May 30th, 2012 at 9:18:15 PM
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Quote:andysifi have been searching the whole internet with no answer.

what question exactly did he solve?

Why 5 pounds of cement won't fit is a 3 pound

bucket has always baffled me. It probably

something harder than that, though..

"It's not enough to succeed, your friends must fail."
Gore Vidal

May 30th, 2012 at 9:23:11 PM
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Ray solved two fundamental particle dynamics theories

which physicists have previously been able to calculate only by using powerful computers.

The first problem is: “Particle-wall collision under Hertzian collision force and linear Damping.”

And the second problem is: “Trajectory of an oblique body thrown in the earth’s gravitational field

and Newton’s law of universal gravitation to calculate Flow resistance.":

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 [the preceding should probably read

"the impedance (acceleration due to friction plus gravity) is"]:

\bigg( x''(t)2 + (y''(t)+g)2 \bigg)(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."

"Here's a forward solution (found by reverse-engineering the answer):

Consider a projectile moving in gravity with quadratic air resistance.

The governing equations are

u' = -a u \sqrt{ u^2 + v^2 }

v' = -a v \sqrt{ u^2 + v^2 } - g

where a is the coefficient of air resistance defined by |F| = ma|v|^2 .

Cross-multiply and rearrange to find

a \sqrt{ u^2 + v^2 } (uv'-vu') = gu'

Substitute v = su and separate variables:

a \sqrt{ 1 + s^2} s' = g\frac {u'}{u^3}

Integrate both sides to get the answer:

\frac g {u^2} + a \left(\frac{v \sqrt{ u^2 + v^2 }}{u^2} + \sinh^{-1}\left|\frac v u\right| \right)= C"

which physicists have previously been able to calculate only by using powerful computers.

The first problem is: “Particle-wall collision under Hertzian collision force and linear Damping.”

And the second problem is: “Trajectory of an oblique body thrown in the earth’s gravitational field

and Newton’s law of universal gravitation to calculate Flow resistance.":

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 [the preceding should probably read

"the impedance (acceleration due to friction plus gravity) is"]:

\bigg( x''(t)2 + (y''(t)+g)2 \bigg)(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."

"Here's a forward solution (found by reverse-engineering the answer):

Consider a projectile moving in gravity with quadratic air resistance.

The governing equations are

u' = -a u \sqrt{ u^2 + v^2 }

v' = -a v \sqrt{ u^2 + v^2 } - g

where a is the coefficient of air resistance defined by |F| = ma|v|^2 .

Cross-multiply and rearrange to find

a \sqrt{ u^2 + v^2 } (uv'-vu') = gu'

Substitute v = su and separate variables:

a \sqrt{ 1 + s^2} s' = g\frac {u'}{u^3}

Integrate both sides to get the answer:

\frac g {u^2} + a \left(\frac{v \sqrt{ u^2 + v^2 }}{u^2} + \sinh^{-1}\left|\frac v u\right| \right)= C"

In a bet, there is a fool and a thief.
- Proverb.

May 30th, 2012 at 10:21:48 PM
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thanks. must be a genius to solve this when it takes this long to just write the question.

May 30th, 2012 at 10:26:21 PM
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1. Find German teen.

2. Teach him to count cards at blackjack.

3. ???

4. Profit!!!

2. Teach him to count cards at blackjack.

3. ???

4. Profit!!!

May 31st, 2012 at 2:04:08 PM
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Quote:konceptum1. Find German teen.

2. Teach him to count cards at blackjack.

3. ???

4. Profit!!!

This is stupid. The mathematical part of card counting is easy. It's everything else that is hard, and I'm guessing the kid's social skills aren't in the same realm as his math skills.

"So drink gamble eat f***, because one day you will be dust." -ontariodealer

May 31st, 2012 at 3:02:08 PM
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Quote:WongBoSort of sad that it will improve the field of ballistics...

Not sad at all. Doping the wind will still be art.

May 31st, 2012 at 3:08:00 PM
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Quote:AcesAndEightsThis is stupid. The mathematical part of card counting is easy.

The movie Rainman was stupid, as are most BJ movies.

Totally out of whack with reality. The latest is on the

TV show Breaking Bad. The main guy makes millions

by cooking meth, and his wife explains the money as

winnings from card counting BJ. Everybody buys it,

even the audience. Nobody on the shows discussion

boards even raised an eyebrow, movies have convinced

them wealth is easily attainable this way.

"It's not enough to succeed, your friends must fail."
Gore Vidal

May 31st, 2012 at 3:44:55 PM
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Quote:rxwineQuote: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.

I'm curious about how long it will take someone to find his error.

May 31st, 2012 at 4:37:20 PM
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Quote:AcesAndEightsThis is stupid. The mathematical part of card counting is easy. It's everything else that is hard, and I'm guessing the kid's social skills aren't in the same realm as his math skills.

That's why the number 3 is in question marks. Figure out the hard part, and the profit is easy to come by.

But, ultimately, I figured more people would recognize the meme. I was wrong.

May 31st, 2012 at 4:52:34 PM
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There's a Nobel Prize in Mathematics, right? Cause this kid should get it. Or the Fields Medal. And when he gets his Ph.D. in two years he'll probably get an awesome job in whatever national defense program Germany has.