Hey, I'm doing more of the Challenge problems in Mr. Patrick's Intro book.  I want to check if my answers and logical reasoning are correct, and alternate methods to solving the problem that would render it easier.

2.27 For how many of the positive integers $24-125$ inclusive is the digital sum a multiple of $7$?

First, I divided the cases into two-digit and three-digit numbers.

I noted that the possible multiples of $7$ for numbers $24-99$ were $7$ and $14$.  Digits satisfying $x+y=7$ were $(2,5), (3,4), (4,3), (5,2), (6, 1), (7,0)$.  Digits satisfying $x+y=14$ were $(5,9), (6,8), (7,7), (8,6), (9,5)$.  There are $12$ such two-digit numbers.

Possible multiples of $7$ for numbers $100-125$ was only $7$ itself.  Digits satisfying $x+y+z=7$ were $(1,0,6), (1,1,5), (1,2,4)$.  There are $3$.

So there are $15$ total.

2.28 Mr. Smith brings home 7 animals for his 7 children.  Each child will adopt a pet to be his or her own.  There are 4 different cats, 2 different dogs, and a goldfish.  Anna and Betty refuse to takecare of the goldfish, and Charlie and Danny insist on having cats.  The other kids will take anything.  In how many ways can Mr. Smith give the children pets?

I started out with the restrictions, saying that Charlie could choose a cat in $4$ ways and Danny then in $3$ ways.  There are now 5 animals left, but Anna and Betty refuse the goldfish.  That means that Anna can also choose from only $4$ and Betty from $3$.  The other kids have 3 animals left between them and can choose between them in $3!$ ways.  The total number is $864$ ways.

2.30 In how many ways can you spell the word $NOON$ below?  You can start on any letter, then on each step you can move one letter in any direction (up, down, left, right, or diagonal).  You cannot visit the same letter twice.

NNNN
NOON
NOON
NNNN

Obviously you have to start on the perimeter to start with an 'N.'  Divide into two cases, starting on the corner, and merely starting on the perimeter.

Starting on the corner, there are is only one way to go to the next O.  Then there are 3 ways to go to the 2nd O.  There are then 5 ways to go the last N.  There are 4 corners, so 3*4*5=60 choices for starting on the corner.

Starting on the any location on a side which is not a corner, there are two choices for the first movement, 3 choices for the second.  The 3 choices themselves make up two subcases.

For subcase one, as usual, start on a non corner side, and there are two possibilities.  If you choose to go to the 'O' bordering the same side you started on (which there is one of), then you have only 4 choices of the last N because you cannot go back to an earlier letter.  2*1*4=8.

For subcase two, start the same way, but go to one of the two 'O's not bordering the side you started on.  This leads to 5 possible end locations, with 2*2*5=20

The total for case 2 is 28 possibilities, and there are 8 starting locations, for 224 possibilities.

224+60=284.

Hmm, counting problems are annoying.  You never know if your logic is correct.  it must be a pain on the AIME.    :D

Thanks for the help, I'll be posting more later.  Don't want my computer to crash right now..

Here's another way to do the NOON one.

Choose two Os, then choose an N for each O. Divide into two cases:

1. The two Os are next to each other.
2. The two Os are diagonal to each other.

Case 1:
There are 5 Ns to choose from for each O, so that's 5*5 = 25 ways. But there are 2 ways to choose the same N twice, which is not allowed, so we must subtract those; in fact there are 25 - 2 = 23 ways. Since there are 4 ways to choose two Os that are next to each other, Case 1 renders 23*4 = 92 total ways.

Case 2:
There are 5 Ns to choose from for each O, so that's 5*5 = 25 ways. There's no way to choose the same N twice, so we don't have to worry about that. And since there are 2 ways to choose two Os diagonal to each other, Case 2 renders 25*2 = 50 total ways.

Add up the two cases to get 142 total paths. But for each path, we can go in two different directions, so we actually need to multiply that by 2, to get our final answer of 284.


Given a vector field (4x+2x^3z)i-y(x^2 +y^2)j -(3x^2z^2 +4y^2z)k which closed surface has the greatest flux. I imagine that the divergence theorem palys a role but I'm not sure.

Don't double post. Your other thread may be moved here, but that doesn't mean you should ask the same question again.

It's ok. I've decided to stop being lazy and think. However it is difficult computationally and any help is welcome.

[quote="jmerry"]Of course it's the divergence theorem- the divergence is a function which is positive on a bounded region and negative outside that region. How would you maximize the integral of a function that looks like that?[/quote]


Let $x, y, z$ positive real numbers such that $x^{2}+y^{2}+z^{2}=1$, and let $n$ be a positive integer. Show that

                     \[\frac{x}{1-x^{2n}}+\frac{y}{1-y^{2n}}+\frac{z}{1-z^{2n}}\geq\frac{(2n+1)^{1+1/2n}}{2n}.\]

[quote="zweig"]"Posted: Today, at 8:36 pm"

"Posted: Today, at 8:45 pm"

....... Are you a genious N.T.TUAN ?  :o[/quote]
No, I am not :P . This is the old problem.  :D


Triangle ABC and point P in the same plane are given. Point P is equidistant from A to B, angle APB is twice angle ACB, and $\overline{AC}$ intersects $\overline{BP}$ at point D. If PB=3 and PD=2, then $AD\cdot CD=$
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

[hide="Hint and answer"]What will happen if you draw circle with center $P$ and radius $PA$? Answer: $5$.[/hide]

[hide="Solution"] Draw the circle with center P. Extend PB and call the intersection with the circle E. EP is 3 because it is a radius. By power of a point, ED * DB = 5 * 1 = 5 = CD * AD, so the answer is 5.[/hide]

[hide]Draw circle with center P that passes through A,B, and C. Using power of a point, $AD\cdot CD=DB\cdot(PD+R)=1\cdot(2+3)=5.$
$\boxed{A: 5}$.[/hide]


Let $x$, $y$ be two positive integers such that $\displaystyle 3x^2+x=4y^2+y$.
Prove that $x-y$ is a perfect square.

TST? oh it's also for iran (95 or 96) see here :
 http://www.mathlinks.ro/Forum/viewtopic.php?p=204325&highlight=

Another solution: if $3x^2+x=4y^2+y$, then $(3x+3y+1)(x-y)=y^2$. Now if $p$ is a prime and $p \mid x-y$, we have $p \mid y^2 \Rightarrow p \mid y \Rightarrow p \mid x-y+y=x$. Now suppose $p \mid x-y, p \mid 3x+3y+1$, then as $p \mid x, p \mid y$ we have $p \mid 3x+3y-1-3x-3y=1$, contradiction, so $(3x+3y+1, x-y)=1$. Thus the result.

Also from the same contest:
Find all numbers x such that both x and x inverse ( x in reverse order) are powers of 2. (same level of difficulty)

[quote="ali"]Also from the same contest[/quote]

Which one?  ;)  Not the french TST...
Please start a new thread whith this new problem.

Pierre.

[quote="Igor"]My (one line  :) ) solution:
[hide]
$\displaystyle 3x^2+x=4y^2+y  \Leftrightarrow  (4y-3x)^2=(x-y)(12x-12y+1)$ and the result follows
[/hide][/quote]Nice solution. Actually we can prove that $x-y = (\gcd (x,y))^2 $ :)

the relation can be developped and so on:
(x-y)(1+3(x+y))=y²
note a=(x-y)
and b=(x+y)
then 4a(1+3b)=(b-a)²
the equation X²-14aX+(a²-4a)=0 has b as a solution
so the small discriminant is a perfect square
this discriminant =48a²+4a=4a(12a+1)
so 4a|a perfect square and gcd((12a+1),4a)=1
so 4a is perfect square

We can rewrite $ 3x^2 \plus{} x \equal{} 4y^2 \plus{} y$ as:

$ (8x \plus{} 1)^2 \equal{} (4x)^2 \plus{} (8y \plus{} 1)^2$

but that is no more than a pythagorean triple.
Exist $ u,v$ positive integers whit some conditions such that:

$ u^2 \plus{} v^2 \equal{} 8x \plus{} 1$
$ 2uv \equal{} 4x$
$ u^2 \minus{} v^2 \equal{} 8y \plus{} 1$

because $ 4x$ is even, while $ 8y \plus{} 1$ is odd.
From the first and the third ecuation: $ 2v^2 \equal{} 8(x \minus{} y)$, then ($ 2 \mid v$ and)
$ (\frac {v}{2})^2 \equal{} x \minus{} y$


geometry
geometric transformation
homothety
incenter
radical axis
power of a point
geometry open

Let $ O_1,O_2,O_3$ be three circle. Let $ I$ be a circle tangent to $ O_1,O_2,O_3$ internally at $ A,B,C$ respectively and $ J$ be a circle tangent to $ O_1,O_2,O_3$ externally at $ A',B',C'$ respectively. Then internal similitude center of circle $ I,J$ is radical center of $ O_1,O_2,O_3$. Also, this point is intersection of $ AA',BB',CC'$. Then how can we say about external similitude center of circle $ I,J$?

We can say nothing about the external similarity center of the circles $ I, J.$ Non-degenerated Apollonius problem has 8 solutions. But when the circles are considered directed (clockwise or counter-clockwise), tangent only when they have the same directions at their tangency point, the problem has only 2 solutions. When circles are involved, directed lines resemble properties of points better then undirected lines.

1. 2 circles can have up to 2 common points, up to 4 common undirected lines (common tangents). But 2 directed circles have only up to 2 common directed lines.

2. Directed angle is dual to a line segment. Only an angular interval of lines passing through a point fill a directed angle. Only linear interval of points on a line fill a line segment. 2 circles form equal angles at their common points. Tangent lengths of 2 directed circles are equal.

2. One circle passes through 3 points, 4 circles are tangent to 3 lines. But only one directed circle is tangent to 3 directed lines.

3. 2 circles have one radical axis, and 2 similarity centers. But 2 directed circles have only one similarity center: External similarity center of undirected circles, if they have the same direction (say both clockwise), internal similarity center of undirected circles, if they have opposite directions.

4. Common points of 2 circles (if any) are on their radical axis. Common directed lines of 2 directed circle (if any) pass through their similarity center.

5. Tangent lengths from a point on the radical axis are equal. Directed line through similarity center forms equal angles with both directed circles.

6. 3 circles have one radical center, common point of their radical axes, but 3 homothety axes, common lines of their similarity centers. But 3 directed circles have only one homothety axis.

etc.

If you consider directed circle $ I$ externally tangent to all 3 directed circles $ O_1, O_2, O_3,$ it means that they have the same direction. Directed circle $ J$ internally tangent to (intersecting) $ O_1, O_2, O_3,$ is the 2nd solution of general Apollonius problem for directed circles and there are no other solutions. Obviously, $ J$ is directed oppositely then $ I,$ therefore their single homothety center is the internal homothety center of undirected circles. These 2 directed circles do not have any other similarity center.

Thank you for kind answer. I'm not sure if I'm understanding about concept of directed circle correctly.But I think [quote="yetti"] tangent only when they have the same directions at their tangency point [/quote] should be tangent externally or internally when they have the same or opposite directions at  their tangency point. Is that right? And what is difference between direction at point on circle and direction of circle?
If I'm understaing correctly, you are saying  that if we consider circle by directly, then circles $ I$ and $ J$ have opposite direction and there is only one similitude center, i.e internal similitude center of circle $ I$ and $ J$. So we can say nothing about the external similarity center of the circles.  But I think  what circle $ I$ and $ J$ have no external similitude center is the case that we consider circle by directly and I can't understand why we must consider circle by directly because what I requested is just property of external similitude center.

Undirected circle is composed of points equally distant from the center. Its radius is always positive.
Directed circle is composed of directed lines equally distant from the center and forming directed angle either $ \plus{}90^circ,$ or $ \minus{}90^circ$ (but not both) with the radius. Its radius can be positive (for all circles directed counter-clockwise) or negative (for all circles directed clockwise).

If 2 directed lines with opposite directions coincide in position, they are still 2 different directed lines.
If 2 directed circles with opposite directions coincide in position, they are still 2 different directed circles.

The following figures are taken from Yaglom's Geometric Transformations.

Figure 77 shows 2 common directed lines of 2 directed circles, (a) with the same directions (both counter-clockwise) and (b) with opposite directions. Note that the directed circles and their common directed lines have [b]the same directions at the tangency point of undirected circles[/b].

Figure 76 shows examples of (a) tangent directed circles, i.e., directed circles [b]with single common directed line[/b]; again, they have the same directions at the tangency point of undirected circles, and (b) directed circle that are not tangent; the first pair has 2 common directed lines and [b]non-zero tangent distance[b] $ AB,$ the second pair has [b]no common directed line[/b]; both pairs have opposite directions at the tangency point of undirected circles.

Thanks a lot. :lol:
I think there are many advantage by definiting directed circle, especially generalizing problem.
Let $ A,B,C,D$ be four points on circle in that order. Then incenter of triangle $ ABD,ACD$, $ C \minus{} excenter$ of triangle $ ABC$,$ B \minus{} excenter$ of triangle $ DBC$ are collinear.
I'm not sure that this problem can be generalized with directed circle or line, but if it is possible, then how can we generalize so that $ A,B,C,D$ are points on circle in any order?


What is the minimum number of people in a group so that you would be certain that 3 of them were born in the same month and on the same day of the week? Explain your answer.

well you could have two people for each day of the week in each month and that's 168 people

so you gotta have one more guy in order to be sure there's three for at least one month and day of the week

1[b]69[/b]


inequalities
quadratics
function
algebra unsolved
algebra

Let $ a,b,c$ be the three sides of a triangle. Determine all possible values of  $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$

hello, it is
$ 1\le\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}<2$ because
$ a^2\plus{}b^2\plus{}c^2<2ab\plus{}2bc\plus{}2ca$ is equivalent with
$ 0<a(b\plus{}c\minus{}a)\plus{}b(a\plus{}c\minus{}b)\plus{}c(b\plus{}a\minus{}c)$ and
the inequality $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$ holds for all real numbers $ a,b,c$, especially when $ a,b,c$ are sides of a triangle. The equal sign holds for $ a\equal{}b\equal{}c\equal{}1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]$ 0 < a(b \plus{} c \minus{} a) \plus{} b(a \plus{} c \minus{} b) \plus{} c(b \plus{} a \minus{} c)$ [/quote]

I don't think this is sharp for $ a,b,c$ sides of a triangle.

But I think, that
$ b<a\plus{}c$
$ a<b\plus{}c$
$ c<a\plus{}b$ and $ a,b,c>0$ so it's obviously sharp.

But all three of those inequalities cannot be sharp at the same time.  What I mean is as $ b\to a\plus{}c$, one of the other inequalities becomes less sharp.

$ a^2\plus{}b^2\plus{}c^2 <k(ab\plus{}ac\plus{}bc)$

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

$ 2x^2\plus{}2y^2\plus{}2z^2\plus{}2(xy\plus{}xz\plus{}yz)<k(x^2\plus{}y^2\plus{}z^2)\plus{}3k(xy\plus{}xz\plus{}yz)$

Now, k=2 gives a triviality, if k<2 then moving everything over to the L.H.S. gives a positive quadratic in x, which is unbounded positively.

I have'nt checked your solution,rofler :-/ now I have another solution for our problem .Hope it will be correct :D
First of all we have  basic ỉnequality :$ ab \plus{} bc \plus{} ca \leq a^2 \plus{} b^2 \plus{} c^2 < 2ab \plus{} 2bc \plus{} 2ca$. This means that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$ must be on interval $ [1;2)$
Consider a triangle with three sides of which are $ n,n$ and 1
and a function $ f(n) \equal{} \frac {n^2 \plus{} n^2 \plus{} 1}{n \plus{} n \plus{} n^2}$
We easily see that f(n) is a continous function and such that $ f(1) \equal{} 1$ and $ lim_{n \minus{} > \plus{} \infty} f(n) \equal{} 2$
Therefore $ f(n)$ get all values between $ f(1)$ and $ f( \plus{} \infty)$
So $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$ get all values in interval $ [1;2)$


Harvard college Yale Princeton MIT Columbia

I'm having an interview for my "safety" college next week, and I'm worried about them asking me the names of the other colleges that I am applying to. This school has been known to reject people who they think are using them as their safety. I'm not exceptional, but I'm also applying to Harvard, Yale, Princeton, MIT, etc, and the disparity between these schools and my safety is painfully obvious. Can I leave some schools out (e.g. Harvard, Yale, Princeton) and simply mention my other safeties and one or two top schools, or would this be dishonest?

Haha, perhaps we should have an entire forum dedicated to ethics issues ...

I don't remember being asked that question by interviewers anywhere, but I mostly had alumni, fairly informal interviews.  I doubt, if that is the type of interview you're having, that they will ask you about it except perhaps "off the record."  You could always say you'd rather not tell, although that would be kind of odd.  I don't think it's fair to only tell them some (but not all) of the schools you are applying to as if it were a complete list, however.

one could phrase it ambiguously, as in: "my list includes [school A], [school B], and [school C]." under pressure, one could say that one is also applying to princeton, yale, etc- but i, for one, would speak of these as true longshots, as "lets see what happens" kind of schools, as not being... *TRULY* serious about one's chances at these. i realize that the application fee for those schools makes it more serious than that, but one can make it sound like one used one's birthday money on these, just as a whim. i am planning to apply to one-two colleges "on a whim" so i dont see it as being that... crazy an idea. just a thought!

If someone asks you, "What other schools are you applying to?" it makes sense to give a straight answer. The old saying is, "Honesty is the best policy."

The much more likely question you will encounter in an interview, and one you had very well better have an answer for, is "Why do you want to go to my college?" That's the question you should be prepared to answer in detail.

You have my sympathies. I know you live in the Northeast, where there are strong private schools that many applicants to the most famous schools use as "safeties," which results in those private schools trying extra hard to figure out who their REAL applicants are. You have to be convincing, by first convincing YOURSELF, that you could fit in and add to the student community at any of the schools to which you apply. Here in the upper Midwest, where I live, anyone can apply to the U of Wisconsin, the U of Minnesota, the U of Iowa, or whatever, and the kind of strong student who typically participates on AoPS can be sure of getting in "by the numbers," with no interview, and no embarrassment about applying to Harvard, Princeton, MIT, or whatever. You are in a more delicate situation, but you won't look better in an interview by being evasive. The application deadlines have mostly passed by now, and the interviewer knows you know where you have already applied.

My interviewer for Columbia asked me where else I was applying. I think I gave him a full list.

Lol, Columbia isn't really a safety, and I wouldn't mind saying the full list for Columbia and up. It is just uncomfortable when you have schools that aren't as prestigous, so to speak, that it sounds blatantly obvious that they are your safety school.

Is it defensible for any university to ask about other
applications?  (It seems even less defensible to utilize
the answer in their admissions deliberations, so why
ask?)

There are obvious reasons the university would be
curious about such information, but curiousity doesn't
necessarily entitle an inquiry.

[quote="JBL"]Many colleges seem to ask once you turn them down, which seems perfectly defensible to me.  I don't remember being asked before having been accepted or rejected.[/quote]

Wait... colleges send you a letter asking you to explain why you turned them down?

To enforce "single-choice" early admission (SCEA) plans, or early decision (ED) plans, colleges have some interest in knowing where else applicants are applying.

i dont think anyone asks where you applied early; at least i wasn't asked that.  If they did, that would hurt you... i.e i applied ED and EA, if my EA schools asked where I applied early, my chances would most likely go down if i told them I applied ED somewhere else.

For ED, schools definitely compare lists with each other to make sure that no one is applying to more than one.  I don't remember if everywhere I didn't go sent me a little postcard asking where I was going, but at least a couple of schools did.  This is after you've already chosen where you're going, mind you, and it's highly voluntary -- I have no idea what the return rate it.  U. Chicago actually also sent me a very nice letter after I turned them down.


Simply state your favorite zelda game(s) and what you found so enrapturing about them!

My favorites are, of course, [b]the legend of the wind waker and the ocarina of time[/b]!

Those two almost top my all time favorite video game list(second only to Halo and call of duty)!

They had beautiful graphics(considering the time they were made in), had wonderful plots, awesome weapons, mind testing puzzles, and AMAZING boss battles. They are, in all aspects, flawless games.

Although FFX is still top on my game list...

Ocarina of Time and Twilight Princess for me!

For the above reasons, Ocarina of Time was perfect.
And I'm only halfway through Twilight right now, but I'm very impressed by the graphics and the not-so-standard-for-LoZ plot.

Windwaker was a bit boring at times (cough gathering the triforce cough)

Well, I've only played 3 LoZ games, but I've played 6 games with Link in them... I've played OoT, Four Swords Adventures, and TP. Twilight Princess was the best by far for me. Do Smash Brothers count as LoZ?

Not really, though it is very fun.  I mean, you wouldn't call the new Soul Caliber a Star Wars game because Yoda was in it :P

(I think Yoda was in it...)

Ocarina of time has stood the test of time...  :P

It is one of the only games that gamespot.com rated a perfect 10/10.(The others were MGS4 and GTA)

I'm going to buy Ocarina of time for the wii, just because it was so awesome...


If 3x-8y=28 and x and y are integers

What is the smallest value of x-y?

[hide]

Let $x=8k+4$ and $y=3k-2$ for some $k$. Then $3x-8y=3(8k+4)-8(3k-2)=28$. The value of $x-y$ then is $8k+4-(3k-2)=5k+6$, so there is no lower bound. $-\infty$?
[/hide]

:huh:
How?
If $lim_{y \rightarrow-\infty}$ then $8y \not=28$, that could not be what you mean though.
Sorry for the bad LaTeX.

[quote="AstroPhys"]:huh:
How?
If ${lim}_{y \rightarrow \infty}$ then $8y \not{=}28$, that could not be what you mean though.
Sorry for the bad LaTeX.[/quote]

Um, maybe you are reading the sign wrong? 3x-8y=28

says 3x-8y... and there is definitely no lower bound

[hide]$3x-8y=28 \Rightarrow y= \frac{3x-28}{8}$

Let $h(x,y)=x-y \Rightarrow h(x)=x-\frac{3x-28}{8}\Rightarrow h(x)= \frac{5}{8}\cdot x+\frac{7}{2}$

$h'(x)=\frac{5}{8}\Rightarrow h(x)$ is increasing  $\forall ~~x$ thus it has a range of $(-\infty,+\infty)$[/hide]

[quote="AstroPhys"][hide]
$3x+8y= 28$
$3(4)+8(2) = 28$
$x-y= 1$
[/hide][/quote]

I'm not sure what you did there.  You just plugged in random numbers and said that 4-2=1.

[quote="AstroPhys"][hide]
$3x+8y= 28$
$3(4)+8(2) = 28$
$x-y= 1$
[/hide][/quote]

how do you know thats the smallest?

[quote="Teki-Teki"]what happens if we restrict them to positive integers?[/quote]

[hide="Restricted to positive integers,"]$x = \frac{8y+28}{3}$

$x-y = \frac{5y+28}{3}$

Clearly, to minimize $x-y$, we must minimize $y$. Let $y=1$. So $x=12$. Thus, $x-y =11$.[/hide]


[quote="kobebryant24"]could someone perhaps post the 2008 national sprint round, the 1995 and 1997 sprint, target, and team rounds or email to asdfasdfsdf12390@gmail.com?[/quote]
It's on the MATHCOUNTS website...

the only thing i see are the 2008 state, chapter, and school rounds


function
search
algebra unsolved
algebra

Let $ f : R\to R$ satisfy $ |f(x)|\leq 1$ for all $ x\in R$
and $ f(x\plus{}\frac{13}{42})\plus{}f(x)\equal{}f(x\plus{}\frac{1}{6})\plus{}f(x\plus{}\frac{1}{7})$
Prove that $ f$ is period function

คุณ apollo ชื่อจริงชื่ออะไรครับ
ช่วยพีเอ็มมาบอกมด้วย
Thank you.  :lol:


superior algebra
superior algebra unsolved

Suppose $ L$ is an algebraically closed field, $ F \subset L$ is a subfield, and $ f \in F[x]$ is a nonconstant polynomial. Prove that $ L$ cfontains a unique subfield extending $ F$ which is a splitting field for $ f$ over $ F$.

The proof could depend a little on your definition of a splitting field.  Hint: if $ f$ splits completely in $ K,K'$, then $ f$ splits completely in $ K\cap K'$.  If you're not given the existence of such a field $ K$, you'll need to prove that.


Can you check me to make sure I'm right....

If two lines are parallel then they lie on the same plane......(my answer FALSE)

If two plances intersect then the intersection is either a point or a line...
(my answer   TRUE)

The first statement is [color=blue]true[/color] and the second [color=red]false.[/color]
 But , please write at the getting start section this kind of question. Probably you are nev here !
[u] Babis[/u]

[quote="stergiu"][b]NOTE[/b]

  As [color=red]Virgil Nicula[/color] pointed out to me, the second question can be taken as '' true'' in the sense that one part of the answer is true: the intersection of two planes is a line (axiomatic definition) and never a point.But the sentense  << ....is a point [color=blue]or[/color] a line >> is true , because it contains the part '' ...a line''.
 Any way , sorry for not giving to the question the necessary attention.

[u] babis [/u][/quote]

ure answer is completely, complete. :wink:


modular arithmetic
number theory unsolved
number theory

Let $ p$ be a prime of the form $ 4k\plus{}1$. Let $ i$ be a positive integer, $ 0<i<p$. Is it true that there is a positvie integer $ n$ st $ 2^n\minus{}i$ is a multiple of $ p$? If it is true can anyone post a proof?

No.  Take, for example, $ p \equal{} 17, i \equal{} 3$.

What about this: does there exist an infinite number of primes $ p$ st for any positive integer $ i$, $ 0<i<p$, there is a positive integer $ n$ st $ 2^n\minus{}i$ is a multiple of $ p$?

This is known as [url=http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots]Artin's conjecture[/url] and is open.  The relevant sequence of primes is [url=http://www.research.att.com/~njas/sequences/A001122]A001122[/url].

What you're asking is whether there are infinitely many primes $ p$ such that $ 2$ is a primitive root (generator) mod $ p$. I am not sure whether it is known or not (there are many results of the smallest primitive root mod $ p$, so probably not). Yet it can be shown that $ 2$ is a primitive for a prime $ p$ such that $ p \equiv 3 \pmod 8$ and $ \frac {p \minus{} 1}{2}$ is also a prime. It is defiantly not known whether there are infinitely many such primes.
[EDIT: as t0rajir0u said, it is open]


inequalities
geometry
geometric transformation
reflection
inequalities proposed

For positive real numbers $a,b$ and $c$,
\[ \sum_{cycl}\frac{ab(a^2+bc)}{b+c} \ge \sqrt{3abc(ab^2+bc^2+ca^2)} \]

\[ \sum_{cycl}\frac{ab(a^2+bc)}{b+c} \ge \sqrt{3abc(ab^2+bc^2+ca^2)}  \]
$<\to (\sum_{cycl}\frac{ab(a^2+bc)}{b+c})^2 \ge 3abc(ab^2+bc^2+ca^2)$
Use the well-known inequality $(x+y+z)^2 \geq 3(xy+yz+zx)$ hence
      $(\sum_{cycl}\frac{ab(a^2+bc)}{b+c})^2 \geq 3\sum \frac{ab(a^2+bc)}{b+c}\frac{bc(b^2+ac)}{a+c}$
 $\sum \frac{ab(a^2+bc)}{b+c}\frac{bc(b^2+ac)}{a+c} \geq abc(ab^2+bc^2+ca^2) <\to \sum_{cycl} b(a+b)(a^2+bc)(b^2+ac) \geq (ab^2+bc^2+ca^2)(a+b)(b+c)(c+a) <\to \sum a^3b^3+2abc(\sum a^3)+3a^2b^2c^2+\sum_{cycl} a^2b^4+abc(\sum_{sym}a^2b)+(b^5c+c^5a+a^5b) \geq 2abc(\sum_{sym}a^2b)+\sum a^3b^3+abc(\sum a^3)+3a^2b^2c^2+\sum_{cycl} a^2b^4 <\to abc(\sum a^3)+\sum_{cycl}b^5c \geq abc(\sum_{sym} a^2b)$
Use AM-GM inequality we have $\sum_{cycl}b^5c \geq 3a^2b^2c^2$
So that $abc(\sum a^3)+\sum_{cycl}b^5c \geq abc(\sum a^3)+3a^2b^2c^2\geq abc(\sum_{sym} a^2b)$ (Shur's inequality)
therefore the inequality is proved :lol:  :lol:  :lol:

or we can just prove that $\sum \frac{ab(a^2+bc)}{b+c}\geq \frac{3abc}{2}+\frac{\sum a^2 b}{2}$ wich yealds the desired inequality with $Am-Gm$
we know that
$\sum \frac{ab(a^2+bc)}{b+c}=abc \sum \frac{b}{b+c}+\sum \frac{a^3 b}{b+c}\geq \frac{3abc}{2}+\sum a^2 b\frac{a}{b+c}$ which yealds what we want with rearrangement.

[quote="Albanian Eagle"]or we can just prove that $\sum \frac{ab(a^2+bc)}{b+c}\geq \frac{3abc}{2}+\frac{\sum a^2 b}{2}$ [/quote]
The above inequality is wrong !!!

[quote="nhat"]
$abc\sum a^3 + \sum_{cycl}b^5c \ge abc\sum_{sym} a^2b$ [/quote]
I think you made a smal mistake in your calculation. Shouldn't it be
$abc\sum_{cycl}a^3 + \sum_{cycl}b^5c \ge 2abc\sum_{cycl}ab^2$

[quote="nhat"]I don't think so,Ductrung.I checked my solution and I didn't find any mistakes  :)  :)[/quote]

I verify all these with a computer algebra system:
\[ X=\sum_{cycl} b(a+b)(a^2+bc)(b^2+ca) =3a^2b^2c^2+2abc\sum_{cycl}(a^3+a^2b)+\sum_{cycl}a^3b^3+\sum_{cycl}a^2b^4+\sum_{cycl}a^5b \]
\[ Y=(a+b)(b+c)(c+a)\sum_{cycl} ab^2 =3a^2b^2c^2+abc\sum_{cycl} a^3+2abc\sum_{cycl}a^2(b+c)+\sum_{cycl}a^3b^3+\sum_{cycl}a^2b^4 \]
and
\[ X-Y=\sum_{cycl}a^5b+abc\sum_{cycl}a^3-2abc\sum_{cycl}ab^2 \]

OK,After i checked againt,I had found my mistakes.So I will prove the inequality:
                    $\sum_{cyclic} a^5b+abc(\sum a^3) \geq 2abc(ab^2+bc^2+ca^2)$
Firstly,Apply AM_GM's inequality we have $\sum a^3 \geq 3abc$
So that:$\sum_{cyclic} a^5b+abc(\sum a^3) \geq \sum_{cyclic} a^5b+3a^2b^2c^2$
we only need to prove:$\sum_{cyclic} a^5b+3a^2b^2c^2 \geq 2abc(ab^2+bc^2+ca^2)$
Apply Shur's inequality we have:
             $\sum_{cyclic} a^5b+3a^2b^2c^2 \geq (abc)^{\frac{1}{3}}(\sum_{sym}a^3b^2)$ (1)
We need to prove $\sum_{sym}a^3b^2 \geq 2(abc)^{\frac{2}{3}}(ab^2+bc^2+ca^2)$(2)
Lemma:For $a,b,c>0$ Prove that:$3(\sum_{sym}a^3b^2) \geq 2(ab+bc+ca)(ab^2+bc^2+ca^2)$ (2')
Solution:(2) equalvilent to $\sum_{sym} a^3b^2+2(\sum_{cycly}a^3b^2 \geq 2abc(a^2+b^2+c^2+ab+bc+ca)$ (*)
Apply AM_GM's inequality we have:  $a^3b^2+a^3c^2+a^3bc \geq 3a^3bc$
                                                     $b^3c^2+b^3a^2+b^3ac \geq 3ab^3c$
                                                     $c^3a^2+c^3b^2+abc^3 \geq 3abc^3$
Sum up these inequalities we have $\sum_{sym}a^3b^2 \geq 2abc(\sum a^2)$ (3)
By similar way we can prove that $\sum_{cyclic} a^3b^2 \geq abc(\sum ab)$ (4)
From (3),(4) we have (*).So (2') is proved
From (2') it follows that $\sum_{sym}a^3b^2 \geq 2(abc)^{\frac{2}{3}}(ab^2+bc^2+ca^2)$
So that this problem is proved.Cheer  :lol:  :lol:

[quote]Apply Shur's inequality we have:
$\sum_{cyclic} a^5b+3a^2b^2c^2 \geq (abc)^{\frac{1}{3}}(\sum_{sym}a^3b^2)$
[/quote]

Hi, Nhat

I cannot see how you get the above.
Following you: using
$x^3+y^3+z^3+3xyz \ge xy(x+y)+yz(y+z)+zx(z+x)$
where $x=\sqrt[3]{a^5b}$, $y=\sqrt[3]{b^5c}$, $z=\sqrt[3]{c^5a}$, right?

Now
$3xyz=3\sqrt[3]{a^6b^6c^6}=3a^2b^2c^2 \ \ \Rightarrow$ OK
$x^2y = \sqrt[3]{(a^5b)^2 b^5c} = \sqrt[3]{a^{10}b^7c} = \sqrt[3]{abc}(a^3b^2) \ \ \Rightarrow$ OK too
$xy^2 = \sqrt[3]{a^5b (b^5c)^2} = \sqrt[3]{a^5b^{11}c^2} \ \ \Rightarrow$???

Could you please explain?

[quote="bomb"]Use generalised Schurs
Bomb[/quote]

Which 'generalised Schurs' do you refer to? Care to explain?

[quote]For positive real numbers $a,b$ and $c$,
\[\sum_{cycl}\frac{ab(a^{2}+bc)}{b+c}\ge \sqrt{3abc(ab^{2}+bc^{2}+ca^{2})}\]
[/quote]

By the well-known inequality $(x+y+z)^{2}\ge 3(xy+yz+zx)$ it suffices to prove that
$\sum_{cycl}\left( ab\frac{a^{2}+bc}{b+c}\cdot bc\frac{b^{2}+ca}{c+a}\right) \ge abc(ab^{2}+bc^{2}+ca^{2})$
or equivalently
$\sum_{cycl}\frac{b(a^{2}+bc)(b^{2}+ca)}{(b+c)(c+a)}\ge ab^{2}+bc^{2}+ca^{2}$
Note that $(a^{2}+bc)(b^{2}+ca)-ab(a+c)(b+c)=c(a+b)(a-b)^{2}\ge 0$, that is,
$\frac{(a^{2}+bc)(b^{2}+ca)}{(a+c)(b+c)}\ge ab$ (*)
Hence
$\frac{b(a^{2}+bc)(b^{2}+ca)}{(a+c)(b+c)}\ge ab^{2}$
Adding this and similar inequalities, the conclusion follows.

[b]Apropos:[/b] (*) implies trivially
$\sqrt{\frac{(a^{3}+abc)(b^{3}+abc)}{(a+c)(b+c)}}\ge ab$
and so
$\sum_{cycl}\sqrt{\frac{(a^{3}+abc)(b^{3}+abc)}{(a+c)(b+c)}}\ge ab+bc+ca$
This together with a well-known inequality by Cezar Lupu yields
$\sum_{cycl}\sqrt{\frac{a^{3}+abc}{b+c}}\ge a+b+c$
In similar manner, we also have
$\sum_{cycl}\sqrt{\frac{b+c}{a^{3}+abc}}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
$\sum_{cycl}\sqrt{\frac{a^{2}+bc}{b+c}}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}$
$\sum_{cycl}\sqrt{\frac{b+c}{a^{2}+bc}}\le \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$

[quote="cefer"][quote="ductrung"]This together with a well-known inequality by Cezar Lupu yields
$\sum_{cycl}\sqrt{\frac{a^{3}+abc}{b+c}}\ge a+b+c$
[/quote]

Which inequality is it? :maybe:[/quote]

Cefer,
The answer is part of the solution. You will know it if you try to solve the problem :-)

Otherwise, look at Problem Corner of the current issue of Math. Reflection:  [url]http://reflections.awesomemath.org/2006_5/2006_5_problems.pdf[/url]


geometry
rectangle
analytic geometry
combinatorics unsolved
combinatorics

On the plane are positioned at most $ \frac {4n}{3}$ rectangles with sides parallel to the coordinate axes. Assume that every rectangle intersects at least $ n$ other rectangles. Show that there exists a rectangle that intersects all other rectangles.

Do you mean that there are at most $ \frac{4n}{3}$ rectangles?

[quote="probability1.01"]Do you mean that there are at most $ \frac {4n}{3}$ rectangles?[/quote]

OK. Edited.

[hide="Probably the right idea"]Suppose we have $ m \le \frac{4n}{3}$ rectangles. Consider the four rectangles with the highest bottom edge, lowest top edge, rightest left edge, and leftest right edge. A rectangle which intersects all four of these rectangles will intersect all the rectangles. Now you play around with the numbers and use pigeonhole.[/hide]

[quote="probability1.01"] Consider the four rectangles with the highest bottom edge, lowest top edge, rightest left edge, and leftest right edge. [/quote]

Are you sure? Look at the following diagram. The big rectangle intersects all the four rectangles you mentioned. Yet it doesn't intersect all rectangles...

P.S. Sorry for my bad drawing

May be i misunderstand the problem, but does this construction satisfies the condition of problem?

@greenvert: no, since each rectangle intersects $ n\equal{}4$ rectangles, there should be most $ 4\times \frac{4}{3}<6$ rectangles, but in your diagram there are $ 6$ rectangles.

Consider the four rectangles: (using coordinates) the one with the less right horizontal edge, the one with the greater left horizontal edge, the one with the less upper vertical edge, and the one with the greater lower vertical edge. It is clear that if there is a rectangle that intersects this four, then it intersects all rectangles. Each of this rectangles intersect at least other $ n.$ Suppose there isn't that rectangle. We need $ n$ intersections from each special rectangle, each of which can be counted at most trice. This means we need at least $ \frac{4n}{3}\plus{}1$ rectangles. Contradiction.

[quote="Carlez Tevos"][quote="probability1.01"] Consider the four rectangles with the highest bottom edge, lowest top edge, rightest left edge, and leftest right edge. [/quote]

Are you sure? Look at the following diagram. The big rectangle intersects all the four rectangles you mentioned. Yet it doesn't intersect all rectangles...[/quote]

I thought rectangles included the area within. Do you think this holds when rectangles are only their boundaries?

[quote="Nemion"]Consider the four rectangles: (using coordinates) the one with the less right horizontal edge, the one with the greater left horizontal edge, the one with the less upper vertical edge, and the one with the greater lower vertical edge. It is clear that if there is a rectangle that intersects this four, then it intersects all rectangles. Each of this rectangles intersect at least other $ n.$ Suppose there isn't that rectangle. We need $ n$ intersections from each special rectangle, each of which can be counted at most trice. This means we need at least $ \frac {4n}{3} \plus{} 1$ rectangles. Contradiction.[/quote]

By my calculations, if there are exactly $ \frac{4n}{3}$ rectangles, it is possible that the four special rectangles all intersect one another, and all other rectangles intersect three of the four special rectangles.

Am I wrong?

[quote="Carlez Tevos"]
By my calculations, if there are exactly $ \frac {4n}{3}$ rectangles, it is possible that the four special rectangles all intersect one another, and all other rectangles intersect three of the four special rectangles.

[/quote]

I think so. Then you have to deal with this case separately.

[quote="discredit"]

I think so. Then you have to deal with this case separately.[/quote]

But how?  :maybe:


$ a,b,c \geq 0,$prove that:

$ \sqrt{\frac{3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\frac{3}{b^3\plus{}(a\plus{}c)^3}}\plus{}\sqrt{\frac{3}{c^3\plus{}(a\plus{}b)^3}} \leq\frac{ab\plus{}ac\plus{}bc}{abc\sqrt{a\plus{}b\plus{}c}}$

[quote="rachid"]$ a,b,c \geq 0,$prove that:

$ \sqrt {\frac {3}{a^3 \plus{} (b \plus{} c)^3}} \plus{} \sqrt {\frac {3}{b^3 \plus{} (a \plus{} c)^3}} \plus{} \sqrt {\frac {3}{c^3 \plus{} (a \plus{} b)^3}} \leq\frac {ab \plus{} ac \plus{} bc}{abc\sqrt {a \plus{} b \plus{} c}}$[/quote]

indeed , it's easy but nice
it sufficies to show the stronger :
$ \sum{\frac{9}{a^3 \plus{} (b \plus{} c)^3}} \leq  \frac{ (ab\plus{}ac\plus{}bc)^2}{ (abc)^2(a\plus{}b\plus{}c)}$
but $ (a\plus{}b)^3\plus{}c^3\equal{} a^3\plus{}b^3\plus{}c^3\plus{}3bc(b\plus{}c) \geq 3abc\plus{}3bc(b\plus{}c) \equal{} 3bc(a\plus{}b\plus{}c)$
 so the inequality is equivalent to :  $ (ab\plus{}ac\plus{}bc)^2\minus{}3abc(a\plus{}b\plus{}c) \geq 0$ which is true  :wink:

[quote="anas"][quote="rachid"]$ a,b,c \geq 0,$prove that:

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

indeed , it's easy but nice
it sufficies to show the stronger :
$ \sum{\frac {9}{a^3 + (b + c)^3}} \leq \frac { (ab + ac + bc)^2}{ (abc)^2(a + b + c)}$
but $ (a + b)^3 + c^3 = a^3 + b^3 + c^3 + 3ab(a + b) \geq 3abc + 3bc(b + c) = 3ab(a + b + c)$
 so the inequality is equivalent to :  $ (ab + ac + bc)^2 - 3abc(a + b + c) \geq 0$ which is true  :wink:[/quote]
My solution is a little similar to yours anas.

[hide="Solution"]
$ \sum{\sqrt {\frac {3}{a^3 + (b + c)^3}} \leq \sum \frac {1}{\sqrt {bc(a + b + c)}} \leq \frac {ab + ac + bc}{abc\sqrt {a + b + c}}}$[/hide]


How many numbers can you get by adding two or more distinct members of the set  $ \{0,1,2,4,8,16\}$ together?

If we convert these numbers to binary, we'll easily see that every positive integer up to $ 11111_2\equal{}\boxed{31}$ can be made.

Or you can just add up the numbers, $ 16+8+4+2+1=\boxed{31} $


[quote="summerBoy"]I know a lot of complex numbers but in my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

PROBLEM 1

Find Re(w) and Im(w) [where $z=x+yi$]:

$w=z+\frac{1}{z}$

So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ($x$ and $y$).

My way would be:

$w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=\frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...$

If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.

However, the final solution is:

$Re(w)=x+\frac{x}{x^{2}+y^{2}}$ and $Im(w)=y-\frac{y}{x^{2}+y^{2}}$

Does anybody know the steps leading to the final solution?

PROBLEM 2

Find all solutions to the equation [where $z=x+yi$]:

$2z^{2}-3\overline{z}^{2}=10i$

It is a quadratic equation and has a conjugate of complex number in it.

Thanks for help in advance.[/quote]

Let

$z=x+iy\Rightarrow z+\frac{1}{z}=x+iy+\frac{1}{x+iy}=x+iy+\frac{x-iy}{(x+iy)(x-iy)}=x+\frac{x}{x^{2}+y^{2}}+i\left(y-\frac{y}{x^{2}+y^{2}}\right)$

$2(x+iy)^{2}-3(x-iy)^{2}=y^{2}-x^{2}+10ixy$. So we have $y^{2}-x^{2}=0,xy=1$ where from we got $x=y=1$ so $z=1+i$

summerBoy: complex numbers $\neq$ complex analysis.  Questions of this sort belong in one of the high school fora (maybe Intermediate or PreOlympiad).

cckek: you should never need to quote the entire previous post.  It is a assumed by default that you are responding to the post immediately preceding yours, and quoting the whole thing just makes a mess.

Also, you made a small error at the end of the second question: $x^{2}= y^{2}$ and $xy = 1$ has two solutions in the real numbers.

Thanks for help, cckek.

JBL, you are right. There are two solutions to the equation.

Sorry, I posted the topic very quickly, so I didn't notice it doesn't belong to this section. Moderator should move it where it has to be.


On Saturday night, one-half of the students in Ms. Schultz's math class went to the dance, one-third of the remainder went to the movies, and the remaining 8 people stayed home. How many students are in her class?

1/2 went to the dance and 1/3 of the remainder(or 1/6) went to the movies. This means that the remainder, 8 people, is $ 1 \minus{} \left ( \frac{1}{2} \plus{} \frac{1}{6} \right ) \equal{} \frac{1}{3}$ of the class, meaning there's 24 people.


The compound $ \text{MON\minus{}0585}$ is a non-toxic, biodrgradable larvicide (highly selective against mosquito larvae). Synthesize this compound.

http://ctd.mdibl.org/detail.go;jsessionid=F8289B9243A077E36E3782881A0F7655?type=chem&acc=C006114

This link shows what the compound looks like.

[hide="Answer"]Here's a possible synthesis from phenol:

1) $ C_6H_5OH \plus{} C_6H_5COCl/pyr \longrightarrow C_6H_5\minus{}O\minus{}CO\minus{}C_6H_5$

2) $ C_6H_5\minus{}O\minus{}CO\minus{}C_6H_5 \plus{} AlCl_3/25^oC \longrightarrow p\minus{}C_6H_5\minus{}CO\minus{}C_6H_4\minus{}OH$

3) $ p\minus{}C_6H_5\minus{}CO\minus{}C_6H_4\minus{}OH \plus{} NaHO/H_2O/Cl_2 \longrightarrow\,\,\,sodium\,\,4\minus{}benzoyl\minus{}2,6\minus{}dichlorophenoxide$

4) $ sodium\,\,4\minus{}benzoyl\minus{}2,6\minus{}dichlorophenoxide \plus{} CH_3I \longrightarrow 4\minus{}benzoyl\minus{}2,6\minus{}dichloroanisole$

5) $ 4\minus{}benzoyl\minus{}2,6\minus{}dichloroanisole \plus{} HOCH_2CH_2OH/H^\plus{} \longrightarrow 4\minus{}benzoyl\minus{}2,6\minus{}dichloroanisole ketal$

6) $ 4\minus{}benzoyl\minus{}2,6\minus{}dichloroanisole ketal \plus{} 2Mg/THF,\,\,then 2CO_2,\,\, then H_3O^\plus{} \longrightarrow 2\minus{}metoxi\minus{}5\minus{}benzoylisophthalic acid$

7) $ 2\minus{}metoxi\minus{}5\minus{}benzoylisophthalic acid \plus{} excess\,Me_3Al \longrightarrow O\minus{}methyl\, MON\minus{}0585$

8) $ O\minus{}methyl\, MON\minus{}0585 \plus{} conc.HI \longrightarrow MON\minus{}0585$.[/hide]

By the way, the question allows you to use benzene as well. The aromatic rings can come from benzene, phenol or both.

I quite dislike the synthesis given in the book. My primary concern is that it doesn't appear, to me, to be phsically applicable (well who am I to say that) as it involved Friedel Craft Alkylation which is hard to control (and I think they are a bit too specific).

I will post it if wanted.

Yes, post the book's synthesis, but hidden, if you don't mind.

[hide="Synthesis"]

Start with benzene:

1. $ \ce{CH(CH3)2Cl}$ and $ \ce{AlCl3}$ to make $ \text{isopropyl benzene}$ (is that the name?)

2. $ \ce{(PhCO2)2}$ and $ \ce{CCl4}$ and $ NBS$ to achieve benzylic bromination

3. $ \ce{FeBr3}$ and phenol to link the previou molecule to the para position of the phenol

4. $ \ce{C(CH3)3Cl}$ and $ \ce{AlCl3}$ to introduce $ \text{tert\minus{}butyl}$ groups to both positions ortho to the hydroxyl group
[/hide]

Wouldn't it be hard to be specific about where you want to introduce those alky branches?

The first step would be tricky, since polyalkylation is an important side reaction.

Exactly, hence my dislike of this method. I am still trying to fully understand your method  :oops: . Anyway thanks for the alternative method.

In general can we control the degree of branching during Friedel-Craft Alkylation?

I don't think so. Anyway, that's not needed, since we have the Friedel-Crafts Acylation reaction as an alternative.


Hi -

I'm trying to get the following equations lined up, so the the equals are all one above the other:

F_i(t) = \mbox{fitness of best individual in population i at time t}

D_i(t) = \left\{
\begin{array}{l l}
  \frac{|F_i(t) - F_i(t - 1)|}{F_i(t - 1)} & \quad \mbox{: F_i(t - 1) > 0}\\
  0.05 & \quad \mbox{: otherwise}\\ \end{array} \right.

Q_i(t) = \left\{
\begin{array}{l l l}
  \alpha \frac{D_i(t)}{\sum_{j} D_j(t)} + (1 - \alpha) & \frac{F_i(t)}{\sum_{j}F_j(t)} & \quad \mbox{: \sum_{j} D_j(t) > 0}\\
  & \frac{F_i(t)}{\sum_{j}F_j(t)} & \quad \mbox{: otherwise}\\ \end{array} \right.

I've tried nesting arrays, so that there's one big array with the equals in one column, but I can't seem to get it to work.

Any ideas? Help much apreciated!
-Asbestos

[P.S. I haven't been able to get the LaTeX to show: the original uses slash-squarebracket to start and end the equations, the forum aparently doesn't like that, nor replacing those with dollar-signs... Couldn't work it out, sorry]

Are you familiar with [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#eqnarray]eqnarray[/url]?

(link is fixed now)

Here is one that works. Use [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#eqnarray]eqnarray[/url]. I know why rruscyzk's doesn't work. He used his local computer name instead of artofproblemsolving.com. e.g. If i type in my computer name(in this case it is server) it would go to dasarathy0.tripod.com


inequalities
trigonometry
inequalities unsolved

prove that if $ a,b,c \ge 2$ satisfy the condition $ a^2\plus{}b^2\plus{}c^2\equal{}abc\plus{}4$
then
$ a \plus{} b \plus{} c \plus{} ac \plus{} bc \ge 2\sqrt {(a \plus{} b \plus{} c \plus{} 3)({a^2} \plus{} {b^2} \plus{} {c^2} \minus{} 3)}$

[quote="Victory.US"]prove that if $ a,b,c \ge 2$ satisfy the condition $ a^2 \plus{} b^2 \plus{} c^2 \equal{} abc \plus{} 4$
then
$ a \plus{} b \plus{} c \plus{} ac \plus{} bc \ge 2\sqrt {(a \plus{} b \plus{} c \plus{} 3)({a^2} \plus{} {b^2} \plus{} {c^2} \minus{} 3)}$[/quote]
it's must be like this
$ \sum a \plus{} \sum ab\ge 2\sqrt {(a \plus{} b \plus{} c \plus{} 3)({a^2} \plus{} {b^2} \plus{} {c^2} \minus{} 3)}$
and this is my solution
assume that c=min{a,b,c}
let $ a \plus{} b \equal{} x$
the $ ab \equal{} \frac {x^2 \plus{} c^2 \minus{} 4}{2 \plus{} c}$
we can rewrite the ineq like this
$ \frac {x^2 \plus{} c^2 \minus{} 4}{2 \plus{} c} \plus{} cx \plus{} x \plus{} c\ge 2\sqrt {(\frac {x^2 \plus{} c^2 \minus{} 4}{2 \plus{} c}*c \plus{} 1)(c \plus{} x \plus{} 3)}$
which is equivalent to
$ \frac {(x^2 \minus{} 4c \minus{} 8)(2 \minus{} c \minus{} c^2 \plus{} x^2)^2}{(c \plus{} 2)^2}\ge 0$
Q.e.d

i post this problem and wnat it to be sloved by trigonometry method? does it have the solution use trigonomẻtry?


red_dog, I'm not sure you can do that without a good explanation.
A direct use of Newton's binomial would be
$ (n+2)^{n}=[(n-k+1)+(k+1)]^{n}=\sum_{i=0}^{n}\binom{n}{i}(n-k+1)^{n-i}(k+1)^{i}$,
since a dummy index should not be a variable already used.

and it's not the same expression, there is an extra $ (n-k+1)$


Suppose $ f: R^\plus{} \rightarrow R^\plus{}$ is a decreasing continuous function such that for all $ x,y \in R^\plus{}$,

$ f(x \plus{} y) \plus{} f(f(x) \plus{} f(y)) \equal{} f(f(x \plus{} f(y))) \plus{} f(y \plus{} f(x))$

Prove that $ f(f(x))\equal{}x$.

[quote="irantst"][size=150]سلام
سوال‌سختیه
خودت‌به‌چیزی‌نرسیدی؟
اینو‌فارسی‌گفتم‌تا‌بقیه‌در‌جریان‌نباشند[/size] :lol:[/quote]

سلام،نه نشد هر كاري كردم يه جايي گير ميكنه[size=150]..
مرسي از جواب سريعت.
اميرحسين[/size]

ليس لدي أي فكرة ما قلتموه

It`s Iran TST 1997 . If you really interested in see a solution I can post it.

[hide="@amparvardi"]asoone ha ! khodet mitooni hal koni. rasti mikhain az in joor harfa bezanid PM bezanid behtare , choon age in ja farsi type koni maro ba ARABaye bi pedar , heivan sefat eshtebah migiran  ...[/hide]


An elevator of mass 2000k.g rises with acceleration of 1m/s.What is the force with which the cable is puling the elevator?

You're correct.  Another way to think of this is to measure the acceleration against the gravational reference, which is accelerating towards the center of the earth at g.  If the elevator was "stationary" relative to the ground, it would have a F equal to the weight of the elevator (sort of the definition of weight).  One then merely has to add the F from the apparent acceleration.  You get the same formula, but this is just a different way to think about why.

Yes the above are correct.All we have to think is that to attain an accleration of $1m/s^2$,we cable of the elevator will have to apply a force to accelerate the elevator by $11m/s^2$(since we need a net acceleratuion of $1m/s^2$)


calculus integration function logarithms derivative algorithm Euler

Can some one show me how to evaluate this integral?

integral(e^[-x^2])dx.

Thanks

Ah, Jimmy. Is it time to let you in on some of our secrets?

Let's start with some terminology. We have certain "building block" functions: the powers of $x$, the exponential and the logarithm, the trigonometric and inverse trigonometric functions. Any function that we can created from these elements using a finite number of steps of addition, subtraction, multiplication, division, and composition, is called an "elementary function." By that definition, $e^{-x^2}$ is clearly an elementary function.

Theorem: the derivative of an elementary function is an elementary function. The proof of this theorem is by algorithm: we know the (elementary) derivatives of the building block functions, and we know how to combine them using the product rule, quotient rule, and chain rule.

The converse of that theorem is utterly false. There exist elementary functions whose antiderivatives (primitives) are not elementary. A notorious example of an elementary function with no elementary primitive is $e^{-x^2}.$ (This is known, but even hinting at the proof would take us too far afield here.)

In fact, your default assumption should be that if an elementary function doesn't obviusly have an elementary primitive, the chances if it having an elementary primitive at all are at best remote. This applies to simple-looking formulas like $(1+x^3)^{1/3}$ and to complicated messes.

But don't read too much into that. Does $e^{-x^2}$ have a primitive? Absolutely yes, as we can claim for all continuous functions. We can simply write that primitive as $\int_0^xe^{-t^2}dt.$ The integral exists because any continuous function is integrable on any finite interval.

Could we write the primitive as a power series? Yes, easily and usefully. Just write the power series for $e^{-x^2}$ and integrate it term by term. Could we estimate any $\int_a^be^{-x^2}dx$ as closely as we please? Certainly, using either the power series or some technique of numerical integration.

Can any definite integrals be computed exactly? In this case yes:

$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}.$

Thanks,
but could you show me where I could find int(e^[-x^2]) worked out analytically? It's driving me crazy

What exactly do you mean by "worked out analytically"? Are you referring to a particular definite integral? If that's what you mean, you should say so. Do you mean as an analytic function (that is, a power series)? If that's what you mean, you should say so. If you're looking for a (closed form) formula for the antiderivative, then you're asking for something that doesn't exist, and you should re-read what I wrote above.

Sorry for being unclear, I meant an analytic function.

But I should be able to figure this out on my own, I wasn't thinking too clearly earlier.
Thanks for your help

hey kent, you said that the primitive of e^(-x^2)dx is square root of (2*pi)?

I used my 89 to solve that primitive from negative infinity to infinity and i got 1.77 which is not the same as square root of (2*pi).

Can you tell me what i did wrong?

Actually there is a typo in Kents post.$\int_{-\infty}^{+\infty}{e^{-x^2}dx}=\sqrt{\pi}$

and how would you get a definite integral of this function?  seems like he just integrated it from -infiniti to infiniti

Dear [b]zubairkhan14[/b] have you read [b]post #7[/b]?

yes i did, but i dont understand it.  if you're given that integral, how do you come up with a definite integral for that.  is there a way to do it by substitution, trig, partial fractions, integ by parts?

Note that it is sufficent to calculate the integral $\int_{0}^{+\infty}{\frac{dx}{(1+x^2)^n}}$.Use trigonometric substitution $x=\tan{z}$ $z\in[0,\frac{\pi}{2}]$.You will come to the integral of type  $\int_{0}^{\frac{\pi}{2}}{{\sin{x}}^n dx}$.I think this one will be in you r calculus coursebook because it is famous integral.

[quote="Myth"][quote="LevonNurbekian"]As I remember the shortest way to count the Euler-Poisson integral is by using that $\lim_{n{\rightarrow}+\infty}{(1+\frac{x^2}{n})^{-n}}=e^{-x^2}$ and so writing
$\lim_{n{\rightarrow}+\infty}{\int_{0}^{+\infty}{(1+\frac{x^2}{n})^{-n}dx}}=\int_{0}^{+\infty}{e^{-x^2}dx}$(of course with testing some conditions).[/quote]

I thought the shortest path is the following:
$I^2=\int_{\mathbb R^2}e^{-x^2-y^2}dxdy=\int_0^{+\infty}\int_{0}^{2\pi}re^{-r^2}dr=-2\pi e^{-r^2}|_0^{+\infty}=2\pi$.  :D[/quote]
The set of your memory is larger than mine [b]Myth[/b].  :D


Hi, I was wondering if it is too late to get your school to sign up for Mandlebrot? I want to take it but it is hard to convince my school to do stuff like that.

im thinking about getting my school to do mandelbrot too. the only problem is that most of my friends already take it at this math program called icae, which i cannot attend. so im not sure if they can continue taking the exam there (which they will want to, given icae's sucess in this contest) or if they will be forced to take it our school. of course, if they can continue with icae, i will be the only one doing the team contest, which means our scores will be dismally low. :D

I'm really excited this year.  I founded "Math Club" at our school and convinced my math teacher to sign up for the Mandelbrot.  Ya.  Just thought I'd let you know.

As for the deadline, just check http://www.mandelbrot.org.  If needbe, you could send them a message using the "Contact Us" page.

P.S. Does anyone know what happened to Mandelbrot Midlevels?  I thought there used to be a competition called that (at http://www.midlevels.org) but it's not there...


Do there exist a function $f: R\to R$ such that for all $x,y\in R$
$f(x^{2}y+f(x+y^{2}))=x^{3}+y^{3}+f(xy)$?

Sorry, I can't understand what you mean.But ,this problem is quite easy.
for x=0:ff(y^2)=y^3+f(0)
for y=0:ff(x)=x^3+f(0)


Determine the units digit of [1020000/(10100 + 3)].

A hint: you might want to use the binomial theorem :D

'Het binomium van Newton' i.e. the expansion of (a + b)n.


the letters in the word $ RANDOM$ are arranged randomly. What is the probability that an arrangement chosen at random ends with a vowel?

[hide="Solution"]Number of ways to have it end with a vowel:
$ 2$ choices for the last letter, $ 5!$ ways to place the rest of the letters.

Total number of arrangements: $ 6!$.

Answer: $ \frac {2\cdot 5!}{6!} \equal{} \frac {2}{6} \equal{} \boxed{\frac {1}{3}}$.[/hide]

Why are there 4! ways to arrange the rest of the letters?

since, other than the last two letters, there are 4 letters left that we can arrange in any random order.

Sorry, I was initially multiplying by $ \binom{5}{1}$ before that, to choose a letter to be a vowel, then I realized that I could just be multiplying by $ 5!$, and then forgot to change it for some reason.

Or you could just realize that there are 6 total letters all different and two vowels, so the chance is automatically 2/6 or 1/3.


Any idea why the [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-tse/-pdf/tse01.pdf]USA TST 2001[/url] had 9 problems instead of 6? Assuming that wasn't a mistake, on the AoPS resources section, there are only 1-6, so have 7-9 been posted (just not added to the resources section)?

Because there was a 3-way tie for the 6th position on the team after the first 6 problems.


I get this asymptote error when running the following script

[code]size(150);
unitsize(1cm);
draw((-7,-7)--(7,7));[/code]

on Windows Vista x64, Asymptote 1.77

Error I get
[code]C:\tmp>asy -verbose temp.asy
Processing temp
temp.asy: 1.1: invalid token 'ï'
temp.asy: 1.2: invalid token '»'
temp.asy: 1.3: invalid token '¿'[/code]

Any ideas?

It seems like your program file has encoding different from the standard ASCII, so what your editor shows as "siz" is perceived as a strange chain of fancy symbols. What editor did you use to create the file?

It's PSPAD. But I'm using normal settings, the file looks the same in Notepad (simple notepad), Texnic center and so on ...

Are you still able to compile other .asy files? Have you tried to rename the file? (i.e., are you absolutely sure that test.asy that you created is the same test.asy as the one being compiled?).

Nevermind you're right. The file was UTF-8 encoded, and it needs to be ANSII ...


geometry
incenter
inradius
trigonometry
geometry solved

This problem is making me some sirious troubles....

Let $I$ be the incentre of $\triangle ABC$, $r$ inradius, $M$ the midpoint of side $AB$ and $N$ the midpoint of $CM$. Prove that if $r=CN-IN$ then $AC=BC$ or $\angle ACB=90^o$.

Who solve it first get's a CANDY....    :)  :)

[quote="lomos_lupin"]Let me say the problem in another word.  :)
Let the $m_a$ be a median of the triangle $ABC$(from $A$ to $BC$).
We know that the circle with diameter $m_a$ is tangent to Incircle.
Prove that $A=90$ or $AB=AC$[/quote]

Well....... that is not so helpful....
But, do you know what wloud be very helpful....

$SOLUTION$  :)  :)  :)  :)  :)  :)

[quote="lomos_lupin"]We know that the circle with diameter $m_a$ is tangent to Incircle.
[/quote]
The  circle with diameter $m_a$ also passes through $D$, the foot of the altitude from $A$ (I'm using the notations of lomos_lupin). If $AB\ne AC$ then $D$ and $M$ are distinct. It is not difficult to see that considering the pencil of circles passing through $D$ and $M$, only one is tangent to the incircle (having the incircle in its interior). But it is well known that the nine point circle is tangent to the incircle (and, of course, passes through $D$ and $M$). Therefore, the circle with diameter $m_a$ coincides with the nine point circle. This means that $A$ is the orthocenter of $ABC$, that is, $\angle A=90^{\circle}$.

Do you see lupin...

that is the $SOLUTION$
tnx

Just kidding, lupin did a lot of deal with his "let's put the problem in another word..."

 :)  :)  :)

m@re

Thank God! This problem killed me. Thanks, Sir Enescu (Bogdan ?) and congratulations !

Actually, thanks are due to lomos_lupin. His rewording of the problem gave me the idea.

nice solution ,enescu.

To M@re:Yup i didnt gave the solution ,but just a simple hint.
The real work is for enescu
I think it was a nice problem.
By the way
[quote="M@re"]

Who solve it first get's a CANDY....    :)  :)[/quote]
,Enescu wants a candy :D ?

Still once congratulations ( :first: - the first candy) for the his solution, to [i]Bogdan Enescu[/i],
and ( :first: - the second candy) for the his thought of the tangent, to my friend
(from this moment) [i]lomos lupin[/i] (and no Arsene, he knows why) !.
Here is (now it is easily) a variation on the same theme (lupin's notes):
$AN-IN=r\Longleftrightarrow I_a N-AN=r_a\Longleftrightarrow b=c\ or\ A=90^{\circ}.$

It is ready ! At last I managed to prove (with the Lupin's notations) that:\[AN-IN=r\Longleftrightarrow I_aN-AN=r_a\Longleftrightarrow \frac {(s-a)}{r}=\frac {2m_a}{a}\Longleftrightarrow\]\[m_a\cdot h_a=s(s-a)\Longleftrightarrow b=c \ \vee\ A=90^{\circ}\][b]Remark.[/b] $S=\frac 12 ah_a=sr=s(s-a)\tan \frac A2 .$

You point out if you should like to present the its solution. All right !


Ειναι το ακολουθο γραμμικα ανεξαρτητο ,ναι η οχι και γιατι:

{$ e^x,e^{2x}$}

Οι συναρτήσεις $ e^x,e^{2x}$ είναι Γ. ανεξάρτητες (ή με άτοπο ή με ορίζουσα...)

Εχω ενα προβλημα με το παραπανω και το παραθετω¨

Εξ ορισμου το παραπανω ειναι γραμ ανεξαρτητο εαν >

για ολα τα a,b,x και $ ae^x \plus{} be^{2x}\equal{}0$,τοτε a=b=0 και σε μορφη ποσοδεικτων εχουμε:

$ \forall a\forall b\forall x[ (a\in\Re\wedge b\in\Re\wedge x\in\Re\wedge ae^x \plus{} be^{2x} \equal{} 0)\Longrightarrow a\equal{}b\equal{}0]$..........................................................................................1

Tωρα ,η αρνηση της (1) [b]η οποια και συνεπαγεται γραμμικη εξαρτηση[/b] ειναι΄

[b] υπαρχουν a,b,x[/b] ετσι ωστε $ ae^x \plus{} be^{2x} \equal{} 0$ [b] KAI [/b] $ a\neq 0,b\neq 0$

[b]Επομενως εαν βαλλουμε : a=2,b=-2 ,x=0,εχουμε $ ae^x \plus{} be^{2x} \equal{}0$ και ετσι ικανοποιουμε την αρνηση της (1) και αρα ΕΧΟΥΜΕ ΓΡΑΜΜΙΚΗ ΕΞΑΡΤΗΣΗ[/b]

[quote="τρικλινο"]
για ολα τα a,b,x και $ ae^x \plus{} be^{2x} \equal{} 0$,τοτε a=b=0 και σε μορφη ποσοδεικτων εχουμε:[/quote]

Η σωστή διατύπωση είναι: οι $ e^x, e^{2x}$ λέγονται γραμμικά ανεξάρτητες τότε και μόνο τότε όταν για κάθε $ a,b\in\mathbb{R}$ τέτοια ώστε $ ae^x \plus{} be^{2x} \equal{} 0\,\forall x\in\mathbb{R}$, ισχύει $ a \equal{} b \equal{} 0$.

Από τη στιγμή που δεν ισχύει $ 2e^x \minus{} 2e^{ 2 x} \equal{} 0\,\forall x\in\mathbb{R}$, το αντιπαράδειγμα που έδωσες δεν είναι σωστό.

Cheerio,

Durandal 1707

[quote="Durandal"][quote="τρικλινο"]
για ολα τα a,b,x και $ ae^x \plus{} be^{2x} \equal{} 0$,τοτε a=b=0 και σε μορφη ποσοδεικτων εχουμε:[/quote]

Η σωστή διατύπωση είναι: οι $ e^x, e^{2x}$ λέγονται γραμμικά ανεξάρτητες τότε και μόνο τότε όταν για κάθε $ a,b\in\mathbb{R}$ τέτοια ώστε $ ae^x \plus{} be^{2x} \equal{} 0\,\forall x\in\mathbb{R}$, ισχύει $ a \equal{} b \equal{} 0$.

[/quote]

Ποια ειναι η αρνηση του ορισμου σου??

ΕΑΝ γραψεις τον ορισμο σου σε συμβολα θα ηταν ποιο ευκολο για την αρνηση

[quote="τρικλινο"]ΣΥΓΝΩΜΗ θελω να πω ,οτι εαν γραψεις τον ορισμο σου σε συμβολα τοτε θα δεις οτι δεν διαφερει απο τον δικο μου ,επομενως η αρνηση σου ειναι λαθος[/quote]

Nope. Δεν αγαπάω ιδιαίτερα τους ποσοδείκτες και κάτα πάσα πιθανότητα δε θα το γράψω σωστά (οπότε Dem, please correct me), αλλά ο ορισμός μου σε σύμβολα είναι:

$ \forall a \forall b\left[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge(ae^x \plus{} be^{2x} \equal{} 0)]] \implies (a \equal{} b \equal{} 0) \right]$

Edit επεξήγησης: αν $ a,b\in\mathbb{R}$ και $ ae^x \plus{} be^{2x} \equal{} 0, \forall x\in\mathbb{R}$ τότε $ a\equal{}b\equal{}0$. Ελπίζω τώρα να καταλαβαίνεις καλύτερα γιατί οι δύο ορισμοί είναι διαφορετικοί...

Χμμμ, δεν μπορώ να πω ότι καταλαβαίνω γιατί επιμένεις τόσο να γράφεις τα πράγματα με ποσοδείκτες, αλλά περί ορέξεως... ;)

Cheerio,

Durandal 1707

ΓΙΑΤΙ ΜΟΝΟ εαν γραψεις  μια προταση σε ποσοδεικτες σωστα ,τοτε μπορεις να αποδειξεις την αρνηση της προτασεως  .Διαφορετικα [b]μπορει[/b] να κανεις λαθος.

Αλλα ας περιμενουμε να δουμε αν συμφωνει και ο Δημητρης και συνεχιζουμε¨.

[quote="τρικλινο"]ΓΙΑΤΙ ΜΟΝΟ εαν γραψεις  μια προταση σε ποσοδεικτες σωστα ,τοτε μπορεις να αποδειξεις την αρνηση της προτασεως  .Διαφορετικα [b]μπορει[/b] να κανεις λαθος.[/quote]

Sure, whatever. Αλλά τουλάχιστον σιγουρέψου ότι ξεκινάς από το σωστό ορισμό πριν τον γράψεις σε ποσοδείκτες... το λέω αυτό γιατί ο ορισμός που έδωσες:

[quote="τρικλινο"]για ολα τα a,b,x και $ ae^x \plus{} be^{2x} \equal{} 0$, τοτε a=b=0 και σε μορφη ποσοδεικτων εχουμε:
[...][/quote]

δεν είναι καν ο σωστός ορισμός της γραμμικής ανεξαρτησίας. Με τον ορισμό αυτό όλες οι συναρτήσεις θα έβγαιναν γραμμικά εξαρτημένες ;)

Τέλος, όταν έχεις μια απορία είναι γενικά καλύτερο για τη λειτουργία του forum να τη διατυπώνεις ως έχει από την αρχή, και να μην περιμένεις posts άλλων για να μετατρέψεις μία ερώτηση ανάλυσης σε ερώτηση μαθηματικού φορμαλισμού. Για να ξέρουμε και ποιά είναι η πραγματική σου ερώτηση στο κάτω-κάτω...

My tuppence anyway,

Durandal 1707

Έστω ότι είναι γρ.εξαρτημένα τότε
Θα υπάρχει $ m\in R$(ανεξάρτητο του x)$ : e^{2x}\equal{}me^x$για κάθε $ x\in R$ ή
$ : e^x\equal{}m$για κάθε $ x\in R$
Παραγωγίζοντας προκύπτει το άτοπο

[quote="τρικλινο"]
Αλλα ας περιμενουμε να δουμε αν συμφωνει και ο Δημητρης και συνεχιζουμε¨.[/quote]

Χωρίς να διεκδικώ κανένα αλάθητο.

Με λόγια: Δυο συναρτήσεις $ f,g: \mathbb{R} \to \mathbb{R}$ είναι γραμμικώς εξαρτημένες αν και μόνο αν υπάρχουν $ a,b \in \mathbb{R}$ που δεν είναι και τα δύο ίσα με 0 ώστε $ af(x) \plus{} bg(x) \equal{} 0$ για κάθε $ x \in \mathbb{R}$.

Με ποσοδείκτες

$ \exists a \in \mathbb{R} \exists b \in \mathbb{R} ((\neg(a\equal{}0 \wedge b\equal{}0)) \wedge \forall x\in \mathbb{R} (af(x) \plus{}bg(x)\equal{}0 ))$

(Ο ορισμός του Παναγιώτη μου φαίνεται ισοδύναμος)

H άρνηση

$ \forall a \in \mathbb{R} \forall b \in \mathbb{R} ((\neg(a\equal{}0 \wedge b\equal{}0) \Rightarrow \exists x \in \mathbb{R} (af(x) \plus{} bg(x) \neq 0))$

Όπου $ \exists x \in \mathbb{R} P(x)$ σημαίνει $ \exists x (x \in \mathbb{R} \wedge P(x))$ και $ \forall x \in \mathbb{R} P(x)$ σημαίνει $ \forall x (x \in \mathbb{R} \Rightarrow P(x))$

Αν το πρόβλημα είναι στα λόγια, να προσθέσω ότι σκεφτόμαστε τις συναρτήσεις σαν σημεία σε διανυσματικό χώρο (pointless space είχε πει κάποτε ο Halmos) και φυσικά σε αυτόν τον χώρο μια συνάρτηση είναι ίση με 0 αν και μόνο αν παίρνει την τιμή 0 σε όλα τα σημεία (και όχι μόνο σε τουλάχιστον ένα) όπου ορίζεται.

[quote="Durandal"][quote="τρικλινο"]ΣΥΓΝΩΜΗ θελω να πω ,οτι εαν γραψεις τον ορισμο σου σε συμβολα τοτε θα δεις οτι δεν διαφερει απο τον δικο μου ,επομενως η αρνηση σου ειναι λαθος[/quote]

Nope. Δεν αγαπάω ιδιαίτερα τους ποσοδείκτες και κάτα πάσα πιθανότητα δε θα το γράψω σωστά (οπότε Dem, please correct me), αλλά ο ορισμός μου σε σύμβολα είναι:

$ \forall a \forall b\left[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge(ae^x \plus{} be^{2x} \equal{} 0)]] \implies (a \equal{} b \equal{} 0) \right]$

Edit επεξήγησης: αν $ a,b\in\mathbb{R}$ και $ ae^x \plus{} be^{2x} \equal{} 0, \forall x\in\mathbb{R}$ τότε $ a \equal{} b \equal{} 0$. Ελπίζω τώρα να καταλαβαίνεις καλύτερα γιατί οι δύο ορισμοί είναι διαφορετικοί...

Χμμμ, δεν μπορώ να πω ότι καταλαβαίνω γιατί επιμένεις τόσο να γράφεις τα πράγματα με ποσοδείκτες, αλλά περί ορέξεως... ;)

Cheerio,

Durandal 1707[/quote]

Mιας και ο ΔΗΜΗΤΡΗΣ θεωρει την φορμουλα σου σωστη, ας παρουμε την αρνηση  [b] της φορμουλα σου[/b] ,ετσι εχουμε.

$ \neg \forall a\forall b[(a\in R)\wedge(b\in R)\wedge[\forall x[x\in R\wedge(ae^x \plus{} be^{2x}\equal{}0)]]\Longrightarrow (a\equal{}b\equal{}0)]$  η οποια ειναι ισοδυναμη με

$ \exists a\exists b${$ \neg[(a\in R)\wedge(b\in R)\wedge[\forall x[x\in R\wedge (ae^x \plus{} be^{2x}\equal{}0)]]\wedge a\neq 0,b\neq 0$}..........................................................................1

Αλλα $ \neg[(a\in R)\wedge(b\in R)\wedge[\forall x[x\in R\wedge (ae^x \plus{} be^{2x}\equal{}0)]]$ ειναι ισοδυναμη με

$ ((a\in R)\wedge(b\in R))\Longrightarrow\neg[\forall x[x\in R\wedge(ae^x \plus{} be^{2x}\equal{}0)]]$...................................................................................................................................2

Αλλα $ \neg[\forall x[x\in R\wedge(ae^x \plus{} be^{2x}\equal{}0)]]$ ειναι ισοδυναμη με

$ \exists x[x\in R\Longrightarrow (ae^x \plus{} be^{2x}\neq 0)]$...............................................................................................................................................................................3

Τωρα εαν αντικαταστησουμε (3) στην (2) και την φορμουλα που θα προκυψει εις την (1) εχουμε.

$ \exists a\exists b[((a\in R)\wedge(b\in R))\Longrightarrow\exists x[x\in R\Longrightarrow(ae^x \plus{} be^{2x}\neq 0)]\wedge (a\neq 0,b\neq 0)]$

ΑΥΤΗ η φορμουλα μας λεει:

Υπαρχουν a,b ετσι ωστε εαν a,b ανοικουν στους πραγματικους αριθμους τοτε ,υπαρχει χ το οποιον ανηκει στους πραγματικους αριθμους,το οποιον συνεπαγεται $ ae^x \plus{} be^{2x}\neq 0$, και $ a\neq 0,b\neq 0$

και ετσι εαν παρουμε a=1 ,b=6 and x=2,θα εχουμε $ e^2\plus{} 6e^4\neq 0$ και $ a\neq 0,b\neq 0$

ΑΡΑ ΕΧΟΥΜΕ γραμ. εξαρτηση

Και ας ΧΡΗΣΙΜΟΠΟΙΗΣΟΥΜΕ την φορμουλα την οποια εγραψα στο ποστ #3 ,για να αποδειξουμε γραμ.ανεξαρτησια>

Εαν $ ae^x \plus{}be^{2x} \equal{}0$ =====> $ e^x(a \plus{}be^x) \equal{}0$ ====> $ a \plus{}be^x \equal{}0$ ====>$ \frac{d( a \plus{} be^x)}{dx} \equal{}0$ ====>$ be^x \equal{} 0$ ====> b=0=a

κια εαν γενικευσουμε $ \forall a\forall b\forall x[ ae^x \plus{} be^{2x} \equal{}0\Longrightarrow a\equal{}b\equal{}0]$

[quote="Durandal"][quote="τρικλινο"]
$ \neg \forall a\forall b[(a\in R)\wedge(b\in R)\wedge[\forall x[x\in R\wedge(ae^x \plus{} be^{2x} \equal{} 0)]]\Longrightarrow (a \equal{} b \equal{} 0)]$  η οποια ειναι ισοδυναμη με

$ \exists a\exists b${$ \neg[(a\in R)\wedge(b\in R)\wedge[\forall x[x\in R\wedge (ae^x \plus{} be^{2x} \equal{} 0)]]\wedge a\neq 0,b\neq 0$}[/quote]

Nope, δεν έβαλες την άρνηση $ \neg$ στο σωστό σημείο. Η άρνηση του ορισμού είναι (εκτός και κάνω κάπου κάποιο typo):

$ \exists a \exists b \big[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge (ae^x \plus{} be^{2x} \equal{} 0)]]\wedge \neg(a \equal{} b \equal{} 0) \big]$

(η οποία συμφωνεί ακριβώς και με την άρνηση που έγραψε ο Dem)

Τελικά ακόμα δεν έχω καταλάβει: έχεις κάποια genuine απορία ή απλώς προσπαθείς να δείξεις ότι υπάρχει κάποιο πρόβλημα στον ορισμό της γραμμικής εξάρτησης συναρτήσεων?

Cheerio,

Durandal 1707[/quote]

ΣΕ ποια γραμμη της αποδειξεως μου εχω κανει λαθος

[quote="Durandal"][quote="τρικλινο"]ΣΕ ποια γραμμη της αποδειξεως μου εχω κανει λαθος[/quote]

Στη δεύτερη. Δεν κοίταξα πιο πέρα γιατί δεν είχε νόημα.

Cheerio,

Durandal 1707[/quote]

  Ποιο ειναι το λαθος

[quote="Durandal"][quote="τρικλινο"]Ποιο ειναι το λαθος[/quote]

Δεν έκανες σωστά την άρνηση της συνεπαγωγής.

Cheerio,

Durandal 1707[/quote]

[b]ΠΟΛΥ ΣΩΣΤΑ[/b].

[b]ΟΜΩΣ[/b] εχουε το εξης [b]ΠΕΡΙΕΡΓΟ[/b]. Η φορμουλα που εγραψα στο ποστ #3 και η οποια ειναι¨

$ \forall a\forall b\forall x[((a\in R)\wedge(b\in R)\wedge(x\in R)\wedge(ae^x \plus{} be^{2x}\equal{}0)) \Longrightarrow a\equal{}b\equal{}0]$ [b] συνεπαγεται[/b] την φορμουλα σου οπως αποδεικνυεται παρακατω.

Υποτεθεστω οτι $ (a\in R)\wedge(b\in R)\wedge\forall x[(x\in R)\wedge(ae^x \plus{} be^{2x} \equal{} 0)]$ .ΑΥΤΟ συνεπαγεται $ (a\in R)\wedge(b\in R)\wedge(x\in R)\wedge(ae^x \plus{} be^{2x} \equal{}0)$

Kαι συμφωνα με την παραπανω φορμουλα (δικη μου) εχουμε  a=b=0.

ΚΑΙ γενικευοντας εχουμε την δικη σου φορμουλα.

$ \forall a\forall b[(a\in R)\wedge(b\in R)\wedge[\forall x[(x\in R)\wedge(ae^x \plus{} be^{2x} \equal{}0 )]]\Longrightarrow (a\equal{}b\equal{}0)]$

[b]ΣΥΝΕΠΩΣ η φορμουλα μου ειναι γενικωτερη της δικη σου[/b]

Η αρνηση ομως της φορμουλας μου ειναι αποδεικτεα και επομενως  [b] συνεπαγεται εξαρτηση των συναρτησεων [/b]  {$ e^x,e^{2x}$ }

[b] Αρα και η αρνηση της φορμουλα σου[/b] θα πρεπει να συνεπαγεται εξαρτηση των συναρτησεων. Απλα  δεν εχουμε βρει την λυση ακομα

Εκτος και εαν καποιος γραψει μια φορμαλ αποδειξη για οποιαδηποτε φορμουλα και επομενως ολα τα αλλα καταρυφθουν

Βρε συ, σκέψου λίγο τι λες λογικά :P

Έχουμε γράψει δύο ορισμούς για τη γραμμική εξάρτηση: τον δικό σου (Α) και τον "δικό" μου (Β).
(BTW, άσκηση: δες ποιές είναι οι συναρτήσεις που ικανοποιούν το δικό σου ορισμό Α... θα δεις ότι είναι πάρα πολύ λίγες)

Μετά, λες πολύ σωστά το εξής: αν δύο συναρτήσεις ικανοποιούν τον (Α), τότε ικανοποιούν και τον (Β). Αντιθέτως, υπάρχουν συναρτήσεις που ικανοποιούν τον (Β) αλλά όχι τον (Α) - οι $ e^x, e^{2x}$ είναι ένα τέτοιο παράδειγμα.
(Εναλλακτικά, έστω $ A$ το σύνολο των (ζευγών) συναρτήσεων που ικανοποιούν τον (Α) και έστω $ B$ το σύνολο αυτών που ικανοποιούν τον (Β). Προφανώς, $ A\subseteq B$ αλλά, το inclusion είναι proper.)

Έτσι, αν δείξεις ότι δύο συναρτήσεις δεν ικανοποιούν τον (Α), δεν έχεις δείξει τίποτα για το αν ικανοποιούν ή όχι τον (Β). Στη βιβλιογραφία, ο ορισμός που κανείς χρησιμοποιεί είναι ο Β και όχι ο δικός σου ο Α. Με βάση αυτόν τον ορισμό (B) δεν υπάρχει κανένα πρόβλημα, ελπίζω πια να το καταλαβαίνεις αυτό.

Σε κάθε περίπτωση, μου φαίνεται ότι διϋλίζεις τον κώνωπα και καταπίνεις την κάμηλο: καλοί οι ποσοδείκτες, δε λέω, αλλά βλέπω ότι η χρήση τους καθόλου δε σε προστατεύει από το να κάνεις σφάλματα λογικής. Γι'αυτό, θα σου πρότεινα πρώτα να βεβαιωθείς ότι είναι σωστή η λογική δομή των επιχειρημάτων σου, και μετά προσπάθησε να τα μεταφέρεις σε ποσοδείκτες.

Cheerio,

Durandal 1707

[quote="Durandal"]

Μετά, λες πολύ σωστά το εξής: αν δύο συναρτήσεις ικανοποιούν τον (Α), τότε ικανοποιούν και τον (Β). Αντιθέτως, υπάρχουν συναρτήσεις που ικανοποιούν τον (Β) αλλά όχι τον (Α) - οι $ e^x, e^{2x}$ είναι ένα τέτοιο παράδειγμα.

Cheerio,

Durandal 1707[/quote]

Για να δουμε

Εστω:

a,b,x ανοικουν στους πραγματικους αριθμους, και $ ae^x \plus{} be^{2x} \equal{}0$,αλλα:

$ ae^x \plus{} be^{2x}\equal{}0\Longrightarrow a \plus{}be^x\equal{}0\Longrightarrow\frac{d(a \plus{} be^x)}{dx} \equal{} 0\Longrightarrow be^x \equal{} 0\Longrightarrow a\equal{}b\equal{}0$

[b]Kαι γενικευοντας εχουμε[/b].

$ \forall a\forall b\forall x[ a,b,x\in R\wedge (ae^x \plus{} be^{2x} \equal{}0)\Longrightarrow a\equal{}b\equal{}0]$

Aρα η φορουλα μου (Α) ικανοποιει την ανεξαρτησια των {$ e^x,e^{2x}$} συναρτησεων

[b]Η υπαρχει λαθος στην παραπανω αποδειξη??[/b]

Τρίκλινο (real name btw?), sorry αλλά έχεις κάνει πολύ σημαντικό λάθος που δεν έχει καθόλου να κάνει με τη χρήση ποσοδεικτών. Επίσης, από εσένα που κυρήττεις τόσο έντονα την αυστηρή λογική και θεμελίωση, θα περίμενα πολύ περισσότερη προσοχή...

[quote]Εστω:

a,b,x ανοικουν στους πραγματικους αριθμους, και $ ae^x \plus{} be^{2x} \equal{} 0$,αλλα:

$ ae^x \plus{} be^{2x} \equal{} 0\Longrightarrow a \plus{} be^x \equal{} 0\Longrightarrow\frac {d(a \plus{} be^x)}{dx} \equal{} 0\Longrightarrow be^x \equal{} 0\Longrightarrow a \equal{} b \equal{} 0$
[/quote]

Το ότι δέχεσαι ότι υπάρχουν κάποια $ a,b,x$ ώστε $ ae^x \plus{} be^{2x} \equal{} 0$, επ'ουδενεί δε σημαίνει ότι αυτή η σχέση ισχύει σε μία περιοχή του $ x$ ώστε να μπορείς την [i]παραγωγίσεις[/i] ώστε να καταλήξεις ότι $ be^x \equal{} 0$.

Δηλαδή, έλεος, σκέψου λίγο και ένα παράδειγμα: αν $ a \equal{} 1, b \equal{} \minus{} 1, x \equal{} 0$, προφανώς έχουμε $ ae^x \plus{} be^{2x} \equal{} 0$, αλλά δεν έχουμε $ be^x \equal{} 0$, πόσο μάλλον $ a \equal{} b \equal{} 0$... αυτό δε σου χτυπάει καμπανάκια για την "απόδειξή" σου? :what?:

Ξαναλέω... κατάλαβε πρώτα την πραγματική λογική και το τι λένε οι έννοιες που χρησιμοποιείς (π.χ. η γρ. ανεξαρτησία), και μετά ασχολήσου με τους ποσοδείκτες και τις "ελλείψεις" της αυστηρής θεμελίωσης της ανάλυσης. Sorry, αλλά αυτό που κάνεις δε νομίζω ότι είναι ιδιαίτερα παραγωγικό...

Cheerio,

Durandal 1707

[quote="Durandal"]

Δηλαδή, έλεος, σκέψου λίγο και ένα παράδειγμα: αν $ a \equal{} 1, b \equal{} \minus{} 1, x \equal{} 0$, προφανώς έχουμε $ ae^x \plus{} be^{2x} \equal{} 0$, αλλά δεν έχουμε $ be^x \equal{} 0$, πόσο μάλλον $ a \equal{} b \equal{} 0$... αυτό δε σου χτυπάει καμπανάκια για την "απόδειξή" σου? :what?:

Cheerio,

Durandal 1707[/quote]

Γιατι, το ιδιο δεν συμβαινει και με την δικη σου φορμουλα??

Η φορμουλα σου ειναι.

$ \forall a \forall b\left[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge(ae^x \plus{} be^{2x} \equal{} 0)]] \implies (a \equal{} b \equal{} 0) \right]$

Εαν λοιπον βαλουμε : a=1 ,b=-1 ,x=0 εχουμε:

$ ae^x \plus{} be^{2x} \equal{} 0$, αλλά δεν έχουμε $ be^x \equal{} 0$, πόσο μάλλον $ a \equal{} b \equal{} 0$

χριστος

[quote="τρικλινο"]Η φορμουλα σου ειναι.
$ \forall a \forall b\left[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge(ae^x \plus{} be^{2x} \equal{} 0)]] \implies (a \equal{} b \equal{} 0) \right]$

Εαν λοιπον βαλουμε : a=1 ,b=-1 ,x=0 εχουμε:

$ ae^x \plus{} be^{2x} \equal{} 0$, αλλά δεν έχουμε $ be^x \equal{} 0$, πόσο μάλλον $ a \equal{} b \equal{} 0$[/quote]

Gaaaah, έλεος, όχι!! Ας πάμε βήμα-βήμα μπας και τελειώσουμε ποτέ:
1. Διαλέγεις α=1. ΟΚ
2. Διαλέγεις β=-1. ΟΚ
3. Μπαίνουμε στην αγκύλη.
4. $ a,b\in\mathbb{R}$, OK.
5. Μετά, για την άρνηση του ορισμού θέλεις: $ ae^x \plus{} be^{2x} \equal{} 0$ [b][i][size=150]ΓΙΑ ΚΑΘΕ[/size][/i][/b] $ x\in\mathbb{R}$. Γι'αυτό έχει μπει σε ολόκληρη αγκύλη αυτό το κομμάτι. Εσύ έχεις $ ae^x \plus{} be^{2x} \equal{} 0$ ΜΟΝΟ για $ x \equal{} 0$. Για $ x \equal{} 1$ δεν βγάζεις 0, άρα δεν ικανοποιείται η άρνηση του ορισμού.

Αν δεν καταλαβαίνεις τη διαφορά, δε νομίζω ότι μπορώ να σε βοηθήσω άλλο.

Cheerio,

Durandal 1707

Η  φορμουλα σου ειναι:

$ \forall a \forall b[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge(ae^x + be^{2x} = 0)]] \implies (a = b = 0) ]$;

Aφου η φορμουλα σου ισχυει για ολα τα  a , b θα ισχυει και για α=1 και b = -1

Αντικαθιστουμε λοιπον στην φορμουλα σου και εχουμε:

$ (1\in R)\wedge(-1\in R)\wedge[\forall x[(x\in R)\wedge(e^x - e^{2x} = 0)]]\implies (1=-1=0)$

Tωρα η δεξια μερια αυτης της συνεπαγωγης ειναι ψευδης.

Αλλα για να ειναι η συνεπαγωγη αληθης πρεπει και η αριστερη μερια να ειναι ψευδης,γιατι ψευδες συνεπαγεται ψευδες ,θα κανει την συνεπαγωγη αληθη

Αλλα $ 1\in R$ EINAI αληθες,επισης $ -1\in R$ ειναι αληθες επομενως το

$ [\forall x[(x\in R)\wedge(e^x - e^{2x} = 0)]]$ πρεπει να ειναι ψευδες

[b]ΠΩΣ ΕΙΝΑΙ ΔΥΝΑΤΟΝ ΤΟ $ [\forall x[(x\in R)\wedge(e^x - e^{2x} = 0)]]$ ΝΑ ΕΙΝΑΙ ΨΕΥΔΕΣ ΓΙΑ ΟΛΑ ΤΑ x ΟΤΑΝ ΓΙΑ x = 0 ΕΧΟΥΜΕ $ e^0 - e^0 =0$ ΤΟ ΟΠΟΙΟΝ ΕΙΝΑΙ ΑΛΗΘΕΣ???[/b]

Επομενως οτι συμβαινει με την φορμουλα μου συμβαινει συμβαινει και με την δικη σου

Ασε την αρνηση εδω δεν ικανοποιειται καν η φορμουλα η ιδια για ορισμενες τιμες

[quote="τρικλινο"][b]ΠΩΣ ΕΙΝΑΙ ΔΥΝΑΤΟΝ ΤΟ $ [\forall x[(x\in R)\wedge(e^x \minus{} e^{2x} \equal{} 0)]]$ ΝΑ ΕΙΝΑΙ ΨΕΥΔΕΣ ΓΙΑ ΟΛΑ ΤΑ x ΟΤΑΝ ΓΙΑ x = 0 ΕΧΟΥΜΕ $ e^0 \minus{} e^0 \equal{} 0$ ΤΟ ΟΠΟΙΟΝ ΕΙΝΑΙ ΑΛΗΘΕΣ???[/b][/quote]

Για να είναι η πρόταση $ \forall x[(x\in\mathbb{R})\wedge(e^x \equal{} e^{2x})]$ ψευδής, αρκεί απλώς να υπάρχει ένα $ x\in\mathbb{R}$ που να μην ικανοποιεί τη σχέση μέσα στην αγκύλη:
$ \neg[\forall x[(x\in\mathbb{R})\wedge(e^x \equal{} e^{2x})]] \equal{} \exists x[(x\in\mathbb{R})\wedge(e^x\neq e^{2x})]$

Νομίζω αυτό το διδάσκουν κάποια στιγμή στο λύκειο ή στο γυμνάσιο (όχι με ποσοδείκτες, νομίζω κοινή λογική το λένε στο σχολείο), αλλά μπορεί και να κάνω λάθος.

Επίσης, όσο benefit of doubt και να κρατήσω, είναι πια προφανές ότι έχεις agenda εδώ πέρα, οπότε sorry, απλώς δεν έχω άλλο χρόνο/υπομονή να ασχοληθώ με αυτό το θέμα. Αν ακόμα δεν καταλαβαίνεις τι γίνεται (που δεν είμαι σίγουρος), το καλύτερο μέρος για να έχεις αυτήν τη συζήτηση βρίσκεται εδώ:

[url=http://annals.princeton.edu/annals/about/cover/cover.html]Annals of Mathematics[/url]

Επίσης, είμαι σίγουρος ότι o Josiah Carberry (καθηγητής στο Brown), θα χαρεί να σε βοηθήσει :cool: :evilgrin:

Cheerio,

Pan

geia sas!syggnomi pou epemvaino,alla an k den eimai foititis akomi thelo n rotiso an mporei n teleiosei i askisi me tin orizousa tou wronski..me voithaei kapoios???efxaristo..

Ναι τελειώνει και με την Wronski

ΝΟΜΙΖΩ οτι οι παραπανω συναρτησεις ειναι ανεξαρτητες για ολες τις τιμες [b]εκτος για x=0[/b].

Γιατι για χ=0 [b]δεν υπαρχει φορμουλα που να ικανοποιει την ανεξαρτησια τους[/b]

[quote="τρικλινο"]ΝΟΜΙΖΩ οτι οι παραπανω συναρτησεις ειναι ανεξαρτητες για ολες τις τιμες [b]εκτος για x=0[/b].[/quote]

Δύο συναρτήσεις είναι είτε γραμμικά ανεξάρτητες, είτε γραμμικά εξαρτημένες, δεν υπάρχει η έννοια της γραμμικής ανεξαρτησίας σε κάποιο [i]σημείο[/i].

Το "ανεξάρτητες για όλες τις τιμές εκτός για $ x \equal{} 0$" που λες είναι δικό σου συμπέρασμα (και δείχνει ότι δεν έχεις καταλάβει την έννοια της γραμμικής ανεξαρτησίας), οπότε μην μπερδεύεις όσους μπορεί να έχουν genuine απορίες πάνω στο ζήτημα. Γιατί αλλιώς, όλες οι συναρτήσεις θα ήταν γραμμικά εξαρτημένες σε κάποιο σημείο τους και δε θα είχε νόημα να μιλάμε για γραμμική ανεξαρτησία.

Αν όντως νιώθεις μπερδεμένος ακόμα, drop it και πάμε σε pm's. Αλλιώς, stop trolling, γιατί έχει κουραστεί το θέμα...

Cheerio,

Durandal 1707

ΤΟΤΕ ΑΣ γραψει καποιος μια φορμουλα που να οριζει την ανεξαρτησια των συναρτησεων

Η δικη σου φορμουλα δεν ισχυει ,οπως σου απεδειξα..

Αλλα το ξαναγραφω εδω για να δουμε ποιος μπερδευεται.

 Η φορμουλα σου ειναι¨

$ \forall a \forall b\left[ (a\in\mathbb{R})\wedge (b\in\mathbb{R})\wedge [\forall x[(x\in\mathbb{R})\wedge(ae^x \plus{} be^{2x} \equal{} 0)]] \implies (a \equal{} b \equal{} 0) \right]$;

Για a=1, b=-1,x=0 η φορμουλα σου γινεται:

$ [((1\in R)\wedge(\minus{}1\in R)\wedge(0\in R)\wedge( e^O \minus{}e^0\equal{}0))\Longrightarrow (1\equal{}\minus{}1\equal{}0]$

Οπως ειπα και στο παραπανω ποστ μου ,

Η ΑΡΙΣΤΕΡΑ ΜΕΡΙΑ ΑΥΤΗΣ της συνεπαγωγης ειναι [b] αληθης και η δεξια ψευδης[/b] και αυτο κανει ολη την συνεπαγωγη[b] ψευδη[/b]

Αρα η φορμουλα σου για χ=0 ,a=1,b=-1 ειναι ψευδης

Π.Χ  η φορμουλα $ \forall x( x^2>0)$ ειναι αληθης για ολα τα χ εκτος χ=0, το ιδιο συμβαινει και με την φορμουλα σου.

Δυο πραγματα συμβαινουν εδω λοιπον:

1) Η παραδεχομεθα την φορμουλα σου οποτε εχουμε ενεξαρτησια ,εκτος για χ=0

2)Η ΔΕΝ παραδεχομεθα την φορμουλα σου οποτε πρεπει να βρουμε μια αλλη φορμουλα να οριζη την ανεξαρτησια των συναρτησεων.

ΚΑΙ ΠΑΨΕ να αντιγραφεις απο αλλα φορουμ βλακειες οπως ( stop trolling) και τετοιες ανοησιες.

ΔΙΕΚΔΙΚΕΙΣ ΤΟ ΑΛΛΑΘΗΤΟ ΤΟΥ ΠΑΠΑ???

[quote="τρικλινο"] ΚΑΙ ΠΑΨΕ να αντιγραφεις απο αλλα φορουμ βλακειες οπως ( stop trolling) και τετοιες ανοησιες.
ΔΙΕΚΔΙΚΕΙΣ ΤΟ ΑΛΛΑΘΗΤΟ ΤΟΥ ΠΑΠΑ???[/quote]

1. Περιμένω άρθρο στα [url=http://annals.princeton.edu/annals/about/cover/cover.html]Annals of Mathematics[/url] που να περιέχει τις ανακαλύψεις σου.
2. Ξύδι για τα νεύρα.
3. Αλάθητο με ένα "Λ" - το λέω σε περίπτωση που χτυπάς πολλά πλήκτρα ταυτόχρονα επειδή έχεις νεύρα και επειδή ταιριάζει η ειρωνεία στο συγκεκριμένο σημείο.

Χαίρομαι που έδειξες μόνος σου ότι δεν αξίζει να ασχολείται το forum μαζί σου. Thatta boy! :)

Cheerio,

Durandal 1707


From Moldova Mathematical Olympiad.

For a,b,c strictly positive numbers and x=(abc)^1/3 prove that

1/(a+b+x) + 1/(b+c+x) + 1/(c+a+x) ? 1/x

PS I think that the symbol ? in the original text means that you must find
     the correct sign of the inequality.

Write a = x, b = y, c = z.

The problem becomes :

1/(x + y + xyz) + 1/(y + z + xyz) + 1/(z + x + xyz) \leq 1/(xyz)

which is exactly USAMO97/5. I think this has been discussed 3 times or so ...


geometry
3D geometry
calculus
integration
number theory proposed
number theory

Solve $a^3+2b^3+4c^3=6abc+1$ in integers $a,b,c$.

Well, here's a guess.

We can factorise the given equation:
\begin{eqnarray*} 1&=&a^3+2b^3+4c^3-6abc\\ &=&(a+\sqrt[3]{2}b+\sqrt[3]{4}c)(a+\omega\sqrt[3]{2}b+\omega^2\sqrt[3]{4}c)(a+\omega^2\sqrt[3]{2}b+\omega\sqrt[3]{4}c) \end{eqnarray*}
where $\omega$ is a complex cube root of 1.  If $a$, $b$ and $c$ are integers, then the solutions $A_n$, $B_n$, $C_n$ to the simultaneous equations
$A_n+\sqrt[3]{2}B_n+\sqrt[3]{4}C_n=(a+\sqrt[3]{2}b+\sqrt[3]{4}c)^n$
$A_n+\omega\sqrt[3]{2}B_n+\omega^2 \sqrt[3]{4}C_n=(a+\omega \sqrt[3]{2}b+\omega^2\sqrt[3]{4}c)^n$
$A_n+\omega^2\sqrt[3]{2} B_n+\omega\sqrt[3]{4}C_n=(a+\omega^2\sqrt[3]{2}b+\omega\sqrt[3]{4} c)^n$
are also integers.  So, proceeding by analogy with the solution of a Pell's equation, we "should" get all solutions from raising a base solution to various powers.

I think the base solution is 19, 15, 12.  This gives second solution 1081, 858, 681.  And so on.

Edit:  factorisation and some other errors corrected, thanks ZetaX.

I think your factorisation is wrong (the last factor should be different), but you are on the right way, my solution is more or less the same.
But note that there is a solution $a=b=c=1$ ;)

I don't have all the integral solutions, but by Dirichlet's Theorem on units in the ring of integers of a number field, they are all of the form $(a_n,b_n,c_n)$ such that $a_n+b_n\sqrt[3]2+c_n\sqrt[3]4=(a_1+b_1\sqrt[3]2+c_1\sqrt[3]4)^n\ (*)$ for some "small" solution $(a_1,b_1,c_1)$. The problem is determining this small solution. For example, $(a,b,c)=(-1,1,0)$ also works, and it cannot be obtained from $(1,1,1)$ by means of $(*)$.

Grobber: you forgot the negative powers of that minimal solution. The solution $(-1,1,0)$ is just the inverse of $(1,1,1)$, so the $-1$-th power of it.


Just started chem. and have a question:

Is there a way to convert same units with diff. exponents?
If so, how do you do it?

EX: How many cm are there in 2 cm^3?
and How many mm are in 5 m^3?

Thanx :)

What you propose is actually impossible.  For example, a m^2 represents is a two-dimensional area, while one m is a one-dimensional length.

You can think about this problem geometrically.  To try to change something with area (for example, a rectangle) into something with length only (e.g. a line segment) one could imagine cutting the rectangle into "thin" strips and laying them end to end.  However, no matter how "thin" you cut these strips, they will always have some finite width--hence no matter how much you try, something that starts with area can never be converted to just a length.

You should note, however, that it is certainly possible to convert between, say, cm^3 and m^3... to do so, recall 100 cm = 1 m, so 100^3 cm^3 = m^3.


let $ a,b,c$ be non-negative numbers, no two of which are zero. prove that :
$ \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2}$


geometry
3D geometry
tetrahedron
sphere
probability

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point P is selected at random inside the circumscribed sphere. The probability that P lies inside one of the five small spheres is closest to

(A) 0
(B) 0.1
(C) 0.2
(D) 0.3
(E) 0.4

[hide]Let $R$ be the radius of the circumscribed sphere, $r$ be the radius of the inscribed sphere, and $s$ be the radius of the tangent spheres. It can be shown that $r = \frac{R}{3}$ (the centers of the circumscribed/inscribed spheres are the centroid of the tetrahedron) and then $s = \frac{1}{2}(R-r) = \frac{R}{3}$ (draw the two radii to the tangent point, the difference is the diameter of the tangent sphere) also, so the probability is

$5\left(\frac{1}{3}\right)^{3}= \frac{5}{27}\approx 0.2$.[/hide]


I believe this will include most trignometry results useful in Olympiads. Any additions are welcome!!!

I don't get the point of this thread.  Can you explain, Rushil?


Knowing that $\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\frac{1}{11}$, calcule $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$


Δίνεται τρίγωνο $ ABC$ και έστω $ A1, B1, C1$ τα μέσα των πλευρών $ BC, CA, AB$ αντίστοιχα. Έστω $ P$ τυχόν σημείο του περίκυκλου και έστω ότι οι $ PA1, PA2, PA3$, τέμνουν τον κύκλο στα $ A', B', C'$ αντίστοιχα ώστε κανένα από τα σημεία αυτά να μην ταυτίζεται με τις κορυφές του τριγώνου και επιπλέον οι φορείς των $ AA', BB', CC'$ να σχηματίζουν τρίγωνο (να τέμνονται ανά δύο και μάλιστα σε διαφορετικά σημεία). Να αποδειχθεί ότι το εμβαδόν του τριγώνου αυτού είναι ανεξάρτητο της επιλογής του $ P$.

Ρε Κώστα, με κατέστρεψες!!!! Σε περίοδο εξετάσεων, χάνω πάνω από μια ώρα για ένα πρόβλημα...

Λοιπόν συνοπτικά τα βήματα της λύσης είναι:
Με Pascal προκύπτει ότι αν το τρίγωνο που σχηματίζεται είναι το $ A''B''C''$, ta $ A'', B'', C''$ βρίσκονται στις πλευρές του τριγώνου $ A'B'C'$ (να' ναι καλά το Cabri που με βοήθησε να το δω αυτό :P ).

Το 2ο βήμα είναι να δείτε ότι το εμβαδόν του $ AB'C$ είναι σταθερό (χωρίς βλάβη της γενικότητα έχω θεωρήσει το $ P$ στο τόξο $ AC$). Επειτα, το εμβαδόν του τριγώνου που ψάχνουμε έχει την γωνία $ B''$ κοινή με το παραπάνω τρίγωνο! Ετσι $ \frac {(A''B''C'')}{(AB'C)} \equal{} \frac {A''B''\cdot C''B''}{AB''\cdot CB''}$. Παρατηρείστε τέλος ότι αυτός ο τελευταίος λόγος είναι ίσος με τη μονάδα αφού:
$ \frac {A''B''}{AB''} \equal{} \frac {A'B''}{AB' \minus{} A'B''} \equal{} \frac {A'B''}{C'B''}$

[quote="Protonios"]To Pascal itan i olh istoria... Xwris ayto den xekinaei (yparxei allh mia lysh alla poly eyfantasth kai de vgainei eykola, an thelei kapoios thn postarw avrio)

Re Mim, ti na diavaseis? Na pw oti exoyme viologia ok... :) haha[/quote]

Εγώ ο βλάκας είδα το Pascal σχεδόν αμέσως κι έφαγα την περισσότερη ώρα να ψάχνω τη συνέχεια :blush:

Και εννοείται πως βιολογία διαβάζω  :P

A! Γράψε και τη λύση σου όταν μπορείς...

IMO Shortlist 2007 . Αν δεις το Pascal μετά αρκεί να παρατηρήσει κανείς ότι τρίγωνα που έχουν κορυφές να κινούνται σε παράλληλη προς τη βάση ευθεία έχουν ίσα εμβαδά .
http://www.mathlinks.ro/viewtopic.php?p=1186790#1186790


calculus integration geometry derivative LaTeX rectangle function

That's not a derivative; it's just algebra.

If you want pretty math, $\LaTeX$ is enabled on this forum. Just put an expression between dollar signs to make it work.

Your integral can be written as
\int_{-4}^3 (3-x)-(x^2-9)\, dx
$\int_{-4}^3 (3-x)-(x^2-9)\, dx$
Most of that should be easy to understand; the \, is just a code for a thin space.

[quote="alltime"]I'm trying to grasp a firm understanding on "Applications of the definite integral"

In my book there is a problem which solves the area between two curves

Here ispart of how the problem is worked out:

A = int(-4 -> 3) [(3-x) - (x2  - 9 ) ] dx

   = int(-4 -> 3) (-x2-x+12) = [ (-x3)/3 - (x2)/2 + 12x] (from -4 -> 3)

Now I do not udnerstand the taking of derivative from

[quote] int(-4 -> 3) [(3-x) - (x2  - 9 ) ] dx [/quote]

to

[quote]int(-4 -> 3) (-x2-x+12) [/quote]

I understand derivatives but for some reason, I'm not seeing this.  Also, how do you do the integral sign on this board?[/quote]

alltime,

I believe the part that might be confusing you is the $dx$ after the $\int$ sign, no?

The $dx$ is not a derivative. It is the width of an infinitesimal bit of distance along the $x$ direction.   The height of a given curve at a point $x$ is $f(x)$, so that the product $f(x)\,dx$ is the area of a very thin rectangle.  The $\int$ sign is just a script "S" meaning sum.  In other words, take the areas of all the thin rectangles, $f(x)\, dx$ and add them up. The result of this addition is the total area under the curve.

It's best to see $dx$ as just part of the $\int$ symbol- it's the indicator to tell you which variable to integrate with respect to, and the integral is badly ambiguous without it. $\int_0^a x^y\, dx\neq\int_0^ax^y\, dy$. The $dx$ also conveniently tells you where the function to be integrated ends. Since leaving it off can only lead to confusion, always include it.


inequalities
triangle inequality
absolute value
geometry solved
geometry

Let triangle $ABC$ is inscribed in the circle with radius 1. Let $P$ is an interior point. Prove that:
$PA\cdot{PB\cdot{PC}}<\frac{32}{27}$.

This is an inequality of Paul Erdös. I am reformulating it in a more canonical way:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle inscribed in a circle k with radius R. Let P be a point in the interior or on the boundary of the triangle ABC. Then, we have the inequality $PA\cdot PB\cdot PC \leq \frac{32}{27}R^3$.[/color]

This theorem is considered to be so difficult only since it requires a slightly unusual strategy: When one sees an inequality such as $PA\cdot PB\cdot PC \leq \frac{32}{27}R^3$, one usually tries to prove it by keeping the triangle ABC fixed and maximizing the term $PA\cdot PB\cdot PC$ when the point P moves around the triangle ABC. For this theorem, exactly the other way round works: One maximizes the term $PA\cdot PB\cdot PC$ for a fixed point P, with the points A, B, C moving around the circle k (with the additional condition that the point P must remain lying in the interior or on the boundary of the triangle ABC).

Here is the [i]proof of Theorem 1[/i] in detail:

Let O be the center of the circle k. If the point P coincides with the point O, then the distances PA, PB, PC are all equal to the radius R of the circle k (since the point P coincides with the center O of the circle k, and the points A, B, C lie on this circle k); thus, $PA\cdot PB\cdot PC=R\cdot R\cdot R=R^3\leq \frac{32}{27}R^3$, and Theorem 1 must hold. Hence, we have proven Theorem 1 for the case when the point P coincides with the point O; thus, in the following, we need only to consider the case when $P\neq O$.

Also, Theorem 1 is trivial if the point P is one of the vertices of triangle ABC (in this case, the product $PA\cdot PB\cdot PC$ is zero, and thus clearly $\leq \frac{32}{27}R^3$). Thus, we can also assume, in the following observations, that the point P is distinct from each of the vertices of the triangle ABC. Since we also know that the point P lies in the interior or on the boundary of the triangle ABC, it follows that the point P lies inside the circle k.

Let the lines AP, BP, CP meet the circle k at the points X, Y, Z (apart from the points A, B, C, respectively). Since the point P lies in the interior or on the boundary of the triangle ABC, the points X, Y, Z lie on the arcs BC, CA, AB of the circle k which don't contain the points A, B, C, respectively.

Now, consider the angles A = < CAB, B = < ABC and C = < BCA of triangle ABC, and the angles X = < ZXY, Y = < XYZ and Z = < YZX of triangle XYZ. Since the sum of the angles of a triangle is always = 180°, we have A + B + C = 180° and X + Y + Z = 180°, and thus X + Y + Z = A + B + C. If all three inequalities X < A, Y < B and Z < C would simultaneously hold, then we would also have X + Y + Z < A + B + C, what would contradict the equation X + Y + Z = A + B + C. Thus, the three inequalities X < A, Y < B and Z < C cannot all simultaneously hold, so that at least one of them must be wrong. WLOG assume that the inequality X < A is wrong; in other words, we have $X\geq A$.

But X = < ZXY, and since the points X, Y, Z and B lie on one circle (namely, on the circle k), we have < ZXY = < ZBY. Thus, X = < ZXY = < ZBY = < ZBP.

On the other hand, A = < CAB, and since the points A, B, C and Z lie on one circle (namely, on the circle k), we have < CAB = < CZB. Thus, A = < CAB = < CZB = < PZB.

Since X = < ZBP and A = < PZB, the inequality $X\geq A$ becomes $\measuredangle ZBP\geq \measuredangle PZB$. But since, in a triangle, the greater angle lies opposite to the longer side, it follows that, in the triangle ZPB, we have $PZ\geq PB$. In other words, $PB\leq PZ$. Consequently, $PA\cdot PB\cdot PC\leq PA\cdot PZ\cdot PC$.

Hence, in order to prove the inequality $PA\cdot PB\cdot PC \leq \frac{32}{27}R^3$, it is enough to verify $PA\cdot PZ\cdot PC \leq \frac{32}{27}R^3$. This certainly must be easier, since the point P lies on the segment ZC.

By the triangle inequality for the (possibly degenerate) triangle APO, we have $PA\leq AO+PO$. Since the point O is the center of the circle k, and the point A lies on the circle k, the distance AO equals the radius R of the circle k; thus, we get $PA\leq R+PO$. On the other hand, since the point P lies on the chord ZC of the circle k, the product $PZ\cdot PC$ is the absolute value of the power of the point P with respect to the circle k, and thus it equals $PZ\cdot PC=R^2-PO^2$ (since O is the center and R is the radius of the circle k). In other words, $PZ\cdot PC=\left(R-PO\right)\cdot\left(R+PO\right)$. This equation and the inequality $PA\leq R+PO$ together imply

$PA\cdot PZ\cdot PC = PA\cdot\left(PZ\cdot PC\right) \leq \left(R+PO\right)\cdot\left(\left(R-PO\right)\cdot\left(R+PO\right)\right)$
$=\left(R+PO\right)^2\cdot\left(R-PO\right)=\frac12\cdot\left(R+PO\right)^2\cdot 2\left(R-PO\right)$.

Since the point P lies inside the circle k, while O is the center and R is the radius of this circle k, we have PO < R; thus, R - PO > 0, so that 2 (R - PO) > 0. Consequently, the three numbers R + PO, R + PO and 2 (R - PO) are all nonnegative; hence, we can apply the AM-GM inequality to these three numbers, and obtain

$\sqrt[3]{\left(R+PO\right)\cdot\left(R+PO\right)\cdot\left(2\left(R-PO\right)\right)}\leq\frac{\left(R+PO\right)+\left(R+PO\right)+\left(2\left(R-PO\right)\right)}{3}$.

This simplifies to

$\sqrt[3]{\left(R+PO\right)^2\cdot 2\left(R-PO\right)}\leq\frac43 R$,

and cubing yields $\left(R+PO\right)^2\cdot 2\left(R-PO\right)\leq\left(\frac43 R\right)^3=\frac{64}{27}R^3$. Thus,

$PA\cdot PZ\cdot PC \leq \frac12\cdot\left(R+PO\right)^2\cdot 2\left(R-PO\right) \leq \frac12\cdot\frac{64}{27} R^3 = \frac{32}{27} R^3$.

This completes our proof of Theorem 1.

Note that a closer look at our proof of Theorem 1 shows that, in the inequality $PA\cdot PB\cdot PC \leq \frac{32}{27}R^3$, equality holds if and only if the triangle ABC is an isosceles triangle with base angle $90^{\circ}-\arccos\frac13$, and the point P is the midpoint of its base. Hence, if we add the additional condition that the point P should lie (strictly!) in the interior of the triangle ABC, then we always have the strict inequality $PA\cdot PB\cdot PC < \frac{32}{27}R^3$.

This solves your problem.

  Darij

Thank you very much, Darij for this very detailed proof.

What can I say, darij, veryvery nice. Thank you very much!!