$ \frac {a^2}{b} \plus{} \frac {b^2}{a}\ge \sqrt {2(a^2 \plus{} b^2})$
is equivalent to
$ a^6\plus{}b^6\plus{}2a^3b^3\geq 2a^4b^2\plus{}2a^2b^4$.
Since
$ 2a^4b^2\plus{}2a^2b^4\minus{}2a^3b^3\equal{}2a^2b^2(a^2\minus{}ab\plus{}b^2)\equal{}2a^2b^2\frac{a^3\plus{}b^3}{a\plus{}b}$,
the original equation is equivalent to
$ (a^6\plus{}b^6)(a\plus{}b)\geq 2a^2b^2(a^3\plus{}b^3)$
However, according to Muirhead, that is correct as
$ (7,0)\succeq (5,2)$ and $ (6,1)\succeq (5,2)$.

Also, we could replace Muirheads, I imagine, by AM-GMs - it shouldn't be too hard as the inequality does have only two variables.

[quote="thanhnam2902"][color=darkblue]Let $ a, b > 0$. Prove that: $ \frac {a^2}{b} + \frac {b^2}{a}\ge \sqrt {2(a^2 + b^2})$[/color][/quote]
${ \frac {a^2}{b} + \frac {b^2}{a}\ge \sqrt[9] {256(a^9 + b^9})}$ is true too, but ${ \frac {a^2}{b} + \frac {b^2}{a}\ge \sqrt[10] {512(a^{10} + b^{10}})}$ is wrong already. :wink:

[color=darkblue][size=150]This is FaiLurE'solution:[/size]
$ \frac {a^2}{b} \plus{} \frac {b^2}{a}\ge \sqrt {2(a^2 \plus{} b^2})$

$ \Longleftrightarrow \frac {a^4}{b^2} \plus{} \frac {b^4}{a^2} \plus{} 2ab\ge2a^2 \plus{} 2b^2$

$ \Longleftrightarrow\frac {a^4}{b^2} \plus{} \frac {b^4}{a^2}\ge2(a^2 \minus{} ab \plus{} b^2)$

$ \Longleftrightarrow(a \plus{} b)(\frac {a^4}{b^2} \plus{} \frac {b^4}{a^2})\ge2(a^3 \plus{} b^3)$

$ \Longleftrightarrow(\frac {a^5}{b^2} \plus{} \frac {b^5}{a^2} \minus{} a^3 \minus{} b^3) \plus{} (\frac {a^4}{b} \plus{} \frac {b^4}{a} \minus{} a^3 \minus{} b^3)\ge0$

As $ (a^5 \minus{} b^5)(a^2 \minus{} b^2)\ge0$

$ \Longleftrightarrow\frac {a^5}{b^2} \plus{} \frac {b^5}{a^2} \minus{} a^3 \minus{} b^3\ge0$, (1)

As $ (a^4 \minus{} b^4)(a \minus{} b)\ge0$

$ \Longleftrightarrow\frac {a^4}{b} \plus{} \frac {b^4}{a} \minus{} a^3 \minus{} b^3\ge0$, (2)

$ (1) \plus{} (2): (\frac {a^5}{b^2} \plus{} \frac {b^5}{a^2} \minus{} a^3 \minus{} b^3) \plus{} (\frac {a^4}{b} \plus{} \frac {b^4}{a} \minus{} a^3 \minus{} b^3)\ge0$
Q.E.D.[/color]

The problem is equivalent to
$ a^6\minus{}2b^2a^4\plus{}2b^3a^3\minus{}2b^4a^2\plus{}b^6\equal{}(a\minus{}b)^2(a^4\plus{}2ba^3\plus{}b^2a^2\plus{}2b^3a\plus{}b^4) \ge 0.$

[quote="dgreenb801"]The problem is equivalent to
$ a^6 \minus{} 2b^2a^4 \plus{} 2b^3a^3 \minus{} 2b^4a^2 \plus{} b^6 \equal{} (a \minus{} b)^2(a^4 \plus{} 2ba^3 \plus{} b^2a^2 \plus{} 2b^3a \plus{} b^4) \ge 0.$[/quote]

[color=darkblue]It's nice solution. Thank you very much.[/color]

[quote="thanhnam2902"][color=darkblue]Let $ a, b > 0$. Prove that: $ \frac {a^2}{b} \plus{} \frac {b^2}{a}\ge \sqrt {2(a^2 \plus{} b^2})$[/color][/quote]

Hello [b]thanhnam2902[/b]!

The following identity is true for all non-negative $ a,b$:

\[ {{\frac {{a}^{2}}{b}}\plus{}{\frac {{b}^{2}}{a}}\minus{}\sqrt {2 ({a}^{2}\plus{}\,{b}^{2})
}\equal{}{\frac { \left( {a}^{4}\plus{}2\,{a}^{3}b\plus{}{a}^{2}{b}^{2}\plus{}2\,a{b}^{3}\plus{}{b}^{
4} \right)  \left( a\minus{}b \right) ^{2}}{ab \left( {a}^{3}\plus{}{b}^{3}\plus{}ab\sqrt
{2 \left({a}^{2}\plus{}\,{b}^{2}\right)} \right) }}}\]

Thank you!

Sergey Markelov
markelov@mccme.ru

[quote="Markelov"][quote="thanhnam2902"][color=darkblue]Let $ a, b > 0$. Prove that: $ \frac {a^2}{b} \plus{} \frac {b^2}{a}\ge \sqrt {2(a^2 \plus{} b^2})$[/color][/quote]

Hello [b]thanhnam2902[/b]!

The following identity is true for all non-negative $ a,b$:
\[ {{\frac {{a}^{2}}{b}} \plus{} {\frac {{b}^{2}}{a}} \minus{} \sqrt {2 ({a}^{2} \plus{} \,{b}^{2}) } \equal{} {\frac {\left( {a}^{4} \plus{} 2\,{a}^{3}b \plus{} {a}^{2}{b}^{2} \plus{} 2\,a{b}^{3} \plus{} {b}^{ 4} \right) \left( a \minus{} b \right) ^{2}}{ab \left( {a}^{3} \plus{} {b}^{3} \plus{} ab\sqrt {2 \left({a}^{2} \plus{} \,{b}^{2}\right)} \right) }}}
\]
Thank you!

Sergey Markelov
markelov@mccme.ru[/quote]

[color=darkblue]Thank you very much.[/color]


linear algebra
matrix
trigonometry
integration
superior algebra
superior algebra solved

Let matrix $A=\left(\begin{array}{cc}\cos \theta+i\sin \theta & \sin \theta \\
\sin \theta & \cos \theta-i\sin \theta\end{array}\right)\  (i=\sqrt{-1})$ and  $A^n\left(\begin{array}{cc}1  \\ 0\end{array}\right)=\left(\begin{array}{cc}p_n(\theta) \\  q_n(\theta)\end{array}\right)\  (n=1,2,\cdots)$.

$\Phi_n(\theta)$ is defined as follows.

\[\Phi_n(\theta)=\int^{\frac{\pi}{2}}_{0} [p_n(\theta)-iq_n(\theta)] d\theta\].

Evaluate\[ \sum_{n=1}^\infty\frac{\Phi_{n+1}(\theta)\Phi_{n}(\theta)}{n}\].

From Cayley Hamilton we have: $A^2=2\cos\theta-\cos^2\theta I_2$

We put $A^n=a_nA-b_nI_2$. Results that
$A^{n+1}=a_n(a_2A-b_2I_2)-b_nA=(a_na_2-b_n)A-a_nb_2I_2$ (where
$a_2=2\cos\theta, b_2=\cos^2\theta$), so $a_{n+1}=a_na_2-b_n$ and
$b_{n+1}=a_nb_2$.

We find that $a_{n+1}=a_na_2-a_{n-1}b_2$ which gives us
$a_n=n\cos^{n-1}\theta$ and $b_n=(n-1)\cos^n\theta$.

Now we have that $A^n=n\cos^{n-1}\theta A-(n-1)\cos^n\theta I_2$.

We can easily find that $p_n(\theta)-iq_n(\theta)=\cos^n\theta$

so
$\displaystyle\Phi_n(\theta)=\int_0^{\frac\pi2}\cos^n\theta
d\theta=\int_0^{\frac\pi2}(\sin\theta)'\cos^{n-1}\theta d\theta =
(n-1)\int_0^{\frac\pi2}(1-\cos^2\theta)\cos^{n-2}\theta d\theta =
(n-1)\Phi_{n-2}(\theta)-(n-1)\Phi_n(\theta)$
results $\displaystyle
\Phi_n(\theta)=\frac{n-1}n\Phi_{n-2}(\theta)$

$\Phi_0(\theta)=\int_0^{\frac\pi2}1d\theta = \frac\pi2$.

$\Phi_1(\theta) = \int_0^{\frac\pi2}\cos\theta d\theta = 1$.

Now for $n$ even:

$\Phi_n(\theta)=\frac{n-1}n\frac{n-3}{n-2}\cdot\ldots\cdot\frac34\frac12\frac\pi2$

$\Phi_{n+1}(\theta)=\frac{n}{n+1}\frac{n-2}{n-1}\cdot\ldots\cdot\frac23$

and for $n$ odd:

$\displaystyle\Phi_n(\theta)=\frac{n-1}n\frac{n-3}{n-2}\cdot\ldots\cdot\frac45\frac23$

$\Phi_{n+1}(\theta)=\frac{n}{n+1}\frac{n-2}{n-1}\cdot\ldots\cdot\frac12\frac\pi2$

Results that $\Phi_n(\theta)\Phi_{n+1}(\theta)=\frac{\pi}{2(n+1)}$

and now our sum will be equal to $\frac\pi2$

Hope I didn't misscalculated smth.

Thank you for your reply, Kuba!  :)  :)  :)

You are right!

Another solution in deriving for $A^n$

[quote="Kuba"]

From Cayley Hamilton we have: $A^2=2\cos\theta-\cos^2\theta\cdot I_2$ rewrites $(A-\cos \theta\cdot I_2)^2=\mathbb{O}$.

Set $B=A-\cos \theta\cdot I_2$, we have $B^2=\mathbb{O}$,yielding $B^m=\mathbb{O}\ (m=2,3,\cdots)$.

In using  [b]binominal theorem[/b],

$A^n=(\cos \theta\cdot I_2+B)^n=\cos ^n\theta\cdot I_2+_nC_1 \cos ^{n-1}\theta\cdot B$

$=\cos ^n\theta\cdot I_2+n\cos ^{n-1}\theta\cdot(A-\cos \theta\cdot I_2)$

$=n\cos ^{n-1}\theta\cdot A-(n-1)\cos ^n\theta\cdot I_2$.

kunny


inequalities
logarithms
function
inequalities theorems

My teacher told me the Huygens inequality is: $a_{i},b_{i}$ ($i=1,2,...,n$) are positive real numbers and $\displaystyle\sum_{i=1}^{n}p_{i}=1$, then
\[\displaystyle\prod_{i=1}^{n}(a_{i}+b_{i})^{p_{i}} \ge \displaystyle\prod_{i=1}^{n}a_{i}^{p_{i}}+\displaystyle\prod_{i=1}^{n}b_{i}^{p_{i}}.\]
However it is different from the one Darij posted in this [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Huygens&t=18846]link[/url]. Moreover I dont know whether $p_{i}$ can be negative. So I want to ask:
(1) Is this inequality true? And can anyone show me a proof?
(2) If it is true, then should I called it the 'Huygens inequality' or the 'Generalized Huygens inequality'?
(3) What is $p_{i}$ there?
Thanks.

Divide both sides by $\prod a_i^{p_i}$ to obtain $\prod (1+x_i)^{p_i}\geq 1+\prod x_i^{p_i}$, where $x_i=b_i/a_i$. Let $x_i=e^{y_i}$, then finding the logarithm your inequality becomes Jensen's inequality for a convex function $\ln(1+e^y)$, $y\in\mathbb R$.

Then do we have this inequality  :? :
If $a_{ij}$ ($i=1,...,m; j=1,...,n$) are positive real numbers and $p_{j}$ ($j=1,...,n$) are real numbers such that $\displaystyle\sum_{j=1}^{n}p_{j}=1$,
\[\displaystyle\prod_{j=1}^{n}(\displaystyle\sum_{i=1}^{m}a_{ij})^{p_{j}} \ge \displaystyle\sum_{i=1}^{m}\displaystyle\prod_{j=1}^{n}a_{ij}^{p_{j}}.\]

Hmm...
We need to apply your first inequality several times here, don't we? :?

:(  :blush:  :(  :blush:  Sorry for always posting some nonsense things...... It seems that I am very stupid recently...


vector
linear algebra
linear algebra unsolved

Let $ V$ be finite dimensional complex space and $ f: V\longrightarrow V$ be linear mapping. Let $ W_f$ be the whole complex vector space which comprise of bilinear form $ b: V*V\longrightarrow \mathbb{C}$ such that $ b(v,\ w) \equal{} b(w,\ v)$ and $ b(f(v),\ w) \equal{} b(v,\ f(w))$ for all $ v,\ w\in{V}$.

(1) Let $ V \equal{} \bigoplus_{i \equal{} 1}^r {V_i}$ be direct sum decomposition of $ V$ by general eigen space $ (\text{eigen space in the wider sence })$.

Prove that $ b\left(\sum_{i \equal{} 1}^r v_i,\ \sum_{i \equal{} 1}^r w_i\right) \equal{} \sum_{i \equal{} 1}^r b(v_i,\ w_i)$ for $ b\in{W_f}, \ v_i,\ w_i\in{V_i}\ (1\leq i\leq r)$.

(2) If dim$ (V) \equal{} 3$, find the possible values of dim$ (W_f)$.

"properly space" doesn't make grammatical sense in English, and the closest term I can think of is "proper subspace", which doesn't really fit mathematically. Can you clarify what is intended?

Thank you for pointing out that.

I exchanged the term in into eigen space.


i believe that to box things we use the code
\ boxed { ........ }
only there would be no spaces and there would be $ \$$s right?

Yes, that is correct.
\boxed{4} becomes $ \boxed{4}$ when you add the dollar signs.

that is exactly what i am doing but it is not working? plz check my file

Add
\usepackage{amssymb,amsmath}
into your preamble (before \begin{document}).

A point on style: put any math, including numerals and equal signs, in math mode, i.e., between dollar signs. Also, have you tried using enumerate?


geometry
circumcircle
inequalities
inradius
geometry proposed

he circumcircle $k$ of triangle $ABC$ has radius $r$. The bisectors of the angles of the triangle intersect the the circle $k$ besides the points $A,B,C$ in the points
$A',B',C'$. If $P$ respectively $Q$ is the area of triangle $ABC$, respectively $A'B'C'$, prove the inequality\[16 \cdot Q^3 \geq 27 \cdot r^4 \cdot P.\]

First show that Q / P = r / 2R, where r is the circumradius and R is the inradius (odd? :D ) This is straightforward by expressing [A'IB'] in terms [AIB], which equals to c/2 * r. Then sum it up to get Q / P. Lot of computation, but easy logic.

Now substitute 4r = abc / P, then use AM-GM.  :)

Equality occurs when a=b=c, or ABC is equilateral.


two stars, and I think about 180-190 posts.

[quote="baseballfreak2"]do the stars relate to the posts?[/quote]

Yes, see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24794

[quote="baseballfreak2"]How many stars do you have, and how many posts?

[i]Moderator note: 5 stars has been added to the poll][/i][/quote]

Either way, the moderators have the same starts as they would have by their posts if they weren't moderators. The Administrators have six (6) stars, and you don't count them. Besides you don't count people in P vs NP that have 1/2 star, and people in Rieman Hypothesis that have two and a half!

On the other hand, the results on the Poll are not as they are in reality. For example, if you count in the member section (up to five minutes ago) You would have (Number of post/Name/Number of stars/Number of members in that cathegory/percentage of members) The data is from some minutes ago, it changes every time, but the persentages should still be almost constant

0 - 19       New Member                      (Zero Stars)  6154 - 90.954%
20 - 49       P versus NP                       (Half Star)  246 - 3.636%
50 - 99       Hodge Conjecture              (One Star) 135 - 1.995%
100 - 249       Poincare Conjecture          (Two Stars) 119 - 1.759%
250 - 499       Riemann's Hypothesis       (Two and Half stars) 48 - 0.709%
500 -  999       Yang Mills Theory              (Three Stars) 32 - 0.473%
1000 - 2499       Navier-Stroke's Equation   (Four Stars) 25 - 0.370%
2500 - $\infty$   Birch & Swinnerton Dyer.   (Five Stars) 7 - 0.104%

I counted inside there the administrators, because I didn't find the way to find them all and substract them from this data in a fast way. but any way, I don't think that there are too many administrators, and probably all of them are in the last two cathegories.

Best regards,

[quote="djimenez"]I counted inside there the administrators, because I didn't find the way to find them all and substract them from this data in a fast way. but any way, I don't think that there are too many administrators, and probably all of them are in the last two cathegories.

Best regards,[/quote]There are 4 main admins, Richard, Mathew, Dave and me. And there's Vanessa Rusczyk, but I think she's not really into math stuff too much :)

[quote="djimenez"][quote="baseballfreak2"]How many stars do you have, and how many posts?

[i]Moderator note: 5 stars has been added to the poll][/i][/quote]

Either way, the moderators have the same starts as they would have by their posts if they weren't moderators. The Administrators have six (6) stars, and you don't count them. Besides you don't count people in P vs NP that have 1/2 star, and people in Rieman Hypothesis that have two and a half!

On the other hand, the results on the Poll are not as they are in reality. For example, if you count in the member section (up to five minutes ago) You would have (Number of post/Name/Number of stars/Number of members in that cathegory/percentage of members) The data is from some minutes ago, it changes every time, but the persentages should still be almost constant

0 - 19       New Member                      (Zero Stars)  6154 - 90.954%
20 - 49       P versus NP                       (Half Star)  246 - 3.636%
50 - 100       Hodge Conjecture              (One Star) 135 - 1.995%
100 - 249       Poincare Conjecture          (Two Stars) 119 - 1.759%
250 - 499       Riemann's Hypothesis       (Two and Half stars) 48 - 0.709%
500 -  999       Yang Mills Theory              (Three Stars) 32 - 0.473%
1000 - 2499       Navier-Stroke's Equation   (Four Stars) 25 - 0.370%
2500 - $\infty$   Birch & Swinnerton Dyer.   (Five Stars) 7 - 0.104%

I counted inside there the administrators, because I didn't find the way to find them all and substract them from this data in a fast way. but any way, I don't think that there are too many administrators, and probably all of them are in the last two cathegories.

Best regards,[/quote]

You made a little mistake, Hodge cnjecture is to 99 not 100 :D

[quote="sheepwarrior"]this may sound sorta stupid, but how do the administrators become the administrators?[/quote]Ahem.... we created (and we upgrade & administrate) the site(s)? :)

[quote="jli"]You made a little mistake, Hodge cnjecture is to 99 not 100 :D[/quote]
Yep, edited and corrected, thanx.

Best,

[quote="shobber"]4 stars now![/quote]
i wish i had 4 stars

[quote="shobber"]i think having stars under your name can urge you post more...
anyway, i have two stars now :D[/quote]

 ;)  im agree with u dear shobber.

now i have about $1500$ and $4$ stars...

which one is the best the one who do math because of love...

and the one who lave math because of stars? ;)

jesus christ, the tug of war lol 2 and a half woo hooo

1 star!!! Just over $2\frac{3}{4}$ posts per day. :D That's my favorite smiley face emoticon thing. :D

i have 2.5, would've had maybe 3 if i had taken advantage of the tug of war.

the stars you never get:

$n/2$ where $n\epsilon N$, $n = 3,7,9,$ or $n>10$


[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {a^2 \plus{} b^2 \minus{} 3c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} 3a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2 \plus{} a^2 \minus{} 3b^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {4(ab \plus{} bc \plus{} ca)}{3(a^2 \plus{} b^2 \plus{} c^2)}\leq\frac {1}{3}$[/quote]

first we prove that

$ \frac {a^2 \plus{} b^2 \minus{} 2c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} 2a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2 \plus{} a^2 \minus{} 2b^2}{c^2 \plus{} ca \plus{} a^2}\leq 0.$

this ineq can be written as

$ \sum_{cyc}{\frac {(b \minus{} c)^2(b \plus{} c)(a \plus{} b \plus{} c)}{(a^2 \plus{} ab \plus{} b^2)(c^2 \plus{} ca \plus{} a^2)}} \ge 0,$

clear true.

so we just need to prove that

$ \frac {4(ab \plus{} bc \plus{} ca)}{3(a^2 \plus{} b^2 \plus{} c^2)}\leq\frac {1}{3} \plus{} \frac {c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \plus{} ca \plus{} a^2},$

by CS, we have

$ \frac {c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \plus{} ca \plus{} a^2}$

$ \ge \frac {(a \plus{} b \plus{} c)^2}{2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca}$

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

$ \equal{} \frac {1}{3} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{3(a^2 \plus{} b^2 \plus{} c^2)},$

so we just need to prove that

$ \frac {4(ab \plus{} bc \plus{} ca)}{3(a^2 \plus{} b^2 \plus{} c^2)}\leq \frac {2}{3} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{3(a^2 \plus{} b^2 \plus{} c^2)}$

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

clear true, done.

[quote="kuing"]
first we prove that
$ \frac {a^2 \plus{} b^2 \minus{} 2c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} 2a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2 \plus{} a^2 \minus{} 2b^2}{c^2 \plus{} ca \plus{} a^2}\leq 0.$
this ineq can be written as
$ \sum_{cyc}{\frac {(b \minus{} c)^2(b \plus{} c)(a \plus{} b \plus{} c)}{(a^2 \plus{} ab \plus{} b^2)(c^2 \plus{} ca \plus{} a^2)}} \ge 0$,
clear true.
[/quote]
We can just say it follows from rearrangement, can't we?  :P

[quote="kuing"][quote="filomen"]If $ a,b,c > 0$, then

$ \frac {a^2 \plus{} b^2 \minus{} 3c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} 3a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2 \plus{} a^2 \minus{} 3b^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {4(ab \plus{} bc \plus{} ca)}{3(a^2 \plus{} b^2 \plus{} c^2)}\leq\frac {1}{3}$[/quote]

first we prove that

$ \frac {a^2 \plus{} b^2 \minus{} 2c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} 2a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2 \plus{} a^2 \minus{} 2b^2}{c^2 \plus{} ca \plus{} a^2}\leq 0.$

this ineq can be written as

$ \sum_{cyc}{\frac {(b \minus{} c)^2(b \plus{} c)(a \plus{} b \plus{} c)}{(a^2 \plus{} ab \plus{} b^2)(c^2 \plus{} ca \plus{} a^2)}} \ge 0,$

clear true.
[/quote]
From this ineq we can prove the stronger:
$ \frac {a^2 \plus{} b^2 \minus{} 3c^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2 \plus{} c^2 \minus{} 3a^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2 \plus{} a^2 \minus{} 3b^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {4}{3}\leq\frac {1}{3}$


At this moment, I cannot compute how many zeros does $ z^4\plus{}z^3\plus{}z^2\plus{}2z\plus{}2$ has in the unit ball $ |z|<1$. Maybe we need to use the Rouche's theorem but how can we apply? Thank you.

Let $ f(z)\equal{}z^{4}\plus{}z^{3}\plus{}z^{2}\plus{}2z\plus{}2\equal{}g(z)\plus{}h(z)$.  Now $ \left|g(z)\right|\equal{}\left|z^{4}\plus{}z^{3}\plus{}z^{2}\right|\leq 3$ and $ \left|h(z)\right|\equal{}\left|2z\plus{}2\right|\leq 4$ on $ C$, where $ C$ is the unit circle.  So $ \left|g(z)\right|<\left|h(z)\right|$ on the unit circle.  Since $ h(z)$ has no zeroes inside $ C$, so does $ f(z)$.

[quote="JRav"]Let $ f(z) \equal{} z^{4} \plus{} z^{3} \plus{} z^{2} \plus{} 2z \plus{} 2 \equal{} g(z) \plus{} h(z)$.  Now $ \left|g(z)\right| \equal{} \left|z^{4} \plus{} z^{3} \plus{} z^{2}\right|\leq 3$ and $ \left|h(z)\right| \equal{} \left|2z \plus{} 2\right|\leq 4$ on $ C$, where $ C$ is the unit circle.  So $ \left|g(z)\right| < \left|h(z)\right|$ on the unit circle.  Since $ h(z)$ has no zeroes inside $ C$, so does $ f(z)$.[/quote]

Why does $ \left|g(z)\right| < \left|h(z)\right|$ on the unit circle friend?

[quote="JRav"]Read the sentence before that one.[/quote]

Oops, at least at $ z\equal{}\minus{}1$, $ |g(z)|\equal{}1>0\equal{}|h(z)|$.

Indeed.  I've forgotten Rouche's theorem!

Just multiply your function by $ z\minus{}1$. ;)

You've used that trick before, I recall. :wink:

$ (z\minus{}1)(z^{4}\plus{}z^{3}\plus{}z^{2}\plus{}2z\plus{}2)\equal{}z^{5}\plus{}z^{4}\plus{}z^{3}\plus{}2z^{2}\plus{}2z\minus{}z^{4}\minus{}z^{3}\minus{}z^{2}\minus{}2z\minus{}2\equal{}z^{5}\plus{}z^{2}\minus{}2$.  This has a single real root, $ z\equal{}1$, but we just supplied that root when we multiplied by $ z\minus{}1$ ($ \left|z^{5}\plus{}z^{2}\right|<1<\left|2\right|$, which has one root).  So the original function doesn't have any.

[quote="JRav"]You've used that trick before, I recall. :wink:

$ (z \minus{} 1)(z^{4} \plus{} z^{3} \plus{} z^{2} \plus{} 2z \plus{} 2) \equal{} z^{5} \plus{} z^{4} \plus{} z^{3} \plus{} 2z^{2} \plus{} 2z \minus{} z^{4} \minus{} z^{3} \minus{} z^{2} \minus{} 2z \minus{} 2 \equal{} z^{5} \plus{} z^{2} \minus{} 2$.  This has a single real root, $ z \equal{} 1$, but we just supplied that root when we multiplied by $ z \minus{} 1$ ($ \left|z^{5} \plus{} z^{2}\right| < 1 < \left|2\right|$, which has one root).  So the original function doesn't have any.[/quote]

Thanks but I still confuse how to prove $ z^5\plus{}z^2\minus{}2$ has a unique solution which is nothing but $ z\equal{}1$. The above argument $ \left|z^{5} \plus{} z^{2}\right| < 1 < \left|2\right|$ cannot be applied to $ z\equal{}1$ which lies on the boundary of $ D(0,1)$.


http://chemed.chem.purdue.edu/genchem/topicreview/bp/1biochem/amino2.html#table

Scroll down a little and you'll see a big list of amino acids and their structures.

Which ones out of those amino acids have only one isomer (no enantiomers are possible)?

I know what an isomer is, but I'm a little confused as how to figure out which ones only have one.  Any help is greatly appreciated. Thanks!
:)

"No enantiomers are possible" means that the molecule can be superimposed on its mirror image. Another word for this is "not chiral."

While there are several way at arriving at the other, chiral, state, the simplest is to have a carbon covalently bonded to four different things. The bonds in such a carbon are in a tetrahedral arrangement. To simulate what is going on, sketch a tetrahedron with the letters A, B, C, and D at its four vertices. Can you verify to your own satisfaction that such a thing cannot be superimposed on its mirror image? But that if the letters were A, B, C, C, then it could be so superimposed?

The usual biologically important amino acids have a central carbon atom with four things bonded to it. These four things are:

1. An amine group, $ \text{NH}_2.$ (Actually $ \text{NH}_3^ \plus{}$ at physiological pH.)

2. An carboxylic acid group, $ \text{COOH}.$ (Actually $ \text{COO}^ \minus{}$ at physiological pH.)

3. A hydrogen.

4. A side chain.

The differences between the different amino acids come entirely from differences between the side chains. And none of the side chains are either a simple amine group or a simple acid group. So that has to be four different items (hence chiral or having enantiomers) with exactly one possible exception: the exception being that the side "chain" might consist only of hydrogen. That amino acid does exist: glycine.

Proline is a little different in that #1 (the amine) and #4 (the side chain) are connected in a loop. But that still means that proline has an asymmetric carbon.

(I might have classified this topic as chemistry rather than biology.)


arithmetic sequence
number theory
prime numbers
number theory proposed

For each positive integer $n \geq 2$, let $\rho(n)$ be the product of the distinct prime divisors of $n$. Let $a_{1}\geq 2$ be an integer and define $a_{n+1}= a_{n}+\rho(a_{n})$. Show that the sequence $\left(a_{n}\right)$ contains arbitrarily long arithmetic progressions.

Let $p_{0}=2,p_{1}=3,....$ primes and \[M_{k}=\prod_{i=0}^{k}p_{i}, S_{k}=\sum_{i=1}^{k}p_{i}.\] Define k(n), suth that $S(k(n))<n\le S(k(n)+1)$.
It is easy to show $a_{n}=M(k(n))*(n+1-S(k(n)))$.
Then your needed result is obviosly.

Hm... what about $a_{1}$?? I mean, the first term is arbitrary, so your formula can't be right?

I give solution for $a_{1}=2$. It can be be modification for $a_{1}>2$. But in this case is difficult give exactly formuls.

Let $a_{n+1}=a_{n}+rad(a_{n})$ (rad(x) is product of different prime divisors), then $rad(a_{n+1})\ge rad(a_{n})$. Let $b_{n}=\frac{a_{n}}{rad(a_{n})}$. If $rad(a_{n})=rad(a_{n+1})=...=rad(a_{n+k-1})<rad(a_{n+k})$ then  $b_{n+i}=b_{n}+i$ and $rad(b_{n}+i)|rad(a_{n}), i=1,2,...,k-1$ and $b_{n+k}=\frac{b_{n}+k}{p}<b_{n}$ if $b_{n}>1$, because prime p>k. It give $b_{n_{1}}> b_{n_{2}}> b_{n_{3}}> ...$ Therefore exist n when $b_{n}=1$. It mean $a_{N}=M_{k}=\prod_{i\le k}p_{i}$. Then sequence defined by
Let $l(n): \ S_{l}-S_{k}\ge n-N<S_{l+1}-S_{k}$, were $S_{m}$ is the sum first m primes. Then $a_{n}=M_{l(n)}(n+1-N-S_{l(n)}+S_{k})$. It is obviosly $a_{n}$ have $p_{k+1}$ terms arithmetic progression, then $p_{k+1}$ terms e.t.c.

[i]Lemma :[/i]  For each prime $p$ there exists $i$ that for $n\geq,p|a_{n}$.
[b]Proof :[/b] If $a_{k}=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{t}^{\alpha_{t}}$ then $a_{k+1}=p_{1}p_{2}\dots p_{t}(p_{1}^{\alpha_{1}-1}p_{2}^{\alpha_{2}-1}\dots p_{t}^{\alpha_{t}-1}+1)$ so there is a $s\geq k$ that $a_{s}$ is Free of Square. Then after some while the least prime number that does not divide $a_{n}$ increases.

If $p_{1},p_{2},\dots,p_{k}$ are the first $k$ prime numbers, then there is an $n$ that $p_{1}p_{2}\dots p_{k}|a_{n}$ and $a_{n}$ is free square. Assume that $a_{n}=p_{1}p_{2}\dots p_{k}q_{1}q_{2}\dots q_{s}$. So for each $1\leq i\leq p_{k}$ you see that $a_{n+i-1}=ip_{1}p_{2}\dots p_{k}q_{1}q_{2}\dots q_{s}$ and we have got a long arithmetic progression.


function
algebra
functional equation
real analysis
real analysis unsolved

fix $ t\in(0,1)$.
find all continuous $ f: \mathbb{R}\to\mathbb{R}$ such that \[ f(tx\plus{}(1\minus{}t)y) \equal{} tf(x)\plus{}(1\minus{}t)f(y)\]

please help me to deal with this problem :(

The function is convex on I, if for every $ x, y\in I; \quad x<y$ we have
\[ f(tx\plus{}(1\minus{}t)y)\leq tf(x)\plus{}(1\minus{}t)f(y).\]
The function is concave on I, if for every $ x, y\in I; \quad x<y$ we have
\[ f(tx\plus{}(1\minus{}t)y)\geq tf(x)\plus{}(1\minus{}t)f(y).\]
Therefore your function is concave and convex on $ \mathbb{R}$.

What can you deduce from that?

[hide]Straight lines are the only solutions. Hence $ f(x)\equal{}cx\plus{}d$, where $ c, d \in \mathbb{R}.$  [/hide]

[quote="tmrfea"]The function is convex on I[/quote]
This is not a direct consequence of the given condition, since $ t$ is fixed. If continuity of $ f$ is omitted, even $ f$ can fail to become convex.

Here is a direct approach:

Let $ f$ be a solution to the given functional equation $ (*)$. Fix any two distinct point $ x < y$ in $ \mathbb{R}$.

[b]Step 1.[/b] For any $ s \in (0, 1)$. we construct a certain sequence $ z_1, z_2, \cdots$ which converges to $ z \equal{} sx \plus{} (1\minus{}s)y$.

Put $ r \equal{} \max \{ t, 1\minus{}t \}$, $ I_1 \equal{} [a_1, b_1] \equal{} [x, y]$ and $ z_1 \equal{} tx \plus{} (1\minus{}t)y$. Ten inductively define $ I_n \equal{} [a_n, b_n]$ and $ z_n$ as follows:

If $ z \in [a_n, z_n]$, put $ I_{n\plus{}1} \equal{} [a_n, z_n]$. Otherwise, put $ I_n \equal{} [z_n, b_n]$. Then define $ z_{n\plus{}1} \equal{} ta_{n\plus{}1} \plus{} (1\minus{}t)b_{n\plus{}1}$.

Clearly $ I_1 \supset I_2 \supset I_3 \supset \cdots$ and $ |b_{n\plus{}1} \minus{} a_{n\plus{}1}| \leq r|b_n \minus{} a_n|$, so $ \textstyle \bigcap_n I_n$ consists of only a single point, which is $ z$ by our construction. So $ z_n \to z$.

[b]Step 2.[/b] Now write $ z_n \equal{} t_n x \plus{} (1\minus{}t_n) y$ for some $ t_n \in (0, 1)$. Then we claim that $ f(z_n) \equal{} t_n f(x) \plus{} (1\minus{}t_n) f(y)$.

The case $ n \equal{} 1$ follows from definition, so suppose our claim holds for all $ k \leq n$. First we assume $ I_{n\plus{}1} \equal{} [a_n, z_n]$.

By our construction, $ a_n$ is either $ x$ or $ z_k$ for some $ k < n$, so that either $ z_{n\plus{}1} \equal{} tx \plus{} (1\minus{}t)z_n$ or $ z_{n\plus{}1} \equal{} tz_k \plus{} (1\minus{}t)z_n$.

Then it is easy to confirm that our claim holds for this case. The case $ I_{n\plus{}1} \equal{} [z_n, b_n]$ also follows in a similar mannar.

[b]Step 3.[/b] Then $ t_n \to s$, so by continuity of $ f$, we have $ f(sx \plus{} (1\minus{}s)y) \equal{} sf(x) \plus{} (1\minus{}s)f(y)$. This shows that the graph of $ f$ is a line.


function
geometry
3D geometry
algebra
polynomial

The numbers $ x, y$ and $ z$ satisfy

$ x\plus{}y\plus{}z\equal{}1$
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}2$
$ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{}3$.

Show that $ yz\plus{}zx\plus{}xy \equal{} \minus{}\frac{1}{2}$.

Show also that $ x^{2}y\plus{}x^{2}z\plus{}y^{2}z\plus{}y^{2}x\plus{}z^{2}x\plus{}z^{2}y \equal{} \minus{}1$, and hence that $ xyz\equal{}\frac{1}{6}$.

Hence find numbers $ a, b$ and $ c$, independent of $ n$, such that

$ S_{n\plus{}1} \equal{} aS_{n} \plus{}bS_{n\minus{}1} \plus{} cS_{n\minus{}2}$,

where $ S_{n} \equal{} x^{n}\plus{}y^{n}\plus{}z^{n}$.

For the first part:
Squaring both sides of the first equation yields:
$ (x\plus{}y)^2\plus{}2z(x\plus{}y)\plus{}z^2\equal{}1$
$ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2xz\plus{}2yz\equal{}1$
From the second equation, $ x^2\plus{}y^2\plus{}z^2\equal{}2$, so subtracting that expression from both sides of the squared first equation gives
$ 2xy\plus{}2xz\plus{}2yz\equal{}\minus{}1$
$ 2(yz\plus{}zx\plus{}xy)\equal{}\minus{}1$
$ yz\plus{}zx\plus{}xy\equal{}\minus{}\frac{1}{2}$

I'm too lazy to type the second part, but I'll give a hint:

Use the same method as before, except cube the first equation. Then use all the info you are given already to prove that statement true.

I suppose I'll try the second part. As gauss1181 mentioned, we start off by cubing both sides of the first equation and subtracting both sides from the third equation:

$ x^3\plus{}3x^2y\plus{}3x^2z\plus{}y^3\plus{}3xy^2\plus{}3y^2z\plus{}z^3\plus{}3xz^2\plus{}3yz^2\plus{}6xyz\equal{}1$

$ 3x^2y\plus{}3x^2z\plus{}3xy^2\plus{}3y^2z\plus{}3xz^2\plus{}3yz^2\plus{}6xyz\equal{}\minus{}2$

$ x^2y\plus{}x^2z\plus{}xy^2\plus{}y^2z\plus{}xz^2\plus{}yz^2\equal{}\minus{}\frac{2}{3}\minus{}2xyz$

Now we start again on the left hand side.

$ x^2y\plus{}x^2z\plus{}y^2z\plus{}y^2x\plus{}z^2x\plus{}z^2y\equal{}x^2y\plus{}xy^2\plus{}xyz\plus{}x^2z\plus{}xyz\plus{}xz^2\plus{}xyz\plus{}y^2z\plus{}yz^2\minus{}3xyz$

$ x^2y\plus{}x^2z\plus{}y^2z\plus{}y^2x\plus{}z^2x\plus{}z^2y\equal{}xy(x\plus{}y\plus{}z)\plus{}xz(x\plus{}y\plus{}z)\plus{}yz(x\plus{}y\plus{}z)\minus{}3xyz$

$ x^2y\plus{}x^2z\plus{}y^2z\plus{}y^2x\plus{}z^2x\plus{}z^2y\equal{}(xy\plus{}xz\plus{}yz)(x\plus{}y\plus{}z)\minus{}3xyz$

$ x^2y\plus{}x^2z\plus{}y^2z\plus{}y^2x\plus{}z^2x\plus{}z^2y\equal{}(\minus{}0.5)(1)\minus{}3xyz$

$ x^2y\plus{}x^2z\plus{}y^2z\plus{}y^2x\plus{}z^2x\plus{}z^2y\equal{}\minus{}\frac{1}{2}\minus{}3xyz$

Now we substitute:

$ \minus{}\frac{2}{3}\minus{}2xyz\equal{}\minus{}\frac{1}{2}\minus{}3xyz$

$ xyz\equal{}\frac{1}{6}$

$ x^2y\plus{}x^2z\plus{}y^2z\plus{}y^2x\plus{}z^2x\plus{}z^2y\equal{}\minus{}1$

The answers follow almost directly from Newton sums:

for a cubic polynomial (wlog let $ a_3 \equal{} 1$) $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0$ we have

$ S_1 \plus{} a_2 \equal{} 0$

$ S_2 \plus{} a_2S_1 \plus{} 2a_1 \equal{} 0$

$ S_3 \plus{} a_2S_2 \plus{} a_1S_1 \plus{} 3a_0 \equal{} 0$

substituting in $ S_1 \equal{} 1,S_2 \equal{} 2,S_3 \equal{} 3$ we have $ a_2 \equal{} \minus{} 1,a_1 \equal{} \minus{} 1/2,a_0 \equal{} \minus{} 1/6$ so the polynomial we seek is
$ x^3 \minus{} x^2 \minus{} \frac12 x \minus{} \frac16\implies xy \plus{} yz \plus{} zx \equal{} a_1 \equal{} \minus{} 1/2,xyz \equal{} \minus{} a_0 \equal{} 1/6$

$ (x \plus{} y \plus{} z)^3 \equal{} S_3 \plus{} 3R \plus{} 6xyz \equal{} 1\implies R \equal{} \minus{} 1$

For $ S_{n \plus{} 1} \equal{} aS_{n} \plus{} bS_{n \minus{} 1} \plus{} cS_{n \minus{} 2}$, this follows immediately from Newton sums ($ a \equal{} \minus{} a_2,b \equal{} \minus{} a_1,c \equal{} \minus{} a_0$


A man can run x feet in y seconds.How many yards can he run in z min.

[hide]In one second, the man runs $\frac{x}{y}$ feet.  In 1 minute, the man runs $\frac{60x}{y}$ feet.  That is the same as $\frac{20x}{y}$ yards.  Therefore, in $z$ minutes, he will run $\frac{20xz}{y}$ yards.[/hide]

Isn't this problem in AOPS volume 1?
[hide]If the man runs $x$ feet in $y$ minutes, his rate is
$\frac xy$ feet per second.

Multiply by $60$ to find minutes so now it is

$\frac{60x}{y}$ feet per minute.

Divide by $3$ to find yards so now it is

$\frac{20x}{y}$ yards per minute.

To find how many yards he can run in $z$ minutes, you use the $d=rt$ formula and plug in:

$d=\frac{20x}{y} \times z$

So he can run $\boxed{\frac{20xz}{y}}$ yards per minute.
[/hide]


inequalities proposed
inequalities
algebra

[quote=sqing]Let $a,b,c$ are non negative number such that $a+b+c=\frac{3}{5} $ Prove that
$$\frac{9}{2} \leq\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 6$$[/quote]

$$\frac 92 \leq\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 6\Leftrightarrow \frac 32 \leq\frac{2a}{1-a}+\frac{2b}{1-b}+\frac{2c}{1-c}\leq 3.$$
We know that $a^2+b^2+c^2\ge \frac{(a+b+c)^2}3=\frac 3{25}$.
By Cauchy-Schwarz, we have $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{a^2}{a-a^2}+\frac{b^2}{b-b^2}+\frac{c^2}{c-c^2}\ge \frac{(a+b+c)^2}{(a+b+c)-(a^2+b^2+c^2)}=\frac{\frac{9}{25}}{\frac 35-(a^2+b^2+c^2)}\ge \frac{\frac{9}{25}}{\frac 35-\frac 3{25}}=\frac 34.$$
Equality occurs when $a=b=c=\frac 15$.

Let $f(a,b,c)=\frac{2a}{1-a}+\frac{2b}{1-b}+\frac{2c}{1-c}$. It is easy to see that $f(x,y,z)\leq f(x+y,0,z)$ where $x+y+z=\frac 35$ and $x,y,z\ge 0$.
Hence, $f(a,b,c)\leq f(a+b,0,c)=f(a+b,c,0)\leq f(a+b+c,0,0)=3$.
Equality occurs when $(a,b,c)=\left(\frac 35,0,0\right)$ and its permutations.

[hide=Thanks.][quote=BarisKoyuncu][quote=sqing]Let $a,b,c$ are non negative number such that $a+b+c=\frac{3}{5} $ Prove that
$$\frac{9}{2} \leq\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 6$$[/quote]

$$\frac 92 \leq\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 6\Leftrightarrow \frac 32 \leq\frac{2a}{1-a}+\frac{2b}{1-b}+\frac{2c}{1-c}\leq 3.$$
We know that $a^2+b^2+c^2\ge \frac{(a+b+c)^2}3=\frac 3{25}$.
By Cauchy-Schwarz, we have $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{a^2}{a-a^2}+\frac{b^2}{b-b^2}+\frac{c^2}{c-c^2}\ge \frac{(a+b+c)^2}{(a+b+c)-(a^2+b^2+c^2)}=\frac{\frac{9}{25}}{\frac 35-(a^2+b^2+c^2)}\ge \frac{\frac{9}{25}}{\frac 35-\frac 3{25}}=\frac 34.$$
Equality occurs when $a=b=c=\frac 15$.

Let $f(a,b,c)=\frac{2a}{1-a}+\frac{2b}{1-b}+\frac{2c}{1-c}$. It is easy to see that $f(x,y,z)\leq f(x+y,0,z)$ where $x+y+z=\frac 35$ and $x,y,z\ge 0$.
Hence, $f(a,b,c)\leq f(a+b,0,c)=f(a+b,c,0)\leq f(a+b+c,0,0)=3$.
Equality occurs when $(a,b,c)=\left(\frac 15,0,0\right)$ and its permutations.[/quote]

[/hide]


function
algebra
domain
induction
functional equation

[quote="boxedexe"]Find all functions $f: \mathbb{N}\rightarrow\mathbb{N}$ such that $f(f(m)+f(n))=m+n$.

Masoud Zargar[/quote]

Solution in Kalva:

$f$ is obviously surjective. If $f(a) = f(b)$, then $c + a = f( f(c) + f(a) ) = f( f(c) + f(b) ) = c + b$, so $a = b$. Hence $f$ is injective. So the inverse $f^{-1}$ exists. The functional equation gives immediately $f(m + n) = f^{-1}(m) + f^{-1}(n)$ and $f^{-1}(m + n) = f(m) + f(n)$.

$f(2n) = f^{-1}(2n-1) + f^{-1}(1) = f(2n-2) + f(1) + f^{-1}(1)$. Iterating, $f(2n) = n f(1) + n f^{-1}(1)$. Similarly, $f(2n+1) = n f(1) + (n+1) f^{-1}(1)$.

So we putting $f(1) = h$, $f^{-1}(1) = k$, we have $f(1) = h$, $f(2) = h + k$, $f(3) = h + 2k$, $f(4) = 2h + 2k$, $f(5) = 2h + 3k$, ... . But $h, k \geq 1$ and $f$ is surjective, so we must have $h = k = 1$. Hence $f(n) = n$ for all n.

[quote="luimichael"][quote="M4RI0"]
Solution in Kalva:

$f$ is obviously surjective. [/quote]
I think it is not that obvious:
What is the pre-image of 1 ?
That is, if $f(n)=1$ then $n = ?$[/quote]

The domain is $\mathbb N$ and range is $\mathbb N$. Isn't that sufficient to know the function is surjective?

Even the domain and the range are the same, the function may be NOT surjective.
An easy understanding example:
$f: \mathbb{N}\rightarrow\mathbb{N}$
Define $f(n) = n + 1$ for all natural numbers $n$.
This function is NOT surjective because you can't find the pre-image of 1.

Anyway, I still think the term "obvious" is NOT suitable.

But we have $f(x) = m + n$ for some $x$ and all $m,n$ so $f$ is clearly surjective because it can map to all integer values.

Told you that the problem arised from $f(n)=1$ already!
Note that $m\geq 1$ and $n\geq 1$. Then $m+n\geq 2$.
We cannot find the pre-image of 1 by assigning values to m and n to the given condition direclty.
Therefore, in my opinion,  the surjective property is NOT obvious.

[quote="Jan"]I don't know if anybody is still interested since there has been a solution posted...
Anyway, here's a more elementary solution to the problem: (It doesn't need surjectivity)

Set $m=n=x$: $f(2f(x))=2x$
Set $m=n=2f(x)$: $f(4x)=4f(x)$
Set $m=n=0$: $f(0)=4f(0) \Rightarrow f(0)=0$
Set $m=0$: $f(f(n))=n$

The function is injective, because if $f(m)=f(n) \Rightarrow m=n$ (take $f$ of both sides)
So \[ f(f(m)+f(n))=m+n=f(f(m+n)) \Rightarrow f(m+n)=f(m)+f(n)  \]

Now it follows that $f(x)=xf(1)$
Take $f$ of both sides, and obtain $x=f(1).f(x)=f(1).x.f(1)$ So it follows that $f(1)=1$ (the possibilty $f(1)=-1$ can be ignored since the range is $\mathbb{N}$)

Induction: $f(0)=0$
$f(x+1)=f(x)+1$

So $f(x)=x$.[/quote]

This might sound stupid but, does 0 belong to $\mathbb N$?

[quote="Jan"]Yes, I'm quite sure $0$ does belong to $\mathbb{N}$. Why else would the frequently used symbol $\mathbb{N}_0$ exist?  :)[/quote]

In America, it is standard although not absolutely universal to define ${\bf N} \equiv \{1, 2, \ldots\}$.  In which case this problem is still unsolved, since as luimichael pointed out the original solution assumes ${\bf N } = \{m + n \mid m, n \in \bf N\}$.

I'm also not sure I've ever seen the notation ${\bf N}_0$ before.

This is very strange.  :huh:  Look on mathworld: http://mathworld.wolfram.com/NaturalNumber.html

[quote][...] there seems to be no general agreement about whether to include it.[/quote]

So we could change $\mathbb N$ to $\mathbb {N}_0$, and the problem would have a nicer solution.

Wether $0\in\mathbb{N}$ is a long arguing between different conventions, different habits, ...

Usually people just write $\mathbb{N}$ for both and let the context decide, and in cases like here where it does matter, you should write
- $\mathbb{N}\cup \{0\}$ for $0,1,2,...$
- $\mathbb{N}\backslash\{0\}$ or $\mathbb{N}^*$ for $1,2,...$
which are universal and cannot be misunderstood.

The problem with the symbol $\mathbb{N}_0$ is exactly that both conventions use it for the contrary of their $\mathbb{N}$ meaning.


[quote="Anthony Cai"]How do you use AM-GM to find the minimum value of $ x^{2}+2x+\frac{24}{x}$[/quote]you can split up that $ \frac{24}{x}$ like this: $ \frac{24}{x}=\frac{8}{x}+\frac{8}{x}+\frac{8}{x}$, so you have as many x's in the numerator as in the denominator. So you have: $ \frac{x^{2}+2x+\frac{8}{x}+\frac{8}{x}+\frac{8}{x}}{5}\geq\sqrt[5]{2\cdot 8^{3}}=4$, equality when $ x=2$.
The difficult part, though, is that you want to find an "arrangement and division" of the terms that will ensure you have a verifiable equality case.


[quote="peeta"]solve for all $x,y,z\in\mathbb R$:

\[x+y+z=6\\
xy+xz+yz=12\][/quote]

suppose $x,y,z$ is the roots of the cubic equation $f(t)=t^3-6t^2+12t+k$ for some $k$ . Since $f'(t)=3(t-2)^2$ , this implies $f(t)$ has three factors of $t-2$ . Hence $f(2)=8+k=0$ so gives us $f(t)=t^3-6t^2+12t-8=(t-2)^3$ . Hence $x=y=z=2$ is the solution .

A clever problem!
There are a lot of ways to do it,
personally I will use inequality in solving...

We have $(x+y+z)^2-3(xy+yz+zx)=6^2-3 \cdot 12=0$
so $(x+y+z)^2=3(xy+yz+zx)$
or $x^2+y^2+z^2-xy-yz-zx=0$
or $(x-y)^2+(x-z)^2+(y-z)^2=0$
so $x=y=z$
and finally $x=y=z=\frac{6}{3}=2$......

By AM-GM-HM, $xyz \le 8$ and $\frac{3xyz}{12} \le \sqrt[3]{xyz} \rightarrow xyz \ge 2$. So $2 \le xyz \le 8$ and the greatest possible value:

$8/z+8/y+8/x=12$.
$1/z+1/y+1/x=3/2$.

And for $xyz=2$,

$2/z+2/y+2/x=12$.
$1/z+1/y+1/x=6$.

So $\frac{3}{2} \le \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le 6$

which isn't useful at all... nevermind :(

i like socrates way, i came up with this problem equally playing around with the chebychev inequalty:

$36=(x+y+z)(y+z+x)\le3(xy+yz+xz)=36$

so the only solutions are when $x=y=z$
pretty cool  :gleam:  i think


Set A: The union of geometric figures, started with a circle of radius 1. A square is inscribed inside the circle, then a circle is inscribed in the square, then a square in that, etc.

Set B: Same thing, started with a circle of radius 1, then an equilateral triangle inscribed in that, then a circle, eq. triangle, etc.

Set B is overlayed over set A such that the circles of radius 1 coincide. Are there any other circles in the union of the two sets that coincide? If so what is the radius of next-largest circle?

Hope that made sense. You might try converting this to algebra terms.

It is clear that inscribing a eq. triangle, then a circle, because of the 30-60-90 triangle apparent, that the circle shrinks by factor 1/2 each time

It is clear that inscribing a square, then a circle, shrinks it by factor 1/sqrt(2).

By symmetry, the origin of all the circles is the same.

Now the result is clear:  the answer is 1/2.

Easier than I thought.  :lol:

Try this one: using the same method, put two more planes on it, with a regular pentagon and regular hexagon, and find the circle common to all planes.


i know a^2 -7(b^2) = 11 has no integer solutions, but how do i show that it doesn't?

You show it by working $ \mod 4$. And this is obviously posted in a wrong forum.


I was asked to post these problems in a separate topic, so here they are ...

We've got a week-long qualification camp in Slovakia and there is an awful lot of problems there, but most of them are from former IOI and CEOI's. If you're interested in some treining material, try those. Our national round had two practical and three theoretical problems, they can be found at http://www.ksp.sk/mop/english/. I think they are only in Slovak, unfortunately. Here is a short description of the practical (that is an IOI-like) contest problems:

1. You get two strings of length N < 1 000 000. (char by char, and simultaneously) Find out whether they can be changed one into the second by repeating this operation: "take two adjacent chars in a string, which are NOT adjacent in the alphabet, and interchange them".
For example, the strings "abraka" and "krabaa" are equivalent (interchangeable), but strings "dabra" and "badar" are not.

2. You get a road map of a land with N < 100 000 towns and M < 500 000 roads between them. The government wants to pave the roads, so they bought a lot of asphalt. Asphalt comes in large barrels, one barrel is enough to pave exactly two of the roads. Once you open a barrel, you should use it all up on two adjacent roads (two roads sharing an endtown). You don't want to end up in a town that has all its roads paved with a half-full barrel of asphalt! Find a way how to pave the roads or output that it is not possible. (a way how to pave the roads = give a list of pairs of adjacent roads that are to be paved with one barrel of asphalt)
For example, if N=5 and M=8, and the roads are between towns:
1 5
1 4
1 3
1 2
3 2
3 4
4 5
4 2

We need four barrels, paving these pairs of roads
5 1 4 = (road from 5 to 1 and then from 1 to 4)
5 4 3
4 2 3
3 1 2

---

The time limits were not specified, but in 1., they said we should use <stdio>, not <iostream> to read the input, because cin is too slow;)

Ok. I have seen the pdf containing solution. And the solution of the first problem given there is faster than mine proposed solution. The idea is same but the checking time is improved!

For the second one, I saw one line of the given solution and that is DFS(). Then after thinking a lot I think I got the solution. Though I can’t compare my solution with the official one because those are in Slovak language.

Anyhow the idea is to start DFS() from any point. And while backtracking pair each adjacently discovered edges. After backtracking if the last edge is undiscovered then keep it in memory and try to discover through another vertices adjacent to that vartice. And thus try to pair like last time. And see if you can pair that unpaired edge with any edge. If it is impossible after completeing the process of a vertice then the procedure is impossible. Run this process until all the vertices are completed. And if completed print Possible. That’s all.

Actually it is easier to code than describing the process. ;)

[hide="First one"]
My solution was this: for each pair of adjacent letters (a and b, b and c, ...) I kept a linked list of excess of those letters in one of the strings, as I passes thorugh them simultaneously. For example for strings

abababxxxx
xxaxxbabab

after processing first 4 chars of both stings, there will be "bab" in the (a and b) list, "bb" in the (b and c) list, and "xxx" in the (x and y) and (w and x) list, with a flag that excess is in the second string.

This way you need to traverse the strings only once and the memory is 2*N worst case.[/hide]

[hide="Second one"]
It is possible to do it if M is even. I forgot to mention that the graph is guaranteed to be interconnected. I did it this way: Construct a DFS tree (well, in fact, any spanning tree is OK), and then, when returning back from the leaves, for each town do this:

* Pair roads to all its neighbors that are marked done.
* If there was an odd number of them, pair the last road with the road to the parent and delete the edge to the parent from the graph.
* Mark this town done.

O(N+M)
[/hide]

My and your solution tot he second one is same.

For the first one, i had the same idea but a bit time costing! :oops:

Anyhow, r u making IOI?

Yes, I'm going. Too bad it's so close, only 50 km from Slovak borders;) Anyway, we ought to meet there ... [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36480]I think many more mathlinkers are going[/url] ... well, I see that we're going to meet in Mexico also, arent' we? ;-)

Aha! I did not noticed that. But anyhow hope to meet u at mexico! :)

click at the link that kubus gave. the problem he gave was probably in the almost last part. and after finding the problem pdf, click the soln pdf!


Εστω η ακολουθεια $a_{n}=a_{n-1}a_{n-2}$ με $n>2$ της οποιας ολοι οι οροι της ειναι θετικοι ακεραιοι $(a_{1}=a$ και $a_{2}=b)$.Νδο:
αν ενας τουλαχιστον απο τους 3 πρωτους ορους της ειναι τελοιο τετραγωνο τοτε εχει απειρους ορους που ειναι τελεια τετραγωνα.Αν παλι κανενας απο τους 3 πρωτους ορους της δεν ειναι τελειο τετρ τοτε κανενας ορος της δεν ειναι τελειο τετραγωνο. :)

Μιας και την εφτιαξα ας βαλλω και την λυση μου...$a_{n}=a_{n-1}a_{n-2}=a_{n-2}^{2}a_{n-3}$.......(ειναι η ιδεα) :wink:


Let $x$, $y$, and $z$ be non-negative real numbers satisfying $x + y + z = 1$.  Show that
\[ x^2 y + y^2 z + z^2 x \leq \frac{4}{27} \]
and find when equality occurs.

[quote="shobber"]Let $ x$, $ y$, and $ z$ be non-negative real numbers satisfying $ x \plus{} y \plus{} z \equal{} 1$.  Show that
\[ x^2 y \plus{} y^2 z \plus{} z^2 x \leq \frac {4}{27}
\]
and find when equality occurs.[/quote]

Really classical on ML:

http://www.mathlinks.ro/Forum/viewtopic.php?t=216
http://www.mathlinks.ro/Forum/viewtopic.php?t=52258 (a generalization to $ n$ variables)
http://www.mathlinks.ro/Forum/viewtopic.php?t=63530
http://www.mathlinks.ro/Forum/viewtopic.php?t=86688 (a generalization)
http://www.mathlinks.ro/Forum/viewtopic.php?t=30843 (a generalization by Hungkhtn)
http://www.mathlinks.ro/Forum/viewtopic.php?t=30337 (a generalization by Sbos)
http://www.mathlinks.ro/Forum/viewtopic.php?t=2549 (a well-known generalization)
http://www.mathlinks.ro/Forum/viewtopic.php?t=24113 (here in a slightly modified version)
http://www.mathlinks.ro/Forum/viewtopic.php?t=76726 (an attempt at a generalization, half-failed)
http://www.mathlinks.ro/Forum/viewtopic.php?t=28680 (two similar problems)
http://www.mathlinks.ro/Forum/viewtopic.php?t=83964 (a wrongly-formulated generalization)
http://www.mathlinks.ro/Forum/viewtopic.php?t=82136 (some related facts)
http://www.mathlinks.ro/Forum/viewtopic.php?t=3090
http://www.mathlinks.ro/Forum/viewtopic.php?t=78790

If I were mod, I would merge some of these topics...

  Darij


$ \mbox{Fie ABCDA'B'C'D' un cub. Notam cu M si N mijloacele muchiilor [BC] si respectiv [DD'], iar cu: } \\
\{P\} \equal{} DM\cup AC, \ \{Q\} \equal{} CN\cup DC'. \mbox{ Demonstrati ca: } PQ\parallel{}(ABC').$
[b]REMARCA[i](M. Miculita)[/i]:[/b] $ \mbox{Concluzia problemei ramane adevarata si in cazul in care ABCDA'B'C'D' este un paralelipiped arbitrar!}$

$ \mbox{Notam cu O si cu O' centrele paralelogramelor ABCD si respectiv CDD'C', iar cu S mijlocul muchiei [CD]. Avem, atunci: }$
$  \begin{array}{c} ABCD - paralelogram \\
\{O\} = AC\cap BD \end{array}\}\Rightarrow|BO| = |DO|\Rightarrow[CO] - mediana \ in \ \triangle{BCD}.\ (1)$
$ \mbox{Pe de alta parte: }[DM] - mediana \ in \ \triangle{BCD} \mbox{ si }[BS] - mediana \ in \ \triangle{BCD}.\ (2) \\
\mbox{Din }(1),(2)\Rightarrow[CO]\cap[DM]\cap[BS] = \{P\}\Rightarrow \\
\Rightarrow\mbox{P - este centrul de greutate in }\triangle{BCD}\Rightarrow\frac {|PS|}{|PB|} = \frac {1}{2}.\ (3) \\
\mbox{In mod analog se arata ca: } \\
\mbox{Q - este centrul de greutate in }\triangle{CDD'}\Rightarrow\frac {|QS|}{|QD'|} = \frac {1}{2}.\ (4) \\
 \begin{array}{c} \mbox{Din: }(3),(4)\Rightarrow\frac {|PS|}{|PB|} = \frac {|QS|}{|QD'|}\Rightarrow PQ\parallel{}BD' \\
BD'\subset(ABC'D')\end{array}\}\Rightarrow PQ\parallel{}(ABC'D').$


parameterization
calculus
trigonometry
Trigonometric Identities
algebra
IMO Longlist

Find values of the parameter $u$ for which the expression
\[y = \frac{ \tan(x-u) + \tan(x) + \tan(x+u)}{ \tan(x-u)\tan(x)\tan(x+u)}\]
does not depend on $x.$

Basically what we need to do is to derivate $ y(x,u)$ in terms of $ x$ and find $ u$ for which the derivative cancels. Before we do that, using the addition formula for tangents, we reduce the expression to the form
\[ y = \frac 12 \cdot \frac { \cos { 2u} + 3 \cos {2x} + 2 }{ \sin (x-u) \sin (x+u) } .\]
Now we can derivate easily, and obtain the equation
\begin{align*}
\cos 2 u \cot(u-x)+3 \cos 2x  \cot(u-x)-\cos 2u\cot(u+x) & \\
-3 \cos 2x \cot (u+x)-6 \sin 2x & = 0.
\end{align*}
Via [url=http://www.wolframalpha.com/input/?i=solve+%28cos%282+u%29+cot%28u-x%29%2B3+cos%282+x%29+cot%28u-x%29-cos%282+u%29+cot%28u%2Bx%29-3+cos%282+x%29+cot%28u%2Bx%29-6+sin%282+x%29%29+%3D+0]Wolfram[/url], we get that the solutions we need are $ u = \dfrac 14 (4\pi  k \pm \pi)$, where $ k\in \mathbb Z$.

i don't know if i'm mistaken, but i obtained that it was $ k\pi\pm\frac {\pi}{3}$, for integers $ k$....

i got also that $ y \equal{} \dfrac{3\cos 2x \plus{} \cos 2u \plus{} 2}{\cos 2u \minus{} \cos 2x} \equal{} \minus{} 3 \plus{} \dfrac{2 \plus{} 4\cos 2u}{\cos 2u \minus{} \cos 2x}$...

so for the expression not to depend on $ x$ we should have that $ 2 \plus{} 4\cos 2u \equal{} 0$, which implies that $ u \equal{} k\pi\pm\frac {\pi}{3}$  :D

I solved the given equation.

    y=(3cos^x - sin^u)/(sin^2x - sin^u) .
Further i am not able simplify it for solution. Could some one please help me to simplify it.

Pardon me for my ignorance as this is my first post and i don't know the rules  of the forum.
Thank you

campos, you're right, your approach is nicer than mine (I failed to see the denominator transformation). Seems your solution is correct. There might have been a typo with wolfram :D


Suppose $ a$ is jointly proportional to $ b$ and $ c$.  If $ a\equal{}4$ when $ b\equal{}8$ and $ c \equal{} 9$, then what is $ a$ when $ b\equal{} 2$ and $ c\equal{}18$?

a is jointly proportional to b, c so a = kbc from some k.

When b=8 and c=9, bc=72 and a=4 so k = 4/72 = 1/18

Then when b=2 and c=18, a = kbc = 2*18/18 = 2.

a = 2

I have no idea what jointly proportionally means, but c doubled from the first to second and b got divided by 4. So the answer, I believe, is $ 4\cdot\frac{1}{2}\equal{}2$

Jointly proportional means, like Waynerius said, for numbers $ x$, $ y$, and $ z$, where $ x$ is jointly proportional with $ y$ and $ z$, $ x \equal{} kyz$, and $ k$ is some constant.


Since $ a^ba^c \equal{} a^{b\plus{}c}$, we have $ 2000^1 \times 2000^{2000}$, or $ \boxed{\textbf{A}}$.

The explanation is in the post above you.

We can rewrite the question as:

$ 2000\cdot 2000^{2000}\equal{}2000^1\cdot 2000^{2000}$.

For any numbers,
$ x^y\cdot x^z\equal{}x^{y\plus{}z}$

So we have:

$ 2000^1\cdot 2000^{2000}\equal{}2000^{2001}$

[quote="karpov352"]how to solve this sum?plz teach me[/quote]
Just to make it easier:
$ 4^2  \cdot  4^6  \equal{}  4^8$
How?
If you have the same base number which is here $ 4$, and there is exponents on the base numbers. We simply put the same number as it is $ 4$ then we add the exponent like here in the example $ 4^{2\plus{}6}$. Which is $ \equal{} 4^8$

Note:- If you have just a number like $ 4$ this is mean $ 4^1$ as well.

Hopes that help  :)

[quote=Darrenthegoat][hide=Solution]$2000(2000^{2000}) = 2000^{1}\times 2000^{2000} = 2000^{2000 + 1} = 2000^{2001}$. Thus, the correct choice would be $(A)$. [/hide][/quote]

You really needed to bump an 8 year old thread by re-stating the solution?

[quote=scrabbler94]
You really needed to bump an 8 year old thread by re-stating the solution?[/quote]

:rotfl:
Isn't there anyone to lock this thread ? ;)

[quote=physicskiddo]The bump makes me rage[/quote]

Then you have a problem

Usually it's not bad -- when it's done for the purpose of furthering someone's knowledge. This user posted a solution that was already posted and explained, not really helping anyone. Yes, we all should have just ignored it and let it go, but these are middle schoolers, c'mon, they have to comment.

[quote=Darrenthegoat][hide=Solution]$2000(2000^{2000}) = 2000^{1}\times 2000^{2000} = 2000^{2000 + 1} = 2000^{2001}$. Thus, the correct choice would be $(A)$. [/hide][/quote]

I will eat someone if there is another 8 year revive :P

[quote=dogsareawesome123]I don't get what's wrong with replying to an old thread. Does it make AoPS slower or something? Not saying it's ok to bump, just asking why it's bad.[/quote]

Bumping really old threads without adding much significant content is generally not considered good forum etiquette (plus it gets in the way with more recent questions). Also the OP hasn't logged on in 4 years.

There are cases where bumping old threads is perfectly fine - typically when adding something meaningful that hasn't been added before.

I'm not sure what the purpose of bumping-year-old posts is, but I know one thing. It's really annoying. This isn't the first time that the bumper has done something like this....

[quote=swagsavage]Bro 8 years actually[/quote]

Bro 9 months really

the irony is that we're actually perpetuating this thread at the same time as complaining about it being bumped... please

Answer is clearly A for millionth time. Someone pls lock this thread for it is very old and question is very ez


[quote="13375P34K43V312"]On question 3a, I filled in the entire the triangle and added the numbers up, which is brute-forceish. Would I lose a point for that? (I got the problem correct)[/quote]

I think you would get full points for [b]that part[/b]

that's what I did.  I'm sure you'll get full credit for that part

The USAMTS doesn't require your solutions to be elegant. They only require that your solution works and is complete.

[quote="K81o7"]If I found, with proof, an exact formula in Part b and then used it in part a, would i have lost credit?[/quote]

I would guess that the USAMTS graders would be flexible with that. That's what I did in fact.

Well, it probably is best to do part a on its own to check the formula...

Well since during part 3a you're not supposed to have an explicit formula yet (although I'm sure they would be fine if you used your formula from b to find the answer), I think they expected you to just add it up.  Besides, part 3a is pretty insignificant, 3b is the important one.  I doubt 3a is worth more than 1 point.

[quote="K81o7"]If I found, with proof, an exact formula in Part b and then used it in part a, would i have lost credit?[/quote]
That's what I did - I said "I'll develop a general formula satisfying part b, and then apply it to find the solution to part a."  Since I gave the correct solution to both parts, I have a complete solution; they wouldn't take credit off for that.

I suspect Part A was there to let students who failed to get a general formula still get [u]some[/u] points.
It really is pretty insignificant.

I bruteforced 3a because I thought that's what they "wanted" you to do, and then i generalized the formula on 3b and checked the answer. I don't think it really matters though as long as you get 3b right.


Here is a nice geometrical problem

$\bigstar$ Show that the perpendicular bisectors of the internal bisectors of the angles of a triangle meet the respective sides of the triangle in trhee collinear points.

Any solution??

[color=darkblue]The bisector ray of the angle $\widehat{BAC}$ meets the opposite side in the point $D$. The tangent $AA$ to the circumcircle meets the opposite side-line in the point $A'$. The triangle $A'AD$ is isosceles because $m(\widehat{A'AD})=m(\widehat{A'DA})=C+\frac{A}{2}$. Therefore, the point $A'$ belongs to the bisector of the segment $[AD]$. From the well-known property - " $A'\in AA\cap BC$, $B'\in BB\cap CA$, $C'\in CC\cap AB$ $\Longrightarrow$ $A'\in B'C'$ " - results the conclusion of the given problem.[/color]


superior algebra
superior algebra unsolved

Show that $ (\mathbb{Z}[\sqrt{2}])^*$ is a direct product of a cyclic group of order 2 and an infinite cyclic group.

Show that the units are of type $ \pm (1\plus{}\sqrt 2)^n$.


Show that all odd numbers can be written as a difference between two perfect squares

[hide=":yinyang:"]If we let two consecutive perfect squares be $n^{2}$ and $(n+1)^{2}$, their difference will be $(n^{2}+2n+1)-n^{2}=2n+1$. Doing this for n=1,2,3... we can find all odd intergers greater than 1.
Not exactly a rigorous proof, but you get the idea.[/hide]

[hide="Solution"]Any given odd number can always be expressed as one more than an even number. (Zero is defined to be even.)

Thus, we can say for a fact that $2n+1$ is always odd, given that N is an integer.

We want to show a way to express this with difference of two perfect squares.

Let the first perfect square be $N^{2}$, for some integer N.
This forces the next perfect square to be $(N+1)^{2}$

Their difference is $(N+1)^{2}-N^{2}$

This can be expanded to $N^{2}+2N+1-N^{2}$

This simplifys to $2N+1$, which is always odd.
Thus any odd number may be expressed as the difference of two squares.
$\mathcal{QED}$.
[/hide]

[hide="different idea"]
$x^{2}-y^{2}=(x+y)(x-y)$
Now let $x-y=$ 1 and $x+y$ equal the number.[/hide]

[hide="Another different idea"]
Every odd number can be shown drawn like this:

*  *  *  *  *  *  *
*
*
*
*
*
*

Now add on some zeroes...

*  *  *  *  *  *  *
*  0  0  0  0  0  0
*  0  0  0  0  0  0
*  0  0  0  0  0  0
*  0  0  0  0  0  0
*  0  0  0  0  0  0
*  0  0  0  0  0  0

The middle thing's a square and so is the outer thing.

$QED$
[/hide]

Follow up question: What numbers cannot be written as the difference of two squares?

[hide="answer"]Numbers that are congruent to 2 mod 4. Now go prove it.[/hide]

[quote="1=2"][hide="problem 1"](n+1)^2-n^2=2n+1[/hide]

[hide="problem 2"]Squares are either 0 or 1 mod 4.
1-0=1
0-0=0
0-1=-1=3mod4
But that's it. So only numbers 0, 1, or 3 mod 4 can be written as the difference of 2 squares.[/hide]

I can't believe my four year old brother lied to me and said that Dad was home![/quote]

You haven't shown that all numbers that are $0, 1, 3 \pmod{4}$ can be written as the difference of two squares, though. To prove this,

[hide="hint"] For what $a, b$ do there exist positive integers $x, y$ such that $x+y = a$ and $x-y = b$?[/hide]

[quote="paladin8"][hide="hint"] For what $a, b$ do there exist positive integers $x, y$ such that $x+y = a$ and $x-y = b$?[/hide][/quote]

[hide="Alternate hint"] Since the difference between two consecutive squares is an odd number, the difference between two squares is the sum of consecutive odd numbers. [/hide]


Let $m^{2}+m=2n^{2}+n$ and $m$, $n$ are positive integers. Prove that $m-n$, $m+n+1$, $2m+2n+1$ are all squares of integers.

[hide="is something wrong with my solution..."]
This factors to $(m-n)(m+n+1)=n^{2}$.
Now, set $m-n=k$ and $m+n+1 = \frac{n^{2}}{k}$ for some positive integer $k$. $m-n$ will always be positive because $m>n$.
Adding the two equations and simplifying gives:
$k^{2}-k(2m+1)+n^{2}=0$. Checking the discriminant , we see that it is

$\sqrt{(2m+2n+1)(2m+2n-1)}$. But this cannot be a perfect square, because the product of two integers whose difference is two is always one less than a perfect square. In addition, $2m+2n-1$ cannot be zero. Therefore, there is no integer $k$, that satisfies the conditions. Thus, the equation has no solutions...
By a computer search,(14,10) was found to be a solution and there are others.
[/hide]

[quote="Leao"]... the discriminant... is $\sqrt{(2m+2n+1)(2m+2n-1)}$
[/quote]

The discriminant is actually $(2m+2n+1)(2m\,\boxed{-}\,2n\,\boxed{+}\,1)$


im having a final in a couple of days, and I wanted to review problems like :

$ f(g(x))$

say that $ f(x)$ = $ 2x \plus{} 7$ and $ g(x)$ = $ 8x$

and the rule was $ f(g(x))$

What would you do next?

Also, on problems like $ (f$0$ g)$ $ (x)$ , and $ f(x) \equal{} 2x$ and $ g(x) \equal{} 3$, how would you solve this? Help!

note: the 0 is supposed to show the function f of g; it's not multiplication

Just plug the value of $ g(x)$ into $ f(x)$. For example, the first one would be $ f(g(x))\equal{}f(8x)\equal{}2(8x)\plus{}7\equal{}16x\plus{}7$.

Expanded:
Using you example $ f(x) \equal{} 2x$ and $ g(x) \equal{} 3$, The value the function $ g(x)$ returns (in this case 3), will be substitute into $ f(x)$ (for your example $ f\circ g\text{ }x\implies2g(x)\equal{}2\cdot3\equal{}6$).

Also $ f\circ g\text{ }x \equal{} f(g(x))$

Let f(x)=2x+7 and g(x)=8x.

If f(x)=2, find g(f(x)) and f(g(x))


1级：内（那）谁啊，咱哥们是惦记不上了。
　　A.--我跟本就不喜欢她。
　　B.--我才没空惦记她那。
　　C.--她瞧不上咱爷们呀。
　　D.--她没准儿早有男朋友啦。

2级：您稍绷会儿。。
　　A.--您的皮筋儿快断了吧。。
　　B.--说你哪！不能笑啊！
　　C.--您等一会儿。
　　D.--在这儿呆着，千万别动啊。

3级：这可是正宗温叉儿P的！
　　A.--您瞧清楚，可不是温叉儿Q的。
　　B.--Windows XP 版的。
　　C.--只有我姓温的这里的叉儿P才正宗啊。
　　D.--我这儿可是老字号哦。

4级：哽儿P着凉大海棠
　　A.--哽儿他们家吃多了海棠着凉啦。
　　B.--海棠吃多了可是要他妈的着凉打嗝放P的。
　　C.--这人完了，4翘的了。
　　D.--大海棠他着凉了，哽儿了。

can you explain how you got your answer? i lived in beijing for almost 10 years but i cant get these questions

1, 有句话叫: 不怕贼偷就怕贼惦记. 所以惦记这个词可以用于形容男孩对女孩的向往. 有调侃的意思.

2, 不解释了...

3, 我在置顶贴里解释过了.

4, 嗝儿屁=死了. 不过我一般把这个短语理解为死了感冒.


inequalities
function
logarithms
inequalities unsolved

Assume that a_1,a_2...a_n are positive real number satisfying the condition[tex]\frac{1}{a_1^2}+\frac{1}{a_2^2}+...\frac{1}{a_n^2}=1[/tex].Prove that:
[tex]\prod_{i=1}^{n}(a_i-1)\geq\ (\sqrt{n}-1)^n[/tex]

Hmmm, this is just a reformulation of a problem of Vasc and me. I think Myth gave a solution using his famous IPT. I will try to find the link.

[quote="Fedor Petrov"]Sorry, using what?[/quote]
Myth's inflection point theorem.

I have a solution, but it looks too easy to be true after you mentioned IPT and such arsenal:
By generalised Holder we have:

$\prod _{i=1}^n (a_i -1) \geq (\sqrt[n]{\prod _{i=1}^n a_i} -1)^n$

so it suffices to prove that $\sqrt[n]{\prod _{i=1}^n a_i} \geq \sqrt{n}$

using given constraint and QM-GM we have:

$\sqrt{\frac{\sum _{i=1}^n \frac{1}{a_i ^2}}{n}} \geq \sqrt[n]{\prod _{i=1}^n \frac{1}{a_i}}$

what after multiplying gives $\sqrt[n]{\prod _{i=1}^n a_i} \geq \sqrt{n}$ as wanted

[quote="zhaobin"][quote="Megus"]

$\prod _{i=1}^n (a_i -1) \geq (\sqrt[n]{\prod _{i=1}^n a_i} + (-1)^n)^n$

[/quote]
sorry,I think it is wrong ;)[/quote]

why? can you explain?

[edit] I've corrected it a bit (I forgot about nth root), is it okey now?

[quote="Vasc"]Try $a_1=1$ and $n$ even.[/quote]

but if $a_1=1$ then LHS of $\prod_{i=1}^{n}(a_i-1)\geq\ (\sqrt{n}-1)^n$ is zero which contradicts our inequality, or am I missing something ?  :?

My idea:
Since [tex]\sum\frac{1}{a_{i}^2}=1[/tex] ,hence:
There exist:
[tex]x_{1},x_{2},..x_{n} [/tex]st;
[tex]a_{i}=\sqrt{\frac{x_{1}+x_{2}+...+x_{n}}{x_{i}}[/tex]
Then given inequality becoms:
[tex]\prod_{i=1}^n[\sqrt{\frac{x_{1}+...+x_{n}}{x_{i}}}-1]\geq\ [\sqrt{n}-1]^n[/tex]
I used Jensen's ineq but it's wrong...
I will try it...
Good Luck! :D

I was serious: it really fails for $n=2$, because inflection point of the function $f(x)=\ln(\sqrt{1/x}-1)$ is $4/9<1/n$ for $n=2$.


At school we had a debate about if people could change their ways or not. I was curious about what people on AoPS think about people changing their ways.
So do you think people can change their ways?

I think people can change, although sometimes not completely. For example, if there was a really mean guy, and he was disciplined into a guy who acts really nice, it may be that that he still has mean ideas but chooses not to act upon them.

When you say "if people can change their ways", what part of a person's "ways" are you referring to? And what would constitute changing those "ways"?

[quote="cyberspace"]I'll assume you're asking about if people can change the type of person they are.  I don't think people can ever change that. A person may be able to reform a way of life or even a way of acting.  However, no matter what actions occur to us, deep down we'll never be able to change how our genetics dictate our traits. If a person's genetics tell that he will always react to a certain situation in some way, then he will, regardless of whether or not he wishes it or even if he thinks he's changed from it.

Of course, if you mean can a person change their actions, then sure.  People can always use their mental will to reform their actions, however deeply ingrained they may be.  My point is that even if a person does this, it may run contrary to his core principles whether he knows it or not.[/quote] Well I personally think that you are a function of nature and nurture: not only your genes define who you are. So yeah, it is possible imo to change your ways.

I think that for the most part everyone's posts make good sense... everyone will always have some feelings that they can't really get rid of, although a mean person obviously can change to be a nice person. However, so far all of this has been to the extent of how much someone can change [i]on their own[/i]. I know that probably most of you don't believe in Christianity and probably only some of you believe in God at all... however, I know from personal experiences (not to mention my own) that however much someone can't change/get rid of something on their own, after becoming Christians they have the power to change from the inside out through the power of Christ. When anyone accepts Christ as their savior, they are putting off the old self and becoming a new person. 2 Corinthians 5:17: "If anyone is in Christ, he is a new creation; The old has gone, the new has come!" Yup, that's my opinion, anyway.  :lol:

[quote="kookamunga"] know from personal experiences (not to mention my own) that however much someone can't change/get rid of something on their own, after becoming Christians they have the power to change from the inside out through the power of Christ. [/quote]

I'm Christian too, but you can change, regardless of religion.  You CHOSE bad behavior and you can CHOOSE to reverse it.

[quote="kookamunga"]after becoming Christians they have the power to change from the inside out through the power of Christ. When anyone accepts Christ as their savior, they are putting off the old self and becoming a new person. 2 Corinthians 5:17: "If anyone is in Christ, he is a new creation; The old has gone, the new has come!" Yup, that's my opinion, anyway.  :lol:[/quote] I actually think that the only way to change yourself is by becoming a non-Christian. When you see that Christ isn't your savior, you can become a new person. (Yes, I am being sarcastic.)

[quote="HilbertThm90"]People are changing all the time whether they want to or not. Everything is in constant change. No millisecond goes by such that you haven't changed. It is the nature of the universe. The question "can people change" is strange to me. The better question is, "Can people not change?" Which I believe the answer to be no.[/quote]

Whoa...That's deep!  :rotfl:

[quote]I'm Christian too, but you can change, regardless of religion. You CHOSE bad behavior and you can CHOOSE to reverse it.
[/quote]
I get what you're saying... to some extent you do indeed choose between good and bad behavior. However, on your own you can't always be good. Paul wrote about this in Romans 7:18-25:
    "I know that nothing good lives in me, that is, in my sinful nature. For I have the desire to do what is good, but I cannot carry it out. For what I do is not the good I want to do; no, the evil I do not want to do - this I keep on doing. Now if I do what I I do not want to do, it is no longer I who do it, but it is sin living in me that does it.
   So I find this law at work: When I want to do good, evil is right there with me. For in my inner being I delight in God's law; but I see another law at work in the members of my body, waging war against the law of my mind and making me a prisoner of the law of sin at work within my members. What a wretched man I am! Who will rescue me from this body of death? Thanks be to God - through Jesus Christ our Lord!"
   Paul finds that however much he desires to have "good behavior", he cannot carry it out due to our sinful nature. He concludes by saying that God, however, has indeed rescued him from this body of death which follows its sinful desires.

As one gets progessively older, it becomes harder to change. This is why youth groups were set up in Hitler Germany, Stalin Russia, and Mao China, to ensure that a group of people would think a certain way. One can change still, but the efforts needed are increasing as one gets older.

A person though, can easily appear to be changed, and not listen to what his or her true feelings dictate.

There is also good and evil in every person. For example, Hitler created the terrible Holocaust, yet he also helped little children in a city (I can't remember the name right now).

[quote="lightning"]
There is also good and evil in every person. For example, Hitler created the terrible Holocaust, yet he also helped little children in a city (I can't remember the name right now).[/quote]

Hitler was completely insane; the Holocaust was classified as "good" for him.

@modularmarc101
i'm not too sure at what you're trying to say....

I was just saying that no one is completely good or bad, that there is some good in every person, and some bad in every person. Hitler did think that the Holocaust was good, but that doesn't make him 100% evil, just very close to it, because he did some generous deeds. For example, "[i]at Berchtesgaden, he loved to see the nieghborhood children and give them ice cream and cake." [/i]Of course, I do not support Hitler, but would like to point out that there is at least a little good in him, which was overshadowed by his horrific Holocaust.

thinking about it, I'm not too sure if a truly evil person is possible. I think a truly evil person would want to hurt everyone, including him/herself.

The quote was taken from an article named "Evil" written by Time magazine. We had to read this article for our English 3 class.

This is an interesting topic. Without putting too much though into it, I'd have to say that believing that you can't change is obviously not the way to go...if you believe that you can't, then you obviously won't. Thus, regardless of what is fact, we should still believe in ourselves to be able to change.

[quote="cyberspace"]I agree to a point.  I think in the early developmental stages, both nature and nurture count.  During that phase, people can change dramatically.  However, once people are out of the "nurture" phase, they're stuck.[/quote]

this is too scientific a view imo

I must disagree however with the notion that as one gets older, it is harder to change. In my younger years, I changed quite a bit, receded for a while, and then am currently changing again. I believe it is the same for everyone.
But generally, people change very little (unless they have a strong desire to) after their developmental phase.

[quote="modularmarc101"]I believe a person can change, if and only if, he/she want to.[/quote]

I think you should make that just an "if." Iff doesn't hold because people unwillingly must change all the time (take in case me). Actually, perhaps that would a direct representation of my true subconscious form, which would be depressing to think about...

[quote="modularmarc101"]I believe a person can change, if and only if, he/she want to.[/quote]

I think you should make that just an "if." Iff doesn't hold because people unwillingly must change all the time (take in case me). Actually, perhaps that would a direct representation of my true subconscious form, which would be depressing to think about...[/quote]

Lol now this is leading back into math (biconditional statements) :P

As a christian I believe that when you become a christian you are a different person. You have a new future a new life, a new attiude and a new wayr of life. After I accepted Christ I felt different I had new goals and plans for my life.  That's why I know people can change.


A computer operator can enter 75 data forms each hour. After the operator has been working for 20 minutes, a second operator starts working also. Two hours later, they have entered 295 forms. How long would it have taken the first operator to do the entire job?

[quote="SnowStorm"][hide]
$R_1(60) = 75$
$R_1 = \frac{5}{4}$
Since he started 20 minutes early he will have x number of forms all ready done:
$\frac{5}{4}(20) = x$
$x = 25$
So the two operators only need to finish $295-25 = 270$ more forms which they do in two hours.
$\frac{5}{4}(120) + R_2(120) = 270$
$120(R_2) = 120$
$R_2 = 1$
So the number of minutes it would've take this operator to do all the forms is:
$1(T) = 295$
$T = 295$ minutes[/hide][/quote]

It's not asking how long it would have taken the second operator alone! It's asking how long it would have taken the first operator if he had done the entire job by himself! ;)

[hide]
It would take the first operator 4 hours and 16 minutes to finish the entire job.

How do we know this? First we have to find out how many forms the operator has to finish. There is the 295 the two operators did together and the 25 he did before. If we add the 295 to 25, we get 320. So, the whole job is to enter 320 forms. If the first operator can finish 75 in an hour, we can do 320/75 x 60 to get 256 minutes, which equals 4 hours, 14 minutes.[/hide]


parameterization
function
real analysis
real analysis theorems

In the derivation of `variation of parameters' the text indicates the following:

$ y'' \plus{} a_1(x)y' \plus{} a_0(x) y \equal{} f(x)$

Assume homogenous  solution is known to be:

$ y_H \equal{} C_1y_1(x) \plus{} C_2y_2(x)$

We look for a particular solution

$ y_p \equal{} u(x)y_1(x) \plus{} v(x)y_2(x)$

Differentiate $ y_p$ to obtain

$ y_p' \equal{} u(x)y'_1(x) \plus{} v(x)y'_2(x) \plus{} \underbrace{u'(x)y_1(x) \plus{} v'(x)y_2(x)}_{Assumed\;Zero}$

Why can we assume $ u(x)'y_1(x) \plus{} v'(x)y_2(x) \equal{} 0 ?$

We artificially introduced two new functions, yet only have a single equation to solve. This gives us an extra degree of freedom, and there are presumably many solutions to our new equation. If we impose some arbitrary equation to restrict that, we'll only be taking away that extra degree of freedom, and should still have a chance of having a solution.

That's it. We arbitrarily impose the restriction $ u'y_1 \plus{} v'y_2 \equal{} 0.$ Why that equation? Because it's convenient; because it works.

The theoretical justification of all of this comes from the back end. There's really only one theorem worth anything in all of differential equations: the existence-uniqueness theorem. So we employ some algebraic method full of arbitrary assumptions and "just because we can" manipulations. In the end, do we have enough solutions that we can solve for all allowable initial conditions? If so, then we have found all solutions. The end justifies the means.

Speaking just in concept then, I would be free to assume $ u'(x)y_1(x) \plus{} v'(x)y_2(x) \equal{} 1 \; or \;g(x)$  as well?  I don't know why it would be necessary, but just in concept?

In concept yes. In practice, it might or might not be effective - the only way to see that would be to try it. This is really all about practicality.


6 students took a test. The test consists of 5 multiple choice questions. Each question has 3 choices (A, B, and C). Each question worths 1 point, thus the maximum total score is 5.

Here are the answers of the first 5 students:

[b]Student 1[/b]:  A B C B A
[b]Student 2[/b]:  B B C A C
[b]Student 3[/b]:  B B B C C
[b]Student 4[/b]:  A B A C B
[b]Student 5[/b]:  C A B C C

The only thing we know is that they all got 2 points (out of 5).

Then, what is the score of this student?

[b]Student 6[/b]: C C C B C

[hide="answer key"]A C C C C[/hide]
[hide="therfore"]he scored three[/hide]

That's correct. :)

Interestingly, there's another possible answer key which still gives 3 points to the student. Can you find it?

Another possible answer key ==> C B A B C
                                                                            Marks
                                     Student 1: A B C B A          2
                                     Student 2: B B C A C          2
                                     Student 3: B B B C C          2
                                     Student 4: A B A C B          2
                                     Student 5: C A B C C          2

                                     Student 6: C C C B C          3


geometry
circumcircle
Euler
cyclic quadrilateral

Let  $ ABCD  $  a cyclic  quadratilateral  and $ K , L $  the orthocenters of triangles  $  ADC  $ and  $BDC$ . Prove that  $ KL = AB $

[i]Babis[/i]

Well , no solution untill now , so I offer it to all friends in this forum.Please  , pay attention to the  basic theorem that we use in this proof.

  Let $ M $ the midpoint of $ CD $ and $ O $ the circumcenter of $ ABCD $. Then
     $ AK = 2OM , BL = 2 OM $ and $ AK // BL $ . Thus  $ AKLB $ is parallelograme
    and $ AB = KL $

 Babis

[quote="agolsme"]how did you get that $ AK = 2OM $ and $ BL = 2OM $ ?[/quote]

This follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 post #4 Theorem 3, applied to the triangles ADC and BDC with their circumcenter O, their orthocenters K and L, and with M being the midpoint of their common side CD.

As for the initial problem, it follows from Theorem 2 in the note "The Euler point of a cyclic quadrilateral" on [url=http://de.geocities.com/darij_grinberg]my website[/url].

  Darij


ratio
geometry
3D geometry
tetrahedron
octahedron

A regular tetrahedron and a regular octahedron have edges of the same length. What is the ratio between their volumes?

i think its...

[hide]1:4[/hide]

i just visualized it in my head.

Solution:

[hide]
$ \frac{\text{area tetrahedron side x}}{\text{area octahedron side x}}$

$ =\frac{\frac{x^{2}\sqrt{3}}{4}x\sqrt{\frac{2}{3}}\frac{1}{3}}{2\frac{1}{3}x^{2}\frac{x}{\sqrt{2}}}=$

$ =\boxed{\frac{1}{4}}$[/hide]


Find the smallest positive value taken by $a^3 + b^3 + c^3 - 3abc$ for positive integers $a$, $b$, $c$ .
 Find all $a$, $b$, $c$ which give the smallest value

Use the identity :

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$

Afterwards,

$\frac{a^2+b^2}{2} \geq ab$
$\frac{a^2+c^2}{2} \geq ac$
$\frac{b^2+c^2}{2} \geq bc$

So, $a^2+b^2+c^2 \geq ab+bc+ab$

So, $a^3+b^3+c^3-3abc \geq 0$ with egality iff $a=b=c$

There is more simple :

$\frac{a^3+b^3+c^3}{3} \geq \sqrt [3] {a^3b^3c^3}$ (AM-GM)

So, $a^3+b^3+c^3-3abc \geq 0$ with equality iff $a=b=c$

That's NOT the point of the question, Labiliau.

The questions wants us to find the smallest positive integer $n$ (where $0$ is not a positive integer) such that $n$ can be written in the form \[ n = a^3 + b^3 + c^3 - 3abc  \] for some positive integers $a$, $b$ and $c$.

Well, Labiliau's post #2 almost answers the problem:

Consider three positive integers a, b, c such that the value of $a^3+b^3+c^3-3abc$ is positive.

Since a, b, c are positive integers, each of them is $\geq 1$; if all of them are = 1, then $a^3+b^3+c^3-3abc=0$, contradicting with the positivity of $a^3+b^3+c^3-3abc$. Hence, at least one of the integers a, b, c must be $\geq 2$. So two of the integers a, b, c are $\geq 1$ and the third one is $\geq 2$; thus, $a+b+c\geq 4$.

On the other hand, the integer $a^2+b^2+c^2-bc-ca-ab$ is nonnegative (as shown by Labiliau in post #2), so it is either = 0 or $\geq 1$; hence, the value of

$a^3+b^3+c^3-3abc=\underbrace{\left(a+b+c\right)}_{\geq 4}\cdot\underbrace{\left(a^2+b^2+c^2-bc-ca-ab\right)}_{=0\text{ or }\geq 1}$

is either = 0 or $\geq 4$. Since we required it to be positive, it thus must be $\geq 4$. And actually, the value 4 is achieved, since for a = 1, b = 1 and c = 2, we have $a^3+b^3+c^3-3abc=4$.

Thus, the smallest positive value of the expression $a^3+b^3+c^3-3abc$ for positive integers a, b, c is 4.

  darij

[quote=Chandrachur]$a^3+b^3+c^3-3abc=\frac {1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$.
Now $a+b+c>=3$ but $a+b+c=3 \implies a=b=c=1 \implies a^3+b^3+c^3-3abc=0$. So $a+b+c>=4$.
$[(a-b)^2+(b-c)^2+(c-a)^2]>=2$ as otherwise it must be $0$ (check why it can't be $1$ ;) )
and there exists a solution $(1,1,2)$ where each of these attain their minimum values so answer is $\boxed{4}$
:D @ as simple as that :rotfl:[/quote]

Your answer is still partially correct even though you have solved it. It has asked for all values of a,b,c
Here a b and c are clearly symmetric. Hence there will be three solutions which will give the smallest value they are
(1,1,2)   (1,2,1)    (2,1,1)


Determine all continuous functions $f,g,h: \mathbb R \to \mathbb R$ satisfying the relation \[f(x+y)=g(x)+h(y)\] for all reals $x$ and $y$.

$ f(x\plus{}y)\equal{}g(x)\plus{}h(y)\equal{}g(y)\plus{}h(x)$;
$ g(x)\plus{}h(y)\equal{}g(y)\plus{}h(x)$;
$ h(x)\minus{}g(x)\equal{}h(0)\minus{}g(0)\equal{}A\equal{}const$;
$ h(x)\equal{}g(x)\plus{}A$ - we will put it to $ f(x\plus{}y)\equal{}g(x)\plus{}h(y)$:

$ f(x\plus{}y)\equal{}g(x)\plus{}g(y)\plus{}A$;
After $ y\equal{}0$: $ f(x)\equal{}g(x)\plus{}B$, $ B\equal{}A\plus{}g(0)$;
$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}2B\plus{}A$;

We will try to find continuous solutions of $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}C$.
After taking substitution $ v(x)\equal{}f(x)\plus{}C$ we will have:
$ v(x\plus{}y)\equal{}v(x)\plus{}v(y)$, where $ v$ is continous.

It's well-known, that $ v(x)\equal{}kx$ is solution of last equation (with $ v$ - continous).
So, $ f(x)\equal{}kx\plus{}b$, $ k, b \in R$.
It's easy to check that this class of functions is don't  break any conditions (for $ f(x)\equal{}kx\plus{}b$ we can take $ g(x)\equal{}kx\plus{}\frac{b}{2}$, $ h(x)\equal{}kx\plus{}\frac{b}{2}$).

[quote="Kirill"]$ f(x \plus{} y) \equal{} g(x) \plus{} h(y) \equal{} g(y) \plus{} h(x)$;
$ g(x) \plus{} h(y) \equal{} g(y) \plus{} h(x)$;
$ h(x) \minus{} g(x) \equal{} h(0) \minus{} g(0) \equal{} A \equal{} const$;
$ h(x) \equal{} g(x) \plus{} A$ - we will put it to $ f(x \plus{} y) \equal{} g(x) \plus{} h(y)$:

$ f(x \plus{} y) \equal{} g(x) \plus{} g(y) \plus{} A$;
After $ y \equal{} 0$: $ f(x) \equal{} g(x) \plus{} B$, $ B \equal{} A \plus{} g(0)$;
[/quote]

Your solution is OK but the beginning may be shorter :
$ y\equal{}0$ $ \implies$ $ g(x)\equal{}f(x)\plus{}a$
$ x\equal{}0$ $ \implies$ $ h(y)\equal{}f(y)\plus{}b$
And so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}c$


prove that $ \forall k>1$ there is a multiple of k like $ i$ such that $ i$ doesnt have more that 4 different  digit and $ k^4>i$

sorry if it posted before... :maybe:


find the natural solution of the equation
$x^{3}-y^{3}=x y+61$

[quote="ashrafmod"]find the natural solution of the equation
$x^{3}-y^{3}=xy+61$[/quote]
[hide]$x=6,\,y=5$

Guess and check. :ninja:[/hide]

WLOG, assume x > y.

$y = x-c$
$x^{3}-(x-c)^{3}= x(x-c)+61$
$3x^{2}c-3xc^{2}= x^{2}-xc+60$
$3xc(x-c) = x(x-c)+60$
$(3xc-x)(x-c)=60$
$x(3c-1)(x-c)=60$

This implies that $x|60$, and through plugging in values of x that satisfy $x|60$, we find that $c=1$, meaning that our answer is $\boxed{(6,5)}$.

@Treething:
1. Where did $c^{3}$ from the LHS go?
2. How $61$ became $60$?

Very similar problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=121446]here[/url], solution repeated for convenience.

[hide="Solution"]Since RHS is positive, the LHS must be too, therefore $x>y\implies x-y\geqslant 1$.

$xy+61=(x-y)(x^{2}+xy+y^{2})=(x-y)[(x-y)^{2}+3xy]$

Substituting $a=x-y,b=xy$ we get

\begin{eqnarray*}a(a^{2}+3b)=b+61\iff a^{3}+3ab=b+61\iff b(3a-1)=61-a^{3}\iff b={61-a^{3}\over 3a-1}\end{eqnarray*}

Since the denominator is positive, the numerator must be too, hence $a\leqslant 3$. Plugging $a=1,2,3$ we get an integer $b$ only for $a=1$, giving $b=30$, and the system $x-y=1, xy=30$ has a unique natural solution $x=6, y=5$.[/hide]


quadratics
algebra
polynomial
sum of roots

Both roots of the quadratic equation $x^2-63x+k=0$ are prime numbers. The number of possible values of $k$ is

\[ \text {(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 4 \qquad \text{(D)} \ \text {more than four} \]

[hide]

Sum of roots is 63. If both primes are odd, than their sum is even, so one must be even; and the only even prime is 2. Thus, the only possible solution is (2,61), and 61 is prime so there is one solution.

[/hide]


vector
function
induction
inequalities
calculus
calculus computations

Also if you define a [b]k-cell[/b] as $ (x_1,...,x_n)$ and $ a \leq x_i \leq b$ $ \forall i$ then a k-cell is also convex. Considering how an ellipsoid is shaped relative to this "hypercube" you can prove the n-dimensional ellipsoid (hyperellipsoid??) is also convex.

To show it is convex take $ i,j,k \in E$ and show $ \lambda i \plus{} (1\minus{}\lambda)j \minus{} k < r$ where $ r \equal{} max(a,b,c)$ Using $ |j\minus{}i|<r$ and $ |k\minus{}j| <r$ you'll be able to do the rest.
What I am showing (or trying to :P) is that if you draw a line between any two [i]fixed[/i] points and then choose a third (all arbitrary) we can draw a line from the third point to any point on the line between the first two points.
This shows its convex.

By any chance could we use the bilinearity condition of the dot product?

Note: if you define $ \|x\|_A^2 \equal{} \sum_{k \equal{} 1}^n\left(\frac {x_k}{a_k}\right)^2,$ and with it the inner product $ \langle x,y\rangle_A \equal{} \sum_{k \equal{} 1}^n\frac {x_ky_k}{a_k^2},$ you have defined $ \mathbb{R}^n$ to be a very slightly nonstandard inner product space. (The example above has $ n \equal{} 3$ and uses $ a,b,c$ instead of $ a_1,a_2,a_3,$ but it's the same thing.) In which case, $ E$ is the closed unit ball in this inner product space.

Claim: the closed unit ball of a real inner product space is convex.

It's not all that hard. Suppose $ x$ and $ y$ are in this unit ball, so that $ \|x\|\le 1$ and $ \|y\|\le 1.$ The line segment between $ x$ and $ y$ is $ \{(1 \minus{} t)x \plus{} ty\mid t\in[0,1]\}.$

$ \|(1 \minus{} t)x \plus{} ty\|^2 \equal{} \langle (1 \minus{} t)x \plus{} ty,(1 \minus{} t)x \plus{} ty\rangle$

$ \equal{} (1 \minus{} t)^2\langle x,x\rangle \plus{} 2(1 \minus{} t)t\langle x,y\rangle \plus{} t^2\langle y,y\rangle$

$ \le (1 \minus{} t)^2\|x\|^2 \plus{} 2(1 \minus{} t)(t)\|x\|\,\|y\| \plus{} 2t^2\|y\|^2$ by Cauchy-Schwarz

$ \equal{} \left((1 \minus{} t)\|x| \plus{} t\|y\|\right)^2\le \left((1 \minus{} t)\cdot 1 \plus{} t\cdot 1\right)^2 \equal{} 1.$

So the line segment is contained in the unit ball. Since $ x$ and $ y$ were arbitrary elements of that unit ball, the ball is convex.

=====

Generalization: prove that the closed  unit ball of any real normed vector space is convex.

(Stating it that way means I had a far easier proof that I could have used above, using only the triangle inequality.)

Can we apply a similar strategy for boxes?

If a box is just a Cartesian product of intervals:$ [a_{1}, b_{1}], \times \cdots \times [a_{m}, b_{m} ]$ in $ R^{m}$, and we're asked to prove that all boxes are convex, we can invoke the dot product to prove this? If so, what would my bounds be?

Just do it component by component. We have $ x$ and $ y$ in the box, and we want to show that $ (1\minus{}t)x\plus{}ty$ is in the box. The $ i$th component of $ (1\minus{}t)x\plus{}ty$ is $ (1\minus{}t)x_i\plus{}ty_i.$ So what we need to prove is this:

If $ a_i\le x_i\le b_i$ and $ a_i\le y_i\le b_i,$ then $ a_i\le (1\minus{}t)x_i\plus{}ty_i\le b_i$ for $ 0\le t\le 1.$

There's no point in saying this has much to do with a dot product. (OK, taking a component amounts to taking a dot product with $ e_i,$ the $ i$th standard basis vector.)

Note also that if the box is centered at the origin, it could function as the unit ball for a norm (a weighted $ \infty$ norm), and hence what I said about the unit ball in any normed vector space applies. The proof of that general statement requires two of the three axioms for a norm: the scaling property and the triangle inequality.

mstrip, try this one:

Let $ v_j,\ 1\le j\le k$ be a collection of nonzero vectors in $ \mathbb{R}^n$ and $ b_j$ a collection of real numbers.

Define $ E \equal{} \{x\in\mathbb{R}^n\mid v_j\cdot x\le b_j\text{ for all }1\le j\le k\}.$

Assume that $ E$ is nonempty. (Not guaranteed unless we insist on it.)

Prove that $ E$ is convex.

(This is a region bounded by hyperplanes - a polyhedron, or whatever you would call the higher dimensional analogue of a polyhedron. Your box is a special case of this.)

Kent,

here's my attempt - though I think I might need help. Since $ v_{j}$ was a vector in n-dimensional Euclidean space ($ \mathbb{R}^{n}$), I made the claim that for $ \parallel{}z_{j}\parallel{}\le b_{j}$ and $ \parallel{} y_{j}\parallel{} \le b_{j}$ we can form a linear combination for $ v_{j}$ written as $ (1\minus{}t)z_{j} \plus{} ty_{j} \equal{} v_{j}$. I was using the dot product to show that $ v_{j} \cdot x$ was going to be less than $ b_{j}$, but treating $ x$ as a scalar and proceeding in that direction didn't bring me very far. I tried induction, but I don't see how to prove definitively my $ k\plus{}1$ case.

Any advice?

[quote="mstrip"]I made the claim that for $ \parallel{}z_{j}\parallel{}\le b_{j}$ and $ \parallel{} y_{j}\parallel{} \le b_{j}$...[/quote]
Not true, and not relevant. Don't use the norms at all. Use the given inequality directly.

And it's not so much an induction on $ k$ as a matter of showing that all $ k$ of these conditions are still true for a point that is a convex combination of two points for which all $ k$ conditions are true.

Could we say then, that for any $ a,b \in E$ that we can draw a line between $ a$ and $ b$ where $ (\frac{1\minus{}t}{x})a \plus{} (\frac{t}{x})b \equal{} v_{j}$ since $ v_{j}$ is a collection of vectors? The denominator of $ x$ just helps to form a convex combination (and at the same time a linear combination since $ v_{j}$ is a vector). Would this do it?

What? $ x$ is a vector. You can't divide by it.

Here's the path:
Choose two generic points in the set- say, $ x$ and $ y$.
Consider the segment $ \{tx\plus{}(1\minus{}t)y| 0\le t\le 1\}$ between them.
For a generic point on this segment, show that it satisfies each of the defining inequalities.


Suppose that the bisector of angle A of triangle ABC intersects BC
at D. Suppose that AB + AD = CD and AC + AD = BC.
Determine the angles B and C.

[color=blue][b]1.[/b][/color] According to the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23232]x angle[/url] in High School, Intermediate Topics:
Let $\triangle ABC$ be a triangle with the angle $\measuredangle B = 80^o$ and the bisector of this angle intersecting the side $AC$ at a point $D$. If $BC + BD = AD$, then the angles at the vertices $A, C$ are $\measuredangle A = 20^o$, $\measuredangle C = 80^o$ and the triangle $\triangle ABC$ is isosceles with $AB = AC$.

[color=blue][b]2.[/b][/color] According to the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25971]Geometry[/url] in High School, Pre-Olympiad:
Let $\triangle ABC$ be a triangle with the angle $\measuredangle B = 100^o$ and the bisector of the angle $\measuredangle A$ intersecting the side $BC$ at a point $K$. If $AK + BK = AC$, then the angles at the vertices $A, C$ are $\measuredangle A = \measuredangle C = 40^o$ and the triangle $\triangle ABC$ is isosceles $AB = BC$.

Let the $\triangle ABC$ have the angles $\measuredangle A = 120^o$, $\measuredangle B = 40^o$, $\measuredangle C = 20^o$. Let $D$ be the intersection of the bisector of the $\measuredangle A$ with the side $BC$ and $E$ another point on the side $BC$, such that the triangle $\triangle ADE$ is isosceles. Since the angle

$\measuredangle ADB = 180^o - \left(\frac{\measuredangle A}{2} + \measuredangle B\right) = 180^o - (60^o + 40^o) = 80^o$

the points $B, E, D, C$ follow on the line $BC$ in this order.

The triangle $\triangle ABE$ has the internal angles

$\measuredangle ABE = \measuredangle B = 40^o$
$\measuredangle AEB = 180^o - \measuredangle AED = 180^o - 80^o = 100^o$
$\measuredangle EAB = 180^o - (\measuredangle ABE + \measuredangle AEB) = 180^o - (40^o + 100^o) = 40^o$

Hence, it is isosceles with $AB = AE$ and we can apply the result of post [color=blue][b](2)[/b][/color]: Let $F$ be the intersection of the bisector of the angle $\measuredangle BAE$ with the side $BE$. Then $AF + EF = AB$.

The triangle $\triangle AEC$ has the internal angles

$\measuredangle ACE = \measuredangle C = 20^o$
$\measuredangle AEC \equiv \measuredangle AED = 80^o$
$\measuredangle EAC = 180^o - (\measuredangle ACE + \measuredangle AEC) = 180^o - (20^o + 80^o) = 80^o$

Hence, it is isosceles with $CA = CE$ and we can apply the result of post [color=blue][b](1)[/b][/color]: Let $G$ be the intersection of the bisector of the angle $\measuredangle CAE$ with the side $CE$. Then $AE + AG = CG$.

The triangles $\triangle ADF \cong \triangle AEG$ are congruent by ASA, because the sides $AD = AE$ are equal, the angles $\measuredangle ADF = \measuredangle AEG$ are equal (the triangle $\triangle ADE$ is isosceles) and the angles $\measuredangle FAD = \measuredangle EAG$ are also equal:

$\measuredangle FAD = \frac{\measuredangle EAB}{2} + 180^o - 2 \measuredangle ADE = 20^o + 180^o - 160^o = 40^o$

$\measuredangle EAG = \frac{\measuredangle EAC}{2} = 40^o$

Consequently, $AF = AG$, $DF = EG$, $\measuredangle AFD = \measuredangle AGE = 180^o - (\measuredangle AFD + \measuredangle FAD) = 180^o - (80^0 + 40^o) = 60^o$ and the triangle $\triangle AFG$ is equilateral. In addition,

$EF = DF - DE = GE - DE = GD$

Calculating the segment lenghts $BC$, $CD$:

$CD = CG + GD = (AG + AE) + GD =$

$= (AF + AD) + EF = (AF + EF) + AD = AB + AD$

where we applied the result of post [color=blue][b](1)[/b] [/color], $AD = AE$, $AF = AG$ and the result of post [color=blue][b](2)[/b] [/color].

$BC = BE + EC = AE + AC = AD + AC$

where we applied $BE = AE$ and $EC = AC$. Hence, the triangle $\triangle ABC$ with the angles $\measuredangle A = 120^o$, $\measuredangle B = 40^o$ and $\measuredangle C = 20^o$ satisfies the conditions of the problem.