AP Calculus Chapter 9: Infinite Series

AP Calculus Chapter 9: Infinite Series

9. Sequences a, a 2, a 3, a 4, a 5,... Sequence: A function whose domain is the set of positive integers n = 2 3 4 a n = a a 2 a 3 a 4 terms of the sequence Begin with the pattern or nth term. What s the sequence of numbers? a n = ( ) n (2n) Graph each sequence on your graphing calculator in SEQ MODE. Use the trace feature to view the terms of the sequence. a n = 2 2 n

Define a sequence recursively. a = 0 a 2 = 3 a 3 = 6 a 4 = 9 a n+ = a n + 3 Begin with a sequence of numbers. What s the pattern or nth term? Finding the nth term of a sequence can sometimes be a little tricky. Recognizing a few common forms can be helpful. For example 2 n = 2, 4,8,6, 2n = 2, 4,6,8,0, 2n =,3,5,7, 2 n =,3, 7,5, ( ) n n =,2, 3, 4, 5,6, n! =,2,6,24,20, ), 2, 4, 8, 6,... a n = 2 n 2) 2, 4 2, 8 6, 6 24, 32 20, 64 720, a n = 3) 3 7, 5 0, 7 3, 9 6, 9,... a n = 4) 3 2, 4 3, 5 4, 6 5, a n = 5) 3 4 5 2, 2 3, 3 4, 4 5, 5 6,... a n = Graph sequence # and #4 above and observe the terms. What s happening with these sequences as n becomes large?

Convergence and Divergence: When a sequence approaches a particular value for large values of n, we say that the sequence converges. More formally, we use the concept of a limit. Limit of a Sequence Let L be a real number. Let f be a function of a real variable such that If a n lim f (x) = L. x { } is a sequence such that f (n) = a n for every positive integer n, then lim x a n = L Graph u(n) = / (2n) using sequence mode. Graph the sequence and trace the points to see the values of the sequence. What is the limit of this sequence based on the graph? lim n Evaluate the limit analytically: 2n When the limit exists, the sequence CONVERGES. When the limit does not exist, the sequence DIVERGES. Determine the convergence or divergence of the sequence.. a n = 3 + ( ) n 2. a n = n 2n 3. a n = n2 2 n 4. a n = ( ) n 4 5. a n = ( )n n 6. a n = 2n n! 7. a n = n3 3 8. a (n 2)! n n = n!

Monotonic Sequence: A sequence is monotonic when its terms are Non-decreasing or non-increasing. See examples at right. If a sequence is monotonic and bounded then it converges! Bounded Sequences:

9.2 Series and Convergence Two basic questions for this chapter: Does a series converge or diverge? If a series converges, what is the sum? Review: A sequence converges if A series is just the sum of a sequence: a n = a + a 2 + a 3 + + a n + Infinite Series A series can be analyzed as a sequence of partial sums: S = a S 2 = a + a 2 S 3 = a + a 2 + a 3 S n = a + a 2 + a 3 + + a n Convergence or Divergence of a Series If the sequence of partial sums, { S n }, converges to S, then the series converges and the sum is S. If the sequence of partial sums diverges, then the series diverges. Find the sum of each infinite series. Examine the sequence of partial sums.. = 2 n 2 + 4 + 8 + 6 + 2. = + + + + Geometrically determine the sum from #4, using the image from Larson, p. 609.

Geometric Series: A geometric series with ratio r: ar n = a + ar + ar 2 + ar 3 + + ar n + a 0 n=0 Convergent: when 0 < r < and converges to S = a r Divergent: when r Determine whether the geometric series converges or diverges, and its sum, if it exists. 3. 0. + 0.0 + 0.00 + 0.000 +... 4. 3 0 + 3 00 + 3,000 + 3 0,000 + 5. 5 00 + 5 0,000 + 5,000,000 + 6. 3 2 n n=0 7. n=0 3 2 n

Telescoping Series: ( b b 2 ) + ( b 2 b 3 ) + ( b 3 b 4 ) + A telescoping series will converge if and only if b n approaches a finite number as n. If the series converges, its sum is S = b lim n (b n+ ). 8. Determine the sum of the telescoping series: n n + = 2 + 2 3 + 3 4 + 9. Use partial fractions to write this series as a telescoping series and determine its sum, if it exists. 2 4n 2

n th Term Test for Divergence If lim n a n 0, then a n diverges, i.e. if the last term does not approach 0, the series is DIVERGENT. If a n converges, then lim a n = 0, i.e. if the series is CONVERGENT, the last term n approaches 0. If the last term approaches 0, the series could be CONVERGENT or DIVERGENT. NOTE: The n th term test is merely a test for DIVERGENCE!! Apply the n th Term Test for Divergence to determine whether or not a series is DIVERGENT. 0. 2 n. n=0 n! 2. 2n! + n

9.3 The Integral Test and p-series Consider the area under the curve from to. Consider the sum of the function heights for integer values from to. Since the area under the curve can be approximated by a series of rectangles having the same height as the function heights, does it make intuitive sense that the convergence or divergence of these two sums would behave in the same way. That is the essence of the Integral Test. Either both sums converge or both sums diverge. 0.4 0.2 2 3 4 5 6 7 8-0.2 The Integral Test If a function, f is positive, continuous and decreasing for x, and a n = f (n), then a n and f (x)dx either both converge or both diverge. THIS DOES NOT IMPLY THAT THEY CONVERGE OR DIVERGE TO THE SAME VALUE ONLY THAT EITHER BOTH CONVERGE OR BOTH DIVERGE.

Example : Apply the Integral Test to the series n. n 2 + 0.4 0.2 The series could be written 2 + 2 5 + 3 0 + 4 7 + 2 3 4 5 6 7 8 Does this series converge or diverge? -0.2 x Does the function f (x) = meet the conditions of the Integral Test? x 2 + a) Is f (x) positive for x >? (Hint: examine the function) b) Is f (x) continuous for x >? (Hint: any undefined values for x? any breaks in the graph?) c) Is f (x) decreasing for x >? (Hint: is the derivative negative for x >?) Because the function meets of the conditions of the integral test, we can apply it. x x 2 + a x dx = lim a x 2 + dx = let u = x 2 + and du = 2x du 2 = u 2 lim ln(x 2 a +) a = 2 lim ln a a2 + ( ) ln 2 ( ) = Since the improper integral is divergent, the series is also divergent.

Example 2: Apply the Integral Test to the series = n 2 + 2 + 5 + 0 + 7 + Does this series converge or diverge? Does the function f (x) = meet the conditions of the Integral Test? x 2 + a) Is f (x) positive for x >? b) Is f (x) continuous for x >? c) Is f (x) decreasing for x >? ***Note that even though the integral converges to π / 4, that does not mean that the series converges to π / 4. It merely means that the series converges by the integral test because the improper integral converges.***

Example 3: For each series below, confirm that the Integral Test can be applied to the series. Use the Integral Test to determine convergence or divergence. 2 3n + 5 a) Is f (x) positive for x >? b) Is f (x) continuous for x >? c) Is f (x) decreasing for x >? n ln n a) Is f (x) positive for x >? b) Is f (x) continuous for x >? c) Is f (x) decreasing for x >?

n n 4 + 2n 2 + a) Is f (x) positive for x >? b) Is f (x) continuous for x >? c) Is f (x) decreasing for x >? 3 + 5 + 7 + 9 +... a) Is f (x) positive for x >? b) Is f (x) continuous for x >? c) Is f (x) decreasing for x >?

Example 4: Explain why the Integral Test does not apply to the series a) Is f (x) positive for x >? b) Is f (x) continuous for x >? c) Is f (x) decreasing for x >? 2 e n cosn.5 0.5 5 0-0.5 p-series and Harmonic Series p-series: n p = + p 2 + +. p p 3.. Converges if p > Diverges if 0 < p Harmonic Series: = n + 2 + 3 +... (special case of the p-series with p =, the only p-series with a special name) Example 5: Determine the convergence or divergence of the p-series. 3 n 5/3 n.04 n 0.99 + 3 4 + 3 9 + 3 6 + 3 25 +... 5 n

9.4 Comparison of Series Direct Comparison Test: This test allows you to compare a series having complicated terms with a simpler series whose convergence or divergence is known. For example: is not a geometric series, but resembles the simpler series which is 2 + 3 n 3 n geometric. The Direct Comparison Test applies only to series with positive terms. For a given series, if its terms are less than a convergent series, then it must also converge. (If the series in question is below a known convergent series it must converge.) For a given series, if its terms are greater than a divergent series, then it must also diverge. (If the series in question is above a known divergent series it must also diverge.) Direct Comparison Test Let 0 < a n b n for all n.. If b n converges, then a n converges. The a n terms are below the b n terms. 2. If a n diverges, then b n diverges. The b n terms are above the a n terms. If the larger series converges, the smaller series must also converge. If the smaller series diverges, the larger series must also diverge.

Example : Determine the convergence or divergence of 3 + 2 n Compare this series to the convergent geometric series 2 n Term by term comparison shows that 0 < 3 + 2 n < 2 n. Since the larger series, 2 n, converges, the smaller series, 3 + 2 n, must also converge. Therefore, by the Direct Comparison Test, the given series converges. 3 + 2 n Example 2: Determine the convergence or divergence of Compare this series to the divergent p-series 2 + n n /2 Term by term comparison shows that 0 < requirements for divergence. 2 + n which does not meet the /2 n Try comparing to another divergent series, the harmonic series n Term by term comparison shows that 0 < for divergence. n 2 + n which does meet the requirements Therefore, by the Direct Comparison Test, the given series diverges. 2 + n

Example 3: Use the Direct Comparison Test to determine the convergence or divergence of the series. 3n 2 + 2 4 n 5 n + 3 n 3 +

9.5 Alternating Series A series with alternating positive and negative terms is called an alternating series. For example: ( ) n n = + 2 3 + 4 5 +... ( ) n+ n = 2 + 3 4 + 5... odd terms are negative or even terms are negative The Exploration below will help you to understand what is meant by the convergence of an alternating series. Consider the alternating harmonic series: ( ) n+ = n 2 + 3 4 +... Find S, S 2, S 3, S 4..., S 0 to two decimal places and plot each on the line below. This line has been started for you. 0.50.00 S 2 S Describe the pattern of the S n. Based on this pattern, do you think the series converges? Where is S? (Write an inequality involving S.) Thanks to Ruth Dover, IMSA for this exploration. We can test for the CONVERGENCE of these series based on the Alternating Series Test.

Alternating Series Test Let a n > 0. The alternating series ( ) n a n and converge if the following two conditions are met.. lim a n = 0 and n ( ) n+! 2. a n+ a n, for all n i.e. The limit of the nth term is zero and the terms of the series are not increasing. Determine the convergence or divergence of the following series: a n ) ( ) n n Are the two conditions met? ) lim n n = 2) Is n + n for all n? Does the series converge? 2) n ( 2) n 3) ( ) n+ (n + ) = 2 n 3 2 + 4 3 5 4 + 6 5...

Approximating the Sum of an Alternating Series Alternating Series Remainder If a convergent alternating series satisfies the condition a n+ a n, then the absolute value of the remainder R N is less than or equal to the first neglected term. That is, S S N = R N a N + In other words, add up a limited number of the first terms in the series and the sum of the series will be different by no more than the value of the next term in the series. Calculator tip: To find the sum of a sequence. 2 nd STAT MATH sum( 2 nd STAT OPS seq( sum( seq( expr, variable, start, end, step) Paste Enter Can use 2 nd Entry to go back to the input above and change parts of it. Approximate the sum of each series below by using the first 6 terms. Use the next term of the series to approximate the error. 5) ( ) n+ n! =! 2! + 3! 4! + 5! 6! +... 6) 4( ) n+ ln(n + )

More on the error in approximating the sum of an Alternating Series The example below answers three essential questions. ) What is the magnitude of the error in calculating the sum of an alternating series based on a given n number of terms? Consider the alternating series ( ) n+ n If we approximate the sum, S, of this alternating series by using n = 5 terms, what will be the magnitude of the error? In other words, what is the value of the n + term, in this case, the 6 th term? Begin by finding R5 = a 6 (i.e., the Remainder for a Sum of 5 terms is the 6 th term) 2) Can we write the sum as an approximation within a particular interval? Find S 5, and write an inequality about S. S 5 R 5 S S 5 + R 5 3) How many terms, n, must we include to make sure that the error is less than a given value? Begin with a n+ 0.0

Absolute and Conditional Convergence NOT EMPHASIZED ON AP CALCULUS BC EXAM If the series converges, then the series converges absolutely. a n a n If the series converges but the series diverges, then the series a n converges conditionally. a n a n Example: a " = ( ) " ) " a " converges, a " diverges. a " converges conditionally Determine whether each series is convergent or divergent. Classify any convergent series as absolutely or conditionally convergent. 7) ( ) n n! = 0! 2 n 2! 0 2 + 2! 2 3! 2 3 +... 2 n=0 8) ( ) n = n + 2 3 + 4... 9) ( ) n(n+)/2 = 3 9 + 27 + 8... 3 n

9.6 The Ratio Test The Ratio Test is particularly useful for series that converge rapidly such as series involving factorials or exponentials. Ratio Test Let a n be a series with nonzero terms.. a n converges absolutely if lim n a n+ a n <. a 2. a n diverges if lim n+ n a n > or if lim n a n+ a n =. 3. The Ratio Test is inconclusive if lim n a n+ a n =. Determine the convergence or divergence of the series: ) n=0 2 n n! 2) n=0 n 2 2 n+ 3 n ( ) n 3) n n +

Strategies for Testing for Convergence or Divergence: You now have 8 tests for determining convergence or divergence. How do you know which to choose??? Courtesy of Lin McMullin Let the following questions guide your decision.. Does the nth term approach 0? If not, the series diverges! 2. Is the series one of the special types? Geometric, p-series, telescoping, or alternating? 3. Can the Integral test (useful when the nth term can be easily integrated), or the Ratio Test (useful with functions that increase rapidly like exponentials or factorials) be applied? 4. Can the series be compared favorably with one of the special types? Let s look at some examples and select an approach.

. n + 3n + 2. π 6 n 3. ne n2 3 4. ( ) n 4n + 5. n! 0 n 6. n! 3 n n 7. n 3 8. n 2 + 9. 3 + n 0. 3n 3n +

9.7 Taylor Polynomials and Approximations Our goal is to find a polynomial function P that approximates another function f. We begin by choosing some number c such that the value of f and P at c are the same, i.e. P(c) = f (c). Not only must that be true, but at point c, P'(c) = f '(c). If this is true, we say that the approximating polynomial P is expanded about or center at c. With these two requirements, we can obtain a simple linear approximation of f using a first degree polynomial function P. Example : For the function f (x) = e x, find a first-degree polynomial function P(x) = a x + a 0, a line, whose value and slope agree with the slope of f at x = 0. Given that P(0) = f (0), determine the constant a 0 in the equation P(x) = a x + a 0 Given that P'(0) = f '(0), determine the slope a in the equation P(x) = a x + a 0. Put it all together in a st degree polynomial approximation: P(x) = The graph of P(x) along with f (x) = e x is below. You can tell that at points near (0, ) the graph of the function f (x) and its approximation P(x) are very close. However, as we move away from (0,), the approximation is not good and the two graphs move further apart.

To improve the approximation, we can further require that the value of the second derivatives of f (x) and P(x) agree at x = 0. Let s develop a 2 nd degree polynomial approximation P 3 (x) = a 2 x 2 + a x + a 0 for f (x) = e x. Given that P(0) = f (0), find a 0 Given that P'(0) = f '(0), find a Given that P''(0) = f ''(0), find a 2 Put it all together in a 2 nd degree polynomial approximation: P 2 (x) = The graphs of P 2 (x) and f (x) = e x are shown below. Not bad! But we can do better! Develop a 3 rd degree polynomial approximation P 3 (x) = a 3 x 3 + a 2 x 2 + a x + a 0 for f (x) = e x. Given that P(0) = f (0), find a 0 Given that P'(0) = f '(0), find a Given that P''(0) = f ''(0), find a 2 Given that P'''(0) = f '''(0), find a 3 Put it all together in a 3 rd degree polynomial approximation: P 3 (x)= The graphs of P 3 (x) and f (x) = e x are shown below. We could continue creating higher order polynomial approximations to get a better fit!

Use your graphing calculator to find numerical values for the function f (x) = e x and its 3 rd degree polynomial approximation P 3 (x) = a 3 x 3 + a 2 x 2 + a x + a 0. x -.0-0.2-0.0 0 0. 0.2 f (x) = e x P 3 (x) What do you notice about the values for f (x) = e x and further away from x = 0? and P 3 (x) as you examine values further It is not necessary that the polynomial be centered at x = 0. It can also be centered at some number c. This work was done by English mathematicians Brook Taylor (685-73) and Colin Maclaurin (698-746). Definition of nth Taylor and Maclaurin Polynomials If f has n derivatives at c, then the polynomial P n (x) = f (c) + f '(c)(x c) + f ''(c) 2! is called the nth Taylor polynomial for f at c. (x c) 2 + f '''(c) 3! (x c) 3 +!+ f (n) (c) (x c) n n! If c = 0, then P n (x) = f (0) + f '(0)x + f ''(0) 2! x 2 + f '''(0) 3! x 3 +!+ f (n) (0) x n n! is called the nth Maclaurin polynomial for f at c. The Maclaurin Polynomial is just a special case of the Taylor Polynomial.

Example 2: Find the Taylor polynomials P, P2, P3, P4 for f (x) = ln x centered at c =. To start, find f () =. Now find: f '(x) = and f '() = f ''(x) = and f ''() = f '''(x) = and f '''() = f (4) (x) = and f (4) () = So now we can write: P (x) = P 2 (x) = P 3 (x)= P 4 (x) =

Example 3: Find the Maclaurin polynomials P0, P2, P4, P6 for f (x) = cos x. Because it is a Maclaurin Polynomial, you will expand around c = 0. To start, find f (0) =. Now find: f '(x) = and f '(0) = f ''(x) = and f ''(0) = f '''(x) = and f '''(0) = f (4) (x) = and f (4) (0) = Repeated differentiation gives the pattern So now we can write: P 0 (x) = P 2 (x) = P 4 (x) = P 6 (x) = Use P 6 (x) to approximate cos 0.2. What is the difference between your approximation for cos 0.2 and what you get using the cosine key on your calculator?

Example 4: Find the Maclaurin polynomials P, P3, P5, P7 for f (x) = sin x. Use the result to approximate sin 0.2. Example 5: Find the 4 th Maclaurin polynomial to approximate the value of ln.. You could use f (x) = ln x but a polynomial centered at x = 0 isn t going to give a great approximation for a value at x =.. So, let s use f (x) = ln(x +) centered at x = 0 and approximate x = 0.. Example 6: Let f (t) be a function that has derivatives for all orders for all real numbers. Given f (0) = 5, f '(0) = 3, f ''(0) =, f '''(0) = 4, write the 3 rd degree Maclaurin polynomial for f (t).

Accuracy and the Lagrange Remainder An approximation technique is of little value unless we have some idea of how accurate it is. To measure the accuracy of a Taylor or Maclaurin polynomial for a function f (x), we use the concept of a remainder R n (x) defined as follows: f (x) = P n (x) + R n (x) exact value = approximate value + remainder It follows that R n (x) = f (x) P n (x) and we call the absolute value of R n (x) the error associated with the approximation. Lagrange form of the Remainder: R n (x) = f (x) P n (x) Taylor s Theorem If a function f (x) is differentiable through order n + in an interval I containing c, then for each x in I, there exists a z between x and c such that : f (x) = f (c) + f '(c)(x c) + f ''(c) 2! (x c) 2 + f '''(c) 3! (x c) 3 +!+ f (n) (c) (x c) n + R n (x) n! exact value = approximate value + remainder R n (x) = f (n+) (z) (x c)n+ (n +)! This is called the Lagrange form of the remainder. This is essentially the next term in the sequence. f(x) = the function that we are trying to model P " x = the Taylor or Maclaurin polynomial n = the degree of the polynomial c = the center of the polynomial x = a given input value z = the mystery variable related to the error c z x or x z c R " (x) = the error for the worst case scenario

Example 7: Use the 3 rd degree Maclaurin polynomial for f (x) = sin x (see example 4) to approximate sin 0.. Evaluate the Lagrange form of the remainder to determine the error in the approximation. Recall: P 3 (x) = x x3 6 P 8 0. = 0. 0. 8 6 is the 3 rd degree Maclaurin polynomial for f x = sin x. = 0.0998333 Using Taylor s Theorem for this problem we can write: f (x) = P n (x) + R n (x). sin x = x x3 6 + f (4) (z) 4! x 4 What is f (4) (z)? What is the Lagrange form of the remainder? To evaluate the Lagrange form of the remainder, we can t simply plug in 0 for z because the entire remainder would be 0, so we must consider the range of values for sin z. If we estimate the maximum possible error, then we can write a range of values that we know will include the actual value for sin 0.. In the interval [0, 0.] ( i.e. the distance between the center and the x-value that we are trying to approximate), the sine function must lie between 0 and. If we let sin x in the Lagrange error equal, we are determining the maximum possible error. For sin z 4! (0.)4 letting sin z = gives R 3 (x) = 4! (0.)4 0.0000046 Therefore, Sin 0. = P 3 (0.) ± R 3 (0.) = 0.099833 ± 0.0000046 And.099829733 < sin 0. <.0998374933 That s quite a small error considering we only used a 3 rd degree approximation!!

Example 8: Find the accuracy of the 4 th Maclaurin polynomial for e 0.4. See Example for the Maclaurin polynomial. e x P 4 (x) = R 4 (x) = < e 0.4 <

9.8 Power Series In section 9.7 we learned to approximate functions by Taylor Polynomials. Taylor Polynomials have a discrete number of terms. The more terms, the more accurate will be the approximation. With Power Series, we approximate functions exactly as an infinite series. f (x) = e x e x + x st degree polynomial e x + x + x2 2! e x + x + x2 2! + x3 3! e x + x + x2 2! + x3 3! + x4 4! e x + x + x2 2! + x3 3! + x4 4! + x5 5! 2nd degree polynomial 3rd degree polynomial 4th degree polynomial 5th degree polynomial e x + x + x2 2! + x3 3! + x4 4! + x5 xn + + + POWER SERIES 5! n! Definition of Power Series If x is a variable, then an infinite series of the form a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 + + a n x n + n=0 is called a power series. In general, an infinite series of the form a n (x c) n = a 0 + a (x c) + a 2 (x c) 2 + a 3 (x c) 3 + + a n (x c) n + n=0 is called a power series centered at c, where c is a constant. **To simplify the notation for power series, we agree that (x c) 0 =, even if x = c. Note: A Geometric Series is really a special case of the power series with c = 0 and a n = a, a constant. Geometric Series: ax n = a + ax + ax 2 + ax 3 + n=0

The main focus of this lesson is to determine the values of x for which a given power series will converge. There are 3 possible ways that a power series will converge.. A power series centered at c will converge only at the point c, in which case the radius of convergence is 0. 2. A power series will converge over an interval that is R units to the right and left of c in which case the radius of convergence is R. We can then talk about an interval of convergence (c R, c + R). The endpoints may or may not be part of the interval of convergence. 3. A power series will converge for all real numbers, in which case the radius of convergence is R =, and the interval of convergence is These three cases are illustrated in 3 examples below. The Ratio Test works well to determine the interval of convergence. As a reminder, the Ratio Test tells us: lim x a n+ a n < then the series is convergent, (,) If lim x a n+ a n > then the series is divergent, lim x a n+ a n = the ratio test is inconclusive.. Find the interval of convergence of the power series: n=0 x 7 n

2. Find the radius and interval of convergence of n=0 n!x n 3. Find the radius and interval of convergence of n=0 3(x 2) n

4. Find the interval of convergence of the power series: n=0 ( 3x) n ( 2n)! 5. Find the radius and interval of convergence of n=0 ( ) n x 2n+ (2n + )!

Differentiation and Integration of Power Series A function defined by a power series can be differentiated and integrated in the same way as any polynomial function by using the power rule for derivatives and integrals. The interval of convergence for the derivative and integral of the series will be the same as for the original series, except possibly at the endpoints. So, CHECK THE ENDPOINTS! J Given the function f (x) = a n (x c) n = a 0 + a (x c) + a 2 (x c) 2 + a 3 (x c) 3 + n=0 The derivative is: f '(x) = na n (x c) n = a + 2a 2 (x c) + 3a 3 (x c) 2 + The integral is: (x c) n+ (x c) f (x)dx = C + 2 (x c) 3 a n = C + a 0 (x c) + a + a 2 n + 2 3 n=0 + In the next section of our textbook, we will use differentiation and integration of known power series to find the power series for a new function. EXAMPLE ( ) n+ (x ) n+ Let = f (x) =. Find the interval of convergence of n + n=0 f (x), f '(x), f ''(x) and function. f (x)dx. Be sure to check the endpoints of the interval for each

9.9 Representation of Functions by Power Series DAY In this lesson we will represent a function as a Power Series. In our earlier lesson geometric series we learned that for the series x n, n=0 we could write the sum of the infinite number of terms as S = x Now, let s suppose we begin with a function in the form of the sum, f (x) = x, you can see that it is a geometric series that is represented by with a, a constant, and r, a function of x In the following examples, we will attempt to rewrite a given expression in the form as a power series and determine the interval of convergence. x Example : Find a power series for f (x) = x + convergence. centered at 0. Determine the interval of

Example 2: Find a power series for f (x) = centered at. Determine the interval of x convergence. Example 3: Find a power series for f (x) = convergence. centered at 0. Determine the interval of 3 + x

Example 4: Find a power series for f (x) = centered at 0. Determine the interval of 2 4x convergence. Example 5: Find a power series for f (x) = convergence. x 7x + centered at 0. Determine the interval of

Example 6: Find a power series for f (x) = convergence. x 2 + 9 centered at 0. Determine the interval of Example 7: Find a power series, centered at 0, for the function f (x) = 3x x 2 (Hint: First separate the function using partial fractions. J )

DAY 2 Operations with Power Series Let f (x) = a n x n and g(x) = b n x n. f (kx) = a n k n x n Multiply within the function by a constant! 2. f x N n=0 ( ) = a n x nn Exponentiate the variable within the function. n=0 3. f (x) ± g(x) = (a ± b)x n Add/subtract functions to form a new function. n=0 Example 8: Find a power series for f (x) = ln x centered at x =. (Hint: Integrate the series that we already know for the function f (x) =. (See example 4) x dx = ln x + C x

Example 9: Find a power series by integration for the function f (x) = arctan x, centered at 0.

9.0 Taylor and Maclaurin Series The Taylor and Maclaurin Series are very similar to the Taylor and Maclaurin polynomials that we generated earlier in this chapter. The series are an infinite set of terms while the polynomials are a particular number of terms with a remainder. Definition: If a function f has derivatives of all orders at x = c, then the series n=0 f (n) (c) (x c) n n! = f (c) + f '(c)(x c) + + f (n) (c) (x c) n + n! is called the Taylor Series for f (x) at c. If c = 0, then the series is called the Maclaurin Series for f. Example : Use the function f (x) = sin x to form the Maclaurin series and determine the interval of convergence. Note: In this example we have shown that the Maclaurin series converges to a sum for all x, but have not demonstrated that it converges necessarily to f(x) = sin x. Demonstrating convergence to the function for all x goes beyond the scope of this course. However, a brief explanation is that if the Lagrange error goes to 0 as n approaches infinity, then the Maclaurin or Taylor series converges to f(x). f (n) (z)(x c) n i.e. If lim R n = lim = 0, then the series converges to the function. n n n!

Taylor and Maclaurin Series for a Composite Function Use a basic list of Taylor and Maclaurin series for elementary functions to find series for composite functions. For instance, if you want to find the Taylor Series for f (x) = cos x 2, write out the Taylor Series for f (x) = cos x and substitute x 2 for x. Or, if you want to find the Taylor Series for f (x) = e x sin x, write out the series for f (x) = e x and for f (x) = sin x, and multiply them together. Below is a list of basic functions from the Larson text. For the AP exam, it would be helpful to memorize the series for ln x, e x, sin x, and cos x. From this list, we can determine power series for other functions by the operations of addition, subtraction, multiplication, division, differentiation, integration, and composition with known power series. This provides a very useful SHORTCUT to finding the coefficients of a Maclaurin or Taylor series directly. J Memorize the Big 4 for the AP Exam: f(x) = ln x f(x) = e x f(x) = sin x f(x) = cos x

Example 2: Find the Maclaurin series for f (x) = sin x 2. Example 3: Find the first five terms of the Maclaurin series for f (x) = e x arctan x

Example 4: Use a power series to approximate e x2 dx. Use the first four terms of the series for your approximation. 0