## Min bok om relativitetsteori

07 november 2017

Einsteins relativitetsteori

Annonser

## Falling through the Earth

20 juli 2017

Falling through planets and stars – a thought experiment

## A not so easy integral

25 mars 2016 $\int_0^{\infty}\int_0^{\infty}\frac{dx\,dy}{1+xy(x+y)}=\frac{1}{3}\Gamma(1/3)^3$

Can you prove it?

## A trigonometric equation

09 januari 2011

tan(x + y) = tanx · tan y  The equation tan(x + y) = tanx * tan y corresponds to the graph in figure 1. It should really consists of closed simple curves, the gaps is due to the computer program. The “bubbles” are symmetric around the line… y = x, they are infinite in number and separated by π in x- and y-direction. This is naturally a direct consequence of the period length π of the tangent function. The second picture is a magnification of the “bubble” going through the origin, in sequel this “bubble” is called B0. The third picture shows tan x versus tan y.

A few challenges:

1. Find some “simple” solution (e g rational multiple of π or maybe a little less simple) corresponding to a point on B0.

2. Solve the equation for y, to get y = f(x) for intervals corresponding to a points on B0.

3. Find the exact coordinates for the point where B0 intersects the line y = x.

## Expected projection

08 januari 2011

Find the mathematical expectation of the area of the projection of a cube with edge of length onto a plane with an isotropically distributed random direction of projection.

(Proposed by Bhargav Gnv/FB/Math)

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Let the cube be positioned in a 3d coordinate system with centre at the origin and edges parallel to axes,and with edge length 1. Let the direction of projection be given by a unit vector n = (cos α, cos β, cos γ) where α, β, γ are the angles of the vector to the coordinate axes. Also, for simplicity we confine ourselves to the first octant. Because of symmetry that doesn’t affect the result.

Now, from any direction exactly three non-parallel faces of the cube can be seen, including the ”degenerate” cases when some face is seen from its own plane. Therefore the area of the projection is cos α + cos β + cos γ. Isotropically distributed random direction of projection means that every position of the tip of vector n on the unit sphere around the origin is equally probable. Since the area element of the sphere is sin θ dθ dφ in polar coordinates the probability density f(θ) for the tip’s position is proportional to sin θ (no φ-dependence by azimuthal symmetry). By normalization f(θ) = (2/π)*sin θ. Thus the expectation value
<A> = ∫∫ (cos α + cos β + cos γ)(2/(π))*sin θ dθ dφ
Σ
where Σ = [0,π/2]x[0,π/2].

To calculate this we must express the direction cosines cos α, cos β, cos γ in polar angles θ, φ. Let’s say that α, β, γ are the angles with the x-, y-, and x-axis respectively. Then the tip of the vector n is at (sin θ cos φ, sin θ sin φ, cos θ). Using the scalar product of n with the unit vetors e_x, e_y, e_z, these components are equal to the direction cosines. Thus,
<A> = ∫∫ (sin θ cos φ + sin θ sin φ + cos θ)(2/π)*sin θ dθ dφ =
= (2/π)*∫∫ sin²θ (cos φ + sin φ) + cos θ sin θ dθ dφ =
= (2/π) * ∫ [θ=0,π/2] sin²θ dθ * ∫ [θ=0,π/2] cos φ + sin φ dφ
+ (2/π) * (π/2) * ∫ [θ=0,π/2] cos θ sin θ dθ =
= (2/π)*(π/4)*2 + 1*1/2*2 = 1 + 1 = 2.

With edge a we get of course <A> = 2a². Seems resonable, since three faces are seen contracted and each face has area a², so <A> must be something between a² and 3a².

Ooppss! ∫ [θ=0,π/2] cos θ sin θ dθ = 1/4*2 = 1/2 so <A> = 3a²/2.

Also, the estimate can be improved. The greatest projected area occurs in the direction (1/√3, 1/√3, 1/√3) and that area is 3a²/√3 = a²√3. So we must have a² < <A> < a²√3, which satisfied by <A> = 3a²/2. – In general the expaction value of a non-constant random variable must be strictly between the smallest and the greatest value.

## Integral distances

08 januari 2011

Find coordinates of a set of eight non-collinear planar points so that each has an integral distance from others.

(Proposed by Bhargav Gnv/FB/Math)

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These eight points all lie on a circle with radius 4225, so no three of them are collinear:

A = [1183, 4056]
B = [1183, -4056]
C = [2975, 3000]
D = [2975, -3000]
E = [-2975, 3000]
F = [-2975, -3000]
G = [-1183, 4056]
H = [-1183, -4056]

The 28 connecting segments all have integer length:

AB = 8112 , AC = 2080 , AD = 7280 , AE = 4290 , AF = 8190 , AG = 2366 ,
AH = 8450
BC = 7280 , BD = 2080 , BE = 8190 , BF = 4290 , BG = 8450 , BH = 2366
CD = 6000 , CE = 5950 , CF = 8450 , CG = 4290 , CH = 8190
DE = 8450 , DF = 5950 , DG = 8190 , DH = 4290
EF = 6000 , EG = 2080 , EH = 7280 , FG = 7280 , FH = 2080
GH = 8112 ## Reinventing the natural logarithm

08 januari 2011

The natural logarithm ln(x) can be defined as ∫(1 to x) 1/t dt. Using this definition, prove that it is a logarithm, i.e. as follows.

The logarithm of base g, log_g, is defined with the biimplication a=g^x ⇔ log_g(a)=x. Prove that there exists a constant e, such that log_e(a)=∫(1 to a) 1/t dt.

(Because this is a reinvention of the natural logarithm, be careful not to use any fact you know from the natural logarithm, including any fact of the constant e.)

(Proposed by Mauri Ericson Sombowadile/FB/Math)