function
conics
parabola
algebra
domain
calculus
calculus computations

Dear all;

can anyone offer view if my explanation is right for the question below:

Given g(x)=max(f,0). Describe the graph.

My Explanation: The function g(x) will not exists for x<0.

Thank You.

Thats true. Also I take it $ f$ means $ f(x)$. There could be undefined points I think, like if $ f(x) \equal{} \tan(x)$ and $ x\equal{}\frac{\pi}{2}$.

I suppose so because the exact question is:

Describe the graph of g in terms of the graph of f if ...... follow by my question.

Thanks. You clear my ambiguity as to whether I should write that in.  :lol:

[quote="Symbian"]My Explanation: The function g(x) will not exists for x<0.[/quote]
That's wrong. $ x < 0$ has nothing to do with anything.

Here's an example to look at: $ f(x) \equal{} 4\minus{}x^2$ for $ x\in\mathbb{R}.$ Of course, the graph of $ f(x)$ is familiar - it's a parabola opening downward with a vertex at $ (0, 4).$

Now, let $ g(x) \equal{} \max(f(x),0).$

That means that for $ x$ such that $ f(x) < 0,$ we will have $ g(x) \equal{} 0,$ and for $ x$ such that $ f(x)\ge 0,$ we will have $ g(x) \equal{} f(x).$ We still have $ g(x)$ defined everywhere $ f(x)$ is defined, which in this case is everywhere.

$ g(x) \equal{} \begin{cases}0 & x < \minus{} 2 \\
4\minus{}x^2 & \minus{} 2\le x\le 2 \\
0 & x > 2\end{cases}.$

The graph is the portion of the $ x$ axis for $ |x|\ge2$ together with a parabolic bump that sticks out above the axis on the interval $ ( \minus{} 2,2).$

General features:

* The domain of $ g$ is the same as the domain of $ f$.

* If $ f$ is continuous at $ a,$ then $ g$ is continuous at $ a;$ or more generally, if $ f$ is continuous on the set $ S,$ then $ g$ is continuous on $ S.$ (In the example above, both $ f$ and $ g$ are continuous on $ \mathbb{R}.$) The converse of this statement is false.

* The range of $ g$ is the intersection of the range of $ f$ with $ [0,\infty).$

* $ g$ need not be differentiable even if $ f$ is. The example above shows this clearly.

* This is simply function composition. Let $ m(x)$ be the piecewise linear continuous function defined by $ m(x) \equal{} \max(x,0).$ Then $ g(x) \equal{} m(f(x)).$

Kent, you have some max vs. min errors you might want to fix, and also your piecewise expression for $ g$ should have $ x^2 \minus{} 4$, not $ x^2$.

Yikes! How did I screw it up that badly?

I just rewrote the whole thing so it's now max instead of min (which makes JBL's comment slightly out of date - but I appreciated it.)

Dear Kent;

What if I assume the function $ f(x)$ to be $ x^2\minus{}1$? Clearly I will achieve a parabola almost like yours but is inverted. Hence from this graph, I describe it as follows:

Since the question says: $ g(x)\equal{}max(f,0)$, the curve below the x-axis will not exists. So from this I conclude that the function will not exists for $ x<0$. Do you think I am on the right track?

Thanks

No, if $ f(x) \equal{} x^{2} \minus{}1$ then

$ g(x) \equal{} \begin{cases}x^{2}\minus{}1 \;  \forall \; x \in (\minus{}\infty,\minus{}1] \cup [1, \infty)  \\
0 \; \text{otherwise} \end{cases}$

Symbian:

Your use of mathematical language is so sloppy that I can't even tell what track you're on.

Let's try this again. Suppose $ f(x)\equal{}x^2\minus{}1$ for $ x\in\mathbb{R}$ and $ g(x)\equal{}\max(f(x),0).$

Then $ g(x)$ exists for all $ x\in\mathbb{R}.$ So try to phrase what you mean to say without using the words "not exist."

And, once again, $ ``x<0"$ is completely irrelevant and completely wrong. I suspect you probably meant to say something involving $ ``y<0"$ but that's not what you said.

To communicate mathematics, you must choose the meanings of your words and symbols carefully. You're not succeeding at that.

Dear Kent;

Thank you for your patience. Yes, my use of mathematical words are not really good. So I always prefer diagrams when I want to explain something. Attached forthwith is the two diagram and my complete answer to the question. Do let me know what is your view and once again, I would like to thanks everyone for their patience in attending my problem.

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Consider the graph, $ f(x) \equal{} x^2\minus{}1$

The graph of $ f(x)$ is like in Figure 1. Since the question implies $ g(x)\equal{}max(f,0)$, the graph of $ g(x)$ will be like the graph as shown in Figure 2, where the curve below the $ x$ axis is cut off such that the curve will not cross below the $ x$ axis. In other words, the functions $ f(x)$ will not exists for $ x<0$.
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Is there anyone had further comment on my republished work?

Yes: you don't appear to understand the difference between $ x < 0$ and $ y < 0$.

Oh dear.. Thank You very much JBL and Kent. I didn't even aware of my big mistake until both of you point it out.. Thank you..... Is there any more comment assume the error had been corrected from $ x < 0$ to $ y < 0$..... Sorry about the mistake. Thanks.


[quote="mathVNpro"]Let $ a,b,c$ be three positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove:
\[ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac {9}{2\left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right )}\]
[/quote]
It's equivalent to $ \sum_{cyc}\frac {1}{1 \plus{} a^2b^2}\geq\frac {9}{2(a \plus{} b \plus{} c)},$ where $ a,$ $ b$ and $ c$ are
non-negatives such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$
Let  $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3$ for the last inequality.
Hence, the last inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is convex function.
Thus, enough to prove your inequality in two cases:
1) for $ c \equal{} 0$ ;
2) for $ b \equal{} c.$

[quote="mathVNpro"]Let $ a,b,c$ be three positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove:
\[ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac {9}{2\left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right )}\][/quote]
Mu poof
By $ Cauchy \minus{} Schwarz$
$ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac{9}{3\plus{}ab\plus{}bc\plus{}ca}$
We have $ a^2\plus{}2\sqrt{a} \ge 3a$
$ \Rightarrow a^2\plus{}b^2\plus{}c^2\plus{}2\left (\sqrt{a} \plus{}\sqrt{b}\plus{}\sqrt{c} \right ) \ge 3\left (a\plus{}b\plus{}c  \right )\equal{}9$
$ \Leftrightarrow \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c} \ge ab\plus{}bc\plus{}ca$
And $ \left (\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}  \right ) ^2 \le 3(a\plus{}b\plus{}c)\equal{}9$
$ \Rightarrow \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c} \le 3$
So
$ \frac{9}{3\plus{}ab\plus{}bc\plus{}ca} \ge \frac{9}{2\left ( \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c} \right )}$

[quote="nguyenvanhuyen"][quote="mathVNpro"]Let $ a,b,c$ be three positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove:
\[ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac {9}{2\left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right )}\]
[/quote]
Mu poof
By $ Cauchy \minus{} Schwarz$
$ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac {9}{3 \plus{} ab \plus{} bc \plus{} ca}$
We have $ a^2 \plus{} 2\sqrt {a} \ge 3a$
$ \Rightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} 2\left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \right ) \ge 3\left (a \plus{} b \plus{} c \right ) \equal{} 9$
$ \Leftrightarrow \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge ab \plus{} bc \plus{} ca$
And $ \left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \right ) ^2 \le 3(a \plus{} b \plus{} c) \equal{} 9$
$ \Rightarrow \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \le 3$
So
$ \frac {9}{3 \plus{} ab \plus{} bc \plus{} ca} \ge \frac {9}{2\left ( \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \right )}$[/quote]
[color=darkblue]
This solution is not true  :)

[b](1)[/b] $ \blacksquare \ \ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge ab \plus{} bc \plus{} ca$ [b]and[/b] $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \le 3 \stackrel{\mbox{not}}{\Longrightarrow} 3 \plus{} ab \plus{} bc \plus{} ca \le 2\left ( \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \right )$

[b](2)[/b] $ \blacksquare \ \ 3 \plus{} ab \plus{} bc \plus{} ca \nleq 2\left ( \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \right )$
try $ (a,b,c) \equal{} (1,5\ \ ; \ \ 1,5 \ \ ; \ \ 0)$[/color]

My Proof  :)  ( I don't think this is the nicest proof)

We have

$ \dfrac{1}{{1 \plus{} ab}} \plus{} \dfrac{1}{{1 \plus{} bc}} \plus{} \dfrac{1}{{1 \plus{} ca}} \ge \dfrac{9}{{2\left( {\sqrt a \plus{} \sqrt b \plus{} \sqrt c } \right)}}$

$ \Leftrightarrow \left( {1 \minus{} \dfrac{1}{{1 \plus{} ab}}} \right) \plus{} \left( {1 \minus{} \dfrac{1}{{1 \plus{} bc}}} \right) \plus{} \left( {1 \minus{} \dfrac{1}{{1 \plus{} ca}}} \right) \le 3 \minus{} \dfrac{9}{{2\left( {\sqrt a \plus{} \sqrt b \plus{} \sqrt c } \right)}}$

  $ \Leftrightarrow \dfrac{{ab}}{{1 \plus{} ab}} \plus{} \dfrac{{bc}}{{1 \plus{} bc}} \plus{} \dfrac{{ca}}{{1 \plus{} ca}} \le 3 \minus{} \dfrac{9}{{2\left( {\sqrt a \plus{} \sqrt b \plus{} \sqrt c } \right)}}$

From AM-GM inequality, we have,

$ \dfrac{{ab}}{{1 \plus{} ab}} \plus{} \dfrac{{bc}}{{1 \plus{} bc}} \plus{} \dfrac{{ca}}{{1 \plus{} ca}} \le \dfrac{{ab}}{{2\sqrt {ab} }} \plus{} \dfrac{{bc}}{{2\sqrt {bc} }} \plus{} \dfrac{{ca}}{{2\sqrt {ca} }} \equal{} \dfrac{1}{2}\left( {\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} } \right)$

It suffices to prove that,

$ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \plus{} \dfrac{9}{{\sqrt a \plus{} \sqrt b \plus{} \sqrt c }} \le 6$

Setting $ \sqrt a \equal{} x,\sqrt b \equal{} y,\sqrt c \equal{} z \Rightarrow {x^2} \plus{} {y^2} \plus{} {z^2} \equal{} 3$

The inequality is written as

 $ xy \plus{} yz \plus{} zx \plus{} \frac {9}{{x \plus{} y \plus{} z}} \le 6$

  $ \Leftrightarrow xy \plus{} yz \plus{} zx \plus{} \dfrac{{\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)\sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} }}{{x \plus{} y \plus{} z}} \le 2\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)$

To prove this inequality, I have 2 proofs

[u][b][i]First Proof[/i][/b][/u]

Due to homogeneity, we may assume that $ x \plus{} y \plus{} z \equal{} 1$

Setting $ xy \plus{} yz \plus{} zx \equal{} A \Rightarrow {x^2} \plus{} {y^2} \plus{} {z^2} \equal{} 1 \minus{} 2A,A \le \dfrac{1}{3} < \dfrac{2}{5}$

The inequality is equivalent to,

$ A \plus{} \frac {{\left( {1 \minus{} 2A} \right)\sqrt {3\left( {1 \minus{} 2A} \right)} }}{1} \le 2\left( {1 \minus{} 2A} \right)$

 $ \Leftrightarrow \left( {1 \minus{} 2A} \right)\sqrt {3\left( {1 \minus{} 2A} \right)} \le 2 \minus{} 5A$

  $ \Leftrightarrow 3{\left( {1 \minus{} 2A} \right)^3} \le {\left( {2 \minus{} 5A} \right)^2}$

  $ \Leftrightarrow 0 \le \left( {1 \minus{} 3A} \right)\left( {1 \plus{} A \minus{} 8{A^2}} \right)$

This inequality is true since $ 1 \minus{} 3A \ge 0,1 \plus{} A \minus{} 8{A^2} > 1 \minus{} 9{A^2} \ge 0$

[u][i][b]Second Proof[/b][/i][/u]

The inequality is equivalent to

$ xy \plus{} yz \plus{} zx \le \left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)\left[ {\dfrac{{2\left( {x \plus{} y \plus{} z} \right) \minus{} \sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} }}{{x \plus{} y \plus{} z}}} \right]$

  $ \Leftrightarrow xy \plus{} yz \plus{} zx \le \left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)\dfrac{{4{{\left( {x \plus{} y \plus{} z} \right)}^2} \minus{} 3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)}}{{\left( {x \plus{} y \plus{} z} \right)\left[ {2\left( {x \plus{} y \plus{} z} \right) \plus{} \sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} } \right]}}$

  $ \Leftrightarrow xy \plus{} yz \plus{} zx \le \left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)\dfrac{{{x^2} \plus{} {y^2} \plus{} {z^2} \plus{} 8\left( {xy \plus{} yz \plus{} zx} \right)}}{{\left( {x \plus{} y \plus{} z} \right)\left[ {2\left( {x \plus{} y \plus{} z} \right) \plus{} \sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} } \right]}}$

It is easy to check 2 following inequalities:

$ 2\left( {x \plus{} y \plus{} z} \right) \plus{} \sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} \le 3\sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)}$

$ {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} \left( {xy \plus{} yz \plus{} zx} \right) \le {\left( {x \plus{} y \plus{} z} \right)^3}$
  $ \Leftrightarrow 27\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)\left( {xy \plus{} yz \plus{} zx} \right)\left( {xy \plus{} yz \plus{} zx} \right) \le {\left( {x \plus{} y \plus{} z} \right)^6}$ (which is true from AM-GM inequality)

Therefore,

$ \left[ {2\left( {x \plus{} y \plus{} z} \right) \plus{} \sqrt {3\left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)} } \right]\left( {xy \plus{} yz \plus{} zx} \right) \le {\left( {x \plus{} y \plus{} z} \right)^3}$

It suffices to prove that

$ {\left( {x \plus{} y \plus{} z} \right)^4} \le \left( {{x^2} \plus{} {y^2} \plus{} {z^2}} \right)\left[ {{x^2} \plus{} {y^2} \plus{} {z^2} \plus{} 8\left( {xy \plus{} yz \plus{} zx} \right)} \right]$
  $ \Leftrightarrow 4\left( {xy \plus{} yz \plus{} zx} \right)\left( {{x^2} \plus{} {y^2} \plus{} {z^2} \minus{} xy \minus{} yz \minus{} zx} \right) \ge 0$ (which is true)

The proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$

[/u][/i][/b]

[quote="mathVNpro"]Let $ a,b,c$ be three positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove:
\[ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac {9}{2\left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right )}\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-36-unknown-author.html

[quote="mathVNpro"]Let $ a,b,c$ be three positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove:
\[ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} bc} \plus{} \frac {1}{1 \plus{} ca}\ge \frac {9}{2\left (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right )}\]
[/quote]

this is my problem

initially I have not solution so I posted to vimf.freeforum.org

and I received a solution very nice from NGUYỄN ANH TUẤN


Given that A.B.C are points of a circle such that arc AB= Arc AC = Arc BC = 120. P is any point of arc AB. Prove that PA+PB=PC.

Can Someone please solve this. Thank You

[hide="Big hint"]If you draw in segments AB, BC and CA, you have yourself a cyclic quadrilateral with diagonals.[/hide]


Let there be $ a_1$ objects of first kind, $ a_2$ objects of second kind and so on up to $ a_n$ objects of $ n$ th kind with both $ n$ and $ a_j,\forall 1\le j\le n$ being finite and $ a_j\ge 2\quad \forall 1\le j\le n$. How many arrangements (when placed on a straight line) are there so at least one neighbour of every object is of same kind?

Provided $ a_1\plus{}\dots \plus{}a_n\equal{}k$, when does the number of arrangements become maximum?

In other words, in a required arrangement, elements of each kind should appear in groups of two or more.
Let $ g_i$ stand for the number of (maximal) groups of the $ i$-th kind appearing in an arrangement. Then the total number of arrangements is given by
\[ \sum_{g_1 \equal{} 0} \binom{a_1 \minus{} g_1 \minus{} 1}{g_1 \minus{} 1} \sum_{g_2 \equal{} 0} \binom{a_2 \minus{} g_2 \minus{} 1}{g_2 \minus{} 1} \dots \sum_{g_n \equal{} 0} \binom{a_n \minus{} g_n \minus{} 1}{g_n \minus{} 1} M(g_1,g_2,\dots,g_n)\]
where each $ \binom{a_i \minus{} g_i \minus{} 1}{g_i \minus{} 1}$ accounts for the number of compositions of $ a_i$ elements into $ g_i$ groups of size at least two each, and $ M(g_1,g_2,\dots,g_n)$ is the number of sequences of groups with $ g_i$ groups of the $ i$-th kind where every two adjacent groups are not of the same kind. The formula for  $ M(g_1,g_2,\dots,g_n)$ is given in the paper:
L. Q. Eifler, K. B. Reid Jr., D. P. Roselle  [url=http://dx.doi.org/10.1007/BF01819761]"Sequences with adjacent elements unequal"[/url]. Aequationes Mathematicae (1971), 6 (2-3), 256-262. (see also a [url=http://www.cse.sc.edu/~maxal/gpscripts/]PARI/GP code[/url] for computing $ M()$)


abstract algebra
advanced fields
advanced fields unsolved

Suppose that $R$ is a commutative ring, and $P = R[x,y]$. Consider $R$ as a $P$-module via the map $P \to R$ by evaluation at $x=0$ and $y = 0$. What is $\text{Ext}^{n}_{P}(R,R)$ for $n \in \mathbb{N}$?

Use an explicit free resolution of the first argument. (This is an easy special case of the Koszul complex.) owk

I just want to make sure I have the right idea before I start the computations.

I consider the free resolution
\[0 \to P \stackrel{f}{\to}P \oplus P \stackrel{g}{\to}P \stackrel{h}{\to}R \to 0 \]
where $f(p) = (Yp, Xp)$, $g(p,q) = Xp-Yq$ and $h(p) = p1 = p(0,0)$. This induces
\[\text{Hom}(P,R) \stackrel{h'}{\to}\text{Hom}(P \oplus P ,R) \stackrel{g'}{\to}\text{Hom}(P,R) \to 0 \]
Then $\text{Ext}_{1}(R,R) = \text{ker}(g')/\text{im}(h')$, $\text{Ext}_{2}(R,R) = \text{Hom}(S,R)/\text{im}(g')$, and $\text{Ext}_{n}(R,R) = 0$ for $n > 2$. Am I on the right track?

Yes, you are. (But your notation appears to be mixed up.) The remaining calculations are really easy. owk

Yes the notation is mixed up. It should be
\[\text{Hom}(P,R) \stackrel{g'}{\to}\text{Hom}(P \oplus P,R) \stackrel{f'}{\to}\text{Hom}(P,R) \to 0, \]
$\text{Ext}_{1}(R,R) = \text{ker}(f')/\text{im}(g')$, and $\text{Ext}_{2}(R,R) = \text{Hom}(P,R)/\text{im}(f')$

Then it was not hard to show that both $f'$ and $g'$ are trivial maps, which implies that
\[\text{Ext}_{1}(R,R) = \text{Hom}(P \oplus P, R) \cong \text{Hom}(P,R) \oplus \text{Hom}(P,R) \cong R \oplus R \]and\[\text{Ext}_{2}(R, R) = \Hom(P,R) \cong R. \]
Does this sound right?

Thank you for all your help these past few days.


linear algebra
matrix
linear algebra unsolved

Let $A$ and $B$ be $n \times n$ matrices. Prove that if $e^{tA}$ is bounded for $t \geq 0$, then $e^{tA+B}$ is also bounded for $t \geq 0$.

I have a question :

What does it mean that a sequence of matrices is bounded ?

Actually, norm-bounded. I just copied the original statement of problem.

Sorry, i've just found it http://www.mathlinks.ro/Forum/viewtopic.php?highlight=e%5E%7BtA%7D&t=43488


logarithms
trigonometry
function
modular arithmetic
LaTeX

I know that
[code]$\text{log}_327=3$[/code] comes out as
$ \text{log}_327\equal{}3$

But is there any other way to write a logarithm without using \text{log} to save time?

Can't you just use \log:

$ \log_327\equal{}3$?

[code]$\log_3 27 = 3$[/code] or [code]$\log_{3} 27 = 3$[/code]

I recommend [b]lower[/b] notation.

Thanks both of you.  So basically
[code]$\log$[/code]
is the same as
[code]$\text{log}$[/code], right?

No they aren't; using \log gives you better spacing, and the font itself is slightly different:

$ \log_327$ --- \log_327

$ \text{log}_327$ --- \text{log}_327

Can't really notice the font difference on the forums, but I can tell on my pdfs.

You can also do this with other 'things':
$ \sin$

$ \cos$

$ \tan$

Note: You can also do "\mod" to make $ \mod$.

i wish i could become a mod like that......

also, many regular text-like functions can easily be inducted that way;

$ \csc$, $ \cot$, $ \sec$ are also examples

Oh yeah:
[code]$\bmod{5}$[/code]
$ \bmod{5}$
[code]$\pmod{5}$[/code]
$ \pmod{5}$


[quote="Sadi"]$\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\geq\frac{3}{a+b}$

-this is easy problem for beginers :)[/quote]

[hide]$\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\ge \frac{(x+y+z)^{2}}{(a+b)(xy+yz+zx)}$ by Cauchy

and the famous $(x+y+z)^{2}\ge 3(xy+yz+zx)$[/hide]

Dear Sadi

This problem is from the series of Thomas Mildorfs Inequalities. If you don't have that book you can't get that searching in google.com. GOOD LUCK!

davron latipov


limit
complex analysis
real analysis
real analysis theorems

If $\left(a_{n}\right)$ is a sequence of non-negative real numbers  with $\lim_{n\to\infty}a_{n}=0$, $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ and $\lim_{x\uparrow{1}}f(z)=a\in\mathbb{C}$ prove that $\sum{a_{n}}=a$.

The full converse of Abel's Theorem is false. Any partial converse - partial, because an additional hypothesis is imposed, typically on $a_{n}$ - is a Tauberian theorem, with the additional condition called the Tauberian condition. The Tauberian condition here is that $a_{n}\ge 0.$

Still that doesn't solve the problem. I was thinking to do something like $a=\lim_{x\uparrow{1}}f(x)=\lim_{x\uparrow{1}}\lim_{k\to\infty}f_{k}(x)=\lim_{k\to\infty}\lim_{x\uparrow{1}}f_{k}(x)=\lim_{k\to\infty}\sum_{i=0}^{k}a_{i}=\sum{a_{n}}$ but I am sure I am missing something since I am not using $\lim_{n\to\infty}a_{n}=0$ and $a_{n}\geq0$

With everything is sight positive, this is not complex analysis - it is real analysis, and I'm moving it.

Two comments: you won't ever need to explicitly use that $a_{n}\to0.$ The theorem could be rephrased as this: if $a_{n}\ge0$ and we consider only $x\ge 0,$ then

$\lim_{x\uparrow 1}\sum_{i=0}^{\infty}a_{i}x^{i}= \sum_{i=0}^{\infty}a_{i}$

The equality sign is meant to convey that if either side is infinite, then the other side must also be infinite.

The second is that this would be false if you ever left the positive numbers behind.

By $x\uparrow1$ I meant as $x$ approaches $1$ from inside the unit disc so there is no such thing as $x\geq0$. I understand the reasoning in the Real case but I cannot yet generalize it to the Complex. Help ??

Then I think this is false, although I don't have an immediate counterexample. Tauber's theorem says it's true if $a_{n}=o\left(\frac1n\right).$ Littlewood's Tauberian theorem says it's true if $a_{n}=O\left(\frac1n\right).$ Just the condition $a_{n}\to0$ shouldn't be good enough.

Take a look at the bottom of the page at http://www.jstor.org/view/0003486x/di961710/96p0039q/0

I believe this proves that the problem is solvable. So, is there anyone able to post a solution or at least a sketch? I sure have tried for these couple of days but my limited knowledge didn't lead me to any close-by result so far.


Suppose that $ K$ is a compact Hausdorff space and $ K\equal{} \cup_{n\equal{}0}^{\infty}A_n$, where $ A_n$ is metrizable and $ A_n \subset A_m$ for $ n<m$. Prove that $ K$ is metrizable.

[i]Z. Balogh[/i]

The idea is to reduce the problem to the case when $A_n$ are closed and afterwards to prove it for this special case.
Denote by $d_m$ the metrict of $A_m\,,m\in \mathbb{N}$. WLOG, we can assume $d_m(x,y)\leq 1, \forall x,y\in A_m$, otherwise we can take $\overline{d_m}=\min\{d_m,1\}$.
EDIT>: Proceeding, will we need that:

$d_{m_1}(x,y) \ge C d_{m_2} (x,y) \,, \forall x,y\in A_{m_1} \,\,\,\,\,\,\, \text{(1)}$

for any $m_2>m_1$, where $C$ is a constant depending only on $m_1$ and $m_2$. It is not guaranteed, but we can ensure it by changing the metrics and  leaving the induced topologies intact. Namely let us define:
\[\overline{d_n}(x,y)=\sum_{m\ge n}2^{n-m}d_m(x,y)\,\, \forall x,y\in A_n  \]
Clearly $\overline{d_n} \ge \frac{1}{2} \overline{d_{n+1}}$. Thus, WLOG we can assume $\text{(1)}$ is true.
>end of EDIT.
Fix $n$. To show that $\overline{A_n}$ is metrizable, it is natural somehow to extend by continuity the metric $d_m$ over $A_m\,,m\geq n$ to a metric over $\overline{A_n}$. The problem is that it can happen it couldn't be done for some set of "bad" points of $\overline{A_n}\setminus A_m$. We will prove that when $m\to\infty$ these "bad" points will eventually vanish.
Let us denote by $B_{n,m}\,\,, n\le m$ the set of these "bad" points, i.e.:

$ B_{n,m}=\left\{x\in \overline{A_n}\mid \exists \varepsilon>0\,,\forall U\in\tau(K),x\in U,\, \exists u_1,u_2\in U\cap A_m\,,\,d_{m}(u_1,u_2)>\varepsilon  \right\} $.

Some properties of $B_{n,m}$
1) $B_{n,m} \subset \overline{A_n}\setminus A_m$
2) $B_{n,m}$ is closed.
3) $B_{n,m_{2}}\subset B_{n,m_{1}}\,,\, m_2>m_1 $

To prove 2) let $b$ is a limit point of $B_{n,m}$, we must show $b\in B_{n,m} $. Clearly $b\in \overline{A_n} $. Take some neighbourhood $U$ of $b$. There exists $b_1\in B_{n,m}$ with $b_1\in U$. Since $K$ is compact and Hausdorff, it is regular (in fact it is normal), thus we can separate $b_1$ from $K\setminus U$, in particular there exists open set $V$, $b_1\in V\,, V\subset U$. Applying the definition of $B_{n,m}$, with respect to $b_1\,,V$ we establish the existence of "not close"point $u_1,u_2\in U$, thus $b\in B_{n,m}$.
3) can be proved in a similar way.
Now, since $B_{n,n}\supset B_{n,n+1}\supset \ldots$ , $B_{n,m}$ are compact and $\bigcap_{m} B_{n,m}=\emptyset$, it follows $B_{n,m}=\emptyset$ for all sufficiently big $m$. Then, we can extend by continuity the metric $d_m$ over $\overline{A_n}$.

It remains to consider the case $A_n$ are closed. The idea is to embed $K$ into the cube $[0,1]^{\mathbb{N}}$ with the product topology.
Since each $A_n$ is a compact metric space, it is separable. (one can consider the finite $1/k$ nets, $k=1,2,\ldots$). Let $a^{(n)}_i\in A_n \,,i\in\mathbb{N}$ be a dense set in $A_n$.
Define functions $f^{(n)}_i :A_n \to [0,1]\,, f^{(n)}_i(x)=d_n(x,a^{(n)}_i )$. They are continuous functions on a closed sets of normal topological spase $K$. Applying the Tietze extension theorem, we can extend $f^{(n)}_i $ to continuous map $f^{(n)}_i : K\to [0,1]$. Let us denote $F=\{f^{(n)}_i \mid n,i \in \mathbb{N}\}$. We consider the map: $\phi : K\to [0,1]^F\,, \phi(x)=\langle f(x) \rangle_{f\in F}$.
One can check straightforwardly that:

1) $\phi$ is continuous. ( considering $[0,1]^F$ with the product topology)
2) $\phi$ is injective.
3) $\phi$ is open mapping. (in $\phi(K)$ with the induced topology)

It follows that $\phi$ is a homeomorphism between $K$ and $\phi(K)$. Since $[0,1]^F$ is metrizable, the result follows.

As a rule, there are always things one should patch in the second glance.
Here, as written "... 3) $B_{n,m_{2}}\subset B_{n,m_{1}}\,,\, m_2>m_1$..." is not generally true. To prove it, we need:

$d_{m_1}(x,y) \ge C d_{m_2} (x,y) \,, \forall x,y\in A_{m_1} \,\,\,\,\,\,\, \text{(1)}$

where $m_2>m_1$ and $C$ is a constant depending only on $m_1$ and $m_2$. It is not guaranteed, but we can ensure it by changing the metrics and  leaving the induced topologies intact. Namely let us define:
\[\overline{d_n}(x,y)=\sum_{m\ge n}2^{n-m}d_m(x,y)\,\, \forall x,y\in A_n  \]
Clearly $\overline{d_n} \ge \frac{1}{2} \overline{d_{n+1}}$. Thus, WLOG we can assume $\text{(1)}$ is true.
I have fixed it in the above post.


geometry
power of a point
radical axis
geometry solved

Excircle of triangle ABC (opposite A) touches side AB, BC, CA at points K, L, M respectively. Incircle of triangle ABC touches sides AB, BC, CA at points P, Q, R respectively. Line AQ meets excircle at two points and  T is  the nearest A. Line AL meets incircle at two points and  S is the neatest L. Prove that lines SQ, TL meet on line joining midpoints RM and PK.

First of all, the line joining the midpts of $RM,PK$ is simplythe radicalaxis of the two circles. Since $A$  is the pole ofan inversion which turns the two circles into each other, it means that $AQ\cdot AT=AL\cdot AS$,which means that $Q,T,L,S$ are concyclic, so $XS\cdot XQ=XT\cdot XL$, where $X=SQ\cap TL$. In turn, this means that $X$ has equal powers wrt the two circles,so it lies on the radical axis, Q.E.D.


For some reasons the problems from 99 and 2000 that are on the official ARML website are very difficult to read because the letters are blurry. Any methods of dealing with this problem.

Also, I don't understand why the ARML problems always state  let T=TNYWR?

[color=cyan]Only the relays have TNYWR.  It means "The Number You Will Receive."  Because someone is going to hand you a number.  And they want you to call it T.

If you print out the ARML questions, I think they come out nicer.  At least, that was my experience last time I tried.[/color]

When I printed the ARML they didn't come out too nice.

can you still read it?  I think when i printed them out, it was still readable, and it was also readable on the computer.  Can you read it fine from the website?

are you zooming in?  if you put your mouse over the picture for a second or two, an icon will appear and you click it and it will get larger so that you can read it.  If you dont do that, you cant read it.


algebra
polynomial
calculus
derivative
floor function
combinatorics unsolved
combinatorics

Each of the subsets $ A_1$, $ A_2$, $ \dots,$ $ A_n$ of a 2009-element set $ X$ contains at least 4 elements. The intersection of every two of these subsets contains at most 2 elements. Prove that in $ X$ there is a 24-element subset $ B$ containing neither of the sets $ A_1$, $ A_2$, $ \dots,$ $ A_n$.

[b]Proof[/b]:

Firstly we can simply assume $ |A_i|\equal{}4$ If not,we can repalce $ A_i$ by a $ 4\minus{}element$ subset of $ A_i$

We randomly choose $ 24$ element from $ X$ call it $ Y\equal{}\{1,2,...24\}$

Of course,$ Y$ may contain some $ A_i$,denote $ m$ by the number of $ A_i$ such that $ A_i$ is contained in $ Y$,($ m$ is obviously finite)

All we need to do is to decrease $ m$ :D

WLOG let $ A_1\equal{}\{1,2,3,4\}$ contained in $ Y$

We delete $ 1$ can add an element from $ \{25,26,...2009\}$

After deleting $ 1$,$ m$ decrease by at least $ 1$,if we add $ i$ into $ Y$,$ \{i,p,q,r\}$($ i \in \{25,26,...2009\},p,q,r \in \{1,2,...24\}$) isn't any of $ A_i$

Then we are done.So for any $ i \in \{25,26,...2009\}$ $ \exists p,q,r,A_j$ so that $ A_j\equal{}\{i,p,q,r\}$

Let us count the possible $ A_j$,It is easily to show there are at most $ \binom{23}{3}$ such $ A_j$(because $ |A_i \cap A_j|\le 2$ and $ |A_j|\equal{}4$)

but from $ \{25,26,...2009\}$ we are allowed to choose $ 1985$ which is bigger than $ \binom{23}{3}$

Contradiction!

QED

This is a new song on an old motif (triplets with at most one point in common).
Take $ | X | \equal{} n \geq 4$, and a maximal cardinality set $ B$ with that property (such sets exist, e.g. any set of at most 3 elements); denote $ | B | \equal{} b$. Then for any $ x \in X \setminus B$ there exists at least one set $ x \in A_i$ such that $ A_i \setminus \{ x \} \subseteq B$ (otherwise we can append $ x$ to $ B$). Since $ | A_i \setminus \{ x \} | \geq 3$, one may associate to $ x$ any triplet $ T_x \subseteq A_i \setminus \{ x \}$. For $ y \neq x$, $ y \in X \setminus B$, with set $ A_j$ associated, one has $ T_x \neq T_y$, or else $ | A_i \cap A_j | \geq 3$. Therefore, by counting all possible triplets in $ B$, one has $ | X \setminus B | \equal{} n \minus{} b \leq \genfrac {(} {)} {0pt} {} {b} {3}$, leading to $ b^3 \minus{} 3b^2 \plus{} 8b \geq 6n$. This is a strictly increasing polynomial in $ b$ (its derivative is strictly positive), and one checks that $ b \geq \lfloor \sqrt [3]{6n} \rfloor \plus{} 1$. For $ n \equal{} 2009$ this yields $ b \geq 24$.


Για το φίλο [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=461337#p461337]jj-the-king[/url] (και για οποιονδήποτε άλλο, φυσικά) :)

Μπορεί να θεωρηθεί σαν προγραμματιστικός γρίφος.

Είναι γνωστός ο τρόπος με τον οποίο κάνουμε "swap" (ανταλλαγή τιμών) σε δύο μεταβλητές Α,Β:

C=A;
A=B;
B=C;

Υπάρχει τρόπος να κάνουμε ανταλλαγή τιμών [u]χωρίς[/u] τη χρήση βοηθητικής μεταβλητής?

A = A xor B
B = B xor A
A = A xor B

Και έχει αντιστραφεί το (A, B)

Οι πολλαπλασιασμοί (πόσο μάλλον οι διαιρέσεις) είναι "βαριές" πράξεις, οπότε συμφέρει μόνο για $x = y = 1$, που είναι ακριβώς ο τρόπος του dionyzi.


superior algebra
superior algebra unsolved

$ F \subset K \subset F(x)$ and $ F\neq K$ where $ F,K$ fields.

prove that $ K\equal{}F(f)$ where $ f \in F(x) \backslash F$

I can reduce it to the case $ K \equal{} F(f,g)$ where $ f \in F(x) \backslash F , g \in F(x)$. does anyone have an idea how to get rid of the $ g$?


What is the units digit of $13^{2003}$?

$\left(A\right) 1$
$\left(B\right) 3$
$\left(C\right) 7$
$\left(D\right) 8$
$\left(E\right) 9$

[hide]C. There's a pattern. :lol: [/hide]

[hide]Since multiplication will yield the same last digit as the product of the last digits, we can simply consider the last digit of $3^{2003}$.

The last digit of $3^{1}$ is obviously $3$. The last digit of $3^{2}$ is obviously $9$. The last digit of $3^{3}$ is $7$. The last digit of $3^{4}$ is $1$. The last digit of $3^{5}$ is $3$, so the pattern is repeating.

Thus, for any integer $n \ge 0$, if:

$n \equiv 0 \bmod{4}$, then the last digit of $3^{n}$ is $1$.

$n \equiv 1 \bmod{4}$, then the last digit of $3^{n}$ is $3$.

$n \equiv 2 \bmod{4}$, then the last digit of $3^{n}$ is $9$.

$n \equiv 3 \bmod{4}$, then the last digit of $3^{n}$ is $7$.

Since $2003 \equiv 3 \bmod{4}$, the last digit of $3^{2003}$ is $7$, and so the last digit of $13^{2003}$ is also $7$. C.

This pattern, of course, doesn't require mods. However, I included the modular explanation for the sake of clarity.
[/hide]


geometry
3D geometry
sphere
arithmetic sequence
algebra
linear equation

Let $s_1$ be the sum of the first $n$ terms of the arithmetic sequence $8,12,\cdots$ and let $s_2$ be the sum of the first $n$ terms of the arithmetic sequence $17,19\cdots$. Assume $n\ne 0$. Then $s_1=s_2$ for:

$\text{(A)} \ \text{no value of n} \qquad \text{(B)} \ \text{one value of n} \qquad \text{(C)} \ \text{two values of n}$
$\text{(D)} \ \text{four values of n} \qquad \text{(E)} \ \text{more than four values of n}$

Yeah you are right, but still, I'm kind of bored sohere is the solving process:
$S_1=\frac{n[16+(n-1)4]}{2}$
$S_2=\frac{n[34+(n-1)2]}{2}$

Setting them equal to each other and canceling out some terms we get:
$16+(n-1)4=34+(n-1)2$
$16+4n-4=34+2n-2$
$2n=20$
$n=10$
Done

[quote="ShoeFactory"][quote="4everwise"][hide]Apply formula for sum of arithmetic series. Setting them equal, we get a linear equation for $n$, so there is clearly only 1 solution.  :) [/hide][/quote]

I didnt take a complete look at the question, but just because its a linear equation doesnt mean theres 1 solution for this problem.

What if n is negative or a fraction? it clearly doesnt work for a series.[/quote]

I think we can assume that n is a positive integer. It is irrevalent, since a fractional/negative of n would not give the correct answer either.

Also, linear equations describe straight lines. Two lines intersect only at one point, so a linear equation only has one solution.

[quote="236factorial"]Two lines intersect only at one point, so a linear equation only has one solution.[/quote]

Two lines can ALSO be parallel, in which case they don't intersect each other, or can be congruent, in which case they have infinitely many points of intersection.

Examples:

The linear equation $2x+2=2x+3$ has no solution, and the equation $3x+5=3x+5$ has infinitely many solutions.


Harley had an 8-gallon keg of wine and a jar. One day, he drew off a jarful of wine and filled up the keg with water. Later on, when the wine and the water got thoroughly mixed, he drew off another jarful and again filled up the keg with water. The keg then contained equal quantities of wine and water. What was the capacity of the jar?

Let the capacity of the jar be $ x$.

After the first time, the keg contained $ 8\minus{}x$ of wine and $ x$ of water and the jar contained $ x$ of wine and then none.

Then he takes $ \frac{x}{2}$ of wine and $ \frac{x}{2}$ of water from the keg, leaving $ 8\minus{}\frac{3x}{2}$ of wine and $ \frac{x}{2}$ of water.

$ 8\minus{}\frac{3x}{2}\equal{}\frac{x}{2}$

$ x\equal{}\frac{8}{3}$.

HOW DO we get this?
CORRECT answer should be 8-4(sqrt 2) Gallons

Sorry orangefronted, I've made a very silly mistake: when he drew off the second jar it was not half wine and half water.  In fact, $ \frac{(8\minus{}x)}{8}$ of the $ x$ gallons in the jar is wine and the other $ \frac{x}{8}$ of the $ x$ gallons in the jar is water. This leaves $ (8\minus{}x) \minus{} \frac{(8\minus{}x)}{8}x$ wine and $ x\minus{}\frac{x^{2}}{8}$ water. He then adds $ x$ gallons of water, so that the equal quantities of wine and water in the keg are given by

$ (8\minus{}x) \minus{} \frac{(8\minus{}x)}{8}x\equal{}x\minus{}\frac{x^{2}}{8} \plus{}x$

Multiplying by $ 8$, we get

$ 64\minus{}8x\minus{}8x\plus{}x^{2} \equal{} 16x\minus{}x^{2}$

$ 2x^{2}\minus{}32x\plus{}64\equal{}0$

$ x^{2} \minus{} 16x \plus{}32\equal{}0$

Applying the quadratic formula, one root is greater than 8 and so the required value of $ x$ is the smaller root: $ 8\minus{}4\sqrt{2}$.

I hope this is OK.


inequalities
trigonometry
geometry proposed
geometry

Prove that $ \boxed {\ \frac pR \plus{} 2\cdot\left(\cos\frac A2 \plus{} \cos\frac B2 \plus{} \cos\frac C2\right)\le 3\sqrt 3\cdot\left(\sin\frac A2 \plus{} \sin\frac B2 \plus{} \sin\frac C2\right)\ }\ ,$

where $ 2p \equal{} a \plus{} b \plus{} c$ and $ R$ is the length of the circumradius.

the inequality is equivalent to

$ \sum \sin A\plus{}2\cos\dfrac{A}{2}\leq 3\sqrt3\sum \sin\dfrac{A}{2}$.

we'll prove a more general result,

$ \sum \sin 2A\plus{}2\sin A\leq 3\sqrt 3\sum \cos A$.

(the first inequality follows letting ABC to be an acute triangle and taking $ (A,B,C)\equal{}\left(90\minus{}\frac{A}{2},90\minus{}\frac{B}{2},90\minus{}\frac{C}{2}\right)$).

recall that
$ \sum \sin 2A\equal{}4\prod\sin A\equal{}32\prod \cos \dfrac{A}{2}\sin\dfrac{A}{2}$,

$ \sum \sin A\equal{}4\prod \cos\dfrac{A}{2}$,

and $ \sum \cos A\equal{}1\plus{}4\prod\sin\dfrac{A}{2}$.

then,

$ \sum \sin 2A\plus{}2\sum\sin A\equal{}32\prod \cos \dfrac{A}{2}\sin\dfrac{A}{2}\plus{}8\prod \cos\dfrac{A}{2}\equal{}8(1\plus{}4\prod\sin\dfrac{A}{2})\prod \cos\dfrac{A}{2}.$

the result follows from the well-known inequality $ 8\prod \cos\dfrac{A}{2}\leq 3\sqrt3$.


[color=darkred]Let $ ABC$ be a triangle for which $ AB\ne AC$ .
Denote the its centroid $ G$ and the its incircle $ C(I,r)$ .
Prove that $ IB\cdot IC \equal{} r\cdot IA$ $ \Longleftrightarrow$ $ IG\perp BC\ .$[/color]

Let $M$ be the midpoint of $BC,$  $D$ the tangency point of the incircle $(I)$ with $BC$ and $P$ the second intersection of $AI$ with the circumcircle $(O)$ of $\triangle ABC.$ It's well-known that $P$ is the circumcenter of $\triangle IBC.$ Thus, $IP,ID$ are isogonals with respect to $\angle BIC$  $\Longrightarrow$ $IB \cdot IC \equal{} 2ID \cdot IP \equal{} 2r \cdot IP \ (\star).$  Obviously, $IG \perp BC$ $ \Longleftrightarrow$ $IG \parallel MP.$  Hence by Thales theorem we claim that $IG \perp BC$  $\Longleftrightarrow$ $\frac {_{AI}}{^{IP}} \equal{} \frac {_{AG}}{^{GM}} \equal{} 2.$  Then, from the expression $(\star),$  it follows that $IG \perp BC$ $\Longleftrightarrow$ $IB \cdot IC \equal{} r \cdot IA.$


Find all couples $ (a,\ b)$ of integers such that $ a^2b^2\plus{}a^2\plus{}b^2\minus{}4a\minus{}4b\plus{}3\leq 0$.

[quote="kunny"]Find all couples $ (a,\ b)$ of integers such that $ a^2b^2 \plus{} a^2 \plus{} b^2 \minus{} 4a \minus{} 4b \plus{} 3\leq 0$.[/quote]
We have $ (a\minus{}2)^2\plus{}(b\minus{}2)^2 \ge 0$ so $ a^2\plus{}b^2\minus{}4a\minus{}4b\plus{}8 \ge 0$.
It follows that $ a^2b^2 \le 5$ or $ a^2b^2\equal{}0,1,4$
If $ a^2b^2\equal{}0$ or $ a^2b^2\equal{}1$,  impossible
If $ a^2b^2\equal{}4$, we have $ (a,b)\equal{}(1,2),(2,1)$

[quote="NguyenDungTN"]
It follows that $ a^2b^2 \le 5$ or $ a^2b^2 \equal{} 0,1,4$
If $ a^2b^2 \equal{} 0$ or $ a^2b^2 \equal{} 1$,  impossible[/quote]

Why?

$ a^2b^2 \plus{} a^2 \plus{} b^2 \minus{} 4a \minus{} 4b \plus{} 3\leq 0 \Leftrightarrow a^2b^2 \plus{} (a \minus{} 2)^2 \plus{} (b \minus{} 2)^2 \leq 5$
So $ \minus{} 2\leq ab, (a \minus{} 2), (b \minus{} 2)\leq 2$, wich implies $ 0\leq a,b \leq2$
From here it is easy to see that the couples that work are $ (0,1)$,$ (0,2)$,$ (1,1)$,$ (1,2)$ and its permutations.

 :)

You can get one more pair without considering permutation.


Hey ,
Check this out -- It's fun....

http://www.actofme.co.uk/bush_speech/bushspeechwriter.html

Go on posting some other interesting websites as well...

the nintendo choir one is hilarious.
i couldn't stop laughing.  :rotfl:
Except... it's somewhat out of tune. but thas ok.
it was still funny XD


Find all triples of positive integers x, y, z with
x ≤ y ≤ z
 such that
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}= 1$.

when

$x=y=z$ i think there is only one solution that is
$x=y=z=3$

$x=1/4,y=1/4,z=1/2$ also works..

Well, I think there would be infinitely many, but I am retarded so I can't really do it.

ya there's probably way more i just thought i would post the obvious one's  :lol:

Actually come to think of it, I think there are only a finite number of triples.

Kurt Godel just posted before me, and yeah I thought wrong initiially.

3 solutions that is a lot.

[hide]Two values of $1/k$ are $1/2$ and $1/3$. Clearly two of the numbers can't be $1/2$, and if one is $1/2$ and the other is $1/3$ we get $1/6$ back. So $1/6$ is our smallest number. $1/5$ obv. doesn't work.

The average term is $1/3$, so unless we have a $1/2$, we must have $1/3,\;1/3,\;1/3$.

If we do have a $1/2$, we have left to make a $1/2$, which can be done with $1/4,\;1/4$ or $1/3,\;1/6$.

Triples: $(3,3,3),\;(2,4,4),\;(2,3,6)$.[/hide]


Since you didn't mention topology, I'm going to assume you mean the linear dual, which is very large for infinite-dimensional spaces.

An example:
Let $ V$ be a space with countable dimension, spanned by vectors $ e_1,e_2,\cdots$. Let $ e_i^*$ be the functional which takes $ e_i$ to 1 and $ e_j$ to zero for all other $ j$. The vector space $ S$ spanned by $ e_i^*$ is a proper subspace of $ V^*$, and annihilates only zero.

Doesn't the vector space spanned by $ e_i^*$ annihilate the nonzero vector $ e_j$, $ j \ne i$, by definition?

I was trying to show that if V cannot be written as a finite union of proper subspaces if and only if for each finite subset S of V (not containing the zero vector) there is a linear functional which does not send any member of S to zero. Kent Merryfield made that claim a while ago but I did not think much of it at the time.

I'm having a hard time proving $ \Leftarrow$ though.

I mean the vector space spanned by [b]all[/b] of the $ e_i^*$.

[quote="§outh§tar"]If V is a linear space over an infinite field and S is a proper subspace of V*, can the subspace annihilated by S be {0}?[/quote]
Let V be the (real) vector space of all bounded sequences of real numbers. For $ x \equal{} (x_1,x_2,x_3,\ldots)\in V$ let $ e_n(x) \equal{} x_n$, and let S be the subspace of V* spanned by the functionals $ e_n$.

Clearly  the subspace annihilated by S is {0}. But S is not the whole of V*. For example, the functional f given by $ f(x) \equal{} \lim_{\mathcal U}x_n$ (where $ \mathcal U$ is a free ultrafilter on $ \mathbb N$) is not in S.

[b]Edit.[/b] In fact, that is just a slightly more concrete version of [b]jmerry[/b]'s example above.

Mine was perfectly concrete- the difference is that you've shifted up to uncountable dimension, and thus can't easily describe a linear functional outside your chosen subspace.

For my example, $ \sum_{i\equal{}1}^{\infty} x_i$ is a linear functional outside the chosen subspace.


Consider $ a,b,c \in \mathbb{R}$ so that $ a\plus{}b\plus{}c\equal{}1$. Prove that at least one of the numbers $ a\minus{}3bc,b\minus{}3ca,c\minus{}3ab$ is nonnegative.

[quote="moldovan"]Consider $ a,b,c \in \mathbb{R}$ so that $ a \plus{} b \plus{} c \equal{} 1$. Prove that at least one of the numbers $ a \minus{} 3bc,b \minus{} 3ca,c \minus{} 3ab$ is nonnegative.[/quote]
$ (a \minus{} 3bc) \plus{} (b \minus{} 3ca) \plus{} (c \minus{} 3ab) \equal{} (a\plus{}b\plus{}c) \minus{} 3(ab\plus{}bc\plus{}ca) \equal{} (a\plus{}b\plus{}c)^2\minus{}3(ab\plus{}bc\plus{}ca) \equal{} \frac{1}{2} ( (a\minus{}b)^2 \plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 ) \ge 0$. The sum is nonnegative and hence: At least one of the numbers is also.


induction
vector
analytic geometry
ratio
LaTeX
inequalities
algebra unsolved

[quote="Lek Huo Tang"]Here is how the problem goes:

An infinite sequence $\{x_n\}$ ($x_n \neq 0$, $n \ge 3$) satisfies the relation \[\ (x^2_1 + x^2_2 + ... + x^2_{n-1})(x^2_2 + x^2_3 + ... + x^2_{n}) = (x_1x_2 + x_2x_3 + ... + x_{n-1}x_n)^2.\] Prove that the sequence $ \{x_n\} $ is a geometric progression.

It is easy to prove for $n=3$ and I suppose we need to use induction for $n\ge 4$. But how should we proceed?

Should we assume that $x_i=aq^{i-1}$ for $1\le i \le n-1$ and then deduce that $x_n=aq^{n-1}$? But that will lead to some messy algebra, I think.[/quote]

this is what we call the case of equality of the Cauchy's inequality.
Applying this result we get that the vectors $u=(x_1,x_2,\ldots,x_{n-1})$ and
$v=(x_2,x_3,\ldots,x_{n})$ are colinear. Then there exists a real $\lambda$
such that $u=\lambda v$.
I think I said enough - just write down coordinate by coordinate to get
the geometric progression.

Fabrice

This is the solution I have, but I don't understand it.

When $n=3$, $x^2_2=x_1\cdot x_3$, and it is clear that these terms form a GP.

Suppose that the statement is true for $n=k$ $(k\ge3)$, $\{x_n\} (n=1, 2, ..., k)$ is a GP with a common ratio of $q$. Then when $n=k+1$, \begin{eqnarray*} (x^2_1+x^2_2+...+x^2_{k-1}+x^2_k)(x^2_2+x^2_3+...+x^2_k+x^2_{k+1}) = (x_1x_2+x_2x_3+...+x_{k-1}x_k+x_kx_{k+1})^2 =2abc=2xyz+ytup\end{eqnarray*}
and if
$x^2_1+x^2_2+...+x^2_{k-1} = a^2 (a\neq0)$,
then $(x_kq-x_{k+1})^2 = 0$, therefore the result holds for $n=k+1$ as well.

I think a lot of steps have been skipped. Please fill up the holes for me.

Sorry those abcxyz are nonsense, I just wanted to try alignment here. Apparently I was not successful.

Can we use LaTex to type equations in Microsoft Word?

[quote="Lek Huo Tang"]This is the solution I have, but I don't understand it.

When $n=3$, $x^2_2=x_1\cdot x_3$, and it is clear that these terms form a GP.

Suppose that the statement is true for $n=k$ .... Please fill up the holes for me.
[/quote]

The first thing you should do is to stop trying to use induction.
This won't work.
One way to tackle those sort of problems should be to study
the inequality of  [b]Cauchy-Scwharz[/b]:

$\sum\limits_{i=1}^{n} a_i b_i  \leq \sqrt{ \sum\limits_{i=1}^{n} {a_i}^2}
\sqrt{ \sum\limits_{i=1}^{n} {b_i}^2}$

and most of all things, the case of equality must be carefully learned.
One good geometric interpretation is to think of this inequality as a scalar
product in an Euclidian space - for example $\mathbb{R}^3$:
$(\vec{u} | \vec{v} ) \leq ||\vec{u} || \cdot ||\vec{v}||$.
The proof is not very difficult, but you need to be carefull
for the equality case, which is equivalent to the fact that
$\vec{u}$ and $\vec{v}$ are colinear - make a little drawing, with $\vec{u}$, $\vec{v}$,
the angle $\theta$  well-defined in the plane spanned by $\vec{u}$ et $\vec{v}$.
The result is then straightforward if you compute the scalar product of $\vec{u}$
 with $\vec{v}$ !
Then reread my proof, everything should be easier.

Fabrice

Thank you, in fact I fully understand your solution which uses Cauchy.

But I wonder why induction won't work in this case.


Show that if a square matrix X satisfies X^3 + 4X^2 − 2X + 7I = 0 then so does t(X) transpose of X, where I is the identity matrix and 0 denotes the zero matrix.

t(AB)= t(B)t(A)...not sure where your clue leads me...but I would appreciate more help. Thanks!

what do you obtain when you transpose the right side and the left side of your equation ?


pigeonhole principle
combinatorics
USAMO

Suppose that in a certain society, each pair of persons can be classified as either [i]amicable [/i]or [i]hostile[/i]. We shall say that each member of an amicable pair is a [i]friend[/i] of the other, and each member of a hostile pair is a  [i]foe[/i] of the other.  Suppose that the society has $\, n \,$ persons and $\, q \,$ amicable pairs, and that for every set of three persons, at least one pair is hostile.  Prove that there is at least one member of the society whose foes include $\, q(1 - 4q/n^2) \,$ or fewer amicable pairs.

Consider the graph with two people joined iff they are friends.

For each person $X$, let $A(X)$ be the set of its friends and $B(X)$ the set of its foes. Note that any edge goes either: from $X$ to $A(X)$ (type 1), from $A(X)$ to $B(X)$ (type 2) or from a point of $B(X)$ to another (type 3), but there's no edge joining two points of $A(X)$ (since they would form a triangle with $X$). Let the number of type 1, type 2, type 3 edges of $X$ be $x_1, x_2, x_3$ respectively, so that $x_1$ is the degree of $X$ and we want to show that for some $X$, we have $x_3 \leq q(1-4q/n^2)$.

Since each edge is of one of those types, we have $x_1+x_2+x_3=q$. Thus
\[ x_3 \leq q(1-4q/n^2) \Longleftrightarrow \\ x_3 \leq q-4q^2/n^2 \Longleftrightarrow \\ q-x_1-x_2 \leq q-4q^2/n^2 \Longleftrightarrow \\ x_1+x_2 \geq 4q^2/n^2. \]
That is, what we want is equivalent to proving that for some vertex $X$, the set of edges touching either $X$ or a vertex joined to $X$ is at least $4q^2/n^2$. Obviously now we'll sum $x_1+x_2$ over all vertices $X$. In the resulting sum, an edge joining $X, Y$ is counted once for each edge that $X, Y$ have, that is, it is counted $D(X)+D(Y)$ times, where $D(X)=x_1$ is the degree of $X$. Thus each vertex $X$ contributes to the overall sum with $D(X)$ for each edge it has, and since it has $D(X)$ edges, it contributes with $D(X)^2$. Thus the considered sum is equal to
\[ \sum_{X \in G} D(X)^2 \geq \frac{(\sum_{X \in G} D(X))^2}{n}=\frac{(2q)^2}{n}=4q^2/n. \]
(That's Cauchy.) Since we are summing over $n$ vertices, one of the summands is at least $4q^2/n^2$ by pigeonhole, which is what we wanted to prove.

Hmm, a less trickier way for less tricky people.
Call $q = E$ and $n = V$ for convenience. The graph is $3$-free if we look at the graph $G$ made by friend relationships, and let people be the edges.
Instead of looking at the subgraph induced by eliminating $A$'s friends, we do something a bit different. Instead, look at each edge, and consider say, edge ${u, v}$. We ask, how many times does a given edge have a vertex which is not connected to either? Clearly, if we sum up all of these, we will be able to finish. This is normally a very difficult task, but observe that the graph is $3$-free, so the vertices that $u$ is adjacent to are disjoint from those adjacent to $v$. Hence, there are $D(u) - 1 + D(v) - 1 = D(u) + D(v) - 2$ vertices adjacent to one of them (which are not $u$ or $v$), so there are $(V-2) - D(u) - D(v) + 2 = V - D(u) - D(v)$ vertices adjacent to neither $u$ nor $v$. We can sum all of these up over all edges, and then note that in order to finish, we will need less than $VE - 4E^2/V$ edges so that piegonhole suffices. Note that $\sum_{X\in V}D(X) = 2E$.
\[ \sum_{{u, v}\in E}V - D(u) - D(v)=EV-\sum_{{u, v}\in E}D(u) + D(v)=EV - \sum_{u \in V} D(u)^2 \le EV - \frac{4E^2}{V} \], (where the last step is Cauchy and $\sum_{X\in V}D(X) = 2E$.) so we are done.

Suppose that it is not possible. Create a graph to represent the scenario and label vertices $v_1,v_2,...,v_n$. Let $d_i$ be the degree of vertex $v_i$ and let $f(v_i)$ be the number of amicable pairs among the foes of $v_i$. Let's call a group of three vertices with one edge between them a "1-triplet" and with two edges between them a "2-triplet". Consider the sum $\sum_{i=1}^{n} d_i(d_i-1)+f(v_i)$. This counts all edges among the foes of $v_i$ each of which is used to make a 1-triplet with $v_i$ once, and the $d_i(d_i-1)$ term counts the number of 2-triples with $v_i$ adjacent to both edges twice. However, if we consider some edge $v_1v_2$ and one of the vertices $v_3,v_4,...,v_n$ then each vertex along with this edge either makes a 1-triplet or a 2-triplet, and 1-triplets are counted once while 2-triplets are counted twice as in the above sum.
Hence $\sum_{i=1}^{n} d_i(d_i-1)+f(v_i)=q(n-2)$. On the other hand $f(v_i)>q(1 - 4q/n^2)$.
Therefore $q(n-2)=\sum_{i=1}^{n} d_i(d_i-1)+f(v_i)>nq-4q^2/n+4q^2/n-2q=q(n-2)$ contradiction, hence there must be some vertex $v_i$ such that $f(v_i)\leq (1 - 4q/n^2)$.

Let $V(x)$ be the set of vertices connected to $x$, and let $V$ be the set of all vertices. If the result were false, then
$$k - \sum_{v \in V(x)} d(v) > k (1 - \frac{4k}{n^2})$$
for each $x \in V$. Thus by counting in two ways the number of triples $(A, B, C)$ where $A, C$ (not necessarily distinct) are both connected to $B$, we find that
$$nk (1 - \frac{4k}{n^2}) < \sum_{x \in V} (k - \sum_{v \in V(x)} d(v)) = nk - \sum_{x \in V, v \in V(x)} d(v) = nk - \sum_{v \in V} d(v)^2 \le nk - \frac{4k^2}{n}$$
by Cauchy-Schwarz, contradiction. Thus the result is true.

Nice and easy!And my first USAMO too :P Here's my solution
Rephrase the problem in graph theoric terms
[quote=USAMO 1995 Rephrased]Consider a $3$free-graph $G$ with $n$ vertices and $q$ edges.Connect two edges if they are amicable.Prove that there is atleast one  vertice $S$ such that the number of edges $e$ between $v$ and $w$ $v,w \in G/N[S]$ is atmost $q\left(1-\dfrac{4q}{n^2}\right)$ [/quote]
$\mathbf{Solution:}$ Assume the contrary that there is no such vertice $S$ satisfying the problem conditions.Then for any vertice $S$ the number of edges in $G/N[S]$ will be $q-d(S)$ but we have included the edges originating from the $N(S)$ also thus we have to subtract $d(S_i)-1
$$\forall S_i \in N(S)$.Note that the graph is $3$-free sotheir is no edge between two vertices $\in N(S)$,
$\therefore$ The number of edges will be $q-d(S)-(\sum_{S_i \in N(S)}d(S_i)-1)=q-(\sum_{S_i \in N(S)}d(S_i)) > \left(q-\dfrac{4q^2}{n^2}\right) \implies \dfrac{4q^2}{n^2} > \sum_{S_i \in N(S)}d(S_i)$.Doing this for all the edges we get $$\dfrac{4q^2}{n}=n\cdot\left(\dfrac{4q^2}{n^2}\right)>\sum_{S\in G}\sum_{S_i\in N(S)}d(S_i)=\sum_{S\in G}d(S)^2 \ge \dfrac{(\sum_{S\in G}d(S))^2}{n}=\dfrac{4q^2}{n}$$ A contradiction !! Hence our assumption is wrong and thus $\exists$ a member whose foes include atmost $q\left(1-\dfrac{4q}{n^2}\right)$ amicable pairs so we are done.$\blacksquare$

NOTE-$N[S]$ represents the closed neighbourhood of $S$ i.e. the set of vertices sharing an edge to $S$ and also including $S$ and $N(S)=N[S]/S$

I am confused about sabbasi's solution. When he says $\sum_{i=1}^{n} d_i(d_i-1)+f(v_i)$
[quote]counts all the edges among the foes of $v_i$ each of which is used to make a 1-triplet with $v_i$ once[/quote]
he is counting the number of edges in the spanning subgraph of the overall graph, consisting only of the foes of $v_i$. If this is the case, then shouldn't the number of edges among the foes of $v_i$ be $\sum_{i=1}^{n} d_i(d_i-1)-f(v_i)$ since two vertices are connected by an edge if they are enemies. Please correct me if I misunderstood something.

Note: I think I am misunderstanding what edges represent in sabbasi's solution.

Take the obvious graph theoretic interpretation.
Label the vertices $v_1,v_2,\dots,v_n.$ For $i\in\{1,2,\dots,n\},$ let $x_i$ be the number of edges adjacent to $v_i$ or a neighbor of $v_i.$\\\\ Since $G$ is triangle-free, $$x_i=\sum_{w\in N(v_i)}\text{deg}(w).$$ Therefore, $$x_1+x_2+\dots+x_n=\sum_{w}\text{deg}(w)^2\ge\frac{(\sum_{w}\text{deg}(w))^2}{n}=\frac{4m^2}{n}.$$ Thus, there exists some $k$ such that $x_k\ge\left(\frac{2m}{n}\right)^2,$ as desired.

We solve the following equivalent problem:
[quote = USAMO 1995/5 Reformulated]
Let $G$ be a triangle-free graph with $n$ vertices and $m$ edges. Show that one can select a vertex $v$ such that deleting vertex $v$ and all its neighbors kills at least $(\tfrac{2m}{n})^2$ edges.
[/quote]

If the neighbors of a vertex $v$ are $u_1,u_2,\dots,u_t$, then the number of edges lost by the specified deletion is
\[\sum_i \deg u_i\]
because there are no triangles. Observe that the expected value of the sum is
\[\sum_{u\in G}\deg u\cdot\frac{\deg u}{n}=\frac{1}{n}\sum_{u\in G}(\deg u)^2\ge \frac{1}{n^2}\left(\sum_{u\in G}\deg u\right)^2=\left(\frac{2m}{n}\right)^2\]
by the Cauchy-Schwarz Inequality, implying the desired.

Let $T_2$ be the number of sets of three people with $2$ amicable pairs, and let $T_1$ be the number of sets of three people with $1$ amicable pair. Since there are no sets of three people with $3$ amicable pairs, we have
\[ 2T_2 + T_1 = q(n-2),\]
by a simple double counting argument on the number of amicable pairs. Now if $A(p)$ is the number of friends of a person $p$, observe that
\[ \sum_{p} \binom{A(p)}{2} \ge n \binom{\sum\limits_p A(p)}{2} = n \binom {\frac{2q}{n}}{2} \]
by Jensen's inequality, so the number of pairs containing a person $p$ and a hostile pair of people who are both friends with $p$ is at least $n \binom {\frac{2q}{n}}{2}$. But this is simply $T_2$, so
\[ T_1 = q(n-2) - 2n \binom {\frac{2q}{n}}{2} = qn - 2q - n \left( \frac{2q}{n} \right) \left( \frac{2q}{n} - 1 \right) = qn - \frac{4q^2}{n}.\]
To finish, notice that there is a one to one correspondence between sets of three people with exactly $1$ amicable pair and pairs containing a person $p$ and a amicable pair of people who are both foes to $p$. Thus, a person $p$ has on average
\[ \frac1{n} \left(qn - \frac{4q^2}{n} \right) = q \left( 1 - \frac{4q}{n^2} \right),\]
amicable pairs in its foes, which implies the conclusion. $\blacksquare$

Construct a graph $G=(V, E)$ with vertex set $V$ being the set of $n$ people and edge set $E$ being the set of $q$ amicable pairs. By the problem statement, $G$ is triangle-free. For each vertex $v$, denote by $\text{FE}(v)$ the set of edges between foes of $v$. The task is to prove that there exists $v\in V$ such that $|\text{FE}(v)|\le q\left(1-\frac{4q}{n^2}\right)$.

We shall count the pairs $(v, e)$ where $v\in V$ is a vertex and $e$ is an edge in $\text{FE}(v)$. Counting from the perspective of $v$'s, number of pairs is $\sum_{v\in V} |\text{FE}(v)|$. Counting from the perspective of edges $uv$'s, since $u$ and $v$ have no common neighbour, the number of pairs is

\[\sum_{uv\in E} \left(n-\deg{u}-\deg{v}\right) = nq-\sum_{v\in V} (\deg{v})^2 \le nq - \frac{4q^2}{n}\]

where the inequality is due to Cauchy-Schwarz. Hence, we have

\[\sum_{v\in V}|\text{FE}(v)|\le nq-\frac{4q^2}{n} = nq\left(1-\frac{4q}{n^2}\right).\]

So, by pigeonhole principle, at least one $v\in V$ satisfies $|\text{FE}(v)|\le q\left(1-\frac{4q}{n^2}\right)$. $\square$

Fairly weak problem; every step is basically forced.

We solve the problem in the following alternative formation: given a triangle-free graph with $n$ vertices and $m$ edges, we show that we can select a vertex $v$ such that deleting $v$ and its neighbors removes at least $\left(\frac{2m}n\right)^2$ from the graph.

Consider pairs $(v, u, e)$ where $e$ is an edge and $v, u$ are vertices with $uv$ an edge such that $e$ is incident with $u$. First, fix $u$, and notice that there are $(\deg u)^2$ ways to pick $v$ and $e$. So the number of pairs is bounded below by $$k = \sum_v (\deg v)^2 \geq \frac{(2m)^2}n$$ by Cauchy-Schwarz.

But there are at most $n$ choices for $v$ and thus there exists a $v$ associated with $\left(\frac{2m}n\right)^2$ such pairs. Because no two neighbors $u_1, u_2$ of $v$ are connected, all such edges $e$ are then distinct for this $v$, which finishes.


Consider the sequence $p_{1},p_{2},...$ of primes such that for each $i\geq2$, either $p_{i}=2p_{i-1}-1$ or $p_{i}=2p_{i-1}+1$. Show that any such sequence has a finite number of terms.

Here are some observations, not a complete proof.  Suppose there is an infinite sequence of such terms.

First, note that 3 can't be one of the primes, since otherwise the sequence would quickly die out.  (Either 3-5-11-23-47 or 3-7-13.)

Second, we can't have a "-1" move followed by a "+1" move.  Otherwise, we would have the sequence p, 2p - 1, 4p - 1.  Looking mod 3, we see that one of these terms must be divisible by 3, contradicting the first observation.

Third, we can't have a "+1" move followed by a "-1" move.  Otherwise, we would have the sequence p, 2p + 1, 4p + 1.  Again looking mod 3, one of these terms must be divisible by 3, contradicting the first observation.

So, we have reduced to the simpler problem where either all the moves are "-1" or all the moves are "+1".

If all signs are + then sequence is  $p,2p+1,4p+3,...,2^{k}(p+1)-1,...$
one of fifth member divisible 5 if $5\not |(p+1)$.
Let q is prime, s.t. exist m: $q|2^{m}-p-1$, then $q|2^{k}(p+1)-1$ for all $k=r(q-1)-m$.
If all signs are - then sequence is $p,2p-1,4p-3,...,2^{k}(p-1)+1,...$
Let q is prime, s.t. exist m $q|2^{m}+p-1$, then for all $k=r(q-1)-m$ we have $q|2^{k}(p-1)+1$.


Very interesting problem, I will try to help but no insight for now... where do you get such problems?

This is very nice!  I solved it working backwards with calculus, but it turns out that you can turn it around forwards and an elementary solution comes out.

[hide="The key idea"]$\sqrt{2}$ is a lie.  Instead, use $\frac{3}{2}$.[/hide]

[hide="Solution"]For $y > 1, x > 0$ define $F(x, y) = \log_{y}\left(y^{x}+\left(\frac{y}{2}\right)^{x}\right)$.  We have:
$F(x, y) = \log_{y}\left(y^{x}\cdot \left(1+\frac{1}{2^{x}}\right)\right) = x+\log_{y}\left(1+\frac{1}{2^{x}}\right)$.  This expression is clearly decreasing with respect to $y$ (that is, when $x$ is held constant), so in particular $F(x, 2) > F(x, 3)$.

But this means $\log_{2}\left(2^{x}+1\right) > \log_{3}\left(3^{x}+\left(\frac{3}{2}\right)^{x}\right)$, and surely $\log_{3}\left(3^{x}+\left(\frac{3}{2}\right)^{x}\right) \geq \log_{3}\left(3^{x}+\sqrt{2}^{x}\right)$, so we're done.[/hide]


geometry
inequalities solved
inequalities

In any triangle $ ABC $ , prove that :
 $ 3a^2 + 2b^2+2c^2 \geq 16S $ , where $ a,b,c $ are 3 sides of triangle $ ABC $ and $ S $ is the area of triangle $ ABC $

Here is a hint : try to prove : $ b^2+c^2 \geq \frac{a^2}{2}+2h_a^2 $ (You can prove this one by drawing an attitude from A )  ;)

Nobody will answer you if you post your problem here. If you intend to post a problem you solved and want others to try it, please post it in the proposed section.

really good problem, jarod! :)
I think I worked it out!

Taking into account your hint we have

$\displaystyle  3a^2 + 2b^2+2c^2= 3a^2 + 2(b^2+c^2)\geq 3a^2+2(\frac{a^2}{2}+2h_a^2 )= 4a^2+4h_a^2\geq 8a \cdot h_a=8 \cdot 2S=16S$

and we are done!!


quadratics
calculus
integration
geometry
algebra
quadratic formula

I'm trying to get x all by itself, but I can't.
Help, please!

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ceaf0a4dd1dce936ed9cd735b43b0859.gif[/img]

edit: Oh, right. Haha. Thanks!

edit: OK. I have a problem again, x can't have more than one answer. How do I prove it has only one answer?

When you moved $a+b$ from the right hand side to the numerator, you forgot to multiply by $2x$. Since the equation is a quadratic, it can have two solutions. Say $a=8$ and $b=6$. Then $x=2$ and $x=12$ both satisfy the equation.

[hide="Quadratic Formula used..."]Multiply both sides by $2x$ and bring everything to $1$ side (by adding or subtracting), and you get:

$2x^{2}-2(a+b)x+ab=0$
Applying the quadratic formula,

$x = \frac{2(a+b)\pm\sqrt{4(a+b)^{2}-8ab}}{4}$

$=\frac{2(a+b)\pm\sqrt{4a^{2}+4b^{2}}}{4}$

$=\frac{(a+b)\pm\sqrt{a^{2}+b^{2}}}{2}$

Let $a=12$, $b=5$. Then $x=2$ or $x=15$. So there could be more than $1$ solution.

As long as $a=2rs$ and $b=r^{2}-s^{2}$, where $r$ and $s$ are integers and $r>s$, $x$ will have 2 integral values, namely $x=r(r+s)$ and $x=s(r-s)$.[/hide]

I'm sorry. I wasn't clear enough. The origin of this problem was actually a geometry problem. I have to prove that x is greater than 0 and is less than a and b.

The only solution that gives us an x within these boundaries is [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a1a52c22b3b95f108e5f501811b0674b.gif[/img].


polynomial
Ring Theory
superior algebra
superior algebra unsolved

I have $ f(x) \equal{} y^2 \minus{} p(x) \in \mathbb{C}[x,y]$, with $ p(x)$ without repeated roots. A factorization would be $ (y \minus{} u)(y \plus{} u)$, where $ u^2 \equal{} p$, but then $ u(x)$ wouldn't be anymore a polynomial. Does this imply that $ f(x)$ is irreducible and why?

Never dealt with irreducibles in two variables, so I have no idea how to use. Besides, I need the irreducibility in order to prove that it forms a prime ideal..

$ \mathbb{C}[x,y]\equal{}R[y]$ with $ R\equal{}\mathbb{C}[x]$, so it's only one indeterminate :) You only need that $ x\minus{}\lambda$ is prime ($ \lambda$ a simple zero of $ p$). owk

You mean $ x \minus{} \lambda$ forms a prime ideal? And to use Eisenstein criterion on it? I'm new to this stuff, so probably don't understand intellingent hints, sorry

He never said that $ x\minus{}\lambda$ generates a prime ideal (inspite this is true), but said the element itself is prime. Now use Eisenstein's criterion to this element.


In order for Mateen to walk a kilometer(1000m) in his rectangular
backyard, he must walk the length 25 times or walk its perimeter 10
times. What is the area of Mateen's backyard in square meters?

We can find the length by doing 1000/25, so the length of the rectangle is 40.
The perimeter is 1000/10, or 100, and so the width is 10.
Area of a rectangle = length*width, so our answer is 40*10, or 400.


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14864[/img]

 :(

Assuming $ \angle A$ is not right. Draw the circumcircle $ (O)$ with the given radius $ R.$ Draw arbitrary diameter, best parallel to one leg of the given angle $ \angle A.$ Mark the diameter endpoints $ A', B.$ Through $ A',$ draw a parallel line to the other leg of the given angle $ \angle A.$ This line cuts the circle $ (O)$ at $ C.$ Perpendicular bisector of $ BC$ cuts the circle $ (O)$ at $ M, N.$ If the angle $ \angle A$ is acute, $ M$ is on the same side of $ BC$ as $ A'.$ If the angle $ \angle A$ is obtuse, $ M$ is on the opposite side of $ BC$ than $ A'.$ Draw a circle $ \mathcal K$ with diameter $ BM.$ Draw a circle $ (B)$ with center $ B$ and radius $ \frac{b \plus{} c}{2}.$ The circles $ \mathcal K, (B)$ meet at $ P$ inside $ (O).$ The angle $ \angle BPM$ is right and $ BP \equal{} \frac{b \plus{} c}{2}.$ By Archimedes theorem, $ P$ bisects the polygonal path $ B \rightarrow A \rightarrow C$ of the $ \triangle ABC.$ Line $ BP$ cuts $ (O)$ again at $ A.$ The $ \triangle ABC$ solves the problem. (Reflection of the $ \triangle ABC$ in the perpendicular bisector $ MN$ of $ BC$ also solves the problem.)

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14875[/img]

 :thumbup:


Let $ X$ be a set containing $ 2k$ elements, $ F$ is a set of subsets of $ X$ consisting of certain $ k$ elements such that any one subset of $ X$ consisting of $ k \minus{} 1$ elements is exactly contained in an element of $ F.$ Show that $ k \plus{} 1$ is a prime number.

It is also Indian TST 1998, apparently:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195941
(I found this by searching for "subsets prime" (without quotes) in the Combinatorics forums.)

Also, the converse is on the forum (but no responses yet):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253058

It is a topic of combinatorial Design Theory. When each $(k-1)$-set is included in one and only one block of the family, it follows that for any value $1\leq a\leq k-1$, each $a$-set is included in a same number of blocks. This is easily computed: there are $\binom {2k-a} {k-1-a}$ $(k-1)$-sets containing a given $a$-set; each block containing the $a$-set contains $\binom {k-a} {k-1-a} = k-a$ of those $(k-1)$-sets, therefore there are exactly $\frac {1} {k-a} \binom {2k-a} {k-1-a}$ such blocks.
In particular, for $k\geq 3$, we have a $(v,b,k,r,\lambda)$-BIB, with $v=2k$, $b = \frac {1} {k+1} \binom {2k} {k}$, $r = \frac {1} {k+1} \binom {2k-1} {k}$, $\lambda = \frac {1} {k+1} \binom {2k-2} {k}$.
It means $k-a$ divides $(2k-a)(2k-a-1)\cdots (k+2)$ for all $1\leq a \leq k-2$, including all the primes $p=k-a$ in the range $2\leq p \leq k-1$, therefore none can divide $k+1$, hence the latter is a prime.
The converse requires the existence of such highly symmetric designs (there is a vast literature related), and this is not the case for all primes. Little is known on which work.


A sequence of numbers is defined as follows:

$ a_0  \equal{} 0$ ;

 $ a_{n\plus{}1} \equal{} 3\cdot a_n \plus{} 1$  $ \forall$ $ n \in \mathbb {W}$.

Show that $ a_{1995}$ is divisible by eleven but $ a_{1994}$ is not.


algebra
polynomial
calculus
calculus computations

Let $ a,\ b,\ c$ be real numbers and $ f(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c.$ Prove that if equations : $ f(x) \equal{} 0$ and $ f'(x) \equal{} 0$ have common root, then the root is real.

If $ f$ and $ f'$ have a common root, it is a double root of $ f$; $ f$ must have either one or three real roots; the two complex roots are in a conjugate pair. So, if the double root $ z$ is complex, it must be self-conjugate and therefore real.

[quote="Nukular"][quote="kunny"]
Could you explain the fact in detail?
[/quote]

Sure... if $ f(x) \equal{} (x \minus{} a)g(x)$, then $ f'(x) \equal{} (x \minus{} a)g'(x) \plus{} g(x) \implies f'(a) \equal{} g(a)$. If $ f'(a) \equal{} 0$, then $ g(a) \equal{} 0$ and so $ (x \minus{} a)$ divides $ g(x)$; therefore $ f(x) \equal{} (x \minus{} a)^2h(x)$ for some $ h$.[/quote]

Thank you for the explanation, Nukular.

It is understandable to Japanese high school students.

The more algebraic generalization has been moved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299583]here.[/url]

[quote="kunny"][quote="Nukular"]If $ f$ and $ f'$ have a common root, it is a double root of $ f$.[/quote]

Could you explain the fact in detail?[/quote]

If $ f$ vanishes at $ t$, then $ f(x) \equal{} (x\minus{}t)g(x)$ for some polynomial $ g$ with $ \deg{g}\plus{}1\equal{}\deg{f}$.  If $ f'$ also vanishes at $ t$, since $ f'(x) \equal{} (x\minus{}t)g'(x)\plus{}g(x)$, we must have $ g(t)\equal{}0$ in order to have $ f'(t)\equal{}0$, and hence $ g(x) \equal{} (x\minus{}t)h(x)$ for some polynomial $ h$ with $ \deg{h}\plus{}1\equal{}\deg{g}$.  Substituting, $ f(x) \equal{} (x\minus{}t)^2h(x)$, and that's that.


[quote="Ji Chen"]Let $x,y,z$ be nonnegative numbers such that $x+y+z=4$. Prove that

$3(x^{4}+y^{4}+z^{4})+118 \geq 13(x^{3}+y^{3}+z^{3})$.[/quote]
It's fourth degree. Hence, by $uvw$ it remains to check two cases only:
1. $y=x$ and $z=4-2x$, where $0\leq x\leq2$, which gives
$9x^4-51x^3+88x^2-48x+9\geq0$, which is true for $0\leq x\leq2$;
2. $z=0$, $y=4-x$, which gives $(x-1)^2(x-3)^2\geq0$. Done!


Alcumus
videos
induction
algebra
binomial theorem
Pascal\u0027s Triangle

Binomial Theorem Part 2

We find a formula and an algebraic explanation for the pattern we found in the previous lesson.  The formula we find for raising a binomial to a positive integer power is called the Binomial Theorem.

I'm too lazy to explain it, so... [url=http://www.artofproblemsolving.com/Wiki/index.php/Pascal's_Identity]Pascal's Identity[/url]

I was also going to ask if Pascal's Triangle and Pascal's Identity are introduced anywhere in the videos.  If not, is there a list anywere of things like that, that are assumed to be known before you start Alcumus?

Pascal's Identity is taught in a video lesson, along with 2 others about Pascal's Triangle.

I did notice the other day that there were 3 Pascal's Triangle videos available in the Teacher's Panel.  But when I look at the videos that are available to me based on the problems I have actually worked on, I don't see the Pascal's Triangle videos, either in my mastered topics or my current topics.  I have the Binomial Theorem videos, but not the Pascal's Triangle videos which are clear prerequisites.  I'd be interested in learning how the available videos are chosen for students.  As a teacher, can I make a video available that the system is not giving to some of my students yet?

Thanks!


If $ a_1,a_2,...,a_n \in \{\minus{}1,1\}$ and $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}0$.
Prove that there exists $ k \in \{1,2,3,...,n\}$ such that $ |a_1\plus{}2a_2\plus{}3a_3\plus{}...\plus{}ka_k|$≤$ [\frac{k}{2}]$.
($ [x]$ denote the greastest integer≤x)

[quote="Timestopper_STG"]I can prove it but only without using combinatoric and wonder how you prove it using combinatoric. :([/quote]
Can I see your solution?I think this is an algebric problem.


MATHCOUNTS geometry perimeter inequalities

How many [b]different[/b] isosceles triangles have integer side lengths and perimeter of 81 units?

We have the equation

\[ 2x \plus{} y \equal{} 81
\]

Find how many solutions there are when $ x$ and $ y$ are positive integers.  That should be your answer.

Remember the triangle inequality, so 2x>y (because x+y is obviously larger than x)


:D  Today was Korean MO and one inequality was proposed.

for all $x,y,z>0$

prove that

$xyz(x+2)(y+2)(z+2)$ <= ${(1+2(xy+yz+zx)/2 )}^{3}$

can you see your problem i can't understand it

modified.

 :D

right side...

$({1+2(xy+yz+zx)/3 })^{3}$

When $x, y, z$ are positive real number, prove this inequality.
$xyz(x+2)(y+2)(z+2) \leq (1+\frac{2(xy+yz+zx)}{3})^{3}$

I didn't solve this problem  :mad:

If you make the change $\sqrt{yz}=a,\; \sqrt{xz}=b; \sqrt{xy}=c$ it becomes equivalent to:

$\sqrt [3]{ abc(\frac{bc}{a}+2)(\frac{ac}{b}+2)(\frac{ab}{c}+2) }\leq 1+\frac{2\sum a^{2}}{3}$

But,

$\sqrt [3]{ abc(\frac{bc}{a}+2)(\frac{ac}{b}+2)(\frac{ab}{c}+2) }\leq \frac{ \sum a(\frac{bc}{a}+2) }{3}=\frac{\sum bc+\sum 2a }{3}$

So it is just left to prove:
$\sum bc+\sum 2a\leq 3+2\sum a^{2}$
$\sum (a^{2}-2a+1)+\sum a^{2}-\sum bc \geq 0$
$\sum (a-1)^{2}+\frac{\sum (a-b)^{2}}{2}\geq 0$

Maybe I'm wrong, my brain starts making me bad jokes, but if I'm not:

Really nice problem!!  :lol: .

here is my soln
 first inq equivalent to (ab+bc+ca)^3+6(ab+bc+ca)^2+12(ab+bc+ca)+8 >= (abc)^2+4(ab+bc+ca)abc+16abc(a+b+c)+64abc

Applying AM-GM inq we get
 12 (ab+bc+ca+a^2b^2c^2)>=48abc
 13/27 .(ab+bc+ca)^3 >=13(abc)^2
 8/27. (ab+bc+ca)^2 +8>= 16abc
 16/3 (ab+bc+ca)^2 >= 16abc(a+b+c)
 2/3 (ab+bc+ca)^2+6(abc)^2 >= 4(ab+bc+ca)abc

add all inq we have result
 may be i'm wrong


To complete the inequality list posted by Manlio:

For $a,b,c$ non-negative numbers, the inequality holds

$\sum\frac{a(b+c)}{b^2+bc+c^2} \ge2$.

Because the inequality form is nice, but it is very hard to find a nice solution. :P

[quote="Vasc"]For $a,b,c$ non-negative numbers, the inequality holds

$\sum\frac{a(b+c)}{b^2+bc+c^2} \ge2$.[/quote]

Well, this one can be proven in a similar way to http://www.mathlinks.ro/Forum/viewtopic.php?t=14577 ; just one needs a little bit more phantasy.

We have to show the inequality

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

(here and in the following, all sums are cyclic).

Now we note that

$\frac{a\left( b+c\right) }{b^{2}+bc+c^{2}}=\frac{2a}{a+b+c}+\frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left( a+b+c\right) \left(b^{2}+bc+c^{2}\right) }$,

so that we can rewrite our inequality as

$\sum \left( \frac{2a}{a+b+c}+\frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left(a+b+c\right) \left( b^{2}+bc+c^{2}\right) }\right) \geq 2$.

This is clearly equivalent to

$\sum \frac{2a}{a+b+c}+\sum \frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left(a+b+c\right) \left( b^{2}+bc+c^{2}\right) }\geq 2$.

But

$\sum \frac{2a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2$;

thus, the inequality in question simplifies to

$2+\sum \frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left( a+b+c\right) \left(b^{2}+bc+c^{2}\right) }\geq 2$.

Upon subtraction of 2, this becomes

$\sum \frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left( a+b+c\right) \left(b^{2}+bc+c^{2}\right) }\geq 0$.

Multiplication with a + b + c simplifies this into the form

$\sum \frac{ab\left(a-b\right)-ca\left(c-a\right)}{b^{2}+bc+c^{2}}\geq 0$.

We can break the sum into two sums:

$\sum \frac{ab\left( a-b\right) }{b^{2}+bc+c^{2}}-\sum \frac{ca\left(c-a\right) }{b^{2}+bc+c^{2}}\geq 0$.

Now, since $\sum \frac{ca\left(c-a\right) }{b^{2}+bc+c^{2}}=\sum \frac{ab\left(a-b\right) }{c^{2}+ca+a^{2}}$, this rewrites as

$\sum \frac{ab\left( a-b\right) }{b^{2}+bc+c^{2}}-\sum \frac{ab\left(a-b\right) }{c^{2}+ca+a^{2}}\geq 0$.

Now, we put the two sums together again:

$\sum \left( \frac{ab\left( a-b\right) }{b^{2}+bc+c^{2}}-\frac{ab\left(a-b\right) }{c^{2}+ca+a^{2}}\right) \geq 0$,

and simplify the term:

$\sum \left( ab\left( a-b\right) \left( \frac{1}{b^{2}+bc+c^{2}}-\frac{1}{c^{2}+ca+a^{2}}\right) \right) \geq 0$.

But this inequality is true, since the left hand side is a sum of three nonnegative numbers, and thus nonnegative. Why is the left hand side a sum of three nonnegative numbers? Well, let's show, for instance, that the number

$ab\left( a-b\right) \left( \frac{1}{b^{2}+bc+c^{2}}-\frac{1}{c^{2}+ca+a^{2}}\right)$

is nonnegative. In fact, we have two cases: either $a\geq b$ or $a\leq b$.

In the case $a\geq b$, we have $a-b\geq 0$; on the other hand, $a\geq b$ yieds $b\leq a$, so that $b^2\leq a^2$ and $bc\leq ac$, and thus $b^2+bc+c^2\leq a^2+ac+c^2=c^2+ca+a^2$, and thus

$\frac{1}{b^2+bc+c^2}\geq\frac{1}{c^2+ca+a^2}$,

and

$\frac{1}{b^2+bc+c^2}-\frac{1}{c^2+ca+a^2}\geq 0$;

multiplying this with $a-b\geq 0$, we get

$\left( a-b\right) \left( \frac{1}{b^{2}+bc+c^{2}}-\frac{1}{c^{2}+ca+a^{2}}\right)\geq 0$,

so that, finally, the number

$ab\left( a-b\right) \left( \frac{1}{b^{2}+bc+c^{2}}-\frac{1}{c^{2}+ca+a^{2}}\right)$

is nonnegative.

In the case $a\leq b$, we can proceed similarly; we then have $a-b\leq 0$ and

$\frac{1}{b^{2}+bc+c^{2}}-\frac{1}{c^{2}+ca+a^{2}}\leq 0$,

and again, the product is $\geq 0$.

So the inequality is proven.

Incidentally, this proof is more or less an elegant formulation of my proof of the same inequality in http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9.

BTW, I will try some of your other inequalities in that thread.

  Darij

Darij, your proof is MAGNIFICENT. Congratulations!  :)  :P  :)

In fact, the following nice and strange (unexpected) identity occurs:

$\displaystyle \sum\frac{a\left( b+c\right) }{b^{2}+bc+c^{2}}= 2+\sum\frac{bc(b-c)^2}{(a^2+ac+c^2)(a^2+ab+b^2)}$.

wow!
both proofs are very very nice!!!!!

A little curiosity about you, Vasc.

How did you discover this

$\displaystyle \displaystyle \sum\frac{a\left( b+c\right) }{b^{2}+bc+c^{2}}= 2+\sum\frac{bc(b-c)^2}{(a^2+ac+c^2)(a^2+ab+b^2)}$ ?

These are the answers. Please, choose one or more answers, OK. :lol:

A) By trial

B) By intuition

C) By computer

D) By a book

E) By a your secret method

F) By a standard well-known method

Manlio, my answer to your question is F). :lol:

And now, here is a similar nice inequality:
If $a,b,c \ge 0$, then

$\displaystyle \displaystyle \sum\frac{ab-bc+ca}{b^2+c^2} \ge \frac{3}{2}$.

I think somebody will find a smart solution.

If it is well known, how is it called?

Sorry for my ignorance :blush:

It is called "expanding method".
Expandind the inequality

$\sum\frac{a(b+c)}{b^2+bc+c^2} \ge 2$,

we get

$\sum{bc(b-c)(b^3-c^3)} \ge 0$,

that is

$\sum{bc(b-c)^2(b^2+bc+b^2)} \ge 0$.

From this, the identity clearly follows. But it also follows from Darij's solution.

May I ask, Vasc, if there is an efficient way to "expand the inequality," as you put it? I am wondering if there is a way to avoid filling pages of calculations.

It is not necessary to fil pages with calculations. In our inequality $A \ge B$, we have

$A=\sum bc(b^4+c^4)+\sum b^2c^2(b^2+c^2)+2\sum b^3c^3+2abc\sum a^3+$
$4abc\sum bc(b+c)+6a^2b^2c^2$

and

$B=2\sum b^2c^2(b^2+c^2)+2\sum b^3c^3+2abc\sum a^3+4abc\sum bc(b+c)+6a^2b^2c^2$.

Consequently

$A-B=\sum bc(b^4+c^4)-\sum b^2c^2(b^2+c^2)=\sum bc(b-c)(b^3-c^3) \ge 0$. ;)

[quote="Vasc"]If $ x,y,z \ge 0$ and $ yz+zx+xy>0$, then

$ \frac {xy - yz + zx}{y^2 + z^2}+\frac {yz - zx + xy}{z^2 + x^2}+\frac {zx - xy + yz}{x^2 + y^2} \ge \frac {3}{2}.$

I think somebody will find a smart solution.[/quote]$ \sum{\frac {xy - yz + zx}{y^2 + z^2}=\frac {3}{2}+\sum{\frac{yz(y-z)^2\left[(y+z-x)^2+x^2+3y^2+3z^2\right]}{4\left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)}}\geq\frac {3}{2}.}$

[quote="Vasc"]To complete the inequality list posted by Manlio:

For $ a,b,c$ non-negative numbers, the inequality holds

$ \sum\frac {a(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \ge2$.[/quote]
$ LHS \equal{} \sum\ \frac{a^2}{\frac{a(b^2\plus{}bc\plus{}c^2)}{b\plus{}c}} \geq\ \frac{(a\plus{}b\plus{}c)^2}{\sum\ \frac{a(b^2\plus{}bc\plus{}c^2)}{b\plus{}c}} \geq\ 2$
$ \Leftrightarrow a^2\plus{}b^2\plus{}c^2 \plus{}2abc\sum\ \frac{1}{a\plus{}b} \geq\ 2(ab\plus{}bc\plus{}ca)$
$ LHS \geq\ a^2\plus{}b^2\plus{}c^2 \plus{} \frac{9abc}{a\plus{}b\plus{}c} \geq\ 2(ab\plus{}bc\plus{}ca)$
$ \Leftrightarrow a^3\plus{}b^3\plus{}c^3\plus{}3abc \geq\ ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)$
 :wink:

[quote="Ji Chen"][quote="Vasc"]If $ x,y,z \ge 0$ and $ yz + zx + xy > 0$, then

$ \frac {xy - yz + zx}{y^2 + z^2} + \frac {yz - zx + xy}{z^2 + x^2} + \frac {zx - xy + yz}{x^2 + y^2} \ge \frac {3}{2}.$

I think somebody will find a smart solution.[/quote]$ \sum{\frac {xy - yz + zx}{y^2 + z^2} = \frac {3}{2} + \sum{\frac {yz(y - z)^2\left[(y + z - x)^2 + x^2 + 3y^2 + 3z^2\right]}{4\left(y^2 + z^2\right)\left(z^2 + x^2\right)\left(x^2 + y^2\right)}}\geq\frac {3}{2}.}$[/quote]

I proved Ji Chen equality this way:

First I proved (easy)

$ \frac {xy - yz + zx}{y^2 + z^2} + \frac {yz - zx + xy}{z^2 + x^2} + \frac {zx - xy + yz}{x^2 + y^2} - \frac {3}{2} = \sum{\frac {(x^2 + y^2)(x - y)^2\left[\sum xy - z^2\right]}{2\left(y^2 + z^2\right)\left(z^2 + x^2\right)\left(x^2 + y^2\right)}}$
then I summed

$ \sum{\frac {(x - y)^2(y - z)(z - x)\left[x^2 + xy + y^2\right]}{2\left(y^2 + z^2\right)\left(z^2 + x^2\right)\left(x^2 + y^2\right)}} = 0$

and all is made.

Nice identity Ji Chen. :)

sdfgsdfgsdgdfhdf (I will write my solution later...)

Dear Mudok I found that:

Rewrite it as

$\sum a^2\frac{b+c}{b+c} +abc\sum \frac{1}{b+c}+abc\sum \frac{1}{c+a} \geq 2\sum ab$

$\sum \frac{a^2b}{b+c} +\sum\frac{b^2a}{c+a}+abc\sum \frac{1}{b+c}+abc\sum \frac{1}{c+a}=
\sum ab\frac {a+c}{b+c} +\sum ab\frac{b+c}{a+c} = \sum ab(\frac {a+c}{b+c} +\frac{b+c}{a+c}) \geq 2\sum ab$

by AM-GM. Is is true?  :blush:

Can you please post your solution? Thank you very much

This is a simple CS proof of

$\sum a^2 +2abc\sum \frac{1}{a+b} \geq 2\sum ab$

Proof:

$2\sum a^2  + 2abc\sum \frac{1}{a+b} \geq (\sum a)^2$

$\sum (2a^2 + 2abc\frac{1}{b+c}) \geq (\sum a)^2$

$2\sum ab \sum \frac{a}{b+c} \geq (\sum a)^2$

$\sum a(b+c)\sum \frac{a}{b+c} \geq (\sum a)^2$

which is true by C-S :)


let $S$ be the set of complex numbers such that $\forall z \in S$ satisfies$\left(\frac{\Re(z)-3}{2}\right)^2+\left(\frac{\Im(z)+7}{5}\right)^2 \le 1$
if $w \in S$, then find the maximum and minimum value of $|w|$

Well, if we put it into a x-y plane, we have:

$(\frac{x-3}{2})^2 + (\frac{y+7}{5})^2 \le 1$

So we got to find the minimum and maximum values of $\sqrt{x^2+y^2}$ with $x$ and $y$ in the ellipse

$(\frac{x-3}{2})^2 + (\frac{y+7}{5})^2 =1$

So that can be solved pretty easily..... and I need to do my English essay that's due tomorrow, so I need to go.

Hint:

[hide]This is the equation of all points inside an ellipse on the complex plane with center , vertical axis , and horizontal axis .[/hide]

just a process rather than working through all the algerbra is necessary, i just chose the numbers randomly (i know what to do, it was just an interesting problem that i thought of


provided 2 events [a] & [b] are non-complementary.
p[a]=0.5  ..what can you say about p[b]?

Um, you really didn't give us any relation between A and B other than they're not complementary, so nothing whatsoever can be determined, not even that $ P[B]\ne 1\minus{}P[A]$, since non-complementary events can still have probabilities adding up to 1.


sqrt(-2)*sqrt(-18) = i*sqrt(2)*i*sqrt(18) = i*i*sqrt(36) = -6

sqrt[(-2)*(-18)] = sqrt(36) = 6

How could it be explained intuitively and/or in terms of abstract algebra ?

$\sqrt{\ }$ is not multiplicative over $\mathbb{C}$ (no matter how you define it), and you proved it yourself.
$\sqrt{\ }$ is simply not a nice function over $\mathbb{C}$. What could be more trivial?


Could someone help me to prove this:

$a,b,c\geq0$

$(a^2+2)(b^2+2)(c^2+2) \geq 3(a+b+c)^2$

Al least an idea? :D

I dont have time to look at it more closely, but looks like a problem where you might be able to use Cauchy?

Very nice proof, argady. It is really strange for me.
And here is my solution:
Let
\[ f(a,b,c)=(a^{2}+2)(b^{2}+2)(c^{2}+2)-3(a+b+c)^{2}\]
We will prove that:
\[ f(a,b,c)-f \left(a,\frac{b+c}{2},\frac{b+c}{2}\right)\ge 0 \]
Without loss of generality, we can assume that $ b\le 1$ and $ c\le 1\to t=\frac{b+c}{2}\le 1$. Therefore:
\[ f(a,b,c)-f \left(a,\frac{b+c}{2},\frac{b+c}{2}\right)\ge 0 \]

\[ \iff (a^{2}+2)(b-c)^{2}[16-(b+c)^{2}-4bc]\ge 0 \]
which is clearly true.
If you want to complete solution, we will prove that the following inequality:
\[ f(a,t,t)\ge 0 \]

\[ \iff (a-t)^{2}(t^{4}+4t^{2}+1)+(1-t^{2})[t^{4}+3t^{2}+8-at(2t+10)]\ge 0 \]
We are done.
A nice inequality. It is stronger than a well-known inequality. This is a inequality in APMO 2004.
\[ (a^{2}+2)(b^{2}+2)(c^{2}+2)\ge 3(a+b+c)^{2}\ge 9(ab+bc+ca)\]
We have a similar to: (created by me and Hungkhtn)
\[ (a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\ge 9(ab+bc+ca)\]
when $ a,b,c\in \mathbb{R}$

a strong one which is a question of an old apmo
$ (a^{2}+2)(b^{2}+2)(c^{2}+2)\ge9(a^{2}+b^{2}+c^{2})$
is also true.

[quote="lasha"]Expanding LHS we get x2y2z2 + 2(x2y2 + y2z2 + z2x2) + 4(x2 + y2 + z2) + 8.

By AM/GM we have 3(x2 + y2 + z2) ≥ 3(xy + yz + zx) and (2x2y2 + 2) + (2y2z2 + 2) + (2z2x2 + 2) ≥ 4(xy + yz + zx). That leaves us needing x2y2z2 + x2 + y2 + z2 + 2 ≥ 2(xy + yz + zx) (*). That is hard, because it is not clear how to deal with the xyz on the lhs.

By AM/GM we have (a+b)(b+c)(c+a) ≥ 8abc. So if (u+v-w), (u-v+w), (-u+v+w) are all non-negative, we can put 2a = -u+v+w, 2b = u-v+w, 2c = u+v-w and get uvw ≥ (-u+v+w)(u-v+w)(u+v-w). If u, v, w are positive, then at most one of (u+v-w), (u-v+w), (-u+v+w) is negative. In that case uvw ≥ (-u+v+w)(u-v+w)(u+v-w) is trivially true. So it holds for all positive u, v, w. Expanding, we get u3 + v3 + w3 + 3uvw ≥ uv(u+v) + vw(v+w) + wu(w+u), which is a fairly well-known inequality. Applying AM/GM to u+v etc we get, u3 + v3 + w3 + 3uvw ≥ 2(uv)3/2 + 2(vw)3/2 + 2(wu)3/2. Finally, putting u = x2/3, v = y2/3, w = z2/3, we get x2 + y2 + z2 + 3(xyz)2/3 ≥ 2(xy + yz + zx) (**).

Thus we have x2y2z2 + x2 + y2 + z2 + 2 ≥ ≥ (x2 + y2 + z2) + 3(xyz)2/3, by applying AM/GM to x2y2z2, 1, 1, and now (**) gives the required (*).[/quote]

[url]http://www.mathlinks.ro/viewtopic.php?t=5261[/url]  :wink:

[quote="M4RI0"]Could someone help me to prove this:

$ a,b,c\geq0$

$ (a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 3(a+b+c)^{2}$

Al least an idea? :D[/quote]
The Ineq finer:$ (a^{2}+2)(b^{2}+2)(c^{2}+2) \geq k(a^{2}+b^{2}+c^{2})+(9-k)(ab+bc+ca)$
k=4 is the greatest value of constant

[quote="ducpbc"]But the Inequalities:$ (a^{2}+2)(b^{2}+2)(c^{2}+2) \ge 9(a^{2}+b^{2}+c^{2})$ is not true.See a=0,b=0.....[/quote]

Yes. Maybe hoang is wrong when supposing that $ a,b\le 1$.

hix I have made mistake :D  my solution for the second problem ... Is it correct?   :maybe: I misundersand why we can not assume $ a,b\le 1$ in this case.

[quote="ducpbc"]The finer inequality :

$ (a^{2} + 2)(b^{2} + 2)(c^{2} + 2) \geq k(a^{2} + b^{2} + c^{2}) + (9 - k)(ab + bc + ca)$

$ k = 4$ is the greatest value of constant.[/quote]$ \prod(a^2 + 2)\geq 4\sum{a^2} + 5\sum{bc}\Longleftrightarrow$

$ 0\leq 8 - 5\sum{bc} + 2\sum{b^2c^2} + a^2b^2c^2$

$ \equiv\frac {8}{9}\left(3 - \sum{bc}\right)^2 + Q(a,b,c)$,

where

$ Q(a,b,c) = a^{2}b^{2}c^{2}+\frac{2}{9}\left(5\sum{b^{2}c^{2}}-8abc\sum{a}\right)+\frac{1}{3}\sum{bc}$

$ = \left(abc + \frac {5}{9}\sum{\frac {bc}{a}} - \frac {8}{9}\sum{a}\right)^2$

$ + \frac {5}{81}\left(\sum{\frac {bc}{a}} - \sum{a}\right)\left(8\sum{a} - 5\sum{\frac {bc}{a}\right)}$

$ + \frac {1}{27}\left[5\sum{a^2\left(\frac {b}{c} + \frac {c}{a}\right)} - 8\sum{a^2} - 2\sum{bc}\right]\geq0$.

[quote="M4RI0"]Could someone help me to prove this:

$ a,b,c\geq0$

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3(a \plus{} b \plus{} c)^2$

Al least an idea? :D[/quote]

We can prove that:
$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \ge 3(a\plus{}b\plus{}c)^2\plus{}(abc\minus{}1)^2$  :blush:

[quote="M4RI0"]$ a,b,c\geq0\Longrightarrow$

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3(a \plus{} b \plus{} c)^2$.[/quote]See also : http://www.mathlinks.ro/viewtopic.php?t=267785[quote="caubetoanhoc94"]We can prove that

$ \left(a^2 \plus{} 2\right)\left(b^2 \plus{} 2\right)\left(c^2 \plus{} 2\right) \ge 3(a \plus{} b \plus{} c)^2 \plus{} (abc \minus{} 1)^2.$[/quote]$ \prod{\left(a^2 \plus{} 2\right)} \minus{} 3\left(\sum{a}\right)^2 \minus{} (abc \minus{} 1)^2$

$ \equal{} \left[\frac {\sqrt {199} \minus{} 3\sqrt {7}}{12}(bc \plus{} ca \plus{} ab) \plus{} \frac {7\sqrt {7} \minus{} \sqrt {199}}{12}(a \plus{} b \plus{} c) \minus{} \sqrt {7}\right]^2$

$ \plus{} \frac {225 \minus{} \sqrt {1393}}{288}\sum{a^2(b \minus{} c)^2} \plus{} \frac {\sqrt {1393} \minus{} 25}{36}\sum{bc(a \minus{} 1)^2}$

$ \plus{} \frac {7\sqrt {7} \minus{} \sqrt {199}}{288}\sum{a\left(\sqrt {199}a \plus{} 12\sqrt {7}\right)(b \plus{} c \minus{} 2)^2}\geq0.$

By the way, caubetoanhoc94 inequality holds for real numbers $ a,b,c$ with $ 3\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab\right) \plus{} 14(a \plus{} b \plus{} c) \plus{} 21\geq0.$

[quote="ducpbc"]
The Ineq finer:$ (a^{2}+2)(b^{2}+2)(c^{2}+2) \geq k(a^{2}+b^{2}+c^{2})+(9-k)(ab+bc+ca)$
k=4 is the greatest value of constant[/quote]
If $b=c=0$ and $a\neq0$ so $k\leq4+\frac{8}{a^2}$.
Hence, for $k>4$ the inequality
$ (a^{2}+2)(b^{2}+2)(c^{2}+2) \geq k(a^{2}+b^{2}+c^{2})+(9-k)(ab+bc+ca)$ is wrong.
We'll prove that for $k=4$ the inequality is true.
Id est, $ (a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 4(a^{2}+b^{2}+c^{2})+5(ab+bc+ca)$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $ (a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 4(a^{2}+b^{2}+c^{2})+5(ab+bc+ca)\Leftrightarrow$
$\Leftrightarrow a^2b^2c^2+8+\sum_{cyc}(2a^2b^2-5ab)\geq0\Leftrightarrow f(u)\geq0$, where
$f(u)=-12uw^3+18v^4-15v^2+w^6+8$, which is linear  function.
Hence, $f$ gets a minimal value, when $u$ gets an extremal value,
which happens when two numbers from $\{a,b,c\}$ are equal.
Let $b=c$. Hence, we need to prove that
$(b^4+4b^2)a^2-10ab+2b^4-5b^2+8\geq0$, for which we need to prove that
$25b^2-(b^4+4b^2)(2b^4-5b^2+8)\leq0$, which is
$b^2(b-1)^2(b+1)^2(2b^2+7)\geq0$, which is obviously true.

[quote="zaizai-hoang"]
We have a similar to: (created by me and Hungkhtn)
\[ (a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\ge 9(ab+bc+ca)\]
when $ a,b,c\in \mathbb{R}$[/quote]
It's wrong! Try $a=0$ and $b=c=-1$ :wink:
But the following nice inequality is true already.
Let $a$, $b$ and $c$ are real numbers such that $a+b+c\geq 3$. Prove that:
\[ (a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\ge 9(ab+bc+ca)\]

Generalization of the inequality : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=396767

Let  $ a^{2}=2(tanx)^{2} $  ect.
inq  become
$ (cosxcosycosz(tanx+tany+tanz))^{2}\leq \frac{4}{3} $
$ x ,y ,z\in(0,\frac{\pi}{2}) $

[quote="arqady"][quote="Emo916math"]This problem belongs in the Inequalities forum.[/quote]
There was something such there:
$ (a^{2}+2)\left(1+\frac{(b+c)^{2}}{2}\right)\geq(a+b+c)^{2}.$
Hence, it remains to prove that $ (b^{2}+2)(c^{2}+2)\geq3\left(1+\frac{(b+c)^{2}}{2}\right),$
which equivalent to $ (b-c)^{2}+2(bc-1)^{2}\geq0.$[/quote]
Let $a,b,c$ be positive real numbers .For all  positive real number $k$ ,prove that\[ (a^2+k)(b^2+k)(c^2+k)\geq \frac{3k^2}{4}(a+b+c)^2. \]
[b]Proof:[/b]
$  (b-c)^{2}+2(bc-k)^{2}\geq0\iff (b^{2}+k)(c^{2}+k)\geq\frac{3k^2}{4}\left(1+\frac{(b+c)^{2}}{k}\right),$
$(a^2+k)(b^2+k)(c^2+k)\geq \frac{3k^2}{4}(a^2+k)\left(1+\frac{(b+c)^{2}}{k}\right)\geq \frac{3k^2}{4}(a+b+c)^2. $
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=1452134]here[/url]

$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \ge 3(a\plus{}b\plus{}c)^2\plus{}(abc\minus{}1)^2$

The following inequality is also true
[img]http://s5.sinaimg.cn/middle/0018zOAxgy70UfIRCug24&690[/img]

[quote=luofangxiang]
The following inequality is also true
[img]http://s5.sinaimg.cn/middle/0018zOAxgy70UfIRCug24&690[/img][/quote]

This is equivalent to $a^2+b^2+c^2+2abc+1 \geq 2(ab+bc+ac)$ which is well-known...


if n=a^2+2mb^2 ,where a,b,m are integers, is a perfect square,find the integer roots of x^2+y^2=n-mb^2


[color=cyan]An 8x8 grid is filled with the numbers 1 to 64, one number per square such that the sums of the numbers in every column, row, and diagonal are the same.  Compute that sum.[/color]

Spoiler:

[hide]Sum of all numbers = 64(64+1)/2 = 2080

Sum per row = 2080/8 = 260[/hide]

nice
didn't think of that....
shoulda tho

Actually, when I wrote this problem, [hide]I didn't realize how badly overdetermined it was.  Most people figured out that you didn't need most of the information in the question, though.[/hide][/hide]

Is this type of problem generalizable?

I mean, given the integers 1-n, and they are placed in a k by p grid, where kp=n, and the sum of all the terms in each row, column, and diagonal are the same, find that common sum?

[color=cyan]Well, squares are easy.  Do it yourself for squares.  Rectangles, I don't know about.  Actually, never mind.  Yes I do.  Figure it out, too -- just try the same logic.[/color]

Yea, squares are very easy,

Rectangles are slightly harder tho,

and another question: would this work for any k,p,n, only given that kp=n?

If you mean it's "slightly harder" to create a magic rectangle, you're right...

[color=cyan]Well, in some cases you clearly can't construct the rectangle at all . . . such as a 1xn.  However, the question is still answerable for these cases, bizarrely enough.  You also have to wonder how you sum a diagonal of a rectangle like that . . .[/color]

[color=cyan]Actually, on further thought, it would probably take a lot of thought to come up with a nice definition for a magic rectangle . . . it should be doable though.  Maybe each of the rows sum to one thing and each of the columns to another?[/color]

But the sum of the rows and columns should be the same (never mind the diagonals, take those out)

How about this?

I'll give you $1000 if you can draw for me a non-square rectangular grid of distinct numbers such that the sums of the entries in all the rows and all the columns are the same.

Not quite

[hide]

How about:

1    2   -3

-1  -2   3

[/hide]

When do I get my $1000?

:shock: This reminds me of a movie...(hint: think about where TripleM lives. Then think about a continent near it.)

Hehe. Teratomato and I were both imagining that they all had to be positive. If some of them don't, that changes things a bit, doesn't it?


The hypotenuse $ c$ and one arm $ a$ of a right triangle are consecutive integers. The square of the second arm is:

$ \textbf{(A)}\ ca \qquad\textbf{(B)}\ \frac {c}{a} \qquad\textbf{(C)}\ c \plus{} a \qquad\textbf{(D)}\ c \minus{} a \qquad\textbf{(E)}\ \text{none of these}$

hello, let $ c\equal{}n\plus{}1$ and $ a\equal{}n$ with $ c>a$ and $ n \in \mathbb{N}$, then we have $ b^2\equal{}c^2\minus{}a^2\equal{}2n\plus{}1\equal{}a\plus{}c$.
So the answer is $ \text{C}$.
Sonnhard.


If $x,y,z$ are positive prove the following nice inequality
$\sum\frac{\sqrt{xy+4yz+4xz}}{x+y} \geq \frac{9}{2}$

It was clear to me when I saw the title... Anyway, it was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=28175 , and yes, the only proof known is by brute-force calculation (don't ask me how).

  Darij

Thanks for the link but it does not help. So we should discuss it here to find a solution i think .the only who has it is the proposer and maybe you Darij. thanks for the hearing. have a nice time with this very difficult problem. thanks

my idea is to prove that

$\sum\frac{\sqrt{xy+4yz+4xz}}{x+y} \geq\ 2(\frac{xy+yz+zx}{(x+y)^2}+\frac{xy+yz+zx}{(y+z)^2}+\frac{xy+yz+zx}{(z+x)^2})$ .

I think that after expanding it holds but i am not sure.

No, it doesn't; the inequality is stronger and not weaker than Iran 1996.

  Darij

[quote="Ji Chen"][quote="darij grinberg"]the inequality is stronger and not weaker than Iran 1996.[/quote]If $ x,y,z$ are positive, then

$ 3\sqrt {\sum{\frac {yz \plus{} zx \plus{} xy}{(y \plus{} z)^2}}}\geq\sum\frac {\sqrt {yz \plus{} 4zx \plus{} 4xy}}{y \plus{} z} \geq3\sqrt [5]{\frac {27}{8}\sum{\frac {yz \plus{} zx \plus{} xy}{(y \plus{} z)^2}}}.$

See also : http://www.mathlinks.ro/viewtopic.php?t=244562[/quote]

WOW  Ji chen you keep amazing me every day  :lol:  , I'd  to see a solution  please  :)


trigonometry
calculus
integration
real analysis
real analysis unsolved

Find:
1. $ \int \frac{sinxcosx}{sin^{4}x\plus{}cos^{4}x}\,dx$
2. $ \int \frac{sin^{2}xcos^{2}x}{sin^{5}x\plus{}cos^{5}x}\,dx$

Setting $ x\equal{}\frac{\pi}{2}\minus{}\theta$ gives the answer $ Constant$ .

[quote="borislav_mirchev"]Find:
1. $ \int \frac {sinxcosx}{sin^{4}x \plus{} cos^{4}x}\,dx$
2. $ \int \frac {sin^{2}xcos^{2}x}{sin^{5}x \plus{} cos^{5}x}\,dx$[/quote]

1. divide by $ \cos^4\,x$ to get

$ I\equal{}\int\;\dfrac{\tan\,x\;\sec^3\,x\;dx}{1\plus{}\tan^4\,x}$

put $ u\equal{}\sec^2\,x$ to get $ I\equal{}\frac{1}{2}\int\dfrac{\sqrt u\;du}{u^2\minus{}2u\plus{}2}$

solving this then gives $ I\equal{}\minus{}\frac{1}{2}\tan^{\minus{}1}\,(\cos\,2x)$

2. doing a similar thing gives, $ I\equal{}\frac{1}{2}\int\dfrac{\sqrt{\;u(u\minus{}1)\;}\;du}{1\plus{}(u\minus{}1)^{\frac{5}{2}}}$

hmmm... this might require some thinking... i need to run now, but i will post the rest of the answer later.


calculus
integration
trigonometry
LaTeX
calculus computations

shouldnt $\frac 1{\sqrt x}=\frac 1{\sqrt{\sin \theta}}$ since $x=\sin \theta$ from the subsitution $\theta=\sin^{-1}x$?

[quote="Carcul"]Yes Jhonny Ruiz, but the conclusion is the same.[/quote]

yeah i understand, but at first look I saw that little mistake, its all, thanks

[quote="sylow_theory"]$\int \frac{1}{\sqrt{x}}\cdot \frac{1}{\sqrt{1-x^{2}}}dx$

So if you let $\theta = \sin^{-1}x$ then $\frac{d\theta}{dx}= \frac{1}{\sqrt{1-x^{2}}}$ and use also have $\frac{1}{\sqrt{\cos \theta}}= \frac{1}{\sqrt{x}}$

So we have, upon substitution,

$\int \frac{1}{\sqrt{\cos \theta}}d\theta$

This happens to be an elliptic integral of the first kind.
http://en.wikipedia.org/wiki/Elliptic_Integral[/quote]
I'm sorry, but in China people cannot open the pages in wiki :oops:

[quote="Twins"]
I'm sorry, but in China people cannot open the pages in wiki :oops:[/quote]

Dont worry, use http://mathworld.wolfram.com/EllipticIntegral.html

Searching for a math topic will usually bring up both the wikipedia article and the mathworld article- and both of those are usually high quality. At least recently, I prefer linking to the wiki version.

[quote="jmerry"]. At least recently, I prefer linking to the wiki version.[/quote]

That because they use $\LaTeX$.  :rotfl: