Consortium of Indian Defence Websites
https://forums.bharat-rakshak.com/
(There is a lot of interesting discussion following those pages)AmberG. wrote:There is $100,000 prize, not to mention fame, (See:http://www.math.unt.edu/~mauldin/beal.html to either find a counter example (or prove it is impossible) for an equation of the kind
x^a+y^b = z^c (a,b,c >2 and certain restrictions on x,y,z - see the above link.
So, Vinaji (
For those who want to try it..the challenge is sponsored by the American Mathematical Society; the rules, including where to submit proposed solutions by email or snail mail, are available at:Want to make a quick million? All you have to do is figure out a little math problem that goes like this: A^x + B^y = C^z. Simple algebra, right?
Oh how deceptively innocuous a few elementary variables can seem. You’re actually looking at something inspired by one of the great mysteries of mathematics, known as Fermat’s Last Theorem and named after the 17th century French lawyer and mathematician Pierre de Fermat. Fermat came up with his own theorem back in 1637, scribbling it in the margins of his copy of the Greek text Arithmetica by Diophantus and surmising that — put your math caps on and buckle up — if n were an integer greater than 2, then the equation Xn + Yn = Zn has no positive integral solutions. The note was discovered after Fermat’s death, and it took over 350 years and untold failed attempts by others for someone to prove the theorem. In 1995, British mathematician Andrew Wiles, who’d been fascinated with the theorem since he was a child, finally got the job done, having puzzled over it in secret for roughly six years.
That’s where Texas billionaire D. Andrew Beal comes in. In 1993, he posited a closely related number theory problem hence dubbed Beal’s Conjecture (that first A-B-C equation above), where the only solution is possible when A, B and C have a common numerical factor and the exponents x, y and z are greater than 2. Beal’s been trying to solve his theorem ever since, reports ABC News, offering cash rewards in steadily increasing amounts — $5,000 in 1997, $100,000 in 2000 – to anyone with the knack to get the job done.
The prize total in 2013: $1 million, which is either a sign of Beal’s magnanimity or his skepticism that it’s actually possible. (Since Beal is worth a reported $8 billion, there’s little need to worry about whether he’ll pay the winner.)
It’s apparently not just about the money for Beal, either: In a statement, he said “I’d like to inspire young people to pursue math and science. Increasing the prize is a good way to draw attention to mathematics generally … I hope many more young people will find themselves drawn into the wonderful world of mathematics.”
)
Chapter 8. Crossing the Atlantic
... Kempe's solution had been shown to be faulty ...
The feeling was also beginning to emerge in some quarters that a solution to the four-colour problem had not been forthcoming because no really good mathematician had worked on it. Indeed a story is told about the distinguished German number-theorist Hermann Minkowski in the first decade of the twentieth century. While lecturing on topology at Gottingen University, he mentioned the four-color problem:
'This theorem has not yet been proved, but that is because only mathematicians of the third rank have occupied themselves with it', Minkowski announced to the class in a rare burst of arrogance. 'I believe I can prove it.'
He began to work out his demonstration on the spot. By the end of the hour he had not finished. The project was carried over to the next meeting of the class. Several weeks passed this way. Finally one rainy morning, Minkowski entered the lecture hall, followed by a crash of thunder. At the rostrum, he turned towards the class with a deeply serious expression on his face.
'Heaven is angered by my arrogance', he announced. 'My proof of the FCT is also defective.' He then took up the lecture on topology at the point where he had dropped it several weeks before.
SaiIK, I am splitting the asnwer into two parts.SaiK wrote:sorry matrimc. the answer I get is yes, but I was wanting to understand more as to why it is so. [1]
regarding markov, question: regarding the state transition to which probability is attributed, the probability would be always one or probability given the condition to trigger, if the event is known ahead right? sorry for asking without reading much about it. [2]
Thanks! Interesting read. (I had a somewhat similar experience with FCT, I may share that story in other post)matrimc wrote:AmberG ji, thanks for cautioning about the $1M prize.
Following is a quote from a popular book on Four Color Theorem (FCT) - worth pondering over by people before proceeding to solve the problem on the spot.
Four Colors Suffice: How the Map Problem Was Solved [Paperback]
Robin Wilson (Author)
"""Print counterexamples to Beal's conjecture.
That is, find positive integers x,m,y,n,z,r such that:
x^m + y^n = z^r and m,n,r > 2 and x,y,z co-prime (pairwise no common factor).
ALGORITHM: Initialize the variables table, pow, bases, powers such that:
pow[z][r] = z**r
table.get(sum) = r if there is a z such that z**r = sum.
bases = [1, 2, ... max_base]
powers = [3, 4, ... max_power]
Then enumerate x,y,m,n, and do table.get(pow[x][m]+pow[y][n]). If we get
something back, report it as a result. We consider all values of x,y,z
in bases, and all values of m,n,r in powers.
"""
def beal(max_base, max_power):
bases, powers, table, pow = initial_data(max_base, max_power)
for x in bases:
powx = pow[x]
if x % 1000 == 0: print 'x =', x
for y in bases:
if y > x or gcd(x,y) > 1: continue
powy = pow[y]
for m in powers:
xm = powx[m]
for n in powers:
sum = xm + powy[n]
r = table.get(sum)
if r: report(x, m, y, n, nth_root(sum, r), r)
def initial_data(max_base, max_power):
bases = range(1, max_base+1)
powers = range(3, max_power+1)
table = {}
pow = [None] * (max_base+1)
for z in bases:
pow[z] = [None] * (max_power+1)
for r in powers:
zr = long(z) ** r
pow[z][r] = zr
table[zr] = r
print 'initialized %d table elements' % len(table)
return bases, powers, table, pow
def report(x, m, y, n, z, r):
x, y, z = map(long, (x, y, z))
assert min(x, y, z) > 0 and min(m, n, r) > 2
assert x ** m + y ** n == z ** r
print '%d ^ %d + %d ^ %d = %d ^ %d = %s' % \
( x, m, y, n, z, r, z**r)
if gcd(x,y) == gcd(x,z) == gcd(y, z) == 1:
raise 'a ruckus: SOLUTION!!'
def gcd(x, y):
while x:
x, y = y % x, x
return y
def nth_root(base, n):
return long(round(base ** (1.0/n)))
import time
def timer(b, p):
start = time.clock()
beal(b, p)
secs = time.clock() - start
return {'secs': secs, 'mins': secs/60, 'hrs': secs/60/60}
Oh, forgot to add the link to the article at ElesieverAlan Turing was also a highly accomplished runner, competing with the Walton Athletic Club in Surrey, UK. According to a blog post by Lubor Ptacek, he finished fourth in the Amateur Athletic Association Championships marathon with a time of 2 hours 46 minutes and 3 seconds – even faster than Lance Armstrong’s time of 2:46:42 in the 2007 New York Marathon. Here, Turing is running a race at the National Physical Laboratory Sports Day in December 1946. (Photo courtesy of the National Physical Laboratory)
You might want to post python code with theAmber G. wrote: Let me add a story abut a 6 year old Isabella who said something to the effect : "I'm good at math. Can I do that and win the prize? Then I can buy that toy I wanted."..I think she (probably with the help of her father) has posted a nice computer program to look for counter examples for the Beal's problem.
Here is the complete python program..They said that anyone can use it, small request, if you indeed win the prize please share some money so that Isabella can buy her toy....snip...
Code: Select all
and [[i]/[/i]code] tags instead of [quote] tags. For one thing, the quote tag destroys indentation, which is very vital to python code.
Her dad (who wrote all the code) is quite well paid by Google and is a well known author in the field of AI research. He can easily afford to buy Isabella any toy she wants :).Narad said to Bhaskara:" A number is 'pavitra' if
1. it is prime. (Let us call it p)
2. If you take half and round it up it is a perfect square. (IOW (p+1)/2 = a^2)
3. If you take half of its square and round it up, it is still a perfect square. (IOW, (p^2+1)/2 = b^2)
O Bhaskara! can you find me pavitra number(s)? ( all, or any, or as many as you can)
Solution - As said, it is simpleAmber G. wrote:Just for fun, if some one is interested, here is some what harder (or may be very simple
You have a 6x6 board (instead of 8x8) and you want to cover it entirely. (with 18 dominoes (of the 2 x 1 size) as described above)
Prove that you can always find a straight line which divides the board into two (not necessarily equal) parts, such that no domino is cut by that line.
)Just saw this. I know that there are multiple solutions (hint: counter example was given by a mathematician with the initials N.E. and he proposed a technique to find infinite solutions to the above equation) and also that the original equation was proposed by Euler. The mathematician N.E. was briefly mentioned in Simon Singh's book, Fermat's Enigma: The Epic Quest to Solve the World's Greatest Mathematical Problem. It was during the time when Andrew Wiles was trying to solve Fermat's last equation and had hit some significant difficulties in the proof and was wondering if he was on the right path or was his proof hitting difficulties because Fermat was wrong after all. Apparently, someone in the math community circulated an e-mail on April Fool's day that N.E. had found a counter-example to Fermat's last theorem (N.E. had become famous when he found the counter example to the above Euler equation a few months earlier) and the news nearly gave Andrew Wiles a heart attack before he realized that it was an April Fool's day joke.Amber G. wrote:Meanwhile, if some one likes to try their hand on computer program, try to find (counter) example for
x^4+y^4+z^4=w^4
(Hint: example exists)
(Many thought, that there are no solution for the above kind of eq for n>3)
Any square board (2nx2n) greater than 6, can be tiled to have non-intersecting cut. No solution for 6x6 or smaller board.matrimc wrote:Excellent.
I have another question to think about.
Does this property hold only for a 6x6 board? So the question would be for what other values of n other than 3 does a 2n x 2n board does admit a domino non-intersecting cut?
No, I do not have a solution may thinking on the lines of either n is a prime or some multiple of 3.
Please post the answers... when you get time..ArmenT wrote:Just saw this. I know that there are multiple solutions (hint: counter example was given by a mathematician with the initials N.E. and he proposed a technique to find infinite solutions to the above equation) and also that the original equation was proposed by Euler. The mathematician N.E. was briefly mentioned in Simon Singh's book, Fermat's Enigma: The Epic Quest to Solve the World's Greatest Mathematical Problem. It was during the time when Andrew Wiles was trying to solve Fermat's last equation and had hit some significant difficulties in the proof and was wondering if he was on the right path or was his proof hitting difficulties because Fermat was wrong after all. Apparently, someone in the math community circulated an e-mail on April Fool's day that N.E. had found a counter-example to Fermat's last theorem (N.E. had become famous when he found the counter example to the above Euler equation a few months earlier) and the news nearly gave Andrew Wiles a heart attack before he realized that it was an April Fool's day joke.Amber G. wrote:Meanwhile, if some one likes to try their hand on computer program, try to find (counter) example for
x^4+y^4+z^4=w^4
(Hint: example exists)
(Many thought, that there are no solution for the above kind of eq for n>3)
I'll post the answer if no one else answers this in a while.
AmberG ji, yes. As soon as I posted (on an iPodAmber G. wrote:...
), I left for work and on the way reached the same conclusion as above. We need n > 0, integer and 4n -2 > n^2, which is true for n = 1, 2, or 3 Just to add to above, yes N.E. (Noam Elkies)'s counter example to Euler equation is quite note worthy but Noam was quite famous before that. At the age of 14 (!!!) he got a gold medal (and a perfect score! (youngest to ever have done so!) in International Math Olympiad. He was also a Putnam fellow (really very big deal) at age of 16 (again perhaps youngest) and PhD at the age of 20 from Harvard. (according to wiki) and youngest full professor later, in the history of Harvard!ArmenT wrote: Just saw this. I know that there are multiple solutions (hint: counter example was given by a mathematician with the initials N.E. and he proposed a technique to find infinite solutions to the above equation) and also that the original equation was proposed by Euler. The mathematician N.E. was briefly mentioned in Simon Singh's book, Fermat's Enigma: The Epic Quest to Solve the World's Greatest Mathematical Problem. It was during the time when Andrew Wiles was trying to solve Fermat's last equation and had hit some significant difficulties in the proof and was wondering if he was on the right path or was his proof hitting difficulties because Fermat was wrong after all. Apparently, someone in the math community circulated an e-mail on April Fool's day that N.E. had found a counter-example to Fermat's last theorem (N.E. had become famous when he found the counter example to the above Euler equation a few months earlier) and the news nearly gave Andrew Wiles a heart attack before he realized that it was an April Fool's day joke.
Answer: Color the board with colors Row 1 = (1,2,1,2,1,2..) Row 2 = (3,4,3,4,3,4,..), All odd rows same as row 1, all even rows - same as row 2.. (see figure below)..Amber G. wrote: Armen T - Thanks for mentioning Singh's book, I have not read it, will do when I get a chance.
Invariant concept is very powerful in Physics too.. Many difficult problems, (including everyday problems) can be solved this way....The trick is to find the right invariant property..
For example, if our dominoes are 4x1 (instead of 2x1) can one cover a 6x10 board with them? (The board is to be fully covered)
Indians did publish treatises on negative numbers, however I don't think anyone dealt with the concept of complex numbers. The Bakshali manuscript from 400 AD shows -ve numbers and Brahmagupta also wrote about it in 620 AD and also established the rules of addition/subtraction/multiplication and division for +ve and -ve number combinations. On the question of square roots, he touches on it briefly and IIRC says that the square root of -ve numbers is unanswered.Nilesh Oak wrote:All Math Gurujan,
Need your help....
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I was reading about complex numbers (in a context of my research- Astronomy) and found that on Wikipedia it lists Gerolamo Cardano as its inventor.
http://en.wikipedia.org/wiki/Gerolamo_Cardano
The Wikipedia entry o Gerolamo Cardano says that complex numbers came into the picture (need for them) in solving certain trigonometric functions/solutions etc.
This entry (on Gerolamo Cardano) as it related to complex numbers.
------------
In his exposition, he acknowledged the existence of what are now called imaginary numbers, although he did not understand their properties (described for the first time by his Italian contemporary Rafael Bombelli, although mathematical field theory was developed centuries later).
--
AmberJi, Matrimc and others..
Do we any evidence for existence, usage or inkling of complex numbers in ancient Indian mathematics/astronomy/trigonometry..etc., anytime before say 1500 CE ?
Appreciate your help,
Some may be interested in:matrimc wrote: ....
Another student of Prof. Leighton (I don't remember the name) was the founder. He was in one of the planes that hit WTC. Looking at his thesis might give some details.
