MATH 0 - A - Spring 009 Representation of Functions as Power Series. Our starting point in this section is the geometric series: x n = + x + x + x 3 + We know this series converges if and only if x <. So this is a power series in x, centred at x = 0, it has radius of convergence R =, and its interval of convergence is the open interval (, ). Furthermore, whithin this interval, we may compute explicitly the sum of the series: x n = + x + x + x 3 + =, for x <. x Let s turn the tables: If we start from the function f(x) =, the reasoning above x tells us that within the interval (, ), we can expand this function as a power series in x, so we may write or in sigma-notation, x = + x + x + x 3 +, for x < ; (*) x = x n, for x <. (**) We will now see how to use these ideas in order to find power series expansions for some other rational functions. Example. Find a power series expansion for the function of convergence. +x, and specify its interval Example. Express as the sum of a power series, and find the interval of convergence. x+ Example 3. Find a power series representation for +x. Example 4. Find a power series representation for x +x.
MATH 0 - A - Spring 009 Theorem (On differentiation and integration of power series). Suppose that the power series c n(x a) n has radius of convergence R > 0. Then the function f(x) = c 0 + c (x a) + c (x a) + c 3 (x a) 3 + = c n (x a) n is differentiable (and therefore also continuous) on the open interval (a R, a + R). Furthermore, within this interval, the function f(x) can be differentiated term-by-term f (x) = c + c (x a) + 3c 3 (x a) + = nc n (x a) n ; n= or integrated term-by-term f(x) dx = C + c 0 (x a) + c (x a) = c n (x a) n+ n + + C. + c (x a) 3 3 + c 3 (x a) 4 4 + The radii of convergence of the power series expansions of f (x) and f(x) dx are also R (the same radius of convergence as the original power series). Remark. The moral of this theorem is that, within their radius of convergence, power series may be treated as polynomials (at least for purposes of differentiation and integration). Remark. Notice the theorem states that the radius of convergence of a power series remains the same after you differentiate it or integrate it (term-by-term). However, this does not mean that the interval of convergence remains the same. There are examples where the original series converges at an endpoint, whereas the differentiated series diverges there; consider for instance the series n= x n n. On the other hand, there are also examples where the original series does not converge at one of the endpoints, whereas the integrated series converges there; consider for instance n= ( )n x n. Example 5. Find a power series expansion for /( x). What is the radius of convergence? Example 6. Find a power series representation for ln( + x), and specify its radius of convergence. Example 7. Find the sum of the series n n=. n Example 8. (a) Evaluate x +x dx as a power series. (b) Use the result of part (a) to approximate / 0 x +x dx correct to five decimal places.
MATH 0 - A - Spring 009 3 Solution of Example. The idea is to relate the given function to the sum of a +x geometric series as in equation (*). We do this by a simple algebraic manipulation: + x = ( x) = + ( x) + ( x) + ( x) 3 + = ( x) n = ( ) n x n. This series converges presisely when x < ; i.e. when < x <, so the interval of convergence is the open interval (, ). Solution of Example. Similarly here, we want to relate the function to the sum x+ of a geometric series. To express this function as something that involves the sum of a geometric series, we do a few simple algebraic manipulations: x + = + x = ( + x/) = ( x/) = ( x/) n = ( ) n x n n = ( ) n x n n+. This series converges presisely when x/ < ; i.e. when < x <, so the interval of convergence is the open interval (, ). Solution of Example 3. Once again, we manipulate algebraicaly the function +x to relate it to the sum of a geometric series: + x = ( x ) = + ( x ) + ( x ) + ( x ) 3 + = ( x ) n = ( ) n x n. This series converges presisely when x < ; i.e. when < x <, so the interval of convergence is the open interval (, ). Solution of Example 4. We now know from the previous example how to expand the function as a power series: +x + x = ( ) n x n, for < x <. Multiplying both sides of this equation by x we get x + x = x ( ) n x n = ( ) n x n+, for < x <. The interval of convergence for this power series expansion is also (, ); Solution of Example 5 (copied shamelessly from page 73 of the textbook). We start from the geometric series: x = + x + x + x 3 + = x n, for x <.
MATH 0 - A - Spring 009 4 Differentiating each side of this equation we get ( x) = + x + 3x + = nx n. For a neater presentation, we may shift the summation index setting k = n, so now we may write the power series starting from zero: ( x) = (k + )x k. k=0 According to the theorem, the radius of convergence of the differentiated series is the same as the radius of convergence of the original series, i.e., R =. Solution of Example 6. n= The idea in the solution of this example is to notice that ln( + x) = + x dx. In Example we obtained a power series expansion for +x : + x = x + x x 3 + = ( ) n x n, for x <. Integrating each side of this equality we get x ln( + x) = dx = x + x + x3 3 x4 4 + = ( ) n xn+ n +. By the theorem, the radius of convergence of the integrated series is the same as the radius of convergence of the original series, namely, R =. { It is worthwhile to note that the power series expansion we obtained for ln( + x) converges also for x =, by the AST. So this is an example where the interval of convergence of the integrated power series includes an endpoint that the original series didn t! } Solution of Example 7. The idea here is to find the function whose power series expansion is n= n x n, and then set x = /. We start from the result obtained in Example 5: ( x) = nx n, n= n= or equivalently, nx n = n= x ( x). Now differentiate both sides of the equation on the right to obtain [ n x n = d ] nx n = d [ ] x = + x dx dx ( x) ( x). 3 Next, multiplying by x we get n= n x n = n= x( + x) ( x) 3.
MATH 0 - A - Spring 009 5 These operations do not change the radius of convergence, so the last power series converges (to the indicated function) for < x <. Setting x = / we get n= n (/)( + (/)) = = 6. n ( (/)) 3 Solution of Example 8. (a) We found in Example 4 that x + x = ( ) n x n+, for < x <. Now we integrate both sides of this equation: x + x dx = ( ) n xn+ + C, for < x <. n + (b) In order to evaluate / x dx we use the antiderivative found in part (a) and the 0 +x FTC (and note it suffices to use the antiderivative with C = 0). We have n + / x 0 + x dx = ( ) n xn+ x=/ x=0 = ( ) n (n + ) n+. The sum s of the infinite series on the right is the exact value of the given definite integral. Since this is an alternating series, we know s = s n + R n, where the remainder can be estimated by R n < b n+, where b n = (n + ) n+. This comes from the theorem about estimation of the remainder in the AST. Since we want accuracy to five decimal places, we want the remainder to satisfy R n < 0 6. From the previous inequality, it suffices to require that b n+ < 0 6, or equivalently Since (n+4) ((n + ) + ) = (n+)+ (n + 4) < n+4 0. 6 <, it suffices to require < n+4 0, or equivalently, 6 4n+7 > 0 6. Upon inspection, we notice that 4 0 = 048576, which is larger than 000000 = 0 6. Hence, for the required accuracy, it suffices to take n + 7 0, i.e., n 3. This means that we may take s 3 as a good approximate value of the definite integral: / 0 x + x dx s 3 = 4 + 4 6 6 8 ; 8 and this approximate value is correct to five decimal places.