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The four Hermite basis functions. The interpolant in each subinterval is a linear combination of these four functions. On the unit interval [,], given a starting point at = and an ending point at = with starting tangent at = and ending tangent at =, the polynomial can be defined by = (+) + (+) + (+) + (), where t ∈ [0, 1].
In mathematics, the degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
The roots, stationary points, inflection point and concavity of a cubic polynomial x 3 − 6x 2 + 9x − 4 (solid black curve) and its first (dashed red) and second (dotted orange) derivatives. The critical points of a cubic function are its stationary points, that is the points where the slope of the function is zero. [2]
One approach is to color the vertices (with two colors, e.g., black and white) and require that adjacent tiles have matching vertices. [32] Another is to use a pattern of circular arcs (as shown above left in green and red) to constrain the placement of tiles: when two tiles share an edge in a tiling, the patterns must match at these edges. [21]
A tiling that cannot be constructed from a single primitive cell is called nonperiodic. If a given set of tiles allows only nonperiodic tilings, then this set of tiles is called aperiodic. [3] The tilings obtained from an aperiodic set of tiles are often called aperiodic tilings, though strictly speaking it is the tiles themselves that are ...
The pattern corresponds to each of the following: symmetrically staggered rows of identical doubly symmetric objects; a checkerboard pattern of two alternating rectangular tiles, of which each, by itself, is doubly symmetric; a checkerboard pattern of alternatingly a 2-fold rotationally symmetric rectangular tile and its mirror image
In the mathematical study of polynomial splines the question of what happens when two knots, say t i and t i+1, are taken to approach one another and become coincident has an easy answer. The polynomial piece P i (t) disappears, and the pieces P i−1 (t) and P i+1 (t) join with the sum of the smoothness losses for t i and t i+1.
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