By Jiwei D., Lee P.K.
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Additional info for 4th-order spline wavelets on a bounded interval
The relation is involutory: (F')' =F. It is not hard to show that all poles of lines through a given point lie on the polar of that point. Conversely, all polars of points of a given line pass through the pole of that line. If S is a plane curve we consider a variable tangent to S. Regarded as a moving polar, this induces a moveable pole that traces out the reciprocal polar curve of S. Treated algebraically, the polars are quite simply obtained. Let C = 0 be the equation of the fixed conic, where C is some polynomial, of second degree in x and y.
As in Fig. 2b we fix two bars X and Y at right angles and make the arrangement of sleeve-slides as shown. If the points A and B are at 1 and a units from the origin, we get a mechanical way of generating the first few powers 1, a, a 2 , a 3 , a 4 , as, ... Alternatively, working in reverse, we can use the power linkage for an approximate simultaneous determination of the quantities a 11 s, a 21 s, a 31 s, a 41 s, given a (positive and fairly close to 1). By using the power linkage together with the angle multisectors (to be described) the reader may wish to design a mechanical method for powers and roots of complex numbers.
More generally, we can also tell whether C lies on a sphere. For, let C be represented in the Cartesian form by its radius vector as a function of s: x = x(s). We produce a sphere of possibly high order of contact with C, its so-called osculating sphere. The equation of the sphere is written in FURTHER TOPICS IN GEOMETRY 40 vector form: (u-c) · (u-c)-r2 =0 so that c is its center and r its radius. We substitute x for u, and differentiate successively with respect to s; using the Frenet-Serret formulas (3) we get the equations (x- c)· (x- c)- r 2 = 0, (x- c)· t = 0, K(x- c)· n + 1 = 0, (x-c) · (K 1 n-K 2 t+KTb)=0.
4th-order spline wavelets on a bounded interval by Jiwei D., Lee P.K.