10.9 Representing Functions as Power Series Contemporary Calculus 1 10.9 REPRESENTING FUNCTIONS AS POWER SERIES Power series define functions, but how are these power series functions related to functions we know about such as sin(x), cos(x), ex, and ln(x)? How can we represent common functions as power series, and why would we want to do so? The next two sections provide partial answers to these questions. In this section we start with a function defined by a geometric series and show how we can obtain power series representations for several related functions. And we look at a few ways in which power series representations of functions are used. The next section examines a more general method for obtaining power series representations for functions. The foundation for the examples in this section is a power series whose sum we know. The power series ∑n=0∞ xn is also a geometric series, with the common ratio r = x, and, for |x| < 1, we know the sum of the series is 11–x . Geometric Series Formula For | x | < 1, ∑n=0∞ xn = 1 + x + x2 + x3 + x4 + ... = 11 – x . One simple but powerful method of obtaining power series for related functions is to replace each "x" with a function of x. Substitution in Power Series Suppose f(x) is defined by a power series f(x) = ∑n=0∞ anxn = a0 + a1x + a2x2 + a3x3 + ... + anxn + ... that converges for –R < x < R. Then f( xp ) = ∑n=0∞ an{ xp }n = a0 + a1xp + a2{ xp }2 + a3{ xp }3 + ... + an{ xp }n + ... = a0 + a1xp + a2x2p + a3x3p + ... + anxnp + ... converges for –R < xp < R, and f( x–c ) = ∑n=0∞ an(x–c)n = a0 + a1(x–c) + a2(x–c)2 + a3(x–c)3 + ... + an(x–c)n + ... converges for –R < x – c < R.10.9 Representing Functions as Power Series Contemporary Calculus 2 We can use this substitution method to obtain power series for some functions related to the Geometric Series Formula 11 – x . Example 1: Find power series for (a) 11 – x2 , (b) 11 + x , and (c) x1 – x . Solution: (a) Substituting "x2" for "x" in the Geometric Series Formula, we get 11 – x2 = 1 + {x2} + {x2}2 + {x2}3 + {x2}4 + .... = ∑n=0∞ ( x2 ) n = 1 + x2 + x4 + x6 + x8 + .... = ∑n=0∞ x2n for –1 < x < 1. (b) Substituting "–x" for "x" in the Geometric Series Formula, 11 + x = 11 – (–x) = 1 + {–x} + {–x}2 + {–x}3 + {–x}4 + .... = ∑n=0∞ (–x) n = 1 – x + x2 – x3 + x4 + .... = ∑n=0∞ (–1) n xn for –1 < x < 1. (c) We need to recognize that x1 – x is a product, x1 – x = x . 11 – x . Then x . 11 – x = x { 1 + x + x2 + x3 + x4 + .... } = x + x2 + x3 + x4 + x5 + .... = ∑n=0∞ xn+1 or, equivalently, ∑n=1∞ xn for –1 < x < 1. Practice 1: Find power series for (a) 11 – x3 , (b) 11 + x2 , and (c) 5x1 + x . One of the features of polynomials that makes them very easy to differentiate and integrate is that we can differentiate and integrate them term–by–term. The same result is true for power series.10.9 Representing Functions as Power Series Contemporary Calculus 3 Term–by–Term Differentiation and Integration of Power Series Suppose f(x) is defined by a power series f(x) = ∑n=0∞ anxn = a0 + a1x + a2x2 + a3x3 + ... + anxn + ... that converges for –R < x < R. Then, (a) the derivative of f is given by the power series obtained by term–by–term differentiation of f: f '(x) = ∑n=1∞ n.an xn–1 = a1 + 2a2x + 3a3x2 + 4a4x3 + ... + n.anxn–1 + ... (b) an antiderivative of f is given by the power series obtained by term–by–term integration of f: ⌡⌠ f(x) dx = C + ∑n=0∞ an xn+1n+1 = C + a0x + a1 x22 + a2 x33 + a3 x44 + ... + an xn+1n+1 + ... The power series for the derivative and antiderivative of f converge for –R < x < R. (The power series for f, f ' and the antiderivative of f may differ in whether or not they converge at the endpoints of the interval of convergence, but they all converge for –R < x < R.) The proof of this result is rather long and technical and is omitted. Like the previous substitution method, term–by–term differentiation and integration can be used to obtain power series for some functions related to the Geometric Series Formula 11 – x . Example 2: Find power series for (a) ln( 1 – x ) , and (b) arctan( x ) . Solution: These two are more challenging than the previous examples because we need to recognize that these two functions are integrals of functions whose power series we already know. (a) ln( 1 – x ) = ⌡⌠ –11 – x dx = – ⌡⌠ { 1 + x + x2 + x3 + x4 + .... } dx = – { x + x22 + x33 + x44 + .... } + C = C – ∑n=1∞ xnn . We can find the value of C by using the fact that ln( 1 ) = 0:10.9 Representing Functions as Power Series Contemporary Calculus 4 for x = 0, 0 = ln( 1 – 0 ) = C – ∑n=1∞ 0nn = C so C = 0 and ln( 1 – x ) = – { x + x22 + x33 + x44 + .... } = – ∑n=1∞ xnn for –1 ≤ x < 1. (b) arctan( x ) = ⌡⌠ 11 + x2 dx = ⌡⌠ { 1 – x2 + x4 – x6 + x8 – .... } dx = C + { x – x33 + x55 – x77 + x99 – ... } = C + ∑n=0∞ (–1) n x2n+12n+1 . We can find the value of C by using the fact that arctan( 0 ) = 0: for x = 0, 0 = arctan( 0 ) = C + ∑n=0∞ (–1) n 02n+12n+1 = C so C = 0 and arctan( x ) = x – x33 + x55 – x77 + x99 – ... = ∑n=0∞ (–1) n x2n+12n+1 for –1 ≤ x ≤ 1. Practice 2: Find a power series for ln( 1 + x ) . Power series can also be used to help us evaluate definite integrals. Example 3: Use the power series for arctan( x ) to represent the definite integral ⌡⌠00.5 arctan(x) dx as a numerical series. Then approximate the value of the integral by …