Calculus Power Series

Power Series We will review the relevant Definitions, Calculus, Strategies, Examples, and Applications. Definition: A Power Series Centered at a has the following form: ∞ X cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + c4 (x − a)4 + . . . n=0 where a is a constant value and the Coefficients cn may depend on n. Here x is the input variable. Note: Convergent for x = a. Definition: A Power Series Centered at a = 0 has the following form: ∞ X cn x n = c0 + c1 x + c2 x 2 + c3 x 3 + c4 x 4 + . . . n=0 where the Coefficients cn may depend on n. Here x is the input variable. Note: Convergent for x = 0. Think: Power Series are Infinite Versions of Polynomials. We will study their Properties and Applications. First, we can think of Power Series as functions and plug in a specific finite Real x-value. Then the Series becomes an Infinite Series, a sum of infinitely many real numbers. For Example, ∞ X xn = 1 + x + x2 + x3 + . . . is a Power Series Centered at 0. n=0 When we plug in x = 2 the Power Series becomes an Infinite Series ∞ X 2n which is a Divergent n=0 Geometric Series |r| = 2 > 1. Hence x = 2 is not in the Domain. ∞  n X 1 1 However, when we plug in x = the Power Series becomes an Infinite Series 4 4 n=0 1 1 which is a Convergent Geometric Series |r| = < 1. Hence, x = is in the Domain. 4 4 Question: What is the Domain of these function-like Power Series? That is, what is the collection of finite Real numbers that can be plugged into the Power Series and keep it finite, meaning Convergent as a series. Definition: The Interval of Convergence is the collection of finite Real numbers at which the Power Series is a finite Infinite Series. That is, the Domain of the Power Series. Definition: The Radius of Convergence is half the length of the full Interval of Convergence. This essentially captures the size of the Domain Interval. 1 Domains We have a full classification of the Domain of a Power Series ∞ X cn (x − a)n . One of the n=0 following Three things must happen: 1. The Power Series converges only at the Center point x = a. I = {a} R=0 2. The Power Series converges for all Real numbers I = (−∞, ∞) R=∞ 3. The Power Series converges on a finite Interval Centered at the Center Point a and possibly one or both of the endpoints. I = (a − R, a + r) or I = (a − R, a + r] or I = [a − R, a + r) or I = [a − R, a + r] Radius = R To find the Domain of a Power Series we have two main options: 1. If the Series is Geometric, use the Geometric Series Test. This is beneficial since the Geometric Series Test automatically gives Divergence at the endpoints of the Domain Interval. You do not need to test endpoints for the Geometric Examples. 2. Run the Ratio Test. Here you must manually check convergence at the endpoints for the finite interval case, since the Ratio Test is Inconclusive at the endpoints, that is, where L = 1. Example: Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. ∞ X (x − 6)n 7n n=0 The Geometric Series Test yields Convergence for x−6 < 1 and Divergence otherwise. 7 Here we need |x − 6| < 7 ⇒ −7 < x − 6 < 7 ⇒ −1 < x < 13 and finally, the Interval of Convergence is given by I = (−1, 13) and the Radius is given by R = 7. Note that the Center of the Domain Interval is indeed the Center of the Power Series, x = 6 here. Cool! 2 We will now present an example of each of the three Domain options, using the Ratio Test. Example (Finite Interval): Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. ∞ X (−1)n (9x − 4)n (n + 1) 5n n=1 Use Ratio Test. L = lim n→∞ = an+1 an   (−1)n+1 (9x − 4)n+1 1 (9x − 4)n+1  n + 1 5n (n + 2) 5n+1 = lim = lim ·  · n+1 n→∞ n→∞ (−1)n (9x − 4)n (9x − 4)n n+2 5 n (n + 1) 5 |9x − 4| 5 The Ratio Test gives convergence for x when |9x − 4| < 1 or |9x − 4| < 5. 5 1 That is −5 < 9x − 4 < 5 =⇒ −1 < 9x < 9 =⇒ − < x < 1 9 Manually test Convergence (where L = 1) at the Endpoints: ∞ X (−1)n (9(1) − 4)n • Take x = 1. The original series becomes (n + 1) 5n n=1 ∞ ∞ X X (−1)n 5n (−1)n = = which is convergent by AST: (n + 1) 5n n+1 n=1 n=1 1. bn = 1 >0 n+1 1 =0 n+1 1 1 1 1 3. bn+1 = < = bn or becau se f (x) = has derivative f 0 (x) = − < n+2 n+1 x+1 (x + 1)2 0 so the terms are decreasing. 2. lim bn = lim n→∞ n→∞ ** Turn the page for the other endpoint check ** 3 Other endpoint: n    1 −4 9 − ∞ (−1) ∞ X X 1 (−1)n (−5)n 9 • Take x = − . The original series becomes = 9 (n + 1) 5n (n + 1) 5n n=1 n=1 n = ∞ X (−1)n (−1)n 5n n=1 (n + 1)5n 1  even * ∞ = ∞ *  2n  X  (−1)  n=1 n+1 = X n=1 ∞ X1 1 ≈ Div. harmonic p-series p = 1. n + 1 n=1 n 1 n 1 = lim = 1 which is finite and non-zero. LCT: lim n + 1 = lim 1 1 n→∞ n + 1 n→∞ n→∞ 1+ n n ∞ X 1 Therefore, is also divergent by LCT. n+1 n=1   1 5 Finally, Interval of Convergence I = − , 1 with Radius of Convergence R = . 9 9 Example (Infinite Interval): Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. ∞ X (−1)n xn (2n + 1)! n=1 Use Ratio Test. (−1)n+1 xn+1   an+1 xn+1 (2n)! (2(n + 1)! L = lim = lim = lim · n→∞ n→∞ n→∞ (−1)n xn an xn (2n + 2)! (2n)! :0  |x|  = lim  ∞ ∞0 < 1 :  :=    n→∞  (2n + 2) (2n  + 1)   So the Series Converges by the Ratio Test for all Real numbers x. Finally, the Interval of Convergence is I = (−∞, ∞) and the Radius is R = ∞ . 4 Example (Collapsed to a (Center) point Interval): Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. ∞ X (3n)!(x − 6)n n=1 Use Ratio Test.   an+1 (3(n + 1))!(x − 6)n+1 (x − 6)n+1 (3n + 3)! L = lim = lim = lim · n→∞ n→∞ n→∞ an (3n)!(x − 6)n (x − 6)n (3n)! : ∞   : ∞  :∞ = lim |x − 6| (3n + 3) (3n (3n   + 2)  + 1) = ∞ > 1 n→∞ So the Series Diverges by the Ratio Test for all Real numbers x UNLESS x − 6 = 0 ⇒ x = 6 (when L = 0 < 1 which yields convergence by the Ratio Test). Finally, the Interval of Convergence is I = {6} and the Radius is R = 0 . Term-by-Term Differentation and Integration of Power Series Next, we can Differentiate or Integrate Power Series Term-by-Term. That is, given a Power ∞ X cn xn with Radius of Convergence R > 0 then we can compute its Derivative or Series n=0 Integral and the Radius remains unchanged. (The convergence may change at the endpoints.) ∞ ∞ X d X cn x n = cn nxn−1 dx n=0 n=0 and Z X ∞ n=0 Example: ∞ ∞ d X (−1)n xn X (−1)n n xn−1 = dx n=0 n! n! n=0 Example: Z X ∞ ∞ X (−1)n x4n+2 (−1)n x4n+3 dx = +C (2n)! (2n)! (4n + 3) n=0 n=0 5 n cn x dx = ∞ X n=0 cn xn+1 +C n+1 Power Series Representations for Functions Question: Does a function have a Power Series Representation? We start with the Geometric Series form: ∞ X 1 = xn = 1 + x + x2 + x3 + . . . 1 − x n=0 valid for |x| < 1 by the Geometric Series Test Next, we can extend this formula to find new Power Series for other functions: ∞ ∞ X X 1 1 Example: = = (−x)n = (−1)n xn = 1 − x + x2 − x3 + x4 + . . . 1+x 1 − (−x) n=0 n=0 valid when | − x| = |x| < 1 using the Geometric Series Test. Here the Radius is R = 1. ∞ Example: ∞ X X
1 1 2 n = = −x (−1)n x2n = 1 − x2 + x4 − x6 + x8 + . . . = 2 2 1+x 1 − (−x ) n=0 n=0 valid when | − x2 | = |x2 | < 1 or |x| < 1 using the Geometric Series Test. Here the Radius is R = 1. ∞ Example: valid when | − 5x| < 1 or |x| < 3 Example: = ∞ X X 1 1 = = (−5x)n = (−1)n 5n xn = 1 − 5x + 52 x2 − 53 x3 + . . . 1 + 5x 1 − (−5x) n=0 n=0 x = x3 6+x  1 1 using the Geometric Series Test. Here the Radius is R = . 5 5 1 6 − (−x) 3  =   3 x  1 x  x = 6 1− − 6 6 ∞  X x n − 6 n=0 ! ∞ ∞ x3 X (−1)n xn X (−1)n xn+3 = 6 n=0 6n 6n+1 n=0 valid when − x < 1 or |x| < 6 using the Geometric Series Test. Here the Radius is R = 6. 6 6 More Power Series Representations for functions using Intergation Next, we explored the Power Series for ln(1 + x) and arctan x. We derived these Series using Integration of a known Power Series (derived from the Geometric Series Formula) Z ln(1 + x) = = ∞ X (−1)n n=0 P.S. Z Z X Z X ∞ ∞ 1 1 n (−1)n xn dx (−x) dx = dx = dx = 1+x 1 − (−x) n=0 n=0 xn+1 x 2 x3 x4 +C =x− + − + ... + C n+1 2 3 4 Expanding the Power Series a few terms and plugging in the Center point value x = 0 we can show that C = 0 here 0 = ln(1 + 0) = 0 − 0 + 0 − 0 + . . . + C → C = 0 Finally, ln(1 + x) = ∞ X n=0 Z Next, arctan x = = (−1)n xn+1 n+1 P.S. Z Z X Z X ∞ ∞ >
1 1 2 n dx = (−1)n x2n dx dx = −x  2 dx = 2) 1 + x 1 − (−x  n=0 n=0 ∞ X x2n+1 x 3 x5 x7 (−1)n +C =x− + − + ... + C 2n + 1 3 5 7 n=0 Expanding the Power Series a few terms and plugging in the Center point value x = 0 we can show that C = 0 here 0 = arctan 0 = 0 − 0 + 0 − 0 + . . . + C → C = 0 ∞ X x2n+1 (−1)n Finally, arctan x = 2n + 1 n=0 Taylor and MacLaurin Series We’ve been able to find Power Series Representations for a restricted class of functions, related to Geometric Series. Now we want to try and generalize. Definition: The Taylor Series for a function f at a is given by ∞ X f (n) (a) n=0 n! (x − a)n = f (a) + f 0 (a) f 00 (a) f 000 (a) (x − a) + (x − a)2 + (x − a)3 + . . . 1! 2! 3! 7 Definition: For the special case when a = 0, the Taylor Series centered at a = 0 is called the MacLaurin Series for a function f at 0. It is given by ∞ X f (n) (0) n=0 n! xn = f (0) + f 0 (0) f 00 (0) 2 f 000 (0) 3 x+ x + x + ... 1! 2! 3! We used this Definition (or Chart Method) to gain Power Series for sin x and ex . Review these computations. We differentiated the P.S. for sin x to find P.S for cos x. Key Fact: If a function f has a Power Series centered at a given a, then the Series must be the Taylor Series (or MacLaurin if a = 0). Here is a chart of the 6 MacLaurin Series that we derived: e x = ∞ X xn n=0 n! =1+x+ x2 x3 x 4 x5 + + + + ... 2! 3! 4! 5! sin x ∞ X (−1)n x2n+1 x 3 x5 x7 x9 =x− + − + + ... = (2n + 1)! 3! 5! 7! 9! n=0 cos x ∞ X (−1)n x2n = (2n)! n=0 1 1−x = arctan x = ∞ X xn =1− x2 x4 x6 x 8 + − + + ... 2! 4! 6! 8! = 1 + x + x2 + x 3 + x4 + x5 + . . . n=0 ln(1 + x) = ∞ X x 3 x5 x7 x9 (−1)n x2n+1 =x− + − + + ... 2n + 1 3 5 7 9 n=0 ∞ X (−1)n xn+1 n+1 n=0 =x− 8 x 2 x3 x4 x5 + − + + ... 2 3 4 5 Helpful Tip: There are 4 Main Options for finding a functions Power Series representation: 1. 2. 3. 4. Using Substitution into a known Power Series Differentiating a known Power Series. Integrating a known Power Series. Sometimes messier with solving for +C. Using the Definition above, sometimes known as the Chart Method. → This option has its limitations if the derivatives are complicated. Think about which options we have used so far for each function. Fun Challenges: • See if you can use 3 options to find the MacLaurin Series for cos x or sin x. // •////// See/// if////// you////// can////// use///// all/// 4/////////// options//// to/////// find////// the///////////////// MacLaurin///////// Series///// for////////// cosh x//// or/////////// sinh x. Applications We investigated several Applications for Power Series. • New Series • New Indefinite Integrals • New Sums • New Estimates using the Alternating Series Estimation Theorem • New Limits New Series, using substitution into a known Series Example: ∞ ∞ 2n+1 2n+1 X X x 32n+1 x2n+6 n3 =x (−1) = x arctan(3x) = x (−1) (−1)n 2n + 1 2n + 1 2n + 1 n=0 n=0 n=0 5 5 ∞ X n (3x) 2n+1 5 1 1 Here need |3x| < 1 or |x| < , so R = . 3 3 x Example: =x 1 + 7x  1 1 − (−7x)  =x ∞ X n (−7x) = x n=0 ∞ X n=0 1 1 Here need | − 7x| < 1 or |x| < , so R = . 7 7 9 ∞ X (−1) 7 x = (−1)n 7n xn+1 n n n n=0 New Indefinite Integrals Z Example: Use Series to compute Z P.S. dx =  2 : 4−x xe Z x 4 ∞ X (−x2 )n n! n=0 = 2 x4 e−x dx Z dx = x 4 ∞ X (−1)n x2n n! n=0 Z X ∞ (−1)n x2n+4 dx = dx n! n=0 ∞ X (−1)n x2n+5 +C (n!)(2n + 5) n=0 Note: Here we can leave the +C since we don’t have a given function on the left side of the equality chain to plug into to solve for +C. New Sums: Pattern matching in reverse ∞ X n ∞ X 2n n ∞ X 2n  π 2n (−1)n (−1) π (−1) π 6 = = n (2n + 1)! 2n (2n + 1)! (36) (6) (2n + 1)! n=0 n=0 n=0  π 2n  π   π 2n+1 n ∞ (−1)n ∞ π  6 1 X 6 6 X (−1) 3 6 6 6   · π = = sin = = = (2n + 1)! π n=0 (2n + 1)! π 6 π 2 π n=0 6 Example:  Example: ∞ ∞ ∞ X 1 X (−1)n (ln 8)n 1X (−1)n (ln 8)n = = 3n+1 n! 3 n=0 3n n! 3 n=0 n=0 − ln 8 3 n! n − ln 8     @ “ A ” 1 1 ln 8− 13 1  −1  1 1 1 1 1 3 = e = e = 8 3 = = = 1 3 3 3 3 83 3 2 6 0 1 ∞ X n 2n (−1) π = 4n (2n)! 2 n=0 √ π  1 2 = cos = √ = 4 2 2 Example: ∞ X n=0 n 2n (−1) π = (22 )2n (2n)! ∞ X n=0 2n (−1) π = 42n (2n)! π3 π5 π7 π9 Example: π − + − + − . . . = sin π = 0 3! 5! 7! 9! 10 n ∞ X n=0 (−1)n  π 2n 4 (2n)! Example: − π2 π4 π6 π8 + − + − . . . = (cos π) − 1 = −1 − 1 = −2 2! 4! 6! 8!   1 1 1 1 1 1 1 1 Example: −1 + − + − + . . . = − 1 − + − + − . . . = − ln(1 + 1)) = − ln 2 2 3 4 5 2 3 4 5 New Estimates for Values of Alternating Series Example: Estimate cos(1) with error less than cos(1) = ∞ X (−1)n (1)2n (2n)! n=0 ≈1− =1− 1 . Justify. 100 1 1 1 1 1 1 + − + ... = 1 − + − + ... 2! 4! 6! 2 24 720 1 1 24 12 1 13 + = − + = 2 24 24 24 24 24 ← estimate Using the Alternating Series Estimation Theorem (ASET), we can approximate the actual 13 sum with only the first three terms as , with error at most the absolute value of the first 24 1 1 1 neglected term, < as desired. . Here 6! 720 100 New Estimates for Definite Integrals 1 Z x3 ln(1 + x3 ) dx with error less than Example: Estimate 0 Z 1 3 3 Z x ln(1 + x ) dx = 0 0 ∞ 1X n=0 x7 x10 x13 = − + − ... 1 · 7 2 · 10 3 · 13 ∞ 1 . Justify. 30 X (−1)n x3n+7 (−1)n x3n+6 dx = n+1 (n + 1)(3n + 7) n=0 1 0 x7 x10 x13 = − + − ... 7 20 39 1 0 1 0 1 1 1 − + − . . . − (0 − 0 + 0 − . . .) 7 20 39 1 1 20 7 13 ≈ − = − = ← estimate 7 20 140 140 140 = Using the Alternating Series Estimation Theorem (ASET), we can approximate the actual sum with only the first two terms, and the error from the actual sum will be at most the 1 1 1 absolute value of the next (first neglected) term, . Here < as desired. 39 39 30 11 Z Example: Estimate 1
x sin x2 dx with error less than 0 Z 1 2 Z 1 x sin(x ) dx = 0 x 0 ∞ 2n+1 X (−1)n (x2 ) (2n + 1)! n=0 ∞ 1X 1 Z x dx = 0 ∞ X 1 . Justify. 1000 ∞ X (−1)n x4n+2 n=0 (2n + 1)! dx 1 (−1)n x4n+3 (−1)n x4n+4 x4 x8 x12 dx = = − + − ... = (2n + 1)!(4n + 4) 0 1 · 4 3! · 8 5! · (12) 0 n=0 (2n + 1)! n=0   1 11 1 1 1 1 = − + − . . . − (0 − 0 + 0 − . . .) ≈ − = ← estimate 4 48 1440 4 48 48 Z 1 0 Using the Alternating Series Estimation Theorem (ASET), we can approximate the actual 11 sum with only the first two terms as , with error at most the absolute value of the first 48 1 1 1 neglected term, < as desired. . Here 1440 1440 1000 New Limits using Series Example: Use series to evaluate the following limit   x2 x4 x6 x 1− + − + ... x cos x 2! 4! 6! lim = lim x→0 sin x x→0 x 3 x5 x7 x− + − + ... 3! 5! 7!   :0   6 x2 x4 x6 x2 x4 x  + − + ... x 1− 1 − + + ... − 2! 4! 6!  2!  4! 6!  = lim = lim  = 1 :0  x→0 x→0 x2 x4 x6 4 6 2 x x x  + − + ... x 1− 1 − + + ... − 3! 5! 7!  3! 5! 7! In closing, there are many more subtopics to study for Power Series, but we have a great base of material to build off of for now. Focus on the • Definitions • Domains • Differentiation/Integration • Derivation of Power Series for Functions, Multiple Methods • Memorize the 6 Derived MacLaurin Series, study how each one was derived. • Many Applications 12