Elementary trigonometry based on a first order differential equation

arXiv:1111.6839v1 [math.HO] 23 Nov 2011 Elementary trigonometry based on a first order differential equation Horia I. Petrache Department of Physics, Indiana University Purdue University Indianapolis, Indianapolis, IN 46202 November 7, 2018 Abstract It is shown that with appropriate boundary conditions, a real function satisfying the differential equation f ′(x) = f(x + a) has all known properties of the sine function. A number of elementary derivations are presented including proofs for periodicity and trigonometric identities. 1 Introduction In elementary analysis, the sine and cosine functions are identified as the two linearly independent solutions of the second order differential equation f ′′ = −f. Therefore, any function f that satisfies this harmonic equation is periodic, bounded, and sat- isfies a number of trigonometric identities. However, many of these trigonometric properties are derived not from f ′′ = −f but from alternative definitions of sine and cosine functions. These alternatives include primarily geometrical definitions based on right triangles and unit circles [1,2,3], power series and complex exponentials [4,5], and integral forms of arc lengths [6,7]. The question then is what properties follow directly from f ′′ = −f. For example, it is quite easy to show by direct differ- entiation that f 2 +f ′2 = const., which then implies that f and f ′ are both bounded. However, other properties are more difficult to prove in abstract form, e.g. the fact 1 that f is periodic. An even more interesting question is whether there is any other property of the sine and cosine functions from which all known properties can be derived. Here it is shown that the differential equation f ′(x) = f(x+a) specifies the sine function given the appropriate boundary conditions. Derivations are carried at an abstract level that does not involve series expansions or complex exponentials. 2 Derivations Consider the differential equation f ′(x) = f(x + a), (1) in which the argument x and the constant a are real. In the following treatment we assume that f is well defined on the entire real axis, is of class C2 or higher, and it is not identically zero. Let us first note that the differential operator changes the parity (symmetry) of an odd or even function. If f is odd then f ′ is even and vice versa. For the above equation to hold, the translation operator that takes f(x) into f(x + a) must perform the same symmetry transformation as the differential operator. From this perspective, we will investigate solutions of Eq. 1 with well defined parity. Proposition 1. f ′(x) = f(x + a) implies that f is periodic. Proof. Use notation g(x) = f ′(x) and assume that f is odd. (The proof for an even function is similar.) If f is odd then g = f ′ is even. We have f(x) Eq. 1 = f ′(x −a) def = g(x −a) sym = g(−x + a) Eq. 1 = f(−x + 2a) sym = −f(x −2a), where the text above the equal sign indicates the property used at each step, and where sym stands for symmetry and def for the definition (notation) g = f ′. We have obtained that f(x) = −f(x −2a) for any x, which implies that f(x −2a) = −f(x −4a), so that 2 f(x) = f(x −4a) for any x. It means that f is periodic of period 4a. Proposition 2. f ′(x) = f(x + a) implies that g′(x) = g(x + a), where g = f ′. Proof. g′(x) = f ′′(x) Eq. 1 = f ′(x + a) = g(x + a). Proposition 3. f ′(x) = f(x + a) implies that g′(x) = −f(x), where g = f ′. Proof. Using Proposition 2, we have g′(x) = g(x + a) sym = g(−x −a) def = f ′(−x − a) Eq. 1 = f(−x −a + a) = f(−x) sym = −f(x). We have in fact proven that f satisfies the second order differential equation f ′′ = −f. (2) Proposition 4. f 2 + g2 = constant. Need to show that the derivative of f 2+g2 is zero. We have ff ′+gg′ = ff ′+f ′(−f) = 0. Because the functions f and g are specified up to an arbitrary scale, we can choose the constant to be 1 without loss of generality. If f is odd, we have f(0) = 0 and therefore g(0) = ±1 and we can choose g(0) = 1. Proposition 5. The functions f and g = f ′ are linearly independent. Proof. Assume that f = cf ′ with c a real constant. We have f ′ = cf ′′ = −cf = −c2f ′, implying that c2 = −1 which contradicts the assumption that c was real. (Alternatively, it can be verified that the Wronskian of f and f ′ is non-zero.) Proposition 6. If f is a solution of f ′′ = −f, then g = f ′ is also a solution and f and g are linearly independent. Proof. g′′ = f ′′′ = −f ′ = −g. Linear independence was proven above. Proposition 7. f ′′ = −f implies that f(x + y) = f ′(y)f(x) + f(y)f ′(x). Proof. First note that f(x + y) is a solution of f ′′ = −f for any y, where the derivative is with respect to the variable x. That means that f(x + y) is a linear 3 combination of f and f ′, which were shown earlier to be independent solutions of Eq. 2. We have f(x + y) = A(y)f(x) + B(y)f ′(x), (3) where the coefficients A and B are functions of y. With f(0) = 0 and f ′(0) = 1, we have f(0 + y) = A(y)f(0) + B(y)f ′(0) (4) f(x −x) = A(−x)f(x) + B(−x)f ′(x), (5) which g

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