Let $f(x)=\sqrt{x^2-1}$. Prove that $f^{(n)}(x)>0$ for odd $n$ and all $x$ and the opposite for all even $n$ and $x>1$.

$f'(x)=\frac{x}{\sqrt{x^2-1}}=\left(1-\frac1{x^2}\right)^{-\frac12}$

$=1-\frac12\left(-\frac1{x^2}\right)+\frac{(-\frac12)(-\frac32)}2 \left(-\frac1{x^2}\right)^2+\cdots$

$=1+\sum_{k-1}^{\infty}c_kx^{-2k}$ where $c_k>0$ for all $k.$

The series converges for $x>1$ and can be differentiated term by term.

So $f''(x)=-\sum_{k=1}^{\infty}(2k)c_kx^{-2k-1)}$

$f'''(x)=\sum_{k=1}^{\infty}(2k)(2k+1)c_kx^{-2k-2}$

In general, $f^{n}=(-1)^{n+1}\sum_{k=1}^{\infty}(2k)(2k+1)\cdots(2k+n-2)c_k ^{-2k-n+1}$

Since all of the terms of each sum are the same sign, the sign of the sum is clear.


What is the largest even integer which cannot be written as the sum of two odd composite numbers? (Recall that a positive integer is said to be composite if it is divisible by at least one positive integer other than 1 and itself.)

Hmm... it just says odd composite positive integers, but not necessarily distinct? If so, you can limit it to numbers of the form $2p$, where $p$ is some prime. You also know that it's $<999$, because it's an AIME problem. :lol:

If the answer is n, then at least (roughly) 1/4 of the numbers below n are primes.  Note that about 1/4 of numbers are not multiples of 2, 3, 5.  I guess that can help you get an estimate (for example you'd guess that the number is probably not over 10).

For this one, you have to start playing around with it. After all, it is pretty tough: #14. Let's see which numbers can be written as, for example, $3(\text{odd})+5(\text{odd})$, i.e. $3(2a+3)+5(2b+3)=2(3a+5b)+24$ for non-negative integers $a$ and $b$.

Which even numbers can be written as $3a+5b$ for non-negative integers $a$ and $b$?
[hide]By experimenting, we find that we can get all numbers larger than and including 8:
8=5+3
9=3+3+3
10=5+5
11=3+3+5
etc.[/hide]

What does that tell us about the answer to the problem?
[hide]By using the $3(\text{odd})+5(\text{odd})$ form above, we can write all even numbers above and including $2(8)+24=40$ as the sum of two odd composites. Thus, the answer is less than 40.[/hide]

So, what [i]is[/i] the answer?
[hide]We need to be checking numbers below 40, so we first try 38:
[b]3[/b]+35
[b]5[/b]+33
[b]7[/b]+[b]31[/b]
9+[b]29[/b]
[b]11[/b]+27
[b]13[/b]+25
15+[b]23[/b]
[b]17[/b]+21
[b]19[/b]+[b]19[/b]
It works! 38 is the answer.[/hide]

So, that works pretty well by just testing numbers up to a point, especially if you can limit it down to a certain set (like $n=2p$, as I did). All in all, a fairly easy #14, but then you have the nagging suspicion, unless you tested every number up to 998, that you missed one that was actually higher, so your method probably works better. :P

It's a reference to Goldbach's conjecture, that every even number more than 2 can be written as the sum of two primes.

I prefer the book's solution. I also came up with a nice one (but more testing) when I did it:

[hide]The first time that three consecutive odd composites appear are 91,93,95. Now we can add three consecutive odds to get 91+7, 93+5, 95+3. Since 7,5,and 3 are all prime, this doesn't work here, but they are the only triplet of primes. Therefore, the answer is in the two digits. Now you can just work down until you get it.[/hide]

[quote="pachao"]i dont really get it....
one question, why did u pick 2a+3 instead of 2a+1 as an odd number?[/quote]
I picked $2a+3$ and $2b+3$ so I could deal with $a$ and $b$ as NON-NEGATIVE integers instead of POSITIVE integers.

I remember getting this problem fairly quickly in 9th grade and being quite proud of it at the time (after all it was a #14 on the AIME). Hence, my approach has stuck with me to this day. I did it almost entirely just by using the fact that every odd multiple of 5 other than 5 itself is composite. Keeping that in mind, you can pretty much come up with a way to cover anything larger than a certain point.


inequalities
Cauchy Inequality
inequalities proposed

Define $ y_i \equal{} 1 \plus{} x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2$. ($ y_0 \equal{} 1$)

$ \frac {x_1 }{y_1 } \plus{} \frac {x_2 }{y_2 } \plus{} .. \plus{} \frac {x_n }{y_n } < \sqrt{n}$

By Cauchy Inequality,

$ (LHS)^2 \leq n( \frac {x_1^2 }{y_1^2 } \plus{} \frac {x_2^2 }{y_2^2 } \plus{} .... \plus{} \frac {x_n^2 }{y_n^2 }$.

To establish $ LHS \leq \sqrt {n}$, it suffices to show that $ RHS \leq n$, or
$ \sum_{i \equal{} 1}^{n} \frac {x_i^2 }{y_i^2 } < 1$.

Since we have $ y_{i \plus{} 1} \minus{} y_i \equal{} x_{i \plus{} 1}^2 \geq 0$, we have that

$ \sum_{i \equal{} 1}^{n} \frac {x_i^2 }{y_i^2 } < \sum_{i \equal{} 1}^{n} \frac {x_i^2 }{y_i y_{i \minus{} 1} }$
$ \equal{} \sum_{i \equal{} 1}^{n} ( \frac {1 }{y_{i \minus{} 1}} \minus{} \frac {1}{y_i} )$ (Partial Fraction)
$ \equal{} 1 \minus{} \frac {1}{y_n} < 1$ (Telescoping Sum)

Just want to say that this problem was at Kazakhstan National Olympiad 2003,don't know whether this problem initally appeared there...


I know some SAT dates are easier than others, is the same true for SAT IIs?  If so, when is the best time to take them?

Really??????

Which SAT I dates are easier?

I think it is an urban legend, not based on fact, that any SAT I test date is easier or harder than any other. There is no reason to expect that there would be such a systematic difference, especially given how the test's standard scores are equated from one administration to the next.

Correspondingly, there is no reason to suppose that any one date is easier for SAT II testing than any other. Take an SAT II test when you have learned the material thoroughly.

[quote="aznphatso"]I actually believe in this urban legend, I guess. I'd say December and January are usually slightly more favorable to strong students.[/quote]
so do you mean that December and January tests are slightly harder?

One of the reasons I don't believe the legend is that any explanation I have ever seen about why that should be so doesn't make sense, based on how the College Board explains the equating of test scores. But, please, explain why one test date should be any easier than any other.

definitely anything I say is a rumor ought to be taken with a grain of salt, but the rationale is this:

October = all the prepared seniors getting their last test in before EA/ED deadlines, most fresh after a summer of prep
November = same thing
December = last minute procrastinating seniors, so the curve is a bit nicer to strong prepared students
January = extreme procrastinating seniors, same reasoning as December.

The SAT is not curved.  It is scaled.  There is a big difference.  Read the College Board site for an explanation.

I think the whole easier/harder test date thing is just because some tests happen to be easier for certain people... something inconsistent like that which may have, by chance, produced some sort of pattern. That said, I certainly scored much higher on my December SAT I than I had expected (even right after taking it).

Hmm.  Okay.  I've just always heard that people tend to do best on the December SAT.  I guess it is purely myth.

[quote="Ravi B"]The SAT is not curved.  It is scaled.  There is a big difference. [/quote]

Doesn't the S in SAT imply that the scaling is itself a form of curving, i.e. adjustment of results relative to some set of people taking the same test?

[b]Edit:[/b] the intended meaning was S as in Standardized (Aptitude Test).

The S used to stand for "Scholastic", although now the letters S-A-T officially stand for nothing.

Your score on an SAT does not depend on the particular students who take the test the same day.  Instead, each SAT question has been given to previous students in an unscored, equating ("experimental") section.  (Students do not know in advance which section is experimental.)  By comparing student performance on the experimental questions and the real questions, the ETS can assign a difficulty rating to each experimental question.  The test assemblers then pool such questions to create a test whose difficulty is known in advance.

So your score is not determined by comparing you with current students.  In a sense, you are being compared with a fixed group of students from decades ago.

I happen to work for a tutoring company that administers old official SATs to its students (at least it did until the SAT changed last year).  Looking at our data, I do not see any correlation between the old test month and test scores.  It's true, as paladin8 mentioned, that an individual student's scores will vary from test to test.  But that seems to have more to do with that student's strengths and weaknesses, and even his or her mood that day, than to systematic differences in test forms.

[quote="Ravi B"]The S used to stand for "Scholastic", although now the letters S-A-T officially stand for nothing. [/quote]

I confused this S with "Standardized", sorry.   The point was that the Scaling or Standardization or other named procedure used by ETS,
involves fitting the results to some population of test-takers (curving)
and then fixing that curve thereafter.  However, even whether the curve is fixed (for given test material) is not clear, as ETS apparently could change it without notice.

One question: is the ETS scoring procedure simply a weighted sum of the correct answers, or are there dependencies between different questions?

For scoring, there are not any dependencies between different questions.  In each category (Math, Critical Reading, Writing Multiple Choice), a raw score is first computed: the number of correct answers minus one-fourth the number of incorrect multiple-choice answers.  Each SAT form then has its own table that converts the raw score to a scaled score from 200 to 800.  For example, on the October 2005 test that I am looking at, a Math raw score of 50 (out of 54 questions) translates to a scaled score of 720.

The Writing composite score is given by a chart with two dimensions: the Writing multiple-choice raw score and the essay score.

english i think starts at 300, not 200 math stars at 200

No, the lowest score is 200 on Math, Reading, and Writing.  You have to miss a few questions to get that low of a score.  If you leave all the questions blank, you get about 220, according to an official chart I am looking at.

(Actually, if you leave all the questions in a section blank, the College Board will treat that as a request to cancel your score.)


[quote="ductrung"]Let $ a,b$ and $ c$ be positive real numbers such that $ a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}\equal{} abc$. Prove that
\[ \sqrt{\frac{b\plus{}c}{1\plus{}a}}\plus{}\sqrt{\frac{c\plus{}a}{1\plus{}b}}\plus{}\sqrt{\frac{a\plus{}b}{1\plus{}c}}\le\frac{3}{\sqrt{2}}\]
[/quote]
$ a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}\equal{} abc\Leftrightarrow\frac{ab}{c}\plus{}\frac{ac}{b}\plus{}\frac{bc}{a}\equal{}1.$
Let $ \frac{bc}{a}\equal{}x^{2},$ $ \frac{ac}{b}\equal{}y^{2}$ and $ \frac{ab}{c}\equal{}z^{2},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}1,$  $ a\equal{}yz,$ $ b\equal{}xz$ and $ c\equal{}yz.$
Id est, we need to prove that $ \sum_{cyc}\sqrt{\frac{z(x\plus{}y)}{1\plus{}xy}}\leq\frac{3}{\sqrt2}.$
But $ \frac{9}{2}\geq(x\plus{}y\plus{}z)\sum_{cyc}\frac{x\plus{}y}{1\plus{}xy}\geq\left(\sum_{cyc}\sqrt{\frac{z(x\plus{}y)}{1\plus{}xy}}\right)^{2}.$

[quote="arqady"]Hence, $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{} 1,$  $ a \equal{} yz,$ $ b \equal{} xz$ and $ c \equal{} yz.$[/quote]
Yes, that's what I did  :)
[quote="arqady"]$ \frac{9}{2}\geq(x\plus{}y\plus{}z)\sum_{cyc}\frac{x\plus{}y}{1\plus{}xy}$[/quote]
Proving the above is interesting by itself. Do you know if it exists here - any link?
Since $ 3(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}xy)\minus{}2(x\plus{}y\plus{}z)(x\plus{}y)\equal{}z^{2}\minus{}xy\plus{}(x\minus{}z)^{2}\plus{}(y\minus{}z)^{2}$
It suffices to show that
\[ \frac{z^{2}\minus{}xy}{1\plus{}xy}\plus{}\frac{y^{2}\minus{}zx}{1\plus{}zx}\plus{}\frac{x^{2}\minus{}yz}{1\plus{}yz}\ge 0\]
Subtracting both sides from 3, this becomes
\[ \frac{(x\plus{}y)^{2}}{1\plus{}xy}\plus{}\frac{(y\plus{}z)^{2}}{1\plus{}yz}\plus{}\frac{(z\plus{}x)^{2}}{1\plus{}zx}\le 3\]
Any nice solution or any link?

[quote="ductrung"]
It suffices to show that
\[ \frac{z^{2}\minus{}xy}{1\plus{}xy}\plus{}\frac{y^{2}\minus{}zx}{1\plus{}zx}\plus{}\frac{x^{2}\minus{}yz}{1\plus{}yz}\ge 0\]
Subtracting both sides from 3, this becomes
\[ \frac{(x\plus{}y)^{2}}{1\plus{}xy}\plus{}\frac{(y\plus{}z)^{2}}{1\plus{}yz}\plus{}\frac{(z\plus{}x)^{2}}{1\plus{}zx}\le 3\]
Any nice solution or any link?[/quote]
My proof the same, but the last step is following:
$ \sum_{cyc}\frac{z^{2}\minus{}xy}{1\plus{}xy}\ge 0\Leftrightarrow\sum_{cyc}\frac{(z\minus{}x)(z\plus{}y)\plus{}(z\minus{}y)(z\plus{}x)}{1\plus{}xy}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(z\minus{}x)(z\plus{}y)}{1\plus{}xy}\minus{}\frac{(y\minus{}z)(z\plus{}x)}{1\plus{}xy}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(x\minus{}y)(x\plus{}z)}{1\plus{}yz}\minus{}\frac{(x\minus{}y)(y\plus{}z)}{1\plus{}xz}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(x\minus{}y)^{2}(1\plus{}xz\plus{}yz\plus{}z^{2})}{(1\plus{}xz)(1\plus{}yz)}\geq0.$

[quote=ductrung]Let $ a,b$ and $ c$ be positive real numbers such that $ a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}\equal{}abc$. Prove that
\[ \sqrt{\frac{b\plus{}c}{1\plus{}a}}\plus{}\sqrt{\frac{c\plus{}a}{1\plus{}b}}\plus{}\sqrt{\frac{a\plus{}b}{1\plus{}c}}\le\frac{3}{\sqrt{2}}\][/quote]
The following inequality is also true
[img]http://s15.sinaimg.cn/middle/006ptkjAzy76rAuzztQce&690[/img]


inequalities
quadratics
inequalities unsolved

I will prove that

$\frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$

holds for all positive reals x, y, z, and that equality holds only for x = y = z.

In fact, after AM-GM (= the inequality between the arithmetic and the geometric mean), we have

$\frac{x^3}{yz}+\frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq 4\root{4}\of{\frac{x^3}{yz}\cdot \frac{x^3}{yz}\cdot \frac{y^3}{zx}\cdot \frac{z^3}{xy}}=4x$,

and similarly

$\frac{y^3}{zx}+\frac{y^3}{zx}+\frac{z^3}{xy}+\frac{x^3}{yz}\geq 4y$

and

$\frac{z^3}{xy}+\frac{z^3}{xy}+\frac{x^3}{yz}+\frac{y^3}{zx}\geq 4z$.

Summing this all together, we get

$4\frac{x^3}{yz}+4\frac{y^3}{zx}+4\frac{z^3}{xy}\geq 4x+4y+4z$,

so that

$\frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$

and our inequality is proven. Now, each time we applied AM-GM, we had equality if and only if $\frac{x^3}{yz}=\frac{y^3}{zx}=\frac{z^3}{xy}$. Therefore, in the resulting inequality, we also have equality if and only if $\frac{x^3}{yz}=\frac{y^3}{zx}=\frac{z^3}{xy}$. But $\frac{x^3}{yz}=\frac{y^3}{zx}=\frac{z^3}{xy}$ is equivalent to $x^4=y^4=z^4$, i. e. to x = y = z. Hence we see that in your original inequality, equality holds only for x = y = z.

  Darij

It also follows directly from Muirhead, as $(3, -1, -1) \succ (1, 0, 0)$.

Or by Chebycheff and AM-GM:

\[
\frac{x^{3}}{yz}+\frac{y^{3}}{zx}+\frac{z^{3}}{xy}\geq \frac{1}{3} (x^3+y^3+z^3) \left(\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy}\right)
\geq (xyz) \left(\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy}\right).\]

[quote="Iris Aliaj"]Show that x^3/(yz) + y^3/(zx) + z^3/(xy) ≥ x + y + z for any positive reals x, y, z. When do we have equality?[/quote]

I may as well provide another proof:

a^4 + b^4 + c^4 >= 3*[(a+b+c)/3]^4 by AM-GM

But [(a+b+c)/3]^3 >= abc again by AM-GM

So:

LHS >= abc (a+b+c), and, upon dividing out abc, we are done.

By Cauchy we have x^4/xyz+y^4/xyz+z^4/xyz :ge: (x^2+y^2+z^2)^2/3xyz Then by AM-GM x^4+y^4+z^4 :ge: xyz(x+y+z) and 2(x^2y^2+x^2z^2+y^2z^2) :ge: 2xyz(x+y+z) That means we have  (x^2+y^2+z^2)^2/3xyz :ge: x+y+z and we are done

[quote="Iris Aliaj"]Show that x^3/(yz) + y^3/(zx) + z^3/(xy) ≥ x + y + z for any positive reals x, y, z. When do we have equality?[/quote]

Just for fun, here is yet another proof (or, better, another version of one of your proofs):

We have to show

$\frac{x^3}{yz}+\frac{y^3}{zx}+\frac{z^3}{xy}\geq x+y+z$.

This is obviously equivalent to

$x^4+y^4+z^4\geq xyz\left( x+y+z\right)$.

Now, we will establish this by proving the [b]two inequalities[/b]

$x^4+y^4+z^4\geq y^2z^2+z^2x^2+x^2y^2$
and $y^2z^2+z^2x^2+x^2y^2\geq xyz\left( x+y+z\right)$.

[b]The first of these two inequalities can be proved[/b]

- by rearrangement:

$x^4+y^4+z^4=x^2\cdot x^2+y^2\cdot y^2+z^2\cdot z^2\geq x^2\cdot y^2+y^2\cdot z^2+z^2\cdot x^2=y^2z^2+z^2x^2+x^2y^2$.

- using Cauchy-Schwarz: you have

$x^4+y^4+z^4=\frac{1}{3}\cdot \left( \left( x^2\right) ^2+\left( y^2\right) ^2+\left( z^2\right) ^2\right) \cdot \left( 1^2+1^2+1^2\right)$
$\geq \frac{1}{3}\left( x^2\cdot 1+y^2\cdot 1+z^2\cdot 1\right) ^2=\frac{1}{3}\cdot \left( x^2+y^2+z^2\right) ^2$
$=\frac{1}{3}\cdot \left( x^4+y^4+z^4+2y^2z^2+2z^2x^2+2x^2y^2\right)$
$=\frac{1}{3}\cdot \left( x^4+y^4+z^4\right) +\frac{2}{3}\cdot \left( y^2z^2+z^2x^2+x^2y^2\right)$,

hence $\frac{2}{3}\cdot \left( x^4+y^4+z^4\right) \geq \frac{2}{3}\cdot \left( y^2z^2+z^2x^2+x^2y^2\right)$, and thus $x^4+y^4+z^4\geq y^2z^2+z^2x^2+x^2y^2$.

- by simple quadratic representation:

$\left( x^4+y^4+z^4\right) -\left( y^2z^2+z^2x^2+x^2y^2\right) =\frac{1}{2}\left( \left( y^2-z^2\right) ^2+\left( z^2-x^2\right) ^2+\left( x^2-y^2\right) ^2\right) \geq 0$.

- using $u^2+v^2\geq 2uv$ three times:

$x^4+y^4+z^4=\frac{1}{2}\cdot \left( \left( y^4+z^4\right) +\left( z^4+x^4\right) +\left( x^4+y^4\right) \right)$
$\geq \frac{1}{2}\cdot \left( 2y^2z^2+2z^2x^2+2x^2y^2\right) =y^2z^2+z^2x^2+x^2y^2$.

- using power means for three variables:

$\root{4}\of{\frac{1}{3}\left( x^4+y^4+z^4\right) }\geq \sqrt{\frac{1}{3}\left( x^2+y^2+z^2\right) }$

yields $3\left( x^4+y^4+z^4\right) \geq \left( x^2+y^2+z^2\right) ^2$, so $3\left( x^4+y^4+z^4\right) \geq x^4+y^4+z^4+2y^2z^2+2z^2x^2+2x^2y^2$, and $2\left( x^4+y^4+z^4\right) \geq 2y^2z^2+2z^2x^2+2x^2y^2$ and finally $x^4+y^4+z^4\geq y^2z^2+z^2x^2+x^2y^2$.

- by Muirhead, since $\left( 4;\;0;\;0\right) \succ \left( 2;\;2;\;0\right)$.

[b]The second of the two inequalities can be proved[/b]

- using $u^2+v^2\geq 2uv$ three times:

$y^2z^2+z^2x^2+x^2y^2=\frac{1}{2}\cdot \left( x^2\left( y^2+z^2\right) +y^2\left( z^2+x^2\right) +z^2\left( x^2+y^2\right) \right)$
$\geq \frac{1}{2}\cdot \left( x^2\cdot 2yz+y^2\cdot 2zx+z^2\cdot 2xy\right) =x^2yz+y^2zx+z^2xy=xyz\left( x+y+z\right)$.

- by simple quadratic representation:

$\left( y^2z^2+z^2x^2+x^2y^2\right) -xyz\left( x+y+z\right) =\frac{1}{2}\cdot \left( x^2\left( y-z\right) ^2+y^2\left( z-x\right) ^2+z^2\left( x-y\right) ^2\right) \geq 0$.

- by rearrangement again: we have

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

- using power means for three variables:

$\sqrt{\frac{1}{3}\left( y^2z^2+z^2x^2+x^2y^2\right) }\geq \frac{1}{3}\left( yz+zx+xy\right)$

yields $3\left( y^2z^2+z^2x^2+x^2y^2\right) \geq \left( yz+zx+xy\right) ^2$, so $3\left( y^2z^2+z^2x^2+x^2y^2\right) \geq y^2z^2+z^2x^2+x^2y^2+2yzzx+2zxxy+2xyyz$, and $2\left( y^2z^2+z^2x^2+x^2y^2\right) \geq 2yzzx+2zxxy+2xyyz$ and finally $y^2z^2+z^2x^2+x^2y^2\geq yzzx+zxxy+xyyz=xyz\left( x+y+z\right)$.

- by Muirhead, since $\left( 2;\;2;\;0\right) \succ \left( 2;\;1;\;1\right)$.

- or finally by applying the first inequality to the variables $\sqrt{yz}$, $\sqrt{zx}$, $\sqrt{xy}$.

  Darij


vector
Putnam
linear algebra
matrix
geometry proposed
geometry

Can $n+2$ points be chosen in $\mathbb{R}^n$ in such a way that each distance between a pair of points is an odd integer? Prove that if $n\not\equiv 6 \mod 8$ then this is impossible.

This is a result from Graham, Straus and Rotschild, which appeared in the A.M.M. (1974).

Pierre.

Thanks, Pierre. I didn't know that. But it's such a nice argument - let's see if one of our group can find it without looking it up in a 30-year-old [i]Monthly.[/i]

First hint: Use vector language. Let one point lie at the origin, and consider the other $n+1$ points to lie at the ends of vectors $u_1,\dots,u_{n+1}.$

Second hint: If $k$ is odd, then $k^2\equiv1\mod8.$

The Putnam problem based on this was the $n=2$ case: can you put 4 points in the plane with all distances odd integers?

We can straightforwardly generalize the solution to the Putnam problem, can't we?

We [i]use vector language. Let one point lie at the origin, and consider the other $n+1$ points to lie at the ends of vectors \[u_{1}, \dots, u_{n+1}.\][/i] Next, embed $\mathbb R^{n}$ into $\mathbb R^{n+1}$. Now consider the matrix $A = \left( u_{1}\ \ u_{2}\ \ \ldots \ \ u_{n+1}\right) .$ We obviously have that $\det A = 0$.

Obviously, \[\left( A^{T}A \right)_{ij}= u_{i}\cdot u_{j}= \frac{u_{i}^{2}+u_{j}^{2}-\left( u_{i}-u_{j}\right)^{2}}2 .\] We also have that $\det \left( A^{T}A \right) = 0$. If we multiply each column by $2$, the determinant $\det \left( u_{i}^{2}+u_{j}^{2}-\left( u_{i}-u_{j}\right)^{2}\right)_{ij}$ obviously equals $0$. Furthermore, all of its entries are integers (according to the hypothesis, $u_{i}^{2}, u_{j}^{2}, \left( u_{i}-u_{j}\right)^{2}$ are all integers). Now reduce the new determinant modulo $8$. [i]If $k$ is odd, then $k^{2}\equiv 1 \mod 8$[/i]. Thus, the determinant will look like this
\[\begin{vmatrix}2 & 1 & \ldots & 1 \\ 1 & 2 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \ldots & 2 \end{vmatrix}. \]
Now add to the first row the rest. It will look like this: $\left( n+2 \ \ n+2 \ \ \ldots \ \ n+2 \right)$. Now pull a $n+2$ in the front of the determinant and from each of the other rows subtract the new first row $\left( 1 \ \ 1 \ \ \ldots \ \ 1 \right)$. So our determinant is $n+2$ modulo $8$, and we have our contradiction right here.

Is the case $n \equiv 6 \mod 8$ studied?
What about the maximum number of points in $\mathbb R^{n}$ such that each distance between a pair of points is an odd integer?


geometry
incenter
circumcircle
geometry proposed

Dear Mathlinkers,
let ABC be a triangle, (1) the incircle of ABC, I the incenter of (1), DEF the contact triangle of ABC, D' the antipôle of D wrt (1), D" the symmetric of D wrt B, and O the center of the circumcircle of ABC.
Prove : D'D" is parallel to OI.
Sincerely
Jean-Louis

To the moderator : can you please delete the above message. Thank you.

Dear Mathlinkers,
let ABC be a triangle, O the center of the circumcircle of ABC, (1) the incircle of ABC, I the incenter of (1), DEF the contact triangle of ABC, D' the antipôle of D wrt (1), D* the meetpoint of OI and BC, and D" the point of contact of the second tangent to (1) issue from D*.
Prove : D'D" is parallel to OI.
Sincerely
Jean-Louis

$ \Delta D''D*I\sim \Delta D''DD'$ then $ \angle D''ID*\equal{}\angle D''D'D\equal{}\angle D'D''I$
We are done!
Is it too easy?  :maybe:

Dear Livetolove,
yes, it is a direct application of Reim's theorem when considering the circle with diameter ID* and (1).
Sincerely
Jean-Louis


Find all functions $f: \Re \to  \Re$ satisfying:
$(x+y)(f(x)-f(y))=(x-y)f(x+y)$

Nice problem!

If we tak $x=-y$ and $x \neq 0$, then $f(0)=0$.
Now we can assume that $f(1)=k$ and $f(2)=a$, where a and k are reals.
Taking $y=1$ and $x=n+2$, where n is a natural, we have $f(n+3)=\frac{n+3}{n+1}(f(n+2)-k)$. And now we can prove by induction that $f(n+3)=\frac{a-2k}{2}n^2+\frac{5a-8k}{2}n+(3a-3k)$, for all integer n such tah $n\geq 0$.
We can extending this result for all integer n. It´s just obtain $f(-1)$ and $f(-2)$ in function of $a$ and $k$, and make a new induction. To obtain $f(-1)$ we can take $y=-1$ and $x=3$ and to obtain $f(-2)$ we can take $y=-2$ and $x=3$.
And now we have to extend this result for all racional n. We can take $x=qy$ and $y=\frac{p}{q}$ and we´ll have $(q+1)(f(p)-f(\frac{p}{q}))=(q-1)f(p+\frac{p}{q})$. And taking now $x=-p$ and $y=p+\frac{p}{q}$ we can calculate $f({p+\frac{p}{q}})$, and finally estend for all racional n.
In this point it´s just verify that $f$ is a continuous function. And this is in fact a merely consequence of it´s definition. We can prove that $f$ is diferenciable just looking to this definition. And soon we can extend that formula for all reals n.


Let \[ p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0, \] where the coefficients $a_i$ are integers. If $p(0)$ and $p(1)$ are both odd, show that $p(x)$ has no integral roots.

Theorem:
If $P(x)$ has integer coefficient and integers $a$ and $b$ are distinct integer numbers, then:
$a-b|P(a)-P(b)$

So assume, that integer roor exist, so $P(c)=0$, where $c$ is integer.$P(0)=2k+1, P(1)=2l+1$. Using theorem we get:
$c|2k+1$
$c-1|2l+1$
One of numbers: $c$ or $c-1$ is even, so one of numbers: $2k+1$ or $2l+1$ is even, which is impossible. Contradiction

The theorem is easy to prove, because of this formula:
$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-2}x+y^{n-1})$

Plus, if $a_{n}$ is odd to, then the polynomial doesn't have rational roots. ;)

[hide="Solution"]
Since $P(0)$ is odd, $a_{0}$ is odd. Mod 2, we have
\[p(x) \equiv 0+0+\cdots+0+a_{0}\equiv 1, \]
if $x$ is even and
\[p(x) \equiv a_{n}+a_{n-1}+\cdots+a_{1}+a_{0}\equiv p(1) \equiv 1, \]
if $x$ is odd. Therefore $p(x)$ can never be even, so it can never be zero. Thus it is impossible for $p(x)$ to have integer roots.
[/hide]

Here is my solution which is different from the above 2 so I am posting it here :)
[b]Proof:[/b]
$\longrightarrow$ Firstly note that $P(x)= \sum_{i=0}^{n} a_ix^i$ where $a_i \in \mathbb{Z}$.

Since $P(0)\equiv 1(mod 2) \implies a_0 \equiv 1\pmod {2}$

Since $P(1) \equiv 1(mod 2) \implies \sum_{i=0}^{n} a_i \equiv 1(mod 2) $

Thus, $ \sum_{i=1}^{n} \equiv 0(mod 2) $.

Now for the sake of contradiction we suppose that there exists an integer $c$ such that $P(c) =0$
.
$\implies \sum_{i=1}^{n} a_ic^i + a_0 = 0 \equiv 0(mod 2)$

$\implies \sum_{i=1}^n a_ic^i \equiv 1(mod 2)$

$\implies c\sum_{i=1}^n a_ic^{i-1} \equiv 1(mod 2)$ $\longrightarrow c\equiv 1(mod 2)$
$\implies a_n(c^{n-1} - 1) + a_{n-1}(c^{n-2}-1) +\dots + a_2(c-1)  + 2k \equiv 1(mod 2)$,where $2k = \sum_{i=1}^n a_i$
This is a contradiction since both the terms in the above expression are even.

Let $r$ be an arbitrary root of $p(x)$.
Suppose that $r \in \mathbb{Z}$.
Since $a_0 \equiv 1 \pmod 2$, it must be that $r \equiv 1 \pmod 2$ by RRT, and thus $r^k \equiv 1 \pmod 2$ ($\spadesuit$) for some positive integer $k$.
Additionally, $\sum_{i=0}^{n} a_i \equiv 1 \pmod 2 \Longleftrightarrow \sum_{i=1}^{n} a_i \equiv 0 \pmod 2$ ($\clubsuit$). (Because $a_0 \equiv 1 \pmod 2$).
Using ($\spadesuit$) and ($\clubsuit$), $\sum_{i=1}^{n} a_i r^i \equiv r\sum_{i=1}^{n} a_i \equiv 0 \pmod 2$. However, $p(r) \equiv \left( r\sum_{i=1}^{n} a_i \right) + a_0 \equiv 1 \pmod 2$, so $p(r) \neq 0$, contradiction. $\blacksquare$

$p(0) =odd \implies a_0 =2k+1$ and $p(1) =odd \implies \sum_{i=1}^{n} a_i=even$
Now let y be the root of $p(x)$
Then we have $y(a_n*y^{n-1} +... +a_1)=-a_0=odd$ clearly $y$ can't be even
For $y=2k+1$ we have $a_n*(2k+1)^{n-1}+.. a_2(2k+1) +a_1\equiv \sum_{i=1}^{n} a_i (mod 2)\equiv p(1) (mod 2) \equiv 0(mod 2)$
So $y$ can't be odd also hence no integer solution!!


geometry
3D geometry
tetrahedron
vector
Pythagorean Theorem
cyclic quadrilateral
area of a triangle

1. If triangle ABC has AB = c, BC = a, CA = b then its area is $S =\sqrt{p(p-a)(p-b)(p-c)}$(Heron formula, $p=(a+b+c)/2$).
  [b]Question[/b]: A tetrahedron with the lengths of the sides are given, can we calculate its volume?
2. Similar question for a convex polygon (area) and a convex polyhedron (volume)

If you give me six side lengths, I might be able to construct different tetrahedra with different volumes with those side lengths, so the volume cannot be determined from just the lengths of the sides.
For example, if it is known that the sides of a tetrahedron have lengths {2,2,2,3,3,3}, it is possible to construct a tetrahedron with an equilateral triangular base with side 2 or a tetrahedron with equilateral triangular base with side 3. Some arithmetic with show that these have different volumes.
However, if you give me a specific tetrahedron and the length of its sides, you can choose a base and find its area, and determine its height somehow (probably through a series of applications of the Pythagorean theorem or maybe some vector calculus). From these, you just apply the volume formula, V=(1/3)Bh. I don't think this is what you're looking for, but I also don't think what you're looking for can be found.

if you have a cyclic quadrilateral, it works almost the same as heron's formula

[url]http://mathworld.wolfram.com/CyclicQuadrilateral.html[/url]

i don't think there's a general formula for noncyclic quadrilaterals, because it depends on their angles and stuff.

Nice counter-example, Bictor!  If you want a counter-example whose counter-exampleness is easy to see without actually calculating, take tetrahedra with three sides of length $s$ and three of length $\frac s{\sqrt3} + \epsilon$ for a very small $\epsilon > 0$.  When we assemble the three sides of length $s$ as an equilateral triangle, the volume of the resulting tetrahedron is very close to 0.  If we assemble it the other way, we get a tetrahedron with some actual volume to it (in fact, volume larger than $\frac{s^3\sqrt{3}}{54}$).

As a matter of fact, the sides 2, 2, 2, 3, 3, 3 can be assembled into a tetrahedron in 3 different ways -- anyone care to figure out the volume of the third, less-symmetric version?  (The tetrahedron I offered can also be assembled in this third manner.)

So, why don't I offer related questions:
1. Find constraints on positive reals $a, b$ such that:
(a) There are three distinct tetrahedra with sides $a, a, a, b, b, b$
(b) There are only two such tetrahedra
(c) There is only one such tetrahedron
(d) There are no such tetrahedra

2. What about if the tetrahedron is to have sides of length $a, a, a, a, b, b$?

3. What are the volumes of all the tetrahedra in question 1.(a) and 2?

I am not sure if you are looking for a forumula like this:

[url=http://mathworld.wolfram.com/Cayley-MengerDeterminant.html]mathworld.wolfram.com/Cayley-MengerDeterminant.html[/url]

actually, there is a general formula for a quadrilateral sides a,b,c,d, and given the sum of the opposite angles, this is in aops vol 2 i believe


Let $ n$ be a positive intger and $ x$ a real number distinct from $ 1$ or $ \minus{}1$.
Prove that

\[ \binom{2n}{n}\left(\dfrac{x^{2n\plus{}2}\minus{}1}{x^2\minus{}1}\right)\ge(x\plus{}1)^{2n}\]

:)

Come on this is a nice and easy inequality. Try it!
 :)

[hide]Write the second multiplicand on the LHS as $ x^{2n} \plus{} x^{2n \minus{} 2} \plus{} \ldots \plus{} x^2 \plus{} 1$ and use the binomial theorem on the RHS to get $ \binom{2n}{n}x^{2n} \plus{} \binom{2n}{n}x^{2n \minus{} 2} \plus{} \ldots \plus{} \binom{2n}{n}x^2 \plus{} \binom{2n}{n}\ge \binom{2n}{1}x^{2n} \plus{}$ $ \binom{2n}{2}x^{2n \minus{} 1} \plus{} \binom{2n}{3}x^{2n \minus{} 2} \plus{} \ldots \plus{} \binom{2n}{2n}$ which can be easily inducted. (Hint: assume wlog that x is positive)[/hide]

EDIT: Oh wait; one can't assume without loss of generality that x is positive because of the x+1 on the RHS - I think it can still be inducted upon by considering the two different cases, though.

EDIT again: ...oh yeah, the case where x is negative is somewhat trivial. Never mind.

[hide="Outline"]$ \binom{2n}{n}\equal{}\sum_{i\equal{}0}^n \binom{n}{i}^2$. Use Cauchy.[/hide]

that is precisly my solution worthawholebean, moreover this inequality was created in order for that to work jiji

:)

Heh; I'll have to remember that identity - my solution was ridiculously ugly compared to that.

now you see why this is almost funny, don't you?  :D

We are all learning here fellow. jiji, besides, long solutions also get a 7.  :P

[quote="worthawholebean"][hide="Outline"]$ \binom{2n}{n} \equal{} \sum_{i \equal{} 0}^n \binom{n}{i}^2$. Use Cauchy.[/hide][/quote]

very nice and elegant solutions!!

 :wacko:

I guess the inequaltity isn't so funn nevertheless very simple.
It is very well known that

$ \binom{2n}{n}\equal{}\sum_{k\equal{}0}^n \binom{n}{k}^2$. On substituting it in the given inequality,
we just get a direct consequence of the Cauchy Scwarz Inequality.
Hence, proved.

[quote="worthawholebean"][hide="Outline"]$ \binom{2n}{n} \equal{} \sum_{i \equal{} 0}^n \binom{n}{i}^2$. Use Cauchy.[/hide][/quote]

sorry, but I don't understand how that can be :oops:

[b]@ ElChapin:[/b] What is funny about this inequality?

[b]@ Attila the Hun:[/b] Suppose you have $ 2n$ people and you want to choose $ n$ of them. There are $ \binom{2n}{n}$ ways of doing so. Alternately, we can count the number of ways to do this as follows:

Divide the $ 2n$ people into two groups of $ n$ people, say group A and group B.
For each $ i \in 0 \ldots n$, we can choose $ n$ people by choosing $ i$ people from A and $ n \minus{} i$ people from B.
You should verify for yourself that there is no overcount or undercount.

Thus, for each $ i$, there are $ \sum_{i \equal{} 0}^n \binom{n}{i} \binom{n}{n \minus{} i} \equal{} \sum_{i \equal{} 0}^n \binom{n}{i} ^2$ ways of selecting a total of $ n$ people using the selection scheme above.

Hence, $ \binom{2n}{n} \equal{} \sum_{i \equal{} 0}^n \binom{n}{i} ^2$

ohhhhh yessss I get it now. :trampoline:  I proved the identity myself a few months ago as an excercise question. I did not use a counting proof though. I used generating functions.
consider the product of the two series:

nC0 + (nC1)x + (nC2)x^2 + .... + (nCn)x^n = (1 + x)^n

nC0 + (nC1)/x + (nC2)/x^2 + ...... + (nCn)/x^n = (1 + 1/x)^n.

the required series on the RHS is the term independent of x in the product of these two series

(1+x)^n * (1+1/x)^n

= (1+x)^(2n) / x^n

and so, the sum of the series on the RHS is C(2n, n).

I did not know it was so well known though. :D

however, nice question and nice identity.  :)


The sequence $ {a_n, n\geq 1}$ is defined in the following way: $ a_1 \equal{} 1, a_n \equal{} a_{n\minus{}1} \plus{} (a_{n\minus{}1} mod n)^2, n\geq 2$, where $ a$ mod $ b$ denotes the remainder of division of $ a$ by $ b$. What is the maximal possible number of consecutive equal members of this sequence?

Could you post solution for your conjecture?


trigonometry inequalities geometry trapezoid quadratics Support calculus

First off, I plead JBL not to move this to the Fun Factory  :maybe: ; I'm looking for serious responses.

Will someone give the purpose(s) of formal education?

If the purpose is being fulfilled successfully in schools, and that purpose is a reasonable one, then that is good; please say if this is your opinion. If there is little purpose in an education, or if modern schools do much more than to fulfill their purpose, should not much less money be spent in that direction? If, on the other hand, schools are a huge failure at what they should be accomplishing, please explain what specific goals schools are failing to attain and what makes those goals so important that education should be mandatory. Is it perhaps difficult to ascertain the purpose of forcing children to go to school, and it is only known that it is a good thing and somehow helps society?

For example, if you believe that the purpose is to develop the knowledge, skill, and character of the population (taken from the first Google hit on "the purpose of education"), what knowledge, skill and character should all people have, why do you think that this is a worthy purpose, is it ethical to impose these causes on people (for example, I believe it is bad for someone to weigh more than 300 kg, but I don't believe it should be mandatory that one weigh less than that number), and how are schools doing in terms of fulfilling these purposes?

And perhaps, specifically, what level of mathematics should the population be familiar with and why?

To me the purpose of formal education is to make life easier and further the standard of living. Before technology and all the advances society has made in the last couple centuries life was hard. Most families had to use physical labor or force physical labor in order to put food on the table, and entertainment such as t.v. and movies didn't exist. Education allows humanity to go forwards, most if not all significant technological accomplishments were made by someone with a formal education. And the hope is that future generations will keep this trend, so that the standard of living will continue to rise and life will even become easier.

One big example of all this would be the field of medicine. In the 1600's the average life expectancy of someone was about 40 years old or greater with a very high infant fatality rate. Whereas the life expectancy now is in the upper 70's. This is because of advances such as the discovery of Penicillin, or anesthesiology which allows for modern surgery.

The public schooling system in America was founded in the late 19th century as a result of the Gilded Age. Many working-class families were extremely poor, as there was little legislation protecting workers' rights (i.e. no minimum wage law). Thus, many families had both the husband and wife going to work all day. The children, left at home with nothing to do, became public nuisances, stealing, breaking, running free in the street, etc. This caused a general havoc, and it became necessary to find a place for these children to stay while their parents were at work, so they would not be a detriment to society. Schooling came about as the foremost option.
Education is actually a secondary goal of schooling, and I believe that because of its secondary status, it has been neglected. Mostly it's the math  :) but a lot of other things tend to be repetitive. I think the key is not money, but teachers who understand the topics being taught. A good teacher can teach anyone with any equipment.

[quote="leoxnlin"]Will someone give the purpose(s) of formal education?
.......................
what makes those goals so important that education should be mandatory. Is it perhaps difficult to ascertain the purpose of forcing children to go to school, and it is only known that it is a good thing and somehow helps society?[/quote]
Maybe the goals should be to acquaint people with rules for life in society, and societal morals, ideas and expectations, and to pass on some knowledge and skills for society to survive and develop. We want Johnny to know how he is expected to behave. We want knowledge to be passed on. We want competent fellow members of society because we will depend on them.
[quote="leoxnlin"]what knowledge, skill and character should all people have, why do you think that this is a worthy purpose, is it ethical to impose these causes on people (for example, I believe it is bad for someone to weigh more than 300 kg, but I don't believe it should be mandatory that one weigh less than that number), and how are schools doing in terms of fulfilling these purposes?

And perhaps, specifically, what level of mathematics should the population be familiar with and why?[/quote]
The population should be able to read, write, balance a checkbook and fill forms.  The rest is open to discussion. However, I think diplomas and grades should mean something. Maybe people should be able to opt out of formal education after attaining a minimum competency, but those who choose to continue their education to a diploma should be able to demonstrate the knowledge that they have supposedly obtained.
Personally, I would like most of the population to have at least some rudimentary understanding of physics, chemistry, biology, algebra and geometry. Maybe adding trigonometry to the list is asking too much.

[quote="worthawholebean"][quote="evilhamster"]A good teacher can teach anyone with any equipment.[/quote]It's really hard to teach, though, when you [i]don't[/i] have the equipment, especially in science classes which require special stuff.[/quote]

bah

teach theory then

that is obviously better anyways

Actually, my point is that one can always find ingenious ways to circumvent problems if all parties are interested in doing so. That, I believe, is the other problem of education: Schools have failed to link education to success in life, thus creating disinterest. Nobody can be taught under force, as learning is an internal process.
By the way, KMST, if you will notice, civilization did exist, and it was no worse than today (barring technological differences), before public schooling. I think education should be compulsory, as then society could very quickly filter itself into good and bad. The people who would become outstanding later in life would attend school, because of their desire to learn, while those without internal incentives would not. Thus, a corporation or government could save much time by not having to pick incompetent workers. This would only be possible in a rather communist country, though, as the costs of schooling only a few students would be high.

I am not doubting or challenging what anybody says, although it may sound like it; I merely am curious as to how people feel.

[quote="evilhamster"]...many families had both the husband and wife going to work all day. The children, left at home with nothing to do, became public nuisances, stealing, breaking, running free in the street, etc. This caused a general havoc, and it became necessary to find a place for these children to stay while their parents were at work, so they would not be a detriment to society.....
Education is actually a secondary goal of schooling, and I believe that because of its secondary status, it has been neglected.[/quote]

[quote]I think education should be compulsory, as then society could very quickly filter itself into good and bad. The people who would become outstanding later in life would attend school, because of their desire to learn, while those without internal incentives would not. Thus, a corporation or government could save much time by not having to pick incompetent workers.[/quote]

(Ah! "Compulsory"! Couldn't find that word...   :oops: ) So, school's primary purpose is to lock up children so they'll only bother and annoy a limited number of the adult population, namely teachers, and its secondary purpose is to sort out the competent from the incompetent at an early age? Is this morally right? How are schools doing so far?

Also, you seem to think that whether people will be "competent" or not largely depends on their school performance. Yet, many people regarded by most as successful in life were not necessarily successful in school. Can you elaborate on this?

[quote="MysticTerminator"][quote="worthawholebean"][quote="evilhamster"]A good teacher can teach anyone with[i][out][/i] any equipment.[/quote]It's really hard to teach, though, when you [i]don't[/i] have the equipment, especially in science classes which require special stuff.[/quote]

bah

teach theory then

that is obviously better anyways[/quote]

(I will now respond seriously to sarcasm.) Is there such a thing as a fundamental difference in learning types between theoretical and practically (hmm... I wish there were a better word for this) -minded people? (It frustrates me especially, to have people assume I like science or engineering or whatnot because I show interest in math... even pure, natural sciences seem too practical-real-world-ish to me... and also those online personality tests and career finders that schools make you take say the same thing; I don't know how many times I've read something to the effect of "You are a very logically-minded person... you should consider becoming an engineer...")

[quote="KMST"]
The population should be able to read, write, balance a checkbook and fill forms.  The rest is open to discussion. However, I think diplomas and grades should mean something. Maybe people should be able to opt out of formal education after attaining a minimum competency,...[/quote]

...which is? "Minimum competency" might have quite varied meanings to different people.

[quote]
...but those who choose to continue their education to a diploma should be able to demonstrate the knowledge that they have supposedly obtained.
Personally, I would like most of the population to have at least some rudimentary understanding of physics, chemistry, biology, algebra and geometry. Maybe adding trigonometry to the list is asking too much.[/quote]

None of those topics are typically classified as humanities... are you perhaps biased?  :wink: Ability in English (no, not the ability to write an essay full of nonsense about literary elements in Dickens - he himself probably couldn't have cared less) is arguably quite a lot more useful in life than being able to balance a chemical equation.

While education is generally useful, one must think in terms of the creators of education, namely politicians and upper-class patriarchs during the Gilded Age, who were the only people at the time with the power to create such a widespread institution. These "robber barons" were only interested in education to stop children from costing them money in repairs. They actually did [i]not[/i] want the general public to become educated, as then their workers would be well-informed and a majority would be able to argue against the upper class, as well as be unsatisfied with blue-collar jobs. An uneducated working class would be content with low-level jobs, as well as being unable to argue against the unfair practices of their superiors.
Survival of the fittest rules this decision. The "fittest" became the upper class, and in keeping with the dog-eat-dog principles of this world, they tried to keep the rest of the population under a firm foot. This principle continues to today, even though the public may complain of inequalities and unethical or unfair distribution of wealth (no reference to Obama). Maybe this is why education continues to lag, despite numerous efforts to improve the quality of education.

There could be other reasons why our education is not rising. In my state of Texas kids never "understand" or never try to and when TAKS (Texas Assessment of Knowledge and Skills) comes they flunk and complain that school work was too hard. Thus they make the program less demanding and higher grades are repeating about 85% of the work done in the previous year! and the other 15% is on new ways of understanding and things we were actually supposed to learn. I'm lucky to be in a Pre-AP class (I'm only in 6th grade :( so i should have been learning seventh grade math but nooooooo i have to learn 6th grade before 7th so I'm spending school days learning about prime factorization. Thank god theres an Aops or i would never learn anything. Back on topic there are many hispanics in my school some of them immigrants. I have nothing against them except the fact that they choose not understand and work hard in school. (not all of them are like this but a good part are) They are pulling the rest of us from learning anything new because all the teachers are too busy helping them then to teach us. (or at least their teachers are) Every year now I'm going to be learning the same thing over and over....... WHOO HOO. I can't w8 to start 8th grade and learn about geometry. (Not being sarcastic)

[quote="mathblitz"]There could be other reasons why our education is not rising. In my state of Texas kids never "understand" or never try to and when TAKS (Texas Assessment of Knowledge and Skills) comes they flunk and complain that school work was too hard. ...[/quote]

When something doesn't seem to go well, we must go back and ask ourselves, "What is the point of doing all this anyways?" That is the topic of this..., well, topic.

And please don't blame your fellow students for holding classes back - it's not their fault, although it may seem to be. What's more, they probably have some skills you lack, that they consider more important and may be frustrated, as you are, that others can "get away" in life without learning those skills.

Which brings us back to the question, "What is the point of education? What makes knowing how to find the area of a trapezoid or the Krebs cycle more important than knowing how to... survive in a wilderness?" (although I'd much prefer, personally, learning the former rather than the latter... it being easier...)

Mathematics is the discipline that requires the least knowledge, it seems.

[quote="leoxnlin"]And please don't blame your fellow students for holding classes back - it's not their fault, although it may seem to be. What's more, they probably have some skills you lack, that they consider more important and may be frustrated, as you are, that others can "get away" in life without learning those skills.[/quote]

Actually in my experiences it is because they lack the motivation to work hard, and make the pace of class slower.

Anyways, back on topic. I think knowledge is good for the sake of knowledge. School is there to familiarize students on all the available options they have when they move on to college. For instance the area of a trapezoid is very useful in architecture, or surveying, while having no importance in medicine and a lot of other fields. We learn about it so we can find where we would like to be society is when we move on to college and to work. So even if we don't use everything we've learned in school, some other classmate could be using that information that you are not to help him/her succeed in whatever field they go into.

Then would it be reasonable, if a student absolutely loathes what s/he is learning, for the student to be allowed to skip that course? If they hate it, they're probably not going to pursue a career in that field.

The question is, how old must a person be before their own judgment can be trusted?

(And about the holding classes back, I believe most educators have a curriculum that they are forced to teach... so it's not really a direct effect of those fellow students sitting right in the same classroom as yourself that the classes seem slow for you.)

Yes, but the curriculum is designed with the students in minds.

Back on topic, that is a good question leoxnlin and that comes down to the decision of the parents and the government. The United States is a country that holds a lot of Greek values, such as the value of being well rounded human beings, rather than isolating oneself to one subject for your entire life. Personally I think its a good idea to let a kid skip a class if he/she knows that's what they are not going to major in, but others think the opposite.

yes but some of the things that you learn in other classes can help you succeed in your profession. For example a good mathematician needs the writing skills he gets in ELA to publish his works. A composer needs some knowledge of math to help make the rhythm stay on track. Although i believe we should be allowed to start majoring in a profession when we reach high school.

I agree with silverlight but I dont think that people should be allowed to major in their professions in highschool because even then someone may not know what they want to be and forcing them to choose is harsh.

mathblitz, there's a difference between allowing students to choose a major and [i]forcing[/i] them.

I think that schools today are putting too much focus on well-roundedness.  A mathematician or scientist will most likely not need to know about the structure of European feudalism or the complexities and nuances of a Shakespearean play.  Likewise, a historian will probably not need to know about how to determine the coefficient of friction between two surfaces or how to solve a quadratic equation.  However, students who want to pursue these careers still need to take these "unnecessary" classes.  This is true for me; I hated my world history class.  If a student knows what he/she wants to do in life, then a school should encourage him/her to take more classes in that area and less in other areas.

Not to say that the swedish education system is any good, but we do choose majors in high school (although broader than in university), with a number of more practical specializations so that you graduate certified to do some job (plumber-type jobs, electrician, carpenter, driving various vehicles, farmer etc). If you want to do natural sciences or engineering at university you need to choose an appropriate 'major' in high school, and if you want to do social studies you need to do another (although I think if you do the natural sciences one you can with little effort take some extra classes to do social studies)

It's not really an issue of debate here that this would force people to choose too early or something

So most here agree that a solution to the students-are-learning-useless-stuff problem is to have early specialization?

Does anyone think that each individual should attain a high level of familiarity with many subjects, a level equal to or more than the current (and by 'current' I am thinking of the US, but hmm... I don't really know about other education systems) level attained (or supposedly attained) by college graduates?

[quote="lingomaniac88"]mathblitz, there's a difference between allowing students to choose a major and [i]forcing[/i] them.[/quote]

By forcing them i meant that they had to choose a major and couldnt go any farther without one.

I guess education is just a place where we can stay when our parents are at work or are busy and be productive with our minds except that education today has become much more then just a place to stay and be productive. Now it's pretty much required if you want a good job (thus it has become competitive) and has become i guess a government "business." The point of education is so that we can rise up to the worlds demands and our own and succeed without too much hardship.

Are the world's demands reasonable? Do GPA's and volunteer hours make better employees?

I think leoxnlin has a point, but it should be elaborated. We have all heard stories of multimillionaires and billionaires who started life on the streets and dropped out of school to support their impoverished family, but, through hard work, rose through the corporate ranks (Andrew Carnegie being a good example). Meanwhile, over 90% of lottery winners go bankrupt in the first five years after they win, because they are unable to handle money well. Monetary value and assets are just one indicator, of course, but the same characteristics and life skills probably can be extended to any field or subject. Maybe schools should try to find the defining characteristics of all great people, who can rise to the top no matter where they start, and focus on those instead of education.

Note that many of the sciences have been pushed to new levels by people who didn't even specialize in that field, or have any high degrees. Copernicus was a minister in a church and Edison dropped out of school.

rich people are not the educated
but the frugal
and the hardworking
EDIT: Or you could always go for the pro sports field

[quote="leoxnlin"]...therefore, education is not only a secondary goal of schooling, but also one with little purpose? So far we have established, with almost no opposition, that schools are pens for keeping children during the day.

Is this our final conclusion?[/quote]
Yep  :D . Mind you, that's because it's the only working alternative to general havoc in the streets. The vast majority of students hate school and view it as a burden, no matter how hard education boards try to excite them. Thus, they resort to making schools boring and pointless.
@funtwo: Rich people [i]are[/i] educated-they're educated in the art of being successful in this world. And generally you pick up some pretty good information along the way. Take AoPS-it has nothing to do with school, as it is completely optional, but at the same time teaches academia (namely, math). And pro sports is not necessarily a good field-the NFL has to hire financial advisors for their players to help them use their money better (and so they will not ask for more.).

What a startling consensus we have here! Is this not something big? Should schools take off the cover of saying that the point of school is to educate, and stop giving homework and other graduation requirements (sometimes I wonder what in the world the US would do were it not for required high school community service hours  :roll: )? Or would that be going back to general havoc in the streets? Would that create poorly disciplined adults? I wonder what those top people in education agencies say the purpose of educating our young people is...

There has to be something good about schools, since we've had them for so long. I suppose it develops people's character. And they couldn't figure how else to do so but by giving useless, random stuff to memorize.

Maybe young people should go to work sooner, to be like apprentices or something, like before.

I feel that parents are way more responsible than the schools for what their kids learn and what behaviors they develop. Some parents are apathetic and take the schooling system for granted, not realizing all the work and money that goes into it from the government.

Mandating education creates a reasonably predictable demand level, making it feasible to create a public education system in the first place. Otherwise, the government would not even be able to efficiently (I use this term loosely) provide a decent education to those who want it.

[quote="leoxnlin"]For example, if you believe that the purpose is to develop the knowledge, skill, and character of the population (taken from the first Google hit on "the purpose of education"), what knowledge, skill and character should all people have, why do you think that this is a worthy purpose[/quote]

Well, here's a few of them --

[b]For everyone:[/b]
1. Literacy and communication - [i]English, foreign language, general environment[/i]
2. Basic interpersonal skills - [i]General environment[/i]
3. A (very) basic outline of US history and government - [i]History[/i]
4. Knowledge of basic quantitative tools and scientific concepts that will streamline and improve performance in nearly every profession and in daily tasks - [i]Math, science[/i]
5. Awareness of basic principles of health, such as the hazards of smoking and drinking, and how not to get a venereal disease

[b]For those who chose:[/b]
1. An experience in the arts, because the arts provide a medium of communication and propagate culture
2. Leadership through student-led initiative opportunities, such as student government, student clubs and organizations
3. Understanding and interest in mathematics and science to drive academic and technological innovation
4. Athletics, because they promote good health and healthy competition
5. Community service, because if taken seriously, it is an eyeopening experience that exposes students to other environments, and also because it promotes good character

[quote]is it ethical to impose these causes on people.[/quote]

It is ethical for the United States to require citizens and voluntary residents to support the nation within the scope of the law and the Constitution.

[quote]for example, I believe it is bad for someone to weigh more than 300 kg, but I don't believe it should be mandatory that one weigh less than that number.[/quote]

The most obvious difference is the effect of this person on voting.

Well, assuming that we do not want to give up the current sciences and technologies completely, what would be the alternative to the mandatory education that would allow the society to support them? The main purpose of low level schooling is giving the students the general ability to learn and forcing children to use (and develop) their brains, which, unlike walking and eating, is not something that comes naturally.

Every society has had "mandatory education" for the youngsters (though what exactly was being taught to whom depended heavily on time and place) just to pass the crafts and general knowledge developed by the previous generations. If we do not teach children the current school curriculum, we'll have to teach them something else (say, how to work in coal mines, etc.) unless we want the electric lights in NY to go out tomorrow.

I think that providing the general education to everybody up to the age of 18 or so is a big step forward compared to teaching the son the craft of his father practiced in the medieval ages but if somebody thoroughly dislikes being taught language, history, math, and other "useless general subjects" (whose main goal is just to train one's brain a bit and make one able to learn virtually any craft in existence later on), I do not see why he should be prohibited from being sent to learn some unskilled labor starting at 7 years of age. Indeed, why to waste money and effort on trying to bring everybody's mental abilities to a reasonable level?

Once upon a time, education was a privilege, later it was a right. Is it now considered a burden? If so, we can go back to the ancient Egyptian model of society where everything concerning the value of education was very straight and clear. One written document of that time starts with "Learn, my son, and you'll not live the life of a slave who is beaten with a wooden stick every day..." (well, the quote is not exact, but I don't think I changed the general sense of it). That was quite an incentive, as you can tell. This incentive is still there except the cause-effect relation may be not so immediate nowdays.

The question of how well the current public education system serves its purpose is a separate one. I agree that it is far from perfection. But still, I consider it a great achievement of the modern civilization, much more important than so much praised "democracy", "equality", "freedom", and many other things on a long run. It is interesting that its development in the West more or less coincides in time with the development of the current technological society (both date back to about 1800). I surmise it is not a pure coincidence ;)

I think you are right about forced education's appearance concurring with the development of "power to the people" ideals in Western society. However, in the U.S., real attention to schooling started with Horace Mann, who didn't push for reform in the educational system until half a century after the Constitution had been drafted. Until then, legislative bodies made only a brush-over of public education.
I believe the perceived burden of school comes from the very fact that it is imposed. School becomes routine and boring, and people don't see the point in doing well. In Third World countries, where the population is poor, and education scant or nonexistent, children jump on opportunities to learn. Routine items tend to get drab and joyless, no matter how appealing (breathing is definitely good, but do you enjoy every breath you take in the knowledge that it is keeping you alive?), and thus students view school as a time-drainer.
In addition, students look at those who are much smarter and think "We don't need to do anything, they'll solve all our problems for us!" They are unable to see the world as a whole, and recognize that the world is in dire need of brilliant minds. These students also tend to not see the everyday geniuses who are solving the world's problems quietly, because it's the pop culture stars that are glamorized in the press, and they (again) are unable to see past what is in front of their face. In poor countries, where the problems of the public are brutally displayed every day, louder than any media, the impetus to help society is especially strong. Thus, they work hard at school, and try to scrape their way to a higher education, in the hope that they will be able to bring back a degree and some measure of prestige, attention to their society's plight, and solutions to the problems of undeveloped countries that plagued them when they were children. These laborers recognize that an education is the best one can get in their situation, while the Western youth does not.
What does everyone else think?

I personally don't think that the educational aspect of school is the most important. Much of school is about developing social skills, which is becoming more and more useful in the world (and in the US, in particular). It doesn't matter how knowledgeable you are, if you lack the ability to communicate effectively. I believe that the American school system actually does a pretty good job of preparing students, in this perspective. I guess what I'm trying to say is that the purpose of school, while it may seem so at first, is not to "educate" people as much as to form strong public skills.

So I've spent some time gathering my thoughts.

[quote="fedja"]

I think that providing the general education to everybody up to the age of 18 or so is a big step forward compared to teaching the son the craft of his father practiced in the medieval ages but if somebody thoroughly dislikes being taught language, history, math, and other "useless general subjects" (whose main goal is just to train one's brain a bit and make one able to learn virtually any craft in existence later on), I do not see why he should be prohibited from being sent to learn some unskilled labor starting at 7 years of age. Indeed, why to waste money and effort on trying to bring everybody's mental abilities to a reasonable level?

[/quote]

Seems reasonable.

[quote]
Once upon a time, education was a privilege, later it was a right. Is it now considered a burden? If so, we can go back to the ancient Egyptian model of society where everything concerning the value of education was very straight and clear. One written document of that time starts with "Learn, my son, and you'll not live the life of a slave who is beaten with a wooden stick every day..." (well, the quote is not exact, but I don't think I changed the general sense of it). That was quite an incentive, as you can tell. This incentive is still there except the cause-effect relation may be not so immediate nowdays.
[/quote]

By why, why should the well-learned not be the slave beaten with a wooden stick? As far as I'm concerned, anything taught in schools that has no relevance to more than a good proportion of professions, and is thus forgotten by most adults, and yet is used to separate those beaten with the stick and those beating with the stick is merely a social construct and does not have much true value. Albeit a very common construct - it occurs in just about every culture - the reason seems to be that in history only those who had money had the time to sit around and study prime numbers and whatnot.

I'm very willing to be convinced that all this stuff learned in school, or supposedly learned in school, helps us. In fact, I want to be convinced, since I would hate living a hypocritical life, as I do plan to remember as much as I can from my education and become what I now am forced to view as a useless intellectual doing nothing to help society.

At the moment though, I just don't see it.

[quote]
The question of how well the current public education system serves its purpose is a separate one. I agree that it is far from perfection. But still, I consider it a great achievement of the modern civilization, much more important than so much praised "democracy", "equality", "freedom", and many other things on a long run. It is interesting that its development in the West more or less coincides in time with the development of the current technological society (both date back to about 1800). I surmise it is not a pure coincidence ;)[/quote]

Yes, one could even say that intellectual works of the intellect last longer than anything else created to humans... that scientists are noble workers paving the eternal path that all will follow in the future...

But then... Arguably, most scientists and mathematician's work, especially nowadays, doesn't do a thing for society, and, contrary to what many would have a student believe, will not in the future either. That is, the work is theoretical and not practical, science and not engineering.

[quote="math154"]I guess what I'm trying to say is that the purpose of school, while it may seem so at first, is not to "educate" people as much as to form strong public skills.[/quote]

So do you think that students are being treated unfairly, given all these assignments? Have many a generation toiled needlessly? Is this not a colossal waste of paper, time, and resources? If this is your true view then would you agree that education is the most horrible thing that has ever happened, that it is the biggest failure of humankind?

Or is it all a disguise to hide from the students the truth? 'If students knew, they'd wreak havoc; these young people have not the experience nor the judgment to do what's best for themselves'.

[quote="leoxnlin"]By why, why should the well-learned not be the slave beaten with a wooden stick?[/quote]
Doing so consumes time and effort, while hindering your ability to perform your duties towards yourself, your family, your friends your society, your nation, and God.

[quote]As far as I'm concerned, anything taught in schools that has no relevance to more than a good proportion of professions, and is thus forgotten by most adults, and yet is used to separate those beaten with the stick and those beating with the stick is merely a social construct and does not have much true value.[/quote]

What about reading, writing and social skills? These would not be picked up that well if parents who did not have the time to home-school their kids (most parents) had the burden of providing these.

What about all the life lessons you learn from being around a large group of other students, who are also learning and developing? Without these, you would have to learn everything for yourself, often the hard way.

What about math and science? It is true that a person might forget, say, techniques for computing integrals, later in life. But they've absorbed the concepts and have gotten a feel for what they are, and can apply them as necessary. Had they not learned them, they would not know what they are doing.

[quote]But then... Arguably, most scientists and mathematician's work, especially nowadays, doesn't do a thing for society, and, contrary to what many would have a student believe, will not in the future either. That is, the work is theoretical and not practical, science and not engineering.[/quote]

This leads to an interesting point of discussion -- does science and engineering really even serve society?

In any case, mathematics and science lay the theoretical groundwork from which applications can be discovered. If the math or science isn't there, no brilliant person can jump out and guess its application. It is really unreasonable to expect an engineer to anticipate all the math or science that might be needed to implement a particular novel idea and then ask for it to be researched. That produces very little progress for a large level of input creativity.

[quote]So do you think that students are being treated unfairly, given all these assignments? Have many a generation toiled needlessly? Is this not a colossal waste of paper, time, and resources?[/quote]

No it is not. Just because forming social skills is a major result of education, does not mean it is the principle goal. It is best viewed as a byproduct. The fact is that lots of those assignments help lots of students learn lots of things, whether they admit it or not. Also, having to complete homework assignments develops many other important life skills, such as time management, study techniques, work ethic, etc., either directly or by compelling you to develop them through the other things you do.

[quote="leoxnlin"]
[quote="math154"]I guess what I'm trying to say is that the purpose of school, while it may seem so at first, is not to "educate" people as much as to form strong public skills.[/quote]

So do you think that students are being treated unfairly, given all these assignments? Have many a generation toiled needlessly? Is this not a colossal waste of paper, time, and resources? If this is your true view then would you agree that education is the most horrible thing that has ever happened, that it is the biggest failure of humankind?

Or is it all a disguise to hide from the students the truth? 'If students knew, they'd wreak havoc; these young people have not the experience nor the judgment to do what's best for themselves'.[/quote]
[quote="TZF"][quote]So do you think that students are being treated unfairly, given all these assignments? Have many a generation toiled needlessly? Is this not a colossal waste of paper, time, and resources?[/quote]

No it is not. Just because forming social skills is a major result of education, does not mean it is the principle goal. It is best viewed as a byproduct. The fact is that lots of those assignments help lots of students learn lots of things, whether they admit it or not. Also, having to complete homework assignments develops many other important life skills, such as time management, study techniques, work ethic, etc., either directly or by compelling you to develop them through the other things you do.[/quote]

I agree with TZF in that students do learn things, and that we need education, which guides our world in, at least, some direction. I just don't really like what we do to educate. For example, being a conservative, I think we waste too many resources (especially paper), and that there are many places we could cut on use or be more efficient in usage or both. I also disagree with the styles of teaching, the areas that are focused on, etc.


[quote="Mark9182"]Here are the 3 problems out of the 25 for my homework that I can not figure out.  Any help would be greatly appreciated.

1.Evaluate to four decimal place using a calculator e2.9

2.Find the equation of the hyperbola satisfying the given condition: Vertices at (9.0) and (-9,0); foci at (11,0) and (-11,0)

3.Find the focus and directrix of the parabola x2 = 12y[/quote]

[hide]for number 1, do you mean e^2.9.
=18.1741 using a calc
[/hide]

[hide="#2"]
\frac{\left( x-h \right)^2}{a^2} - \frac{\left( y-k \right)^2}{b^2} = 1
c=121
c^2={a^2}+{b^2}
$121=81+{40^{2}}$

b=\sqrt{40}
it is centered at 0 i think, so just plug in the values[/hide]

Thank you.  The square root is exactly what I meant.  Any help on the other two would be greatly appreciated.

[url=try this link]http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_conics_directrix.xml[/url]

weird site, but has stuff for the 3rd problem. i pretty much forgot all of the parabola stuff, and i don't want to give wrong answers


calculus
integration
trigonometry
function
algebra
rational function
calculus computations

Let $|a|<\frac{\pi}{2}.$ Evaluate the following definite integral.

\[\int_{0}^{\frac{\pi}{2}}\frac{dx}{\{\sin (a+x)+\cos x\}^{2}}\]

[quote="sylow_theory"]Maybe this is the stupid way to approach it but maybe expand $\sin (x+a)$ as a sine angle formula, to obtain a rational function of sines and cosines. Then use a Weierstrass substitution.[/quote]

I think that's not a bad approach. I also expanded $\sin(x+a)$ using angle formula, and then substitued $t = \tan x$.

Is the answer:

[hide]$\frac{2}{1+\text{sin a}}$[/hide]

Ohhh...I typed it wrong :blush:
I also get $\frac{\sec a}{1+\sin a}$ too. :P


On a twelve-hour digital clock which shows hours and minutes, what time is it when the sum of the digits displayed is the greatest?

The greatest possible sum for hours is 9;
The greatest possible number for minutes (tens) is 5;
The greatest possible number for minutes (ones) is 9;

the time must be 9:59.

Note: Don't say 12:59, because even though 12>9, 1+2+3<9.

[color=blue]Fixed your typo. I'm afraid $ 1 \neq 2 \neq 3 < 9$.  :P  -Izzy[/color]


inequalities
real analysis
function
integration
triangle inequality
real analysis unsolved

Let $ 0 < p < 1$ and $ f, g \in L_p$. Show that there exists $ C > 0$ such that
\[ \|f\plus{}g\|_{L_p} \leq C(\|f\|_{L_p} \plus{} \|g\|_{L_p}).\]
What is the best constant $ C$?

Could you explain what $ L_p$ is and what the norm is defined to be there (or should this work for any metric in that space)?

Let $ L_p$ be the space of all Lebesgue measurable functions defined over $ [0,1]$ such that
\[ \|f\|_{L_p} \equal{} \left(\int_0^1 |f|^p\right)^{1/p} < \infty.\]
Function $ \| \cdot \|_{L_p}$ is not a norm when $ 0 < p < 1$ but it is a quasi-norm (i.e. in the triangle inequality there is additional constant $ C > 1$).

Sharp constant is $ 2^{\frac{1}{p}\minus{}1}$. I have no time to reproduce the prove right now, just take a look here http://www.math.ucla.edu/~tao/247a.1.06f/ ,Notes 1 , you can find a skech of the prove and nice generalization.

[quote="leshik"]Sharp constant is $ 2^{\frac {1}{p} \minus{} 1}$..[/quote]
The example that proves that you can't do better than that: Let $ A$ and $ B$ be disjoint sets, each of measure $ a>0$ and let $ f\equal{}\mathbf{1}_A,g\equal{}\mathbf{1}_B.$

Then we have $ \|f\|_p\equal{}\|g\|_p\equal{}a^{1/p}$ and $ \|f\plus{}g\|_p\equal{}(2a)^{1/p}.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Does there exist a finite subgroup of a profinite group with infinite closure?

No. Such a subgroup is compact (because it is finite) and therefore closed (because profinite groups are T_2).


Which compound is more acidic and why: p-methoxyphenol or o-methoxyphenol?

I think p-methoxyphenol>o-methoxyphenol in acidity.
In ortho there are 2 complementary effects -I and +R whereas in para only +R so it is more acidic than ortho.

I think ortho is more acidic. Maybe due to ortho effect?  :maybe:  :maybe:

i think para is more acidic, as intramolecular hydrogen bonding can take place in ortho. also ortho effect is not applicable here, madness, it is applicable in benzoic acids.
and chemrock, +R effect decreases acidity and -I increases acidity

So is max -I at ortho posn the reason?  :maybe:  :maybe:

[quote="madness"]How can there be H bonding in that?  :huh:  :huh: . I dont think there is any H bonding in anisole sub compds  :maybe:[/quote]
may i know the reason for that? you an acidic h attached to O and another O adjacent to it, and chelation is possible. :roll:

Well i thought that the O attached to the $ \ce{CH3}$ grp might not be electronegative enough for H bonding  :maybe:  :maybe:

Yes sorry my reason was correct and so ortho>para in acidity. :oops:
Probably varun u should compare the stabilities of the respective phenoxide ions :maybe:


For any given triangle, find a point in its plane such that the sum of its distances from the vertices of the triangle be minimum.

 :arrow: This point is is called the [b]$ Steiner$ $ Point$[/b].

You mean th Fermat Point? http://en.wikipedia.org/wiki/Fermat_point
I know many books call it Steiner point but i don't understand that..

BTW there is a geometry subforum, where it would be more appropriate to post these questions.


vector
linear algebra
matrix
linear algebra unsolved

Dear friends,

I wonder if there is a closed form solution for this!

Let $\mathbf{A}$ be an $n \times n$ full rank matrix and all its entries are complex numbers. Is it possible to find the linear combination of the columns of this matrix such that it results in a real vector?

I mean can we find solution for the matrix \[\mathbf{A}\mathbf{x}=\mathbf{r}\] where $\mathbf{r}$ is a real vector!

$\mathbf A^{-1}\mathbf r$, where $\mathbf r$ is a real vector of your choice.


Let $ a,b,c$ be trhee non-negative real numbers. Prove that
$ \sum\frac{a(b\plus{}c\minus{}a)}{a^2\plus{}2bc}\geq0$

[quote="lipako"]Let $ a,b,c$ be trhee non-negative real numbers. Prove that
$ \sum\frac {a(b \plus{} c \minus{} a)}{a^2 \plus{} 2bc}\geq0$[/quote]

the inequality is  equivalent  to
$ 2 b^4 a^2 \plus{} 7 a^2 b c^3 \plus{} 7 a^3 b^2 c \plus{} 7 c b^3 a^2 \plus{} 7 c^2 a^3 b \minus{} 3 a^2 b^2 c^2 \minus{} 4 c a^4 b \plus{} 7 a b^3 c^2 \plus{} 7 a c^3 b^2 \plus{} 2 c^4 a^2 \plus{} 2 b^2 c^4 \plus{} 2 a^4 b^2 \minus{} 4 c b^4 a \plus{} 2 a^4 c^2 \plus{} 2 c^2 b^4 \minus{} 4 a^3 b^3 \minus{} 4 b^3 c^3 \minus{} 4 c^3 a^3 \minus{} 4 a b c^4\geq 0$
$ < \equal{} > 2(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{} 3abc\sum{(b \plus{} c)(a \minus{} b)(a \minus{} c)} \plus{} 27a^2b^2c^2\geq 0$

easy  to prove $ 3abc\sum{(b \plus{} c)(a \minus{} b)(a \minus{} c)}\geq 0$  by schur  inequality.


trigonometry
geometry
vector
inequalities
calculus
derivative
geometry solved

3. In acute-angled triangle ABC point D is the perpendicular projection of C on the side AB. Point E is the perpendicular projection of D on the side BC. Point F lies on the side DE and:
$\frac{EF}{FD}=\frac{AD}{DB}$
Prove that $CF \bot AE$

This bloody problem duz really make me angry!!! Anyway here is a halfways nice solution, found by me in an hour. My nerves won't suffice to find a synthetic one for this cursed problem.

Let the parallel to the line AE through the point F meet the line AB at a point T. Then, by Thales, $\frac{AT}{TD}=\frac{EF}{FD}$, and since $\frac{EF}{FD}=\frac{AD}{DB}$, we have $\frac{AT}{TD}=\frac{AD}{DB}$. Thus,

$\frac{AD}{TD}=\frac{AT+TD}{TD}=\frac{AT}{TD}+1=\frac{AD}{DB}+1=\frac{AD+DB}{DB}=\frac{AB}{DB}$,

and hence,

$TD=AD\diagup \frac{AB}{DB}=\frac{AD\cdot DB}{AB}=\frac{b\cos A\cdot a\cos B}{c}$.

Therefore,

$TD\diagup CD=\frac{b\cos A\cdot a\cos B}{c}\diagup \left( b\sin A\right) =\frac{b\cos A\cdot a\cos B}{c\cdot b\sin A}=\frac{a\cos A\cos B}{c\sin A}=\frac{a\cos A\cos B}{a\sin C}=\frac{\cos A\cos B}{\sin C}$,

where we have used the Sine Law in the form c sin A = a sin C.

On the other hand, if X is the orthogonal projection of the point A on the line DE, then $AX\perp DE$, and since $DE\perp BC$, we have AX || BC. Thus, by Thales,

$\frac{AX}{BE}=\frac{AD}{BD}$.

And hence,

$AX=BE\cdot \frac{AD}{BD}=\frac{BE}{BD}\cdot AD=\cos B\cdot AD=\cos B\cdot b\cos A$.

On the other hand, since AX || BC, the segment XE, being the distance from the point X to the line BC, must equal the distance from the point A to the line BC, i. e. it must equal the A-altitude of triangle ABC. Hence, XE = b sin C. Thus,

$AX\diagup XE=\left( \cos B\cdot b\cos A\right) \diagup \left( b\sin C\right) =\frac{\cos A\cos B}{\sin C}$.

Comparing with the above, we see that AX / XE = TD / CD. But AX / XE = tan < XEA and TD / CD = tan < DCT. Thus, tan < XEA = tan < DCT, so that < XEA = < DCT. Since FT || AE, we have < XEA = < DFT, and thus we have < DFT = < DCT. Thus, the points D, T, C and F lie on one circle. Since < TDC = 90°, the segment DC is a diameter of this circle. Thus, also < TFC = 90°, so that $FT \perp CF$. Since FT || AE, we thus get $AE \perp CF$. And we are done.

  Darij

heh  :D , analytic geometry is the best way here [for me , at least]

Using vectors can also solve it quickly. Here's my solution:

Let $B=(0,0)$, $A=(a_{1},a_{2})$, $C=(c,0)$, $D=(ka_{1},ka_{2})$, $E=(ka_{1},0)$, $F=(ka_{1},(1-k)ka_{2})$.

Since $AB$ is perpendicular to CD, so

$(a_{1},a_{2})dot(ka_{1}-c,ka_{2})=0\Rightarrow$

$ka_{1}^{2}-a_{1}c+ka_{2}^{2}=0$

Since the dot product of vector CF and AE is:

$((1-k)a_{1},a_{2})dot(ka_{1}-c,(1-k)ka_{2})$

$=(1-k)[ka_{1}^{2}-ca_{1}+ka_{2}^{2}]$

$=0$

Therefore CF is perpendicular to AE.

[quote="pestich"]If you construct a point D' isotomic congugate of D, draw a line thru it
    parallel to BC to intersect with A-altitude at K, then it is easy to see
    that  B,F,K are collinear, < AED = < ECF = < KAE and this places point X
   (AE x CF)  on the circle with diameter AC.[/quote]

Maybe I am blind but I don't understand why < AED = < ECF = < KAE. Or, more precisely, I understand why < AED = < KAE but it's not clear to me how to involve < ECF here. Could you please give a better explanation? I find your idea really nice, but this point is still obscure to me...

  Darij

[quote="pestich"]Darij,

    I may be wrong (nothing new here),   but it looks like the < AED is measured
    as difference of  corresponding arcs on circle with diameter AC, both feet of
    altitudes from A  and C are on this circle and DE is parallel to A-altitude.

    Equal inscribed angles  put the intersection point of AE and CF on the circle.

    Pestich.

 Also <AEF= <ECF makes triangles CEF and EXF (X is intersection of AE and CF)
similar.[/quote]

This doesn't seem to make things clear. You write that "< AED is measured as difference of  corresponding arcs on circle with diameter AC". Why? The point E doesn't lie on that circle, and the line AE, which is a side of angle < AED, is not directly related to that circle. Also, it may sound banal, but you must use the fact that the points D and D' are isotomically conjugate with respect to BC somewhere in your proof. Do you?

  Darij

[quote="pestich"]Hello Darij,

   The isotomic property of D' is used in collinearity of points K,F,B.

   The fact that E is not on the circle is the reason it is measured by difference of
   arcs, and the point M (intersection of KD' and DE ) is on the circle.

   Altitude AG is definitely parallel to DE.

   Pestich.

  P.S.   Man, what happened to you? You used to love angle chasing so much ![/quote]

Hello Pestich,

The AM-GM inequality shows us that each of the terms is majorized by a product of similar terms.

The fact that the derivatives are not positive is the reason why they are all equal, and the Fundamental Theorem of Arithmetics holds.

$n^2+2n+1$ is definitively divisible by $n+1$.

  Darij.

PS. Don't you see that this all is closely related to the FLT ???

I am sorry, but this is what I get when trying to imitate your style. None of your sentences seems to be a consequence of the previous one. I am not claiming that your proof is wrong, but I am claiming that it is explained in such a way that I am not able to say whether it is correct since I am not able to follow your argumentation. It is so untangible! Each time I think I have caught it I see it disappearing in front of my eyes and hiding again in nowhere! You say that "The isotomic property of D' is used in collinearity of points K,F,B". Okay, it is used there, but then I ask you how do you later use the point K. You then write "The fact that E is not on the circle is the reason it is measured by difference of arcs,". Which arcs? Then: "and the point M (intersection of KD' and DE ) is on the circle." It actually does. But how do you show this? And finally: "Altitude AG is definitely parallel to DE.". Okay, I do know this. But what does it yield?

  Darij

[quote="pestich"]  Two right-angled triangles CDE and ABG (G is a foot of A-altitude) are similar.

  Lines AE and CF are similar lines in similar triangles
  similar sides of which are orthogonal.[/quote]

Pestich, CONGRATULATIONS! And now I'm serious. CONGRATULATIONS for this beautiful proof! You have really seen the point of the problem. And, more importantly, you have finally found a mathematically correct and strict way of reasoning that still occupies just 4 lines. I was either blind when I wrote my solution, or it was too late at night, anyway I made the problem much more complicated than it is. Congratulations and thanks a lot for this proof (and let's forget the ideas from your previous messages in this thread - even if they yield a solution, this solution would be superfluously complicated compared with your new one).

[b]EDIT:[/b] Two other solutions of the initial problem can be found in [url=http://groups.yahoo.com/group/Hyacinthos/message/9964]Hyacinthos message #9964[/url] and [url=http://groups.yahoo.com/group/Hyacinthos/message/11046?expand=1]Hyacinthos message #11046[/url].

  Thanks!
  Darij

I know an old problem which is one particular case of this problem when $CA=CB$ . The proposed problem is an easy extension.
[quote="Megus"]In an acute-angled triangle $ABC$ denote the point $D\AB$ so that $CD\perp AB$ , the point $E\in BC$

so that $DE\perp BC$ and the point $F\in (DE)$ so that $\frac{FE}{FD}=\frac{DA}{DB}$ . Prove that $CF \bot AE$ .[/quote] [b][u]Proof[/u].[/b] Denote $G\in BC$ so that $AG\perp BC$ . Observe that $AG\parallel DE$ and $\frac {EG}{EB}=\frac {DA}{DB}$ , i.e. $\frac {EG}{EB}=\frac {FE}{FD}$ . Since

$\triangle CDE\sim\triangle ABG$ obtain that the points $F\in  DE$ and $E\in BG$ are omologously in  similarity. Thus, $\widehat {FCD}\equiv\widehat {EAB}$ ,

i.e. the quadrilateral $ACLD$ is cyclically where $L\in CF\cap AE$ . In conclusion,  $CD\perp AB\implies  CF\perp AE$ .

Let`s take $E^{\prime}\in BC$ such that $DE^{\prime}\parallel AE$. Using the informations given and the  Thales` theorem we obtain: \[\frac{EF}{FD}=\frac{AD}{DB}=\frac{EE^{\prime}}{BE^{\prime}}\Longrightarrow FE^{\prime}\parallel AB\]. Now it`s easy to see that $E^{\prime}F\perp CD,\ DF\perp CE^{\prime}$, so $F$ is the ortocenter of the $\triangle CDE^{\prime}$, thus $CF\perp DE^{\prime}$, q. e. d.


calculus
derivative
trigonometry
real analysis
real analysis unsolved

Hello!
What can I do with this:

Find the number of solutions
8 arctan sqrt(x) = x^2 + 6
2 arctan sqrt(x) = sqrt(1+x)
arcsin x = x sqrt 2

I've tried different things (some derivatives etc) and nothing works...
Thanks in advance and sorry for my English

$ x  \equal{} \tan^2 \frac{x^2\plus{}6}{8}\equal{}\tan^2 \frac{\sqrt{1\plus{}x}}{2}\equal{}\sin (x\sqrt 2)$

as a system, the two outsides have only one solution, which isnt consistent with the middle, so zero solutions..

unless you meant find the number of solutions for EACH?

Yes, for each of course...

I've dealed with the last one
Take it as f(x), calculate first derivative - find f(x)' = 0 points. It gives two points x1 = 1/2 and x2 = -1/2
We now know that in this points f(x) changes its direction (growing or not).
Calculating f(x) for theese two and for the boundaries of arguments (which are -1 and 1) we get f(1)(-) f(1/2)(+) f(-1/2)(-) and f(1)(+)
From Cauchy we know that it means we have 3 f(x) = 0 points -> 3 solutions of equation.

But in first and second derivative is like 8*1/(x+1)*1/(2sqrt(x))-2x and i have no idea what can i do with this?


abstract algebra
complex numbers
absolute value

Let $ z_1,z_2,...z_n$ complex numbers with the same module, so that \[ z_3 \cdot z_4 \cdot...\cdot z_n \plus{} z_1 \cdot z_4 \cdot ... \cdot z_n \plus{}...\plus{} z_1 \cdot z_2 \cdot ... \cdot z_{n\minus{}2} \equal{}0\] Prove that \[ z_1z_2\plus{}z_2z_3\plus{}...\plus{}z_{n\minus{}1}z_n\equal{}0\]

If the common absolute value's 0, it's trivial. Otherwise, let $ a$ be the square of the common absolute value so $ \overline{z_k}\equal{}\frac{a}{z_k}$ for all $ k$. Now conjugate the first equation, clear denominators, and simplify: the second equation results.


geometry
trapezoid
analytic geometry
quadratics
USAMTS
algebra

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6$, $BC=5=DA$, and $CD=4$. Draw circles of radius 3 centered at $A$ and $B$, and circles of radius 2 centered at $C$ and $D$. A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p$, where $k$, $m$, $n$, and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k+m+n+p$.

Pretty easy for a #12, don't you think?  :P

[hide]Let the radius of the center circle be $r$ and its center be denoted as $O$. The center obviously has an x-coordinate of $3$.

So line $AO$ passes through the point of tangency of circle $A$ and circle $O$. Now let $y$ be height from the base of trapezoid to O. Now, from Pythagoreum Theorem,
$3^2+y^2=(r+3)^2 \rightarrow y=\sqrt{r^2+6r}$.

We use a similar argument with the line $AD$, and find the height from the top of the trapezoid to $O$, $z$, to be $z=\sqrt{r^2+4r}$.

Now $y+z =5$, so we solve the equation $\sqrt{r^2+4r}+\sqrt{r^2+6r}=5$

Solving this, we get $r=\frac{-60+48\sqrt{3}}{23}$

So $k+m+n+p= 60+48+3+23= \fbox{134}$.[/hide]

Sorry to drudge up an old thread, but how exactly can we conclude that the height of the trapezoid is 5?  Isn't the height actually $\sqrt{24}$?

[quote="deej21"]He got the right answer, but you're right about the height.  I don't know how he got that to work out right... maybe I'm misreading his solution.

I did pretty much the same thing, found two equations with $y$ and $r$ and solved for $r$.  That quadratic equation is ridiculous though... the discriminant is huge, although easily factorable.  I imagine there is an easier way...

And yeah that was pretty easy for a 12...  It happens.[/quote]
Ah, yeah, you're right - I had worked out the equation and gotten a different solution and then couldn't figure out where this solution was coming from.  Running the equation with the proper height through Mathematica yields the right solution, so I must just have made an algebra mistake somewhere along the way.

Also, I doubt there's an easier way, since the solution is in the form of a quadratic, so I don't see any other way you could end up with the right solution without solving that equation.

Yeah, good point... didn't realize that.

[quote="1234567890"]
Solving this, we get $ r \equal{} \frac { \minus{} 60 \plus{} 48\sqrt {3}}{23}$[/quote]

Sorry to revive an old topic, but how did you solve this? I solved the problem in the same way that 1234567890 did, but upon getting rid of the square roots, I ended up with $ 3r^2\plus{}23r^3\minus{}44r^2\minus{}384r\plus{}576\equal{}0$.

buzzer, I think you expanded incorrectly, you can not get a cubic from expanding that.

Also I had a question, the center of tangent circles are collinear, right? so why aren't the centers of the circle at C, O(the middle circle), and A collinear?

thank you!

I assumed that the tangency points of the big circles with the small one in the middle form a square... Is this true?
Could someone also explain why the center of the little circle "obviously" has a $y$-coordinate of $3$?

It's the $x$-coordinate that has the value of $3$ ; the center is equidistant from $A,B$ and $C,D$ .

1234567890's answer was right but the equation should be $ \sqrt{r^2+4r}+\sqrt{r^2+6r}=\sqrt {24} $ .


Hello; I recently found out that the terms "joint" and "Double" honours mean something very different, I was planning to do a Joint Honours in Mathematics and Physics, But Im concerned that in taking this degree I'll be well-rounded on both subjects but not particularly an expert at either, like as if I took a double honours in both of these subjects. Should I do a Double honours or a joint Honours?

The exact meaning of these words is likely to be local. That is, they might both be meaningful words with different meanings at one university, while the same is not true (perhaps because at least one of the words has no meaning) at another university.

As for what to do: you may have a better idea than anyone else as to what you are capable of, but there's a balance to be struck between being ambitious and being foolhardy. In particular, before committing yourself to two different undergraduate research or undergraduate thesis projects (if that is what you're talking about), ask yourself if you would be able to give both projects the attention they deserve, while at the same time keeping up with all of your other courses.

How much you learn about each topic depends on how many classes you take.  If you take the bare minimum for each, then yes, you'll be a jack-of-all-trades, master-of-none.  If you continue to take upper level classes, then you'll be a master-of-both.


quadratics
algebra
polynomial
algebra unsolved

Let $p(x) = x^2 +ax +b$ be a quadratic polynomial with $a,b \in \mathbb{Z}$. Given any integer $n$ , show that there is an integer $M$ such that $p(n) p(n+1) = p(M)$.

I also did it by converting the products to the polynomial.
Well, I am more interested in knowing  any property of integer polynomial that makes this statement true. Can anyone tell any connection ( through graphs or else) ? I hope I am clear to everyone.

[quote="pbornsztein"]Direct checking leads to $p(n)p(n+1) = p(an+b+n(n+1)).$

Pierre.[/quote]
Isn't this solution complete?

There are two possible value of $M$
$$p(n)p(n+1) = p(n^2+(a+1)n+b) = p(-n^2-(a+1)n-a-b)$$
This can be easily solved by expanding $p(n)p(n+1)$ and then plugging $M=xn^2+yn+z$. On comparing we would get the desirable value of x,y and z.

Let $f(x)=x^2+ax+b$
Let $\alpha,\beta $ be the  roots of $f(x)$
$x^2+ax+b=(x-\alpha)(x-\beta)$
$f(n)=(n-\alpha)(n-\beta)$
$f(n+1)=(n+1-\alpha)(n+1\beta)$
$f(n)f(n+1)=(n-\alpha)(n-\beta)(n+1-\alpha)(n+1\beta)$
$f(n)f(n+1)=(n(n+1)-n(\alpha +\beta)-\alpha +\alpha \beta)(n(n+1)-n(\alpha +\beta)-\beta +\alpha \beta)=(n(n+1)+an+b-\alpha )(n(n+1)+an+b-\beta)=(m-\alpha)(m-\beta)$
$m=n(n+1)+an+b$


inequalities
blogs
inequalities theorems

Can someone please explain what "Langrange multipliers " are and how to use them?

Please give an example how to use it on olympiad inequalities:

for example: $a^{2}+b^{2}+c^{2}+abc =4$ find the minimum of $a+b+c$. ($a,b,c$ are real numbers. )

by the http://www.mathlinks.ro/weblog_entry.php?p=910888#910888

Let $ f\equal{}a\plus{}b\plus{}c$ and $ g\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}abc\minus{}4$ and $ L\equal{}f\minus{}\mu g\equal{}a\plus{}b\plus{}c\minus{}\mu\left(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}abc\minus{}4\right)$

$ \frac{\partial L}{\partial a}\equal{}1\minus{}\mu a\minus{}\mu bc\equal{}0\Longrightarrow\mu\equal{}\frac{1}{2a\plus{}bc}$
$ \frac{\partial L}{\partial b}\equal{}1\minus{}\mu b\minus{}\mu ca\equal{}0\Longrightarrow\mu\equal{}\frac{1}{2b\plus{}ca}$
$ \frac{\partial L}{\partial c}\equal{}1\minus{}\mu a\minus{}\mu ab\equal{}0\Longrightarrow\mu\equal{}\frac{1}{2c\plus{}ab}$

or $ \mu\equal{}\frac{1}{2a\plus{}bc}\equal{}\frac{1}{2b\plus{}ca}\equal{}\frac{1}{2c\plus{}ab}$

$ \frac{1}{2a\plus{}bc}\equal{}\frac{1}{2b\plus{}ca}\Longrightarrow 2a\plus{}bc\equal{}2b\plus{}ca\Longrightarrow 2(a\minus{}b)\equal{}c(a\minus{}b)\;\;\;(\bigstar1)$

similarly we get :
$ 2(b\minus{}c)\equal{}a(b\minus{}c)\;\;\;(\bigstar2)$
$ 2(c\minus{}a)\equal{}b(c\minus{}a)\;\;\;(\bigstar3)$

by solving these three equations we get $ a\equal{}b\equal{}c\;\;\text{(why?)}$ and by the condition $ g$ given in the problem we must have: $ 3a^{2}\plus{}a^{3}\equal{}4$ or $ a^{3}\plus{}3a^{2}\minus{}4\equal{}(a\minus{}1)\left(a^{2}\plus{}4a\plus{}4\right)\equal{}0$ so we have $ a\equal{}1$ or $ a\equal{}\minus{}2$

and $ \min\left\{a\plus{}b\plus{}c\right\}\equal{}3\times\minus{}2\equal{}\minus{}6$
 :)

[quote="sinajackson"]I have posted about the Lagrange Multipliers Method in my Blog, i hope it can be helpful for you in solving the inequality: http://www.mathlinks.ro/weblog_entry.php?p=910888#910888[/quote]

Where is the proof of this method ? .  :lol:

The proof is really easy :lol:
Well you can see a nice proof of this method in [b]PlanetMath.Org[/b]:
http://planetmath.org/encyclopedia/ProofOfLagrangeMultiplierMethod2.html