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4d+ line average distance from a point, Started 10/16/2024

Per dimension: Q^2+x0*x1/3-x0*Q+x1*x1/3+x0*x0/3-Q*x1

Where Q is point to compare with, x0 is one point on line's dimension, and x1 is other point on line's dimension.

Here is how I got it:

Using end, a line point, minus start, Q. x0+t(x1-x0) is one part in line. Set up integral as this. Integral 1 to 0, (x0+t(x1-x0)-Q)(x0+t(x1-x0)-Q) dt Distribute with part in integral. x0^2+Q^2+2t*x0*x1-2t*x0^2-2x0*Q+t*t*x1*x1-t*t*x1*x0-t*t*x1*x0+t*t*x0*x0-2t*Q*x1+2t*Q*x0 Collect. x0^2+Q^2+2t*x0*x1-2t*x0^2-2x0*Q+t*t*x1*x1-2t*t*x1*x0+t*t*x0*x0-2t*Q*x1+2t*Q*x0 Power rule of t. t*x0*x0+t*Q*Q+t*t*x0*x1-t*t*x0*x0-2x0*Q*t+t*t*t*x1*x1/3-2t*t*t*x1*x0/3+t*t*t*x0*x0/3-t*t*Q*x1+t*t*Q*x0 This variable t is either 0 and sets all to 0 or 1. Use this to solve. x0^2+Q^2+x0*x1-x0^2-2x0*Q+x1*x1/3-2x1*x0/3+x0*x0/3-Q*x1+Q*x0 Collect and we have this. Per dimension: Q^2+x0*x1/3-x0*Q+x1*x1/3+x0*x0/3-Q*x1

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