Taking it to another level

I'm posting a solution by our friend Steven Chase, who did not get the problem right at first due to a misunderstanding based on the wording of the problem.

I will post my own solution below.

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Alright, I modified the code to generate points uniformly on the "surface" of the 4-sphere. I used the algorithm from this link (with x1-x4):

http://mathworld.wolfram.com/HyperspherePointPicking.html

Taking the ratio of (# points on 4-sphere AND in cube) / (# points on 4-sphere) yields a percentage of 58.35, which matches your result.

35 import math import random

N = 10**7

numcube = 0 numsphere = 0

for j in range(0,N):

# Generate coordinates for unit 4-sphere and then multiply by 5

x1 = random.uniform(-1.0,1.0)   # "Seed values"
x2 = random.uniform(-1.0,1.0)
x3 = random.uniform(-1.0,1.0)
x4 = random.uniform(-1.0,1.0)

S1 = x1**2.0 + x2**2.0
S2 = x3**2.0 + x4**2.0

if (S1 < 1.0) and (S2 < 1.0):

    x = 5.0 * x1                # 5 times coordinates for unit 4-sphere
    y = 5.0 * x2                # Coordinates uniformly picked on hypersurface 
    z = 5.0 * x3 * math.sqrt((1.0-S1)/S2)
    w = 5.0 * x4 * math.sqrt((1.0-S1)/S2)

    numsphere = numsphere + 1   # Count number of points on 4-sphere

    if math.fabs(x) < 4.0 and math.fabs(y) < 4.0 and math.fabs(z) < 4.0 and math.fabs(w) < 4.0:

        numcube = numcube + 1   # Count number of points on 4-sphere AND within 4-cube

print 100.0 * float(numcube) / float(numsphere) # Ratio gives fraction of 4-sphere "volume" inside 4-cube Steven Chase - 13 hours ago 1 Reply

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By symmetry, we can at first restrict our attention to the "upper half" of the hypersphere, $H$ , where $w\geq 0$ . We can interpret $H$ as the graph of the function $f(x,y,z)=w=\sqrt{25-x^2-y^2-z^2}$ defined on the ball $B_5$ in $\mathbb{R}^3$ given by $x^2+y^2+z^2\leq 25$ . The scaling factor (for volume) of this parameterization of $H$ is $\frac{5}{\sqrt{25-x^2-y^2-z^2}}$ . Thus the volume of $H$ is $\int\int\int_{B_5} \frac{5}{\sqrt{25-x^2-y^2-z^2}}dV=4\pi\int_0^5 \frac{5}{\sqrt{25-\rho^2}}\rho^2 d\rho=125\pi^2$ and the volume of the whole hypersphere is $V_{total}=250\pi^2$ .

This hypersphere has eight "caps" outside the hypercube, where the absolute value of one of the variables is $\geq 4$ . Luckily, the intersection of any two of these caps is empty, since $4^2+4^2>5^2$ . By symmetry, it suffices to find the volume of one of those caps; let's look at $w\geq 4$ . Now, on the hypersphere, $w\geq 4$ means that $\rho\leq 3$ , so that $V_{cap}=\int\int\int_{B_3} \frac{5}{\sqrt{25-x^2-y^2-z^2}} \ dV=4\pi \int_0^3 \frac{5}{\sqrt{25-\rho^2}}\rho^2 d\rho =250\pi\arcsin\left(\frac{3}{5}\right)-120\pi$ The required proportion is $\frac{V_{total}-8V_{cap}}{V_{total}}\approx 0.584\approx \boxed{58}\text{\%}$

I did the two- and threedimensional cases first, to warm up. They gave percentages 18% and 40%. The procedure is analogous.

The 4-dimensional case (4D hypersphere and tesseract):

The hypersphere is defined by $\sum_{i=1}^{4} x_i^2=R^2$ In order to find the volume, we are going to integrate over and angle $φ$ that runs from $0$ to $π$ . We set $x_4=R \cos φ$ For a fixed value of $φ$ , the intersection of the hypersphere and the hyperplane $x_4=R \cos φ$ is defined by $\sum_{i=1}^{3} x_i^2 = R^2\sin^2 φ$
This is a (3D-)sphere with radius $R\sin φ$ and area $4πR^2\sin^2 φ$ .

Integrating the area associated with each value for $φ$ , with jacobian R gives the total volume of the (boundary of) the hypersphere:

$V_{hypersphere}=\int_0^π 4πR^2\sin^2 φ Rdφ= 2πR^3 \int_{0}^{π} (1 - \cos 2φ) dφ =2πR^3 (φ-\frac{1}{2}\sin2φ) \Bigr|_{0}^{π} = 2π^2R^3$

Letting $φ$ run from $0$ to $\arccos \frac{4}{5}$ , we find the volume sticking out one of the 3D-'faces' of  the tesseract:

$v_{bulge}=2πR^3 (φ-\frac{1}{2}\sin2φ) \Bigr|_{0}^{\arccos \frac{4}{5}} = V_{hypersphere}×\frac{1}{π} (\arccos \frac{4}{5}-\frac{1}{2}\sin(2\arccos \frac{4}{5}))$ .

Now a tesseract has 8 'faces' (two for each cartesian axis), so 8 bulges are sticking out and a fraction $\frac{8}{π} (\arccos \frac{4}{5}-\frac{1}{2}\sin(2\arccos \frac{4}{5})) \text{ } (41.6...\%)$ of the hypersphere is outside the tesseract, hence a rounded $\boxed{  58\%}$ is inside the tesseract.

$\frac{2 \left(48 \pi -50 \pi \sin ^{-1}\left(\frac{3}{5}\right)+25 \pi \sin ^{-1}\left(\frac{7}{25}\right)\right)}{25 \pi ^2}$ is about 58.36%.