inequalities
function
calculus
derivative
algebra
polynomial
inequalities proposed

If a,b,c is positive numbers with abc=1, then \[ \frac{ab}{1\plus{}c^4}\plus{}\frac{bc}{1\plus{}a^4}\plus{}\frac{ca}{1\plus{}b^4}\ge \frac{3}{2}\]

After using Am-Gm and Cauchy Schwart, it's becomes:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{} \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \geq\ 2$ (always trues)
 :)

$ 2^{nd}$ proof:
Let $ x\equal{} \frac{a}{b}$...
By Cauchy Schwart, we only need to prove:
$ \frac{2}{3} (a^3\plus{}b^3\plus{}c^3)^2 \geq\ \sum\ ab(a^4\plus{}b^4)$
There are many ways to prove it  :)

[quote="nguoivn"]$ 2^{nd}$ proof:
Let $ x \equal{} \frac {a}{b}$...
By Cauchy Schwart, we only need to prove:
$ \frac {2}{3} (a^3 \plus{} b^3 \plus{} c^3)^2 \geq\ \sum\ ab(a^4 \plus{} b^4)$
There are many ways to prove it  :)[/quote]
But it's wrong. :(  Try $ a \equal{} 2$ and $ b \equal{} c \equal{} 1.$ :wink:

With same conditions $ \frac {ab}{1 \plus{} c^7} \plus{} \frac {bc}{1 \plus{} a^7} \plus{} \frac {ca}{1 \plus{} b^7}\ge \frac {3}{2}$ is true too.
It seems that $ \frac {ab}{1 \plus{} c^{14}} \plus{} \frac {bc}{1 \plus{} a^{14}} \plus{} \frac {ca}{1 \plus{} b^{14}}\ge \frac {3}{2}$ is true .
But $ \frac {ab}{1 \plus{} c^{15}} \plus{} \frac {bc}{1 \plus{} a^{15}} \plus{} \frac {ca}{1 \plus{} b^{15}}\ge \frac {3}{2}$ is wrong already: $ a \equal{} \frac {1}{1.15^2}$ and $ b \equal{} c \equal{} 1.15.$

Sorry, arquady. The $ 2^{nd}$ proof was false.
I 'll post full first proof:
Lenma: $ 1 \plus{} x^4 \geq\ x(x^2 \plus{} 1)$
$ LHS \equal{} ab \plus{} bc \plus{} ca \minus{} \sum\ \frac {c^3}{1 \plus{} c^4} \geq\ ab \plus{} bc \plus{} ca \minus{} \sum\ \frac {c^2}{c^2 \plus{} 1} \geq\ \frac {3}{2}$
$ \Leftrightarrow ab \plus{} bc \plus{} ca \plus{} \sum\ \frac {1}{1 \plus{} a^2} \geq\ \frac {9}{2}$
Let $ a \equal{} \frac {x}{z}$...
It's becomes: $ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \plus{} \sum\ \frac {x^2}{x^2 \plus{} y^2} \geq\ \frac {9}{2}$
Using Cauchy Schwart:
$ LHS \geq\ \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {(x \plus{} y \plus{} z)^2}{2(x^2 \plus{} y^2 \plus{} z^2)} \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {xy \plus{} yz \plus{} zx}{x^2 \plus{} y^2 \plus{} z^2 } \plus{} \frac {5}{2} \geq\ \frac {9}{2}$

[quote="Nguyen Ngoc Linh"]Your proofs are really magnificent  :rotfl: . Remember, not every mathlinkers is as smart as you are.[/quote]
No, I am only a lazy man  :)

[quote="arqady"]
With same conditions $ \frac {ab}{1 \plus{} c^7} \plus{} \frac {bc}{1 \plus{} a^7} \plus{} \frac {ca}{1 \plus{} b^7}\ge \frac {3}{2}$ is true too.
It seems that $ \frac {ab}{1 \plus{} c^{14}} \plus{} \frac {bc}{1 \plus{} a^{14}} \plus{} \frac {ca}{1 \plus{} b^{14}}\ge \frac {3}{2}$ is true .
But $ \frac {ab}{1 \plus{} c^{15}} \plus{} \frac {bc}{1 \plus{} a^{15}} \plus{} \frac {ca}{1 \plus{} b^{15}}\ge \frac {3}{2}$ is wrong already: $ a \equal{} \frac {1}{1.15^2}$ and $ b \equal{} c \equal{} 1.15.$[/quote]

We can use the following way to prove  :maybe:
Besides, with $ k \equal{} 7$ , if using Cauchy Schwart, we 'll have to prove:
$ \frac {2}{3}(a^4 \plus{} b^4 \plus{} c^4)^2 \geq\  \sum\ ab(a^6 \plus{} b^6)$
Is it trues ?  :wink:

[quote="nguoivn"]
We can use the following way to prove  :maybe:
Besides, with $ k \equal{} 7$ , if using Cauchy Schwart, we 'll have to prove:
$ \frac {2}{3}(a^4 \plus{} b^4 \plus{} c^4)^2 \geq\ \sum\ ab(a^6 \plus{} b^6)$
Is it trues ?  :wink:[/quote]
Wrong. Try $ a \equal{} 2$ and $ b \equal{} c \equal{} 1.$
[quote="Vasc"]This is also true.

If a,b,c are positive real numbers with abc=1, then
\[ \frac {ab}{2 \plus{} c^4} \plus{} \frac {bc}{2 \plus{} a^4} \plus{} \frac {ca}{2 \plus{} b^4}\ge 1
\]
[/quote]
I have proved that with same conditions and for all $ k\geq0$ holds:
\[ \frac {ab}{k \plus{} c^4} \plus{} \frac {bc}{k \plus{} a^4} \plus{} \frac {ca}{k \plus{} b^4}\ge \frac{3}{k\plus{}1}
\]
 :maybe:

[quote="arqady"]
I have proved that with same conditions and for all $ k\geq0$ holds:
\[ \frac {ab}{k \plus{} c^4} \plus{} \frac {bc}{k \plus{} a^4} \plus{} \frac {ca}{k \plus{} b^4}\ge \frac {3}{k \plus{} 1}
\]
:maybe:[/quote]
Me,too. But for $ k \equal{} 2$ there is a nicer solution. :)

[quote="Vasc"]This is also true.

If a,b,c are positive real numbers with abc=1, then
\[ \frac {ab}{2 \plus{} c^4} \plus{} \frac {bc}{2 \plus{} a^4} \plus{} \frac {ca}{2 \plus{} b^4}\ge 1
\]
[/quote]
hi Vasc  :)
your inequality is equivalent to : $ \sum{\frac{x^3}{2x^2\plus{}y^2z^2}} \geq 1$ ( with the same conditions!)
now by cauchy-shwarz : $ LHS (\sum{x(2x^2\plus{}y^2z^2)}) \geq (x^2\plus{}y^2\plus{}z^2)^2$
hence it sufficies to prove that: $ (x^2\plus{}y^2\plus{}z^2)^2 \geq 2(x^3\plus{}y^3\plus{}z^3) \plus{}xyz(x\plus{}y\plus{}z)$
or : $ (x^2\plus{}y^2\plus{}z^2)^2 \geq 2(x^3\plus{}y^3\plus{}z^3)(xyz)^{1/3} \plus{} xyz(x\plus{}y\plus{}z)$
it sufficies to show the stronger :  $ (x^2\plus{}y^2\plus{}z^2)^2 \geq  \frac{2(x^3\plus{}y^3\plus{}z^3)(x\plus{}y\plus{}z)}{3} \plus{} x^2y^2\plus{}x^2z^2\plus{}y^2z^2$
that is : $ \frac{(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}xz)^2}{3} \geq 0$ , which is obvious
for the general case : notice that the inequality is equivalent to : $ \sum{\frac{x^5}{kx^4\plus{}1} \minus{} \frac{3}{k\plus{}1}} \geq 0$ , we put $ x\equal{}e^{a}$ ,and Jensen finishes the proof  :)
I hope that I didn't any mistake  :(

[quote="Vasc"][quote="anas"] your inequality is equivalent to : $ \sum{\frac {x^3}{2x^2 \plus{} y^2z^2}} \geq 1$ ( with the same conditions!)([/quote]
Why ?

And why Jensen, since $ f$ is not convex ?[/quote]

for the first one , put $ x\equal{}ab$ etc.......
for the second one , after putting $ x\equal{}e^{a}$ , consider the function $ f(a)\equal{}\frac{e^{5a}}{ke^{4a}\plus{}1} \minus{} \frac{1}{k\plus{}1}$  the second derivative is : $ \frac{e^{5a}(k^2(e^{4a})^2\minus{}6ke^{4a}\plus{}25)}{(ke^{4a}\plus{}1)^3}$  :wink:

Indeed, you are right Anas. The inequality is trivial by Jensen. :wink:

But we can't apply Jensen for the following :

$ \frac {ab}{2 \plus{} c^5} \plus{} \frac {bc}{2 \plus{} a^5} \plus{} \frac {ca}{2 \plus{} b^5}\ge 1$,

where $ abc\equal{}1$.

[quote="anas"][quote="Vasc"]This is also true.

If a,b,c are positive real numbers with abc=1, then
\[ \frac {ab}{2 \plus{} c^4} \plus{} \frac {bc}{2 \plus{} a^4} \plus{} \frac {ca}{2 \plus{} b^4}\ge 1
\]
[/quote]
hi Vasc  :)
your inequality is equivalent to : $ \sum{\frac {x^3}{2x^2 \plus{} y^2z^2}} \geq 1$ ( with the same conditions!)
now by cauchy-shwarz : $ LHS (\sum{x(2x^2 \plus{} y^2z^2)}) \geq (x^2 \plus{} y^2 \plus{} z^2)^2$
[quote]hence it sufficies to prove that: $ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq 2(x^3 \plus{} y^3 \plus{} z^3) \plus{} xyz(x \plus{} y \plus{} z)$[/quote]
or : $ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq 2(x^3 \plus{} y^3 \plus{} z^3)(xyz)^{1/3} \plus{} xyz(x \plus{} y \plus{} z)$
it sufficies to show the stronger :  $ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq \frac {2(x^3 \plus{} y^3 \plus{} z^3)(x \plus{} y \plus{} z)}{3} \plus{} x^2y^2 \plus{} x^2z^2 \plus{} y^2z^2$
that is : $ \frac {(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} xz)^2}{3} \geq 0$ , which is obvious
for the general case : notice that the inequality is equivalent to : $ \sum{\frac {x^5}{kx^4 \plus{} 1} \minus{} \frac {3}{k \plus{} 1}} \geq 0$ , we put $ x \equal{} e^{a}$ ,and Jensen finishes the proof  :)
I hope that I didn't any mistake  :([/quote]

[quote]hence it sufficies to prove that: $ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq 2(x^3 \plus{} y^3 \plus{} z^3) \plus{} xyz(x \plus{} y \plus{} z)$[/quote]

 Why?

I made a stupid error  :blush:  ,I'm always hurry  :mad:  , I've corrected my proof :

your inequality is equivalent to : $ \sum{\frac {x^3}{2x^2 \plus{} y^2z^2}} \geq 1$ ( with the same conditions!)
now by cauchy-shwarz : $ LHS (\sum{x(2x^2 \plus{} y^2z^2)}) \geq (x^2 \plus{} y^2 \plus{} z^2)^2$
hence it sufficies to prove that: $ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq 2(x^3 \plus{} y^3 \plus{} z^3) \plus{} yz\plus{}xz\plus{}xy$  ( since $ xyz\equal{}1$)
it sufficies to show the stronger : $ (x^2\plus{}y^2\plus{}z^2)^2 \geq  2(x^3\plus{}y^3\plus{}z^3)\frac{x\plus{}y\plus{}z}{3} \plus{} (yz\plus{}xz\plus{}xy)\frac{ (x\plus{}y\plus{}z)^2}{9}$ , which is easy to prove by mixing variables , because  $ f(x,y,z)\minus{} f(x ,\frac{y\plus{}z}{2} , \frac{y\plus{}z}{2}) \geq 0$ is true for $ x\equal{}max(x,y,z)$

[quote="Vasc"]But we can't apply Jensen for the following :

$ \frac {ab}{2 \plus{} c^5} \plus{} \frac {bc}{2 \plus{} a^5} \plus{} \frac {ca}{2 \plus{} b^5}\ge 1$,

where $ abc \equal{} 1$.[/quote]
The Cauchy Schwarz still works here. here is it :)

Using anas' method, we shall prove :

if $ x,y,z$ be positive numbers such that $ xyz \equal{} 1$ and  $ k\geq 0$, then

$ \frac {yz}{k \plus{} x^p} \plus{} \frac {zx}{k \plus{} y^p} \plus{} \frac {xy}{k \plus{} z^p}\geq \frac {3}{k \plus{} 1}$

holds for $ 2 \minus{} 2\sqrt {2}\leq p\leq 2 \plus{} 2\sqrt {2}$.
[hide="Proof."]Substitute $ x \equal{} e^a,y \equal{} e^b,z \equal{} e^c$, the original inequality is equivalent to

$ \frac {3}{k \plus{} 1}\leq \sum{\frac {e^{b \plus{} c}}{k \plus{} e^{pa}}} \equal{} \sum{\frac {e^{ \minus{} a}}{k \plus{} e^{pa}}}\equiv \sum{f(a)}$

with $ a \plus{} b \plus{} c \equal{} 0$. It's trivial by Jensen inequality since

$ f''(a) \equal{} \frac {k^2 \minus{} \left(p^2 \minus{} 2p \minus{} 2\right)ke^{pa} \plus{} (p \plus{} 1)^2e^{2pa}}{e^a\left(k \plus{} e^{pa}\right)^3}$

$ \equal{} \frac {\left[2k \minus{} \left(p^2 \minus{} 2p \minus{} 2\right)e^{pa}\right]^2 \plus{} p^2\left(4 \plus{} 4p \minus{} p^2\right)e^{2pa}}{4e^a\left(k \plus{} e^{pa}\right)^3}\geq 0$

when $ 2 \minus{} 2\sqrt {2}\leq p\leq 2 \plus{} 2\sqrt {2}$.[/hide]
Let $ x,y,z$ be positive numbers such that $ xyz \equal{} 1$ and $ p\geq 2 \minus{} 2\sqrt {2}$, prove or disprove :

$ \frac {yz}{k \plus{} x^p} \plus{} \frac {zx}{k \plus{} y^p} \plus{} \frac {xy}{k \plus{} z^p}\geq \frac {3}{k \plus{} 1}$

holds if and only if $ 0\leq k\leq \frac {1}{2}$.

If $ x,y,z$ be positive numbers such that $ xyz \equal{} 1$, then

$ \frac {yz}{k \plus{} x^5} \plus{} \frac {zx}{k \plus{} y^5} \plus{} \frac {xy}{k \plus{} z^5}\geq \frac {3}{k \plus{} 1}$

holds if and only if $ 0\leq k \leq k_1\equal{}3.8193725239\cdots$ or $ k\geq k_2\equal{}9.6722267758\cdots$,

where $ k_1$ and $ k_2$ are two roots of the irreducible polynomial

$ 32000000000000000k^{33}\minus{}4164907500000000000k^{32}$

$ \plus{}108670789214853515625k^{31}\minus{}1256797627930842690625k^{30}$

$ \plus{}7493064794466948225000k^{29}\minus{}25179783180321850391250k^{28}$

$ \plus{}69998100809896006842966k^{27}\minus{}166282369027994424494619k^{26}$

$ \plus{}248104421159487398937774k^{25}\minus{}400710378866130892958873k^{24}$

$ \plus{}396257666090594064385257k^{23}\minus{}509491530874008328479924k^{22}$

$ \minus{}346738950771159102749972k^{21}\minus{}2408825170503114333783615k^{20}$

$ \minus{}4189941102747476571660963k^{19}\minus{}6765634615405559909195838k^{18}$

$ \minus{}8362769665280304962816235k^{17}\minus{}8896513381195451506877313k^{16}$

$ \minus{}7632962071922521639315144k^{15}\minus{}4837463042034085759524150k^{14}$

$ \minus{}1695616682343142863021315k^{13}\plus{}395117094548260011146381k^{12}$

$ \plus{}927599047555212875891916k^{11}\plus{}634580917305149169698544k^{10}$

$ \minus{}24826238628013847693808k^9\minus{}1021031023301135926285248k^8$

$ \minus{}1866256679324080330596864k^7\minus{}1987337463714132235990272k^6$

$ \minus{}1436664498839673863823360k^5\minus{}735887569963796015874048k^4$

$ \minus{}267609651620487080214528k^3\minus{}66701786894907139031040k^2$

$ \minus{}10326857903621496373248k\minus{}755623749045475344384$.


[quote="thanhnam2902"]I think you can use Drichlet to done. I will try it and post soon.[/quote]

I have a feeling this will be tough...  :P If I'm not getting this wrong, the opening post says:
[quote="Yuriy Solovyov"]$ 100$ positive real numbers
(...)
$ \textbf{(2)}\; x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{100} < 100$
(...)
Prove that there exist numbers $ x_i,\; x_j,\; x_k$ such that $ x_i \plus{} x_j \plus{} x_k > 100$ for some $ 1\leq i,\;j,\;k\leq 100$.[/quote]

However, that's not really possible, since all x's are positive and, if $ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{100} < 100$ then $ x_i\plus{}x_j\plus{}x_k<x_1\plus{}x_2\plus{}...\plus{}x_{100}<100$ (if $ i$, $ j$ and $ k$ are different), so that's a contradiction. If $ i$ and $ j$, for example, can be equal, I can still find a counterexample:
$ x_1\equal{}x_2\equal{}...\equal{}x_{97}\equal{}10^{\minus{}2}$
$ x_{98}\equal{}31$
$ x_{99}\equal{}32$
$ x_{100}\equal{}33$.
As we can see, both conditions are satisfied and the maximum sum of three numbers (not even necessarily different) is $ x_{100}\plus{}x_{100}\plus{}x_{100}\equal{}99<100$. All other sums are smaller than this one. Contradiction.

I'm sorry.... :blush:  :blush:  :blush:
Here sholud be $ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{100} < 300$
I'll edit it.

Sorry, I'm stupid, it's $ 10000$ in the first condition, not $ 1000$. Therefore, my counter-example doesn't hold. Sorry once again. I'll try to solve the problem.


trigonometry
limit
LaTeX
real analysis
real analysis unsolved

Prove that
\begin{eqnarray*} \sum_{n = 1}^{\infty} \left( \frac{2 + \sin ( n )}{3} \right)^{n} & = & + \infty \end{eqnarray*}
[Edit: Corrected a "minor" ;) slip]

[quote="{x}"]Prove that
\begin{eqnarray*} \sum_{n = 1}^{\infty} \left( \frac{2 + \sin ( n )}{3} \right) & = & + \infty \end{eqnarray*}[/quote]
What do the brackets $\left( \right)$ mean here ? If this $a_n =\frac{2 + \sin ( n )}{3}$ is just the $n$'th term of the series $\sum_{n=1}^\infty a_n$ then the series surely doesn't convergce, because $\lim a_n$ isn't only $0$, but even doesn't exist.

The form of this sum reminds me of a much tougher series in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093[/url] that we had proved convergence for earlier.

Do you have a link to the Efnet #math problem?

eugene: Correct, my argument was more complicated, and used the trivial fact that $\frac{2+sin(n)}{3} \geq sin(n)$.
RaviB: Yes, this was the original problem. The author of the problem (landen) suggested that he created this problem by substituing $(-1)^n$ by $sin(n)$ in a divergent serie.  Here is a link, http://www.efnet-math.org/w/POTD#Friday.2C_Apr._21 . The problem was not solved, by #math-ers.

The problem that {x} linked to is the same as the problem that Ravi linked to, and is not the problem that is written here.

[quote="{x}"]Prove that
\begin{eqnarray*} \sum_{n = 1}^{\infty} \left( \frac{2 + \sin ( n )}{3} \right) & = & + \infty \end{eqnarray*}[/quote]

Just because $\frac{2+\sin{(n)}}{3}\geq{\frac{1}{3}}$?

Actually what made my problem absolutely dull, is that I forgot the "exponent n" over the bracker :)

Similar problem: if http://www.mathlinks.ro/Forum/viewtopic.php?t=22093

Anyway, I believe this problem can be solved by just observing that the limit of the terms isn't zero... but you need some sort of approximation on for which n the fraction is close to 1.

Could someone please show that the limit of the terms for the corrected series does not exist? I tried using continued fractions for $\pi$ but I didn't succeed...

You may want to look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90103]here[/url] and, if you still cannot do it, [url=http://www.artofproblemsolving.com/Wiki/index.php/Rational_approximation]here[/url] ;)

Thanks. I have thought through the details and can do it now :). I learned several things I did not know!

Very nice ! I'm happy this problem turned your attention :) I have a somewhat different solutions, which I will hopefully post tomorrow.
TeX Note: Did you type this document in TeXmacs ?

No, I write LaTeX directly as code with Emacs (aucTeX mode).

[quote="Kalle"]Seeing as it turned out to be a lot of work I wrote a long document showing that the limit does not exist: http://lutfisk.se/grans.pdf . I suppose I did it both to practice LaTeX and for my own satisfaction.

The first two lemmata are two theorems from the second page you linked me to, fedja (I suppose only reworded in the way that I understand them). The third lemmata is a rigourised version of what you suggested in the first link (as far as I can tell your argument is not complete). The fourth lemmata and the theorem extend the first three lemmata to show that the limit doesn't exist.

As of now I have not included any references. If this was a "real document" would you reference the proofs? Right now it only says they are well known. It seems that in books there usually aren't references when it comes to theorems, yet I doubt the authors come up with a lot of proofs on their own.

edit: Yes, I realize that this is incredibly pretentious.[/quote]

Kalle, where is that document  :idea:


[size=125]Solve the equation! [/size]

[size=125]$ x\cdot y\plus{}x\plus{}y\plus{}1\equal{}4\cdot\sqrt {xy}$

if  $ x\ge 0, y\ge 0$[/size]

we have $ xy\plus{}x\plus{}y\plus{}1 \equal{}4\sqrt{xy}$
$ \Leftrightarrow (x\plus{}1)(y\plus{}1) \equal{}4\sqrt{xy}$
but by AM-GM $ x\plus{}1 \geq 2\sqrt{xy}$ and $ y\plus{}1\geq \sqrt{y} \rightarrow (x\plus{}1)(y\plus{}1)\geq 4\sqrt{xy}$
then $ x\equal{}y\equal{}1$ because it's the case of equality of AM-GM
Finally $ S\equal{}\{1,1\}$

Not quite, when you multiply those 2 inequalities together, you do not get what was desired.

$ (\sqrt {xy} \minus{} 1)^2 \plus{} (\sqrt {x} \minus{} \sqrt {y})^2 \equal{} 0$

So therefore $ x \equal{} y \equal{} 1$

[hide="solution"]$ x, y \neq 0$ as if so the R.H.S. will be 0, which is impossible

So by AM-GM , $ (xy\plus{}1)\plus{}(x\plus{}y) \ge 2 \sqrt{xy}\plus{}2 \sqrt{xy}\equal{}4 \sqrt{xy}$

equality holds when $ x\equal{}y$, $ xy\equal{}1$, i.e. $ x\equal{}y\equal{}1$[/hide]

[b][color=blue]السلام عليكم ورحمة الله وبركاته[/color][/b]

[quote="peine"]we have $ xy \plus{} x \plus{} y \plus{} 1 \equal{} 4\sqrt {xy}$
$ \Leftrightarrow (x \plus{} 1)(y \plus{} 1) \equal{} 4\sqrt {xy}$
but by AM-GM $ x \plus{} 1 \geq 2\sqrt {xy}$ and $ y \plus{} 1\geq \sqrt {y} \rightarrow (x \plus{} 1)(y \plus{} 1)\geq 4\sqrt {xy}$
then $ x \equal{} y \equal{} 1$ because it's the case of equality of AM-GM
Finally $ S \equal{} \{1,1\}$[/quote]

Do you mean $ x\plus{}1 \ge 2\sqrt x$ and the same for $ y$ ?


function
integration
limit
real analysis
calculus
real analysis unsolved

Let
\[I \subset R^{n}\]
be a closed bounded rectangle. Prove the following statement:

Let
\[\{ f_{j}\}_{j \ge 0}\]
be a sequence of integrable functions I -> R. Suppose that this sequence converges uniformly to a function f: I -> R. Then f also is integrable, and
\[\int_{I}{f = \mathop{\lim }_{j \to \infty }}\int_{I}{f_{j}}\]

Riemann or Lebesgue integration? I assume it's Riemann. For any partition of $I$ and any choice of sample points, the Riemann sums you get for $f$ and $f_{j}$ are different by at most $\mathrm{vol}(I)\sup_{x\in I}|f_{j}(x)-f(x)|$, which tends to $0$.

R is for all reals and it is riemann integration


A positive integer which contains no prime factors except 2 and 3 can be divided
by exactly 24 different numbers and its square can be divided by exactly 77
different numbers. What is the integer?

I have no idea where to even begin on thins problem without using my calculator to multiply 100 numbers.

Let the number be $ 2^x \cdot 3^y$.

Thus:

$ (x\plus{}1)(y\plus{}1)\equal{}24$
$ (2x\plus{}1)(2y\plus{}1)\equal{}77$

Expanding gives

$ xy\plus{}x\plus{}y\equal{}23$
$ 4xy\plus{}2x\plus{}2y\equal{}76 \implies 2xy\plus{}x\plus{}y\equal{}38$

Thus, $ xy\equal{}15$.

This gives $ x\plus{}y\equal{}8$. Clearly, $ x,y\equal{}5,3$ or $ 3,5$.

So there are 2 different possibilites. $ 2^5 \cdot 3^3 \equal{} 32 \cdot 27 \equal{} 864$, or $ 2^3 \cdot 3^5 \equal{} 8 \cdot 243 \equal{} 1944$.

What you need here is the [i]number of divisors[/i] function, commonly denoted $ d(n)$.

If $ n \equal{} \prod p_i^{k_i}$, we have $ d(n) \equal{} \prod (k_i \plus{} 1)$ (Why?).

This fancy notation doesn't really make sense until you see an example, so here it is:
$ d(12) \equal{} d(2^23^1) \equal{} (2 \plus{} 1)(1 \plus{} 1) \equal{} 3\cdot 2 \equal{} 6$. These are $ 1,2,3,4,6,12$.

Edit: Oh, also, try the High School Basics forum next time, this problem belongs there.

Edit: Sorry, used ∑ instead of ∏  :huh: And yes, each $ p_i$ is a prime (I thought that was obvious but maybe not)

[quote="grn_trtle"]What you need here is the [i]number of divisors[/i] function, commonly denoted $ d(n)$.

If $ n \equal{} \prod p_i^{k_i}$, we have $ d(n) \equal{} \sum (k_i \plus{} 1)$ (Why?).

This fancy notation doesn't really make sense until you see an example, so here it is:
$ d(12) \equal{} d(2^23^1) \equal{} (2 \plus{} 1)(1 \plus{} 1) \equal{} 3\cdot 2 \equal{} 6$. These are $ 1,2,3,4,6,12$.

Edit: Oh, also, try the High School Basics forum next time, this problem belongs there.[/quote]

Should that summation after the d(n) be a product?
Also, i know it is probably assumed, but you might want to say that the p_i's are primes :wink:


U have function $f: \mathbb{N}\times \mathbb{N}\to \mathbb{N}$; $f(m,n)=m+\frac{(m+n)(m+n+1)}{2}$      (bijection).

How to find $(m,n) \in \mathbb{N}\times \mathbb{N}$, which $f(m,n)=6666$?

Can be this solved with some algoritm :?:

goska

Hmmm, what you have there is a bijection from the rationals to the integers. Look at the first few values and see if you can spot a pattern...

[hide]
Let $S=m+n$. Then $f(m,n)=m+\frac{S(S+1)}{2}$. But $0\leq m\leq S$ so $\frac{S(S+1)}{2}\leq f(m,n)\leq \frac{(S+1)(S+2)}{2}-1$. Wow, that proves that $f$ is a bijection and gives an easy way to find $S$ and $m$.

In order for $f(m,n)$ to equal 6666, $S$ must be the greatest integer that satisfies $\frac{S(S+1)}{2}\leq 6666$, or $S(S+1)\leq 13332$. $S$ will be somewhere between 110 and 120, so write $S=110+a$. Then
\[S(S+1)=(110+a)(111+a)=12210+221a+a^{2}\]
This is less than or equal to 13332 when $a=4$, but not when $a=5$, so $S=114$. In this case, $\frac{S(S+1)}{2}=6555$, so $m=111$ and $n=3$ as a result: $\boxed{(m,n)=(111,3)}$.
[/hide]


logarithms
calculus
calculus computations

Determine a closed form for the sum $f(y)=\sum_{k=1}^{\infty}\frac1{y^{2}+k^{2}}$.

[quote="didilica"]I think this is the formula you are looking for:

$\pi\frac{e^{\pi x}+e^{-\pi x}}{e^{\pi x}-e^{-\pi x}}=\frac{1}{x}+2x\sum_{k=1}^{\infty}\frac{1}{x^{2}+k^{2}}$.[/quote]

Yes, brilliant! I had done this before, I used:

\[f(y)=\sum_{k=1}^{\infty}\frac1{y^{2}+k^{2}}= \frac1{2y}\sum_{k=1}^{\infty}\frac{2y}{y^{2}+k^{2}}= \frac1{2y}\sum_{k=1}^{\infty}\frac{d}{dy}\left[ \log \left(y^{2}+k^{2}\right)-\log k^{2}\right]\]

and this is premissable because $\frac{d}{dy}\log k^{2}=0$, so we have

\[f(y) =\frac1{2y}\sum_{k=1}^{\infty}\frac{d}{dy}\log\left(1+\frac{y^{2}}{k^{2}}\right) =\frac1{2y}\frac{d}{dy}\log\prod_{k=1}^{\infty}\left(1+\frac{y^{2}}{k^{2}}\right)\]

and since $\mbox{sinh}z = z\prod_{k=1}^{\infty}\left(1+\frac{z^{2}}{\pi^{2}k^{2}}\right)$ we have

\[f(y)=\frac1{2y}\frac{d}{dy}\log\left(\frac{\mbox{sinh}\left(\pi y\right)}{\pi y}\right) =\frac1{2y}\cdot \frac{\pi y}{\mbox{sinh}\left(\pi y\right)}\left(\frac{\pi^{2}y\mbox{cosh}\left(\pi y\right)-\pi\mbox{sinh}\left(\pi y\right)}{\pi^{2}y^{2}}\right)\]
\[=\frac1{2y^{2}}\left(\pi y\mbox{coth}\left(\pi y\right)-1\right)\]


What is the positive difference, in inches, between the greatest cabin
height and the least cabin height of the following business jets?
\[ \begin{tabular}{|c|c|} \hline
\textbf{Business Jet} & \textbf{Cabin Height} \\ \hline
Learjet 31A & 4 feet, 5 inches \\ \hline
Learjet 60 & 5 feet, 8 inches \\ \hline
Cessna Citation V & 4 feet, 7 inches \\ \hline
Gulfstream V & 6 feet, 2 inches \\ \hline
\end{tabular}\]

The smallest height is 4 feet, 5 inches, which = 53 inches. The largest height (I crave a better word) is 6 feet, 2 inches, which is 74 inches. 74 - 53 =  [b]21[/b].
qed


[quote="hydro"]find all positive integers $x,y,z$ s.t. $x+y+z=xyz$[/quote]

I think ordering helps here.

[hide] Assume wlog that $x \geq y \geq z$
Clearly if x = 1 then the equation is false
           if x = 2 then the LHS = 4,5 or 6. However the RHS = 2,4 or 8. (the RHS is 4 when y=2 z=1, but the LHS when y=1)  Hence no solutions.
           if x = 3 then  (3,2,1) is a solution.
                               If y=3 then the RHS $\geq 9$ but the LHS is < 9 unless z=3 (and no solutions here)
                               If y=2 then we only need condsider z=2 which does not yield a solution
                               If y=1 then z=1 and we do not have a solution
           if x > 3 then: if y>2 , the LHS $\geq 3x$ (equality only occurring when z=1) and the LHS $\leq 3x$ (equality only occuring when x=y=z)
                               Therefore y=2 or 1
                               if y=2: The LHS $\leq x+4$ (equality only when z=2) and the RHS $\geq 2x$ but since x is at least 4 this is at least x+4, and equality only occurs when z=1, so y cannot equal 2
                               if y=1 then z=1 and we have x+2=x which is of course absurd

The only solution is x=3 y=2 z=1. (then also y=3 x=2 z=1 and all other permutations)[/hide]

Hopefully there is a neater solution

There you go :
http://www.mathlinks.ro/Forum/viewtopic.php?t=52572 .
And here for the generalization :
http://www.mathlinks.ro/Forum/viewtopic.php?t=63671 .


Ok, game 18 will start as soon as I send out roles and stuff.

People Playing:
Peter VDD
Klebian
SuperNova
LadyKn1ght
lyra
lasagnaman
perfectnumber628
Ars
Captain Sugar
samath
Adunakhor
daine
chesspro
Melissa
bubka
Ignite168
GoBraves
chuckzeduck
slamchess
b-flat
randomdragoon
matt276eagles
dts
Sly Si
poetryinmath

Roles (this is a Mafia vs. Werewolf):
4 Mafia
4 Werewolves
1 Vigilante

1 Inspector
1 Angel
1 Seer
1 Sorcerer
1 Fairy

3 Mason (have the ability to induct on one night)
2 Miller
4 Civilian

1 Assassin
1 Mathematician (for one night, he/she can choose 3 or more people to inspect, and will get a return saying how many of the group are mafia or werewolf. he/she must specify mafia or werewolf)

Accusations are a follows: Accusation and Second do not count as aye. Once seconded, person on block has a maximum of 24 hours to give a defense. Votes begin as soon as defense is given, or 24 hours is up.

Anyone who seems inactive (no post for more than 5 days), and has not given a prior reason as to why, will get one and only one warning and will be subject to replacement if they do not respond.

I'm sending roles out now. DAY BEGINS:

May I be the first to remark that while this game certainly looks fun (Mafia v Werewolves always is), 25 people may be a bit too much.
Watch for around 8 people not posting at all. Anyway, it's strange that we have 2 millers, but whatever. 3 Mason will be interesting, makes for a lot more behind the scenes action.

Anyway, I hope that this post won't get me killed for being the first to speak. If it does, so be it, that's how most Mafia games turn out anyway -_-

Note that I am making no implications in this post, this line being a message to the Mafia, Werewolves, Innocents, and other.

[quote]Watch for around 8 people not posting at all. [/quote]
Not me.

Oh, and can someone give me a quick rundown of how the roles work? Specifically the ones that didn't exist last game, because I read through that one.

mafia vs werewolves! should be fun.....

and besides, if there are 8 inactives, that makes 8 easy lynches right? =P so we don't have to play the "most suspicious" game like we did in 17....

This topic should help:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38004

But I'll explain some too.

Werewolves are similar to mafia, but basically they have a different set of corresponding roles.

Werewolves --- Mafia
Seer --- Inspector
Sorcerer --- Angel (Note that both can save from vigilante kills)

Other than that, it's fairly basic, or was in the last game.
Mathmetician is explained also.

I'm here. My first Mafia/Werewolves game.  :)

I may not be very active (but I'll try to be) until Thursday, since I have a test then.

[quote="poetryinmath"]ok, just showing i'm alive[/quote]
Same here.

i'm here.

minor detail: i will be gone this friday-sunday

posting to show I am here but Randomdragon wait until everbodyy posts before we start random accusations

I'm back from camp finally, but you can replace me if you want.

I hereby declare that I am no longer playing in this game.

I think we're just gonna have to end this game now, or maybe postpone it until summer is over...

how about this:

We restart the whole game in October. That's the start of fall. I'll sign up now.

Signup list:

nutz
akalra1

build off this

okeydokey. I'll pm the past modder of this game and tell him that we are making a signup list.

Hey don't start spamming here now. I'll tell pulak to make a final effort. No registration is open, so there's no ppint in signing up. Each player must be pmed individually, and a huge new set must be taken in fro mthe signup thread. So sign up there if you like.

We are not starting this game over, this is Pulak's game and it's up to him.

ok. If you're talking pianoforte, I pmed him.

Yeah I am now back from Russia so I can rejoin the game

How about this:

we get a whole new signup for game 18(including moderator) So we can actually play.

Is that okay?

How about this: just stop reviving this thread and get over it... this one failed.


Prove that for nonnegative reals $x, y, z,$

\[\sum \sqrt{\frac{2x}{y+z}} \geq 2\sqrt{2}.\]

you correct it very quickly :D
I'll post my solution.
we prove $\sum \sqrt{\frac{x}{y+z}} \geq 2$.
because  $\sqrt{ \frac{x}{y+z}}=\frac{x}{\sqrt {x(y+z)}} \geq \frac{2x}{x+y+z}$
we have proved :)

wt about positive x,y,z, is the minimum 3?

for $x=y=1,z \longrightarrow 0$
$\sqrt{\frac{2x}{y+z}} \longrightarrow 2\sqrt{2}$.

[quote="zhaobin"]you correct it very quickly :D
I'll post my solution.
we prove $\sum \sqrt{\frac{x}{y+z}}\geq 2$.
because  $\sqrt{ \frac{x}{y+z}}=\frac{x}{\sqrt{x(y+z)}}\geq \frac{2x}{x+y+z}$
we have proved :)[/quote]

$\sqrt{ \frac{x}{y+z}}=\frac{x}{\sqrt{x(y+z)}}\geq \frac{2x}{x+y+z}$
What inequality does this come from?
edit: oh wait nv I see it know, it is just AM-GM (but why the wierd form you wrote it in?)

$\frac{x}{\sqrt{x(y+z)}}\geq \frac{2x}{x+y+z}\leftrightarrow \frac{x+(y+z)}{2}\geq \sqrt{x(y+z)}$ or AM-GM on x, y+z

Posted some day ago, http://www.mathlinks.ro/Forum/viewtopic.php?t=108180


ARML MIT college Stanford Princeton CMU Putnam

I'm a couple days early with the X vs Y threads still going on, but maybe some last minute people can get ideas from here.

I'm 95% likely going to Harvey Mudd. Is anyone else? My main concern is that it's so far from home and anyone I know...

Anyway, it's been fun, guys. Not that anyone knows me because I either lurk or spam southeast state boards. I think I'm burnt out on high school math, but maybe I'll head back to the College Playground sometime if my interest returns.

Oh yeah... there's still ARML and Mu Alpha Theta. It never ends!!

I'm going to Caltech as well.

e: Also, I'm from Chicago, so I understand a little bit of how you're feeling about going so far away from home.  In the end, I just decided I had to go to the school I liked the best, and use this as a chance to learn to deal with living far away from home.  And of course, you'll meet new people at HM and make new friends.  It'll probably be fine.

That is funny, I just wanted to open up that thread to figure out which college you guys are finally joining as I think for some colleges the deadlines is on 1st May. Maybe you can also give the main reasons what made you accept the offer, e.g. I want to take courses X, Y, Z and thus thought it is the curriculum that fits me best, I want to take courses under professor A, Stanford has the best weather, most of my friends are going to Caltech, MIT made the most successful advertisement on AoPS, Princeton hold the best college preview session, Berkeley has the best statistics department, at CMU I can happily play with robots etc.

I picked HMC because the only other school to accept me and give significant money was University of Alabama. UA is not so great with math and CS, and I think I'll fit in better at Mudd.

Cool things about HMC: Putnam & pizza night, undergrad focus, lots of research opportunities and Clinics, which let you do real word problem solving in your field.

Princeton for me. Strong analysis program and the undergraduate focus were what attracted me.

Stanford - cs program/classes, people, job opportunities, weather, campus, location, etc.

Stanford- Math + {Physics, Econ, or CS} (i.e., double major)

University of Chicago! Gonna be a huge change in climate, but I'm looking forward to this awesome, new experience! :) Going there because.. they're the best school that accepted me? :) And I really enjoyed my time when I visited, and their staff were some of the friendliest out of the universities I visited.

[quote="Cicatriz"]II'm 95% likely going to Harvey Mudd. Is anyone else? My main concern is that it's so far from home and anyone I know...[/quote].  That's where I'm going.  There were 3 schools I really wanted to go to, but Caltech rejected me and MIT put me on the waitlist, so that leaves Harvey Mudd (and they offered a chunk of money, which is always nice).

But it was one of my favorite schools in the first place because its a great math and science oriented community with an incredible math department surrounded by liberal arts colleges for a nice balance.  I'll be able to do research as an undergrad, and even if I end up in engineering or chemistry, I'll still get to do a lot of math along the way.  I get the feeling I'm going to get a great education in a great environment (I'm talking about the people and the climate, I need to get away from northeastern weather).

Princeton.  They offered more aid than MIT.

I applied to two colleges, Michigan and MIT. I got into both. While Michigan offered me a scholarship, MIT's financial aid was about the same amount. So I chose MIT.

I ended up choosing Harvard. MIT and Harvard's financial offers were about the same. Both are great schools. I loved Harvard's pre-frosh and did cool stuff. I was scared at first when I saw quite the number of drunks Friday evening (including my freshman host friend and his buddies). I didn't attend MIT's CPW because of interview conflicts, so didn't get to see MIT firsthand.

I chose Harvard mostly because of the strong alumni (they had two dinner gatherings for us-one in December and one in April) and close friends there. Pretty sweet.

hey mathfanatic where you at that scholarship dinner during the caltech preview weekend?

I'll be at Princeton next year... for math/finance

MIT, probably Course 6 (Electrical Engineering and Computer Science).

[quote="1234567890"][quote="d343seven"]I'm probably the only one ever from this forum lol but i chose Purdue (I actually live in the same town).
It was between there and Caltech; there were some things I didn't like about Caltech, but overall I liked it ok.
So it was mostly because of the money.

Anyone going to or know anyone else going to purdue too?[/quote]

I think I met some people from West Layafette HS at the Caltech Prefrosh Weekend who were also deciding between Caltech and Purdue.[/quote]

Then that was me and 2 of my friends from here haha.  Whats your name?  I might remember you.  The other two are going to Caltech.  Are you going to Caltech?

I'm Saurabh (I think I met you guys playing mini golf at Ruddock). And, yup, I'm going to Caltech.

Although the one campus which has enrolled the largest number of Southern California ARML team members over our history has been Caltech, this year was an exception: no new Caltech students. There will be SoCal ARML team members as freshmen this year at MIT, Yale, Cornell, Princeton, Penn, UC Berkeley, Stanford, USC, and UCI.

[quote=peelybonehead]Massive bump but I just have to say that these people probably have kids now[/quote]

for real

Bro every single person in this thread go into a t-10 uni,

[quote=Turtle09]everyone pretty much went to either princeton, standford, MIT, or caltech[/quote]

or harvery mudd

[quote name="peelybonehead" url="https://artofproblemsolving.com/community/p25726575"]Massive bump but I just have to say that these people probably have kids now[/quote]

We're like the new generation of AoPS-ers

One day we'll have kids and we'll stop being active here and then the kids of the next generation will bump our threads just like you just did

[quote=ktm]UChicago =) where the squirrels rule -.-[/quote]

Nah mate Vandy is where the squirrels rule :rotfl:


logarithms
integration
calculus
calculus computations

Just note that $ \sum_{n\equal{}0}^{\infty} \frac{(\minus{}1)^n}{3n\plus{}1} \equal{} \int_{0}^{1} \sum_{n\equal{}0}^{\infty} (\minus{}1)^n x^{3n} \, dx \equal{} \int_{0}^{1} \frac{dx}{1\plus{}x^3}$, which is justified by Abel's theorem. The remaining is just a calculation involving partial fraction technique, so we have the answer.

p.s. But the answer is $ \frac{\pi}{9}(\sqrt{3}\pi \plus{} 3 \log 2)$.  :lol:

[quote="sos440"]p.s. But the answer is $ \frac {\pi}{9}(\sqrt {3}\pi \plus{} 3 \log 2)$.  :lol:[/quote]Actually it's $ \frac {1}{9}\left(\pi \sqrt {3} \plus{} 3\log 2\right)$. ;)

[quote="JoeBlow"]Actually it's $ \frac {1}{9}\left(\pi \sqrt {3} \plus{} 3\log 2\right)$. ;)[/quote]

Sorry for typo :blush:


Find all limit points of the sequence $ x_n\equal{}cosn$.

It should be all of $ [\minus{}1,1]$ by the Weyl equidistribution theorem, since $ 2\pi$ is irrational.

Dear blahblahblah thanks for your help.Can you give me information about "Weyl equidistribution theorem"!


Answer is,In nth row there will be n-1 black squares so in 37 th row there will be 36 rows

[hide=Solution 1]So in each row the number of blocks is twice the row number minus 1 so there will will 73 blocks in it. The number of blacks is always half of one less than the number of the blocks. So 73-1 = 72 and 72/2 =36.[/hide]
[hide=Solution 2]An alternative and easier solution is that the number of blacks is always 1 less than the row number. This is because in each row the number of blocks is twice the row number minus 1. The number of blacks is always half of one less than the number of the blocks. Applying this there will be 36 blacks in the 37th row [/hide]

My second solution is easier however the first is more logic based.


Do there exist two coprime posivite integers a,b such that the sequence:
L(0)=a,L(1)=b and L(n+2)=3*L(n+1)-L(n) for n>=0 does not contain any prime?

But is there any solution for this one? If someone has any, then please share it with us!

The answer is yes.
Set  $\lambda  = \frac {3 + \sqrt {5}} {2} $
$ F_n $ = $ \lambda ^ {n} + \lambda ^ {-n} $

$ L_n $ = $ \frac{ \lambda ^ {n} - \lambda ^ {-n} } {\sqrt{5} } $

It's easy to see that both $ F_n $ and  $ L_n $ are integers and satisfy the reccurence of the problem. Moreover, it's easy to see that $ L_n $ and $ L_{n+1} $ are coprime for all positive integers n.
We have $ F_n ^ 2 - 5 L_n ^ 2 = 4 $
Now suppose $ L_n $ is a prime.
Thus $ ( F_n - 2 ) ( F_n + 2 ) = 5 p ^ 2  $ with $ p \in P $
It's easy to see that this equation can't happen for sufficently large n, so there exist a $ N_0 $ such that $ L_n $ isn't a prime for all $ n \geq N_0 $.
Choose $ a= L_{N_0} , b = L_{N_0 + 1 } $ and we are done.

After such a bad day, thanks for this magnificient proof. Now I'm happy I've seen a truly wonderful problem and solution.


geometry
3D geometry
dodecahedron
sphere
vector
symmetry
geometry unsolved

A regular dodecahedron is inscribed in a unit sphere and vectors are drawn from the centre of the sphere to the vertices. What is the sum of these vectors?

By symmetry, each vector has a vector pointing the other way, so the sum is 0. Hope this is right :)


in the figure below we have AP=2PB, find the angule BOP

http://img194.imageshack.us/img194/3549/imagemdwk.png

sorry,but I couldn´t post the image here,

[quote="wanderson"]in the figure below we have AP=2PB, find the angle BOP
[/quote]

Are you referring to the arcs $ AP$ and $ PB$ or the line segments $ AP$ and $ PB$ ?

probably  the line segments AP and PB(I just copied the question),but  can be the arc AP and PB too,
I do not know


inequalities
Pythagorean Theorem
geometry

Let $ ABCD$ be a quadrilateral inscribed in a semicircle with diameter $ AD\equal{}x$. If $ AB\equal{}a$, $ BC\equal{}b$, and $ CD\equal{}c$, then prove \[ x^3\minus{}(a^2\plus{}b^2\plus{}c^2)x\minus{}2abc\equal{}0\]

Anybody solve this problem?

Hint:[hide]ptolemy[/hide]

[hide="solution"] Draw in the other semicircle so ABCD is inscribed in a circle. Notice that, as AD is a diameter, angles ABD and ACD are right.

As that implies ABCD is cyclic, we use Ptolemy's Inequality and the Pythagorean Theorem to get
$ ac \plus{} bx \equal{} \sqrt {(x^2 \minus{} a^2)(x^2 \minus{} c^2)}$.
$ \implies{(ac)^2 \plus{} (bx)^2 \plus{} 2abcx \equal{} x^4 \minus{} (cx)^2 \minus{} (ax)^2 \plus{} (ac)^2}$
$ \implies{x^4 \minus{} (a^2 \plus{} b^2 \plus{} c^2)x^2 \minus{} 2abcx \equal{} 0}$.

As $ x > 0$, we may safely divide by x to get the desired result.[/hide]

A bit easy for a #10? I assume the problems are supposed to fairly increase in difficulty the farther you go...

I am not sure. But this is an interesting problem. A few questions to evaluate what happened:

First, it seems a bit strange to be that when you have the equation for ptolemy, it is clear that $ a$ and $ c$ are interchangeable. But it seems that you couldn't switch say $ a$ and $ b$ without changing the expression. But infact when you expand, the equation is "symmetric" in $ a$, $ b$, $ c$, (that is you can switch them around and the question is unchanged). Can you give a geometric reason for why this is so. As a side note, it is kind of cool that a geometrical argument can give rise to such an algebraic oddity.

Second, geometrically, what does the $ x\equal{}0$ solution correspond to? Why did it appear in our equation?


How many ordered pairs of positive integers $ (x,y)$ are solutions of $ 5x\plus{}3y\equal{}29$?

The pair of numbers with the largest x is $ (4, 3)$.  Then, we can decrease the x by multiples of three.  The only other pair using this method would be $ (1, 8)$.  Therefore, there are $ \boxed{2}$ pairs.


1) find the base 2 decimal expansion of: 11/16 and 4/7

2) find the base 3 decimal expansion of: 11/81 and 7/39

3) in base 10, 9-1=2 :^3: , 9 2 -1 = 2:^4: *5 ,
9 :^4: -1=2 :^3:*5*41
using the above, determine the length of the period in the base [i]nine [/i]expansion of each of the fractions 1/5, 1/(2 :^5: ) and 1/(5*14) in which all of the digits are base 9

[quote="random_mathematician"]... 9 :^3: -1=2 :^3:*5*41 ...
[/quote]
Are you sure that is right? 9 :^3: = 729   :?

oops!! :blush:

yeh fixed it now

its meant to be 9 :^4: not 9:^3:!!

already edited 4 ur convenience...

Quote:1) find the base 2 decimal expansion of: 11/16 and 4/7

[hide]

Note: an underline denotes a repeating cycle of digits.

Since 11 = 8 + 2 + 1 = 10112

dividing by 16 = 24 moves the (binary?) point 4 places to the left.

Answer: 11/16 = 0.10112

Since 7 = 1112, I used long division in base 2 and found that 1/7 = 0.0012

Multiplying by 4 = 22 moves the point 2 places to the right.

Answer: 4/7 = 0.1002[/hide]

I got half of #2 . . .

Quote:2) Find the base 3 decimal expansion of: 11/81

[hide]Since 11 = 9 + 2, we have: 11/81 = 9/81 + 2/81

 which translates to: 1/9 + 2/81 =  0/31 +1/32 + 0/33 +  2/34

Therefore: 11/81 = 0.01023[/hide]


[quote="cezar lupu"]Oh my God that's TUPAC.Please keep it,or if you don't want it just give it to me.I'm a huge $2Pac$ fan. :coolspeak:[/quote]
what? Tupac?

[quote="cezar lupu"]Oh my God that's TUPAC.Please keep it,or if you don't want it just give it to me.I'm a huge $2Pac$ fan. :coolspeak:[/quote]

If you want a tupac avatar, I could find you one, or just look for it on Google Images.

Yes,but I don't know how to find one that little to put it in my profile image.Can you please send me one? :help:

[quote="EUCLA"]it seems to be in a rap club...can i join it?
i was just wondering about my avatar...what do you think about an EMINEM avatar? :lol:

i like TUPAC[/quote]

Sure, get an Eminem one

[quote="Palytoxin"][quote="Iversonfan2005"][quote="cezar lupu"]Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D  :D  :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned   :starwars:[/quote]

True, True, Biggie and Eminem weren't far behind though[/quote]
[quote="G-UNIT"][quote="Iversonfan2005"]Like it? Or was the old one better?[/quote]

 my god, tupac was the most annoying person to ever enter rap, all he did was cry about how african americans were being treated unfairly, when he himslef, killed numerous african americans (excluding the hundreds of cops, he planned on eliminating).  all this person did was cry for attention.  And to think people thought of him as a visionary, more like mockery.  In my opinion, stick with the nba snapshot with iverson or whoever he is.[/quote]

hey cool it off people. you dont have to make the other person what he believes and you two should cool it. otherwise i have to pm the admins to lock this topic[/quote]

  Palytoxin, please use complete and logical sentences when speaking, otherwise; no one understands what you are trying to say.

I don't think the conversation is heated enough to ask the administrators to close this topic.  They've only posted their opinions, strong as they might be.  They're not trying to indoctrinate the other person. ;)

[quote="G-UNIT"][/quote][quote="Palytoxin"][quote="Iversonfan2005"][quote="cezar lupu"]Hey,G-UNIT like it or not $2PAC$ was,is and will be the greatest rapper of all time. :D  :D  :D I think you don't know nothing about Tupac's life and about Rap.... Anyway if you don't know at least one of Pac's songs it means that you know nothing about Rap Music as far as I'm concerned   :starwars:[/quote]

True, True, Biggie and Eminem weren't far behind though[/quote]
[quote="G-UNIT"][quote="Iversonfan2005"]Like it? Or was the old one better?[/quote]

 my god, tupac was the most annoying person to ever enter rap, all he did was cry about how african americans were being treated unfairly, when he himslef, killed numerous african americans (excluding the hundreds of cops, he planned on eliminating).  all this person did was cry for attention.  And to think people thought of him as a visionary, more like mockery.  In my opinion, stick with the nba snapshot with iverson or whoever he is.[/quote]

hey cool it off people. you dont have to make the other person what he believes and you two should cool it. otherwise i have to pm the admins to lock this topic[/quote][quote="G-UNIT"]

  Palytoxin, please use complete and logical sentences when speaking, otherwise; no one understands what you are trying to say.[/quote]

I don't really see the non-logicality of what he wrote. :?

The admins shoot you, fool
Better believe it, they rule

tupac no more :o  :?:

now you have 50 :oops: not bad :!:  ;)

[quote="Iversonfan2005"]Yeah, I'll just switch between Iverson, Tupac, and 50 Cent, like every week or something :)[/quote]

I don't like a lot of the mainstream rappers because all they only sing about "hot girls" (candy shop? wtf? worst song, [b]ever[/b]) to make themselves seem cool. tupac is okay (was he the one that dedicated the song to his mom?) and so are a few others...

there are some underground rappers that manage to sound nice and have meaningful lyrics at the same time... can't recall their names right now, though.

[quote="xdl615x"]Jay Chow is chinese pop not rap!!!!

On a side note, I don't like rap. No offense to any rapper homey's.
I liked your iverson one better.[/quote]

lol I thought it was Jay Chou??

[quote="shobber"][quote="elvenchamp777"]chinese rap is better! go Jay Chow  :D[/quote]
Jay Chow is rap+pop. Anyway, I don't quite like him.

I got my first avatar, how's it?  :P[/quote]
i like it, exept for the pink, i don't like pink :D

[quote="mathclass"]LOL IT'S JAY CHOU. obviously you are not chinese. :P[/quote]
Well I am a Chinese. I just do not know how to spell his lastname. We call his Chinese name all Jay in our daily chats. And the first time I heard of the name Jay I thought of Gay...

[quote="mathclass"]I don't like a lot of the mainstream rappers because all they only sing about "hot girls" (candy shop? wtf? worst song, [b]ever[/b]) to make themselves seem cool. .[/quote]
This is why Eminem is, and will always be, the greatest rapper ever. Every one of his songs has meaning for himself or someone close to him. Unlike meaningless crap such as... "the sweat drips from my balls...".


trigonometry
calculus
calculus computations

I do get the basic aspects of it, but when it comes converting, I am a bit 'shaky'.
1.lim x->0 (x-sinx/tan x-x). I know it is a indeterminate prob. and it is a 0/0 problem. I tried to split into two different equation -sin x/-x and something else to get the right answer. The answer is 1/2 by the way.
2. lim x->pi  (3 sec x/2+tan x) the answer is 3. What I did was I put sec x/tan x on one, 3/1 on the other and needed one more for it to make sense. those would equal 1 and 3 respectively. But it is the 2 that is the problem. Please help.

Stop trying to split up fractions, especially when the addition or subtraction is in the denominator.

On the first: there's a lot of cancellation, and using L'Hopital's rule will require more than one application. It is still the only way you hknow that will work (if you haven't learned power series yet).

On the second: $ \frac{3\sec x}{2+\tan x}=\frac{3}{2\cos x+\sin x}$. It's not indeterminate anymore.


Let $ ABCD$ be a square. Denote $ O\in AC\cap BD$ . For a point $ G\in [OB]$ define

the points $ E\in AG\cap BC$ and $ F\in[CD]$ so that $ GF\perp \overline {AGE}$ . Prove that :

$ 1\blacktriangleright$ For any $ G\in [OB]$ exists the point $ K\in [GF]$ for which $ AK\equal{}EF$  and $ m(\widehat {EKF})\equal{}135^{\circ}$ .

$ 2\blacktriangleright$ The lines $ BD$ , $ AF$ , $ EK$ are concurrently ( [b]I added this point ![/b] ).

This is my solution (sorry, my English is bad)
Note H is the intersection of AF and BD. I'll prove that K is the intersection of GF and HE.
We have ADFG is cyclic so <GAF=<GDF=45. then AGF is right and isosceles, so AG=GF. That causes triangle AGK'= triangle FGE (K' is the intersection of GF and EH, we need prove K= K').
so AK'=EF, we prove EHA is right in H( simple) and have K' is orthocentre of AFE
then <EK'F=180-<EAF= 135. So K'=K.
We have BD, EK, AF is concurrent, too.

1) Call $ H: FG \cap AB$. We have that the quad DAGF and GHBE are cyclic and so $ \angle FAG \equal{} \angle AFG \equal{} \angle GHE \equal{} \angle GEH \equal{} 45^\circ$ then also AHEF is cyclic and so $ AH\equal{}EF$. Then K is clearly the symmetric of H w.r.t. G
2) on the triangle AEF the F-altitude meet the circumcircle at H, then the symmetric of H w.r.t. AE it's the orthocenter of AEF, so K it's the orthocenter of AEF and EK the E-altitude. Call $ P: AF \cap DB$, we have to prove that $ \angle PGA \equal{} \angle AFE$ but $ \angle PGA \equal{} \angle EGB \equal{} \angle EHB \equal{} \angle AFE$


algebra
polynomial
calculus
analytic geometry
graphing lines
slope
trigonometry

That's an interesting thought. As far as I am aware, Taylor Approximations are inherently based on tangency (which is normally a calculus topics) and approximations. However, I suppose that if you did approximations of the tangent slopes that technically you could get Taylor looking stuff from using just reasoning and not much calculus. Just a thought, maybe Kent or someone else could help you more.

Well, I tried something, but it didn't work as well as I had hoped. Rookworm said, "I am thinking of something akin to Archimedes' or ancient Chinese computations of pi." The heart of those computations lies with angle bisection. So consider the following idea:

$\cos x=\sum_{k=0}^{\infty}a_kx^{2x}.$

We limited this to even powers becuase we know that the cosine is an even function. It is also clear that $a_0=1.$

Now we want angle bisection. The cosine satisfies the following identity:

$\cos^2 x=\frac12(1+\cos2x).$

$\sum_{k=0}^{\infty}\left(\sum_{j=0}^ka_ja_{k-j}\right)x^{2x}=1+ \frac12\sum_{k=1}^{\infty}2^{2k}a_kx^{2k}$

So for $k\ge1,\,\sum_{j=0}^ka_ja_{k-j}=2^{2k-1}a_k.$

This gives us the following chain of equations:

$2a_1=2a_1$
$8a_2=2a_2+a_1^2$
$32a_3=2a_3+2a_1a_2$
$128a_4=2a_4+2a_1a_3+a_2^2$

... and so on. The problem with this is that there's no equation that tells us what $a_1$ is. We can successively solve this chain of equations for all the subsequent coefficients, finding:

$a_2=\frac16a_1^2$
$a_3=\frac1{90}a_1^3$
$a_4=\frac1{2520}a_1^4$

If we can determine by some other means that $a_1=-\frac12$ then this would get us that $a_2=\frac1{24}$, $a_3=-\frac1{720}$ and $a_4=\frac1{40320},$ which is what we expect.

But we now need another method for determining $a_1.$

[quote="rookworm"]I'm just wondering-- can anyone think of a nice geometric way to find the Taylor polynomials of sin, cos, etc. [b]without[/b] using calculus (explicity)?. Is there some sequence of polygons that corresponds to these polynomials? I am thinking of something akin to Archimedes' or ancient Chinese computations of pi.[/quote]

I once read a book "Strength in Numbers" by Stein. In it, there is a discussion on how an Indian sage in the 6th century (or some time around that) arrived at the Taylor series for $\tan^{-1} (x)$, using geometrical reasoning. You can try to get the book and see how it is done. Yes, all this before calculus was invented, pretty amazing!


inequalities
geometry
perimeter
algebra proposed
algebra

[color=darkred][b]Theorem.[/b] If $n\in N$, $n>1$, $x_k\in R$, $k\in\overline {1,n}$, $\sum x_k=na$, $\sum x^2_k=nb$
then for any $k\in \overline {1,n}$, $|x_k-a|\le \sqrt{(n-1)(b-a^2)}$.[/color][hide="Some aplications."] $1^{\circ}\blacktriangleright \sum x_k=\frac{n+1}{2}\ ,\ \sum x^2_k=1+\left(\frac{n-1}{2}\right)^2\Longrightarrow$$\left(\forall\right) k\in \overline {1,n}\ ,\ |2nx_k-(n+1)|\le (n-1)^2\ .$

$2^{\circ}\blacktriangleright 2+\sum x_k=\sum x^2_k=n\Longrightarrow \left(\forall\right) k\in \overline {1,n}\ ,\ x_k\in \left[-1\ ,\ 3-\frac 4n\right]\ .$

$3^{\circ}\blacktriangleright$ In any triangle $ABC$ there is the relations

$|r_a+r_b-2r_c|\le \sqrt{(4R+r)^2-3p^2}\ ;\ \max_{x\in \{a,b,c\}} \left|p-\frac 32 \cdot x\right|\le \sqrt{p^2-3r(4R+r)}\ .$[/hide]

For 1:

Assume $(\forall k)|2nx_k-(n+1)|>(n-1)^2$, then square it and expand it, sum up all the inequalities, substitute the given sums and get a contradiction.

For 2:

Assume $(\forall k)(x_k+1)(x_k-3+{4\over n})>0$. Expand it, sum up all the inequalities, substitute the given sums and get a contradiction.

For 3:

What's $p$? Perimeter? Semiperimeter?

The $p$ is the semiperimeter and $P$ is the perimeter (in Romania) !. At other times $p\equiv s$.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

What is the difference between the quotient field of Q[x] and the quotient field of Z[x]?

I dont just want an answer though.. I want someone to help explain the whole reasoning and thought process behind it so that next time I'll be able to reason through it...

Thanks!!!

look at the definitions, these things are just defined differently.
but there is a canonical isomorphism. more generally, if $ R$ is an integral domain, we may identity $ Q(Q(R)[X])$ with $ Q(R[X])$ by taking a common denominator.


Evaluate $ (a \plus{} b)(a^2 \minus{} ab \plus{} b^2)$ for $ a \equal{} 5$ and $ b \equal{} 4$.

We have two solutions - we could substitute $ 5$ for $ a$ and $ 4$ for $ b$ to get the value, but that takes a while, and of course we want to do this quickly. Note that $ (a \plus{} b)(a^2 \minus{} ab \plus{} b^2)$ is the factorized form of $ a^3 \plus{} b^3$. Therefore, all we need to do is compute $ 5^3 \plus{} 4^3 \equal{} 125 \plus{} 64 \equal{} \boxed{189}$.


Can the second reference (that one that's not a math teacher) be a parent? I remember reading that it could be a parent, but I just want to make sure.

Also, do I mail the application test and personal information stuff separately or together?

[quote="zacacox"]We'd strongly prefer the second reference [i]not[/i] come from a parent. If you need a little extra time to solicit a second reference from someone else, we can work with you--send us an e-mail.

If your parents would like to write to us about you, they may submit a parent's statement (details near the bottom of the [url=http://www.awesomemath.org/application.shtml]application page[/url].)

Oh, and mail the stuff together. No sense in using two envelopes.

-Zac[/quote]

im assuming the "stuff" would just be the personal info and the parent letter, correct?

because i think it says for the recommendations to be mailed/emailed separately.

Correct. And the test, too, if you're sending it right at the deadline.


I'd like to introduce some easy AIME to the people who are new to AMC so they can see what AIMEs are like.

It would be solved by me first, of course.

What is the sum of all two-digit numbers that each is divisible by the both of its digits?

Changed the wording from Kalva, who changed wording from the Official AMC Problem. :P

What is the sum of all two digit numbers where each is divisible by both of its digits?
[hide]Well, we obviously have all the double-digits: 11+22+33+44+55+66+77+88+99.
Then there are the even number where the second digit is double the first: 12, 24, 36, 48 (3*4, 3*8, 3*12, 3*16). Finally there is 15 . 11+22+33+44+55+66+77+88+99+12+24+36+48+15=630. So my answer is 630.[/hide]


Two six-sided dice are fair in the sense that each face is equally likely to turn up. However, one of the dice has the $ 4$ replaced by $ 3$ and the other die has the $ 3$ replaced by $ 4$. When these dice are rolled, what is the probability that the sum is an odd number?

$ \textbf{(A)}\ \frac{1}{3}\qquad
\textbf{(B)}\ \frac{4}{9}\qquad
\textbf{(C)}\ \frac{1}{2}\qquad
\textbf{(D)}\ \frac{5}{9}\qquad
\textbf{(E)}\ \frac{11}{18}$

For the first dice the sides are $ {1,2,3,3,5,6}$.
For the second dice the sides are $ {1,2,4,4,5,6}$.
For the sum to be odd, one of the numbers had to be even and the other has to be odd. So, we have two cases.

Case 1 (The first dice rolls to be odd and the other to be even)
$ \frac{4*4}{6*6}\equal{}\frac{4}{9}$
Case 2 (The first dice rolls to be even and the other to be odd)
$ \frac{2*2}{6*6}\equal{}\frac{1}{9}$

Adding these up we get $ \frac{5}{9}$


calculus
derivative
function
vector
calculus computations

For the function $ f(x,y,z) \equal{} x^2 \plus{} y^2 \plus{} z^3 \minus{} 3z$, find $ \nabla f$.

(i) What is the rate of change of $ f(x,y,z)$ in the direction normal to the cylinder $ x^2 \plus{} y^2 \equal{} 25$ at the point $ (3, \minus{}4, 4)$?
(ii) At which points does $ \nabla f$ have no component in the $ z$-direction?
(iii) Find and classify the stationary points of $ f$.
(iv) Sketch the contours of $ f$ and add to the sketch a few arrows showing the directions of $ \nabla f$.

I'll get you started:
$ \nabla f\equal{}\left(f_{x},f_{y},f_{z}\right)\equal{}(2x,2y,3z^{2}\minus{}3)$.  Then a stationary point would be where $ \nabla f\equal{}(0,0,0)$, so that $ 2x\equal{}2y\equal{}3z^{2}\minus{}3\equal{}0$.  Then $ (x,y,z)\equal{}(0,0,\pm 1)$.  If you want to classify them, you can use a Hessian matrix.  The second derivatives are easy to compute.

Okay, this looks like a natural extension of the two-variable case. But what does "normal to the cylinder" mean? I thought we can only find the direction in which the rate of change is the greatest, which is $ \nabla f$

I'm assuming normal to the cylinder means perpendicular to it.  The gradient gives you the tangent vector, from which you can find the normal vector (do you remember how to find normal lines?).

I see, in that case is the normal vector $ <3, \minus{}4, 0>$? Then the answer to the first part would be $ <3, \minus{}4, 0> \cdot <6, \minus{}8, 45>$ for a value of $ 50$. But the how do you interpret this value?

Part (iv) doesn't seem possible because it's like asking us to plot a 4D contour..

No, it's in $ \mathbb{R}^{3}$.  You hold one value fixed and see what the function does for other values.  [url=http://www.math.hmc.edu/calculus/tutorials/surfacecontours/]Here's[/url] a little tutorial.

I can't understand "The gradient gives you the tangent vector, from which you can find the normal vector", but for (i) we have to project $ f$'s gradient into the normal of the cylinder. whose versor is $ ({3\over5},\minus{}{4\over5}).$

Yes, I forgot to normalize there  :maybe:


Let $n$ be a positive integer. Consider all permutations of $\{1,2,...,n\}$. Let $A$ denote the set of those permutations such that each number is either greater than all of the numbers to its left or less than all of the numbers to its right. Let $B$ denote the set of those permutations $a_{1},a_{2},...,a_{n}$ such that for $1\leq i\leq n-1$, there is a $j>i$ with $|a_{j}-a_{i}|=1$. Show that $|A|=|B|$.

[quote="JBL"]I do not think it is true:

For $n = 3$, $A$ consists of all permutations except $\{3, 2, 1\}$, but $B$ consists of all permutations except $\{2, 3, 1\}$ and $\{2, 1, 3\}$.  Did I misunderstand or make a mistake?[/quote]

When $n=3$ then $A=\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2)\}$ and $B=\{(1,2,3),(1,3,2),(3,1,2),(3,2,1)\}$ therefore $5=|A|\not = |B|=4$ !!!!  , problem is false! I think so!

This is a problem in ''A Path To Combinatorics For Undergraduates'' of Tittu and Zum.


WAIT!!!!

How exactly does a FTW tourney work?????

Please tell me Where, When, Who, and How.

...people play against each other as scheduled, results are recorded, a winner is decided.

Yes, on FTW.  Hence the name [b]FTW tourney[/b].

In the future, please read the stickies. One of them pointed to a thread that described in great detail the workings of an FTW tournament.

Locked.


trigonometry
inequalities proposed
inequalities

Prove that for all real numbers $\alpha_1,\alpha_2,\ldots,\alpha_n$ we have \[ \sum_{i=1}^n \sum_{j=1}^n ij \cos {(\alpha_i - \alpha_j )} \geq 0. \]
[i]Octavian Stanasila[/i]

[i]Dear V.V., [/i]this Stanasila's problem is a amusement:

$ S\equiv \sum\limits_{i=1}^n \sum\limits_{j=1}^n ij\cos(x_1-x_j )=\sum\limits_{i=1}^n \sum\limits_{j=1}^n ij(\cos x_i\cos x_j +\sin x_i\sin x_j)= $
$ \sum\limits_{i=1}^n i\cos x_i\cdot\sum\limits_{j=1}^n j\cos x_j+\sum\limits_{i=1}^n i\sin x_i\cdot \sum\limits_{j=1}^n j\sin x_j=\left( \sum\limits_{i=1}^n i\cos x_i \right)^2+ $ $ \left( \sum\limits_{i=1}^n i\sin x_i \right)^2\ge 0. $

[b][u]Remark.[/u] Two interesting problems:[/b]

[b]1. When[/b] $ S=0 $ ?

[b]2. Ascertain sum[/b] $ S\ if\ \left(\forall\right)k\in \overline {1,n},\ x_{k+1}=x_k+r $ [b](arithmetical progression).[/b]

[quote="levi"][i]Dear V.V., [/i]this Stanasila's problem is a amusement:[/quote]It is easy, but it's not the only easy problem given at a Romanian TST :)


We have been given a number $ a\equal{}152855965207715 / 248623746255151$. Determine the rational numbers $ b\equal{}n_1/m_1, c\equal{}n_2/m_2$ such that $ 0<m_1,m_2<10^{15}$ and $ a\minus{}b<0$ and $ a\minus{}c>0$ are as close to $ 0$ as possible but they are not $ 0$.


just by looking at the answer i think its wrong because your answer is not over 145 so the average can't be 145.

Oops... I meant [hide="this"]$154$[/hide]

[hide] 139 + 143 + 144= 426,  An average of 145 on 4 games would bring a total of 580.  580-426=[b]154[/b][/hide]

[hide]She needs 580 points and she currently only has 426 pts. So she needs 154 points. [/hide]

[quote="math92"]In three bowling games, Alice scores 139, 143, and 144.  What score will alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]

[hide]

Add them up. Get 426. 145*4=580. Then, subtract 426 from 580, to get $\boxed{154}$.[/hide]

[hide] you add $139,143,144$. Then, you multiply $145$ by 4. subtract the two sums to get 154. The answer.
$\boxed{154}$
[/hide]

[quote="math92"]In three bowling games, Alice scores 139, 143, and 144.  What score will alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]

[hide]139 is 6 under 145. 143 is 2 less than 145. 144 is one less than 145. 6+2+1=9 which means that Alice is 9 less than where she needs to be. 145+9=154.

$154$[/hide]

[hide][quote="math92"]In three bowling games, Alice scores 139, 143, and 144.  What score will alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]

145*4-139-143-144
580-282-144
580-426
154[/hide]

please don't revive month old topics.  :wink:  :wink:  :wink:


[hide]

$x_{1}+y_{1}<x_{2}+y_{2}<...<x_{n}+y_{n}$
$x_{1}<x_{2}<...<x_{n}$

Just subtract the equations

$y_{1}<y_{2}<...<y_{n}$

And the only possilbe order for the $y_{i}$ is the same as the order of the $x_{i}$

[/hide]

[quote="tjhance"]$x_{1}+y_{1}<x_{2}+y_{2}<...<x_{n}+y_{n}$
$x_{1}<x_{2}<...<x_{n}$

Just subtract the equations

$y_{1}<y_{2}<...<y_{n}$

And the only possilbe order for the $y_{i}$ is the same as the order of the $x_{i}$
[/quote]
You are wrong! $7<8$ and $1<8$ but $7-1<8-8$ is FALSE! :D

The First Mistake in Inequalities, presented by tjhance :P

drat i thought that seemed to easy  :blush:

[hide="click to reveal"]Assume $x_{1}\ne y_{1}$. Consider inequalities of the form $g+x_{j}< x_{j}+y_{k}$ where $j \ne 1$, $g \in S = \{x_{1},x_{2},...,x_{n}\}$, and $y_{k}\in S-\{x_{j}\}$. Start with $g = x_{1}$ and choose some $j \ne 1$. If $y_{k}= g$, we get $g+x_{j}< x_{j}+g$, contradiction. If $y_{k}< g$, we get a contradiction since $x_{1}< x_{2}< ... < x_{n}$. If $y_{k}> g$ set $g$ to $x_{j}$, $x_{j}$ to $y_{k}$, and choose a different value for $y_{k}$ (This is where the permutations kick in). Eventually, if you never run into the $y_{k}\le g$ condition, you will have $g = x_{n}$, $\Rightarrow x_{n}+x_{j}< x_{j}+y_{k}$, contradiction since $x_{n}$ is the largest real according to our given. So if $y_{1}= x_{j}$ with $j \ne 1$, we get a contradiction. Therefore, we must have $x_{1}= y_{1}$. You'll notice that for each determined value of $y_{k}$, $y_{k+1}$ must take a value from the ever-shrinking set $S-\{y_{k}\}$ until we've completely exhausted $S$. So if we assume $x_{i}\ne y_{i}$ for the remainder of the $i$'s, using a similar argument as above we do get $x_{i}= y_{i}$  $\forall$  $i = 1,2,...,n$, $Q.E.D.$[/hide]

Nice Problem, here's my solution :D:
[hide]
Let $A={x_{1},x_{2},x_{3}...x_{n}}$
Consider $S={x_{1}+y_{1},x_{2}+y_{2},...x_{n}+y_{n}}={x_{1}+x_{\in A},x_{2}+x_{\in A},...x_{n}+x_{\in A}}$.
We wish to pick $x_{1}$ and $y_{1}$ so that
$x_{1}+y_{1}<x_{2}+y_{2}<....x_{n}+y_{n}$.

We make a diagram of all the posible $(x_{n},y_{n})$ we can pick for elements of $S$ in order starting with $x_{1}$. From each row we pick at least 1.

$(x_{1},x_{1}) (x_{1},x_{2}) (x_{1},x_{3})...(x_{1},x_{n})$
$(x_{2},x_{2}) (x_{2},x_{3}) ...(x_{2},x_{n})$
$(x_{3},x_{3}) ... (x_{1},x_{n})$
$............$
$(x_{n},x_{n})$

When anything other than $(x_{k},x_{k})$ is picked from each row,
$(x_{k+1},x_{k+1})$ in the next row is eliminated. (There's only 2 identical of $x_{k}$.)

For every row if you don't pick $(x_{k},x_{k})$ , $(x_{n},x_{n})$  will be elimiated before you get to the bottom row, so the number of elements of set $S$ cannot ever reach $n$.
[/hide]


calculus
integration
analytic geometry
geometry
3D geometry
sphere
calculus computations

Write down and calculate an iterated double integral in
rectangular coordinates that represents the volume of the solid that
is bounded below by the xy-plane and bounded above by the sphere
$ z \equal{} 9\minus{}x^2\minus{}y^2$
Convert the above integral to polar coordinates (I think it's a lot easier this way),and then calculate it.

It doesn't affect working the problem, but there is one word in that description that should be changed. The surface $ z\equal{}9\minus{}x^2\minus{}y^2$ is not a sphere or a portion of a sphere. It is a paraboloid.


Find all function  ( $R-\to R$ ) satisfy :
 $f(x)-2f( \frac{x}{2})+f( \frac{x}{4})=x^{2}$
( :maybe: )
[i]Louis Latin and Vicky[/i]

Let $f(x)=\frac{16}{9}x^{2}+g(x)$. Then $g(x)-2g(\frac{x}{2})+g(\frac{x}{4})=0$. Because equation is linear we can find $g(x)=\sum_{k}a_{k}g_{k}(x)$, were $g_{k}(x)=x^{c_{k}}$. It give ${1-2z+z^{2}=0, z=(\frac{1}{2})^{c_{k}}}$, therefore $g_{k}(x)=1$. Continiusly solution is $\frac{16}{9}x^{2}+const$.

Thanks !
But I don't understand " Why you find how to  Let $f(x)=\frac{16}{9}x^{2}+g(x)$ ?
Can teach me ?

Rust,

you still did not answer the "why" you got that certain function.  Where is your motivation for it is what is wanted to be known.  You showed the "what" you did with it.  You need to derive it, if I interpreted right.


trigonometry
complex numbers
algebra unsolved
algebra

Given that \[ \frac{\cos(x) \plus{} \cos(y) \plus{} \cos(z)}{\cos(x\plus{}y\plus{}z)} \equal{} \frac{\sin(x)\plus{} \sin(y) \plus{} \sin(z)}{\sin(x \plus{} y \plus{} z)} \equal{} a,\] show that \[ \cos(y\plus{}z) \plus{} \cos(z\plus{}x) \plus{} \cos(x\plus{}y) \equal{} a.\]

$ u \equal{} e^{ix}, v \equal{} e^{iy}, w \equal{} e^{iz}$ satisfy $ u \plus{} v \plus{} w \equal{} auvw$ and we are asked to show that $ \Re(uv \plus{} vw \plus{} wu) \equal{} a$.  But division gives

$ \frac{1}{vw} \plus{} \frac{1}{uw} \plus{} \frac{1}{uv} \equal{} a$

and we know that for $ z$ on the unit circle we have $ \frac{1}{z} \equal{} \bar{z}$, two complex numbers with the same real part.  The conclusion follows.  (Was this really a Longlist problem?)  We also conclude that $ \sin (x \plus{} y) \plus{} \sin (y \plus{} z) \plus{} \sin (z \plus{} x) \equal{} 0$.

if we multiply the first two expressions it follows that $ \sin (x \plus{} y) \plus{} \sin(y \plus{} z) \plus{} \sin(z \plus{} x) \equal{} 0$...

finally, note that

$ a\equal{}\sum\frac{\cos x}{\cos(x \plus{} y \plus{} z)} \equal{} \sum\frac{\cos(x \plus{} y \plus{} z)\cos(y \plus{} z) \plus{} \sin(x \plus{} y \plus{} z)\sin (y \plus{} z)}{\cos (x \plus{} y \plus{} z)} \equal{} \sum \cos(y \plus{} z)$

and we're done  :D

We have ;\[\frac{(\cos (x+y+z))(\sum_{cyc}\cos (x+y))}{\cos^2(x+y+z)}=\frac{(\sin (x+y+z)(\sum_{cyc}\sin (x+y))}{\sin^2(x+y+z)}=a\]
Now summing up the numerators and the denominators we get
\[a=\cos (x+y+z)\sum_{cyc}\cos (y+z)+\sin (x+y+z)\sum_{cyc}\sin (y+z)=\sum_{cyc}\cos (y+z)\]


calculus
integration
function
limit
derivative
calculus computations

Let $f(x)$ be the function defined for $x\geq 0$ which satisfies the following conditions.
(a) $f(x)=\begin{cases}x \ \ \ \ \ \ \ \ ( 0\leq x<1) \\ 2-x \ \ \ (1\leq x <2) \end{cases}$

(b) $f(x+2n)=f(x) \ (n=1,2,\cdots)$

Find $\lim_{n\to\infty}\int_{0}^{2n}f(x)e^{-x}\ dx.$

More generally, what is the Laplace transform of $f?$

That is, $F(s)=\int_{0}^{\infty}e^{-st}f(t)\,dt.$

In this case, this converges for $s>0.$

kunny's original question asks for the value $F(1).$

Hint: one can make useful connections between Laplace transforms of periodic functions and Fourier series.

Well, this definately isn't the best way, but here's what I did: \[I=\lim_{n\to\infty}\int_{0}^{2n}e^{-x}f(x) \ dx\]  \[=\sum_{j=0}^\infty\left(\int_{2j}^{2j+1}xe^{-x}\ dx+\int_{2j+1}^{2j+2}(2-x)e^{-x}\ dx\right)\]  \[=\sum_{j=0}^\infty \left(2j(e-1)^{2}+e^{2}-2e+1\right)e^{-2(j+1)}\]  \[=\sum_{j=0}^\infty (-4j-2)e^{-2j-1}+\sum_{j=0}^\infty (2j+1)e^{-2j-2}+\sum_{j=0}^\infty (2j+1)e^{-2j}\]  \[=\frac{e^{2}+1}{(e+1)^{2}}\] Edit:  I'm not too sure about my answer because when I tried the solution Kent was referring this is what I got:

For a periodic function $f(x)$ with period $p$, we have $F(s) = \frac{1}{1-e^{-ps}}\int_{0}^{p}f(t)e^{-st}\ dt$, so in our case we have \[F(s) = \frac{1}{1-e^{-2s}}\left(\int_{0}^{1}te^{-st}\ dt+\int_{1}^{2}(2-t)e^{-ts}\ dt\right) = \frac{\tanh{s/2}}{s^{2}}\] and then $F(1) = \tanh 1/2 \not = \frac{e^{2}+1}{(e+1)^{2}}$

You're better off taking the derivative first before doing the Laplace transform, and using that

$\mathcal{L}(f')(s)=sF(s)-f(0).$

Working that way, I get that $F(s)=\frac{e^{s}-1}{s^{2}(e^{s}+1)}$ meaning the integral in this problem is

$F(1)=\frac{e-1}{e+1}.$

OK, Jimmy and I agree with each other - or at least I agree with his second answer.

Any idea where I went wrong in the first one?


integration
calculus
calculus computations

Find the whole portion of:

$A=\sum_{n=1}^{10^{9}}n^{\frac{-2}{ 3}}$

If by "whole portion" you mean "integer part", the answer is 2997.

This sum can be compared to an integral. The comparison is quite visual - draw boxes both above and below the curve $y=x^{-2/3}.$ We can show that

$\int_1^{m+1}x^{-2/3}dx<\sum_{n=1}^mn^{-2/3}<1+\int_1^mx^{-2/3}dx$

$3(m+1)^{1/3}-3<\sum_{n=1}^mn^{-2/3}<3m^{1/3}-2.$

For $m=10^9$ this clearly places the sum between 2997 and 2998.


hi ! $ x,y,z\geq0$ prove that
[img]http://www.texify.com/img/%5CLARGE%5C%21%28xy%2Byz%2Bxz%29%5Csqrt%5B%5D%7B%28x%2By%2Bz%29%5E2%2B4xyz-%28x%2By%2Bz%29%2B1%7D%5Cgeq6xyz.gif[/img]

let $ r\equal{}\sqrt[3]{xyz}$ its obious that $ xy\plus{}yz\plus{}zx\geq 3r^2$ and by AM-GM $ \dfrac{(x\plus{}y\plus{}z)^2}{4}\plus{}1\geq (x\plus{}y\plus{}z)$ and $ \dfrac{3(x\plus{}y\plus{}z)^2}{4}\plus{}4xyz\geq \dfrac{27r^2}{4}\plus{}4r^3$ thus $ (xy\plus{}yz\plus{}zx)\sqrt{(x\plus{}y\plus{}z)^2\plus{}4xyz\minus{}(x\plus{}y\plus{}z)\plus{}1}\geq 3r^2\sqrt{\dfrac{27r^2}{4} \plus{}4r^3}\geq (3r^2)(\sqrt{4r^2})\equal{}6r^3\equal{}6xyz$


Let $ n$ be a positive integer. Given a sequence of $ n$ elements, each 1 or -1. In one step it is allowed to transform sequence in to another one such that every element becomes equal to the product of his neighbours (each element have 2, except first and last element having 1 neighbour). Find all $ n$ such that for every possible arrangement of 1 and -1 in first sequence after finite number of steps we will get sequence $ (1,1,1,\ldots,1)$ (every element is 1).

Example of transformation:
\[ (1, \minus{} 1,1, \minus{} 1,1) \mapsto ( \minus{} 1, 1, 1, 1, \minus{} 1) .
\]
Sorry for my BAD English.  :blush:

Sort of a sketch: Let us go backwards. From $ (1,1,1,1,\dots,1,1,1,1)$ we may reach either $ (1,1,1,1,\dots,1,1,1,1)$ itself or $ ( \minus{} 1,1, \minus{} 1,1, \minus{} 1,\dots,1, \minus{} 1,1, \minus{} 1)$. We consider the latter case. Here the string $ \minus{} 1,1$ appears $ \frac {n \plus{} 1}{2}$ times (in the last appearance a $ 1$ is missing). Now from $ ( \minus{} 1,1, \minus{} 1,1, \minus{} 1,\dots,1, \minus{} 1,1, \minus{} 1)$ we can reach $ (x, \minus{} 1, x, 1, x, \minus{} 1, x,\dots, x, \minus{} 1, x)$, where $ x$ can be replaced with either $ 1$ or $ \minus{} 1$. Note that the string $ x, \minus{} 1,x,1$ appears $ \frac {n \plus{} 1}{4}$ times (in the last appearance a $ 1$ is again missing). Again, from here we can go to $ (y, x, \minus{} y, 1, \minus{} y, x, y, 1, \dots, \minus{} y, x, y)$, where $ x,y\in\{1, \minus{} 1\}$. Here the string $ y, x, \minus{} y, 1, \minus{} y, x, y, 1$ appears $ \frac {n \plus{} 1}{8}$ times (in the last appearance of the string a $ 1$ is again missing). We have something like the following:

. . . . . . . . . . . .
. . . . . . . . . . . .
y, x, -y, 1, -y, x, y, 1, . . ., -y, x, y
x, -1, x, 1, x, -1, x, 1, . . ., x, -1, x
-1, 1, -1, 1, -1, 1, -1, . . ., -1, 1, -1
1, 1, 1, 1, 1, 1, 1, . . ., 1, 1, 1

Proceeding on like this finally we are left with a single string, where the last $ 1$ will be missing, and we have freedom of choice for all the other terms of the string. This implies $ \frac {n \plus{} 1}{2^k} \equal{} 1$ for some $ k$, and thus $ n \equal{} 2^k \minus{} 1$.

But we have a little problem. If there are some $ 1$'s or $ \minus{} 1$'s in the last string, for which we cannot apply our freedom of choice, we can yet go backwards until they are deleted and reach a state where all of the terms can take arbitrary values.

A second approach: Since before reaching $ (1,1,1,1,\dots,1,1,1,1)$ we must reach $ (\minus{}1,1,\minus{}1,1,\dots,\minus{}1,1,\minus{}1)$, we must have $ n\equal{}2n_1\plus{}1$ for some $ n_1$. Now we delete all the $ \minus{}1$'s in our last sequence to get a new sequence of $ n_1$ $ 1$'s. Applying the same argument for $ n_1$ we find that $ n_1\equal{}2n_2\plus{}1$ for some $ n_2$. If we keep going like this we will eventually have $ n_k\equal{}0$ for some $ k$. That means $ n\equal{}2^k\minus{}1$.


Find all functions $f,g,h: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x) - g(y) = (x-y) \cdot h(x+y)$ for $x,y \in \mathbb{R}.$

Take $x=y$ and you find $f=g$. Next, take $y=0$ and you find $ xh(x)=f(x)-f(0)$. Put $ t(x)=f(x)-f(0)$. Then $ (x-y)t(x+y)=(x+y)(t(x)-t(y))$. Now, take $ y=1$ and you find $ t(x+1)=\frac{x+1}{x-1} (t(x)-t(1))$. Substitute here $x+1$ instead of $x$ and obtain $ t(x+2)=\frac{x+2}{x} (\frac{x+1}{X-1}(t(x)-t(1))-t(1))$. But $ t(x+2)=\frac{x+2}{x-2} (t(x)-t(2))$. From the two relations we find that $t$ is of the form $ ax+b$ and since $ t(0)=0$ we have $t(x)=ax$ and then $ f(x)=ax+b$.

Thats not all the possible functions, note for example $ h(x) \equal{} ax$, $ f(x) \equal{} g(x) \equal{} ax^2 \plus{} b$

[quote="harazi"] From the two relations we find that $ t$ is of the form $ ax \plus{} b$ [/quote]

In fact from that tow little guys we get $ t(x)\equal{}x^2\times (\frac{1}{2} f(2)\minus{}f(1))\plus{} x (2 f(1) \minus{} \frac{1}{2} f(2) )$ for $ x\not\equal{} 0, 1, 2$ :D

Make the substitucion $ a\equal{}\frac{1}{2} f(2)\minus{}f(1) \wedge b\equal{}( 2f(1) \minus{} \frac{1}{2} f(2) \Leftrightarrow f(2)\equal{}4a\plus{}b \wedge f(1)\equal{}a\plus{}b$ and we get $ t(x)\equal{}ax^2\plus{}bx$ for all reals, because it is true for $ x\equal{}0$ by defenition of $ t(x)$, and $ f(2)\equal{}4a\plus{}b\equal{}a2^2\plus{}b2$, $ f(1)\equal{}a\plus{}b\equal{}a1^2\plus{}1b$

Recall $ f(0)\equal{}c$ and we get $ f(x)\equal{}ax^2\plus{}bx\plus{}c\equal{}g(x)$ and $ h(x)\equal{}ax\plus{}b$

Now check that $ ax^2\plus{}bx\plus{}c\minus{}ay^2\minus{}by\minus{}c\equal{}(x\minus{}y) \times (a(x\plus{}y) \plus{} b)$ occur for $ x, y$ reals.

Let $P(x,y)$ be the assertion and let $g(0)=c,kf(0)=k$.Then $P(x,0) \Rightarrow f(x)-c=xh(x) \Rightarrow f(x)=xh(x)+c$.$P(0,y) \Rightarrow k-g(y)=-yh(y) \Rightarrow g(y)=k+yh(y)$.Substituting these in the original equation we get $xh(x)-yh(y)+f=(x-y)h(x+y)$ where $f=c-k$.$y=0 \Rightarrow f=0 \Rightarrow c=k$.So the equation may be rewritten as $xh(x)-yh(y)=(x-y)h(x+y)$
Now $P(x,-x) \Rightarrow h(x)+h(-x)=2h(0) \forall x \neq 0$.Substituting $y=-y$ we get
$xh(x)+yh(-y)=(x+y)h(x-y)$
$\Rightarrow xh(x)+y(2h(0)-h(y))=(x+y)h(x-y)$
$\Rightarrow (x-y)h(x+y)+2h(0)y=(x+y)h(x-y)$

Substituting $y=(x-1)$ in this equation yeilds $h(2x-1)=(2x-1)(h(1)-2h(0))$.In other words $h(x)=dx$ for some $d \in \mathbb{R}$.
We may hence conclude that
$f(x)=g(x)=dx^2+c$ for some $d \in \mathbb{R}$

These system clearly satisfies the original FE.


graph theory
combinatorics solved
combinatorics

An infinite tree graph is given in the plane such that every lattice point is a vertex of the graph, and every edge of the graph connects two lattice points whose Euclidean distance is at most 1998.
Prove that there exist two points in the plane, a unit distance apart, such that the path which  connects them in the tree is at least 10^1998 long !

So many Combinatorics Problems. And so few solutions. Where do you find them , ORL? Do you have any solutions for them?

Hi Dust,

at first I cannot refer each problem I posed to a certain olympiad. Sometimes I found it in any journal. I have some of the solutions but I want to make the forum members try to solve it. I cannot evaluate whether the forum members dislike my Combinatorics, especially graph theory, problems (meaning they are not sufficiently appealing) or maybe they regard them as challenging and did not find any approach yet.

I solved it, but all combinatoric problems are very hard to expound in foreign language. It is the main reason why many combinatoric problems haven't solution on this forum, I think.

I think there are only few native English speakers on the forum who might complain. Nobody should try to raise his pedagogical index finger in order to indicate that one's English is superb. As long as the forum members know what you try to explain I feel there is no problem about it. Nobody will kill you ! Simply try !!!  :)

Suppose it is false.

Let T be the tree. Let us appoint (0,0) as the root of T.
Let l be 1998, L be 101998.

Consider square Q with center (0,0) and side-lenght l*L.
Let V be the set of vertex on border of Q.
At initial moment V has 4l*L points with unit distance between neighboring vertexes, and if X,Y are

neighbor then |PathInTree(X,Y)|<L.

Two points X and Y (X,Y \in V) is called neighboring if there are no points on border of Q between X and Y which contains in V.

Step 1) For each vertex in V consider the min path from root to that point. It will be level of the point. Let us choose point A with maximal level.

Step 2) Let B,C \in V and B,C are neighbor of A on border of Q. We know |PathInTree(A,B)|<L and |PathInTree(A,C)|<L.

Step 3) MAIN TRICK: It is fact |PathInTree(B,C)|<L because A has max level!

Step 4) Now we exclude A from V.

Step 5) Let us notice if X,Y \in V and X and Y are neighbor vertex on border of Q then property |PathInTree(X,Y)|<L is hold. |V| -> |V|-1.

Step 6) Go to Step 1).

We will stop process 1)-6) when set V will first time have two neighboring vertex X and Y whose Euclidian distance is at least l*L ==> |PathInTree(X,Y)| \geq l*L/l (because each rib (branch) of graph  is at most l) = L !!!! Contradictory with property |PathInTree(X,Y)|<L.

This is my second proof for that problem. Somebody must verify what I wrote.
Orlando, what do you think about it?


In LaTeX, I want to define a new symbol ${{{\rm M}\atop \substack{1\leq i\leq m}}}$ in my paper, however the spacing between the ${\rm M}$ and $\substack{1\leq i\leq m}$ is too wide. Ideally, I want to be something looks like ${\lim_{n\to \infty}a_{n}}$. Anyone can help me to solve this problem?  :oops:

One way is to use \underset ensuring that you have the amsmath package loaded (which should always be the case) so \underset{1\leq i\leq m}{\text{M}} gives $\underset{1\leq i\leq m}{\text{M}}$.


Which is larger, cos (sin x) or sin( cos x)? Give proof

This one is really classic and can be found here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=807
http://www.mathlinks.ro/Forum/viewtopic.php?t=39441
http://www.mathlinks.ro/Forum/viewtopic.php?t=69100

:)


group theory
abstract algebra
topology
advanced fields
advanced fields unsolved

Hi everyone, I hope you can help me with this two questions :

1) Let $G$ be a compact group and $a \in G$ which is not the identity.  If the cyclic subgroup $<a>$ is dense in $G$, then $G$ is abelian.

2) Let $H$ be a closed subgroup of a compact group $G$ such that $H \neq G$.  Then there exists a character $f: G \to S$ such that $f \not \equiv 1$ and $f(H)=1$, where $S=\{z : |z|=1\}$.

Remembering that a character is just a continuous homomorphism $f: G \to S$

the first one you have to use the fact that a cyclic group is abelian and the operation $G \times G \to G : (x,y) \mapsto xy$ is continuous.

FS

In 2) do you suppose that $H$ is normal? If the conjugates of $H$ generate $G$ (e.g., $G=S_{3}$ and $H=\{e,(12)\}$), then there is no such $f$.

If $H$ is normal, then compose the canonical homomorphism $G\to G/H$ with a non-principal character of $G/H$.

THANKS FOR YOUR HELP.

just two things:

1) $H$ is always normal, since by definition $G$ is Hausdorff (it is a topological group, well i am working with that definition  :P ) , so $H$ is, and by compactness of $G$ , $H$ is compact , then $H$ is a compact hausdorff space, therefore normal  :) .

2)sorry , but I am not very familiar with characters,  so why can i suppose that there exists such a non-principal character? .

Adriana

:) I meant to ask if $H$ is a normal subgroup (as in $x\in H\implies gxg^{-1}\in H$ for any $g\in G$), not a normal topological space.
Non-principal characters exist on any locally compact [b]abelian[/b] group; in fact the characters separate the points of the group by the [url=http://en.wikipedia.org/wiki/Pontryagin_duality]Pontryagin duality theorem[/url]. Using Pontryagin duality may be an overkill, but I just can't think of a simple direct proof of the existence of non-principal characters. :(  For a non-abelian group you take its quotient by the commutant and hope that the quotient is nontrivial.

well thank u so much.   i really appreacite ur help

Adriana

[quote="mlok"]In 2) do you suppose that $H$ is normal? If the conjugates of $H$ generate $G$ (e.g., $G=S_{3}$ and $H=\{e,(12)\}$), then there is no such $f$.

If $H$ is normal, then compose the canonical homomorphism $G\to G/H$ with a non-principal character of $G/ H$.[/quote]

I was thinking about this, and it's not necessary that $H$ be normal, we can consider the left (right) coset $G/H$ which is a group and under the canonical epimorphism is a topological space.

[quote="mlok"]I don't see how $G/H$ is a group... And consider the above example with $S_{3}$: if a homomorphism $f\colon S_{3}\to \mathbb C$ sends $H$ to $1$, then $f$ is identically equal to $1$.[/quote]

Yes, you are right, it was a stupid mistake, of course $H$ has to be normal, otherwise the operation, in $G/H$ , is not well defined.

Adriana


conics
hyperbola
geometry
geometric transformation
reflection

Prove that for every positive integer n , the difference $ s_n = (\sum_{k=1}^{n}[\frac{n}{k}]) - [\sqrt{n}] $ is an even integer , where [x] denotes the integer part of x

Let $A = \{1,2, \cdots , n \}$ and $S = \sum_{k=1}^n [\frac n k ].$

For each $k$, the number $[\frac n k]$ denotes the number of multiples of $k$ in $A$. Therefore, each element $i$ from $A$ contributes for $1$ in $S$ exactly as many times as its number of positive divisors, say $d(i).$
It follows that $S = \sum_{k=1}^n d(k).$
But, it is well known that $d(k)$ is odd if and only if $k$ is a perfect square. Thus $S$ has the same parity than the number of squares in $A$, which is $[\sqrt n ]$, and the desired result now follows easily.

Pierre.

I thought of this geometrically (but in essence it's the same solution as pbornsztein's).

The summation in the problem is the number of lattice points in the first quadrant that are on or below the hyperbola $xy=n$.  Of these lattice points, $[\sqrt{n}]$ of them lie on the line $y=x$.  The remaining lattice points can be paired by reflection across the line $y=x$, so there are an even number of them.

Yakko


geometry
geometric transformation
reflection
geometry unsolved

this is 220 exercise [b]a ra rb [/b]of Prof Luís Lopes Thesis (Handbook of triangles construction)

i made  [b]hc= 2rarb/ (ra+rb)[/b] and i have the line of c side

but 220 exercise says that  ra rb hc is a datum we have hc so the problem is now 164  with [b]a hc ra[/b]  but  this is a mistake because 164 problem says
sin(gama)=[b]hb/a[/b] isn't [b]hc[/b]

i want the A point construction with i've got now

 :(

It is very easy to finish. $ F \in c$ is foot of the C-altitude $ h_c.$ Having the sidelines $ a, c$ interseting at $ B,$ construct the internal and external bisectors $ BI_b$ and $ BI_a$ of the $ \angle B.$ (The excenters $ I_b, I_a$ are unknown, but you can draw the angle bisectors.) Draw 2 parallels $ p, q$ to $ a$: $ p$ at the distance $ r_a$ from $ a$ on the opposite side of $ a$ than $ F,$ $ q$ at the distance $ r_b$ from $ a$ on the same side of $ a$ as $ F.$ The parallel line $ p$ cuts the external bisector $ BI_a$ of the angle $ \angle B$ at $ I_a,$ the parallel line $ q$ cuts the internal bisector $ BI_b$ of the $ \angle B$ at $ BI_b$ at $ I_b.$ Draw the excircles $ (I_a), (I_b)$ tangent to $ a.$ The center line $ I_aI_b$ should cut $ a$ at $ C.$ Reflection of $ a$ in the line $ I_aI_b$ is the common internal tangent of $ (I_a), (I_b)$ other than $ a,$ i.e., the sideline $ b$ cutting the sideline $ c$ at $ A.$


Three machines P, Q, R work together and can do a job in x hours. P needs an extra 6 hours when working alone; Q needs an extra 1 hour when working alone; R needs an extra x hours when working alone. Find x.

Blah, anyone know how to do this? I hate these sort of problems.

remember that their individual rates sum up to the total rate.

I believe there are a couple of these in AoPS 1 Chapter 3 if you want practice.


Show that $a^2\equiv1\pmod8$ for any odd $a$.

Show that $a\equiv{b}\pmod{m}$ and $0<a<m$ and $0<b<m$ imply $a=b$. Try to give a simple, [b]direct[/b] proof.

For the first part, the similar problem is discussed in this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51266

BUT, above thread only asked for answer and since this one asks for "proof," I'm fine with it.  :)

[quote="fourierseries"]We would have:

$m \times k_1 +a = m\times k_2 +b$

Since a and b are restricted,

$k_1=k_2=0$ and therefore $a=b$[/quote]This isn't very rigorous. You basically just restate the conditions and say you're done..

[quote="chess64"][hide="1"]

Since $a$ is odd, $a \equiv 1, 3, 5, \text { or } 7 \pmod {8}.$ Squaring, we see $a^2 \equiv 1, 9, 25, 49 \equiv 1, 1, 1, 1, \equiv 1 \pmod {8} \ \ \ \ \ \mathbb{W}^5$[/hide]

[hide="2"]

$a \equiv b \pmod {m} \implies 0 < a=mk+b < m$
$mk+b < m$
$m(k-1)<-b$
$m(-1)(1-k) < -b$
$m(1-k) > b$
This implies that because $m,b > 0$, $1-k>0 \Rightarrow k<1 \Rightarrow k=0$ because it must be a nonnegative integer.

Therefore $a=m(0)+b \implies a=b \ \ \ \ \mathbb{W}^5$[/hide][/quote]Looks good. Here's my solution for 2.
[hide="2"]Let $a=b+km$. Then $b+km<m\Longrightarrow{k}m<m-b<m\Longrightarrow{k}<1.$ Also $a-km<m\Longrightarrow-km<m-a<m\Longrightarrow-m<km\Longrightarrow{k}>-1.$ Therefore $-1<k<1$ and so $k=0$.[/hide]


Where can I find the questions and solutions of saint peterzbourg's olympiad?

Hi every body
No body know where I can find problems and questions of  Saint petersburg olympiads.

So far, I haven't seen any post about Saint Petersburg Olympiad. Just wait for the Mathlinkers to get it for you. I wish to see them, too. :) In due time, there are many other olympiads worth trying.

- [url=http://www.mathpropress.com/books/LMO/]Leningrad Mathematical Olympiads 1987-1991 by Dmitry Fomin and Alexey Kirichenko (English problems with solutions).[/url]

- [url=http://www.pdmi.ras.ru/~olymp/]Saint-Petersburg Mathematical Olympiad 1998-2008 (Russian problems without solutions).[/url]


analytic geometry
trigonometry
conics
parabola

How would you write $y = x$ in polar form?

$r\sin x = r\cos x$

How would you isolate the $r$?

Polar equations don't always have to have r, just like cartesian equations don't always have both y and x.  In this case, either you can just draw the solution and see, or do what you did, and solve for x to get the equation to be $\theta = \frac{\pi}4$ to be the equation.

Do you mean coefficient, exponent, or constant? Since you added an exponent and a constant I'm guessing you used the wrong term. The equation you gave (a parabola), being a conic, would be graphed in polar coordinates following the equation of a conic in polar. Problem is, conics in polar form need to have the focus at the pole (as far as I know) to be written in this form.

Hmm, so you can't write a polar form of an equation with a coefficient? In coefficient I mean an $x^0$ term like a 1 or a -2 in the equation.

[quote="breez"]So not all $f(x,y)$ can be converted to polar form?

Or would that denote a line going through $45^\circ$?[/quote]

Just as not all $f(x, y)$ can be written explicitly in $y =$ form, so all polar $p(r, \theta)$ cannot be written explicitly in $r =$ form, but all $f(x, y)$ can be written in some sort of implicit polar form by simple substitution.

Yes, the equation $\theta = \frac{\pi}{4}$ denotes the set of all points $(r, \theta)$ such that $\theta = \frac{\pi}{4}$, hence $r$ ranges across the reals.

[quote="breez"]btw, how do I convert the equation into polar form if there is a coefficient? Like $y=x^2 +1$[/quote]

$r \sin \theta = r^2 \cos^2 \theta + 1$
$r^2 \cos^2 \theta - r \sin \theta + 1 = 0$
$r = \frac{\sin \theta \pm \sqrt{\sin^2 \theta - 4 \cos^2 \theta}}{2 \cos^2 \theta}$

How exactly did you substitute the 1 into that polar equation?

Is there a generalized method that works for all coefficients?

You simply add one; no need to convert it.

To what was said earlier -- conics of the form $\frac{ep}{1\pm p\sin\theta}$ (and the corresponding one with $\cos$) have to have a focus at the center, but there are conics of other forms which do not have to have a focus at the center.

[quote="breez"]Isn't the 1 adding to the length of r? In $x^2 + 1$, the 1 adds to the y-axis value. If you added 1 in the polar equation, it would lengthen the r and not increase the y values by 1.

$r \sin \theta$ must increase by 1.[/quote]

I didn't add $1$ to the polar equation, I merely kept the $1$ that was already there.  If you plug my solution (both signs) into your calculator it should produce the same graph.  I don't understand what you're confused about  :huh:

Edit:  As has already been said, [b]not all polar graphs are explicitly $r =$ form.[/b]  Any implicit Cartesian graph $f(x, y) = 0$ can be transformed into an implicit polar graph $p(r, \theta) = 0$ by using $p(r, \theta) = f(r \cos \theta, r \sin \theta)$.

Examples:

$y = x^3 + x^2 + 1 \implies r \sin \theta = r^3 \cos^3 \theta + r^2 \cos^2 \theta + 1$

$x^2 y + y^2 x + 3 = 0 \implies r^3 \cos^2 \theta \sin \theta + r^3 \sin^2 \theta \cos \theta + 3= 0$

$y = e^x + 17 \implies r \sin \theta = e^{r \cos \theta} + 17$


let  ABC a right-angled triangle in C and I medium of [ BC ].  show that sin(<IAB)<=1/3? :oops:

$x=m(\widehat {IAB}),\ m_a=\frac 12\sqrt{a^2+4b^2},\ \cos x=\frac{a^2+2b^2}{\sqrt{(a^2+b^2)(a^2+4b^2)}}.$
$\sin x=\sqrt{1-\cos^2 x}=\frac{ab}{\sqrt{(a^2+b^2)(a^2+4b^2)}}.$
$\sin x\le \frac 13\Longleftrightarrow 9a^2b^2\le (a^2+b^2)(a^2+4b^2)\Longleftrightarrow (a^2-2b^2)^2\ge 0$,
what is true.

[u]A generalization.[/u] $m(\widehat {ACB})\ge 90^{\circ}\Longrightarrow \sin (\widehat {IAB})\le \frac 13$,
with equality iff ${\frac{a}{\sqrt 2}}=\frac b1=\frac{c}{\sqrt 3}.$
(I used the relations $c^2\ge a^2+b^2$ and $\cos C\le 0$.)

A synthetic solution:

[color=blue][b]Problem.[/b] Let ABC be a right-angled triangle with C = 90°, and let I be the midpoint of its catet BC. Show that $\sin\measuredangle IAB\leq\frac13$.[/color]

[i]Solution.[/i] Let M be the midpoint of the hypotenuse AB of the right-angled triangle ABC. Then, AM = MB. On the other hand, since the points I and M are the midpoints of the sides BC and AB of triangle ABC, we have IM || CA, so that < MIB = < ACB = C = 90°. Thus, the point I lies on the circle with diameter MB. The center of this circle is obviously the midpoint N of the segment MB, and the radius of this circle is $\frac12\cdot MB$. Thus, $NI=\frac12\cdot MB$. On the other hand, since N is the midpoint of the segment MB, we have $MN=\frac{MB}{2}$, and thus $AN=AM+MN=MB+\frac{MB}{2}=\frac32\cdot MB$. Now, by the sine law in triangle ANI, we have

$\frac{\sin\measuredangle IAN}{\sin\measuredangle AIN}=\frac{NI}{AN}=\frac{\frac12\cdot MB}{\frac32\cdot MB}=\frac13$.

Hence, $\sin\measuredangle IAN=\frac13\cdot\sin\measuredangle AIN\leq\frac13$ (since sines are always $\leq 1$). In other words, $\sin\measuredangle IAB\leq\frac13$, and the problem is solved.

  darij


linear algebra
matrix
function
logarithms

Let $ M_2(\mathbb{C})$ be the group of [i]all[/i] 2 x 2 matrices with complex entries. Let $ GL(2,\mathbb{C})$ be the group of all invertible matrices with complex entries.
Prove that the matrix exponential map $ \exp$ maps $ M_2(\mathbb{C})$ surjectively into $ GL(2,\mathbb{C})$.

(hint: use jordan cannonical form)

$ \mathcal{M}_2(\mathbb{C})$ isn't a group.

The matrix exponential respects similarity.  What does it do to a diagonal matrix?  What does it do to a $ 2 \times 2$ Jordan block?

[quote="t0rajir0u"]$ \mathcal{M}_2(\mathbb{C})$ isn't a group.[/quote]
It is with addition as the group operation - and with this being the exponential function, that is the right context to be thinking of. (Although the exponential will not be a homomorphism - we only have $ e^{A\plus{}B}\equal{}e^Ae^B$ if $ AB\equal{}BA.$)

It' s true for $ GL_n(\mathbb{C})$ and it isn't difficult.
Let $ A\in{GL_n(\mathbb{C})}$; we seek $ X$ s.t. $ e^X\equal{}A$; $ spectrum(A)\equal{}\{\lambda_1,\cdots,\lambda_n\}\subset\mathbb{C}^*$; there exists a determination of the complex logarithm s.t. the $ (log(\lambda_i))_i$  are defined.
$ A\equal{}P^{\minus{}1}diag(\lambda_1({I_{n_1}}\plus{}J_{n_1}),\cdots,\lambda_k({I_{n_k}}\plus{}J_{n_k}))P$ where $ J_p$ is the nilpotent Jordan block of dimension $ p$. We take $ X\equal{}P^{\minus{}1}diag(Y_1,\cdots,Y_k)P$ where $ Y_p\equal{}log(\lambda_p){I_{n_p}}\plus{}J_{n_p}\minus{}(1/2)J_{n_p}^2\plus{}\cdots\plus{}((\minus{}1)^{n}/(n\minus{}1))J_{n_p}^{n\minus{}1}$.


Evaluate $ \cos36^{\circ} \minus{} \cos72^{\circ}$ without utilizing their individual values.

[hide="Solution"]$ \sin 3x \equal{} Im[(\cos x \plus{} i \sin x)^3] \equal{} 3 \cos^2 x \sin x \minus{} \sin^3 x$
$ \equal{} 3 (1 \minus{} \sin^2 x) \sin x \minus{} \sin^3 x \equal{} 3 \sin x \minus{} 4 \sin^3 x$.
$ \cos 36^\circ \minus{} \cos 72^\circ \equal{} \sin 54^\circ \minus{} \sin 18^\circ \equal{} 3 \sin 18^\circ \minus{} 4 \sin^3 18^\circ \minus{} \sin 18^\circ \equal{}$
$ \equal{} 2 \sin 18^\circ \minus{} 4 \sin^3 18^\circ \equal{} 2 \sin 18^\circ ( 1 \minus{} 2\sin^2 18^\circ) \equal{} 2 \sin 18^\circ \cos 36^\circ$
$ \equal{} \dfrac{2 \sin 18^\circ \cos 18^\circ}{\cos 18^\circ} \cdot \dfrac{2 \sin 36^\circ \cos 36^\circ}{2\sin 36^\circ} \equal{} \dfrac{\sin 36^\circ}{\cos 18^\circ} \cdot \dfrac{\sin 72^\circ}{2 \sin 36^\circ} \equal{} \dfrac{\sin 72^\circ}{2 \cos 18^\circ}$
$ \equal{} \dfrac{\cos 18^\circ}{2\cos 18^\circ} \equal{} \dfrac{1}{2}$[/hide]

[quote="J.Mayers"]Evaluate $ \cos36^{\circ} - \cos72^{\circ}$ without utilizing their individual values.[/quote]
[hide="It may be a simple way to solve"]

By using $ \cos x - \cos y = -2 \sin {x+y\over 2} \sin {x-y\over 2}\cdots(*)$ simply.

 $ \therefore \cos36^{\circ} - \cos72^{\circ}$

 $ = -2 \sin 54 ^{\circ}\sin (-18^{\circ})$

 $ =2 \sin 54 ^{\circ}\sin 18^{\circ}$

 $ = \sin 54 ^{\circ}\cdot {2\sin 18^{\circ}\cos 18^{\circ} \over \cos 18^{\circ}}$

 $ ={\sin 54 ^{\circ}\sin 36^{\circ}\over \cos 18 ^{\circ}}$

 $ ={2\sin 54 ^{\circ}\sin 36^{\circ}\over 2\cos 18 ^{\circ}}$

   Using (*) again,

 $ ={\cos 18 ^{\circ}-\cos 90^{\circ}\over 2\cos 18 ^{\circ}}$

  $ =\frac 1 2$

[/hide]

$ \boxed {E\equiv\cos 36\minus{}\cos 72}\implies 2\sin 36\cdot E\equal{}$ $ 2\sin 36\cos 36\minus{}2\sin 36\cos 72\equal{}$

$ \sin 72\minus{}(\sin 108\minus{}\sin 36)\equal{}$ $ \sin 72\minus{}\sin 108\plus{}\sin 36\equal{}$ $ \sin 72\minus{}\sin 72\plus{}\sin 36\equal{}$ $ \sin 36\implies$

$ \boxed{E\equal{}\frac 12}$ . Exists and a simple geometrical proof of it (you can easily find !).