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Conformal Geometric Algebra in Stochastic Optimization Problems ...

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98 CHAPTER 3. THE CONFORMAL GEOMETRIC ALGEBRA<br />

Fig. 3.4: Inner product between a sphere and a po<strong>in</strong>t <strong>in</strong>side of it. The right side<br />

depicts a cross-section only.<br />

Po<strong>in</strong>t - Sphere<br />

Consider a po<strong>in</strong>t p and a sphere S = �s + 1<br />

2 (�s2 − r 2 s)e + eo. Assume that p lies<br />

outside S. Then the <strong>in</strong>ner product p · S =: α r 2 p, with α = −0.5, gives the radius<br />

rp of a sphere Sp centered at p <strong>in</strong> such a way that Sp is orthogonal 8 to S, i.e.<br />

Sp · S = 0. The sphere Sp can be built by<br />

To see this, compare<br />

Sp = p + (p · S)e. (3.28)<br />

Sp · S = (p + (p · S)e) · S = p · S − p · S = 0.<br />

Note that r 2 p is the power of po<strong>in</strong>t p with respect to the sphere S if p lies outside<br />

S, cf. [111, 19, 107]. Thus given a chord pass<strong>in</strong>g through the po<strong>in</strong>ts a, b and p,<br />

where a and b are assumed to lie on S, it holds that<br />

d [p,a] d [p,b]<br />

const<br />

= r 2 p.<br />

If p lies <strong>in</strong>side S (see below), the <strong>in</strong>ner product becomes positive such that it must<br />

be considered that r 2 p < 0. This implies that the sphere Sp becomes imag<strong>in</strong>ary.<br />

Thus by look<strong>in</strong>g at the sign of the <strong>in</strong>ner product it can be determ<strong>in</strong>ed whether a<br />

po<strong>in</strong>t is <strong>in</strong>side or outside a sphere<br />

p · S < 0<br />

S 2 >0<br />

⇐⇒ po<strong>in</strong>t p lies outside S.<br />

This relationship is exploited by Banarer [9] who utilizes CGA spheres <strong>in</strong> a neural<br />

architecture to def<strong>in</strong>e so-called hypersphere neurones so as to achieve a non-l<strong>in</strong>ear,<br />

that is a hyperspherical, separation <strong>in</strong> classification problems.<br />

Now let p be a po<strong>in</strong>t <strong>in</strong> the <strong>in</strong>side of the sphere S := s − 1<br />

2r2 se. The scenario is<br />

illustrated <strong>in</strong> figure 3.4. Accord<strong>in</strong>g to equation (3.14) the <strong>in</strong>ner product is<br />

p · S = 1<br />

2 (r2 s − d 2 1<br />

[p,s] ) =: 2r2 p,<br />

where <strong>in</strong> this case rp denotes the radius of circle Cp, which is depicted on the left<br />

side of the figure.<br />

8 Recall that this means that the respective surfaces are perpendicular.

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