calculus
integration
logarithms
trigonometry
limit
calculus computations

Evaluate $\int_{0}^{\frac{\pi}{4}}\frac{\ln(\cot\theta)(\sin 2\theta)^{n-1}}{(\sin^{n}\theta+\cos^{n}\theta)^{2}}d\theta$.

$n$ is an integer greater than 1.


You have five pieces of paper. You pick one or more of them and cut each of them into five smaller pieces. Now you take one or more of the pieces from this lot and cut each of these into five smaller pieces. And so on. Prove that you will never have 2003 pieces.

Every time you cut a piece, you create $4$ new pieces.
Since you started with $5$ pieces, the total ammount of pieces, say $x$, will always have this property:
\[ x \equiv 1 (mod 4) \]
Since $2003 \equiv 3 (mod 4)$, you can never have $2003$ pieces of paper.


Let $ a,b,c,d$ be positive real numbers such that $ abcd\equal{}1$, prove that
\[ \frac{1}{4a\plus{}1}\plus{}\frac{1}{4b\plus{}1}\plus{}\frac{1}{4c\plus{}1}\plus{}\frac{1}{4d\plus{}1} \plus{}\frac{1}{a\plus{}b\plus{}c\plus{}d\plus{}1} \ge 1\]
:)
P/s: Please use the elementary methods in you proof!

[quote="Cai Ni"]Can-hang2007:Are you sure this inequality has an easy solution?
Do you have a simple solution?(If you really has a nice solution,please tell us later.Let me think it over.Thank you.)
I think my way is right.But ……"abcd=1"I can't use it completely.
We need to prove:$ \sum\[\[\frac {1 \plus{} 4b}{1 \plus{} 4a} \plus{} \frac {1 \plus{} 4c}{1 \plus{} 4a} \plus{} \frac {1 \plus{} 4d}{1 \plus{} 4a}] \plus{} 4 \ge 4(a \plus{} b \plus{} c \plus{} d).$
This one must be right,but  I don't know how to use" abcd=1"I tried many ways,but failed.
Is my thought  right?[/quote]
Dear Cai Ni,
I am sure that my solution is quite elementary but I am not sure that it is nice since the "beauty" of inequality depends on each person's idea!
I will post my solution here but right now, I am waiting for some other approaches. :)

Dear can_hang2007:
I tried an ugly way,even though I think it's hard. :ninja:
We need to prove following inequality:When abcd=1.
$ 4a^3b^3c^3d^3 + 8\sum\[a^9bcd + 16\sum\[a^4b^8 + 112\sum\[a^4b^4c^4 + 48\sum\[a^5b^5cd\ge 241\sum\[a^6b^62c^2d^2$
It seems to be difficult if we only use AM-GM. :(
I expert your nice solution.I guess you will use Cauchy.I think I will learn many things from you.  :)

Dear Can_hang2007:
Could you please post your solution?I'm waiting for it.
I want to learn something from you.Thank you.

[quote="can_hang2007"]Let $ a,b,c,d$ be positive real numbers such that $ abcd \equal{} 1$, prove that
\[ \frac {1}{4a \plus{} 1} \plus{} \frac {1}{4b \plus{} 1} \plus{} \frac {1}{4c \plus{} 1} \plus{} \frac {1}{4d \plus{} 1} \plus{} \frac {1}{a \plus{} b \plus{} c \plus{} d \plus{} 1} \ge 1
\]
:)
P/s: Please use the elementary methods in you proof![/quote]
Ok, now here is my solution. :)
Firstly, we will take notice at the following 2 interesting lemma (but very easy to prove)
Lemma 1. If $ x,y>0$, then
\[ \frac{1}{x\plus{}1}\plus{}\frac{1}{y\plus{}1} \ge \min\left\{ 1,\frac{2}{\sqrt{xy}\plus{}1}\right\}\]
Lemma 2. If $ x,y,z>0,xy \ge 1,yz \ge 1, zx \ge 1$, then
\[ \frac{1}{x\plus{}1}\plus{}\frac{1}{y\plus{}1}\plus{}\frac{1}{z\plus{}1} \ge \frac{3}{\sqrt[3]{xyz}\plus{}1}\]
Now, turning back to our inequality, assume that $ a \ge b \ge c \ge d$, then if $ (4c) \cdot (4d) \le 1$, according to lemma 1, we will get the result since
\[ \frac{1}{4c\plus{}1}\plus{}\frac{1}{4d\plus{}1} \ge 1\]
If $ (4a) \cdot (4b) \ge (4a) \cdot (4c)\ge (4b) \cdot (4c)  \ge (4b) \cdot (4d) \ge \ge (4c) \cdot (4d) \ge 1$, then we have the following cases
Case 1. If $ a \ge b \ge c \ge 1$, then since $ (4a ) \cdot (4b) \ge (4a) \cdot (4c) \ge (4b) \cdot (4c) \ge 1$, according to lemma 2, we have
\[ \frac{1}{4a\plus{}1}\plus{}\frac{1}{4b\plus{}1} \plus{}\frac{1}{4c\plus{}1} \ge \frac{3}{4\sqrt[3]{abc}\plus{}1}\]
and
\[ a\plus{}b\plus{}c \equal{}\minus{}(1\minus{}b)(1\minus{}c)\minus{}(1\minus{}a)(1\minus{}bc) \plus{}2\plus{}abc \le 2\plus{}abc\]
Put $ t\equal{}\sqrt[3]{abc}\Rightarrow d\equal{}\frac{1}{t^3}$, we have to prove
\[ \frac{3}{4t\plus{}1}\plus{}\frac{1}{4d\plus{}1}\plus{}\frac{1}{t^3\plus{}d\plus{}3} \ge 1\]
\[ \Leftrightarrow \frac{3}{4t\plus{}1}\plus{}\frac{t^3}{t^3\plus{}4} \plus{}\frac{t^3}{t^6\plus{}3t^3\plus{}1} \ge 1\]
\[ \Leftrightarrow (t\minus{}1)^2(3t^7\plus{}6t^6\minus{}3t^5\plus{}6t^4\plus{}15t^3\minus{}8t^2\plus{}8) \ge 0\]
which is true for all $ t>0$.
Case 2. If $ a \ge b \ge 1 \ge c \ge d$, then from the lemma 1, we have
\[ \frac{1}{4a\plus{}1}\plus{}\frac{1}{4b\plus{}1} \ge \frac{2}{4\sqrt{ab}\plus{}1}\]
\[ \frac{1}{4c\plus{}1}\plus{}\frac{1}{4d\plus{}1} \ge \frac{2}{4\sqrt{cd}\plus{}1}\]
and
\[ a\plus{}b\plus{}c\plus{}d\equal{}\minus{}(1\minus{}a)(1\minus{}b)\minus{}(1\minus{}c)(1\minus{}d) \plus{}2\plus{}ab\plus{}cd \le 2\plus{}ab\plus{}cd\]
Put $ u\equal{}\sqrt{ab}\Rightarrow \sqrt{cd}\equal{}\frac{1}{u}$, we have to prove
\[ \frac{2}{4u\plus{}1}\plus{}\frac{2}{\frac{4}{u}\plus{}1}\plus{}\frac{1}{u^2\plus{}\frac{1}{u^2}\plus{}3} \ge 1\]
\[ \Leftrightarrow \frac{2}{4u\plus{}1}\plus{}\frac{2u}{u\plus{}4} \plus{}\frac{u^2}{u^4\plus{}3u^2\plus{}1} \ge 1\]
\[ \Leftrightarrow (u\minus{}1)^2(4u^4\minus{}5u^3\plus{}6u^2\minus{}5u\plus{}4) \ge 0\]
which is true for all $ u>0$.
Case 3. If $ 1 \ge b \ge c \ge d$, we can prove this case quite the same as case 1.
Therefore, our inequality is proved. :)


1.it is known that the equivalent mass of $ S$ in $ SCl_{2}$ is $ 16 ~gm/ eq$ what is it's equivalent mass in $ S_{2}Cl_{2}$.
2.Sulphuric acid and phosphoric acid combine seperately with equal masses of a alkali and form sulphates and dihydrogen phospahtes seperately .what is the mass ratio of sulphuric acid to phosphoric acid?

What is the meaning of "equivalent mass"?

Thanks you very much. But equivalent to what?
I must confess that I had nerver heard before of "equivalent mass", but I made a search of the concept and all I found was equivalent mass in redox reactions and equivalent mass in acid-base reactions. Any clue?

well atleast the text books our country this is quite a common concept and used to solve stochiometry,acid-Base neutralisation,Volumetric... Problems.
i mysefly confused with the way it is presented in most books.
what the definition says is $ 1 eq$ of an susbtance is a collection of that much amount whch displaces or reacts with $ 1 gm$ $ H$.


Let $ a,b > 0$. Prove that:
\[ [5a] \plus{} [5b]\ge [3a \plus{} b] \plus{} [3b \plus{} a]\plus{}[a]\plus{}[b]
\]
Notice that:
\[ x \equal{} [x] \plus{} \left\{ x \right\}
\]

Not hard.
Use $ [5x]\equal{}[x]\plus{}[x\plus{}\frac{1}{5}]\plus{}[x\plus{}\frac{2}{5}]\plus{}[x\plus{}\frac{3}{5}]\plus{}[x\plus{}\frac{4}{5}]$ and consider some case.


logarithms
function
inequalities
inequalities unsolved

[quote="maus"]$ ab/(a^2 \plus{} 3b^2) \plus{} bc/(b^2 \plus{} 3c^2) \plus{} ca/(c^2 \plus{} 3a^2) < \equal{} 3/4$[/quote]
We can assume that $ a,$ $ b$ and $ c$ are positive numbers.
Let $ \frac {a}{b} \equal{} x,$ $ \frac {b}{c} \equal{} y$ and $ \frac {c}{a} \equal{} z.$
Hence, we need to prove that $ \sum_{cyc}\frac {x}{x^2 \plus{} 3}\leq\frac {3}{4},$ where $ xyz \equal{} 1.$
But $ \sum_{cyc}\frac {x}{x^2 \plus{} 3}\leq\frac {3}{4}\Leftrightarrow\sum_{cyc}\left(\frac {1}{4} \minus{} \frac {x}{x^2 \plus{} 3} \plus{} \frac {\ln x}{8}\right)\geq0.$
Let $ f(x) \equal{} \frac {1}{4} \minus{} \frac {x}{x^2 \plus{} 3} \plus{} \frac {\ln x}{8}.$
Hence, $ f'(x) \equal{} \frac {(x \minus{} 1)(x^3 \plus{} 9x^2 \plus{} 15x \minus{} 9)}{8x(x^2 \plus{} 3)^2}.$
It gives that $ f(x) \equal{} 0$ only for $ x_{min} \equal{} 1$ and for $ x \equal{} x_1 \equal{} 0.280...$
and we see that $ f$ is an increasing function on $ (0,x_2],$ where $ f'(x_2) \equal{} 0$ and $ x_1\in(0,x_2].$
Thus, for all $ x\geq x_1$ we obtain: $ f(x)\geq f(1) \equal{} 0.$
Hence, your inequality is proven when $ \min\{x,y,z\}\geq0.281.$

Now let $ g(x) \equal{} \frac {x}{x^2 \plus{} 3}$ and $ \min\{x,y,z\} < 0.281.$
We obtain: $ g(x) \equal{} \frac {1}{x \plus{} \frac {3}{x}}\leq\frac {1}{2\sqrt3}$ and $ g$ is an increasing function on $ (0,\sqrt3].$
Id est, $ \sum_{cyc}g(x)\leq g(0.281) \plus{} \frac {1}{\sqrt3} < 0.75.$ Done!

let $ f(x)\equal{} \frac{1\plus{}x}{1\plus{}3x}$
$ f(x) \le 1$ for positive real $ x$
$ \sum_{}^{}  \frac{ab}{a^2\plus{}3b^2} \le  \frac{1}{4}  \sum_{}^{}  \frac{a^2\plus{}b^2}{a^2\plus{}3b^2}  \equal{} \frac{1}{4} \sum_{}^{} f( \frac{b^2}{a^2}) \le  \frac{3}{4}f( \frac{ \sum_{}^{}  \frac{b^2}{a^2} }{3} ) \le  \frac{3}{4}$ :D  (I used Jensen's inequality of course)

[quote="Dumel"]let $ f(x) \equal{} \frac {1 \plus{} x}{1 \plus{} 3x}$
$ f(x) \le 1$ for positive real $ x$
$ \sum_{}^{} \frac {ab}{a^2 \plus{} 3b^2} \le \frac {1}{4} \sum_{}^{} \frac {a^2 \plus{} b^2}{a^2 \plus{} 3b^2}$[/quote]
Shouldn't 1/4 be 1/2?

[quote="Dumel"]let $ f(x) \equal{} \frac {1 \plus{} x}{1 \plus{} 3x}$
$ f(x) \le 1$ for positive real $ x$
$ \sum_{}^{} \frac {ab}{a^2 \plus{} 3b^2} \le \frac {1}{4} \sum_{}^{} \frac {a^2 \plus{} b^2}{a^2 \plus{} 3b^2} \equal{} \frac {1}{4} \sum_{}^{} f( \frac {b^2}{a^2}) \le \frac {3}{4}f( \frac {\sum_{}^{} \frac {b^2}{a^2} }{3} ) \le \frac {3}{4}$ :D  (I used Jensen's inequality of course)[/quote]
I think, your way is wrong because $ f$ is convex function. :wink:

i noticed 1/4 should be1/2 and i thought the good solution would be:
$ f(x) \equal{} \frac {1 \plus{} x}{1 \plus{} 3x}$
$ f(x) \le \frac {1}{2}$ for positive real $ x \ge  1$
$ \sum_{}^{}\frac {ab}{a^{2} \plus{} 3b^{2}}\le\frac {1}{2}\sum_{}^{}\frac {a^{2} \plus{} b^{2}}{a^{2} \plus{} 3b^{2}} \equal{} \frac {1}{2}\sum_{}^{}f(\frac {b^{2}}{a^{2}})\le\frac {3}{2}f(\frac {\sum_{}^{}\frac {b^{2}}{a^{2}}}{3})\le\frac {3}{4}$
arqady-thx

A similar but easier one:

$ ab/(a^{2} + 2b^{2}) + bc/(b^{2} + 2c^{2}) + ca/(c^{2} + 2a^{2})\leq 1$.

$ Proof$. Let $ \frac {a}{b} = x, \frac {b}{c} = y, \frac {c}{a} = z$.

We can get$ \sum_{cyc}\frac {x}{x^{2} + 2}\leq1$ and $ xyz = 1$.

But because of AM-GM, we can get $ 2x\leq x^{2} + 1$.

So, $ \sum_{cyc}\frac {x}{x^{2} + 2}\leq\sum_{cyc}\frac {x}{2x + 1}$.

So, all we have to do is to prove $ \sum_{cyc}\frac {x}{2x + 1}\leq 1$.

But, if we multiply $ (2x + 1)(2y + 1)(2z + 1)$ on this inequality, and a little bit of calcluation, we can get

$ 3\leqq + y + z$. Since this is obvious because of $ xyz = 1$ and AM-GM, this completes the proof.

[quote="maus"]ab/(a^2+3b^2) + bc/(b^2+3c^2) + ca/(c^2+3a^2) <= 3/4[/quote]
You can see my proof here  :maybe:
http://www.mathlinks.ro/viewtopic.php?t=224237


Let $ x,y,z \in R$. Prove that:
$ \frac {|x \minus{} y|}{1 \plus{} |x \minus{} y|} \le \frac {|x \minus{} z|}{1 \plus{} |y \minus{} z|} \plus{} \frac {|z \minus{} y|}{1 \plus{} |z \minus{} y|}$

Set $ a \equal{} x \minus{} z$ and $ b \equal{} z \minus{} y$
The inequality is equivalent to:
$ \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|} \le \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}$

First we will prove that:
$ \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|} \le \frac {|a| \plus{} |b|}{1 \plus{} |a| \plus{} |b|} \Leftrightarrow 1 \plus{} \frac {1}{|a| \plus{} |b|} \le \frac {1}{|a \plus{} b|} \Leftrightarrow |a \plus{} b|\le|a| \plus{} |b|$ (Triangle's inequality)

Now we want to proove that:
$ \frac {|a| \plus{} |b|}{1 \plus{} |a| \plus{} |b|}\le \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}$, so we can suppose that $ a, b\ge0$, multiplying by $ 1 \plus{} a \plus{} b$:
$ a \plus{} b\le a \plus{} b \plus{} ab(\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b})$

We have equality if and only if $ ab \equal{} 0$, that means $ z \equal{} y$ or $ z \equal{} x$.

[hide="Remark"]
This can be easily extended to:
$ \frac{|\sum{x_i}|}{1\plus{}|\sum{x_i}|}\le\frac{\sum{|x_i|}}{1\plus{}\sum{|x_i|}}\le \sum{\frac{|x_i|}{1\plus{}|x_i|}}$
[/hide]

 :)

Wait, was the $ |y\minus{}z|$ in the denominator of the second term a typo? Maybe that's why I had so much trouble... I thought the problem was trying to be tricky with asymmetry.

mmm... you are right the truth is that I originally misread the inequality, but this causes not much trouble:

$ \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|}\le\frac {|a| \plus{} |b|}{1 \plus{} |a| \plus{} |b|}\le\frac {|a| \plus{} |b|}{1 \plus{} |b|}$ since $ |a|\ge0$.

but now the equality happens when $ a \equal{} 0$, $ x \equal{} z$ only.

 :)

[hide="Overkill"]As in ElChapin's solution, put $ a \equal{} x \minus{} y$, $ b \equal{} y \minus{} z$ and $ f(a) \equal{} \frac {|a|}{1 \plus{} |a|}$ which restates the inequality as $ f(a) \plus{} f(b)\ge f(a \plus{} b) \plus{} f(0)$. Since $ f'(a) \equal{} \mp\frac {1}{(1 \plus{} |a|)^2}$ $ f$ is decreasing in $ \mathbb{R}^ \minus{}$ and increasing in $ \mathbb{R}^ \plus{}$. In addition, we have $ f''(a) \equal{} \minus{} \frac {2}{(1 \plus{} |a|)^3} < 0$ so $ f$ is concave in $ \mathbb{R}^ \minus{}$ and $ \mathbb{R}^ \plus{}$.

[b]Case 1:[/b] $ ab\ge 0$
Since $ a,b,a \plus{} b$ all have the same sign the inequality is just Petrovic's inequality.

[b]Case 2:[/b] $ ab < 0$
WLOG $ |a|\ge|b|$. The $ |a \plus{} b|\le|a|$and $ a$ and $ a \plus{} b$ both have the same sign. The inequality then follows from $ f(a)\ge f(a \plus{} b)$ and $ f(b)\ge 0$.

[hide="Remark"]For every increasing and concave function $ f: \mathbb{R}^\plus{}\mapsto\mathbb{R}^\plus{}$ with $ f(0)\equal{}0$ the inequality $ f(|a|)\plus{}f(|b|)\ge f(|a\plus{}b|)$ holds for any $ a,b\in\mathbb{R}$.[/hide][/hide]

Nice solutions, bertram and ElChapin! My solution is the same as ElChapin's solution .


Which is greater $ \frac{2}{201}$ or $ \ln \frac {101}{100}$?

Note: ln is the natural logarithm to the base e

$ ln{\frac{101}{100}}$ is just slightly greater.

I considered Taylor expansion.

Instead of retyping the solution will give a link outside this forum where I have typed in the soln. Could someone tell me, is my argument right?

Link: http://www.goiit.com/posts/list/algebra-can-anyone-help-71299.htm

Exponentiate both sides. Using the series definition of the exponential:

$ e^{2/201} \equal{} 1 \plus{} \frac {2}{201} \plus{} \frac {\left( \frac {2}{201} \right)^2 }{2} \plus{} \frac {\left( \frac {2}{201} \right)^3}{6} \plus{} \ldots$

$ < 1 \plus{} \frac {2}{201} \plus{} \frac {\left( \frac {2}{201} \right)^2 }{2} \left( \frac {1}{1 \minus{} \frac {2}{201} \cdot \frac {1}{3}} \right)$

$ \equal{} \frac {24605411}{24361803} < \frac {101}{100}$

where, in the first inequality, I used the fact that $ \frac {\frac {x^{n \plus{} 1}}{(n \plus{} 1)!}}{\frac {x^n}{n!}} \equal{} \frac {x}{n \plus{} 1} < \frac {x}{3}$ for $ n > 2$, and thus compared the exponential to a slightly larger geometric series.

[b]Note:[/b] While your argument is meritorious, it is not rigorous to assert that an infinite number of terms are negligible. Instead, it is better to somehow place a numerical bound on how large the quantity can be, and then show that any number within your bounds satisfies your claim.

I'm sure that some type of bounding approach can be used with the natural log series; I chose to use the exponential series because all the coefficients in the expansion of $ e^x$ are positive, so its less tricky to rigorously bound the sum.


inequalities
algebra
polynomial
Divisibility Theory

Find all pairs of positive integers $m, n \ge 3$ for which there exist infinitely many positive integers $a$ such that \[\frac{a^{m}+a-1}{a^{n}+a^{2}-1}\] is itself an integer.

Let us assume that there are infinitely many $ a$ for $ m, n,$ then clearly $ m > n.$
So we may let $ m \equal{} n \plus{} k$ for some $ k > 0.$ Since $ a^m \plus{} a \minus{} 1 \equal{} (a^n \plus{} a^2 \minus{} 1)a^k \minus{} (a^{k \plus{} 2} \minus{} a^k \minus{} a \plus{} 1)$ there are infinitely many $ a$ such that $ a^n \plus{} a^2 \minus{} 1 | a^{k \plus{} 2} \minus{} a^k \minus{} a \plus{} 1.$ Let $ f(x) \equal{} x^n \plus{} x^2 \minus{} 1, g(x) \equal{} x^{k \plus{} 1} \plus{} x^k \minus{} 1, h(x) \equal{} (x \minus{} 1)g(x).$
We have $ \frac {h(x)}{f(x)} \equal{} g(x) \plus{} \frac {r(x)}{f(x)}$ with $ \text{deg}(r) < \text{deg}(f).$ As there are infinitely many $ a$ for which $ r(a)\equal{}0$, we get that $ r$ is identically $ 0,$ or in other words, $ f|h.$
This in turn yields that $ f|g.$

Now, it is clear that $ f$ has a root $ b \in ]0, 1[,$ and then $ b$ is a root of $ g,$ too. So we have $ b^n \plus{} b^2 \minus{} 1 \equal{} 0 \equal{} b^{k \plus{} 1} \plus{} b^k \minus{} 1.$
So we get (consider the degrees) that $ k \geq 2,$ and so $ b^n \geq b^{k \plus{} 1}$ and $ b^2 \geq b^k.$ Now, since $ 1 \equal{} b^n \plus{} b^2 \geq b^{k \plus{} 1} \plus{} b^k \equal{} 1,$ we must actually have $ k \equal{} 2, n \equal{} 3, m \equal{} 5.$
And finally the identity $ a^5 \plus{} a \minus{} 1 \equal{} (a^2 \minus{} a \plus{} 1)(a^3 \plus{} a^2 \minus{} 1)$ shows that the unique solution is the pair $ (m, n) \equal{} (5, 3).$

[quote="Peter"][quote="mathmanman"]So we get (consider the degrees) that $ k \geq 2,$ and so $ b^n \geq b^{k \plus{} 1}$[/quote]Can you explain this?[/quote]
Of course : $ k\plus{}1 \equal{} \deg(g) \geq \deg(f) \equal{} n \geq 3.$ (as for the inequality, remember that $ b \in ]0, 1[.$)

Ok, I see now. Let me reword your first paragraph a bit.

As the polynomial $ x^{m} \plus{} x \minus{} 1 \mod x^{n} \plus{} x^{2} \minus{} 1$ is zero for infinitely many $ x$'s, it is the zero polynomial. So $ x^{n} \plus{} x^{2} \minus{} 1|x^{m} \plus{} x \minus{} 1 \equal{} (x^n \plus{} x^2 \minus{} 1)x^k \minus{} (x \minus{} 1)(x^{k \plus{} 1} \plus{} x^k \minus{} 1)$, and since $ x \equal{} 1$ is not a root of $ x^{n} \plus{} x^{2} \minus{} 1$, it follows that $ x^{n} \plus{} x^{2} \minus{} 1|x^{k \plus{} 1} \plus{} x^k \minus{} 1$ (denote this $ f|g$).

And then we continue[quote="mathmanman"]Now, it is clear that $ f$ has a root $ b \in ]0, 1[,$ and then $ b$ is a root of $ g,$ too. So we have $ b^n \plus{} b^2 \minus{} 1 \equal{} 0 \equal{} b^{k \plus{} 1} \plus{} b^k \minus{} 1.$
So we get (consider the degrees) that $ k \geq 2,$ and so $ b^n \geq b^{k \plus{} 1}$ and $ b^2 \geq b^k.$ Now, since $ 1 \equal{} b^n \plus{} b^2 \geq b^{k \plus{} 1} \plus{} b^k \equal{} 1,$ we must actually have $ k \equal{} 2, n \equal{} 3, m \equal{} 5.$
And finally the identity $ a^5 \plus{} a \minus{} 1 \equal{} (a^2 \minus{} a \plus{} 1)(a^3 \plus{} a^2 \minus{} 1)$ shows that the unique solution is the pair $ (m, n) \equal{} (5, 3).$[/quote]


Let  $ \sum_{i\equal{}1}^n{x_i}\equal{}\sum_{i\equal{}1}^n{x_i^2}\geq{n},n\geq2,x_i\geq0$,prove that $ \sum_{i\equal{}1}^n{x_i^3}\leq {n}$.

Sorry, obviously I'm in the wrong here - could somebody point me to my mistake:

We can WLOG assume $ x_1\geq x_2\geq x_3 ... \geq x_n$. So, also $ x_1^2\geq x_2^2\geq x_3^2 ... \geq x_n^2$.
Per Chebyshev's Inequality
$ \sum_{i \equal{} 1}^n{x_i^3}\geq \frac {\sum_{i \equal{} 1}^n{x_i^2}\cdot \sum_{i \equal{} 1}^n{x_i}}{n}$
However, since $ \frac {\sum_{i \equal{} 1}^n{x_i^2}\cdot \sum_{i \equal{} 1}^n{x_i}}{n}\geq \frac {n\cdot n}{n} \equal{} n$, we have
$ \sum_{i \equal{} 1}^n{x_i^3}\geq n$, which is the opposite of our claim.

I'm not using the condition that $ \sum_{i \equal{} 1}^n{x_i} \equal{} \sum_{i \equal{} 1}^n{x_i^2}$, which is also bizarre. So, where am I wrong?

By Cauchy-Schwarz Inequality, $ \sum x_i \sum x_i^3 \ge \left(\sum x_i^2 \right)^2 \Rightarrow \sum x_i^3 \ge n$

[quote="primoz2"]Maybe the condition is needed and probably numbers arent necessary possitive, so this contition is wrong and than you cant use Chebyshev.[/quote]

Nah, they are positive (nonnegative at least) - it says $ x_i\geq 0$ - if $ t$ of them are zero, we can just treat it as a case for $ n \minus{} t$ non-zero variables (it's even stronger then).* It seems, per hsbhatt, that I was indeed correct and that the problem claim is wrong.

*Wouldn't Chebyshev also be valid for negative numbers too? At least "Old and New Inequalities", p.122 makes no mention of necessary positivity.

Sorry,it is

Let  $ \sum_{i \equal{} 1}^n{x_i} \equal{} \sum_{i \equal{} 1}^n{x_i^2}\geq{n},x_i>0$,prove that $ \sum_{i \equal{} 1}^n{x_i^3}\equal{}n$.

Exactly equal to $ n$? I must say I doubt it's true. I'll try to find a counter-example.

Maybe it really is true.

EDIT: Here you go. This doesn't seem right, but OK. If it's OK, then the claim is true, no matter how unintuitive it looks.

$ \sum_{i \equal{} 1}^n{x_i^2} \geq \frac {{(\sum_{i \equal{} 1}^n{x_i})}^2}{n}$ (Power Mean Inequality)
This is equivalent to $ nt \geq t^2$, which is equivalent to $ t\leq n$ ($ t \equal{} \sum_{i \equal{} 1}^n{x_i^2} \equal{} \sum_{i \equal{} 1}^n{x_i}$). However, $ t\geq n$, so $ t \equal{} n$. Now, since the equality in Power Mean Inequality is achieved only when all members are equal, we conclude that $ x_1 \equal{} x_2 \equal{} ... \equal{} x_n \equal{} 1$, in which case $ \sum_{i \equal{} 1}^n{x_i^3} \equal{} n$ and we have proven the claim.


Adam rolled two non-standard six-sided dice once.  What is the probability that he did not roll a prime number on either die? Express your answer as a common fraction.

First find the possible outcomes. 6*6=36
Now find the non prime numbers. 1,4 and 6
Since there are 3 non prime numbers for each die 3*3=9
9/36=[size=150]1/4[/size]

Or also you could just find the probability of finding a non prime number on one die 1/2, and multiply it by itself.
1/2*1/2=[size=150]1/4[/size]


A 100 meter dash is run on a track in the direction of the vector v = 4i + 7j. The wind velocity w is 5i + j km/hr. The rules say that a legal wind speed measured in the direction of the dash must not exceed 5 km/hr.

Find the component of w which is parallel to v.

it's parallel to one another if it's angle between them is 0 right?

so therefore using the geometric of vectors:

v . w = |v| * |w|, right?

so what I am confused here is does the question asks for a new wind velocity that's parallel to v? or what is it?

[quote="JRav"]The sprint vector $ \vec{v}$ is moving some direction.  The wind vector $ \vec{w}$ is moving in some other direction.  The problem is asking you do break down $ \vec{w}$ into components, and then find the magnitude of the component that is parallel to $ \vec{v}$.  So determine the angles made by $ \vec{v}$ and $ \vec{w}$, and go from there.[/quote]

after I got the angle between the  $ \vec{v}$ and $ \vec{w}$ what do I do?

$ \vec{v}$ and $ \vec{w}$ are parallel if they angle between them is 0 or 180 right?

and if the question asks for the magnitude then why in the answer field it has i and j

Almost
 The scalar product of the vectors gives the component of one in the direction of the other, multiplied by the "others" modulus.


[hide="solutie"]pentru inceput vom gasi la ambele ecuatii solutia $(1,m)$ cu $m\in \mathbb{N}$, verificare imediata. deci consideram $n\geq 2$.
apoi, la prima ecuatie :
impartim prin $n^{2m}$ si obtinem
$\sum_{i=1}^{n}\left(\frac{2i-1}{n^{2}}\right)^{m}=1$.
notam cu $f(m)$ expresia din membrul stang si observam ca aceasta este suma de functii strict descrescatoare, deoarece $n^{2}>2n-1>2i-1$, deoarece $n>2$ si $1\leq i\leq n$. deci $f$ este injectiva, si cum $f(1)=1$, rezulta ca $m=1$ este unica solutie. pentru $m=1$ obtinem chiar identitatea
$\sum_{i=1}^{n} (2i-1)=n^{2}$.
pentru a doua ecuatie
notam cu $Q$ membrul stang.
fie $f(x)=(n+x)^{m}+(n-x)^{m}$. aratam ca $f(x)\geq 2n^{m}$.
calculam $f(x)-2n^{m}$, si vom obtine :
$x\Bigg(\sum_{k=0}^{m-1}n^{m-1-k}\Big((n+x)^{k}-(n-x)^{k}\Big)\Bigg)$.
pentru $k$ fixat, functia $g(y)=y^{k}$ este crescatoare, deci $\big(f(y)-f(z)\big)\big(y-z\big)\geq 0$. (*) mai mult, putem avea egalitate doar pt $k=0$. pentru $y=n+x,\ z=n-x$, obtinem ca $2x\big((a+x)^{k}-(a-x)^{k}\big)\geq 0$, deci, conform celor de mai sus, $f(x)-2n^{m}\geq 0$.
vom face apoi suma dupa toti $x$ de la $1-n$ la $n-1$, astfel incat $n+x$ este impar. acest fapt implica si $n-x$ impar, si, mai mult, acesti $x$ sunt in numar de $n$ (deoarece vom obtine numerele impare de la $1$ la $2n-1$). vom obtine :
$\sum_{x}\Big((n+x)^{m} + (n-x)^{m}\Big)\geq n\cdot 2n^{m}$, unde suma din membrul stang se face dupa $x$ asa cum am zis mai sus. dar aceasta suma este egala cu $2Q$, deci inegalitatea de mai sus devine
$Q\geq n^{m+1}$.
deci pentru a avea solutie, trebuie sa avem egalitate mai sus, deci implicit egalitate in pasul (*) de mai sus. cum acest fapt implica $k=0$, insa cum suma se facea dupa $k$ de la $0$ la $m-1$, rezutla ca $m-1=0$ si $m=1$.
din nou, pentru $m=1$, se verifica imediat solutia $(n,1)$.
deci am obtinut solutiile $(1,m)$, $(n,1)$.[/hide]


calculus
derivative
quadratics
function
algebra
domain
linear algebra

Determine the nature of the critical point(s) of
[tex] f(x,y,z) = x^2 + y^2 + z^2 + xy [/tex]

the only critical point i found was (0,0,0).  is that correct?  i don't know how to classify it - like i know how to find if its a maximum or minimum or a saddle point for 2 variables  using the second derivative test, but i don't know how to do it for 3 variables.

Yes, $(0,0,0)$ is the only critical point.

The second derivative test classifies critical points by looking at the quadratic form generated by the second order terms of the Taylor polynomial. This particular case already is a quadratic form; there's not that much point in going to the derivatives and back to find that the best quadratic approximation to this function is itself.

Here's the general second derivative test for classifying critical points in several dimensions, assuming smooth functions and critical points located in the interior of the domain of the function.

Find the $n\times n$ matrix $H$ whose entries are $h_{jk}=\frac{\partial^2f}{\partial x_j\partial x_k}.$

Evaluate the functions in this matrix at the critical point, so that it is a constant symmetric matrix.

Determine the signs of the eigenvalues of this matrix. (See below for a way to do this.)

If all of the eigenvalues are positive, the matrix is positive definite, and the critical point is a local minimum.

If all of the eigenvalues are negative, the matrix is negative definite, and the critical point is a local maximum.

If there is at least one positive eigenvalue and at least one negative eigenvalue, then the critical point is a "saddle point" (if you're OK with using that word for more than 2 dimensions).

If all the eigenvalues are nonnegative but at least one is zero, or if they are all nonpositive but at least one is zero, then the second derivative test is inconclusive.

For this particular function, the matrix is:

$H=\begin{bmatrix}2&1&0 \\ 1&2&0 \\ 0&0&2\end{bmatrix}.$

This matrix has eigenvalues 1, 2, and 3, and is positive definite. The critical point is a minimum.

There is a test that can be used for positive/negative definite. Find the determinants of all of the square submatrices in the upper left corner, starting with the $1\times1$ case and ending with the $n\times n$ case. If all of these determinants are positive, the matrix is positive definite and the critical point is a minimum. If they have a strictly alternating sign pattern, starting with negative for the $1\times1$ case, the matrix is negative definite and the critical point is a minimum. If there is some other sign pattern, but with no zeros, then the matrix is indefinite and the critical point is a saddle point. It gets more complicated if there are zeros.

In this case:

$|1|=1>0.$

$\left|\begin{array}{cc}2&1 \\ 1&2\end{array}\right|=3>0.$

$\left|\begin{array}{ccc}2&1&0 \\ 1&2&0 \\ 0&0&2\end{array}\right|=6>0.$

Hence positive definite, hence minimum.

This procedure includes as a special case the two-dimensional formulas given in your average American calculus textbook.


What is the order of $GL_{n}(\mathbb{Z}_{p})$?

The problem has appeared before on this site.

The required order is $\prod_{k=1}^{n}(p^{n}-p^{k-1}).$

The idea: count the number of ways we can choose the columns of a matrix to be independent vectors. For each new column, we have $p^{n}$ choices less the span of the preceding $k-1$ columns. That span has $p^{k-1}$ elements.


algebra
polynomial
limit
number theory proposed
number theory

That problem appeared in M&Y Magazine (Vietnam) ,for junior students. :D

[quote="grobber"]We use the fact that $m-n|P(m)-P(n)$.

For $n$ large enough, the sequence $x_n$ becomes strictly increasing. Given $n$ large enough, there exists $m>n$ s.t. $x_n-x_{n-1}|x_m\Rightarrow x_n-x_{n-1}|x_{m-1}\Rightarrow \cdots\Rightarrow x_n-x_{n-1}|x_n\Rightarrow x_n-x_{n-1}|x_{n-1}\Rightarrow \nolinebreak x_n\le \nolinebreak 2x_{n-1}$, so there are infinitely many $n$ for which $P(n)\le 2n$, so $P$ has degree $1\Rightarrow P(x)=ax+b$.

By an argument similr to this one, it's easy to show tht $P(1)=2$, so $a+b=2$. We use this rgument one more time to find that $P(2)-2|2$, so $P(2)\in \{3,4\}$, so $2a+b\in \{3,4\}$. This means that $P(x)=ax+b\in \{x+1,2x\}$. From now on it's easy.

Is it Ok?[/quote]

After almost 10 years,I fear that your solution is wrong.How can how get that the first line implies the second line implies ...... ??

@ sayantanchakraborty , I suppose you mean $x_{n+1}-x_n |x_m \Rightarrow x_{n+1}-x_n|x_{m-1}\Rightarrow .... $
     Because of the property for all $N$ there exists $x_i$ divisible by $N$ , then there exists $m$ such that $x_{n+1}-x_n|x_m$ . Using $a-b|P(a)-P(b)$ , we get $x_{n+1}-x_n|x_m-x_{m-1}$ for all $m>n$ . Combining them together we get $x_{n+1}-x_n|x_m$ if $m\geq n$ .

[quote="grobber"]We use the fact that $m-n\mid P(m)-P(n)$.

For $n$ large enough, the sequence $x_n$ becomes strictly increasing. Given $n$ large enough, there exists $m>n$ s.t. $x_n-x_{n-1}\mid x_m\Rightarrow x_n-x_{n-1}\mid x_{m-1}\Rightarrow \cdots\Rightarrow x_n-x_{n-1}\mid x_n\Rightarrow x_n-x_{n-1}\mid x_{n-1}$, so $x_n\le 2x_{n-1}$, so there are infinitely many $n$ for which $P(n)\le 2n$, so $P$ has degree $1\Rightarrow P(x)=ax+b$.

By an argument similar to this one, it's easy to show that $P(1)=2$, so $a+b=2$. We use this argument one more time to find that $P(2)-2\mid 2$, so $P(2)\in \{3,4\}$, so $2a+b\in \{3,4\}$. This means that $P(x)=ax+b\in \{x+1,2x\}$. From now on it's easy.

Is it Ok?[/quote]
For $n$ large enough, the sequence $x_n$ becomes strictly increasing. Given $n$ large enough, there exists $m>n$ s.t. $x_n-x_{n-1}|x_m$ Why Why??
I can explain as follows: $\mathop {\lim }\limits_{x \to  + \infty } ( {P( {x} ) - 2x =  + \infty  \Rightarrow }$ For $n$ large enough, we have $P( {{x_{n - 1}}} ) - P( {{x_{n - 2}}} ) > P( {{x_{n - 2}}} ) \Leftrightarrow {x_n} - {x_{n - 1}} > {x_{n - 1}}$.
So hypothetically exist $m$ s.t: ${x_n} - {x_{n - 1}}|{x_m} \Rightarrow {x_m} \ge {x_n} - {x_{n - 1}} > {x_{n - 1}} \Rightarrow m \ge n$ $ \Rightarrow {x_n} - {x_{n - 1}}|{x_m} - {x_{m - 1}} \Rightarrow {x_n} - {x_{n - 1}}|{x_{m - 1}}$ and continue as above solution.

[hide="Solution"]
Observe that $p(1)-1>0$, As $p^i(1)\equiv 1\left ( \mathrm{mod}\  (p(1)-1) \right ),\ \forall i\in \mathbb{N}^\star$, by the hypothesis, we get that $p(1)-1=1\Leftrightarrow p(1)=2$.

We'll prove by induction that $p(n)=n+1,\ \forall n\in \mathbb{N}^\star$, which implies the problem. Let $p^k(x)=p\left (p^{k-1}(x)\right )$. Suppose we have proven that $p(k)=k+1,\ \forall 1\le k\le n-1$.

By the induction hypothesis, the sequence $\left (x_i\right )_{i\ge 1}$ is actually $2,3,...\ ,n,p(n),p^2(n),...\ ,p^i(n),...$

As we can find a multiple of $p(n)-1$ in this sequence, we have either one of the following two situations:

[i]i)[/i] $p(n)-1|k$ with $2\le k\le n$

If $k\le n-1$, $p(n)-1\le k\le n-1\Leftrightarrow p(n)\le n$, contradiction. Hence $p(n)-1|n\Rightarrow p(n)\le n+1$. As $p(n)>n$, we infer that $p(n)=n+1$.

[i]ii)[/i] $p(n)-1|p^k(n)$ for some $k\in \mathbb{N}^\star$

 We see that the sequence $p^i(n)$ is periodic modulo $p(n)-1$. Indeed,

$$p^i(n)\equiv p^{i-1}(1)=p^{i-2}(2)=...=p^{i-n}(n)\ \left ( \mathrm{mod}\ (p(n)-1) \right )$$

So we may suppose wlog that $1\le k\le n$.

If $1\le k\le n-1$, by observing that

$$p^k(n)\equiv p^{k-1}(1)=p^{k-2}(2)=...=p(k-1)=k \left (\mathrm{mod}\ (p(n)-1) \right )$$

we get the same contradiction as in the previous case.

Hence $p(n)-1|p^n(n)$, but in the same manner we see that $p^n(n)\equiv n \left (\mathrm{mod}\ (p(n)-1) \right ) $, hence $p(n)-1|n$, which yields $p(n)=n+1$, thus proving our claim.

[/hide]

Obviously the sequence ain't bounded as that would brake the divisibility by taking a sufficiently large number.Supposes $\deg P\geq 2$.This would imply that after a certain point $P(x)>2x$ for all $x>M$.Now notice by Bezout's we have :$x_i-x_{i-1}|x_{i+1}-x_i$ and hence by choosing $m:=x_i-x_{i-1}$ the sequence $x_i$ in $\mathbb{Z}_{m}$ is onwardly  constant .Notice that $x_i-x_{i-1}>x_{i-1}$ for a large $i$  but $x_i-x_{i-1}\mid x_{i-1}$ or less a contradiction.Hence $P$ is linear.If it's monic simply take $m$ to be its free term.Else $x_i=a^{n}+b\frac{a^n-1}{a-1}$ now by taking $m:=a$ we have $a\mid b$ and hence by scalling down the whole sequence by $a$ we obtain a monic linear poly,impossible.$\blacksquare$

[quote=anantmudgal09][quote=Omid Hatami]$p$ is a polynomial with integer coefficients and for every natural $n$ we have $p(n)>n$. $x_k $ is a sequence that: $x_1=1, x_{i+1}=p(x_i)$ for every $N$ one of $x_i$ is divisible by $N.$ Prove that $p(x)=x+1$[/quote]

[hide=solution] Set $N \overset{\text{def}}{:=} x_k-1$ for $k$ adequately large. Suppose that $N \mid x_{\ell}$ for $\ell \geqslant 1$. Take $\ell$ to be the minimum such number.

Note that $x_k-1 \geqslant x_{k-1}$ so $\ell \geqslant k-1$. If $\ell>k-1$ Then $\ell>k$ as well since $P(x_k) \equiv P(1) \pmod{x_k-1}$.

So, we have $$x_{\ell}=P^{\ell-k}\left(P^k(1)\right)=P^{\ell-k}(x_k) \equiv P^{\ell-k}(1)=x_{\ell-k} \pmod{N},$$ contradicting the minimality of $\ell$; so clearly $x_k=x_{k-1}+1$ for all sufficiently large $k$. Consequently, $P(x)=x+1$ vanishes for $x=x_{k-1}$ so it has infinitely many roots so $P(x) \equiv x+1$ as desired.

[/hide][/quote]
What does $\text{def}$ mean ?

@above $\text{def}$ just stands for definition. Basically, it means we are just defining something right now and it wasn't defined before (so that the reader doesn't waste time thinking was this defined before or not).

We show by induction on $m \ge 1$ that $x_i = i ~ \forall ~ 1 \le i \le m$. Base case $x_1 = 1$ is given. Assume assertion is true for $m = k-1$ (with $k \ge 2$), i.e. $x_i = i ~ \forall ~ 1 \le i \le k-1$. Now $x_1 < x_2 < \cdots$. So $x_k := C > x_{k-1} = k-1$. If $C > k$, then modulo $C-1$ the sequence is just
$$ 1,2,\ldots,k-1 ; 1,2,\ldots,k-1 ; \ldots $$
so no number is divisible by $C-1$, contradiction. Hence $C = k$, completing the induction step. $\blacksquare$

Observe that $1 \equiv P(1) \equiv P^2(1) \equiv ... \pmod{P(1)-1}$, so from the problem condition, we have that there exists an $N$ such that $P(1)-1|P^N(1)$, so $P(1)-1|1$, thus $P(1)-1=1$, since $P(1)>1$. Thus, $P(1)=2$. Now, assume by induction that $P(k)=k+1$ for all $k=1,2,...,n$. Therefore, $P(1)=2,P^2(1)=P(2)=3,...,P^n(1)=P(n)=n+1$. Observe that the chain $1 \mapsto P(1) \mapsto P^2(1) \mapsto ...$ is equals to $1 \mapsto 2 \mapsto 3 \mapsto ... \mapsto n \mapsto n+1 \mapsto 1 \pmod{P^{n+1}(1)-1=P(n+1)-1}$. From the problem condition, there is exists an $N$ such that $P^{n+1}(1)-1=P(n+1)-1|P^N(1)$, and $P^N(1)$ belongs to $1,2,...,n+1 \pmod{P(n+1)-1}$. Hence, $P(n+1)-1|k$ for some $1 \leq k \leq n+1 \implies n< P(n+1)-1 \leq k \leq n+1 \implies P(n+1)-1=n+1 \implies P(n+1)=n+2$, for all $n \in \mathbb{Z}_{\geq 0}$. This implies that $P(X) \equiv X+1$, as desired.

$\blacksquare$

Nice problem.
Take some positive integer $n$ and plugging $N=x_n-x_{n-1}$ gives that there exists $i>n$ such that $x_n-x_{n-1} \mid x_i$ and since $x_n-x_{n-1} \mid x_i-x_{i-1}$, we get that $x_n-x_{n-1} \mid x_{i-1}$ and so on until we get that $x_n-x_{n-1} \mid x_{n-1}$ , hence $x_n \leq 2x_{n-1}$ (if $i<n$, since it is given that the sequence is increasing, $x_i \leq x_{n-1}$ and we directly obtain the above mentioned result). Since this holds for all large $n$, we get that the polynomial $P$ grows too slowly ($P(x) \leq 2x$ holds for infinitely many $x$), hence it is linear. If $P(x)=ax+b$, then $P^k(1)=x_k=a^{k}+b\frac{a^k-1}{a-1}$ and let WLOG $(a,b)=1$ (otherwise divide everything by the $gcd$). If $a=1$, then $P^k(1)=kb+1$, so plug $N=b$ to see there are no multiples of $b$. Otherwise, plug $N=a$ to see $a \mid b$, contradiction with the $gcd$ assumption.

Let $P(n) = p(n)$. We have $(x_i) = \{1, P(1), P^2(1), \ldots, \}$ and $x_i = P^{i-1}(1)$.

We induct to show that $x_i = i$ for all $i$.

Base cases: $x_1 = 1$ is obvious. Now we show $x_2 = 2$, which is the same as $P(1) = 2$. Suppose not and there existed a prime $p\mid P(1) - 1$. We see that $\{1, P(1), \ldots, \}$ is always $1\pmod p$, so it can't be divisible by $p$, contradiction. Therefore $P(1) = 2$.

Inductive step: Suppose it was true for all $i \le  n$, for some $n\ge 2$.

Then we have $P^0(1) =1,  P(1) = 2, P^2(1) = 3, \ldots, P^{n-1}(1) = n$. Notice that the sequence $\{1, P(1), P^2(1), \ldots, \}$ is periodic modulo $P^n(1) - 1$ with period dividing $n$, which means it must be $0\pmod{P^n(1) - 1}$ at some point. Since $1,\ldots, n-1$ are not $0$ modulo $P^n(1) - 1$, we have $P^n(1) \equiv 0\pmod{P^n(1) - 1}$ or $P^n(1) - 1 \mid n$. The former is clearly impossible since $P^n(1) > n \ge 2$. Therefore, $P^n(1) - 1\mid n$. Combined with the fact that $P^n(1) > n$, we have $P^n(1) = n + 1$, so $x_{n+1} = n +1$, which completes the induction.

only p6 i can solve

[b][color=pink]Claim:[/color][/b] $P(1) = 2$.
[i]Proof.[/i] FTSOC suppose that $p \mid P(1) - 1$. Then $P(1) \equiv 1 \pmod{p}$ so no $x_i$ is divisible by $p$, contradiction. $\blacksquare$
Now assume $P(n) = n + 1$. Then once again there must be no primes dividing $P(n + 1) - (n+1)$, so $P(n+1) = n+2$. Repeating this gives the result for all integers.

As $P(x) \in \mathbb{Z}[x]$ , so $x_{2}-x_{1}|P(x_{2})-P(x_{1})=x_{3}-x_{2}|\cdots |x_{n}-x_{n-1}|x_{t}-x_{t-1} \implies x_{\ell}-x_{\ell-1}|x_{t}-x_{t-1}$ $\forall$ $ t \geqslant \ell$. Also, there exists a term divisible by $N$ in the sequence , set $N:=x_{n}-x_{n-1}$ as $P(n)>n \implies x_{1}<x_{2}<x_{3}<\cdots x_{n} \cdots < x_{t} \cdots $\\

Since $\exists$ a term divisible by $x_{n}-x_{n-1}$ in the sequence  it would be larger than it  (since each term is positive), consider $x_{m}$ is that term.\\

so $x_{n}-x_{n-1}|x_{m}$ and since $x_{n}-x_{n-1}|x_{m}-x_{m-1} \implies x_{n}-x_{n-1}|x_{m-1}$ continuing in this manner we can descend to this sequence $m \mapsto m-1 \mapsto m-2 \cdots n$ , so $x_{n}-x_{n-1}|x_{n} \implies x_{n}-x_{n-1}|x_{n-1}  \implies x_{n}-x_{n-1} \leqslant x_{n-1} \implies x_{n} \leqslant 2x_{n-1} \implies P(x_{n-1}) \leqslant 2x_{n-1}$\\

which in turns gives $x<P(x) \leqslant 2x$ $\forall$ $x \in \mathbb{Z}^{+}$, now \\

 $\textcolor{red}{\mathrm{Claim:-}}$ $P(x)$ is linear.\\

$\textcolor{blue}{\mathrm{Pf:-}}$ FTSOC , consider $\deg(P(x))>1$ , so sinc we have $x< P(x) \leqslant 2x \implies \lim _{x \rightarrow \infty} \frac{x}{x^{\deg(P(x))}} < \lim_{x \rightarrow \infty} \frac{P(x)}{x^{deg(P(x))}} \leqslant  \lim_{x \rightarrow \infty} \frac{2x}{x^{\deg(P(x))}}  \implies 0< \text{leading \hspace{0.1cm} term} \leqslant 0$\\

 by squeeze theorem we get leading term of $P(x)$ is $0$ which is a contradiction , hence $\deg(P(x))$ is $1$. $\square$\\

Now it implies $P(x)=ax+b$ for some integers $a$ and $b$, we know that $x_{2}-x_{1}|x_{1} \implies x_{2}=2$ , so $P(1)=2 \implies a+b=2$ and $P(2)-2|2 \implies P(2)=3$ or $P(2)=4$ as $P(2)>2$ , so we get that $P(x)$ is $x+1$ or $2x$ , now since for every $N \in \mathbb{Z}^{+}$ there exist a term in the sequence divisible by $N$ but if we choose $N$ to be some odd prime , then we get a contradiction for $P(x)=2x$ , hence $P(x)=\boxed{x+1}$. $\blacksquare$

Assume contradiction. If their exists some $i$ such that $x_i > i$, then we take the smallest $i$ satisfying this condition. Then, by minimality, none of the first $i$ terms are divisible by $x_i-1$. (because $x_{i-1} \leq x_i-1$ and if equality held, then $i-1$ would have been a smaller such $i$.) Now,  note that $x_i \equiv x_1 \equiv 1 \pmod{x_i-1}$, so we cycle back modulo $x_i-1$. Now obviously, $i \geq 2$, so $x_i-1 > 1$, so we get a contradiction. Therefore, there is no such $i$ and since we obviously have $x_i \geq i$ for all $i$, since $P(n) > n$ we have $x_i = i$ for all $i$, thus $P(i) = i+1$ for all positive integers $i$, and we get $P(x)-x-1$ has infinite roots and therefore is identically $0$. So, $P(x) = x+1$.


What is the minimum value of the expression $\frac{2}{abc}-(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) $ when $ a,b,c>\frac{1}{5} $ and $ a+b+c=1 $ ?

Let [tex] f(a,b,c)=\frac{2}{abc}-\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2} [/tex]
So we have:
[tex] 25 \leq f(a,b,c) \leq 27 [/tex] where [tex] a,b,c\geq \frac15 [/tex]
Suppose $ a \leq b \leq c $.We will prove :[tex] f(a,b,c) \geq f(a,\frac{b+c}{2},\frac{b+c}{2})[/tex]
If I'm not wrong:
[tex] f(a,b,c)-f(a,\frac{b+c}{2} , \frac{b+c}{2} )=
\frac{(b-c)^2(2bc-ac-ab)}{b^2c^2(b+c)}\geq0[/tex]
So all we need to do is find the least value of:
[tex]f(a,d,d)=\frac{2}{ad^2}-\frac{1}{a^2}-\frac{1}{d^2}[/tex] where $a+2d=1$
Since [tex]a\geq\frac15[/tex] ,[tex]d\leq\frac25[/tex]
We have:
[tex]f(a,d,d)=\frac{-9d+4}{4d^3-4d^2+d}[/tex] since $ a=1-2d $
And:
[tex]f(a,d,d)-25\geq0 \Leftrightarrow (5d-2)(10d^2-6d+1)\leq 0[/tex] which is right.
The equality holds when [tex] a=\frac15,b=c=\frac25[/tex]
 :)

This was also my approach. Anyway, I find it interesting due to the values for which the equality holds.

This inequality is for my little brain a big nightmare so I implore your help.

Please, a complete solution as I am in loop ( very confused).

Thank you very very very much.


I have had TeXnicCenter set up on my machine (Windows XP) together with MicTex for a while now and have produced several LaTeX documents (as PDFs) so I am satisfied that there is no problem with these packages.

Today I installed Asymptote and ran a few simple command lines to produce some simple outputs in GSView. No problems here.

I then read the Art of Problem Solving guide to Asymptote by Maria Monks. In particular the "Using TeXnicCenter for Asymptote" section in the guide to "Interactive Mode".

I followed the instructions word for word on how to create a new tool in TeXnicCenter, entering the three fields exactly as it said. When I try to perform the subsequent task of hitting my hotkey after entering the prescribed code all I get is a flicker of the windows command prompt then... nothing.

I even went back and entered the arguament field in the tools menu through "browse" to ensure that this was correct.

I don't understand the other two fields so naturally I enterd them exactly as the guide said.

Can anyone offer any advice please?

Regards.

This is what I have. Check that yours agrees and that Ctrl+Shift+A is on the menu next to Asymptote.

My "Arguaments" field was different - I  had:

-batchView -tex "latex" %tc

I changed it to the one you have but exactly the same thing happens.

Although the command prompt (the Asymptote one - I said windows before but must have been mistaken) only appears for a split second, by pressing ctr+alt+A repeatedly I have managed to read a message which appears before it dies:

error: could not load module "test.asy"

or something very similar.

Did you follow this instruction in Using TeXnicCenter for Asymptote?
[quote]To do so, create a new file in TeXnicCenter and save it as test.asy. Type the following code into the document:

draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$\$$P$\$$",(50,50),S);[/quote]

I assume test.asy is open in TexnicCenter when you press Ctr+Alt+A or whatever? When no file is open and you press the hotkey do you get a DOS window with

Welcome to Asymptote version 1.20 (to view the manual, type help)
>

Finally, make sure test.asy is saved in a directory with no spaces in it. Many latex type programs don't like filenames with spaces in them.

Yes it is open and yes that is indeed what happens when I use my hot key whilst no file is open.

I created the directory C:\nospaces and saved it in there but still no luck.

Open a command window (Run, cmd) and then type:

"c:\program files\asymptote\asy.exe" c:\nospaces\test.asy

(changing this to your circumstances). After a pause, does this give an image in GView?

If that works then Asymptote is working fine, but for some reason it can't find test.asy when told by TexnicCenter.

That doesn't work in fact. The old message:

error: could not load module "test.asy"

just comes up when I execute that command.

That means the problem is not with the Asymptote setup in TexnicCenter. Are you [i]absolutely[/i] sure that test.asy exists and you haven't saved it as test.asy.tex as suggested at http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help It's also worth checking the rest of the help on that page.

Yey!!!

It works!!!

It seems that my problem was exactly the second problem in the list. The problem was that even though it looked like an .asy, IT WASN'T.

I feel a bit silly now for not realising that that help page existed. But at least I have a bit of insight into how the process works.

Thank you so much for your help   :D


Harvard
college
Euler
quadratics
number theory
Pythagorean Theorem
geometry

Do someone know any thing about [b]Taniyama_Shimura_Weil Conjecture[/b]
that leads to the final siege agianst the fermat last Theory.(After 300 years of
honorable battle( :starwars:  :starwars:  )

does anyone know another beautiful struggle of mathematicions that lead to a great victory ?

It is not a conjecture anymore. It is a theorem. For a proof, see for example http://abel.math.harvard.edu/~rtaylor/cdt.ps .

Link is fixed now.

I think harvard has deleted the page(i think they dont like this kind of theorem  :lol: , if you click on the link it says
page is not found.

You can see this, but it's very long :o ....
http://math.stanford.edu/~lekheng/flt/wiles.pdf

jesus christ , holy moder of ...
I think this a real example(a bit more than a real example :lol: )of
advanced math.
Thanks puuhikki and leibniz but i tried to read it and i just did understand nothing.  :(

Does anyone know the proof of the more general case $a^{\phi(n)-1}=1 (mod$ $n)$

That's the Euler theorem (easy to prove), generalisation of the FLT (Fermat's little theorem), while these guys were actualling talking about Fermat's LAST theorem!!

Google for a proof. You shouldn't post that kind of questions.

By the way, we need $(a, n) = 1$ and then we have $a^{\phi(n)} \equiv 1\;(\textrm{mod n})$ and not $a^{\phi(n) - 1} \equiv 1\;(\textrm{mod n})$ (which is what you posted).

Hi, I have googled, but have not found.

I would like an algebraic proof that x^3 + y^3 = z^3
has no integer solutions.

(flt case = 3)

ta.

I have not checked, but it looks like there is one on the website http://mathforum.org/library/drmath/view/51546.html

the case $x^3+t^3=z^3$ is a classic
when you are reading about unique factorisation in quadratic extensions of the integers, it becomes pretty easy

You can find it in hardy and wright

the idea is simple, if you call $\rho=e^{\frac{2\pi I}{3}}$ you can rewrite

$z^3=(x+y)(x+\rho y)(x+\rho^2 y)$
when you work in an extension ring of the normal integers that contains $\rho$

such a simple problem... such an ugly complex solution.
I have a theory: just the same way we have simple proofs for things like the pythagorean theorem but struggle when the powers and variables are increased, there is a simple geometric proof to $x^2+y^3=z^3$ if you lived in the fourth dimension...


inequalities
triangle inequality
geometry proposed
geometry

[u][b]The author of this posting is : nttu[/b][/u]
____________________________________________________________________

Let ABC be a triangle, and let M be a point inside the triangle ABC. Prove that

$\frac{MA}{a} + \frac{MB}{b} + \frac{MC}{c} \geq \sqrt3$

[u][b]The author of this posting is : Megus[/b][/u]
____________________________________________________________________

using van Aubel's theorem, rewriting our inequality and after whole bunch of computations [which I'm scared of ;) ] we should get what we wanted

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

[quote="Megus"]using van Aubel's theorem, rewriting our inequality and after whole bunch of computations [which I'm scared of ;) ] we should get what we wanted[/quote]

I don't know what is van Aubel's theorem. Please, clarify it.

[u][b]The author of this posting is : Megus[/b][/u]
____________________________________________________________________

Well I found out that it probably has different name in English, but here goes what I meant:

Consider triangle ABC with arbitrary cevians which intersects each other at point M and AM intersects BC at Y, BM intersects AC at Z and, CM intersect AB at X. Then:

$\frac{AM}{MY} = \frac{AX}{XB} + \frac{AZ}{ZC}$

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

@Megus. Oh yes! It is immediate consequence from center-of-mass approach.

[u][b]The author of this posting is : Megus[/b][/u]
____________________________________________________________________

Do you know how it is called in English because what I found as van Aubel's Theorem is: http://mathworld.wolfram.com/AubelsTheorem.html ?

[u][b]The author of this posting is : manlio[/b][/u]
____________________________________________________________________

http://www.maths.gla.ac.uk/~wws/cabripages/klein/ceva0.html

It is called van Obel's theorem. ;)

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

[quote="Megus"]Do you know how it is called in English because what I found as van Aubel's Theorem is: http://mathworld.wolfram.com/AubelsTheorem.html ?[/quote]

I remembered that russian variant is "van Obel".

[u][b]The author of this posting is : darij grinberg[/b][/u]
____________________________________________________________________

This is not the first time the van Aubel theorem (which I have seen referred to as [b]van Aubel[/b], [b]Aubel[/b], [b]van Obel[/b] and [b]Obel[/b] theorem in different places - there doesn't seem to be "the right" spelling, although van Aubel is my personal favorite) is being discussed on MathLinks, and the thread http://www.mathlinks.ro/Forum/viewtopic.php?p=108079 gives two more links with proofs.

Nevertheless it's not at all clear to me how to do that inequality using van Aubel. But maybe I'm just stupid.

  Darij

[u][b]The author of this posting is : nttu[/b][/u]
____________________________________________________________________

Megus , can you show me how to use Van Obel 's theorem to prove my ineq ?

[u][b]The author of this posting is : Stun[/b][/u]
____________________________________________________________________

I've never heard about Aubel theorem but I can solve nttu's ineq in another way:

$\sum_{\mbox{sym}} \frac{MA}{a} = \sum_{\mbox{sym}} \frac{MA\cdot GA}{a\cdot GA} \geq \sum_{\mbox{sym}} \frac{GA^2}{a\cdot GA}$ (well-known)
$\geq \frac{\sum_{\mbox{sym}} \left(GA^2\right)}{\max\left(a\cdot GA, b\cdot GB, c\cdot GC\right)}$
Let ${a\cdot GA = \max\left(a\cdot GA, b\cdot GB, c\cdot GC\right)}$ so we have
$\frac{\sum_{\mbox{sym}} \left(GA^2\right)}{\max\left(a\cdot GA, b\cdot GB, c\cdot GC\right)}$
$\geq \sqrt3$ (by Cauchy)

[u][b]The author of this posting is : treegoner[/b][/u]
____________________________________________________________________

Well, you can see that in "Hinh hoc cua tam giac" of Hoang Chung. It is a simple theory.

[quote="Myth"][quote="Stun"]$\sum_{\mbox{sym}} \frac{MA\cdot GA}{a\cdot GA} \geq \sum_{\mbox{sym}} \frac{GA^2}{a\cdot GA}$ (well-known)[/quote]

???[/quote]

As always, Myth is completely right with his ???: The inequality $\sum_{\mbox{sym}} \frac{MA\cdot GA}{a\cdot GA} \geq \sum_{\mbox{sym}} \frac{GA^2}{a\cdot GA}$ is not only not well-known, it's actually generally wrong, and Stun's solution is incorrect. The problem is more difficult than it looks like. Here is my solution, which uses a nice idea applied by Anh Cuong in his proof of another inequality posted in http://www.mathlinks.ro/Forum/viewtopic.php?p=136249 .

I will prove a slightly extended form of the problem; actually, the point M needs not lie inside the triangle ABC in order for the inequality to hold. We will prove:

[b]Theorem 1.[/b] If M is an arbitrary point in the plane of a triangle ABC, then

$\frac{MA}{a} + \frac{MB}{b} + \frac{MC}{c} \geq \sqrt3$.

For the proof, I will use two lemmas:

[b]Lemma 2.[/b] We have

$\frac{MB\cdot MC}{bc} + \frac{MC\cdot MA}{ca} + \frac{MA\cdot MB}{ab} \geq 1$.

[b]Lemma 3.[/b] For any three positive numbers x, y, z, if $yz+zx+xy \geq 1$, then $x+y+z \geq \sqrt3$.

From these lemmas, Theorem 1 quickly follows, since Lemma 2 yields $yz+zx+xy \geq 1$, where $x=\frac{MA}{a}$, $y=\frac{MB}{b}$ and $z=\frac{MC}{c}$, and thus by Lemma 3 we have $x+y+z \geq \sqrt3$, so that $\frac{MA}{a} + \frac{MB}{b} + \frac{MC}{c} \geq \sqrt3$, and Theorem 1 is proven.

Remains to proves Lemmas 2 and 3. Well, Lemma 3 is almost evident, since $yz+zx+xy \geq 1$ implies $x^2+y^2+z^2 \geq 1$ (because of the famous inequality $x^2+y^2+z^2 \geq yz+zx+xy$), and thus

$\left(x+y+z\right)^2 = x^2+y^2+z^2 + 2\cdot\left(yz+zx+xy\right) \geq 1 + 2\cdot 1 = 3$

and hence $x+y+z\geq \sqrt3$.

So it remains to show Lemma 2. The most easy way to do it is to work in the complex plane, letting M be the origin and assigning to the points A, B, C the affixes u, v, w. Then,

$\frac{MB\cdot MC}{bc} = \frac{MB\cdot MC}{CA\cdot AB} = \frac{\left|v-0\right|\cdot \left|w-0\right|}{\left|w-u\right|\cdot\left|u-v\right|} = \frac{\left|v\right|\cdot \left|w\right|}{\left|w-u\right|\cdot\left|v-u\right|} = \left|\frac{vw}{\left(w-u\right)\left(v-u\right)}\right|$.

Similarly,

$\frac{MC\cdot MA}{ca} = \left|\frac{wu}{\left(u-v\right)\left(w-v\right)}\right|$;
$\frac{MA\cdot MB}{ab} = \left|\frac{uv}{\left(v-w\right)\left(u-w\right)}\right|$.

Thus, by the triangle inequality,

$\frac{MB\cdot MC}{bc} + \frac{MC\cdot MA}{ca} + \frac{MA\cdot MB}{ab}$
$= \left|\frac{vw}{\left(w-u\right)\left(v-u\right)}\right| + \left|\frac{wu}{\left(u-v\right)\left(w-v\right)}\right| + \left|\frac{uv}{\left(v-w\right)\left(u-w\right)}\right|$
$\geq \left|\frac{vw}{\left(w-u\right)\left(v-u\right)} + \frac{wu}{\left(u-v\right)\left(w-v\right)} + \frac{uv}{\left(v-w\right)\left(u-w\right)}\right|$.

But since

$\frac{vw}{\left(w-u\right)\left(v-u\right)} + \frac{wu}{\left(u-v\right)\left(w-v\right)} + \frac{uv}{\left(v-w\right)\left(u-w\right)} = 1$,

we thus get

$\frac{MB\cdot MC}{bc} + \frac{MC\cdot MA}{ca} + \frac{MA\cdot MB}{ab} \geq \left|1\right| = 1$,

and Lemma 2 is proven. Another proof of Lemma 2 can be found at http://www.mathlinks.ro/Forum/viewtopic.php?t=6037 .

  Darij

Use wellknown lemma: Let triangle $ABC$ with centroid $G$, then: $\sum GA^{2}\leq\sum MA.GA$ with any point $M$ on plane. We have:
$\sum\frac{MA}{a}=\sum\frac{MA.GA}{a.GA}=\sum\frac{3MA.GA}{2a.m_{a}}=\sum\frac{3MA.GA}{a.\sqrt{2b^{2}+2c^{2}-a^{2}}}$
$\geq\sum\frac{3\sqrt{3}.MA.GA}{a^{2}+b^{2}+c^{2}}\geq\frac{3\sqrt{3}.\sum GA^{2}}{a^{2}+b^{2}+c^{2}}=\sqrt{3}$

[b]Remark:[/b]
$\sum\frac{MA}{a}\geq\sqrt{3}\implies\sum\frac{GA}{a}\geq\sqrt{3}(M\equiv G)$
$\implies\boxed{\sum\frac{m_{a}}{a}\geq\frac{3\sqrt{3}}{2}}\implies\boxed{\sum\frac{a}{m_{a}}\geq 2\sqrt{3}}$

my solution
by 96 CMO problem, we have $\sum_{cyc}\frac{PA*PB}{AC*BC}\ge\ 1$,
so we have $\sum_{cyc}\frac{PA}{BC}\ge\sqrt{3}$


[quote="thanhnam2902"][color=darkblue][size=150]Let equation: $ (x \plus{} 1).lnx \minus{} x.ln(x \plus{} 1) \equal{} 0$.

Prove that this equation have only one root?[/size][/color]

--------------------------------------------------------------------------------------------------------------------------
Sory every body. Because I posted this problem at forum Algebra Unsolved Problems.[/quote]There was no need to repost the problem, if the moderators of that forum think it should be moved, they will move it. This one is considered as spam because:

1) Problem reposted twice in AoPS
2) Problem difficulty beyond the intended level of HSB forum

Just don't repost again, so again DO NOT REPOST. Moderators/admins will do their job.

sory, I see, I will try post at right laster.

[quote="JZambrano201"]10000th User, this isn't HSB, this is intermediate topicss.[/quote]It was in HSB if you didn't know, and please stay on topic.

Please continue discussion on the link below since there is no hints/solutions posted in here:
[url]http://www.mathlinks.ro/viewtopic.php?t=201509[/url]


Prove that in any triangle ABC the following inequality holds:
$ \frac{cos^2A}{cos^2B}\plus{}\frac{cos^2B}{cos^2C}\plus{}\frac{cos^2C}{cos^2A}\plus{}8cosAcosBcosC \ge 4$ :)

hello, what is in the case if $ C\equal{}\frac{\pi}{2}$ for example?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, no, in this case makes our inequality no sense.
Sonnhard.[/quote]

Actually, we may be able to work around it by saying if cos C = 0, then the LHS is technically Positive Infinity, which is greater than 4.  But this is very iffy.

[quote="quykhtn-qa1"]Prove that in any triangle ABC the following inequality holds:
$ \frac {cos^2A}{cos^2B} \plus{} \frac {cos^2B}{cos^2C} \plus{} \frac {cos^2C}{cos^2A} \plus{} 8cosAcosBcosC \ge 4$ :)[/quote]
Try also $ a\equal{}b\equal{}5$ and $ c\equal{}9.$ :wink:

[quote="CircleSquared"]if cos C = 0, then the LHS is technically Positive Infinity, which is greater than 4.  But this is very iffy.[/quote]

The LHS isn't technically positive infinity. It's undefined. Division by 0 isn't allowed.

I think there should be an additional restriction about $ A, B, C$ like:
In a non-right angled triangle,....... :maybe:
Of course, the concept $ \frac {x}{0} = + \infty$ is incorrect, it is NOT DEFINED, but that does not $ \Longrightarrow$ that the value of something divided by zero is infinity as infinity and "undefined" are two different terms.


I need some hints
thanks

Runaround numbers are integers with unique digits, none of which is zero (e.g., 81362) that also have an interesting property, exemplified by this demonstration:

If you start at the left digit (8 in our number) and count that number of digits to the right (wrapping back to the first digit when no digits on the right are available), you'll end up at a new digit (a number which does not end up at a new digit is not a Runaround Number). Consider: 8 1 3 6 2 which cycles through eight digits: 1 3 6 2 8 1 3 6 so the next digit is 6.
Repeat this cycle (this time for the six counts designed by the `6') and you should end on a new digit: 2 8 1 3 6 2, namely 2.
Repeat again (two digits this time): 8 1
Continue again (one digit this time): 3
One more time: 6 2 8 and you have ended up back where you started, after touching each digit once. If you don't end up back where you started after touching each digit once, your number is not a Runaround number.
Given a number M (that has anywhere from 1 through 9 digits), find and print the next runaround number higher than M, which will always fit into an unsigned long integer for the given test data.

well it is a brute force problem. u just need to keep a flag for each digit. flag=0 if that digit isn't visited and 1 if that is visited. Now start from the first digit and continue running. when u will reach such a digit for which flag=1 then check whether ur running number is = the number of digits.

OK? :roll:

Sorry, I actually have a solution but it's too slow. For case 5 it has a value of 1000000 and my program runs in 1.13 sec

The way I do it is starting from 1000000 and keep increasing and check if the number is run around or not.

I know that's not too efficient, but is there another way of doing it?

here what official soln says:

The trick to this problem is noticing that since runaround numbers must have unique digits, they must be at most 9 digits long. There are only 9! = 362880 nine-digit numbers with unique digits, so there are fewer than 9*362880 numbers with up to 9 unique digits. Further, they are easy to generate, so we generate all of them in increasing order, test to see if they are runaround, and then take the first one bigger than our input.

I also did something where there can't be a digit that is a multiple of the number of digits in the number.

Great!
Thanks!

Can I have a hint on MoneySystem also?
>_<

As you can see, I'm weak in this section...

Hmm, money systems is a DP problem
nways[n][k]----number of ways to make value n using first k coins
value[k]-----values of coin number k
nways[n][k]=nways[n][k-1]+nways[n-value[k]][k]

Use it wisely  :roll:

how did you guys know how to solve DP problems?
They seem quite difficult to solve...

How did you come up with that relation in the first place?

DP problems is usually very difficult problems, if you can't make them... And very easy if you see the solution directly. :D

That's why IOI always has at least 1 DP problem (usually 2 or more).

The only way to be good at solving them is practicing... "Find DP problems, solve them, find some more, etc..."

DP is usually pretty much like induction. If you know a[0], a[1], ..., a[n], you can calculate a[n + 1]. But it's not always so easy to see that...

Before you start coding "pure" DP solutions, you can try to find memorisation solutions instead. They are a little bit slower, but should be OK for most problems... :D

Weeblie:[quote]memorisation[/quote]

Hmm. I guess it was memoization :D

Weeblie is right. To understand DP you've to see a lot of examples.

A DP solution is of course the more advanced and efficient, but you can also find a way to recursively generate the numbers that fit the digit specifications (i.e. ones that have no zeros, and at most one occurrence of each 1-9 digit) and then write a function to check each of the numbers you generate.  Not as cool, but for those of us to whom DP is still a bit difficult to grasp (i.e. me), it's an alternative that doesn't (or at least didn't for me) time out.

how do you generate all the numbers in descending order?
>_< :?:

i'm worse than i thought...

Use an array or vector (or arraylist in Java...whatever container type you want can work) and generate/incriment each individual digit (you can do it recursively). Use a boolean array to keep track of digits used.  When you have to incriment in such a way that alters the size of the number (ie. 9 to 10) it gets a little trickier, but it can be done!  Let me know if you need some more help, but fool around a bit with that concept if you can.

For some reason, my mind really blanks out when using recursion to do this problem

how exactly would i use recursion to generate all these numbers?

>_< the only question that's left in this section....grr...

actually i am very busy with my last two practical exams and IMO. so plz don't look at me! :D

I actually just solved this one yesterday, but in a different way. I just incremented the input and checked if it was runaround each time. It's pretty inefficient I guess but you can just test for the digit 0 and repeated digits each time and you don't need to check for runaround always. Every time I ran into a 0 I made that 0 a 1 instantly so I didn't need to check tons of numbers with that same 0. I got it to run on each test case fast enough so I guess it works.

Thanks!

Finally can move on to next section :D:D


A florist that is launched upwards, with initial speed $v_{0}$, has a speed of $v < v_{0}$ when he returns to the take-off site. If $|a_{u}|$ is the absolute value of the acceleration during the ascent, $a_{c}$ is the acceleration during the descent and $g$ is the acceleration due to gravity, then we have that:
(a) $|a_{u}| = a_{c}$;
(b) $|a_{u}| > g$;
(c) $|a_{u}| < g$;
(d) $|a_{u}| < a_{c}$.

$(b)$ and $(c)$ are correct. Use Third Equation of motion.
 displacement $S$ is same in both upward and downward motion..

Only one answer is correct. But I must admit that I got the same answer as you. I don't know how to pick the right one (I have the "official solutions", but they give only the answer, without any additional explanations).

If u ignore air resistance(not sure if the right word), the florist will be continously under the influce of acceleration of g in downward direction. That is, $a_{u}=a_{c}=g$. (and $v=v_{0}$, u get this with little math ;) )

Hey, it is stated in the hypothesis that $v < v_{0}$.
The correct answer should be $b$, but I don't quite understand why $c$ is not good.

The problem is mis-stated then... It should say "dont ignore air resistance". Then when the florist goes up, its both affected by gravity and air resistance in downward direction and when it goes down its affected by both again but this time air resistance has upward direction. So thats why $a_{u}>g>a_{c}$.


1) $ a,b,c \geq 0$
Proof
$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}}\geq2$

2) $ a,b,c \geq 0$ and $ abc\equal{}1$
Proof
$ \frac{1}{a^3(b\plus{}c)}\plus{}\frac{1}{b^3(c\plus{}a)}\plus{}\frac{1}{c^3(a\plus{}b)}\geq\frac{3}{2}$

[b]1,[/b] Using the AM - GM Inequality
  $ \sqrt {\frac {a}{{b \plus{} c}}} \equal{} \frac {a}{{\sqrt {a(b \plus{} c)} }} \le \frac {{2a}}{{a \plus{} b \plus{} c}}$

Similarly.
$ \sqrt {\frac {b}{{a \plus{} c}}} \le \frac {{2b}}{{a \plus{} b \plus{} c}}$ and  $ \sqrt {\frac {c}{{a \plus{} b}}} \le \frac {{2c}}{{a \plus{} b \plus{} c}}$

Adding up these three inequalities, we have the desired inequality
[b]2,[/b]
We set $ a \equal{} \frac {1}{x},b \equal{} \frac {1}{y}$ and $ c \equal{} \frac {1}{z}$
$ \Rightarrow xyz \equal{} 1$

The beginning inequality becomes
\[ \frac {{x^2 }}{{y \plus{} z}} \plus{} \frac {{y^2 }}{{z \plus{} x}} \plus{} \frac {{z^2 }}{{x \plus{} y}} \ge \frac {3}{2}\]
Using the Cauchy- Schuwarz Inequality, we have
\[ \frac {{x^2 }}{{y \plus{} z}} \plus{} \frac {{y^2 }}{{z \plus{} x}} \plus{} \frac {{z^2 }}{{x \plus{} y}} \ge \frac {{x \plus{} y \plus{} z}}{2}\]
Which is true from the AM- GM Inequality and  the condition $ xyz \equal{} 1$

[quote="daigiaga1994"][b]1,[/b] Using the AM - GM Inequality
  $ \sqrt {\frac {a}{{b \plus{} c}}} \equal{} \frac {a}{{\sqrt {a(b \plus{} c)} }} \le \frac {{2a}}{{a \plus{} b \plus{} c}}$

Similarly.
$ \sqrt {\frac {b}{{a \plus{} c}}} \le \frac {{2b}}{{a \plus{} b \plus{} c}}$ and  $ \sqrt {\frac {c}{{a \plus{} b}}} \le \frac {{2c}}{{a \plus{} b \plus{} c}}$

Adding up these three inequalities, we have the desired inequality
[/quote]Signs inversed - apart from that, it's correct  :lol:
$ \frac {1}{{\sqrt {a(b \plus{} c)} }} \ge \frac {{2}}{{a \plus{} b \plus{} c}}$ of course


Prove the following are equivalent:
(a) $ \forall V \subseteq Y, V$ is an open subset of $ Y$ iff $ p^{ \minus{} 1}(V)$ is an open subset of $ X$
(b) $ F$ a closed subset of $ Y$ iff $ p^{ \minus{} 1}(F)$ is a closed subset of $ X$.

Could you clarify? In part (a) you make a claim that one condition is stronger than continuity, and you also want us to prove that this condition is equivalent to the condition in part (b). Do you want us to prove that the first claim is stronger than continuity as well?

In the mean time, I will prove what I think you want me to prove, the equivalence of statements (a) and (b).

Assume statement (a) is true, and let $ F$ be closed in $ Y$. $ Y \minus{} F$ is open in $ Y$, so that $ p^{ \minus{} 1}(Y \minus{} F)$ is open in $ X$. Thus $ X \minus{} p^{ \minus{} 1}(Y \minus{} F) \equal{} X \minus{} (p^{ \minus{} 1}(Y) \minus{} p^{ \minus{} 1}(F)) \equal{} X \minus{} (X \minus{} p^{ \minus{} 1}(F)) \equal{} p^{ \minus{} 1}(F)$ is closed in $ X$. Every statement made works when starting with $ p^{ \minus{} 1}(F)$ closed in $ X$, so (a)$ \Rightarrow$(b).

Assume statement (b) is true, and let $ V$ be closed in $ Y$. $ Y \minus{} V$ is closed in $ Y$, so that $ p^{ \minus{} 1}(Y \minus{} V)$ is closed in $ X$. Thus $ X \minus{} p^{ \minus{} 1}(Y \minus{} V) \equal{} X \minus{} (p^{ \minus{} 1}(Y) \minus{} p^{ \minus{} 1}(V)) \equal{} X \minus{} (X \minus{} p^{ \minus{} 1}(V)) \equal{} p^{ \minus{} 1}(V)$ is open in $ X$. Every statement made works when starting with $ p^{ \minus{} 1}(V)$ open in $ X$, so (b)$ \Rightarrow$(a).

Thus, (a)$ \Leftrightarrow$(b), as was to be shown.

[color=Red][b]Correction:[/b][/color]

[b] Definition-[/b] $ X, Y$ spaces.  Let $ p: X \rightarrow Y$ be a surjective map.  $ P$ is a quotient map iff $ \forall$$ V \subseteq Y$,  $ V$ is an open subset of $ Y$ iff $ p^{\minus{}1}(V)$ is an open subset of $ X$.

Prove the following are equivalent:
(a) $ \forall V \subseteq Y, V$ is an open subset of $ Y$ iff $ p^{\minus{}1}(V)$ is an open subset of $ X$
(b) $ F$ a closed subset of $ Y$ iff $ p^{\minus{}1}(F)$ is a closed subset of $ X$.

So you interpreted it correctly.  Thanks for your helpful post!


geometry
quadratics
algebra
quadratic formula
special factorizations

Answer and explain, plz

1. Jeff bicycles 160 miles at the rate of "r" mph. The same trip would have taken 2 hours less if he had decreased his speed by 4 mph. Find r.

2. A bus company has 3000 passengers daily, paying a 25 cent fare. For each 5 cent increase in the fair, the company estimates that it will lose 80 passengers. What increase in fare will produce a 994 dollars daily revenue?

3. A box with no top is to be made by cutting a 2-inch square from each corner of the square sheet of metal shown in Illustration 4. After bending up the sides, the volume of the box is to be 200 cubic inches. How large should the piece of metal be?

the 1st one made noooo sense.
If you decrease mph, you slow down don't you?

you could start q.2 by solving-
        (3000-80x)(25+5x) = 994

[quote="returnofkiladi"]3. A box with no top is to be made by cutting a 2-inch square from each corner of the square sheet of metal shown in Illustration 4. After bending up the sides, the volume of the box is to be 200 cubic inches. How large should the piece of metal be?[/quote]

Okay, since the volume of the box is $ 200$, and the height is $ 2$, the area of the base must be $ 100$, making the side length of the square $ 10$.  Since we took a $ 2 \times 2$ square out of each corner, we took away a length of $ 4$.  $ (10 \plus{} 4)^2 \equal{} \boxed{144}$

[quote="returnofkiladi"]For each 5 cent increase in the fair, the company estimates that it will lose 80 passengers.[/quote]

Hopefully this isn't quadratic or anything, but I assume it's linear, right?

[quote="returnofkiladi"]The same trip would have taken 2 hours less if he had decreased his speed by 4 mph.[/quote]

You mean $ 2$ hours more or increased his speed?

[quote="returnofkiladi"]2. A bus company has $ 3000$ passengers daily, paying a $ 25$ cent fare. For each $ 5$ cent increase in the fair, the company estimates that it will lose $ 80$ passengers. What increase in fare will produce a $ 994$ dollars daily revenue?[/quote]

Let $ x$ be the amount of times you increase the fair.
First make an equation: $ (3000 \minus{} 80x) \cdot (25 \plus{} 5x) \equal{} 994$
Then, everything else is straightforward. Just multiply using FOIL, and solve for $ x$.

[hide="Just in case you're still stuck..."]$ (3000\minus{}80x)(25 \plus{} 5x) \equal{} 994$

$ 7500 \plus{} 13000x \minus{} 400x^2 \equal{} 994$

Set one side of the equation equal to zero.

$ \minus{}400x^2 \plus{} 13000x \plus{} 6506 \equal{} 0$

Quadratic formula, or if you really want to, completing the square (:D).

$ x \equal{} \frac{\minus{}b \pm \sqrt{b^2\minus{}4ac}}{2a}$

$ x \equal{} \frac{\minus{}13000 \pm \sqrt{169000000 \minus{} 4(\minus{}400)(6506)}}{2(\minus{}400)}$

$ x \equal{} \frac{\minus{}13000 \pm \sqrt{169000000 \plus{} 1600 \cdot 6506}}{\minus{}800}$

$ x \equal{} \frac{\minus{}13000 \pm \sqrt{179409600}}{\minus{}800}$

$ x \equal{} \frac{\minus{}13000 \pm 120 \sqrt{12459}}{\minus{}800}$

$ x \equal{} \frac{65}{4} \pm \frac{3\sqrt{12459}}{20}$

$ x \equal{} \frac{325 \pm 3\sqrt{12459}}{20}$

Something like that.  However, we must toss out the negative value, because it's negative. :D  So our answer is $ \boxed{\approx 32.992}$.[/hide]

Actually, for question 3, shouldn't the answer be 196?

How? i think it's 144..  :(
could you give ur explanation?

Ernies reasoning was right, its just that $ (10 \plus{} 4)^2 \equal{} 196$, not 144.  :P

EDIT: also, if the area of the base must be 100 and we subtracted 4 from each side , then $ (s \minus{} 4)(s \minus{} 4) \equal{} 100$. $ (s \minus{} 4)^2 \equal{} 100$. $ (s \minus{} 4) \equal{} 10$, $ s \equal{} 14$, $ 14^2 \equal{} 196$.

Note that we didnt subtract 2 from each side, which is what you probably did to get the side length of 12, cgyao.

[quote="Thunder365"]Ernies reasoning was right, its just that $ (10 \plus{} 4)^2 \equal{} 196$, not 144.  :P[/quote]

D'oh!  That's why I failed MC. :P


Calculate without calculator:
1+2+3...3033


An $n\times n$ table is filled with numbers $-1, 0, 1$ in such a manner that every
row and column contain exactly one $1$ and one $-1$. Prove that the rows
and columns can be reordered so that in the resulting table each number
has been replaced with its negative.

Fix one cell with $1$. Now go to the $-1$ in this row. Then go to the $1$ in this column and  so on. In the end we must come to the starting $1$, so we will got a cycle. It's easy to see that we can rearrange rows and columns belonging to this cycle to get sufficient result.


group theory
abstract algebra
superior algebra
superior algebra unsolved

All groups G are finite, k(G) denotes the number of conjugacy classes in G.

1. Prove that if k(G) is even, then |G| is even.
2. Prove that if G is nonabelian, then k(G)>|Z(G)|+1, where Z(G) is the center of G
3. Let H be a subgroup of G of index 2, let C(h) be the centralizer of h in G. Prove that if for every nontrivial h in H C(h) is in H, then elements from G\H (i.e. elements of G which are not in H) form one conjugacy class.
4. G is a p-group (i.e. it has p^n elements, p -prime) and H is its nontrivial normal subgroup. Let Z(G) be the center of G. Prove that the intersection of H and Z(G) is nontrivial.

1. By the Burnside's lemma, the number of conjugacy classes in $ G$ is

\[ \sum_{g \in G} \frac{|N(g)|}{|G|}\]

where $ N(g)$ is the group of all elements that commute with $ g$. Thus

\[ \sum_{g \in G} |N(g)| \equal{} 2k|G|\]

since there are an even number of conjugacy classes. Suppose that there are $ m$ elements in $ G$, and $ m$ is odd. Then $ |N(g)|$ is odd by Lagrange's theorem. Thus we have the sum of an odd number of odd terms on the left, and an even number on the right. This is a contradiction.

2. Each element in the center is in its own conjugacy class since $ az \equal{} za \iff aza^{\minus{}1} \equal{} z$. Since $ G$ is non-Abelian, there is also at least one more element $ g$ in the group. The claim is that there must exist at least one more element not in the center and also not in the conjugacy class of $ g$. Note that the number of conjugates of $ g$ is at most $ \frac{|G|}{|Z(G)|}$, by the orbit-stabilizer theorem. Furthermore $ |Z(G)| \plus{} \frac{|G|}{|Z(G)|} \leq \frac{|G|}{2}\plus{}\frac{|G|}{2} \leq |G|$ if $ Z(G)$ is non-trivial unless $ |Z(G)| \equal{} 2, |G| \equal{} 4$ (in which case we know the group is Abelian so this is impossible). Thus we may assume $ |Z(G)|$ is trivial. Then we are trying to show that there are more than $ 2$ conjugacy classes in $ G$. We have one...the identity. If any element has non-trivial stabilizer then we would be done. But every non-identity element has non-trivial stabilizer since it commutes with itself. So we are done.

Okay...the second one may be a bit unclear. I essentially use that the number of conjugates of $ g$ is $ \frac{|G|}{|N(g)|}$ many times. That is also essentially how I got the equation at the beginning of the first proof.


In a row of 10 seats, in how many ways can I choose two seats
that are not next to each other?

[hide] There are $ \binom{10}{2}\equal{}45$ ways to choose two seats. Of these, exactly, $ 9$ are next to each other, so the answer is $ 45\minus{}9\equal{}\boxed{36}$.[/hide]


function
search
calculus
calculus computations

For $ n\geq 3,$ prove that $ a_n>a_{n\plus{}1}.$

Where does the continuous function $ f(x)\equal{}x^{1/x}$ have a maximum? Even though that won't be an an integer, it will limit your search to the to integers closest to that point.

actually how do i find which is greater between 2^ (1/2) and 3^(1/3).

stephencheng had that answer - read his post.

Starting from $ 2^{1/2}\stackrel{?}{>}3^{1/3},$ take the 6th power of both sides to see what you have.

Let $ a_n \equal{} \sqrt [n]{n}\ (n \equal{} 1,\ 2,\ 3,\ )$, we prove that $ a_n > a_{n \plus{} 1}$ for $ n\geq 3.$

Proof: $ a_n > a_{n \plus{} 1}\Longleftrightarrow \sqrt [n]{n} > \sqrt [n \plus{} 1]{n \plus{} 1}\Longleftrightarrow n > \left(\frac {n \plus{} 1}{n}\right)^{n}$

$ 1^\circ$ For $ n \equal{} 3,$ by $ 3 > \left(\frac {4}{3}\right)^3 \equal{} 2\frac {10}{27}$, the claim is true.

$ 2^\circ$ For $ n \equal{} k\geq 3$, suppose that the clame is true, we have $ k > \left(\frac {k \plus{} 1}{k}\right)^{k}$ multiplying the both sides by $ \frac {k \plus{} 1}{k} > 0$,

since $ \frac {k \plus{} 1}{k} > \frac {k \plus{} 2}{k \plus{} 1}$, yielding $ k \plus{} 1 > \left(\frac {k \plus{} 1}{k}\right)^{k \plus{} 1} > \left(\frac {k \plus{} 2}{k \plus{} 1}\right)^{k \plus{} 1}$, which shows the claim is also true for $ n \equal{} k \plus{} 1.$

Therefore the calim is true by mathematical induction.

Now from $ a_1 \equal{} 1,\ a_2 \equal{} \sqrt {2},\ a_3 \equal{} \sqrt [3]{3}$, since $ 2^3 < 3^2\Longleftrightarrow \sqrt {2} < \sqrt [3]{3}$ and the above result we have $ a_3 > a_4 > a_5 > \cdots > 1$.

the desired maximum value is $ a_3 \equal{} \sqrt [3]{3}.$

This problem is easy :D .
My solution .
Let $ f(x) \equal{} x^{\frac1{x}}$,$ x\geq3$
We have $ lnf(x) \equal{} \frac1{x}lnx$
$ \Rightarrow$ $ \frac{f'(x)}{f(x)}\equal{}( \frac1{x}lnx)'\equal{}\minus{}lnx.\frac1{x^2}\plus{}\frac1{x^2}$
$ \Rightarrow$ $ f'(x)\equal{}f(x).\frac1{x^2}.(1\minus{}lnx)\leq 0$ because $ lnx \geq ln3 \geq1$
So $ f(x)$ is decrease function $ \forall x\geq3$
So $ f(n)\leq f(3)$.Because $ f(3)\geq f(2)$
$ \Rightarrow f(3)$ is maximum.


Gauss
topology
group theory
abstract algebra
advanced fields
advanced fields unsolved

What is the universal covering space of the orientable genus $ g$ surface, $ M_g$?

the cases $ g\equal{}0,1$ are pretty easy. so from now on $ g\ge 2$.

since you have a constant (gauss) curvature metric on $ M_g$, and by Gauss-Bonnet it has to be negative, its universal cover has to be (as a riemannian manifold) the hyperbolic plane (by the classification theorem), so topologically it's $ \mathbb{R}^2$.

otherwise, i think there's a more elementary way, such like finding a discrete subgroup of $ SL_2(\mathbb{R})$ isomorphic to $ \pi_i(M_g)$, which would guarantee the result.

even more elementarily, you can use classification theorem for noncompact surfaces, which tells you that the only simply connected surfaces are $ S^2$ and the plane: since the universal covering of $ M_g$ can't be compact, it's the plane.

probably, there's a more natural way to do this, but i can't see how to construct a covering map from the plane to $ M_g$.


calculus
derivative
calculus computations

Find the derivative $ f'(x)$ if $ f(x) \equal{} \frac{1}{\sqrt{x \minus{} 1}}$ using the [b]derivative definition of a limit[/b].

I must be missing something here because this problem looks extremely difficult without derivative rules.

[hide="hint"]
In your (simplified with no fractions in the numerator) fraction, try multiplying top and bottom by $ \sqrt{x\minus{}1}\plus{}\sqrt{x\plus{}\delta\minus{}1}$. The deltas should cancel :) and then it should be easy (if I didn't make some strange error)

and it should be the [b]limit definition of the derivative[/b]  :wink:
[/hide]


AMC AIME geometry articles Harvard college MIT

Hello, I am looking for a resource that teaches AIME-level geometry, and maybe goes above that. I find my geometry to be very weak. A book or online article or problems from the AIME would be great. Thank you.

AIME=[b]A[/b]merican [b]I[/b]nvitational [b]M[/b]athematics [b]E[/b]xamination ;)

Challenging Probs in Geo... a great book.

I find the AIME level standard of difficulty fuzzy. If you've mastered the basics, AIME is for the most part an application of the basics in all the major subjects in competition math. (The last few are harder, but you simply apply more advanced concepts - not necessarily more advanced because it depends where you are coming from. i.e. The first problem might deal with Pythagorean Theorem but the last problem could deal with Ceva or Simson Line and the like).

Thus, it would be also beneficial not only to look directly for problems on the AIME level, but to just practice familiarity with main geometric principles. Even proofs can help toward your understanding. I agree that Salkind's book is good, but you could also try other books (I'm aware Zuming Feng has a huge collection of geometry problems - much of it is on a higher level (pre-olympiad, olympiad) but if you really want to sharpen your geomtry skills try those).

Well the book has a detailed solutions section at the end, which is actually what comprises most of the book itself. Solutions are a good teacher.

well, the aops book 1 is good for basic geometry, but you have to do all the problems and read all the examples

Try the geometry events from the Harvard-MIT or Stanford Math Tournaments.  I used them last spring to train for ARML, so they'll probably work for AIME.

HMMT: [url=http://web.mit.edu/hmmt/www/datafiles/problems/]web.mit.edu/hmmt/www/datafiles/problems/[/url]
Stanford: [url=http://sumo.stanford.edu/smt/]sumo.stanford.edu/smt/[/url]


Let $ t = x^{2}-3x+1 = \frac{3+\sqrt{5+4x}}{2}$.  Then $ t$ is a root of

$ t^{2}-3t+(1-x) = 0$

In other words, let $ f(x) = x^{2}-3x+1$.  Then we are trying to solve the polynomial $ f(f(x)) = x$, which clearly has the same roots as $ f(x) = x$, and in this way we have a motivation for factorization.


$ a,b,c\geq 0$
$ \sum_{cyc} \frac{\sqrt{bc\plus{}4a(b\plus{}c)}}{b\plus{}c}\geq \frac{9}{2}$
:lol:

[quote="Rose_joker"]$ a,b,c\geq 0$
$ \sum_{cyc} \frac {\sqrt {bc \plus{} 4a(b \plus{} c)}}{b \plus{} c}\geq \frac {9}{2}$
:lol:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3769
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28175


induction
geometry
integration
algebra unsolved
algebra

To proov that,if $ n$ is a natural number,than:
$ \frac{1}{\sqrt{1}   }\plus{}\frac{1}{\sqrt{2}  }\plus{} \frac{1}{\sqrt{3}  }\plus{}...\plus{}\frac{1}{ \sqrt{n} }  \ge \sqrt{n}$

Induction!

$ n = 1\Longrightarrow L.H.S. = \frac {1}{\sqrt {1}},\ R.H.S. = \sqrt {1} = 1$.

Suppose for $ n = k\geq 2$ the claim is true, since $ \sqrt {k + 1} > \sqrt {k}$, so we have

$ \frac {1}{\sqrt {1}} + \frac {1}{\sqrt {2}} + \cdots + \frac {1}{\sqrt {k}} + \frac {1}{\sqrt {k + 1}}\geq \sqrt {k} + \frac {1}{\sqrt {k + 1}}\geq \sqrt {k + 1}$.

Alternatiive Solution:

Comparison the area of  some figure and that of the rectangular expressed by $ \frac {1}{\sqrt {n}}$ times $ (n + 1 - 1)$.

${ \boxed{{\sum_{k = 1}^n \frac {1}{\sqrt {k}}} > \sum_{k = 1}^n \int_k^{k + 1} \frac {1}{\sqrt {n}}\ dx\ (n = 1,\ 2,\ \cdots)}}$.

$ \sqrt {k} \plus{} \frac {1}{\sqrt {k \plus{} 1}}\geq\sqrt {k \plus{} 1}$---This is not right.It schould be like this:$ \sqrt{k}\plus{}\frac{1}{\sqrt{k\plus{}1}}>\sqrt{k\plus{}1}$,because $ \sqrt {k} \plus{} \frac {1}{\sqrt {k \plus{} 1}} \equal{} \sqrt {k \plus{} 1}$,when $ k \equal{} 4/3$.

Yes,so it must be like that: $ \sqrt{k}\plus{}\frac{1}{\sqrt{k\plus{}1}}>\sqrt{k\plus{}1}$

hello, we have
$ \frac{1}{\sqrt{1}}\plus{}\frac{1}{\sqrt{2}}\plus{}\frac{1}{\sqrt{3}}\plus{}\cdots\plus{}\frac{1}{\sqrt{n}}\geq n\frac{1}{\sqrt{n}}\equal{}\sqrt{n}$
Sonnhard.

Yes,when $ k\ge 2$ this is not right:$ \sqrt{k}\plus{}\frac{1}{\sqrt{k\plus{}1}}\geq\sqrt{k\plus{}1}$


Really far.  They're both rational.  ;)

[hide="Idea 1"] $\sin (\pi-x) = \sin x$
$\cos (\pi-x) =-\cos x$ [/hide]
[hide="Idea 2a"] Complex numbers.  Or... [/hide]
[hide="Idea 2b"] What polynomials have those sets of roots?  What are their leading coefficients?  What are their constant terms? [/hide]

[quote="NeverOddOrEven"]ughh ugly trig. how far are we supposed to simplify these? this is what i have so far. not sure about 2 though.

[hide="1"]
It's easier to type in degrees.

Using $\sin(\theta) = \sin(180-\theta)$,

\begin{eqnarray*}\sin(18)\sin(36)\sin(54)\sin(72) &=& \sin^{2}(18)\sin^{2}(36)\\ &=& 4\sin^{4}(18)\cos^{2}(18)\\ &=& \boxed{4\sin^{4}(18)-4\sin^{2}(18)}\end{eqnarray*}
[/hide]

[hide="2"]
Ah man can't use degrees here  :mad:

\begin{eqnarray*}\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{3\pi}{7}\right) &=& \cos\left(\frac{\pi}{7}\right)\cdot \left[2\cos^{2}\left(\frac{\pi}{7}\right)-1\right]\cdot \left[4\cos^{3}\left(\frac{\pi}{7}\right)-3\cos\left(\frac{\pi}{7}\right)\right] \\ &=& \boxed{8\cos^{6}\left(\frac{\pi}{7}\right)-10\cos^{4}\left(\frac{\pi}{7}\right)+3\cos^{2}\left(\frac{\pi}{7}\right)}\end{eqnarray*}

[/hide][/quote]

You totally missed the point of the problem. The point is to get a [b]rational answer[/b].

EDIT: Sorry NeverOddOrEven. I tried to delete that post, but t0ra posted too quickly.

vishalarul, you've missed the intent of a question more than once.  Give him a break.

[url=http://imageshack.us][img]http://img165.imageshack.us/img165/2690/trgonoca9.jpg[/img][/url]
Shot at 2007-06-26

Im stupid so I actually found the real values of them...

[hide="1"]the first is $(\sin^{2}{\frac{\pi}{5})(\sin^{2}{\frac{2\pi}{5})}}$ -since $\sin{\alpha}=\sin({\pi-\alpha})$.
$\cos{\frac{\pi}{5}}$ gives an exact value (this can be done by using triangles or you can just memorize it) of $\frac{\sqrt{5}+1}{4}$.  Now from $\cos^{2}\alpha+\sin^{2}\alpha=1$, we have $\sin^{2}{\frac{\pi}{5}={\frac{10-2\sqrt{5}}{16}}}$.  We can then use $\sin{2\alpha}=2\sin{\alpha}\cos{\alpha}$.  Since $\sin^{2}{\frac{2\pi}{5}= [2(\cos{\frac{\pi}{5})(\sin{\frac{\pi}{5})]^{2}}}}$ we can plug in our aforementioned values for them and get that to be $\frac{5-\sqrt{5}}{8}$.  W multiply the two values together-- $(\frac{5-\sqrt{5}}{8})\cdot({\frac{10-2\sqrt{5}}{16}})=\frac{20}{64}=\frac{5}{16}$[/hide]

can someone tell me the exact value for $\cos\frac{\pi}{7}$ or is there none?

[quote="vishalarul"][quote="cincodemayo5590"][hide="2"]
Again, let the product equal $P$.

Now, use the formula $\cos \left( \frac{k \pi}{7}\right) = \frac{ n^{k}-n^{-k}}{2}$ where $n = e^{ \frac{i \pi}{7}}$. Similar to problem 1, this means that $n^{7}=1$, so that we can reduce powers, and $n^{6}+n^{5}+n^{4}+n^{3}+n^{2}+n=-1$. We have that:
\[8P = (n^{1}-n^{-1})(n^{2}-n^{-2})(n^{3}-n^{-3}) \]

\[8P=(n^{6}+n)(n^{5}+n^{2})(n^{4}+n^{3}) = n^{6}+n^{5}+n^{4}+n^{3}+n^{2}+n+2 = 1 \]
and thus $\boxed{P=\frac{1}{8}}$.[/hide]

(edited at 4:55 to add #2. edited at 5:00 for typesetting and typo corrections.)[/quote]

Wait, but that's cosine! The identity is:
$\cos x = \frac{e^{ix}+e^{-ix}}{2}$,

not

$\cos x = \frac{e^{ix}-e^{-ix}}{2}$[/quote]
Oops, typo. You can see in later lines that it is. :) (It's fixed now. Thanks.)

Here's a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149204]similar problem[/url].

Refer to Altheman's post for the complex number method.

[quote="shawn1"]can someone tell me the exact value for $\cos\frac{\pi}{7}$ or is there none?[/quote]

Click [url=http://jimvb.home.mindspring.com/TheSeventhRootofUnity.pdf]here[/url].

Hope that helps (even though it is not practical for the problem).

heres a fun method for 2:

note that cos(3pi/7) = -cos(4pi/7)

x = -cos(pi/7)cos(2pi/7)cos(4pi/7)
8xsin(pi/7) = -8sin(pi/7)cos(pi/7)cos(2pi/7)cos(4pi/7)
8xsin(pi/7) = -4sin(2pi/7)cos(2pi/7)cos(4pi/7)
8xsin(pi/7) = -2sin(4pi/7)cos(4pi/7)
8xsin(pi/7) = -sin(8pi/7) = -sin(pi/7)
x = - 1/8

[quote="da ban man"]If its something like adding and subtracting like

cos(pi/7)-cos(2pi/7)+cos(3pi/7) would it be the same idea?[/quote]

[hide="Yup."]
I'm going to use $\omega=e^{ \frac{i \pi}{7}}$ as the variable instead. We have $\cos \left( \frac{k \pi}{7}\right) = \frac{ \omega^{k}+\omega^{-k}}{2}$ and $\omega^{7}=-1$.

So, our sum $S$ is equal to:
\[S= \frac{ w^{1}+w^{-1}-(w^{2}+w^{-2})+w^{3}+w^{-3}}{2}= \frac{ \left(-\omega^{6}+\omega^{5}-\omega^{4}+\omega^{3}-\omega^{2}+\omega-1 \right)+1 }{2}\]\[S= \frac{-\left(\frac{w^{7}+1}{w+1}\right)+1}{2}= \boxed{\frac{1}{2}}\]
[/hide]

[u]Generalizations:[/u]
\[\boxed{ \sum_{k=1}^{n}(-1)^{k+1}\cos \left( \frac{k \pi}{2n+1}\right) = \frac{1}{2}\ \ \ \forall n \in \mathbb{Z}^{+}}\]\[\boxed{ \sum_{k=1}^{n}\sin \left( \frac{k \pi}{2n+1}\right) = \frac{1+\omega}{2i-2i\omega}\ \ \ \forall n \in \mathbb{Z}^{+}, \ \ \omega = e^{ \frac{i \pi}{2n+1}}}\]\[\boxed{ \prod_{k=1}^{n}\cos \left( \frac{k \pi}{2n+1}\right) = \frac{1}{2^{n+1}}\ \ \ \forall n \in \mathbb{Z}^{+}}\] \[\boxed{ \prod_{k=1}^{2n}\sin \left( \frac{k \pi}{2n+1}\right) = \frac{2n+1}{(-4)^{n}}\ \ \ \forall n \in \mathbb{Z}^{+}}\]
(Can we think of any more? :) )


integration
logarithms
trigonometry
calculus
calculus computations

I have some problems. Maybe they are easy for you, but I can solve them.
$\int_{0}^{1/2\,\pi }\!\ln \left( \sin \left( x \right) \right){dx}$
$\int_{0}^{\infty }\! \left( 1+{x}^{\alpha}\left( \sin \left( x \right) \right)^{2}\right)^{-1}{dx}$
$\int_{0}^{1}\!{\frac{\arcsin \left( x \right) }{x}}{dx}$

Let`s start with number 3). Integrate by parts and you should get that:

$\int_{0}^{1}\frac{\arcsin x}{x}dx=\arcsin x \ln x|^{1}_{0}-\int_{0}^{1}\frac{\ln x}{\sqrt{1-x^{2}}}dx$
$=-\int_{0}^{1}\frac{\ln x}{\sqrt{1-x^{2}}}dx$.

Use the substitution $x=\sin t$ and you should get that the integral is equal to:

$-\int_{0}^{\frac{\pi}{2}}\ln \sin t dt=\frac{\pi \ln 2}{2}$.

2) I think that for number 2) you only need to study the convergency.

You need to study the convergency of $I(\alpha)$  according to the values of $\alpha$.

If $\alpha=0$ the integral diverges.

If $\alpha=-\beta$ where $\beta >0$ then the integral diverges since

${\int_{0}^{\infty}\frac{dx}{1+x^{\alpha}\sin^{2}x}}=\int_{0}^{\infty}\frac{x^{\beta}}{x^{\beta}+(\sin^{2}x)^{2}}\geq \int_{0}^{\infty}\frac{x^{\beta}}{x^{\beta}+1}dx=\infty.$
The only case you need to study is when $\alpha>0$.

[quote="didilica"]Definitely we need to study the convergency of your integral for the case when $\alpha >0$, since otherwise the integral diverges.

Now observe that:

$\int_{0}^{\infty}\frac{dx}{1+x^{\alpha}(\sin(x))^{2}}>\int_{0}^{\infty}\frac{dx}{1+x^{\alpha}}=\infty$ if $\alpha <1$. We need to study convergency for $\alpha \geq 1$.
Claim that the integral converges for $\alpha > 2$ and diverges for $0<\alpha \leq 2$.

First observe that $\frac{2}{\pi}x<\sin x <x$ for $0<x<\frac{\pi}{2}$.

I will use from now on the notation $a\approx b$ which mean that the two quantities are the same or have the same nature (in case of integrals).

A calculation shows that

$I=\int_{o}^{\infty}\frac{dx}{1+x^{\alpha}(\sin x)^{2}}=\sum_{k=0}^{\infty}\int_{0}^{\pi}\frac{dx}{1+(k\pi+x)^{\alpha}(\sin(x))^{2}}$

Or

$I=\sum_{k=0}^{\infty}\int_{0}^{\pi/2}\frac{dx}{1+(k\pi+x)^{\alpha}(\sin (x))^{2}}+\int_{0}^{\pi/2}\frac{dx}{1+(k\pi+\pi-x)^{\alpha}(\sin(x))^{2}}$
Now
$\int_{o}^{\pi/2}\frac{dx}{1+(k\pi+x)^{\alpha}(\sin(x))^{2}}\approx \int_{0}^{\pi/2}\frac{dx}{1+k^{\alpha}ay^{2}}\approx \frac{\arctan k^{\alpha/2}}{k^{\alpha/2}}$
The second integral can be estimated in the same way.
Therefore your integral behaves the same as

$S=\sum\frac{\arctan k^{\alpha/2}}{k^{\alpha/2}}<\sum\frac{\pi/2}{k^{\alpha/2}}<\infty$ if $\alpha >2$
If $\alpha/2\leq 1$ then $S>\sum_{k=1}^{\infty}\frac{\pi/4}{k^{\alpha/2}}=\infty$.[/quote]

[quote="Kent Merryfield"][quote="didilica"]
$\int_{o}^{\pi/2}\frac{dx}{1+(k\pi+x)^{\alpha}(\sin(x))^{2}}\approx \int_{0}^{\pi/2}\frac{dx}{1+k^{\alpha}ay^{2}}\approx \frac{\arctan k^{\alpha/2}}{k^{\alpha/2}}$
The second integral can be estimated in the same way.
Therefore your integral behaves the same as

$S=\sum\frac{\arctan k^{\alpha/2}}{k^{\alpha/2}}<\sum\frac{\pi/2}{k^{\alpha/2}}<\infty$ if $\alpha >2$
If $\alpha/2\leq 1$ then $S>\sum_{k=1}^{\infty}\frac{\pi/4}{k^{\alpha/2}}=\infty$.[/quote][/quote]
Can you explain it more clearly? :(


5.  For any positive integer n, let t(n) be the (non-negative) difference between the digit sums of n and n+2. For example t(199) = |19 - 3| = 16. How many possible values t(n) are less than 2000?

Wouldn't it be like 9999999999...
For the difference of the sums of digits to get beyond 2000, one would have to have a pretty large number.

I don't understand why the answer is only 223.

it is asking for the possible values of $t(n)$, not $n$, that are less than $2000$

Well, here's why. For most numbers, the difference will be simply $2$, since there's no carry-over in adding (e.g. sum of the digits of $14$ is $5$, sum of the digits of $14+2$ is $7$ and $7-5=2$).

It's easier if we explore the numbers $n$ and $n-2$ (makes no difference). The problem arises with $n$'s that end up in zeroes or in zeroes with exactly one $1$, since when you subtract $2$, you get a lot of nines and possibly one eight and the sum of digits is not straightforward anymore. So assume we have such a number that ends up in $k$ zeroes:

$xxx\dots xxp\underbrace{00\dots 0}_{k}$

(here $x$'s denote unimportant digits and $p$ is an important digit which is at least $1$)

If we subtract $2$ from it, we'll get

$xxx\dots xx(p-1)\underbrace{99\dots 9}_{k-1}8$

since carry-over will go all the way up to $p$ and reduce it by one. Since all the other digits are the same, we can calculate the difference only for the digits from $p$ rightwards. In the first number the sum is $p+k\cdot 0=p$, and in the second it's $p-1+9(k-1)+8=p+9(k-1)+7$. The difference between these two is $9(k-1)+7.$ The same result is obtained for a number of the form $xxx\dots xxp\underbrace{00\dots 0}_{k-1}1$, the proof of which I leave to you as an exercise.

Now, we've seen that the desired differences can be only of the form $9(k-1)+7$, and we want to see how many of them is less than $2000.$ Simple:

\begin{eqnarray*}9(k-1)+7<2000\implies 9(k-1)<1993\implies k-1<221\frac{4}{9}\implies k-1\leqslant 221\implies k\leqslant 222\end{eqnarray*}

Therefore, there are $222$ such values...

No, wait! We almost forgot that precious little $2$ from the beginning :) Therefore the right answer is $223$.

Not really, that would be a question for U.S. guys :)

[quote="lightfire22000"]Thanks.  Do you know some good resources and guides, online for preparing for the AIME.  I'm in AP Calc, but new to these kinds of questions.[/quote]

[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]Practice tests.[/url]
[url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]AoPS Classes.[/url]


what is
$ \frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}\frac{1}{16}\plus{}...\plus{}\frac{1}{\infty}$
:ddr:  :jump:  :juggle:

[hide="answer"]$ 1$
its just an infinite geometric series
[/hide]

$ \sum_{i \equal{} 1}^\infty{\left(\frac {1}{2}\right)^i} \equal{} \frac {1}{1 \minus{} \frac {1}{2}} \minus{} 1 \equal{} 2 \minus{} 1 \equal{} 1$

This is just a convergent geometric series.

$ \sum_{i \equal{} 0}^\infty{(a \cdot r^i)} \equal{} \frac {a}{1 \minus{} r}$ for all $ 0 < r < 1$
Therefore,
$ \sum_{i \equal{} 0}^\infty{(a \cdot r^i)} \equal{} \frac {a}{1 \minus{} r} \minus{} 1$

Note:$ \frac{1}{2}\plus{} \frac{1}{4}\plus{}...\equal{} \frac{1}{2}\plus{} \frac{1}{2}( \frac{1}{2}\plus{} \frac{1}{4}\plus{}...)$

If we let $ \frac{1}{2}\plus{} \frac{1}{4}\plus{}...\equal{}x$
Then we get:
$ \frac{1}{2}\plus{}\frac{1}{2}x\equal{}x$
Solution is $ x\equal{}1$
Therefore the sum is 1.

[quote="NapoleonXIV"]Note:$ \frac {1}{2} + \frac {1}{4} + ... = \frac {1}{2} + \frac {1}{2}( \frac {1}{2} + \frac {1}{4} + ...)$

If we let $ \frac {1}{2} + \frac {1}{4} + ... = x$
Then we get:
$ \frac {1}{2} + \frac {1}{2}x = x$
Solution is $ x = 1$
Therefore the sum is 1.[/quote]

Your solution reminds me of the infinite square root problem...

$ \sqrt{1 + \sqrt {1 + \sqrt {1 + \sqrt {1 + \sqrt {1 + \sqrt {1 + \sqrt {1 + ...}}}}}}} = x$

Then we have
\begin{align*} x^2 & = x + 1 \\
x^2 - x - 1 & = 0 \\
x & = \frac {1 + \sqrt {5}}{2} \end{align*}

[quote="mathking123"]napolean, your logic is correct, but sometimes that logic of setting up an algebra equation leads to a contradiction, for example:

let n=1+2+4+8+16+....2^n
as n goes to infinity....
1+2+4+8.....=1+2(1+2+4+8+....)
so n=1+2n
which implies that n=-1 which of course is not true..
see if you can see whats wrong with this "proof"[/quote]

The sum is unbounded anyways (as in it doesn't converge), so this whole "proof" doesn't work.

I think, you cant use this method if the limit doesn't exist for this series (or something like that  :?: ).

watch the notation, how would you define $ \frac{1}{\infty}$?

yeah, the sum $ 1\plus{}2\plus{}4\plus{}8\plus{}...$ is divergent, so the proof breaks down.

note that just because the sequence $ \frac{1}{2}$, $ \frac{1}{4}$, $ \frac{1}{8}$, ... is convergent, doesn't mean the series with those terms is.  here, it doesn't converge, but just saying.

[quote="alanchou"]watch the notation, how would you define $ \frac {1}{\infty}$?[/quote]

I was wondering about that, too.

[quote]note that just because the sequence $ \frac {1}{2}$, $ \frac {1}{4}$, $ \frac {1}{8}$, ... is convergent, doesn't mean the series with those terms is.  here, it doesn't converge, but just saying.[/quote]

But the converse is true. If the series converges, then the terms will converge to zero. :D

[quote="Yongyi781"]

But the converse is true. If the series converges, then the terms will converge to zero. :D[/quote]

unfortunately, some people have gotten the idea of proving the converse of a statement instead of the original one


group theory
abstract algebra
superior algebra
superior algebra solved

let $\phi:G\rightarrow G^{\prime}$ be a homomorphism and $H^{\prime}\triangleleft G^{\prime}$. Show that $H=\phi^{-1}(H^{\prime})\triangleleft G$.

It is clear that the associativity holds in $H$.

$e'\in H'$ then $e\in H$. $H$ has identity.

For $a\in H$ and $b\in H$, $\phi(a)\in H'$ and $\phi(b)\in H'$, thus $\phi(ab)=\phi(a)\phi(b)\in H'$. Hence $ab\in H$. We know that $H$ is closed under multiplication.

For $a\in H$, $e'=\phi(aa^{-1})=\phi(a)\phi(a^{-1})$. $\phi(a)\in H'$, hence $\phi(a^{-1})=(\phi(a))^{-1}\in H'$, thus $a^{-1}\in H$.

Hence $H$ is a group. Now we prove that it is normal.

For $x\in G$ and $h\in H$, $\phi(xhx^{-1})=\phi(x)\phi(h)\phi(x^{-1})=\phi(x)\phi(h)(\phi(x))^{-1}\in H'$ because $H'$ is normal in $G'$. Hence $xhx^{-1}\in H$. It follows that $H$ is normal.