Abstract
The derivatives of a quadratic trinomial F = ax 2 + bx + c at its roots x 1, x 2 add up to zero (Vieta’s theorem). Alternatively, we can write this as 1/F′(x 1) + 1/F′(x 2) = 0 or, equivalently, 1/(x 1 − x 2) + 1/(x 2 − x 1) = 0. Also, there is an obvious identity: x 1/(x 1 − x 2) + x 2/(x 2 − x 1) = 1. C.G.J. Jacobi’s concern with questions of dynamics and geometry led him to derive far-reaching generalizations of these identities for polynomials of arbitrary degree, which enabled him to solve several complex problems. For example, he squared an ellipsoid’s surface, determined the geodesics on this surface, and described the dynamics with two motionless centers of gravity. In this chapter, readers will become acquainted with the Jacobi identities and related formulas and their close ties with complex analysis; in addition, they will learn about some less traditional applications of the Jacobi identities (to linear differential equations). The chapter also contains notes about further generalizations and applications and a short guide to the literature.
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Notes
- 1.
The proof in section E12.39 is done for the complex field ℂ. However, it is applicable for any infinite field because a finitely generated extension of the rational field is embeddable in ℂ. (Readers should complete all details.)
- 2.
The fundamental solution possesses n − 2 continuous derivatives, so convolving it with the right-hand side brings n − 1 continuous derivatives, which may be established proceeding step by step as follows:
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1.
Let the functions f and g be identically equal to zero on, respectively, {t < 0} and {t < a} and continuous on, respectively, {t ≥ 0} and {t ≥ a}, where a ≥ 0; show that their convolution h = f*g equals zero identically on t < a and \( h(t)=\int\limits_a^t {f(t-s)g(s)ds} =\int\limits_0^{t-a } {f(s)g(t-s)ds} \) for t ≥ a.
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2.
In addition, let g be continuously differentiable on {t ≥ a}; prove that h is continuously differentiable on {t ≥ a}, with \( {h}^{\prime}(t)=g(a+0)f(t-a)+\int\limits_a^t {f(t-s){g}^{\prime}(s)ds} \). (Use integration by parts.)
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3.
Show by induction on m that if f(t) is an m ≥ 0 times continuously differentiable function with f(0) = … = f [m](0) = 0, then the function \( F(t)=\left\{ {\begin{array} {llll}{\int\limits_a^t {f(t-s)g(s)ds},\;\mathrm{ for}\;t\geqslant a,} \\{0,\quad \quad \quad \quad \mathrm{ for}\;t<a} \\\end{array}} \right. \), with g continuous on {t ≥ a}, is m times continuously differentiable.
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4.
Combining the results of steps 2 and 3 (applied for g′ in place of g), deduce by induction that if, under the conditions of step 2, f is m times continuously differentiable on (−∞,∞) [so that f(0) = … = f [m](0) = 0] and g is continuously differentiable on {t ≥ a}, then f*g is m + 1 time continuously differentiable.
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1.
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Roytvarf, A.A. (2013). Jacobi Identities and Related Combinatorial Formulas. In: Thinking in Problems. Birkhäuser, Boston. https://doi.org/10.1007/978-0-8176-8406-8_1
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