Member: Apfelkuchen

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+104

ax²+bx+c = f(x)

f(0) = a0² + b0 + c = 10 => c = 10
f(1) = a1² + b1 + c = = a + b + 10 = 10.65 => a = 10.65-10-b = 0.65-b
f(2) = a2² + b2 + c = 4a + 2b + 10 = 11.34225
f(3) = a3² + b3 + c = 9a + 3b + 10 = 12.079496

Insert a into f(2):
4(0.65-b) + 2b + 10 = 11.34225 = 2.6 - 4b + 2b + 10 = 2.6 - 2b + 10
=> 2b = 12.6 - 11.34225 = 1.25775
=> b = 1.25775/2 = 0.628875

Insert b into f(3):
9a + 3(0.628875) + 10 = 9a + 1.886625 + 10 = 12.079496
=> 9a = 12.079496 - 10 - 1.886625 = 0.192871
=> a = 0.192871/9 = 0.0214301111

f(x) = 0.02x^2 + 0.63x + 10

ApfelkuchenAug 31, 2013

+104

abs(6x+8) = |6x+8| has to be 50.
Positive case: Replace || with ()
(6x+8) = 50 [-8]
6x = 42 [/6]
x = 7

Negative case: Replace || with -()
-(6x+8) = -6x - 8 = 50 [+8]
-6x = 58 [/-6]
x = -58/6 = -29/3

ApfelkuchenAug 31, 2013

+104

a/a = 1
=> (a/a+1) = 2
=> (1/a + 1)*(a/a + 1) = (1/a + 1)*2 = 2/a + 2

ApfelkuchenAug 31, 2013

+104

Inversely proportional means: F ~ 1/x^2
x is the distance in meters (m). F is the force in Newton (N).
Futhermore F = m*a (1N = kg*m/s^2)

1/m² is not equal to kg*m/s², so we have to add a const. value named n (as desired) with the unit kg*m^3/s^2.
Remember 1T = 1kg/(As^2)
=> kg*m^3/s^2 = T*m^3*A

So we have an equation like: F = n/x^2 with n = B*I*x^3 => F = B*I*x (have a look at the magnetic flux density)
In your case n would be 36Tm^3A

ApfelkuchenAug 31, 2013

+104

i=1: (3-6)*2 = -6
i=2: (8-6)*2 = 4
i=3: (1-6)*2 = -10
i=4: (5-6)*2 = -2
i=5: (4-6)*2 = -4
i=6: (3-6)*2 = -6

=> -6 + 4 -10 -2 -4 -6 = -24

ApfelkuchenAug 31, 2013

+104

You know, this is some kind of discrete optimization problem, because we know neither a, b or c. If we neglect the given characteristics of each parameter, there would be infinite possibilities.
e. g.
a + b + c = 10 with a = 3
So we know b + c = 10-3 = 7
This equation is satisfied with:
b=1,c=6: 1+6=7
b=6,c=1: 6+1=7
b=2,c=5: 2+5=7
and just to exaggerate (not discrete)
b=0.9,c=6.1: 0.9+6.1=7
b=0.99,c=6.01: 0.99+6.01=7
etc.
So using some methods like intelligent water drops or particle swarm optimization we could expect good results.

Here we have a quite large range of possibilities. Your only option is to insert and increment the numbers systematically.
At least you could try to simplify the problem.

If we set a to 1 (first possible value) we'll get:
b^2 + c^2 = d^2 - 1 with d = 427
=> b^2 + c^2 = 427^2 - 1 = 182329 - 1 = 182328
Futhermore:
Max. value of b^2 and c^2: 575^2 = 330625
Min. value of b^2 and c^2: 24^2 = 576
Watch out: 330625 > 182328. So the real max. value of b^2 and c^2 would be 182328-576 = 181752
-> You have to subtract 576 (=24^2) because b or c has to be at least 24.
So max. value out of the range for b and c: sqrt(181752) = 426 (discrete!)
With 426^2 = 181476 < 181752, the min value of b and c has to be 25.

Even though we don't have to go through 24 to 575 anymore, there is no guaranteed solution.

Well, at least the possible values for you:
a=6; b=58; c=423
a=6; b=87; c=418
a=6; b=418; c=87
a=6; b=423; c=58
a=18; b=103; c=414
a=18; b=166; c=393
a=18; b=274; c=327
a=18; b=327; c=274
a=18; b=393; c=166
a=18; b=414; c=103
a=21; b=168; c=392
a=21; b=392; c=168
a=22; b=54; c=423
a=22; b=297; c=306
a=22; b=306; c=297
a=22; b=423; c=54
a=23; b=102; c=414
a=23; b=270; c=330
a=23; b=330; c=270
a=23; b=414; c=102

ApfelkuchenAug 30, 2013

+104

Physics solution.png

ApfelkuchenAug 30, 2013

+104

Heat capacity of water: c = 4,182 kJ/(kg*K)
Heat of fusion of ice: q = 334 kJ/kg

ice water:
Q1 = m*q = 4,0 kg * 334 kJ/kg (= 1336 kJ) = 1,3 MJ

0 °C 10 °C (Diff. 10 K)
Q2 = c*m*T = 4,182 kJ/(kg*K) * 4,0 kg * 10 K (= 167,28 kJ) = 0,17 MJ

Q = Q1+Q2 = 1,3 MJ + 0,17 MJ (= 1,47 MJ) = 1,5 MJ

ApfelkuchenAug 26, 2013

+104

200 km of an orbit are mostly a straight line, therefore it's extraneous.

Station A: (0, 0)
Station B: (200, 0)

Build two linear equations (rem. f(x) = mx + t)
m = dy/dx = tan(n)

=> First lin. equation of Station A: f(x) = tan(23°)*x
Second lin. equation of Station B: g(x) = tan(55°)*(x-200)

Solve f = g for x
tan(23°)*x = tan(55°)*(x-200)
=> tan(23°)/tan(55°) = (x-200)/x = 1-200/x
=> x = 200/[1-tan(23°)/tan(55°)]

Insert x into f(x):
f(200/[1-tan(23°)/tan(55°)]) = tan(23°)*200/[1-tan(23°)/tan(55°)] = 120.8 km

This would result into:

Another solution would occur, if you set the angle of g(x) to 180-55°. Station B would face the opposite direction.
=> f(200/[1-tan(23°)/tan(125°)]) = tan(23°)*200/[1-tan(23°)/tan(125°)] = 65.44 km

Which would lead to:

BUT a satellite in about 65 km height is unrealistic, so it has to be the first one.

ApfelkuchenAug 25, 2013

#12

+104

Zahlen jenseits 64 bit sind in vielen Programmiersprachen nicht Standard und nur durch zusätzliche Bibliotheken möglich.

In diesem Fall stimmts aber schon, dass der Rechenaufwand sehr gering ist. Wenn die Implementation intelligent genug ist, so würde nur (2.4)^(2.4*10^6) berechnet werden. Mit dem diskreten Logarithmus folgt ja bereits, dass (10^6)^(2.4*10^6) = 10^(1.44*10^7) aus 14400000 Nullen besteht. Die eigentliche Zahl hat damit 912507 Stellen, die in knapp 30 Iterationen berechnet werden kann.

Ohne mir die Datentypen in Haskell anzuschauen, würde mich doch interessieren, mit welcher Blockgröße gerechnet wurde.

ApfelkuchenAug 25, 2013

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