4(\beta)cos^4(\gamma) \ge 3 cos^2(\alpha)cos^2(\gamma)cos^2(\beta)$ and  I don't think it's easy.


calculus
derivative
function
logarithms
calculus computations

I'm trying to prove a definition for a problem i'm working on but I have to find the order of a growth of a logarithmic function, i.e. the derivative.  I'm givin the logarithm with no base (so I assume that's the same as the natural log?), anyways...I can't quite remember how to solve this f(x)'.

f(x) = (2x)log x - 15x + 12
f(x)' = ??? - 15

*I realise that 15n becomes 15, and the constant 12 becomes 0...

Any help would be great, thanks.  Also, if the answer is posted, could you please post the equation so I get a sense of "how to" solve the equation.  THANKS!

You should use the product rule $ (uv)'\equal{}u'v\plus{}uv'$ for the first term. [url=http://library.wolfram.com/webMathematica/MSP/Explore/Education/WalkD]This website[/url] gives step-by-step explanations.


conics
parabola
geometry
geometric transformation
reflection
geometry proposed

Prove that three tangents to a parabola always form the sides of a triangle whose altitudes intersect on the directrix of the parabola.

Let $ \mathcal{P}$ be a parabola with focus $ F$ and directrix $ p.$ $ \mathcal{L}_1, \mathcal{L}_2, \mathcal{L}_3$ are three arbitrary tangents of $ \mathcal{P}.$ Define the intersections $ A \equiv \mathcal{L}_1 \cap \mathcal{L}_2,$ $ B \equiv \mathcal{L}_1 \cap \mathcal{L}_3,$ $ C \equiv \mathcal{L}_2 \cap \mathcal{L}_3.$ $ A',B',C'$ are the orthogonal projections of $ F$ on $ \mathcal{L}_1,$ $ \mathcal{L}_2,$ $ \mathcal{L}_3.$ Reflections of focus $ F$ across $ A',B',C'$ lie on $d,$ thus $ A',B',C'$ are collinear $ \Longrightarrow$ $ F \in  \odot(ABC).$ Therefore, the directrix $ p,$ which is the image of $\overline{ A'B'C'}$ under the dilatation with center $ F$ and coefficient $ 2,$ becomes Steiner line with pole $ F$ WRT $\triangle ABC$ $ \Longrightarrow$ $ p$ passes through the orthocenter of $\triangle ABC.$


Let $p(x)$ be a polynomial of degree $n$, not necessarily with integer coefficients. For how many consecutive integer values of $x$ must $p(x)$ be an integer in order to guarantee that $p(x)$ is an integer for all integers $x$?

[hide]
$n+1$

If we have $n+1$ consecutive integral values, then the $n$th level of finite differences, the constant level, must be integral, which means that all integral inputs will produce an integral output.

If you have fewer than $n+1$, then for all you know, the constant level might not be integral, and the polynomial wouldn't give integral outputs for all integral inputs.

[/hide]

I don't think your second part is quite rigorous.

for the second part of his:

[hide]
consider

P(x) = sqrt(2) (x - 1)(x - 2)...(x - n)

Thats a polynomial of degree n where 1 through n give out integers (zero), and no other integer values give out integers
 [/hide]

Or you can choose any non-integer for the $n+1$st point and use Lagrange Interpolation to construct the polynomial.


We all know what the graviton orbital looks like in the third dimension, a swirling torus around the nucleus but in the fourth dimension you have to imagine that the earth is flattened into a zero-mass plate.  Now it is very important that we realize stellar bodies have no mass in the fourth-D because of what I will explain later.
     There is an article in Wired magazine about semi-conductor pseudo-atom formation.  The semi-conductor is shaped like the prongs of a fork all pointing into a central location spaced a few atoms apart.  This allows the formation of a specific number of electrons to form together in a pseudo-atom without a nucleus.  Now, with the absence of a nucleus to make the atomic structure unstable they are able to create huge atomic structures that will interact with other elements much the same as real atoms.  The mass-less atoms are able to support these atomic "super-structures"; by comparison the graviton orbitals are able to remain stable in the 4th-D because 3rd-D stellar bodies also have no relative mass.
     Electron spin cannot be rectified mathematically unless they pass in and out of the 4th-D, therefore the connection between the graviton and the electron must be bound together, proving super-string theory.  The electron passes in and out of the 3rd-D and 4th-D interacting strongly in the 3rd-D and the graviton passes in and out of the 3rd-D and 4th-D interacting weakly with the 3rd-D but strongly with the 4th-D.
     A graviton passing through the Earth would do so instantaneously because of the fact that it is flat in the 4th-D however it would take some time to complete the rest of it's orbit.  A magnetic flux is the fastest known force to us meaning that gravitons must be smaller than a photon to satisfy the Special-Theory of Relativity.

It might be helpful to add more information regarding general relativity.  I admit I understood very little of that.

The basis of gravitons being smaller than photons being that the theory of relativity states that photons are the smallest known particle therefore they travel the fastest.  But in the 4th-D a graviton must have a smaller relative mass.

I always thought that photons moved the speed they did because they had no mass...


What are the last two digits of $57^{2006}$? Please show your work.

[hide="simpler way"]
Look at the pattern of the last 2 digits
$57^{1}-57$
$57^{2}-49$
$57^{3}-93$
$57^{4}-01$
It repeats from here
2006 is congruent to 2 in mod 4,
so $49$
[/hide]

[quote="mathcrazed"]57^2006 is congruent to 57^6 mod 100, so find 57^6 (mod 100)...
p.s. is discussing mods in mathcounts not appropriate?[/quote]

its perfectly fine, as far as i know...

[quote="mathcrazed"]57^2006 is congruent to 57^6 mod 100, so find 57^6 (mod 100)...
p.s. is discussing mods in mathcounts not appropriate?[/quote]

But Euler's Totient isn't the most efficient, notice that it repeats every 4, not every 40 (as Totient gives)

My method is the same as vamathletes', but is there any simpler way to do it? And why $57^{2006}\equiv57^{6}\, (\text{mod}\,100)$?


Write down from $1$ to $999$; we got the number $a=123456789101112...998999$. What's the remainder if $a$ is divided by $11$?

1-11

remainder dividing by 11

from 1-11

remainders go

1,2,3,4,5,6,7,8,9,10

1+10=11

2+9=11
2+8=11
3+7=11
4+6=11
5=5

now since the remainders are cyclic mod11

the remainder shall be roughly 5 * 999/11 where 999/11 is rounded down

11 divides 990 90 times

90*5 = 450

450/11 = 40 remainder 10

remainder of 1-990/11 = 10

remainder 991-999 = 1+2+3+4+5+6+7+8+9

= 1mod11

10+1mod11=mod11

i think this is correct though im tired and have possibly have made a few mistakes soif someone could check through id be greatful


In an acute triangle $ ABC,AC > BC$ and $ M$ is the midpoint of $ AB$ Let altitude $ AP$ and $ BQ$ meet at $ H$ and let lines $ AB$ and $ PQ$ meet at R prove that the two lines $ RH$ and $ CM$ are perpendicular.

$ RH$ is the polar of $ C$ respect to the circle of radius $ AM$ and center $ M$.


$\text{You accidentally drop a glass rod and it shatters into 3 pieces. What is the probability the three pieces form a triangle?}$

There's a really nice solution which uses geometry. Yes, $\mathsf{geometry}$.

[hide="Hint - What Geometric Technique To Use"]Viviani's Theorem[/hide]

[hide="Not computed but method"]Well, one way you could do it is find the area bounded by the equations
x+y+z=3
x,y,z>0
x<z+y, z<x+y, y<x+z[/hide]

Now, what if it breaks into 4 pieces? Probability the four pieces form a quadrilateral?

And, what about n pieces? Probability to form a n-agon?

 :D

What a coincidence.. I was just discussing the same problem a few days ago...
I couldn't solve my generalizations yet though... Too tedious actually :/

[quote="Centy"]Viviani's Theorem[/quote]

Don't think I've ever heard of this :huh: which is strange, since I happen to know tons of geometry theorems ;)

Well :
http://www.cut-the-knot.org/Curriculum/Geometry/Viviani.shtml
http://mathworld.wolfram.com/VivianisTheorem.html
http://www.jimloy.com/cindy/equi.htm

I also didn't know this one... but this is not a surprise, since, unlike you, I know nothing in geometry. ;)

Hm, I see. Well, I don't really call that kind of things a "theorem". It's trivial, after all...

[hide="My Solution"]
We can form a triangle iff all three pieces are less than half the length of the whole rod. Let the altitude of an equilateral triangle be the length of the rod. Pick any point in the triangle and let the length of the three perpendiculars to the sides represent the broken lengths of rod. We can see that to form a triangle, our point needs to lie with the medial triangle, which is $\frac{1}{4}$ the area of our original triangle...and hence our answer.
[/hide]

Here's a sort of like a logical-ish answer:

Call the length of the rod $r$.  The first piece, call it length $s$, has to be less than $\frac{r}{2}$, and the chances of that are

$\frac{\frac{r}{2}}{r}=\frac{1}{2}$.

The second piece has to be of length greater than $\frac{r}{2}-s$, but less than $\frac{r}{2}$.  The chances of that are
$\frac{\frac{r}{2}-(\frac{r}{2}-s)}{r-s}=\frac{s}{r-s}$,
Which ranges evenly from $0$ to $1$... meaning the average probability of the second event is $\frac{1}{2}$, and the chances of the first two pieces (and therefore the third) meeting the conditions equals $\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$.


Let $ F_i$ denote the $ i$-th Fibonacci number and $ p$ be a prime of the form $ 10k\pm 1$. If $ p\mid F_n$ and $ p>n$ for some $ n$, then prove that $ p\equal{}mn\plus{}1$ for some positive integer $ m$.

minimal period for fibonaccci numbers is divisor $ p\minus{}(\frac{p}{5})$. In these case $ p\minus{}1$. Therefore $ n_0|p\minus{}1$ if $ n_0$ is minimal period.
But if $ n$ is not minimal period $ n_0$, it may be false if $ n_0<\frac{p\minus{}1}{2}$.

Or, more explicitly, take $ n \equal{} 33$ and $ p \equal{} 89$.

$ 89 \equal{} F_{11}|F_{33}$

Sorry about that. But I think my problem is correct if $ n$ is prime. :)


[color=green]Find the number of positive integers not exceeding 1000 that are not divisble by 3 or 5.[/color]

Thanks for your help.  :lol:

$1000-\left[\frac{1000}3\right]-\left[\frac{1000}5\right]+\left[\frac{1000}{15}\right]$.

Has school started in Canada?

[hide="info about the method"]This is called PIE (property of inclusion and exclusion). It works because by finding $\left[\frac{1000}3\right]$ you are finding the number of numbers divisible by 3, With 5 you do the same thing. But you count the factors of 15 twice. So you have to add them back again.[/hide]

You find the number not in the list and then subtract. You do the following: $1000-(\frac{1000}{3})-(\frac{1000}{5})+(\frac{1000}{15})$. This is called PIE.


I have posted already my proof for the following inequality:

If: $ a_i\geq0,b_i>0,i\equal{}1,2,\dots,k,$ and $ k,n\in N,k\geq1$, then:
$ \frac{{a_1}^{n\plus{}1}}{b_1^n}\plus{}\frac{a_2^{n\plus{}1}}{b_2^n}\plus{}\cdots\plus{}\frac{{a_k}^{n\plus{}1}}{b_k^n}\geq\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_k}{b_1\plus{}b_2\plus{}\cdots\plus{}b_k}\cdot\left(\frac{{a_1}^n}{b_1^{n\minus{}1}}\plus{}\frac{{a_2}^n}{b_2^{n\minus{}1}}\plus{}\cdots\plus{}\frac{{a_k}^n}{b_k^{n\minus{}1}}\right)$

I used this result (named [u]my lemma [/u]) for proving my first proposed inequality on MathLinks (and you can see also my proof of this lemma there), but I'm convinced that this inequality deserves a separate disscusion. Especially because we can use it for solving some well known problems.
[b]can_hang2007[/b] gave me a different idea for proving it. I'm eager to find out new proofs, new results or new remarks related to this inequality. I haven't seen it before anywhere, and I think that it's a new result.

Marin Bancoș - ROMANIA

By Holder's inequality, $ (b_1 \plus{} b_2 \plus{} \dots \plus{} b_k)^{\frac {n}{n \plus{} 1}}\left(\frac {{a_{1}}^{n \plus{} 1}}{b_{1}^{n}} \plus{} \frac {a_{2}^{n \plus{} 1}}{b_{2}^{n}} \plus{} \cdots \plus{} \frac {{a_{k}}^{n \plus{} 1}}{b_{k}^{n}}\right)^{\frac {1}{n \plus{} 1}}\geq a_1 \plus{} a_2 \plus{} \dots \plus{} a_k$, and
$ (b_1 \plus{} b_2 \plus{} \dots \plus{} b_k)^{\frac {1}{n \plus{} 1}}\left(\frac {{a_{1}}^{n \plus{} 1}}{b_{1}^{n}} \plus{} \frac {a_{2}^{n \plus{} 1}}{b_{2}^{n}} \plus{} \cdots \plus{} \frac {{a_{k}}^{n \plus{} 1}}{b_{k}^{n}}\right)^{\frac {n}{n \plus{} 1}}\geq \frac {{a_{1}}^{n}}{b_{1}^{n \minus{} 1}} \plus{} \frac {{a_{2}}^{n}}{b_{2}^{n \minus{} 1}} \plus{} \cdots \plus{} \frac {{a_{k}}^{n}}{b_{k}^{n \minus{} 1}}$.
If we multiply these together we get the lemma. Note: in this proof, $ n$ can be any positive real number.

Thank you for your proof, [b]bakerbakura[/b]. Your remark is a valuable one.
I'd like to say that in my proof I didn't use any important classical inequality! Maybe this is the main advantage of it. I'm sure you've noticed that.
[b]Vasc[/b] already made your remark at http://www.mathlinks.ro/viewtopic.php?t=310721. He gave me another solution to this lemma. Making a comparison between his proof and mine, it's obvious that his solution is simplier. Also, [b]can_hang2007[/b] gave me a similar solution to Vasc's proof in a previous topic (related to another problem).
I'm really preocupied with finding mathematical "tools" which can be used for solving difficult problems. I'm sure that this lemma could be considered one of them.


geometry
geometric transformation
rotation

There are 8 members in a a bridge committee (committee for making bridges).  Of these 8 members, 3 are chosen to be in special "approval" committee with 1 of 3 members being the "boss."  In how many ways can this happen?

There are five normal members, two approval committee members, and one boss.

$ \frac{8!}{5!\cdot 2!\cdot 1!} \equal{} \boxed{168}$

there we go

There is another way to approach this problem.

[hide="Hint"]
Use the fact of $ \binom{n}{k}$ and "rotation" of $ k$ [/hide]

Anyone wants to solve it this way?

There is $ \binom {8}{3}$ ways to chose committee, and each of three members can be boss so

$ 3*\binom {8}{3}\equal{}168$

[color=green]$8.7.6.3$ ways bcoz 8c3 is normal arrangement and again 3 have equal chance to be boss and so can interchange positions[/color]

8C3 * 3
Coz 3 members can be chosen from 8 by 8C3 ways......and of the three any 1 can be the boss
Now 8C3*3=168


linear algebra
matrix
linear algebra unsolved

Can anyone show me the general form of the inverse of an upper triangular matrix

${A=\left[\begin{array}{cccc}A_{1,1}&A_{1,2}&\ldots&A_{1,n}\\0&A_{2,2}&\ldots&A_{2,n}\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&A_{n,n}\end{array}\right] }$    :?:

bchui,

The inverse of an upper triangular matrix is an upper triangular matrix.

[quote="bchui"]Of course it is an upper triangular matrix but I need the general expression in terms of the elements $A_{1,1},A_{1,2}\ldots$ :roll:[/quote]You can't expect more than a matrix product, just like in the case where the entries are scalars and not matrices.

[quote="loup blanc"]For an (n,n) upper-triangular matrix A=[Aij] over k there exists an easy  recurrence formula for the rows L1,...,Ln of the inverse of A.
Let {E1,...,En} the canonic basis of k^n written in rows.
Ln=(1/Ann)En, and for i<n
Li=(1/Aii)(Ei-A(i,i+1)Li+1-...-AinLn).[/quote]

Hold on,

${ E_{1}=\left[\begin{array}{c}1\\ 0\\ \vdots\\ 0\end{array}\right],\ldots E_{n}=\left[\begin{array}{c}0\\ \vdots\\ 0 \\ 1 \end{array}\right]~,~L_{n}=\left[\begin{array}{c}0\\ \vdots\\ 0\\{A_{n,n}}^{-1}\end{array}\right] }$

What's the explicit expression of $L_{i}$?

1)Li and Ei are not columns but rows.
2) I can't do better.


Here is my updated list

Chapter Round

2000 Sprint Round Only Answers
2001 Sprint Round  Only Answers
2001 Target Round No Answers or Solutions
2001 Team Round No Answers or solutions
2002 Sprint Round With Solutions and answers
2002 Target Round With Solutions and answers
2002 Team Round With Solutions and answers
2003 Sprint Round With Solutions and answers
2003 Target Round With Solutions and answers
2003 Team Round With Solutions and asnwers
2004 Sprint Round With Solutions and answers
2004 Target Round With Solutions and answers
2004 Team Round With Solutions and answers
2005 Sprint Round With Solutions and answers
2005 Target Round With Solutions and answers
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2005 Countdown Round Only Answers
2004 Countdown Round Only Answers
2003 Countdown Round Only Answers
2002 Countdown Round Only Answers
2001 Countdown Round No Answers

State Round

2001 Sprint Round Only Answers
2001 Target Round Only Answers
2001 Team Round Only Answers
2002 Sprint Round With Solutions and answers
2002 Target Round With Solutions and answers
2002 Team Round With Solutions and answers
2003 Sprint Round With Solutions and answers
2003 Target Round With Solutions and answers
2003 Team Round With Solutions and answers
2004 Sprint Round With Solutions and answers
2004 Target Round With Solutions and answers
2004 Team Round With Solutions and answers
2005 Sprint Round With Solutions and answers
2005 Target Round With Solutions and answers
2006 Sprint Round With Solutions and answers
2006 Target Round With Solutions and answers
2006 Team Round With Solutions and answers
2007 Sprint Round With Solutions and answers
2007 Target Round With Solutions and answers
2007 Team Round With Solutions and answers
2008 Sprint Round With Solutions and answers
2008 Target Round With Solutions and answers
2008 Team Round With Solutions and answers
2009 Sprint Round With Solutions and answers
2009 Target Round With Solutions and answers
2009 Team Round With Solutions and answers
2009 Countdown Round Only Answers
2008 Countdown Round Only Answers
2007 Countdown Round Only Answers
2006 Countdown Round Only Answers
2005 Countdown Round Only Answers
2004 Countdown Round Only Answers
2003 Countdown Round Only Answers
2002 Countdown Round Only Answers
2001 Countdown Round Only Answers Maybe

School

2006 Sprint Round With Solutions and answers
2006 Target Round With Solutions and answers
2006 Team Round With Solutions and answers
2006 Countdown Round Only answers
2007 Sprint Round With Solutions and answers
2007 Target Round With Solutions and answers
2007 Team Round With Solutions and answers
2007 Countdown Round Only answers
2008 Sprint Round With Solutions and answers
2008 Target Round With Solutions and answers
2008 Team Round With Solutions and answers
2008 Countdown Round Only Answers
2009 Sprint Round With Solutions and answers
2009 Target Round With Solutions and answers
2009 Team Round With Solutions and answers
2009 Countdown With Solutions and Answers
1999 Sprint Round No Answers
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1999 Team Round No Answers
1999 Countdown Round No Answers

Please email me @ dwade884@gmail.com if you are intersted in trading with me.....

Thanks.

Please move this to the Mathcounts resources forum. You might get more people willing to trade.

Consider this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=133189


Let $ ABC$ be a triangle. Denote by $ M$, $ N$, $ P$ the midpoints of its sides $ AB$, $ BC$, $ CA$. Let $ O$ be the center of the circumscribed circle of triangle $ ABC$. Prove that the circumscribed circles of the triangles $ AON$, $ BOP$ and $ COM$ meet in a point different from $ O$.

I've already read in a book this problem. It is a proposed problem for individual study. The book was about the inversion in geometry. So ... I think that the problem is correct, isn't it?

It is true, and inversion is the key... but I have not the time now, can you wait?

[b]EDIT:[/b] I solved your problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=19720 .

  darij

[color=darkblue]Denote : the circumcircle $w$ of the triangle $ABC$ ; the middlepoints $M$, $N$, $P$ of the sides $[BC]$, $[CA]$ and $[AB]$ respectively ; the intersections $X\in BC\cap AA$, $Y\in CA\cap BB$, $Z\in AB\cap CC$ and the projection $S$ of the center $O$ to the line $d\equiv\overline{MNP}\ ($ the points $M$, $N$, $P$ are collinearly from a well-known property and $d\cap w=\emptyset\ ).$ The quadrilaterals $OMXA$, $ONYB$, $OPZC$ are concyclically and the point $S$ belongs to all their  circumcircles. It is abviously that $S\not\equiv O\ .$[/color]

Inversion in the circumcircle (O) carries the circumcircles $\mathcal C_{A},\ \mathcal C_{B},\ \mathcal C_{C}$ of $\triangle AON,\ \triangle BOP,\ \triangle COM$ passing through the inversin center O into 3 lines passing through A, B, C, respectively. All we have to show that  these 3 lines are concurrent. But these 3 lines are symmedians of $\triangle ABC$ intersecting at its symmedian point. For example, the inversion of $\mathcal C_{A}$ in (O) passes through the inversion N' of N in (O). Since N is the midpoint of BC, N' is the intersection of tangents $t_{B},\ t_{C}$ of (O) at B, C. Thus the inversion of $\mathcal C_{A}$ in (O) is the line AN', the A-symmedian of $\triangle ABC.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ G$ is a group which has only finitely many sub-groups in all,, prove that $ G$ must be finite.

If $ G$ contains an element of infinite order, it is easy to construct infinitely many subgroups.

Thus, assume $ G$ is infinite, but contains no elements of infinite order.  Since every element of $ G$ is contained in a cyclic subgroup with finitely many elements, it follows that there are infinitely many cyclic subgroups.


inequalities
rearrangement inequality
inequalities proposed

If $ x,y,z$ are real numbers then prove the following inequality:

         $ \frac{y^2\minus{}x^2}{2x^2\plus{}1}\plus{}\frac{z^2\minus{}y^2}{2y^2\plus{}1}\plus{}\frac{x^2\minus{}z^2}{2z^2\plus{}1} \geq 0$.

I am not too familiar with rearrangement inequality so I avoided such a solution,raja.
Instead i considered the substitution $ a\equal{}2x^2\plus{}1,b\equal{}2y^2\plus{}1,c\equal{}2z^2\plus{}1$ and then the problem becomes pretty trivial(just a simple use of AM-GM,after that substitution).
Good job fjwxcsl for this nice generalization :)


greatest common divisor
algorithm
number theory
Euclidean algorithm

Well I was experimenting with different numbers today, and I think I came up with a conjecture.

For any prime number, two positive integers that add up to it have a Greatest Common Factor of 1.

For example, with 101, 1+100,2+99,3+98,4+97,5+96,6+95... for all of them, their GCF is 1.

If you can prove or disprove this conjecture please tell how.

-Thanks

[hide="Proof"]Let the prime be $ p$ and the two numbers that add to it be $ x$ and $ p\minus{}x$, where $ x < p$ is a positive integer. Since $ p$ is prime, $ \gcd(x,p) > 1$ iff $ p \mid x$, but since $ p > x$, $ p \nmid x$ and $ \gcd(x,p) \equal{} 1$. By the Euclidean Algorithm, $ \gcd(x,p\minus{}x) \equal{} 1$, as desired.[/hide]

has any1 ever come up with this before me?

Definitely. I've seen this before, but it was awhile ago and I have no idea where.

well if their GCF was >1, then their sum would be divisible by that GCF, therefore making it not prime.

This is a fairly straightforward consequence of the definition of a prime number and could be strengthened to "$ n$ is a prime number if and only if any two positive integers $ a, b$ such that $ a \plus{} b \equal{} n$ satisfy $ (a, b) \equal{} 1$."  If you want to learn more about primes, consult [url=http://mathworld.wolfram.com/topics/PrimeNumbers.html]Mathworld[/url].  :)

For example, here is an alternate characterization of a prime number:

[b]Definition:[/b]  A positive integer $ p$ is prime iff for any two positive integers $ a, b$ such that $ p | ab$ it is true that either $ p | a$ or $ p | b$.

Can you prove that this is equivalent to the usual definition of having no divisors other than $ 1$ and itself?