Power series and inverse functionsIn the section of the notes on the Inverse Function Theorem (Section II.3), there was anassertion that if a function y = f (x) has a convergent power series expansion at x = a and f0(a) 6= 0,then the inverse function x = g(y) has a convergent power series expansion at y = b = f(a). Thepurpose of this document is to include some additional information about this; the statements ofthe results only require concepts from first year calculus, but the discussion of the proofs is moreadvanced and requires material from the theory of functions of a complex variable (Mathematics165A–B).The main resultsThe first 21 pages of the following online document provide a fairly complete summary of basicfacts about infinite power series of the form∞Xk=0ck(x − a)kand we shall use the results stated in that (part of the) document:http://www.grossmount.edu/carylee/Ma280/PowerPoint/Power%20Series.pptWe shall also need the following result:THEOREM. Suppose that we are given functions f(x) and g(x) such that f(0) = g(0) = 0 andboth have convergent power series in intervals (−A, A) and (−B, B). Then the composite functionh(x) = gf(x)also has a convergent power series representation on some interval (−C, C).The conditions f(0) = g(0) = 0 were added for the sake of computational simplicity. Theresult holds more generally if f has a power series representation on (a − A, a + A) and g has apower series representation onf(a) −B, f(a) +B, and the conclusion is that h has a power seriesrepresentation on some interval of the formh(a) − C, h(a) + C.In the theorem we can find the power series coefficients for h by several methods. For example,we can perform a direct substitution using f(x) =Ppkxkand g(x) =Pqkxk, obtaining a messylooking expression of the formXkpk Xmqmxm!k.If we group together all terms in this expression which are constants times xnfor some n, we obtainsomething of the formPnrnxn, and this turns out to be the power series expansion for h thatwe want. Alternatively, we know that the coefficients for this series are expressible in terms of thehigher order derivatives of the composite function h, and we can use standard calculus identitiesto express the latter in terms of the higher order derivatives of f and g; this process will also yieldthe power series expansion for h.We also need the following fact (not in the PowerPoint document explicitly):UNIQUENESS OF POWER SERIES EXPRESSIONS. If f(x) is represented by two powerseries expressions over the same interval, then the corresponding coefficients of these series are allequal.Application to inverse functionsTHEOREM. Suppose now that f and g are inverse functions with f(0) = g(0) = 0 and f0(0) 6= 0,and suppose that f(x) is given by a convergent power series over some interval (−A, A). Then thereis an interval (−B, B) such that B ≤ A and g(x) has a convergent power series expansion over theinterval (−B, B).If f(x) =Ppkxkand g(x) =Pqkxk, then as above one can solve for the qkin terms of thepkby comparing coefficients in the expressionx =Xkpk Xmqmxm!k=Xnrnxnwhere as before the third expression is obtained from the second by combining like terms; inparticular, by uniqueness of power series representations we must have r1= 1 and rn= 0 otherwise.Note that p0= q0= 0 and p16= 0 for the examples we are considering.There is also an identity called the Lagrange Inversion Formula that can be applied to find thecoefficients qk; here is an online site which discusses this formula:http://en.wikipedia.org/wiki/Lagrangeinversion theoremOne difficulty with the power series for the inverse function is that the interval of convergence(−B, B) is often considerably smaller than one might expect. In particular, the functions x+ex−2and x5+ x3+ x satisfy the conditions of the theorem, the functions have positive derivativeseverywhere, they have power series expansions at 0 which are valid for all x, and their limits asx → ±∞ are equal to ±∞ — conditions which imply that the inverse functions can be defined forall values of x. However, it turns out that the power series expansions for the inverse functions areonly valid on a bounded interval (−B, B) and not for all real values of x. This is a consequenceof the following result:THEOREM X. Suppose that f(x) is a function which satisfies f(0) = 0, f0(x) > 0 everywhere,f(x) has a convergent power series expansion which is valid for all real values of x, andlimx→±∞f(x) = ±∞ .Let g be the inverse function to x, and suppose that g(x) also has a convergent power seriesexpansion which is valid for all real values of x. Then f(x) = kx for some positive constant k.The proof of this result requires a number of concepts from the theory of functions of a complexvariable and is beyond the scope of an elementary differential geometry course. However, since thefunctions x + ex− 2 and x5+ x3+ x satisfy the conditions of the theorem, it follows that theconvergent power series expansion for the inverse function g(x) CANNOT be valid for all real valuesof x and hence the interval of convergence must be bounded.Proof of Theorem XSuppose that we are given f(x) and g(x) as above, and assume further that g(x) also has apower series expansion which is valid for all x. We need to show that f(x) and g(x) must be firstdegree polynomials.First of all, basic results about analytic functions show that if f(x) has a power series whichconverges for all real values of x, then the power series f(z) also converges for all COMPLEX numberz; of course the same holds for g(z). Since f and g are inverses of each other and f (0) = 0 = g(0)it follows that f(g(x)) = x = g(f(x)) for all real x in some small interval of the form (−h, h).Since the points where two nonconstant analytic functions agree are isolated from each other, theobservation in the previous sentence implies that we must have f(g(z)) = z = g(f(z)) for allcomplex z. In other words, it follows that the functions f and g are inverse to each other as entireanalytic functions.By the preceding paragraph, the proof of Theorem X reduces to verifying the following result:LEMMA. Let f be an entire analytic function which has an entire analytic inverse. Then f is afirst degree polynomial.Proof of the Lemma. It will suffice to prove the result in the case where f(0) = 0, for if g isan arbitrary function satisfying the conditions in the lemma, then f(z) = g(z) − g(0) still satisfiesthe conditions in the lemma