January 01, 1970
The Lagrange Inversion Theorem provides a formula for the power series of the inverse of an analytic function. We present a straightforward extension to functions with a finite number of derivatives.
The Lagrange Inversion Theorem provides a formula for the power series of the inverse of an analytic function. The book [3] has a historical discussion and a proof using the Cauchy integral formula. The Theorem is stated in various references, e.g. [1,5]. It is useful to be able to apply the formula more generally to functions of limited smoothness. The purpose of this note is to provide a straightforward extension to functions with a finite number of derivatives, since we have not been able to find such a version available. Related results extending a different version to nonanalytic functions are given in [2,4], and there are results for formal power series.
The Lagrange Theorem is stated and proved in [3] as Theorem 2.3.1. We begin with a paraphrase of that statement:
Theorem 1. Suppose \(f\) is an analytic function of \(z \in \mathbb{C}\) in a neighborhood of \(z = a\), \(f(a) = b\), and \(f'(a) \neq 0\). Then there are sufficiently small open disks \(B_1\) about \(a\) and \(B_2\) about \(b\) so that for each \(w \in B_2\) there is a unique root \(z \in B_1\) of \(f(z) = w\). Set \(g(w) = z\). Then the inverse function \(g\) is analytic and \[g(w) = a + \sum_{n=1}^\infty \frac{(w-b)^n}{n!} D^{n-1} \left[\phi(z)^n \right]_{z=a}\] where \(D = d/dz\) and \(\phi\) is the analytic function \[\phi(z) = \frac{z-a}{f(z)-b}\,,\quad z \neq 0 \,, \qquad \phi(a) = \frac{1}{f'(a)}\]
To obtain this statement from Thm. 2.3.1 in [3], in (2.20) we define \(\phi\) as above, choose \(t = w - b\), and set \(\psi(z) = z\). Then for \(z \neq a\) with \(z\) near \(a\), (2.20) in [3] is equivalent to \(w = f(z)\), whereas for \(z = a\) it is \(w = b = f(a)\). The formula (1)-(2) is equivalent to those in [1] and the NIST Digital Library of Mathematical Functions. We will use this theorem, as well as the usual Inverse Function Theorem, to verify the following version for real-valued functions that are smooth rather than analytic:
Theorem 2. Suppose that \(f\) is a real-valued \(C^{N+1}\) function of \(x \in \mathbb{R}\), with \(N \geq 1\), defined near \(x = a\), with \(f(a) = b\) and \(f'(a) \neq 0\). Then there are open intervals \(I\) about \(a\) and \(J\) about \(b\), sufficiently small, so that for each \(y \in J\) there is a unique \(x \in I\) with \(f(x) = y\). Set \(g(y) = x\). Then \(f: g(J) \to J\) and \(g: J \to g(J)\) are inverses and \[g(y) = a + \sum_{n=1}^N \frac{(y-b)^n}{n!} D^{n-1} \left[\phi(x)^n \right]_{x=a} \,+\, O(|y-b|^{N+1})\] where \(D = d/dx\) and \(\phi\) is the \(C^N\) function \[\phi(x) = \frac{x-a}{f(x)-b}\,, \quad x \neq 0\,, \qquad \phi(a) = \frac{1}{f'(a)}\]
To begin the proof, we may assume that \(a = 0\) and \(b = 0\), since we can translate \(x\) by \(a\) and \(f\) by \(b\). With \(f \in C^{N+1}\) and \(f(0) = 0\), it is evident from \(f(x)/x = \int_0^1 f'(tx)\,dt\) that \(f(x)/x\) is \(C^N\), and since \(f'(0) \neq 0\), the same is true for \(\phi(x) = x/f(x)\). Now let \(p\) be the Taylor polynomial of order \(N\), so that \(f(x) = p(x) + O(|x|^{N+1})\). We will apply the above theorem to \(p\), since it is analytic. By the usual Inverse Function Theorem, \(f\) and \(p\) both have \(C^{N+1}\) inverses for \(y\) in a sufficiently small interval \(J\). From Theorem 1 we have \[p^{-1}(y) = \sum_{n=1}^N \frac{y^n}{n!} D^{n-1} \left[\psi(x)^n \right]_{x=0} + O(|y|^{N+1})\,, \qquad \psi(x) = \frac{x}{p(x)}\] and our task is to verify that \[f^{-1}(y) = \sum_{n=1}^N \frac{y^n}{n!} D^{n-1} \left[\phi(x)^n \right]_{x=0} + O(|y|^{N+1})\,, \qquad \phi(x) = \frac{x}{f(x)}\]
We first show that \[f^{-1}(y) - p^{-1}(y) = O(|y|^{N+1}) \quad as \quad y \to 0\] Given \(y \in J\), let \(x = f^{-1}(y)\) and \({\tilde{x}}= p^{-1}(y)\). Then \[0 = f(x) - p({\tilde{x}}) = f(x) - f({\tilde{x}}) + O(|{\tilde{x}}|^{N+1})\] Since \(x = 0\) at \(y = 0\), \(|x| \leq M|y|\) and \(|{\tilde{x}}| \leq M|y|\) for \(y\) in some interval about \(0\) and some \(M\). Then for \(|y|\) small enough, \(|x|\) and \(|{\tilde{x}}|\) are small, so that \(|f'(\xi)| \geq |f'(0)|/2\) for \(\xi\) between \(x\) and \({\tilde{x}}\). Then \(|f(x) - f({\tilde{x}})| \geq |f'(0)| |x - {\tilde{x}}|/2\). Since \(f'(0) \neq 0\), it follows from (8) that \(|x - {\tilde{x}}| = O(|{\tilde{x}}|^{N+1})\) and also \(|x - {\tilde{x}}| = O(|y|^{N+1})\) since \({\tilde{x}}\) is bounded by \(y\). Thus (7) holds.
The functions \(f(x)/x\) and \(p(x)/x\) are \(C^N\) and have the same Taylor expansion to \(N-1\), with remainder \(O(|x|^N)\). The same is true for \(\phi(x) = x/f(x)\) and \(\psi(x) = x/p(x)\), according to standard formulas for series of reciprocals. When we multiply series, the coefficient of \(x^n\) in the product depends only on the coefficients of \(x^m\) for \(m \leq n\) in the factors. Thus, since the expansions for \(\phi(x)\) and \(\psi(x)\) match to order \(N-1\), the same is true for \(\phi(x)^n\) and \(\psi(x)^n\). Then \(\phi(x)^n - \psi(x)^n\) is \(O(|x|^N)\), and its Taylor polynomial of order \(N-1\) is zero. Therefore \(D^{n-1}\phi^n - D^{n-1}\psi^n = 0\) at \(x=0\) for \(n \leq N\). Together with (7) we have now verified that (6) is equivalent to (5).
References
1. G. E. Andrews, R. Askey, and R. Roy, Special Functions, Encyclopedia of Mathematics and its Applications, Vol. 71, Cambridge Univ. Press, 1999.
2. N. Grossman, A \(C^\infty\) Lagrange inversion theorem, Amer. Math. Monthly 112 (2005), 512–514.
3. S. G. Krantz and H. R. Parks, The Implicit Function Theorem, History, Theory, and Applications, Birkhauser, 2002.
4. S. G. Krantz and H. R. Parks, The Lagrange inversion theorem in the smooth case, J. Math. Anal. Appl. 340 (2008), 1263–1270.
5. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, 4th ed., Cambridge Univ. Press, 1927.