Wikipedia mathematics suggests that the volume of a four-dimensional hypercube (tesseract) of side length "a" is 8 a^4.
On the other hand, the B matrix chains predict the volume of a four-dimensional hypercube with side length "a" (tesseract) as simply a^4.
The question arises which answer is true, if any?
Let us first discretize the simple geometry cube considered into 27 equidistant nodes as shown in Fig.1
Fig.1 A 3D cube discretized into 27 equally spaced nodes
It is clear that nodes 1,2,3. . ,. ,. 26.27
W1,W2,. . . . . . . . . . . . . . ,W26,W27
are located in Cartesian space like,
(0,0,0), (h,0,0), (2h,0,0), (0,h,0), (h,h,0), (2h,h,0), (0,2h,0),(h,2h,0),(2h,2h,0)
(0,0,h), (h,0,h), (2h,0,h), (0,h,h), (h,h,h), (2h,h,h) , (0,2h,h), (h,2h,h), (2h,2h,h),
(0,0,2h), (h,0,2h), (2h,0,2h), (0,h,2h), (h,h,2h), (2h,h,2h), (0,2h,2h), (h,2h,2h), (2h,2h,2h).
Now the values of the function W1,W2, . . . . ,W26,W27 for the case W=x+y+z are,
(0 , 1 , 2 , 1 , 2 , 3 , 2 , 3 , 4
(1 , 2 , 3 , 2 , 3, 4 , 3 , 4 , 5
(2 , 3 , 4 , 3 , 4 , 5 , 4 , 5 6 )
And the Statistical weights SW1,SW2, . . . . .SW26,SW27 are given by [1],
0.710215569 0.972846687 0.710215569 0.972846687 1.31404436 0.972846568 0.710215569 0.972846568 0.710215569
0.972846568 1.31404424 0.972846568 1.31404436 1.75985277 1.31404436 0.972846568 1.31404436 0.972846508
0.710215569 0.972846508 0.710215688 0.972846568 1.31404436 0.972846508 0.710215688 0.972846568 0.710215688
The numerical value of the triple integration I will be given by the simple formula,
I=∑ Wi . SWi . . . . . . (1)
Which is calculated as 3h^4.
Should be compared to the analytic form for I which is h^4(x^2+y^2+z^2)/2 . . . . . . (2)
Over the specified limits which is exactly 6 h^4,
back to the hypercube via Eq 1 with all Wi equal to 1, we arrive at,
I=1.0000 h^4