translating a Basic program; stuck on ^(-1/2) and ^(3/2) - basic

Question

I am trying to translate a BASIC program. It has been many decades since I did any BASIC programming. :)
I am having a problem with 2 lines of code:
360 D=D+((X(I)-X(J))^2+(Y(I)-Y(J))^2+(Z(I)-Z(J))^2)^(-1/2)
510 F=((X(I)-X(J))^2+(Y(I)-Y(J))^2+(Z(I)-Z(J))^2)^(3/2)
x(i) etc form (x,y,z) coordinates; so line 360 is - I think - calculating the distance between 2 points; that would work if ^(-1/2) = square-root.
Line 510 is very similar, but the ^(3/2) has me stumped.
Is it sensible to raise a number to the power of 1.5?
I remember that ^2 means square (aka raise to the power 2).
So, can someone please tell me what ^(-1/2) and ^(3/2) mean in BASIC!?
Thanks for you help.
Steve.

Answer 1

Is it sensible to raise a number to the power of 1.5?
Yes. There is an ordinary arithmetical meaning to raising any number to any power. In this case it is equivalent to cubing then square-rooting (or square-rooting then cubing, as multiplication is commutative).
^(-1/2) = square-root.
Not quite. Raising a number to a negative power is equivalent to finding the reciprocal of having raised it to a positive power, so that is a "1 divided by square root X" operation

Answer 2

In BASIC the following is true:
360 D = D + ((X(I) - X(J)) ^ 2 + (Y(I) - Y(J)) ^ 2 + (Z(I) - Z(J)) ^ 2) ^ (-1 / 2)
361 D = D + 1 / SQR((X(I) - X(J)) ^ 2 + (Y(I) - Y(J)) ^ 2 + (Z(I) - Z(J)) ^ 2)
510 F = ((X(I) - X(J)) ^ 2 + (Y(I) - Y(J)) ^ 2 + (Z(I) - Z(J)) ^ 2) ^ (3 / 2)
511 F = SQR(((X(I) - X(J)) ^ 2 + (Y(I) - Y(J)) ^ 2 + (Z(I) - Z(J)) ^ 2) ^ 3)

Answer 3

These functions should report arrays of equal values:
DIM x(10) AS INTEGER, y(10) AS INTEGER, z(10) AS INTEGER
FOR x = 1 TO 10
x(x) = x: y(x) = x: z(x) = x
NEXT
I = INT(RND * 10 + 1): J = INT(RND * 10 + 1)
D1 = ((x(I) - x(J)) ^ 2 + (y(I) - y(J)) ^ 2 + (z(I) - z(J)) ^ 2) ^ (-1 / 2)
D2 = 1 / SQR((x(I) - x(J)) ^ 2 + (y(I) - y(J)) ^ 2 + (z(I) - z(J)) ^ 2)
F1 = ((x(I) - x(J)) ^ 2 + (y(I) - y(J)) ^ 2 + (z(I) - z(J)) ^ 2) ^ (3 / 2)
F2 = SQR(((x(I) - x(J)) ^ 2 + (y(I) - y(J)) ^ 2 + (z(I) - z(J)) ^ 2) ^ 3)
IF D1 = D2 THEN PRINT "values equal to:"; D1
IF F1 = F2 THEN PRINT "values equal to:"; F1
END

Answer 4

Three-dimensional coordinate systems from google:
The distance formula states that the distance between two points in xyz-space is
the square root of the sum of the squares of the differences between corresponding
coordinates. That is, given P1 = (x1,y1,z1) and P2 = (x2,y2,z2), the distance
between P1 and P2 is given by d(P1,P2) = sqr((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)
So, apparently, you can get the distance between 2 xyz points..