German teen solves 300-year-old mathematical riddle posed by Sir Isaac Newton - Chicagoland Sportbikes
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post #1 of 18 (permalink) Old 05-28-2012, 01:05 AM Thread Starter
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German teen solves 300-year-old mathematical riddle posed by Sir Isaac Newton

http://www.foxnews.com/world/2012/05...-isaac-newton/


DRESDEN, Germany A German 16-year-old has become the first person to solve a mathematical problem posed by Sir Isaac Newton more than 300 years ago.
Shouryya Ray worked out how to calculate exactly the path of a projectile under gravity and subject to air resistance, The (London) Sunday Times reported.
The Indian-born teen said he solved the problem that had stumped mathematicians for centuries while working on a school project.
Ray won a research award for his efforts and has been labeled a genius by the German media, but he put it down to "curiosity and schoolboy naivety."
"When it was explained to us that the problems had no solutions, I thought to myself, 'well, there's no harm in trying,'" he said.
Ray's family moved to Germany when he was 12 after his engineer father got a job at a technical college. He said his father instilled in him a "hunger for mathematics" and taught him calculus at the age of six.
Ray's father, Subhashis, said his son's mathematical prowess quickly outstripped his own considerable knowledge.
"He never discussed his project with me before it was finished and the mathematics he used are far beyond my reach," he said.
Despite not speaking a word of German when he arrived, Ray will this week sit Germany's high school leaving exams, two years ahead of his peers.
Newton posed the problem, relating to the movement of projectiles through the air, in the 17th century. Mathematicians had only been able to offer partial solutions until now.
If that wasn't enough of an achievement, Ray has also solved a second problem, dealing with the collision of a body with a wall, that was posed in the 19th century.
Both problems Ray resolved are from the field of dynamics and his solutions are expected to contribute to greater precision in areas such as ballistics.


Read more: http://www.foxnews.com/world/2012/05...#ixzz1w8qWFR5A

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post #2 of 18 (permalink) Old 05-28-2012, 01:13 AM
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Nein! He's Indian, not German.

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post #3 of 18 (permalink) Old 05-28-2012, 09:11 AM Thread Starter
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Yeah, but I'm not one to correct someone's article. My own are bad enough.

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post #4 of 18 (permalink) Old 05-28-2012, 09:41 AM
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post #5 of 18 (permalink) Old 05-28-2012, 09:57 AM
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Lol

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post #6 of 18 (permalink) Old 05-28-2012, 11:08 AM
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Someone should change the title of this thread to "40 year old virgin".

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post #7 of 18 (permalink) Old 05-28-2012, 11:20 PM
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As long as it makes the next version on modern warfare that much more realistic, I'm all for it.
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post #8 of 18 (permalink) Old 05-29-2012, 02:39 AM
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+1 to that lol

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post #9 of 18 (permalink) Old 05-29-2012, 07:29 AM
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For the nerds out there that were actually curious about the problem he solved, here's the solution:

Quote:
The problem he solved is as follows:
Let (x(t),y(t)) be the position of a particle at time t. Let g be the acceleration due to gravity and c the constant of friction. Solve the differential equation:
(x''(t)2 + (y''(t)+g)2 )1/2 = c*(x'(t)2 + y'(t)2 )
subject to the constraint that (x''(t),y''(t)+g) is always opposite in direction to (x'(t),y'(t)).

Finding the general solution to this differential equation will find the general solution for the path of a particle which has drag proportional to the square of the velocity (and opposite in direction). Here's an explanation how this differential equation encodes the motion of such a particle:

The square of the velocity is:
x'(t)2 + y'(t)2

The total acceleraton is:
( x''(t)2 + y''(t)2 )1/2

The acceleration due to gravity is g in the negative y direction.
Thus the drag (acceleration due only to friction) is:
( x''(t)2 + (y''(t)+g)2 )1/2

Thus path of such a particle satisfies the differential equation:
( x''(t)2 + (y''(t)+g)2 )1/2 = c*(x'(t)2 + y'(t)2 )

Of course, we also require the direction of the drag (x''(t),y''(t)+g) to be opposite to the direction of the velocity (x'(t),y'(t)). Once we find the intial position and velocity of the particle, uniqueness theorems tell us its path is uniquely determined.

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post #10 of 18 (permalink) Old 05-29-2012, 07:40 AM
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This one is easier to understand, unless you have a strong understanding of differential equations *shutters*

Quote:
Consider a projectile moving in gravity with quadratic air resistance. The governing equations are
u' = -a * u * sqrt( u2 + v2 )
v' = -a * v * sqrt( u2 + v2 ) - g

where a is the coefficient of air resistance defined by |F| = ma|v|2 .
Cross-multiply and rearrange to find
a * sqrt( u2 + v2 ) * (uv'-vu') = gu'

Substitute v = su and separate variables:
a * sqrt( 1 + s2 ) * s' = g*u'/u3

Integrate both sides to get the answer:
g/u2 + a(v * sqrt( u2 + v2 )/u2 + arcsinh|v/u|) = const

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post #11 of 18 (permalink) Old 05-29-2012, 07:54 AM
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could you translate this in German?

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post #12 of 18 (permalink) Old 05-29-2012, 07:57 AM
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Sure.

Quote:
Betrachten wir ein Projektil bewegt sich in der Schwerkraft mit quadratischem Luftwiderstand. Die Grundgleichungen sind
u '=-a * u * sqrt (u2 + v2)
v '=-a * v * sqrt (u2 + v2) - g

wobei a die Koeffizienten der Luftwiderstand durch | F | = ma | V | 2.
Kreuz-Multiplikations-und neu anordnen, um zu finden
a * sqrt (u2 + v2) * (uv'-vu ') = gu'

Ersetzen Sie v = su und separate Variablen:
a * sqrt (1 + s2) * s '= g * u' / U3

Integrieren Sie beide Seiten auf die Antwort zu erhalten:
g/u2 + a (v * sqrt (u2 + v2) / U2 + arcsinh | v / u |) = const

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post #13 of 18 (permalink) Old 05-29-2012, 08:02 AM
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post #14 of 18 (permalink) Old 05-29-2012, 09:14 AM
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post #15 of 18 (permalink) Old 05-29-2012, 09:55 AM
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I knew all that but no one asked.
You cheated, you went to school with Newton.

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post #16 of 18 (permalink) Old 05-29-2012, 10:40 AM
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You cheated, you went to school with Newton.
He courted Newton's mom. Get your facts straight.

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post #17 of 18 (permalink) Old 05-29-2012, 10:46 AM
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Elder abuse. You should be ashamed.

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post #18 of 18 (permalink) Old 05-29-2012, 11:06 AM
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I knew all this already, but I didn't make a big deal about it. Just another attention whore.
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