Inverse Functions As discussed in the note on grazing goats, a differentiable function y(x) can be inverted at any given point x0, y0 to give a power series for the function x(y). Writing this power series as it’s clear that the coefficients cn are related to the derivatives of x with respect to y by Hence in order to determine the coefficients of the power series of this inverted function we need only determine the derivatives of x with respect to y. Our task is to determine these derivatives from the known derivatives of y with respect to x. We have y0 = y(x0). Letting positive integer subscripts on either of these variables denote derivatives with respect to the other variable, we saw in the previous note that the derivatives are related by Using this formula we can determine the expressions and so on. For a simple example, suppose we wish to determine the power series for the inverse of the exponential function y = ex. In this case yn = ex for all n, and substitution into the preceding expressions gives and so on. We choose to expand the series about the point x0 = 0, y0 = 1, so we evaluate these derivatives at x0 = 0, and we find that Substituting these expressions into the equations for the coefficients, we get the series which of course is the power series for the natural log, the inverse of the exponential function. Notice that the numerators of the expression for xn contains a term for each partition of m = 2(n-1) into exactly (n-1) parts.  Also, the overall signs alternate for successive n, so we can write the general coefficient of the inverse function in the form where the summation is evaluated over all sets of non-negative integers a1, a2, …, an such that The “C” coefficients in the preceding expression are related to the multinomial and binomial coefficients. Specifically, we have To illustrate, we will use these formulas to determine the general expression for the 6th order coefficient in the power series of the inverse function. The seven partitions of m = 2(n-1) = 10 into exactly (n-1) = 5 parts are listed below, along with the corresponding C coefficients. To illustrate how the coefficients are computed, the coefficient corresponding to the second of these partitions is Notice that the sum of the coefficients for n = 6 is (n-1)! = 120, as required. From these results we can now write the sixth order coefficient in the power series expansion of the general inverse function: Naturally these results are consistent with the “reversion of series” formulas, which apply in cases when the original function is given as a power series and we wish to find the power series of the inverse function. We can always shift the variables so that we are expanding about the point x0 = y0 = 0, and as always we require that the derivative of y(x) be non-zero, so we are given a series of the form and we wish to determine the coefficients of the reverted series Inserting the values of the derivatives y1 = h1, y2 = 2h2, y3 = 6h3, etc., into the previous formulas, we get and so on.  In these terms the coefficients in the numerators sum to unity. In the note on grazing goats we considered the function y = sin(x) – x cos(x) at the point x0 = p/2, y0 = 1, so we could just as well have proceeded by re-writing the function in terms of the re-scaled parameters X = x – p/2 and Y = y - 1, which gives Then application of the preceding formula for series reversion gives in agreement with the result found by evaluating the derivatives. It’s interesting that the condition that must be met by a function y(x) in order for the third derivative of x(y) to vanish is y1y3 – (1/3)y22 = 0 which, if we set s0 = y1 and differentiate again, gives This is of the same general form as the ubiquitous “separation equation” in physics. (See the note on inertial and gravitational separations.) This is also reminiscent of the question about whether an accelerating charge radiates, a question that hinges on whether radiation is associated with the second or the third derivative of position with respect to time. Return to MathPages Main Menu