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The Factorial Function

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The Factorial Function The factorial function 1 Geometric series and some calculus In this lecture we’re going to use induction in other ways to establish some facts from calculus. 1.1 Induction and geometric series Proposition 1.1. For every integer n ≥ 0, and for any real number x, the following identity holds: n X 1 − xn+1 xi = 1 + x + ::: + xn = 1 − x i=0 holds. Proof. We establish this by induction. For n = 0, the left hand side is simply the number 1. The (1−x) right hand side is (1−x) = 1 as well. Assume that statement holds for a given value k. In that case, k+1 k X X xi = ( xi) + xk+1: i=0 i=0 1−xk+1 By the induction hypothesis, the first sum is 1−x . Thus, using the induction hypothesis we get: k X 1 − xk+1 ( xi) + xk+1 = + xk+1: 1 − x i=0 Taking a common denominator on the right hand side and doing some algebra yields: 1 − xk+1 1 − xk+1 (1 − x)xk+1 + xk+1 = + 1 − x 1 − x 1 − x 1 − xk+1 + xk+1 − xk+2 = 1 − x 1 − xk+2 = ; 1 − x but this is precisely what we wanted to show. 1 2 1.1 Induction and geometric series Remark 1.2. The statement we made above is a priori a statement about equalities of real numbers. However, x was arbitrary in the statement, so in fact we know more. The left hand side is a poly- nomial function in x, in particular a continuous function of x. The right hand side is a continuous function of x except when x = 1 since the expression on the right hand side is ill-defined there. Nevertheless, if we simply define a new function piecewise by taking the value of the sum on the left for x = 1 (i.e., n + 1) when x = 1, then the equality above is an equality of continuous func- tions. This kind of observation about limits allows us to extend the validity of the above statement yet further. Proposition 1.3. If jxj < 1, then 1 X 1 xi = : 1 − x i=0 Proof. The sum on the left is defined as a limit: n X lim xi: n!1 i=0 Now, the preceding proposition allows us to write this as: 1 − xn+1 1 xn+1 lim = − lim : n!1 1 − x 1 − x n!1 1 − x xn+1 xn+1 Now, we claim that limn!1 1−x = 0 if jxj < 1. To this end, it suffices to show that limn!1 j 1−x j = 0. We will show that jxn+1j is a decreasing sequence that tends to 0. Indeed, jxj ≥ 0 by definition. If x = 0, there is nothing to show, so suppose x 6= 0. In that case, 0 < jxj < 1 by assumption. We claim that jxn+2j < jxn+1j. We can establish this fact using the first principle of mathematical induction. For n = −1, this is the statement that jx1j < jx0j = j1j and is therefore true by assumption on x, i.e., jxj < 1. Now assume that jxk+1j < jxkj for a fixed k. In that case, jxk+2j = jxjjxk+1j. The equality jxj < 1 is preserved by multiplication by a real number (this follows from the fact that the real numbers are ordered). Therefore, jxjjxk+1j < j1jjxk+1j. In other words, jxk+2j < jxk+1j, which is what we wanted to show. By induction, we conclude that jxn+2j < jxn+1j. By transitivity of inequality, we conclude that the sequence is a decreasing sequence of real numbers that is bounded below by 0, and such a sequence automatically converges by the “least 1 upper bound” property of real numbers. Since 1−x does not depend on n, the limit of the right hand side is 0 and we’re done. Remark 1.4. Now that we have functions, we can use various properties of continuity and differ- d 1 −1 entiability to analyze these expressions. For example, dx ( 1−x ) = (1−x)2 . Differentiating the left hand side of the expression above gives n 1 1 d X X X lim ( xi) = ixi−1 = ixi−1: n!1 dx i=0 i=0 i=1 3 1.2 Induction and limits 1.2 Induction and limits Proposition 1.5. The following formula holds: lim xne−x = 0 x!1 Proof. We use induction and L’Hopital’s rule. The statement for n = 0 is the assertion that −x limx!1 e = 0, which is known from calculus. The form of L’Hopital’s rule we will need states f that if we are given a limit of a fraction g , that is an “indeterminate form” (here 1=1), then: f f 0 lim = lim : x!1 g x!1 g0 Now, assume the statement for a given value of k, i.e., xk lim = 0: x!1 ex In that case, xk+1 (k + 1)xk xk lim = lim = (k + 1) lim = 0 x!1 ex x!1 ex x!1 ex where the first equality follows from L’Hopital’s rule, and the second by the induction hypothesis. The assertion then follows from the first principle of mathematical induction. 1.3 The gamma function Definition 1.6. Assume x is a real number. We define Z 1 Γ(x) := e−ttx−1dt: 0 We can compute this for various values of x, e.g., x = 1. In that case, we are computing Z 1 Z a Γ(1) = e−tdt = lim a ! 1 e−tdt: 0 0 We evaluate the RHS by the fundamental theorem of calculus as e−t has anti-derivative −e−t. Thus, we see: Z a lim a ! 1 e−tdt = lim (−e−a) − (−e−0) = lim a ! 11 − e−a: 0 a!1 −a Since lima!1 e = 0, we conclude that Γ(1) = 1. You can compute Γ(2) by integration by parts. Z 1 Γ(2) = te−tdt: 0 Take u = t and dv = e−tdt. Then du = dt and v = −e−t, so Z 1 Z a −t −t a −t te dt = lim −te j0 + e dt: 0 a!1 0 The limit we computed above shows that the first term is 0, while the previous computation shows that the other term is 1. In other words, Γ(2) = 1. The integration by parts trick can be turned into an induction arugment. 4 1.3 The gamma function Proposition 1.7. For every integer n ≥ 0, the formula Γ(n + 1) = n! holds. Proof. For n = 0, this is simply the computation we made above that Γ(1) = 1 = 0!. Therefore assume the statement holds for a fixed k. Let us analyze Γ(k + 1). In that case, by definition: Z 1 Γ(k + 1) = tk+1−1e−tdt: 0 Integration by parts with u = tk and dv = e−tdt gives du = (k)tk−1dt, v = −e−t. Then Z 1 Z 1 k −t k −t a k −t t e dt = lim −t e j0 + (k)t e dt: 0 a!1 0 The first limit is 0 by the proposition we proved above, and the second is kΓ(k) by definition. Now, assume inductively that Γ(k) = (k − 1)!. In that case, Γ(k + 1) = kΓ(k). By the induction hypothesis, this is k(k − 1)! = k! by definition, which is precisely what we wanted to show. Remark 1.8. We will define x! = Γ(x − 1) for any real number x, and the above function can be x viewed as givingp meaing to factorials for values of that are not integers. E.g., one may show that Γ(1=2) = π..
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