Kroner

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15 Iterative Newton’s method solution of <strong>Kroner</strong>’s cubic Rewrite Eq. 211 as: x 3 + ax 2 − bx − c = 0 (272) where x = µ ∗ . It turns out that it is easier to use Newton’s method rather than the formulas for solving a cubic equation. To this end, we can write an exact Taylor’s expansion in the form: f(x + ∆x) = f(x) + f ′ (x)∆x + 1 2 f ′′ (x)(∆x) 2 + 1 6 f ′′′ (x)(∆x) 3 (273) since all derivatives greater than third order vanish. Now apply Newton’s method to first, second, or third order, as follows. Write: f(x n+1 ) = f(x n ) + f ′ (x n )∆x + 1 2 f ′′ (x n )(∆x) 2 + 1 6 f ′′′ (x n )(∆x) 3 (274) x n+1 = x n + ∆x (275) Newton’s method is usually based only on a first order expansion, solving for ∆x assuming that the first derivative does not vanish: f(x n+1 ) = 0 ≈ f(x n ) + f ′ (x n )∆x (276) ∆x = −f(x n) f ′ (x n ) (277) We consider the case where the first derivative is too small by going to a second order approximate solution, and if both first and second derivatives are too small, by going to a third order solution. Details follow. ”Too small” means that the normalized value of the derivative is less than 10 −10 . If the first derivative is too small, then we check the second derivative. If it is also too small, then we use the third order approximation, and the Taylor’s expansion reduces to: f n+1 = f(x n+1 ) = 0 = f n + 1 6 f ′′′ (x n )(∆x) 3 (278) But from Eq.272, f ′′′ (x n ) = 6, so that we can solve for ∆x: f n = f(x n ) (279) ∆x = (−f n ) 1 3 (280) If the first derivative is too small, but the second derivative is not too small, then from the second order approximation, we can solve for ∆x provided that < 0, so that: f n f ′′ n ∆x = √ −2 f n f ′′ n (281)
otherwise there is no solution (More below). If the first derivative is sufficiently large, then we can consider again the second order approximation: f(x n+1 ) = 0 ≈ f(x n ) + f ′ (x n )∆x + 1 2 f ′′ (x n )(∆x) 2 (282) Checking the discriminant D: D = (f ′ n) 2 − 2f ′′ nf n (283) If this discriminant is negative, then we revert to the first order solution Eq. 277. If the discriminant is positive but the second derivative is too small, the we also revert to the first order solution Eq. 277. Finally, if the discriminant is positive and the second derivative is sufficiently large, then we get the second order solution for ∆x: ∆x = −f ′ n + √ D f ′′ n (284) We choose the root with the ”+” sign because as f n ⇒ 0, ∆x ⇒ 0. Note that all cases reported here fell in this second order solution, where both first and second derivatives were large and the discriminant was strictly positive. The case above without solution is interesting. This is the case where f n ′ is have the same sign. If they are both positive, this means that locally (about x n ), f(x) opens upward but does not cross the x-axis. Similarly, if they are both negative, f(x) opens downward but does not cross the x-axis. In both of these cases, there is clearly no solution. too small and both f n and f ′′ n
Page 1 and 2: Derivation and role of Kroner’s c
Page 3 and 4: 1 Abstract The pioneering work of H
Page 5 and 6: Such a tensor can be decomposed as
Page 7 and 8: For a constant point force F acting
Page 9 and 10: or, defining: we can write: ∫ u i
Page 11 and 12: 7 Symmetry properties and model of
Page 13 and 14: where: [k] = k 2 − k 1 (105) [µ]
Page 15 and 16: Markov now describes the critical c
Page 17 and 18: 11 Rotational averages Before proce
Page 19 and 20: Note that from Eqs. 10, 11, and 174
Page 21 and 22: Since I ′ and I ′′ are orthog
Page 23 and 24: value at the arithmetic average of
Page 25 and 26: 14 Major symmetries of tensor D Wri
Page 27: Finally, ω represents any D ijkl t
Page 31: [15] P. Sisodia and M.P. Verma. She

Kroner

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