group theory
abstract algebra
algorithm
linear algebra
matrix
geometry
geometric transformation

Find the smaller order of a group $ G\subset \mathrm{GL}_n(\mathbb{C})$, s.t that $ \mathrm{span}(G) \equal{} \mathcal{M}_n(\mathbb{C})$.

Start with the group of cyclic permutations and allow each nonzero entry to be either $ 1$ or $ \minus{} 1$; that's a group of size $ 2^n n$ (although it might suffice to take a subgroup) and it's not hard to see you can get every $ E_{ij}$ this way.  (Does anyone know what this construction is called?  It's some kind of product of $ (\mathbb{Z}/2\mathbb{Z})^n$ and $ \mathbb{Z}/n\mathbb{Z}$; is it semidirect or wreath?)

Of course, the lower bound on the order of the group is $ n^2$, so this is far from optimal.

Edit:  Anyway, for practical applications it's not so important that $ G$ be small; we want it to be easy to multiply in $ G$.  [url=http://www.cs.caltech.edu/~umans/papers/CKSU05.pdf]Cohn et al.[/url] have remarked that a certain natural conjecture about groups satisfying a certain property would imply that there exists an $ O(n^2)$ algorithm for matrix multiplication.

For $ n\equal{}2$, I can nail down the minimum. The group must be nonabelian (any finite commuting set of diagonalizable matrices can be simultaneously diagonalized), so that's a minimum order of $ 6$.
This minimum can be achieved. The matrices of a $ 120^\circ$ rotation and a reflection generate a spanning dihedral group.

If $ n$ is prime the minimum must be strictly larger than $ n^2$ since a group of order $ n^2$ is abelian.  For $ n \equal{} 3$ this implies that the smallest group has order at least $ 10$.  Now, the only nonabelian group of order $ 10$ is $ D_5$, which doesn't have an irreducible $ 3$-dimensional representation, so the minimum for $ n \equal{} 3$ is a group of order $ 12$.

Now the subgroup of the group I described with an even number of $ \minus{} 1$s (in other words, with the property that the determinant of each element is the sign of the corresponding permutation) obtains this minimum, and this construction works for all $ n > 2$.  This gives a group of size $ 32$ when $ n \equal{} 4$ as opposed to the minimum $ 16$; can someone describe a tricky subgroup in this case?

[b]Edit:[/b]  A few more remarks.

The minimum $ n^2$ is never attainable.  As I remarked, the corresponding representation has to be irreducible, but a group with an irreducible representation of order $ n$ has order at least $ n^2 \plus{} 1$ by the orthogonality relations since it has at least one other representation (the trivial representation).

[b]Edit #2:[/b]  Okay, so I forgot that the dimension of an irreducible representation of a group has to divide its order, which means the lower bound is $ n^2 \plus{} n$.  I'll be using this to simplify the proofs below.

For $ n \equal{} 4$ this gives the minimum $ 20$.  According to [url=http://shell.cas.usf.edu/~eclark/algctlg/small_groups.html]this site[/url], one candidate in this case is the Frobenius group of order $ 20$, which embeds into $ S_5$ and therefore does have a faithful $ 4$-dimensional representation.  I haven't checked more thoroughly, though.

For $ n \equal{} 5$ this gives the minimum $ 30$, but the only nonabelian groups of order $ 30$ that aren't $ D_{15}$ are $ D_5 \times C_3$ and $ D_3 \times C_5$, both of which have more than $ 5$ one-dimensional characters.  A group of order $ 35$ has both Sylow subgroups normal, so it's abelian.  Hence we need order at least $ 40$.

[quote="t0rajir0u"]... can someone describe a tricky subgroup in this case?[/quote]No, because every index-2 subgroup contains $ \minus{} I$. While it may be possible to do better than $ 32$, it certainly requires a different structure.

[Edit- when I initially posted, the sentence quoted was at the end of the preceding post.]

[b]Edit #3:[/b]  I forgot something simple.  Other post has been edited to reflect this.

One more bit: if there's a nontrivial center, that center consists of multiples of the identity, and we're giving away that much by multiplicity.

Edit: You were too quick to dismiss order 20. Consider that Frobenius group, acting on the five vertices $ v_1,v_2,v_3,v_4,v_5$ of a simplex centered at the origin. By averaging powers of $ s$, we can send $ v_1$ to itself and each other $ v_i$ to $ \frac14(v_2\plus{}v_3\plus{}v_4\plus{}v_5)$, which means we send the orthogonal complement of $ v_1$ to zero. Multiplying by powers of $ t$, we can send $ v_1$ to each $ v_i$ while still sending its orthogonal complement to zero; averaging these, we can send $ v_1$ anywhere. Repeat at the other vertices by similarity, and we can send four basis vectors to any four vectors. That's the whole matrix space.

This is essentially the same idea as my $ n\equal{}2$ example. It should work and give the minimum $ n^2\plus{}n$ whenever $ n\plus{}1$ is prime.

I guess I've edited that post too many times now anyway  :oops:  Yes, I just realized my argument was wrong when you posted that.  If $ n(n \plus{} 1)$ is a tight lower bound infinitely often I wonder if $ 2^{n \minus{} 1} n$ is ever a tight upper bound (although I doubt it).  It's likely this question has been investigated before, so I'll check the references in Cohn's paper.

[b]Edit:[/b]  Cohn and Umans are performing a different type of embedding.  I should've realized this, but you can't get any improvement in matrix multiplication by making it isomorphic to a group algebra because multiplying in an irreducible representation isn't any easier than multiplying matrices.

[quote="jmerry"]It should work and give the minimum $ n^2 \plus{} n$ whenever $ n \plus{} 1$ is prime.[/quote]
The group of invertible affine transformations on $ \mathbb{F}_q$ is Frobenius of order $ q(q \minus{} 1)$; is this what you meant?  This should work whenever $ n \plus{} 1$ is a power of a prime, which gives equality for $ n \equal{} 6, 7, 8$, so the small cases left are $ n \equal{} 5, 9$.  I think it should be possible to get a polynomial upper bound on the size of $ G$; $ 2^{n \minus{} 1} n$ seems way too large.

[b]Edit #2:[/b]  Start with the group of cyclic permutations and allow the nonzero entries, from top to bottom, to read $ \zeta_n^k, \zeta_n^{k \plus{} a}, \zeta_n^{k \plus{} 2a}, ... \zeta_n^{k \plus{} (n \minus{} 1)a}$ for some $ k, a \in \mathbb{Z}/n\mathbb{Z}$.  This gives a group of order $ n^3$ with a center of order $ n$.

[b]Edit #3:[/b]  When $ n$ is even and greater than $ 2$ the quotient of the above group by $ \minus{} I$ gives a group of order $ \frac {n^3}{2}$.  Here's another construction:  $ PSL_2(q)$ acts on the projective line over $ \mathbb{F}_q$, which has $ q \plus{} 1$ elements, and subtracting the trivial representation we obtain an irreducible representation of dimension $ q$ (at least, I think).  So when $ n$ is a power of an odd prime we have a group of order $ \frac {n(n^2 \minus{} 1)}{2}$.  This gives the bounds $ 60, 360$ on $ n \equal{} 5, 9$ as opposed to the bounds $ 125, 729$ by the second method and $ 80, 2304$ by the first.  The smallest $ n$ such that $ n \plus{} 1$ isn't a prime power, $ n$ isn't an odd prime power, and $ n$ is odd is $ 21$.

[b]Edit #4:[/b]  Apparently $ PSL_2(q)$ has an irreducible representation of dimension $ q \plus{} 1$, but I don't know what it looks like.  Anyone want to describe it?  In a few cases, that gives us a slightly better $ O(n^3)$ group.


For any positive integer n, $ \binom{n}{0}\equal{}1$. So the answer is just $ \boxed{1}$.

$ \binom{8}{0} \equal{} \frac{8!}{(8\minus{}0)!0!} \equal{} \frac{8!}{8!0!} \equal{} \boxed{1}$

OR

Think about it logically. How many ways are there to choose $ 0$ objects out of $ 8$ objects? Obviously just one way, by choosing none.

OR

Think about Pascal's triangle. All combinations in form of $ \binom{n}{0} \equal{} \binom{n}{n}\equal{} 1$. They are all the ones you see surrounding the triangle.

(It might also help if you've seen the binomial theorem before)

[quote="gauss1181"]For any positive integer n, $ \binom{n}{0} \equal{} 1$. So the answer is just $ \boxed{1}$.[/quote]
Isn't this just a combination problem?

I don't get this...why is it 0 and not 1?

EDIT: Oops the answer was 1

logic

how many ways can we make 0 out of 8? obv 1 which is choosing nothing,  and 0 = 0

[quote=GameBot]Compute $ \dbinom{8}{0}$[/quote]

[hide=Sol]
1. Since any positive number n is $ \binom{n}{0}=1$, replace 8 for n and the answer is $ \boxed{1}$.
2. $ \binom{n}{k}$= $(n!)/[k!(n-k)!]$ so it's just $ \boxed{1}$.
[/hide]


geometry
perimeter
rotation
combinatorics unsolved
combinatorics

ok...here si the problem:
a new planet has recently been discovered. So now there are 10 planets circling around the sun. prove that in every moment there exists a point on the sun from where no more then 4 planets can be seen.
the only known thing is that the perimeter of every planet is smaller then the perimeter of the sun

Assume the opposite. Than, for any point, there are 5 planets seen. But then, for every 2 diametrically opposite pts on the sun, 10 planets are seen. The space  that is not senn from either point must then contain no planets. But if you take the points on the sun right below that space, I am pretty sure it is easy to prove a contradiction.

we will show in 2D, then it's same idea in 3D.
choose one diameter of the sun, say $AB$. now you draw 2 tangents through A ($p$) and B ($q$) and then it is $p$II$q$. we can see anything above $p$ and below $q$. you rotate $p$ and $q$ so that one planet is between them (it can be between them 'cause the diameter of sun is larger then diameter of every planet). now $A$ or $B$ is our point because if there are at least 5 planets above $p$ then there are at most 4 planets below $q$, or opposite.


Each of the five numbers 1,4,7,10, and 13 is placed in one of the
five squares so that the sum of the three numbers in the horizontal
row equals the sum of the three numbers in the vertical column.
What is the largest possible value for the horizontal or vertical sum?
[asy]/* AMC8 1999 #11 Problem */
size(1inch,1inch);
draw((0,1)--(3,1)--(3,2)--(0,2)--cycle);
draw((1,0)--(2,0)--(2,3)--(1,3)--cycle);[/asy]

We place 7 in the middle, and 1 and 13 in the row, and 4 and 10 in the column. This gives us the maximum sum of $ \boxed{21}$.

Hi;

Put the 13 in the middle and the 10 and 1 in the (same row or column) and you make 24.


A 25 foot ladder is placed against a vertical wall of a building. The foot of the ladder is 7 feet from the base of the building. If the top of the ladder slips 4 feet, then the foot of the ladder will slide:

(A)4 ft (B) 5 ft (C) 8 ft (D) 9 ft (E) 15 ft

Okay, so right now, we would have a $ 7$-$ \sqrt{25^2 \minus{} 7^2}$-$ 25$ right triangle, or a $ 7$-$ 24$-$ 25$.  If the height slips by four, we have an $ x$-$ 20$-$ 25$ right triangle.  Solving for $ x$, we have $ \sqrt{25^2\minus{}20^2} \implies 15$.

$ 15 \minus{} 7 \equal{} \boxed{8}$

[hide="Answer"]Since 7-24-25 is a Pythagorean triple, the top of the ladder must be 24 feet up. If it slides 4 feet, it will be 20 feet up. Since 3-4-5 is a Pythagorean triple, the bottom of the ladder is now 15 feet away, which means it slid 15-7 feet. So (C).[/hide]


trigonometry
inequalities proposed
inequalities

Let the three sides of $\triangle ABC$ be $a,b,c$. Prove that
$\displaystyle \frac{\sin^2A}{a}+\frac{\sin^2B}{b}+\frac{\sin^2C}{c} \le \frac{S^2}{abc}$
where $\displaystyle S=\frac{a+b+c}{2}$. Find the case where equality holds.

Using sin law we got equivalent form:

$\frac{abc}{R^2} \leq (a+b+c)$

now using $[ABC]=\frac{abc}{4R}=\sqrt{S(S-a)(S-b)(S-c)}$ we arrive at:

$8(S-a)(S-b)(S-c) \leq abc$ using substitution $x=b+c-a$, $y=c+a-b$, $z=a+b-c$ we have to prove:

$xyz \leq \frac{1}{8} \cdot (x+y)(y+z)(z+x)$ which is obvious by AM-GM

Equality holds when $a=b=c$


abstract algebra
calculus
algorithm
Princeton
college
articles
Ramsey Theory

Story about [url=http://www.theaustralian.news.com.au/story/0,25197,22216398-5012694,00.html]Terry, Nigel and Trevor Tao[/url] who all went to [url=http://www.imo-official.org/country_individual_r.aspx?code=AUS]IMO for Australia.[/url] This can be seen as a follow-up of Richard Rusczyk's previous [url=http://www.mathlinks.ro/weblog_entry.php?t=237957]Web log entry[/url] on early college.

[hide="Report"]
[b]Beautiful minds from August 11, 2007[/b]

[i]What's it like to raise a family of geniuses? With three highly gifted sons, Billy and Grace Tao have learnt to ignore the advice of experts, writes Richard Guilliatt.[/i]

If you were looking for a place to study the Big Questions about human intelligence -- Is genius innate? Why are brilliant minds so often troubled? -- the Tao household in Adelaide's southern suburbs might seem an unlikely choice.

Built on a hill overlooking Flinders University, the house is a standard, four-bedroom, brick-veneer suburban home with a carport. The occupants, Dr Billy Tao and his wife, Grace, are model Chinese immigrants who embraced their new surroundings so enthusiastically that they christened their sons with the Anglo-Saxon names Terence, Trevor and Nigel.

Yet it was here, during the early ’80s, that the Taos came to realise that some fluke of brain chemistry had bestowed unusual gifts on their children. Terry, the eldest, was a bewilderingly smart tot who mastered most of the primary school maths curriculum while still in kindergarten. At seven he could discuss Boolean algebra, Abelian groups and other advanced mathematical concepts, and by nine he was attending physics and maths lectures at the nearby university. Professor Julian Stanley, a US expert on gifted children, pronounced him the most advanced mathematics prodigy he had ever seen.

Terry’s two brothers were similarly gifted. The youngest, Nigel, had an IQ of 180 and as a teenager won bronze medals in two International Mathematical Olympiads. Middle brother Trevor was Australian Junior Chess Champion at 14, won multiple prizes for his classical music compositions, became a gifted pianist and earned a diploma in mathematical sciences. The fact that he is autistic made his story all the more remarkable.

On a recent winter afternoon their proud father reviewed these achievements in the study of his home as Trevor sat playing Rachmaninoff preludes on a grand piano. Dr Billy Tao is a diminutive and voluble 61-year-old with greying, combed-over hair and square-framed spectacles. A pediatrician by training, he has developed an understandably avid interest in the subject of gifted children; before I arrived at the house he had been sitting at his computer, dressed in ugg boots and multiple fleecy sweaters, bashing out his latest thoughts on the subject.

A lot of what Billy Tao has to say runs counter to the accepted wisdom of “gifted education” experts, with their emphasis on IQ ratings and accelerated learning. “It’s not helpful to learn what they have done,” he says cheerfully. “It’s helpful to avoid their mistakes. I have seen too many situations where the parents did the wrong thing.” A brilliant mind, he says, is not just a cluster of neurons crunching numbers but a deep pool of creativity, originality, experience and imagination. “This is the difference between genius and people who are just bright. The genius will look at things, try things, do things, totally unexpectedly. It’s higher-order thinking. Genius is beyond talent. It’s something very original, very hard to fathom.”

What makes Billy Tao’s musings on this subject more pertinent than most is that he, unlike most parents of gifted children, has helped raise a bona-fide genius. When Terry Tao was awarded the Fields Medal (often described as the Nobel Prize of mathematics) last year at the age of 31, he was vaulted to the pinnacle of the mathematics world. Since he became a professor at the University of California, Los Angeles (UCLA), seven years ago, the Taos’ eldest son has dazzled his colleagues with an unstoppable flow of breakthroughs and innovations that some have likened to the musical output of history’s greatest prodigy, Mozart.

“Terry Tao is definitely, in my view, a genius,” says Professor Allan Snyder, director of the Centre for the Mind at the University of Sydney. “I would rarely use that term, but he is. Because he’s not done incremental things; he’s made leaps.”

For Snyder, the Tao family raises fascinating questions about the way a brilliant mind works. Contrary to popular belief, he points out, it’s rare for a child prodigy to blossom into a fully fledged genius; in fact, Snyder ventures to argue that an average kid is much more likely to become an exceptional adult (see box on page 31). What made the Taos an exception to this rule?

For years, Snyder has been researching whether some forms of mental illness open up hidden recesses of the mind which enable great intellectual leaps. Why are autistic people sometimes capable of extraordinary feats of memory or mathematical calculation? Why do bipolar sufferers often produce great art and music?

“Terry has some deep insight that I personally think is facilitated by his type of mind,” Snyder says, choosing his words carefully. “I don’t know what type of mind it is but in his family there’s a history of different types of mind. He’s married, which almost certainly means he’s not autistic. But I do think it’s important that he comes from this kind of a family.”

* * *

For Billy and Grace Tao, life changed irrevocably one afternoon in 1977 when they were visiting friends and looked over to see their two-year-old son sitting on the floor teaching a group of five-year-olds how to spell and add. Asked how he had learnt these skills, Terry replied that he had been watching Sesame Street. By the time he enrolled in primary school three years later, he could sit for hours reading high school calculus textbooks. That same year, 1980, the Taos were told that Trevor, then aged two, was autistic.

“At the time it was a bit like someone throwing a ball to you and saying ‘Catch!’, then throwing another ball right after it,” recalls Billy, with characteristic sangfroid. “It’s not like you have time to wonder whether the balls will be too heavy; you just catch them. You don’t think, ‘I’ve got one child who is gifted and one who is autistic’, you just sit down and work out what to do.”

Autism scrambles the brain signals that govern human communication, and Trevor showed many classic symptoms. He avoided eye contact, didn’t speak, exhibited repetitive behaviours such as endlessly twirling a ball, and would become enraged if his routines were broken. The prognosis for him appeared to be a life of social isolation. But with the help of the Autistic Children’s Association of South Australia (now Autism SA), the Taos hired an instructor, Jean Bryant, who began teaching him behaviour modification techniques.

“When we originally looked at him we thought the outlook was pretty bleak,” recalls Billy. “In those days, if an autistic person could be taught some basic living skills – to make your bed, do the dishes, make a cup of tea – this would be considered good. To finish high school was a bit of a dream. To finish university was a fantasy.”

Jean Bryant later wrote a book about her experience of teaching Trevor Tao, and it’s difficult to connect the five-year-old she describes – who needed cue cards to learn how to dress and speak – with the 30-year-old who sits at the piano playing Rachmaninoff. Tall, gangly and bespectacled, with a high-domed forehead and thatch of spiked black hair, Trevor still speaks with the slightly flattened tonality of many autistic people, and exhibits quirky habits such as clasping his hands behind his back. He is, however, a droll and witty presence. Looking at a scrapbook which features him smiling goofily from the pages of Woman’s Day at the age of 10, he says: “Hmmm. I must have been in my autism mode.”

These days Trevor works as a research scientist at the Defence Science and Technology Organisation in Adelaide, specialising in image analysis. Two years ago, he completed a PhD thesis entitled An Extended Mumford-Shah Model and an Improved Region Merging Algorithm for Image Segmentation (don’t ask). He also writes anti-smoking songs set to tunes such as Supercalifragilisticexpialidocious (“Smoking causes heart disease and osteoporosis …”) The day I visited he played me this tune on the guitar and jokingly cited it as evidence that he is himself a genius. “That’s not genius,” advised his dad. “That’s obsession.”

Music was the key that first unlocked Trevor’s potential. As anyone who has seen the Dustin Hoffman film Rain Man will know, autistic savants have an almost paranormal ability to divine the intricate patterns within maths and musical notation. Allan Snyder likens it to a brain that receives information in ultra-high contrast: in social situations this causes overload and confusion, as the mind fixates on myriad irrelevant details, whereas the static data in a musical score or mathematical equation can be absorbed at a glance. Before he could even speak properly, Trevor could memorise music almost instantly and play it back note-perfect.

For the Tao brothers, it was a happy accident of birth that their father was a pediatrician and their mother a teacher with an honours degree in mathematics. While Billy Tao researched gifted education and autism, his wife quit her job in order to devote herself to the full-time challenge of rearing their sons. (Grace Tao is an exceptionally gifted mathematician, although she deflects almost all questions to her husband). Ironically, the boys also benefited from the fact that special school programs for prodigies and autistic children were hard to find. In a field still taking shape, the Taos devised their own approach, even taking the family to the United States to tour the major centres of gifted-education research.

“People say Terry is lucky he got the right teachers, the right school, the right mentors,” says Billy Tao. “Well, we had to go out and find those people.” With the co-operation of his primary school principal, Terry became the first gifted student in Australia to be “radically accelerated” to high school in maths and physics, while remaining at Bellevue Heights Primary for other subjects. After sitting the Year 12 maths exams at the age of eight (he scored 90 per cent), he began studying maths at Flinders University.

By 1984, the media publicity surrounding this “boy genius” had attracted the attention of international academics. Dr Julian Stanley, who ran a program for maths prodigies at the Johns Hopkins University in Baltimore, sent the Taos a copy of the US College Board’s admission test and was duly astonished by the youngster’s score, the highest he had seen in 14 years of research. Meanwhile, Trevor had become the youngest autistic child ever integrated into a mainstream primary school in South Australia.

Billy Tao recalls that early on he and his wife were helped by the Gifted and Talented Children’s Association of South Australia, but they became uncomfortable with the fixation some parents in the association had with their children’s achievements and IQ ratings. The Taos had already seen the pitfalls of this approach during a trip to the US when they met Jay Luo, a prodigy who had earned a university science degree at age 12, but later dropped out during his PhD studies.

“Many parents of gifted children tend to overestimate their children’s ability, they want to maximise speed,” says Billy. “One thing I disagree about with the gifted-children movement is the emphasis on acceleration. Many gifted-education people, particularly teachers who have diplomas in gifted education, are all brainwashed with this idea of acceleration, acceleration, acceleration. What about lateral thinking? What about creativity?”

In contrast to the effusive praise other parents heaped on their little Mozarts, the Taos avoided excessive flattery and downplayed the importance of winning. It was a policy they put in place partly to deal with the challenge presented by their youngest son, Nigel. A prodigy at chess and maths – his IQ qualifies him as profoundly gifted – Nigel faced the difficulty of being merely exceptional in a family where extraordinary was the norm. At 14 he a won bronze medal in the International Mathematical Olympiad in Hong Kong. By then, however, Trevor had become an international chess player, met prime minister Bob Hawke and been made the subject of a book, a PhD thesis and countless media appearances. Terry, meanwhile, had written his first maths textbook at 15, earned a masters degree at 17 and was starting a doctoral thesis on harmonic analysis at Princeton University.

Miraca Gross, a professor of gifted education at the University of NSW who has known the Taos since 1984, is among the many who note that one of the family’s most remarkable qualities is the absence of egotism and rivalry. “There’s an enormous, deep affection between the three boys,” says Gross. “By the time Terry was 10 years old he was actively looking for ways he could help his brothers on their paths.”

Today, Nigel is 28. After studying maths and economics at the Australian National University he seemed destined for a career as an academic but took a sideways step and now works as a software programmer in the more relaxed environs of the Google headquarters in Sydney. Emerging from his office there recently, he was barefoot and dressed like a teenager in camouflage pants and a blue hoodie.

Talking to Nigel is a reminder that the Tao boys are Aussies. While they may have inherited a genetic gift and a strong work ethic from their Chinese parents, the Australian egalitarian ethos rubbed off along the way and probably helped them avoid the pitfalls of the “troubled genius”.

“I skipped Year 5,” Nigel notes, “but I don’t try and make a big deal of it and I don’t let that define me. Maybe that’s an Australian thing, where you admire someone for being a good bloke rather than something they’ve achieved. If we’d grown up in the US it might have been different. In Australia you can be tremendously good at something but if you’re a whacker, people still aren’t going to like you.

“To me, Terry is just a brother, rather than the towering figure of later years. And Trevor, he’s kind of kooky, but as far as autism goes it’s relatively mild – he’s keeping down a job and once you get past his quirks he’s genuinely friendly and likes making jokes, so he’s fun to be around. So they weren’t outrageously special to me.”

Asked if he sees any parallels between Trevor’s autistic quirks and his own, Nigel doesn’t seem in the least offended. “Well, I think there is a correlation between autistic behaviour and maths and music – that’s quite frequently commented on. And even myself and Terry, we do quite like our clever little puns and crosswords and games. If there are shades of stereotypical autistic behaviour, I’m sure I’ve got fractions of it. The little intellectual patterns in maths and music, I’m quite happy to amuse myself with those things, more than the average man, I think. It’s just a different mindset.”

* * *

On professor Snyder’s office wall at the University of Sydney is a framed sketch of a charging stallion, drawn with perfect perspective and the skill of a masterful illustrator. “That’s by a three-and-a-half-year-old girl who’s mentally impaired, can’t tell her mother from the nurse, but who draws like Leonardo da Vinci,” he says. Such feats of automatic drawing fascinate Snyder, suggesting as they do that savants can tap into the brain’s deeper reservoirs of ability at will. Much of his research involves trying to work out how that occurs.

“If you put into a box all the people who have bipolar disorder, you would find very few of them are creative in any way. But if you take all the people who are creative and put them in a box, you will find a large number with bipolar disorder. Now, why is that?” he asks. “Maybe a lot of geniuses are people who switch uncontrollably between our normal state and that bewildered autistic state where they are overwhelmed by details without interpretation.”

Not everyone in the Tao family is persuaded by this theory. Snyder once wrote to Terry Tao and asked, delicately, whether the mathematician felt he had access to some “privileged” mental ability. Terry replied that mathematics is mainly hard work, pointing out that it took two years of relentless study for him to achieve an acclaimed breakthrough in prime number research. (Asked to comment for this story, Terry demurred; his wife Laura explained that he’s “embarrassed” by the focus on his achievements).

Nor is Billy Tao amenable to Allan Snyder’s speculations. “There is no connection between the savant and the genius,” he asserts. “Allan Snyder seems to have a theory that we all have this hidden potential. But at the end of the day it depends on the child, the time and facilities available to the parents, and a certain serendipity or luck.”

These days, Billy Tao sometimes talks about his oldest son as if he were describing a distant satellite, shot into space by his parents and now freely orbiting. As a young professor at UCLA, Terry works in fields of abstract thinking – combinatorics, harmonic analysis, wave maps, Kakeya conjecture – that his father admits are completely beyond his comprehension. Terry was married five years ago and the Taos have been to the US to meet their daughter-in-law and grandson, William. But their communication is sporadic. “We don’t talk,” says Billy. “We exchange emails.”

If there is a genetic gift of brilliance, however, it may already have passed to the next generation of Taos. A couple of years ago, Terry’s son was eating dirt when his mother told him to stop or he would get germs. “It’s okay,” the three-year-old replied. “The white blood cells will help me fight them off.”

* * *

WHAT BECOMES OF THE GIFTED?

Albert Einstein and Charles Darwin, the two towering scientific geniuses of the past 150 years, were pretty average students. And therein lies a paradox, because most child prodigies grow up to be pretty ordinary adults.

“If I had a child who was six, seven years old and considered a prodigy, I would be sincerely, extremely worried,” says Professor Allan Snyder, director of the Centre for the Mind at the University of Sydney. “The chances of their being able to do something extraordinary are probably far less than someone else coming from a home with a normal background.”

Snyder’s rather harsh summary is backed up by decades of research. In 1921, the American psychologist Lewis Terman began tracking the progress of 1500 intellectually advanced children from California. By the late ’50s, only one had made an enduring impact – Jess Oppenheimer, creator of the sitcom I Love Lucy.

Terman’s results were mirrored in the early ’90s by a retrospective survey of 210 people who had been high-IQ students at Hunter College in New York in the years 1948-60. “Contrary to the expectations associated with the label of ‘genius’,” the study noted, “they tended to hold modest goals for themselves …”

Miraca Gross, a professor of gifted education at the University of NSW, has been following the progress of 40 high-IQ students in Australia since 1989. One of these students is Terry Tao, whose success is outstanding, but the others have had mixed outcomes. Several took successful corporate jobs – actuary, strategic consultant, hedge fund manager – which don’t require a towering intellect. One has become a high-level physicist. One became a disability worker, another failed to get a job in public relations and went back to study law. Several dropped out of their courses, and one developed depression.

Professor Gross says it’s too early to judge the achievements of many of her subjects, the oldest of whom are still in their 30s. But she says her research shows that the students who had the most satisfactory outcomes were those who had the benefits of accelerated learning.

To Allan Snyder, however, a genius is someone who changes the way humanity thinks, a path that requires not just a high IQ but the ability to withstand adversity and take on conventional wisdom. Steve Jobs and Bill Gates, he points out, were college dropouts who changed the world by creating the personal computer revolution.

“It’s a tall order, being a genius,” he concludes.

[/hide]

To be clear, I'm not an advocate of early college as it's typically practiced (and anyone who wants to point at Terry Tao as evidence that they're doing the right thing when they put their 14 year old in college will have to convince me that the 14 year old is Terry Tao first, and that early college will offer the education Terry received -- as I understand, he was identified as a star very early on and was receiving attention that most kids won't get in "early college".)

That said, I am a strong advocate of finding alternatives to the typical school curriculum.  It's just that I'm not conviced early college qualifies as that at many colleges.  (And I haven't even started to touch the social issues, or the importance of learning how to forge your own path, which I think is a far better alternative to early college for most students.)

Richard, I did not want to suggest that you advocate early college. Actually you warned of the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]Calculus Trap[/url] and assume you are in favour to develop and hone one's problem-problem skills via math olympiads, online classes, summer research camps, research with a local university professor, research competitions etc. before you advance to more technical university math.

But I thought it is a good idea to discuss his case. And if you read the article in the attachment you know that his parents wondered whether it is a good idea to send him to college early, his mental capabilities were years ahead but his physical development actually lagged his true age. And as we discussed before, though he might be humble and friendly, it was not clear he fits in with his peers. But he already exhausted more or less the opportunity of middle and high-school math competitions.

If anybody has further ideas and contributions you are very welcome to add them here.  :)

It honestly doesn't matter.

They were just people who's brains were better capable Turing machines than others. Unfortunately, for the rest of our lower-functioning Turing machines, we're unable to preform at such benchmarks. However, perhaps in the near future we'll be able to chemically modify our brains, and all become high-level Turing machines.

Plus I dislike the stance that autism is a disorder. All social interaction is is just an algorithm pre-programmed into our systems in order to aid the survival of our genes to ensure that we reproduce the next generation. Like love and family relationships, it's often given meaning to a phenomenon that's quite unremarkable. Autistic people are just people who have a very trivial algorithm malfunction. Not only that, but there's absolutely no empirical evidence to suggest autism is real (as in, a pathology exists).

I think "Terry Tao" has fit in with his colleagues quite well, he is described as a very friendly person, but the point I would like to share my opinion on is: Terry's talent.

If anyone in this community heard of 'Susan Polgar', I ask them to read her biography, and you would surprised that she was trained meticulously, intensely in chess by her father  at a very young age. Her father sought to prove to the world that : Genius is nurture rather nature.

Susan went on to become a chess champion and her younger sister Judith Polgar is also a consummate chess player; These people managed to reach lofty heights in chess largely due to the fact they were 'trained' by their father from a very young age, and most of the times the training was intense, well-planned.

I feel anyone can become very accomplished in a field  provided they are imparted the necessary knowledge, training from a very young age (when the development of the brain is in the incipient phase).

If you observe any of those so-called 'prodigies' you can see that they have been practicing Math (and their other related fields ) very intensely from a very young age.

I sometimes believe that Terry Tao doesn't possess a wildly superior intellect, his achievements in Math are largely a result of training that he imparted upon himself  from  a very young age, his determination, dedication to his work, and an efficient means of thinking.

I apologize if I offended anyone with these views of mine, but I feel at least some of them are coherent.

i agree totally. hard work is 90%, the fuel, while the talent is just the push that gets the rocket off the ground.

[quote="Gen8"]
I sometimes believe that Terry Tao doesn't possess a wildly superior intellect, his achievements in Math are largely a result of training that he imparted upon himself  from  a very young age, his determination, dedication to his work, and an efficient means of thinking.
[/quote]

I think the quotation of Thomas Edison is kind of famous: "Genius is 1% inspiration, and 99% perspiration.". Though not sure how close this statement to the truth is. I get the impression there is quite some survivorship bias involved. I feel every person has a possible scale for a certain type intellectual domain/skill. And training, mentoring, availability of resources etc. only get you so far, that is tell you where you will be on your personal scale. But it does not mean that you can reach any level of competency beyond your personal scale. There seems to be a personal limit on the speed and level of mental processing capacity.

Today I was attending an event on poetry and math. And I was thinking about the right level of creativity and ingenuity. In a thread that started out as a [url=http://www.mathlinks.ro/viewtopic.php?t=228877]discussion about the aftermath to wall street jobs for math major[/url] we were discussing about people with the key ideas in the fields and the people thought of motar mixers. How much do novel ideas depend on experience? [url=http://www.math.ohio-state.edu/~nevai/MYMATH/rota_ams_notices_01_97.html]In "Ten Lessons I Wish I Had Been Taught"[/url] Gian-Carlo Rota asserts "Every Mathematician Has Only a Few Tricks". Any many contributions that got the Fields medal in the recent past seem be to contributions that re-unite math in the way that they apply a techique from field A and apply it for the first time in field B in a possibly adapted way. Another interesting idea to think about creativity is to consider it as intermediate step between simple, homogeneous states of type I and complex structures of type IV as [url=http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life]Conway's game of life.[/url] [url=http://everything2.com/e2node/Wolfram%2527s%2520classes%2520of%2520cellular%2520automata]Here[/url] is a short overview on Stephen Wolfram's research on cellular automata. Here you can find his article from 1985: [url=http://www.stephenwolfram.com/publications/articles/ca/85-two/]Two-Dimensional Cellular Automata by Stephen Wolfram.[/url]

Thanks for the interesting links, orl!

I cannot help starting with a nitpick on dr. Snyder:
[quote]
“If you put into a box all the people who have bipolar disorder, you would find very few of them are creative in any way. But if you take all the people who are creative and put them in a box, you will find a large number with bipolar disorder. Now, why is that?
[/quote]
I have always been taught that $ |B\cap C|\equal{}|C\cap B|$ if you are talking about absolute numbers and $ \frac{|B\cap C|/|B|}{|C|/|P|}\equal{}\frac{|C\cap B|/|C|}{|B|/|P|}$ if you are talking about percentages above or below the group average in a subgroup. I always find it amusing how social scientists make self-contradictory claims and then ask "Why is that?" :P

More seriously, I believe that it is not "details without interpretation" but self-contradictory and illogical details that shortcut the brain circuits. "Normal" people just do not try to relate things in any way, their world is highly compartmentalized, and they are completely comfortable believing plenty of contradictory things (they just believe them one at a time and switch whenever the circumstances change).

The basis of all creativity is just relating the things that nobody suspected to be related before. Of course, the more things you know, the more opportunities you have, so experience helps. But a lot of people take their experience as "the patterns to follow", not as "the patterns to change and combine" and resemble that Neil brother of Remy in the Ratatouille cartoon at their worst and that female cook that Linguini falls in love with (what's her name? I cannot recall it now :blush:) at their best.

To say that "every mathematician has only a few tricks" is pretty much the same as to say that "every English writer uses just 26 letters". The real assortment of tricks never makes it to the print. When you start solving a problem, the first question is "which thing that I or other people know can be used here?" and the real trick is in how you find the right thing and how you modify it to suit your needs. Creating some new ingredient is next to impossible; if you do it just once in your life, you may be proud of yourself, and if you do it [i]a few[i] times like Hilbert, you'll go into history forever. Fortunately, combining old ingredients in new way and making new dishes out of old ingredients is also valuable and requires a lot of "creativity". As far as that famous Green-Tao paper is concerned, they said themselves that the main part of what they did was analysing the 4 existing proofs of Szemeredi's theorem and putting together the ideas that were already there with suitable modifications. So, this paper contains no tricks at all in Rota's sense but it is still extremely creative according to the highest standards.

As to "hard work", perseverance is the key to all successes, of course. Given enough time and patience, almost anyone can reach any level. The only little problem is that life is rather short :(. So, some people just walk faster than others (or even run) and can cover greater distances in shorter times.

As to the analogy with "cellular automata", our brains are, indeed, something similar to them but the devil, as always, is in the details. We can make many analogies and bring up various associations but, in absence of proven neurosurgery techniques improving creativity, the main question is whether it can be taught and, if so, how exactly to teach it.

I accede with what Mr Fedja asserts, but given some effort, perseverance a bright person can be become more creative.

As for my gentle request to Mr Fedja is, to re-send a private message he has earlier sent to me as an answer to a query I had asked him; Accidentally I deleted the reply, so I ask him to send it again; I ask him to check his inbox and send that(Dated: 17th Nov 2008 or so, somewhere after 15th November,2008)message to me. ( I would feel happy).

It has been more than a month since I had asked him.

I read only the report from the first post and I think that it is very interesting. I watched "A Beautiful Mind", a movie in honor of John Nash, three weeks ago and there was something about schizophrenia. Maybe the unusual polarity of the brain, and sometimes schizophrenia are the most common causes of a beautiful mind.

"Today I was attending an event on poetry and math. "
what event?
EDIT: directed at Orl

[quote="fedja"]

To say that "every mathematician has only a few tricks" is pretty much the same as to say that "every English writer uses just 26 letters". The real assortment of tricks never makes it to the print. When you start solving a problem, the first question is "which thing that I or other people know can be used here?" and the real trick is in how you find the right thing and how you modify it to suit your needs. Creating some new ingredient is next to impossible; if you do it just once in your life, you may be proud of yourself, and if you do it [i]a few[/i] times like Hilbert, you'll go into history forever.
[/quote]

I wanted to note that Timothy Gowers plans on setting up a [url=http://gowers.wordpress.com/2008/11/28/very-brief-tricki-update/]mathematical trick wiki[/url] in the very near future after it has been postponed a few times. [url=http://gowers.wordpress.com/2007/09/11/what-might-an-expository-mathematical-wiki-be-like/]Here[/url] is an earlier article.


Quite simple.

Let f(x)=   :sqrt: (x+5) + (x+1) :sqrt: (x+6) + ...

we are looking for f(1).

my claim is that f(x)= (x+3).

Proof.
Let an = 1 :sqrt: 6+2 :sqrt: 7+3 :sqrt: 8+... + n :sqrt: n+5
Note that  n + (n-4)(n-1) = (n-2)^2

So 4=  :sqrt:  6 + 2  :sqrt: 25 =  :sqrt: 6 + 2 :sqrt: 7 + 3 :sqrt: 36 =...
and thus we can prove by induction that 4>an for all n.
indeed one can show that x+3> fn(x) for all x and all n.

To prove that the sequence actually converges to x+3 one has to work slightly more, but
I think the main point is discussed. I will post the next part if nessasary.
-A

here is a simple way to show that fn(x) indeed converges to x+3.
Note fn(x)=  :sqrt: x+5 + :sqrt: .... +(x+n-1) :sqrt: x+n+4

Now note that if we add  :varepsilon: = (x+n)(x+n+3)
to the last term in the serie, we get that the new sum equals 4.
Now we know that  :sqrt: (a+nx)  <=  :sqrt: n  :sqrt: (a+ x)
So the new sum S' satisfies:

fn(x) <= S'=4 <= ([(x+n)(x+n+3)]^(1/2)^n )fn(x)
but for large enough n, the right hand side expression gets as close to fn(x) as we wish.
In other words fn(x) does converge to 4.
There are of course many other ways of doing this, but i think this is the easiest way.
-Ali

[quote]fn(x)  S'=4  ([(x+n)(x+n+3)]^(1/2)^n )fn(x)
but for large enough n, the right hand side expression gets as close to fn(x) as we wish.
In other words fn(x) does converge to 4. [/quote]

the 4's shoukd be replaced by x+3 ofcourse ;)


group theory
abstract algebra
superior algebra
superior algebra solved

$G$ is a finite group. $H$ is a normal subgroup of $G$ with $|H|$, $[G:H]$ coprime. Show that for every $g\in G$ with $ord(g)||H|$ already follows $g\in H$.

Have much fun!

Misha

P.S. It was the only problem of our exam which I didn't see before it. :D

Counterexample:
$G=S_3$, $H$ has order 2.
Since there are four elements of order dividing 2, they aren't all in $H$.

I think you need $H$ to be normal, or some other condition stronger than "is a subgroup".

Sorry!!! H is really normal. I will correct it.

Thanks Jmerry

Let $f:G\to G/H$ be the projection $x\mapsto xH$. We have $|f(<g>)|\ |\ |<g>|\ |\ |H|$. At the same time, $|f(<g>)|\ |\ |G/H|$, and from here we find $|f(<g>)|=1$, which is precisely what we want.

Is it Ok?

Very nice! It is the best solution I know.

Here is my solution:

We consider $m=[<g>H:H]=\frac{|<g>H|}{|H|}=\frac{ord(g)|H|}{|H||H\cap <g>|}=\frac{ord(g)}{|H\cap <g>|}||H|$. On the other side $\frac{[G:H]}{m}=[G:<g>H]\in\Bbb N$. So $m$ is a divisor of $|H|$ and $[G:H]$. We have $m=1$ and $ord(g)=|H\cap <g>|$ which means $g\in H$.

Misha


calculus
integration
trigonometry
geometry
rectangle
real analysis
real analysis unsolved

I have some troble with this integrals, help me, plz

 [tex]\int_{-\infty}^{+\infty}e^{tx}\frac{e^{-x}}{(1+e^{-x})^2}dx [/tex]

[tex]\int_{-\infty}^{+\infty}e^{tx}\frac{1-cos(ax)}{\pi ax^2}dx [/tex]

Hey artada:

I was looking at your integrals and these particular integrals are difficult to arrive at a general form.  If you sub in consecutive values of t (t=0,1,2,3.......,t) you'll arrive at a pattern, thus giving you a series for the general solution.
i.e for #1. if t=0 you get 1/(1+e^-x); for t=1: 1/(e^x+1)+ln(e^x+1); for t=2: e^x-2ln(e^x+1)-1/(e^x+1);
for t=3: (e^x)^2/2-2e^x+3ln(e^x+1)+1/(e^x+1); t=4:(e^x)/3-(e^x)^2+3e^x-4ln(e^x+1)-1/(e^x+1), and so on. #2 probably plays on the same sort of theme.  There's a simple pattern here which will give you a series for your answer, which I will leave to you to work out. Maybe someone with more expertise than me can give you a better answer.

Hope this helps a little,
Cody aka fredoniahead

If I haven't made a stupid mistake somewhere, the first integral equals $\frac {\pi t}{\sin \pi t}$ for $t\in (-1,1)$. It diverges for $|t|\geq 1$. The easiest way to prove it is to use contour integration. Condider the rectangle contour $\Gamma$ consisting of the interval $[-R,R]$ followed by $[R,R+2\pi i]$ followed by $[R+2\pi i, -R+2\pi i]$ followed by $[-R+2\pi i,-R]$. On one hand, the integral over the contour is $2\pi i$ times the residue at $\pi i$, which can be shown to be $-te^{\pi it}$. On the other hand, the integrals over vertical sides tend to $0$ as $R\to\infty$ and the integrals over the horizontal sides are almost $I$ and $-e^{2\pi i t}I$ where $I$ is the integral to find. Thus, $-2\pi ite^{\pi i t}=(1-e^{2\pi i t})I$, from where the answer follows immediately. This reasoning is valid for $t\neq 0$ but, by continuity, we may conclude that the integral for $t=0$ is $1$, which is easy to check directly as well.

I am confused a bit about the second integral: it converges only for $t=0$ (unless we want to consider purely imaginary $t$ as well). Are you sure it is the one you want to find?


hi!! this is a very simple question, could anybody explain me the difference between the sty and cls files?? both are used for creating the style of a document, no??
I have to write  a resume and in many examples that I've seen both files are used. I guess the sty file is the general one where all the needed features are defined, and the cls is used for customizing the style of some features of the sty file that you need to change. Am I right??

thanks in advance, and sorry for such a silly question!!

ok, I've just realised that the sty files are the old styles that were used in LaTeX 2.09, while the cls files are the new styles used in LaTeX2e.

It's not quite like that. In 2.09 .sty files were both style and class files, but they have now been separated into the two types.

Kopka & Daly's excellent book A Guide to LaTeX gives the details:

[quote].cls. This is a [i]class[/i] file in LaTeX2e, defining the overall format of the document. It is read by the \documentclass command.

.sty This is a [i]package[/i] file in LaTeX2e, to be loaded with \usepackage. A package contains additional LaTeX commands to define new features or alter existing ones [...]

In LaTeX 2.09 the .sty files were [i]main style[/i] files (equivalent to classes) as well as [i]style option[/i] files (packages). The two have quite different properties although they both have the same extension[/quote]
So .sty files are still very much alive!


It has been found that when the chemical bond between chlorine atoms in Cl_2 is formed, 328 kJ is released per mole of Cl_2 formed.  What is the wavelength of light that would be required to break chemical bonds between chlorine atoms?

we haven't done anything with bond strength in class...so there must be a slicker solution?

Hint:
[hide]For a bond to break, it must be hit by a photon with at least the energy of that bond.[/hide]

E = h*c*(1/lambda)
lambda = hc/E

so the wavelength is just planck's constant times the speed of light divided by the energy given in the problem?

Not quite.  What is the energy of a bond in ONE $\text{Cl}_2$ [b]MOLECULE[/b]?

Hint (if you're still stuck):
[hide]Look closely at the units for bond strength.[/hide]

so you have to multiply that energy by 2 mol to cancel out the mol in kJ/mol since there are 2 chlorine atoms (2 mols of chlorine) in 1 mol of Cl_2?  sorry, i'm not good at chem! lol.

You need to find out how many total molecules there are (equivalently, how many bonds there are) in one mole of $\text{Cl}_2$.

yeah, i figured out how to do it last night a while after i had posted that, but was tired and didn't post again.  thanks for your help!


function
trigonometry
algebra unsolved
algebra

Let $f: N\to R$ satify: $f(n+1)+f(n-1)=k.f(n)$ for all natural number $n$.
Finl all values of $k$ so that $f$ ia period function

I think $f(n)=0$ is only  special case for problem with all k belong to R.  Problem will be harder if $f$ is general function that satified codition above.

oh , it is $f: N^*\to R$
solution $|k|\leq 2$

Dear Dieuhuynh, Can you post cpmletely solution?

It's a sequense problem in fact.
$a_{n+1}=ka_n-a_{n-1}$
If $|k|>2$,easy to find when $n$ infinite big,we can find a such sequence let $|a_{n+1}|>a_n$.So it's not period.
When $|k|<2$ ,$k=2\cos\alpha$,
Then we know the roots of $x^2-kx+1=0$ is $\cos\alpha+\mbox{i}\sin\alpha,\cos\alpha-\mbox{i}\sin\alpha$
So $a_n=A(\cos\alpha+\mbox{i}\sin\alpha)^n+B(\cos\alpha-\mbox{i}\sin\alpha)^n=A(\cos n\alpha+\mbox{i}\sin n\alpha)+B(\cos n\alpha-\mbox{i}\sin n\alpha)$ is real.
So $A=B$.$a_n=2A\cos n\alpha$.
So $\{a_n\}$ is period if and only if for some $m$,$\cos (m+k)\alpha =\cos k\alpha$ is hold for all $k$.
Then we know $\sin m\alpha=0$.

[quote="jin"]It's a sequense problem in fact.
$a_{n+1}=ka_{n}-a_{n-1}$
If $|k|>2$,easy to find when $n$ infinite big,we can find a such sequence let $|a_{n+1}|>a_{n}$.So it's not period.
[/quote]
Why do you derive the fact?


analytic geometry
graphing lines
slope
trigonometry

I am a programmer and my math skills are weak.

I want to draw an arrow. I know the start and end points of the arrow.

I need to draw the arrow head which will be made up of three points
Point 1 is the end point which I already know

Point 2 and 3 would be a 45 degree angles a certain distance away from the end point. What is the magical math formula to calculate what points 2 and 3 are?

I would try to help you if you could reword the question.  Can you explain a little more clearly what you are asking?  I fail to comprehend you.

imagine an arrow

 *
/ \
 |
 |
 |
 *

I want to draw an arrow.  I know the two points that are drawn as an asterisk.  The end of the arrow and the top point.  I'm trying to draw the head of the arrow.  The triangle that rests on top of the line.  I know one of the points of the triangle because it represents the top of the arrow.  I need the other two which are lines that go out at a 45 degree angle.

Knowing the two points what equation could I use to get the other two points?

So, it would be like an arrow

    /|\
  /  |  \
/    |    \

And given the middle line, you want to know the equations for the side lines?
Well, if you have the 2 points for the middle line, you can calculate the equation for that middle line. Then use slopes and trig to find the slopes of the 2 side points. And then use the point and these slopes to find the 2 equations for the side lines.

[i]For future reference, topics of this level belong in the "High School Basics" forum.[/i]

Here's the set-up I'll use:
- The tail-end of the arrow is $ (0,0)$. If your first endpoint is not $ (0,0)$, you'll have to adjust the formula appropriately.
- The head-end of the arrow is $ (a,b)$
- The length of the arrow is $ L \equal{} \sqrt {a^2 \plus{} b^2}$
- The length of the ticks is $ \ell \sqrt {2}$ (divide to get the desired $ \ell$)

What you basically want to do is to backtrack a distance $ \ell$ along the direction of the arrow and move out a distance $ \ell$ in a perpendicular direction to get the second endpoint of each tick. There are two perpendicular directions (they are opposite), thus two tickmarks.

Thus, the two points you're interested in are:

$ \left(a \minus{} \frac {\ell}{L} a \plus{} \frac {\ell}{L} b \; , \; \; b \minus{} \frac {\ell}{L} b \minus{} \frac {\ell}{L} a \right)$
$ \left(a \minus{} \frac {\ell}{L} a \minus{} \frac {\ell}{L} b \; , \; \; b \minus{} \frac {\ell}{L} b \plus{} \frac {\ell}{L} a \right)$

Use the case $ (a,b) \equal{} (0,2)$ and $ \ell \equal{} 1$ to get a visual idea of what's going on here.

Good luck programming!


[quote="shobber"]$F_1$, $F_2$ are the focuses of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b>0)$. If there exists a point $P$ such that $\angle{F_1PF_2}=120^o$, find $e$.[/quote]
Allways exist  point P such that $\angle{F_1PF_2}=120^o.$
Id est $0<e<1$  :lol:

[quote="shobber"] $P$ should be on the ellipse.[/quote]
OK
$PF_1^2+PF_2^2+PF_1\cdot PF_2=4c^2,$ where $c^2=a^2-b^2.$
$PF_1+PF_2=2a.$
Hance, $4a^2-PF_1\cdot PF_2=4c^2.$
Hence, $PF_1$ and $PF_2$ are roots of equation: $z^2-2az+4(a^2-c^2)=0.$
Thence, $a^2-4(a^2-c^2)\geq0\Leftrightarrow\frac{c}{a}\geq\frac{\sqrt3}{2}.$
Id est $\frac{\sqrt3}{2}\leq e<1.$ :)


[quote="Treething"]Billy Bob's funeral is from 9 AM to 5 PM on the same day (a Wednesday.) His wife has a 25% chance of coming, his son has a 10% chance of coming, and Arvind_sn has a 90% chance of coming. What is the probability that exactly two people come to his funeral, assuming that everybody else has a 0% chance of coming to the funeral? Express your answer as a common fraction.[/quote]

[hide]$(1/4)(1/10)(1/10)+(3/4)(1/10)(9/10)+(1/4)(9/10)(9/10)=\frac{109}{400}$[/hide]

[hide]There are 3 cases:

Wife and Son come, but not Arvind_sn
$\frac14\cdot\frac1{10}\cdot\frac{1}{10}=\frac1{400}$

Wife and Arvind_sn, but not Son
$\frac14\cdot\frac{9}{10}\cdot\frac9{10}=\frac{81}{400}$

Son and Arvind_sn, but not Wife
$\frac1{10}\cdot\frac9{10}\cdot\frac34=\frac{27}{400}$

Add them up to get:
$\frac{109}{400}$[/hide]

[hide="answer"]I used same method as B-Flat to get [color=red]109/400[/color][/hide]


Can anyone refer me to some webpages that contain the expressions and derivations of Resolving power of microscope,telescope and simple triangular glass prism?I am not having a book and simple google search is not providing results needed.

What I want is the RESOLVING POWER FORMULA for a triangular prism,a microscope and a telescope.As far as I know,the theories for the latter two use the concept of diffraction through a circular aperture.

Please help me with it.

Thank you in advance.

I got this in internet.Please anyone can help how to prove the formula for resolving power of a prism?

http://ece-www.colorado.edu/~ecen5606/Labs/spec040504.pdf

Another link is this one below this sentence:
[url]http://en.wikipedia.org/wiki/Resolving_power[/url]
HTH  :)


logarithms
group theory
superior algebra
superior algebra unsolved

can we have a group that every finite group is all able to embed in that group ? Are these groups unique up to isomorphism ?

Take the sum or product of all finite groups for example. And trivially such groups are not isomorphic, you can always make that group bigger.

Of course, it is a rather natural construction!
But could you embed it into a group similar to symmetric group ? Because I'm
interested in symmetric very much !

You can also embed it into the group $S_{M}$ of bijections on some arbitrary infinite set $M$: As $M$ is infinite, every $S_{n}$ can be embedded in $S_{M}$, and every finite group can be embedded in some $S_{n}$.

@ZetaX: Your first example needs some additional care since the class of all finite groups is not a set. But, of course, it is sufficient to take something like all groups whose underlying set is a subset of $\mathbb N$ (this is pretty close to your second example, of course). owk

I'm aware of this, but if I speak of groups, I mean them always up to isomorphism.

[quote="sylow_theory"]What is the caridnality of all finite groups (up to isomorphism) ?
Surly it is more than $\aleph_{0}$ is it $\aleph_{1}$?[/quote]

Why is it more than $\aleph_{0}$ ? For any integer $n$, there are only finitely many groups of order $n$ up to isomorphism (in fact, at most $n^{n^{2}}$ such groups, since any group of order $n$ is determined by its group table which has $n^{2}$ entries which can take $n$ different values), and the set of integers is countable, so the set of all finite groups up to isomorphism is a countable union of finite sets, and thus is countable...

  darij

Are there any better bounds for the number of isomorphism classes of groups of order $n$ (except trivial improvements like $(n-1)!^{n-1}$)? owk

There is a paper of Holt which proves the following upper bound:
\[n^{(1/6+o(1))\log_{2}^{2}n}. \]
The paper also shows that this bound is tight up to the constant in the exponent. Here is the reference: Holt, Enumerating perfect groups, J. London Math. Soc. (2) 39, no. 1, pp. 67-78, 1989.
http://jlms.oxfordjournals.org/cgi/reprint/s2-39/1/67


geometry
circumcircle
similar triangles
power of a point
geometry theorems

If $X,Y$ are isogonal conjugates in a triangle $ABC$,
then its projections in the sides (let it $X_{a},Y_{a},X_{b},Y_{b},X_{c},Y_{c}$)
are concyclic.

(well, if X is the circumcenter and Y is
P.S.: the reciprocal is true, in some sense?

This is easy, but also very nice.

The center of the sought circle ($\mathcal C$) is the midpoint of $XY$. You can prove that every $4$ of the $6$ points lie on the circle $\mathcal C$. To do that you can use similar triangles and the power of a point theorem. (actually its converse)

The converse should be obviously true. You just the other implication. Or maybe you meant some other converse??

[quote="Johann Peter Dirichlet"]If $ X,Y$ are isogonal conjugates in a triangle $ ABC$,
then its projections in the sides (let it $ X_{a},Y_{a},X_{b},Y_{b},X_{c},Y_{c}$)
are concyclic.[/quote]

This is the [b]pedal circle theorem[/b], and proofs are given on [url=http://www.cut-the-knot.org/Curriculum/Geometry/IsoSimpson.shtml]the Cut-the-Knot site[/url] and in the paper "Isogonal conjugation with respect to a triangle", which can be found as an attachment at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=569871#569871]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 post #6[/url] and on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url] (Theorem 7 [b]b)[/b]).

So let me change the topic to a generalization of this result, which is less known and harder to prove:

[color=blue][b]$ \phi$-pedal circle theorem.[/b] We work with directed angles modulo 180°. Let P and Q be two isogonally conjugate points with respect to a triangle ABC. Let X, Y, Z be points on the lines BC, CA, AB such that $ \measuredangle\left(PX;\;BC\right) \equal{} \measuredangle\left(PY;\;CA\right) \equal{} \measuredangle\left(PZ;\;AB\right) \equal{} \phi$, and let U, V, W be points on the lines BC, CA, AB such that $ \measuredangle\left(QU;\;BC\right) \equal{} \measuredangle\left(QV;\;CA\right) \equal{} \measuredangle\left(QW;\;AB\right) \equal{} \minus{} \phi$. Then, the points X, Y, Z, U, V, W lie on one circle. The center M of this circle satisfies $ \measuredangle MPQ \equal{} \measuredangle PQM \equal{} 90^{\circ} \plus{} \phi$.[/color]

  Darij


Cine credeti ca va castiga O.N.M.2006? :)

Cand am zis cine va castiga O.N.M.2006 probabil iti dai si tu seama ca m-am referit la Romania.(fiid si in sectiunea pentru limba romana).Asa ca te rog frumos sa citesti mai atent mesajele de acum incolo.Multimesc de intelegere!

Marius sunt si eu curios ce solutie ai dat la problema 4 de la nationala.
 Vezi postul http://www.mathlinks.ro/Forum/viewtopic.php?p=486661&highlight=#486661 de la Real Analisys-Unsolved & Proposed Problems. ;)

Eu nu-i stiu, dar sunt sigur ca problema 1 si problema 3 apartin lui Marius Cavachi. :)

Si la ce ora se afiseaza rezultatele ca banuiesc ca sunteti la negruzzi...

Eu, nu. Sunt in Bucuresti. Am treaba pe aici. Dar maine la Baraje o sa fiu in Iasi. :)


geometry
circumcircle
power of a point
radical axis
cyclic quadrilateral
geometry unsolved

Let $ABC$ be a non-isosceles triangle. The incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$. The line $AD$ cuts the incircle at a point $P$ different from $D$. The perpendicular to the line $AD$ at the point $P$ is drawn. The line $EF$ intersects this perpendicular at $Q$. The line $QA$ is drawn. $DE$ and $DF$ cut $QA$ at $R$ and $S$ respectively. Prove that $A$ is the midpoint of $RS$.

Let the normal to AD at P meet the incircle (I) again at X. As $\angle DPX$ is right, DX is a diameter of (I), $I \in DX,$ and $DX \perp BC.$ Let (J) be a circle with diameter AX. Since the angle $\angle APX$ on this diameter is right, $P \in (J)$ and $PX \equiv PQ$ is the radical axis of (I), (J). Let AQ meet (J) again at Y, $\angle AYX$ on the diameter AX is also right. Since Q is on the radical axis of (I), (J), $QY \cdot QA = QF \cdot QE,$ which means that quadrilateral AEFY is cyclic with a circumcircle (K). (K) is the circumcircle of the isosceles $\triangle AEF,$ which in turn is the circumcircle of the cyclic quadrilateral (kite) AEIF (2 right angles $\angle AEI,\ \angle AFI$ on the diameter AI), so that $I \in (K),$ the angle $\angle AYI$ is right and I, X, Y collinear. Since the angle $\angle DPX$ is right, DX is a diameter of (I), $I \in DX$ and $DY \equiv DX \equiv DI \perp BC.$ As a result, $AQ \equiv AY \parallel BC$ (both perpendicular to $DY \equiv DX \equiv DI$). As EA is a tangent of (I) at E, $\angle AEF = \angle FDE \equiv \angle D,$

$\angle AER = 180^\circ-\angle (AEF+\angle FED) = 180^\circ-(\angle D+\angle E) = \angle F.$

As $AR \parallel BC,$ $\angle EAR = \angle BCA = 180^\circ-2 \angle F$ (from the isosceles triangle $\triangle CDE$ with sides CD = CE tangent to (I)). It follows that $\angle ARE = \angle F$ as well and $\triangle AER$ is isosceles with AE = AR. Similarly, $\triangle AFS$ is isosceles with AF = AS. Since AE = AF are tangent lengths form A to (I), we get AR = AS. Note that DA is the D-symmedian of the contact triangle $\triangle DEF.$

$AI$ meets $EF$ at $M$
We have $AI*AM=AF^{2}=AP*AD$
so $P,M,I,D$ are cyclic
so $90-\angle PMA=90-\angle IDA.$ Then $\angle PMF=\angle ADB$ $(1)$
Otherwise $\angle APQ=\angle AMQ=90$ so $A,P,M,Q$ are cyclic
so $\angle PMF=\angle PAQ$ $(2)$
from (1)(2) we have $\angle BDA=\angle QAD$ so $AQ//BC$
Thus$\angle ARE=\angle EDC=\angle DEC=\angle AER$
so triangle $AER$ is isosceles.Then $AE=AR$
similarly $AF=AS$
but $AE=AF$ so $AR=AS$ $q.e.d$

[quote][color=darkred][b]For the incircle $w$ of a non-isosceles $\triangle ABC$ denote $\{\begin{array}{c}D\in BC\cap w\\\ E\in CA\cap w\\\ F\in AB\cap w\end{array}$. Define $\{\begin{array}{c}\{D,P\}\subset AD\cap w\\\ Q\in EF\ ,\ QP\perp AD\\\ R\in DE\cap AQ\ ,\ S\in DF\cap AQ\end{array}$. Prove that $AR=AS$.[/b][/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] I"ll shorten a bit the above [b]nice Huyền Vũ's and Yetty's proofs[/b] of this [b]easy and good-looking problem.[/b]

Denote $\{\begin{array}{c}M\in EF\cap AI\\\ L\in EF\cap BC\\\ N\in AI\cap BC\end{array}$. I"ll show that $AQ\parallel BC$ ([u]see lower remark[/u]). Indeed,

$\{\begin{array}{ccccc}\{\begin{array}{c}PA\perp PQ\\\ MA\perp MQ\end{array}\| & \implies & \mathrm{the\ quad.\ APMQ\ is\ cyclically}& \implies & \widehat{AQM}\equiv\widehat{DPM}\\\\ \{\begin{array}{c}AP\cdot AD=AE^{2}\\\ AI\cdot AM=AE^{2}\end{array}\| & \implies & \mathrm{the\ quad.\ PMID\ is\ cyclically}& \implies & \widehat{DPM}\equiv\widehat{DIN}\\\\ \{\begin{array}{c}DL\perp DI\\\ ML\perp MI\end{array}\| & \implies & \mathrm{the\ quad.\ MIDL\ is\ cyclically}& \implies & \widehat{DIN}\equiv\widehat{DLM}\end{array}\|$ $\implies$ $\{\begin{array}{c}\widehat{AQM}\equiv\widehat{DLM}\\\ M\in QL\end{array}\|$ $\implies$ $AQ\parallel BC$.

The division $(B,C,D,L)$ is harmonically $\implies$ the fascicle $A(B,C,D,L)$ is harmonically $\implies$ the division $(F,E,K,L)$ is harmonically, where $K\in AD\cap EF$ $\implies$

the fascicle $D(F,E,K,L)$ is harmonically and $\overline{SARQ}\parallel DL$ $\implies$ the division $(S,R,A,\infty )$ is harmonically $\implies$ $AS=AR$.

[b][u]Remark.[/u][/b] Let $\triangle ABC$ and $d$ be a line for wich $A\in d$. For the points $\{\begin{array}{c}D\in BC\ ,\ E\in CA\ ,\ F\in AB\\\ X\in AD\cap BE\cap CF\end{array}$ denote $R\in DE\cap d$ and $S\in DF\cap d$.

Then $\boxed{\ d\parallel BC\Longleftrightarrow AR=AS\ }$. Particularly (an well-known property), $AD\perp BC$ and $X\in AD$ $\Longrightarrow$ $\widehat{SDA}\equiv\widehat{RDA}$.[/color]

I have another proof. I have posted it here:http://www.mathlinks.ro/Forum/viewtopic.php?t=145686

Remark: After AQ || BC, see that $ \triangle ARE \sim  \triangle CDE and  \triangle ASF \sim  \triangle BFD$, but the triangles CDE and BFD are isosceles, and so are ARE and ASF; as AE = AF, we get the desired.

Best regards,
sunken rock


[quote="Quevvy"][quote="mitohne"]Hi All:

If someone had 89.50 and has to pay a tax of 8.625 percent on any product purchased, what is the maximum amount of money able to be spent when applying the tax?

How would you go about calculating this? Try to be as thorough as possible with the answer and explanation.[/quote]
Do you mean finding the price of the object before tax?[/quote]

Even if you're given the maximum amount, 82.39 dollars, it's still a bit confusing what formula you would use to come to that answer.
Sorry about the confusion. It's a poorly worded problem. Thanks for your reply, though.

[quote="Quevvy"][hide]If the price of the object is $\$x$, then you would get this equation: $x+x\frac{8.625}{100}=89.50$
Simplifying: $\frac{108625x}{100000}=89.5$
$108625x=8950000$
$x\approx \$82.3936...$
The price would be 82.39 dollars and then tax is 7.11 which totals to 89.50 exactly.[/hide][/quote]

Thanks so much! That makes perfect sense.


Θα ήθελα να ελέξετε την ορθότητα αυτής της απόδειξης ,

διότι στο σχολικό βιβλίο δεν υπάρχει τέτοιος τύπος για την γωνία δυο ευθειών .

[img]http://img295.imageshack.us/img295/1926/newtk5.jpg[/img]

Βασικά επειδή δεν πολυκατάλαβα σε μερικά σημεία και τους χαρακτήρες που χρησιμοποιείς, πώς πας από το:

$\frac{AA'+BB'}{\sqrt{A^{2}+B^{2}}+\sqrt{A'^{2}+B'^{2}}}$

στο

$\frac{AA'}{BB'}+1$ ;

PS: Αν θες καθαρόγραψέ τη (αν δεν μπορείς σε $\LaTeX$) στο Mathtype π.χ. και ξαναποστάρισέ την εδώ.

Καπου εκει πρεπει να εχει λαθος mostel.Oυτε εγω σε εκεινο το σημειο το καταλαβα. :maybe:

(an katalava apo ayta pu grafi..)dieri me $BB'$ to opio 8eori oti ine diaforo tu midenos epidi i e1 e2 den ine katakoryfes....

 :D

δεν ειναι αυτο που γραφεις στελιο.κοιτα καλυτερα...εχει γραμμη κλασματος απο κατω  :)

Ooops! Σωστά, έχει μία γραμμή (?) κλάσματος που δεν την αντελείφθην από την αρχή. Γι αυτό πιστεύω πως καλύτερα θα ήταν να την έγραφε και να την ξαναποστάριζε ως ένα jpg γραμμένο σε mathtype για να μπορούμε να βγάλουμε άκρη..

Μία πρόχειρη Latexia του παραπάνω (και αν κατάλαβα καλά):

Δίνονται οι μη κατακόρυφες ευθείες:

$\epsilon_{1}: \quad Ax+By+C=0$ με $\lambda_{1}=-\frac{A}{B}$
και
$\epsilon_{2}: \quad A'x+B'y+C'=0$ με $\lambda_{2}=-\frac{A'}{B'}$

Θεωρούμε τα διανύσματα:

$\overrightarrow{\delta_{1}}=(-B,A)\parallel\epsilon_{1}$
και
$\overrightarrow{\delta_{2}}=(-B',A')\parallel\epsilon_{2}$

Από το εσωτερικό γινόμενο έχουμε:

$\overrightarrow{\delta_{1}}\cdot\overrightarrow{\delta_{2}}=BB'+AA'$

Επίσης για τα μέτρα των διανυσμάτων ισχύει:

$\left|\overrightarrow{\delta_{1}}\right|=\sqrt{A^{2}+B^{2}}$
και
$\left|\overrightarrow{\delta_{2}}\right|=\sqrt{A'^{2}+B'^{2}}$

Δηλαδή θα έχουμε αντίστοιχα για τη γωνία των δύο διανυσμάτων:

$\sigma\upsilon\nu\left(\overrightarrow{\delta_{1}}\stackrel{\hat{}}{,}\overrightarrow{\delta_{2}}\right)=\frac{\overrightarrow{\delta_{1}}\cdot\overrightarrow{\delta_{2}}}{\left|\overrightarrow{\delta_{1}}\right|\cdot\left|\overrightarrow{\delta_{2}}\right|}=\frac{AA'+BB'}{\sqrt{A^{2}+B^{2}}\sqrt{A'^{2}+B'^{2}}}=\frac{\frac{AA'}{BB'}+\frac{BB'}{BB'}}{\frac{\sqrt{A^{2}+B^{2}}}{B}\cdot\frac{\sqrt{A'^{2}+B'^{2}}}{B'}}=\frac{\frac{A}{B}\cdot\frac{A'}{B'}+1}{\frac{\sqrt{A^{2}+B^{2}}}{\sqrt{B^{2}}}\cdot\frac{\sqrt{A'^{2}+B'^{2}}}{\sqrt{B'^{2}}}}=\frac{\lambda_{1}\lambda_{2}+1}{\sqrt{\frac{A^{2}}{B^{2}}+\frac{B^{2}}{B^{2}}}\cdot\sqrt{\frac{A'^{2}}{B'^{2}}+\frac{B'^{2}}{B'^{2}}}}=\frac{\lambda_{1}\lambda_{2}+1}{\sqrt{\lambda_{1}^{2}+1}\sqrt{\lambda_{2}^{2}+1}}$

[u]Μία σημείωση[/u]:

Σύμφωνα με αυτά που γράφεις, έχουμε: $B=\sqrt{B^{2}}$ και $B'=\sqrt{B'^{2}}$. Αυτό όμως δεν υφίσταται, αφού ισχύει: $a=\left(\sqrt{a}\right)^{2}$ και $\sqrt{a^{2}}=\left|a\right|$. Καλή η σκέψη, αλλά νομίζω ότι εδώ έχουμε να κάνουμε με "προσανατολισμένες" ευθείες και ειδικές περιπτώσεις.

[quote="kostas"]ναι σωστα απλα πρεπει Β,Β' ΝΑ ΕΙΝΑΙ ΘΕΤΙΚΑ? (χαρα στο κουραγιο σου στελιο να τα γραφεις σε λατεξ τετοια ωρα) :P[/quote]

Ευχαριστώ Kostas, ειδικά αν σκεφτείς ότι μου πήρε περίπου ένα τέταρτο το γράψιμο, μιας και δεν ήταν τόσο απλός ο κώδικας.

Σωστά, αυτό προφανώς ισχύει μόνο για τα θετικά, για αυτό και έκανα αναφορά για "προσανατολισμένες" ευθείες. Νομίζω τέτοιοι τύποι μάλιστα υπήρχαν σε παλαιότερα βιβλία του ΟΕΔΒ, αλλά ο ορισμός της θετικής φοράς ήταν διαφορετικός  :wink:

Σε Ευχαριστώ πολύ mostel :oops:  για την διόρθωση της λάθος σκέψης που έκανα ,αν και αναρωτιόμουν χρησιμοποιώντας αυτόν τον τύπο ποια γωνία υπολογίζουμε ( την οξεία ή την αμβλεία γωνία των ευθειών? :idea: )

Στο σχολείο μας είπε έναν τύπο για την γωνία των ευθειών (αλλά δεν τον απομνήμευσα :P , και μάλλον θα πρέπει να ήταν αυτός) , και αυτό μου έδωσε αφορμή για την απόδειξή του. Άραγε υπάρχει τύπος που να δίνει την γωνία των ευθειών
χωρίς κάθε φορά να παίρνουμε διάνυσμα παράλληλο σε ευθεία?

P.S. : Τώρα αρχίζω να μαθαίνω Latex... :oops:

Όταν δύο ευθείες τέμνονται, η γωνία μεταξύ τους περιγράφεται ως η γωνία που θα πρέπει να μετατοπιστεί μία από αυτές για να συμπέσει πάνω στην άλλη. Έστω $\epsilon_{1}$ και $\epsilon_{2}$ οι ευθείες μας και έστω $\phi$ η γωνία των δύο ευθειών. Ακόμη, έστω $a_{2}$ η γωνία που σχηματίζει η ευθεία $\epsilon_{1}$ με τον $x'x$ και $a_{1}$ η γωνία που σχηματίζει η ευθεία $\epsilon_{2}$ με τον $x'x$, με $a_{2}>a_{1}$. Δηλαδή: $a_{2}=a_{1}+\phi\implies\phi=a_{2}-a_{1}$

Έτσι θα έχουμε:

$\tan\phi=\tan(a_{2}-a_{1})=\frac{\tan a_{2}-\tan a_{1}}{1+\tan a_{1}\tan a_{2}}$

'Η αν $\lambda_{1}$ είναι ο συντελεστής διεύθυνσης της $\epsilon_{1}$ και $\lambda_{2}$ είναι ο συντελεστής διεύθυνσης της $\epsilon_{2}$ θα έχουμε:

$\tan a_{1}=\lambda_{1}\quad\kappa\alpha\iota\quad\tan a_{2}=\lambda_{2}$ (με $\lambda_{2}>\lambda_{1}$)

'Η αλλιώς:

$\tan\phi=\frac{\lambda_{2}-\lambda_{1}}{1+\lambda_{1}\lambda_{2}}$

[quote="RAKAR"]na rotiso kati...?
(an kai exun yiloskuriasi ta dianismata pu iksera..)
ta B,B' den mporume na ta epileksume oste na ine 8etika...?
p.x $2x-3y+4=0$ na tin grayume san $-2x+3y-4=0$ gia na isxii o tipos tu giannis..sas rotao gt den exo kratimeno to vivlio tis  b likiu,+ipa mipos lei tipota mesa san leptomeria ,iposimiosi...

 :D

edit:mipos kalitera mpori na dosi kapiois ena antiparadigma pu den isxii o tipos tu giannis,,? :wink:[/quote]

Σωστά RAKAR, αυτό που λες γίνεται σίγουρα στις ευθείες. Αυτό που έκρινα εγώ πάντως ήταν η μετατροπή του $B$ σε $\sqrt{B^{2}}$ χωρίς περαιτέρω εξήγηση, το οποίο δεν ισχύει. Έτσι όπως το θέτεις ισχύει. Θα το ψάξω λίγο αν είναι, και αν βρω αντιπαράδειγμα καλώς.

PS: Όποιος δεν κατάλαβε τι έκανα στον τύπο με τις εφαπτόμενες, μπορώ να κάνω και το σχήμα, όπου θα φανεί καλύτερα. Απλώς ας το ζητήσει.

Ας έχουμε ως δεδομένο ότι: $\tan\phi=\frac{\lambda_{2}-\lambda_{1}}{1+\lambda_{1}\lambda_{2}}$ και ας γυρίσουμε λίγο πίσω, στην ύλη της πρώτης λυκείου.

Από την απόδειξη που υπάρχει και στο σχολικό βιβλίο, ισχύει:

$\cos^{2}\phi=\frac{1}{1+\tan^{2}\phi}$

'H $\frac{1}{1+\tan^{2}\phi}=\frac{1}{\frac{(1+\lambda_{1}\lambda_{2})^{2}+(\lambda_{2}-\lambda_{1})^{2}}{(1+\lambda_{1}\lambda_{2})^{2}}}=\frac{1}{\frac{(\lambda_{1}^{2}+1)(\lambda_{2}^{2}+1)}{(1+\lambda_{1}\lambda_{2})^{2}}}=\frac{(1+\lambda_{1}\lambda_{2})^{2}}{(\lambda_{1}^{2}+1)(\lambda_{2}^{2}+1)}$

Δηλαδή:

$\cos\phi=\pm\frac{\lambda_{1}\lambda_{2}+1}{\sqrt{(\lambda_{1}^{2}+1)(\lambda_{2}^{2}+1)}}$

Σωστός mostel. Πιο συγκεκριμένα:

Αν $BB'>0$ τότε

$\sigma\upsilon\nu\left(\overrightarrow{\epsilon_{1}}\stackrel{\hat{}}{,}\overrightarrow{\epsilon_{2}}\right)=\frac{\lambda_{1}\lambda_{2}+1}{\sqrt{\lambda_{1}^{2}+1}\sqrt{\lambda_{2}^{2}+1}}$

Αν $BB'<0$ τότε

$\sigma\upsilon\nu\left(\overrightarrow{\epsilon_{1}}\stackrel{\hat{}}{,}\overrightarrow{\epsilon_{2}}\right)=-\frac{\lambda_{1}\lambda_{2}+1}{\sqrt{\lambda_{1}^{2}+1}\sqrt{\lambda_{2}^{2}+1}}$

Και στις δύο περιπτώσεις υπολογίζουμε το συνημίτονο της οξείας γωνίας των δύο ευθειών αν $AA'+BB'>0$, ενώ υπολογίζουμε το συνημίτονο της αμβλείας γωνίας αν $AA'+BB'<0$.

Ράλλη πιστεύω τώρα να λύθηκε και η απορία σου. :wink:

Στέλιος

Stedes μου το χάλασες!  :mad: Περίμενα να δω ποιος θα το εξηγήσει, δηλαδή γιατί στη μία περίπτωση είναι $-$ και γιατί στην άλλη $+$, αλλά το πρόλαβες!  :P

Πάντως απορώ γιατί στο σχολικό βιβλίο δεν αναφέρεται ο τύπος  :huh:

θα επρεπε να υπαρχει αυτος ο τυπος στο σχολικο βιβλιο, τουλαχιστον υπαρχει εδω  :)

Σωστά , η απορία λύθηκε.
 :roll: Τα βιβλία των δέσμεων πρέπει να ήταν πιο καλογραμμένα από αυτά που χρησιμοποιούμε τώρα .

P.S. Στο http://www.pi-schools.gr έχει βιβλία από Δέσμες? :?:

[quote="ambros"]mostel Rakar ki stedes μπράβο σας πολύ καλά όλα αυτά θα ήθελα πάντως μια σύνοψη των τύπν μπάς κι μου μείνουν οκ???
ευχαριστώ εκ τω προτέρων
mostel σε παρακαλώ εξήγησε λίγο πιο αναλυτικά τον τύπο της εφαπτομένης[/quote]

Δες εδώ Αμβρόσιε:

http://www.mostel.gr/line.pdf

Απλώς είναι στα αγγλικά επειδή μου ζήτησε την απόδειξη και ένας φίλος από το εξωτερικό. Πρόσεξε το σχήμα και θα το καταλάβεις !

[u]Οι τύποι εν συντομία[/u]:

$\cos\phi=\pm\frac{\lambda_{1}\lambda_{2}+1}{\sqrt{(\lambda_{1}^{2}+1)(\lambda_{2}^{2}+1)}}$

$\tan\phi=\frac{\lambda_{2}-\lambda_{1}}{1+\lambda_{1}\lambda_{2}}$

Παιδιά, ΧΡΟΝΙΑ ΠΟΛΛΑ !

[i] Με χαρά διάβασα μόλις τώρα τις παραπάνω αναζητήσεις.Λοιπόν , εκ του προχείρου ,  εκφράζω μερικές παρατηρήσεις :

[/i] [b]  α)[/b] Μπορούμε πάντα να γράψουμε

     $\frac{\sqrt{a^{2}+b^{2}}}{b}= \sqrt{\frac{a^{2}+b^{2}}{b^{2}}}$, με b θετικό .Αν το $b$ είναι αρνητικό , αλλάζουμε τα πρόσημα  στην εξίσωση της ευθείας και το $b$ γίνεται θετικό( thanks to mostel for pointing me out an error !)
 [b]   β) [/b]Δεν έκανα και εγώ τις πράξεις , αλλά με φευγαλέα ματιά όλα δείχνουν σωστά.Την ορθότητα των πράξεων μπορείτε να τη δείτε και εσεις ξανά.
   [b]  γ)[/b] Δυο τεμνόμενες ευθείες σχηματίζουν δύο γωνίες.Η γωνία ευθειών ορίζεται (στις δέσμες αλλά και αλλού) ως προσανατολισμένη. Έτσι οι τύποι που έχουμε για τη γωνία , αν δεν δούμε σχήμα και δεν βάλουμε κατάλληλα πρόσημα , μας δίνουν το συνημίτονο της μιας από τις δύο γωνίες.Αν θέλουμε την  οξεία γωνία , τότε λύνουμε $cos \phi = |a|$.
[b] δ)[/b] Το θέμα δεν παρουσιάζει κανένα ενδιαφέρον - απλά αντιμετωπίζει τη γενική περίπτωση και αυτό πάλι θέλει περιπτώσεις.[color=red]Στην πράξη [/color], η οξεία γωνία δύο τεμνομένων ευθειών βρίσκεται , υπολογίζοντας την οξεία γωνία δύο διανυσμάτων  που είναι παράλληλα στις ευθείες ( ή των δύο αντίσοιχων κάθετων διανυσμάτων που είναι ακόμα πιο εύκολο !).Καλό είναι σε θέματα γωνιών να μη μπλέκετε με συντελεστές διεύθυνσης , αφού δεν ορίζονται πάντα , ενώ γωνία ευθειών ορίζεται πάντα.
 [b] ε).[/b]Οταν διδάσκαμε στις δέσμες τη γωνία ευθειών και αποδεικνύαμε θεωρητικά θέματα , χαράσσαμε σχήματα , παίρναμε περιπτώσεις και άλλα ...  τραγικά πράγματα , που εισήχθησαν από τη γαλλική μαθηματική σχολή ! Δεν είναι τυχαίο που στο βιβλίο Κατ. της Β' τα πράγματα αυτά παραλήφθηκαν.Η ύλη της Β΄κατ. είναι πιο δύσκολη από αυτήν της Γ΄και πιο δυσνόητη.Οι ασκήσεις δε του σχολ βιβλίου είναι οι περισσότερες ακατάλληλες για τον μέσο μαθητή. Θα το δείτε και στις κωνικές.
[b] στ) [/b]Εσεις που ασχολείστε με τις Ολυμπιάδες , πρέπει ωστόσο να ξέρετε να υπολογίζετε γωνία ευθειών με τον τρόπο που σας περιέγραψα. Είναι ο απλούστερος !Αλλά και οι τελευταίοι τύποι δεν είναι καθόλου άσχημοι !!! Το αντίθετο μάλιστα !! Μπορείτε να τους χρησιμοποιείτε χωρίς απόδειξη , αφού προκύπτουν εύκολα.
                    ΜΠΑΜΠΗΣ

[quote="mostel"][quote="stergiu"]
[b]  α)[/b] Πάντα μπορούμε να γράψουμε

     $\frac{\sqrt{a^{2}+b^{2}}}{b}= \sqrt{\frac{a^{2}+b^{2}}{b^{2}}}$, με b μη μηδενικό.Η αντίστροφη γραφή θέλει προσοχή !

[/quote]

Αν θέσουμε $a=3$, $b=-4$, θα έχουμε:
$\frac{5}{-4}=\frac{5}{4}$

Τι βγαίνει λάθος;  :maybe:[/quote]

 ΑΣΤΟ!!! ΘΕΤΙΚΟ  $b$ είχα και εγώ στο μυαλό μου ή κάτι άλλο  , που κάνουμε στα όρια , αλλά μου ξέφυγε. Τέλος πάντων , κανένα πρόβλημα δεν υπάρχει , αφού με μια αλλαγή προσήμων στην εξίσωση , το $b$ γίνεται θετικό. Θα το αλλάξω και στο κείμενο .
 Ευχαριστώ για την επισήμανση.Καλά να πάθω όμως :(  , γιατί αντί να καθήσω στο κρεββάτι να περάσει η ίοση και ο πονοκέφαλος , τρέχω στο κομπιούτερ και στην αλληλογραφία !!!

Ευχαριστώ mostel για την εξυπηρέτηση που μου προσέφερες κι όσον αφορά την απορία σου το λάθοσ είναι ότι όταν β<0   ΤΟΤΕ - β=ρίζαβ^2
χαιρετώ τον Κύριο Στεργίου κι περαστικά κι ευχές για καλή χρονιά

Και γω με τη σειρά μου να ευχηθώ περαστικά στον κ. Στεργίου.

@Ambros: Ναι, σωστά Αμβρόσιε. Αν παρατήρησες το έγραψα στην αρχή, όπου: $\sqrt{a^{2}}=|a|$

 :roll:


This question asks for each of the following tables of data determine if an exponential model fits the data, where x is the independent variable. Justify your answer.

Table 1
X 1 2 3 4 5
Y 7 -4 8 64

Table 2
X 2 3 4 5 6
Y 3 9 27 81 243

Table 3
X 5 6 7 8 9
Y 18 13 8 3 -2

Table 1
X 1 2 3 4 5
Y 7 -4 8 64

Table 2
X 2 3 4 5 6
Y 3 9 27 81 243

Table 3
X 5 6 7 8 9
Y 18 13 8 3 -2

Looks like Table 2 is the only one with an exponential model data.
$y = 3^{x-1}$

The last one is definitely some sort of arithmetic model, and the first one doesnt even have all the y's needed. I only see 4 responding variable to 5 indepenednet veriables.

Off this goes to Basics.

Also, lillyiggie, if you want anyone to help, show some work so everyone knows they're not just doing your homework for you. :)

Sorry I just dont know how to do this problem to begin with can someone explain how to do this?

Graph each set of data and see if it looks like an exponential graph.

why this topic named 'stupid questions'?


Prove that with $a,b,c,d > 0$ and $n \geq 1$ then:

$\frac{a^n}{b+c+d} + \frac{b^n}{a+c+d} + \frac{c^n}{a+b+d} + \frac{d^n}{a+b+c} \geq \frac{a^{n-1} + b^{n-1} + c^{n-1} + d^{n-1}}{3}$

[hide]
using chebyserv inequaity

$\sum a^n\geq \frac{1}{4}(\sum a^{n-1})(\sum a)$ and

$\sum \frac{a^n}{b+c+d}\geq \frac{1}{4}(\sum a^n)(\sum \frac{1}{a+b+c})$

then AM-HM we have $\sum\frac{1}{a+b+c}\geq \frac{16}{3(\sum a)}$ . Combine all three inequality to get the result
[/hide]


calculus
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linear algebra

By setting the derivative to zero, find the value of $ b_1$ that minimizes

$ \parallel{}b_1 \sin{x} \minus{} \cos{x}\parallel{}^2 \equal{} \displaystyle\int^{2\pi}_0 (b_1 \sin{x} \minus{} \cos{x})^2\,dx$

I took the derivative of the $ (b_1 \sin{x} \minus{} \cos{x})^2$ inside the integral first (with respect to x), and integrated from 0 to $ 2\pi$, but got zero. How can I find this $ b_1$? Can anyone confirm quickly that I did the integral right?

THanks

$ I\equal{}\int_{0}^{2\pi}{(b_1\sin x\minus{}\cos x)^2 dx}$
$ \frac{dI}{db_1}\equal{}\int_{0}^{2\pi}{2\sin x (b_1\sin x\minus{}\cos x) dx}\equal{}2\pi b_1\equal{}0$
So, $ b_1\equal{}0$ minimizes the integral $ I$.
$ I_{min}\equal{}\pi$.

You posted this in the linear algebra forum. Very well, then, let's talk linear algebra.

Suppose we have a (real) inner product space and $ u$ and $ v$ are two nonzero vectors. Our problem is to minimize $ \|b_1u \minus{} v\|^2,$

$ \|b_1u \minus{} v\|^2 \equal{} \langle(b_1u \minus{} v),(b_1v \minus{} u)\rangle \equal{} b_1^2\langle u,u\rangle \minus{} 2b_1\langle u,v\rangle \plus{} \langle v,v\rangle$

$ \equal{} b_1^2\|u\|^2 \minus{} 2b_1\langle u,v\rangle \plus{} \|v\|^2$

We simply complete the square. (We did assume that $ u$ is not the zero vector.)

$ \equal{} \|u\|^2\left(b_1^2 \minus{} 2\,\frac {b_1\langle u,v \rangle}{\|u\|^2} \plus{} \frac {\langle u,v\rangle^2}{\|u\|^4}\right) \plus{} \|v\|^2 \minus{} \frac {(\langle u,v\rangle^2}{\|u\|^2}$

$ \equal{} \|u\|^2\left(b_1 \minus{} \frac {\langle u,v\rangle}{\|u\|^2}\right)^2 \plus{} \frac {\|u\|^2\|v\|^2 \minus{} \langle u,v\rangle^2}{\|u\|^2}$

This algebraic form makes it clear that we minimize this quantity by choosing $ b_1 \equal{} \frac {\langle u,v\rangle}{\|u\|^2}.$

(This is also the proof of the Cauchy-Schwarz inequality, as we note that even this minimum quantity is the the square of a norm and must be nonnegative.)

In this particular case,

$ \langle u,v\rangle \equal{} \int_0^{2\pi}\sin x\cos x\,dx \equal{} 0.$

Hence the quantity is minimized by letting $ b_1 \equal{} 0.$ Had we noted at the the start that $ \sin x$ is orthogonal to $ \cos x,$ we could have made this derivation much shorter.

---

You have to work a little harder and be a little more attentive to detail in a complex inner product space. With the proper interpretation, we can still carry this through.


[b]Problem 1.[/b]

Prove that between any 10 consecutive positive integers($ N$), there can be found one which is relatively prime with all other choosen integers.

[b]Problem 2.[/b]

Number 2003 is written as sum of at least 2 consecutive integers ($ Z$). Determine all possibilities.

[b]Problem 3.[/b] Equation $ 4x^3 \plus{}12x^2 \plus{}9x\equal{}a$ is given, where $ a$ is a real number. How much real solutions does the equation has?

[hide="Problem 2 Solution"][b]Odd number of consecutive integers[/b]: We can write them as $ x \minus{} k, x \minus{} (k\minus{}1), ..., x \plus{} (k\minus{}1), x \plus{} k$, where there are $ 2k\plus{}1$ numbers. These have a sum of $ (2k\plus{}1)x \equal{} 2003$. Notice that $ 2003$ is prime, so $ x \equal{}  1$ and $ k \equal{}  1002$.

The numbers are $ \boxed{\minus{}1000, \minus{}999, ..., 1002}$.

[b]Even number of consecutive integers[/b]: We can write the numbers as $ x \minus{} k, x \minus{} (k \minus{} 1),  ..., x \plus{} (k\minus{}2), x \plus{} (k\minus{}1)$, where there are $ 2k$ numbers. These have a sum of $ 2kx \minus{} k \equal{} k(2x \minus{} 1) \equal{} 2003$. So $ k \equal{}  1$ and $ x \equal{}  1002$, or $ k \equal{}  2003$ and $ x \equal{}  1$

The numbers would be $ \boxed{1001, 1002}$ or $ \boxed{\minus{}2002, \minus{}2001, ..., 2002, 2003}$.[/hide]

[hide="attempt at problem 1"]
Let the sequence be $ o_1,e_1,o_2,e_2,o_3,e_3,o_4,e_4,o_5,e_5.$
where $ o$ represent odd and e represent even numbers. The number we need to select clearly has to be odd.

consider consecutive odd integers:
$ o_1,o_2,o_3,o_4,o_5$

Difference between two consecutive odd multiples of $ 3$ is $ 6$.

So, in $ 5$ consecutive [b]odd[/b] numbers, there are either two numbers which are multiples of 3 or only one(the middle one).
similarly, in $ 5$ consecutive [b]odd[/b] numbers there is only one multiple of $ 5$ and only one multiple of $ 7$.

So, from the preceding conditions, it always possible to choose from 5 consecutive odd integers an odd number which:
1)is not a multiple of $ 3$
2)not a multiple of $ 5$
3)not a multiple of $ 7$

This number will be co-prime with all the other $ 9$ numbers.

similarly for a sequence starting with an even number.[/hide]

[hide="attempt at 3"]consider $ y \equal{} 4x^{3} \plus{} 12x^{2} \plus{} 9x$
It has a maxima at $ \minus{} \frac {3}{2}$ and a minima at $ \minus{} \frac {1}{2}$
It takes a value of  $ 0$ at maxima and $ \minus{} 2$ at minima.

So, $ 4x^{3} \plus{} 12x^{2} \plus{} 9x \plus{} k \equal{} 0$

has one real solution for $ k < 0$ or $ k > 2$,three real distinct solutions for $ 0 < k < 2$, (with one pair of repeating  solutions and the other distinct at  $ 0$ and $ 2$)[/hide]


Help! I don't know where to start on this question:

Show that the product of any sixth root of 1 and any twelfth root of 1 is also a root of 1. What kind of root of 1 is this product?

Let x be one of the sixth roots of 1 and y be one of the twelfth roots of 1.

We have x=cis(60m) and y=cis(30n), where m and n are integers. (Measures are in degrees)

So the product xy=cis(60m)cis(30n)=cis(60m+30n)=cis(30(2m+n))

Since m and n are integers, 2m+n is also an integer, and thus we can rewrite the product as cis(30k), where k is an integer.

Therefore, the product is another twelfth root of 1.


vector
rotation
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hey ppl
here's a problem which has got me really puzzled
here goes
in the planeare given several vectors the sum of whose lengths is 1.show that one can divide them into 3 groups (possibly empty) so  that the sum of  the lengths of the sum of the vectors in each group is atleast [3 \sqrt 3]/2 \pi
please help
best regards

Hello, galois! I'm glad to hear you!

Consider three rays on complex plane: t, e2i\pi /3t, e-2i\pi /3t where t\in R+. These rays divide plane onto 3 angles, and for each angle we consider all vectors which lays in it. So we obtain 3 groups of vectors.
Now we will rotate our rays counterclockwise (groups will be change during rotation). I state that for some angle of rotation the property "the sum of the lengths of the sum of the vectors in each group is atleast \sqrt 27/2\pi" is true.
Indeed, consider projection of vector v to bisector of angle where v lays.  It is easy to see the length of sum of vectors in each group is at least the sum of such projections. The average length of projection of v is |v|3\sqrt 3/2\pi. So average sum of all projections is 2\sqrt 3/2\pi .

I wrote solution 11 month ago. Maybe someone check it?  ;)
And then we can move it to solved section!

Maybe could you give some details about your computations of average projections to make it clearer.

Pierre.

[quote="pbornsztein"]Maybe could you give some details about your computations of average projections to make it clearer.[/quote]
It is just simple integral $\frac{1}{2\pi/3}\int_0^{2\pi/3}\cos t dt$, nothing more.

This problem reminds me of the following result:  Given vectors in the plane whose lengths add up to 1, there is a subset of the vectors whose sum has length at least $1/\pi$.  That result has a proof similar to Myth's.  That result is sometimes useful in complex analysis.

????
There must be something wrong with that. If the vectors add up to 1 it is clear that if we choose all of them the sum will be greater than $\frac 1 {\pi}$....

Pierre.

The vector lengths add up to 1, not the vectors themselves.


function
geometry
Asymptote
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real analysis
real analysis unsolved

function $ f$ is monotonic increasing function. Find a $ x$ that makes $ sum$ $ of$ $ field$ $ a$ and $ field$ $ b$ minimum. I think this is a theorem. Which book i should read to solve this problem and look for which theorem.

look at the attachment to see the field a field b

Just use your common sense. The problem is equivalent to that of maximizing the area of the triangle ABC:
[asy]size(200);
D((0,10)--(0,0)--(8,0),black);
pair A=MP("A",D((1,2))), B=MP("B",D((4,7))), C=MP("C",D((7,8)));
D(MC("f",A..B..C), green);
path P=A--B--C--cycle;
fill(P,yellow);
D(P); D(B);[/asy]
Now the base $ AC$ is fixed, so... By the way, monotonicity is not enough to make the problem meaningful unless you want to talk about oriented areas: you need concavity down to make the regions appear as you drew them. At last, we have Asymptote on this forum for pictures (I didn't bother to reproduce the entire layout from your pdf file but, believe me, it is easy and not time-consuming at all if you invest some time into learning it)

fedja thank you for reply. It is clear that minimizing the he sum of areas of fields means maximizing the area of $ ABC$ but how? And monotonicity is not enough but increasing monotonocity is enough for this problem and convexity and concavity is not matter.

[asy]size(200);
D((0,10)--(0,0)--(8,0),black);
pair A=MP("A",D((1,3))), B=MP("B",D((4,4))), C=MP("C",D((7,8)));
D(MC("f",A..B..C), green);
path P=A--B--C--cycle;
fill(P,yellow);
D(P); D(B);[/asy]

Since the base is fixed, you need to maximize the height, i.e., the distance from $ B$ to $ AC$. You can measure this distance along the perpendicular or along the vertical line, it doesn't matter. If you choose the latter option, you just have to maximize the difference between $ f(x)$ and the linear function with the same values as $ f$ at the endpoints. Now, this is a standard calculus optimization problem that I leave to you.

line $ AC$ is $ \frac{y\minus{}f(c)}{f(c)\minus{}f(a)}\equal{}\frac{x\minus{}c}{c\minus{}a}$ then $ y\equal{}\frac{f(c)\minus{}f(a)}{c\minus{}a}x \minus{} \frac{c}{c\minus{}a}\plus{}f(c)$ for the maximization of the triangle first derivative of this line must be equal to $ f'(b)$ so the solution should be
$ f'(b)\equal{}\frac{f(c)\minus{}f(a)}{c\minus{}a}$
is this solution true and is there any other solution?


Αν η ακολουθία α(ν)>0 για κάθε ν και η σειρά Σα(ν)=(άπειρο) , τότε να αποδείξετε ότι υπάρχει ακολουθία β(ν),με β(ν)>0 για καθε ν  και  β(ν)->0 , τέτοια ώστε η σειρά Σ[α(ν)β(ν)]  να απειρίζεται θετικά.

Αν μπορέσει κάποιος να δώσει μια λύση γιατί την παλεύω εδώ και καιρο και δεν έχω καταφέρει κάτι.....

Μπορούμε να θέσουμε $b_{n}=\frac{1}{\sum_{i=1}^{n}{a_{i}}}$, οπότε η $b_{n}$ τείνει στο $0$, αφού η σειρά $\sum_{i=1}^{n}{a_{i}}$ τείνει στο $+\infty$.
Στη συνέχεια εύκολα βγαίνει ότι η σειρά $\sum_{i=1}^{n}{a_{i}b_{i}}$ τείνει επίσης στο $+\infty$ (...αν θέλει κάποιος μπορώ να το αποδείξω).
Η δυσκολία του προβλήματος είναι πιο πολύ στο να βρούμε την κατάλληλη $b_{n}$.

Μπράβο stedes γιατί δεν μπορούσα να βρω κάτι....
Θα ήθελα να συμπληρώσω ότι το πιο δύσκολο είναι να συνδιάσεις την εύρεση της κατάλληλης ακολουθίας με την εύρεση του κατάλληλου θεωρήματος που θα σε κάνει να μπορείς να το αποδείξεις με τη συγκεκριμένη ακολουθία ...


Παιδιά μια ωραία που βρήκα για Ά Λυκείου ...

Αν $ a>x,  b>y,  c>z,  d>k$,  όπου α,b,c,d,x,y,z,k > 0 , να δ.ο.

$ 8(abcd\plus{}xyzk) > (a\plus{}x)(b\plus{}y)(c\plus{}z)(d\plus{}k)$  ...:)

exw
$ (a\plus{}x)(b\plus{}y)(c\plus{}z)(d\plus{}k)>\equal{}2*riza (ax) *2 riza(by)*2 * riza(cz)*2*riza(dk) \equal{} 16riza(abcdxyzw)$
poy isxyei logw tis anisotitas $ x \plus{} y >\equal{} 2* riza (xy)$..
 ara arkei
$ 8(abcd \plus{} xyzk) >\equal{}16 riza(abcdxyzw)$ <=> $ abcd \plus{} xyzk > 2 riza(abcdxyzw)$,poy isxyei logw tis idias basikhs anisotitas $ x \plus{} y >\equal{} 2* riza (xy)$.
P.S. mipws mporeis na m peis kapoios ti pataw gia tn riza stin latex? eyxaristw  :)

gia thn riza sth latex patas \sqrt . Για περισσότερες πληροφορίες πατα στο μεγαλο εικονιδιο που γραφει LATEX HELP

Κατ' αρχάς η ρίζα γράφεται \sqrt{x}, ($ \sqrt{x}$). Ειδικότερα αν θέλεις ν-οστή ρίζα γράφεις \sqrt[n]{x}.

Kαι τώρα κάτι σημεντικότερο... κάνεις ένα πολύ βασικό λάθος στη λύση σου :!:
Θες να αποδείξεις ότι $ a>b$ και κάνεις το εξής. Αποδεικνύεις ότι $ b>c$ και ότι $ a>c$ κι από αυτά βγάζεις το δεδομένο $ a>b$. Aυτό είναι ένα συχνό λάθος...μπορείς να πάρεις κι ένα παράδειγμα για να δεις ότι αυτή η λογική είναι λάθος.

exeis dikio kai sta dio post sou.signwmi gia ta lathi...

par'ola ayta omws vrika mia alli lysi elpizw swsti an k oxi toso wraia :blush:
apo ta dedomena exw
$ a>x, b>y, c>z ,d>k$
ara kai $ ab> xy ,ac>xz, cd>kz, bd>yk, ad>xk, bc>yz$
alla kai $ bcd>yzk, abc > xyz, abd>xyk, acd>xzk$
anaptysontas to deytero melos tis anisotitas k fernontas to sto prwto melos me paragontopoiisi prokyptei:
$ (bcd \minus{} yzk)(a\minus{}x) \plus{} (abc \minus{} xyz)(d\minus{}k) \plus{} (abd \minus{}xyk)(c\minus{}z) \plus{} (acd\minus{}xzk)(c\minus{}z) \plus{} (acd\minus{}xzk)(b\minus{}y) \plus{}(ab \minus{} xy)(cd \minus{} kz) \plus{} (ac \minus{} xz)(bd \minus{} yk) > 0$ , to opoio isxyei ws athroisma thetikwn


This should be a very interesting question. Which school of thought in economics do you belong to and why?

Even leg, who has a strong knowledge of econ says it is difficult to decide who is right and who is wrong, sorry to put you on the spot leg.

It seems to me that free market economics is morally making more sense then anything else but I dont know. Like imposing taxes on people that dont want to pay, especially pay more then others may not seem moral.

Well thanks :blush:  I think there's a lot of well-thought out ideas
that come from opposing schools of thought. Theories of consumption seem perfectly plausible, and then you realise they're completely contrasting. Does the interest rate have an effect on consumption? Consumption proportional to income (Keynes) or consumption smoothed out over life (Modigliani, Friedman). It is contrasting ideas like these (both logical) that have inspired so many jokes about economists and have lead many untrained in economics to believe that they too can speak sensibly about the subjects (who needs theory if its contrasting).

People choose their economic stance to fit their political preference. I'm glad i don't care much for politics.
There's a lot to be said for free-marketism but there are also many market failures where regulation and taxes are warranted. eg Public goods, pollution and other externalities.

On the macro side, its less clear. We can explain a lot of things, but predict very little.
Nice to see some interest here

Cook, P. Lesley

There are many theories because each is based on different assumptions about the world; it is their relevance rather than their logic which is in dispute.

Economics is not a pure science. It relies more on arguments than proofs. If u think about free market and controlled market u wouldn't be able to decide which one is the best. Because u have not yet seen a pure free market or pure controlled market. I personally dont believe in controlled market, but i dont think that free market is the only option left. Both these two are extremely vulnerable. It is impossible for controlled economy to survive and if there were pure free market(Which actually does not exist) it would be impossible for people to survive.

 :?:

Why would it be impossible to survive free markets?

I believe it would be possible to survive in pure free markets, but there could be some people that don't survive.