SOLVED PROBLEMS ON TAYLOR AND MACLAURIN SERIES

SOLVED PROBLEMS ON TAYLOR AND MACLAURIN SERIES

TAYLOR AND MACLAURIN SERIES Taylor Series of a function f at x = a is ( f k )( a) ( x a) k k! It is a Power Series centered at a. Maclaurin Series of a function f is a Taylor Series at x = 0.

BASIC MACLAURIN SERIES e x = 1 + x + x2 2! + x3 3! + = x k k! sin( x) = x x3 3! + x5 5! cos x = 1 x2 2! + x4 4! = 1 = 1 ( ) ( 1 + x) p = 1 + px + p p 1 2! ( ) k x 2k+1 ( ) k x 2k x 2 + ( 2k + 1)! ( 2k)!

USE TAYLOR SERIES 1 To estimate values of functions on an interval. 2 To compute limits of functions. 3 To approximate integrals. 4 To study properties of the function in question.

FINDING TAYLOR SERIES To find Taylor series of functions, we may: 1 Use substitution. 2 Differentiate known series term by term. 3 Integrate known series term by term. 4 Add, divide, and multiply known series.

TAYLOR & MACLAURIN SERIES For future reference, we collect some important Maclaurin series that we have derived in this section and Section 11.9, in the following table.

IMPORTANT MACLAURIN SERIES Table 1 1 1 x n0 n 2 3 x 1 x x x R 1 e x n 2 3 x x x x 1 R n! 1! 2! 3! n0

IMPORTANT MACLAURIN SERIES Table 1 2n1 3 5 7 n x x x x sin x( 1) x R (2 n 1)! 3! 5! 7! n0 2n 2 4 6 n x x x x cos x( 1) 1 R (2 n)! 2! 4! 6! n0 2n1 3 5 7 1 n tan x ( 1) x R1 n0 x x x x 2n 1 3 5 7

IMPORTANT MACLAURIN SERIES Table 1 k k n kk ( 1) (1 x) x 1kx x n0 n 2! kk ( 1)( k2) 3 x R 3! 2 1

USES OF TAYLOR SERIES One reason Taylor series are important is that they enable us to integrate functions that we couldn t previously handle.

USES OF TAYLOR SERIES In fact, in the introduction to this chapter, we mentioned that Newton often integrated functions by first expressing them as power series and then integrating the series term by term.

USES OF TAYLOR SERIES The function f(x) = e x2 can t be integrated by techniques discussed so far. Its antiderivative is not an elementary function (see Section 7.5). In the following example, we use Newton s idea to integrate this function.

USES OF TAYLOR SERIES Example 10 a. Evaluate e -x2 dx as an infinite series. b. Evaluate e dx correct to within an error of 0.001 1 0 x2

USES OF TAYLOR SERIES Example 10 a First, we find the Maclaurin series for f(x) = e -x2 It is possible to use the direct method. However, let s find it simply by replacing x with x 2 in the series for e x given in Table 1.

USES OF TAYLOR SERIES Thus, for all values of x, Example 10 a e x 2 n0 n0 2 ( x ) n! ( 1) n 2n n! 2 4 6 x x x 1... 1! 2! 3! n x

USES OF TAYLOR SERIES Now, we integrate term by term: Example 10 a 2 4 6 2n 2 x x x x n x e dx 1 ( 1) dx 1! 2! 3! n! 3 5 7 x x x Cx 31! 52! 7 3! 2n1 n x ( 1) (2 n 1) n! This series converges for all x because the original series for e -x2 converges for all x.

USES OF TAYLOR SERIES The FTC gives: Example 10 b 1 0 3 5 7 9 2 x x x x x x dx x 31! 52! 7 3! 94! 1 1 1 1 1 3 10 42 216 1 0.7475 1 1 1 1 3 10 42 216 1 0

USES OF TAYLOR SERIES Example 10 b The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 1 1 115! 1320 0.001

USES OF TAYLOR SERIES Another use of Taylor series is illustrated in the next example. The limit could be found with l Hospital s Rule. Instead, we use a series.

USES OF TAYLOR SERIES Evaluate lim x0 e x 1 x 2 Example 11 x Using the Maclaurin series for e x, we have the following result.

USES OF TAYLOR SERIES lim e x 2 3 x x x 1 1 1x 1! 2! 3! lim x x 2 3 4 x x x lim 2! 3! 4! x0 2 x x0 2 x0 2 Example 11 2 3 1 x x x 1 lim x0 2 3! 4! 5! 2 This is because power series are continuous functions. x

MULTIPLICATION AND DIVISION OF POWER SERIES If power series are added or subtracted, they behave like polynomials. Theorem 8 in Section 11.2 shows this. In fact, as the following example shows, they can also be multiplied and divided like polynomials.

MULTIPLICATION AND DIVISION OF POWER SERIES In the example, we find only the first few terms. The calculations for the later terms become tedious. The initial terms are the most important ones.

MULTIPLICATION AND DIVISION Example 12 Find the first three nonzero terms in the Maclaurin series for: a. e x sin x b. tan x

MULTIPLICATION AND DIVISION Example 12 a Using the Maclaurin series for e x and sin x in Table 1, we have: 2 3 3 x x x x x e sin x1 x 1! 2! 3! 3!

MULTIPLICATION AND DIVISION Example 12 a We multiply these expressions, collecting like terms just as for polynomials: 1 2 1 3 2 6 1x x x x 1 x 3 6 2 1 3 1 4 2 6 xx x x 1 x 3 1 x 4 6 6 2 1 3 3 xx x

MULTIPLICATION AND DIVISION Thus, Example 12 a x 2 1 3 3 e sin xxx x

MULTIPLICATION AND DIVISION Example 12 b Using the Maclaurin series in Table 1, we have: tan x 3 5 x x x sin x 3! 5! cos x x 2 x 4 1 2! 4!

MULTIPLICATION AND DIVISION Example 12 b We use a procedure like long division: 1 3 2 5 3 15 x x x 1 x 2 1 x 4 x 1 x 3 1 x 5 2 24 6 120 1 1 3 1 5 2 24 1 x 3 1 x 5 3 30 x x x 1 x 3 1 x 5 3 6 1 x 5 6

MULTIPLICATION AND DIVISION Thus, Example 12 b 1 3 2 5 3 15 tan xx x x

MULTIPLICATION AND DIVISION Although we have not attempted to justify the formal manipulations used in Example 12, they are legitimate.

MULTIPLICATION AND DIVISION There is a theorem that states the following: Suppose both f(x) = Σc n x n and g(x) = Σb n x n converge for x < R and the series are multiplied as if they were polynomials. Then, the resulting series also converges for x < R and represents f(x)g(x).

MULTIPLICATION AND DIVISION For division, we require b 0 0. The resulting series converges for sufficiently small x.

OVERVIEW OF PROBLEMS Find the Maclaurin Series of the following functions. sin( x 2 ) sin( x) 1 2 3 arctan x x ( ) 4 cos 2 ( x) 5 x 2 e x 6 1 x 3 7 sinh x e x ( ) 8 9 x2 arctan( x 3 ) 1 x

OVERVIEW OF PROBLEMS Find the Taylor Series of the following functions at the given value of a. 10 x x3 at a = 2 11 1 x at a = 2 12 e 2x at a = 12 13 sin x ( ) at a = 4 14 10 x at a = 1 15 ln 1 + x ( ) at a = 2

Find the Maclaurin Series of the following functions.

MACLAURIN SERIES Problem 1 f ( x) = sin( x 2 ) Solution Substitute x by x 2 in the Maclaurin Series of sine. Hence sin x 2 ( x 2 ) 2k+1 ( ) = ( 1) k = 1 ( 2k + 1)! ( ) k x 4k+2 ( 2k + 1)!

MACLAURIN SERIES Problem 2 Solution f ( ) ( x) = sin x x Divide the Maclaurin Series of sine by x. Hence, ( ) sin x x = 1 x ( 1) k x 2k+1 = 1 ( 2k + 1)! ( ) k x 2k ( 2k + 1)!

MACLAURIN SERIES Problem 3 f ( x) = arctan( x) Solution Observe that f '( x) = Series of f ' x Power Series formula. 1. To find the Maclaurin 2 1 + x ( ) substitute x 2 for x in Basic

MACLAURIN SERIES Solution(cont d) Hence f '( x) = 1 1 + x = ( ) k 2 x2 = ( 1) k x 2k. By integrating both sides, we obtain f ( x) = ( 1) k x 2k dx = 1 ( ) k x 2k+1 = 1 + C. 2k + 1 ( ) k x 2k dx

MACLAURIN SERIES Solution(cont d) 0 is in the interval of convergence. Therefore we can insert x = 0 to find that the integration constant c = 0. Hence the Maclaurin series of arctan x ( ) is arctan x ( ) = ( 1) k x 2k+1. 2k + 1

MACLAURIN SERIES Problem 4 f ( x) = cos 2 ( x) Solution By the trigonometric identity, cos 2 ( x) = 1 + cos ( 2x) ( ) 2. Therefore we start with the Maclaurin Series of cosine.

MACLAURIN SERIES Solution(cont d) Substitute x by 2x incos x Thus cos 2x ( ( ) = ( 1) k 2x) 2k and dividing by 2, we obtain ( ) = ( 1) k x 2k. ( 2k)!. After adding 1 ( 2k)!

MACLAURIN SERIES Solution(cont d) cos 2 ( x) = 1 2 1 + 1 = 1 2 ( ( ) k 2x) 2k ( 2x) 2 1 + 1 k=1 =1+ 1 2! ( ) k 2 2k1 ( 2k)! ( + 2x ) 4 4! ( 2k)! x2k

MACLAURIN SERIES Problem 5 f ( x) = x 2 e x Solution Multiply the Maclaurin Seris of e x by x 2. x k Hence, x 2 e x = x 2 = k! x k+2 k!.

MACLAURIN SERIES Problem 6 f ( x) = 1 x 3 Solution By rewriting f ( ( )) 12. By substituting x ( x) = 1 + x 3 by -x 3 in the binomial formula with p = 12 we obtain, 1 x 3 = 1 1 2 x3 1 8 x6

MACLAURIN SERIES Problem 7 f ( x) = sinh( x) Solution By rewriting f ( x) = ex e x 2. Substitute x by x in the Maclaurin Series of e x = 1 + x + x2 2 + = x k, k!

MACLAURIN SERIES Solution(cont d) e x = ( x) k k! Thus when we add e x = 1 x + x2 2! and e x, the terms with odd power are canceled and the terms with even power are doubled. After dividing by 2, we obtain sinh( x) = 1 + x2 2! + x4 4! + = x 2k ( 2k)!

MACLAURIN SERIES Problem 8 f ( x) = ex 1 x Solution We have e x = 1 + x + x2 2! + and 1 1 x = 1 + x + x2 + To find the Maclaurin Series of f ( x), we multiply these series and group the terms with the same degree.

MACLAURIN SERIES Solution(cont d) 1 + x + x2 2! + 1 + x + x2 + ( ) = 1 + 2x + 1 + 1 + 1 2! x2 + higher degree terms = 1 + 2x + 5 2 x2 + higher degree terms

MACLAURIN SERIES Problem 9 f ( x) = x 2 arctan( x 3 ) Solution We have calculated the Maclaurin Series of arctan x arctan x ( ) = ( 1) k x 2k+1. 2k + 1 ( ) Substituting x by x 3 in the above formula, we obtain

MACLAURIN SERIES Solution(cont d) arctan x 3 ( x 3 ) 2k+1 ( ) = ( 1) k = 1. 2k + 1 2k + 1 ( ) k x 6k+3 Multiplying by x 2 gives the desired Maclaurin Series x 2 arctan x 3 ( ) = x 2 ( 1) k x 6k+3 = 1 2k + 1 ( ) k x 6k+4 2k + 1

Find the Taylor Series of the following functions at given a.

TAYLOR SERIES Problem 10 f ( x) = x x 3 at a = 2 Solution Taylor Series of f f ( ( 2)+ f 1 ) + f ( 3 ) ( 2) 3! ( x) = x x 3 at a = 2 is of the form ( )( 2) ( 2) ( x + 2)+ f 2 ( x + 2) 3 + f ( 4 ) 2! ( 2) 4! ( x + 2) 2 ( x + 2) 4 +

TAYLOR SERIES Solution(cont d) Since f is a polynominal function of degree 3, its derivatives of order higher than 3 is 0. Thus Taylor Series is of the form ( )( 2) f ( ( 2)+ f 1 ) ( 2) ( x + 2)+ f 2 2! ( x + 2) 2 + f ( 3 ) ( 2) 3! ( x + 2) 3

TAYLOR SERIES Solution(cont d) By direct computation, f ( 2) ( = 6, f 1 )( 2) = 11, f 2 ( ) ( 2) ( = 12, f 3 ) ( 2) = 6 So the Taylor Series of x x 3 at a = 2 is 6-11( x + 2)+ 6( x + 2) 2 ( x + 2) 3

TAYLOR SERIES Problem 11 f ( x) = 1 x at a = 2 Solution Taylor Series of f x ( f k ) ( 2) ( ) k ( ) = 1 x at a = 2 is of the form x 2. We need to find the general k! expression of the k th derivative of 1 x.

TAYLOR SERIES Solution(cont d) We derive 1 x until a pattern is found. f ( x) ( = 1 x = x 1, f 1 )( x) = ( 1)x 2 ( ) f 2 ( ) ( x) = ( 1) ( 2)x 3, f 3 In general, f k ( f k ) ( ) ( 2) = ( 1) k ( k+1) k!2. ( x) = 1 ( x) = ( 1) k k+1 k!x ( )( 2) 3 ( )x 4 ( ). Therefore

TAYLOR SERIES Solution(cont d) After inserting the general expression of the k th derivative evaluated at 2 we obtain, ( f k )( 2) ( x 2) k 1 = ( k! k! 1 ) k ( k+1) k!2 x 2 ( ) k Hence, the the Taylor Series of 1 x is ( 1) k x 2 ( ) 2 k+1 ( ) k.

TAYLOR SERIES Problem 12 f ( x) = e 2x at a = 12 Solution Taylor Series of f x ( f k ) ( 12) k! x 1 2 ( ) = e 2x at a = 12 is of the form k. We need to find the general expression of the k th derivative of e 2x.

TAYLOR SERIES Solution(cont d) We derive e 2x ( ) = e 2x, f 1 f x ( ) ( ) In general, f k until a pattern is found. ( x) ( = 2e 2x, f 2 ) x ( x) = ( 1) k 2 k e 2x. ( ) = 2 2e 2x ( Therefore f k ) ( 12) = 1 ( ) k 2 k e 2 1 2 = 1 ( ) k 2 k e.

TAYLOR SERIES Solution(cont d) After inserting the general expression of the k th derivative evaluated at 1 2 we obtain, ( ) f k ( 12) x 1 k k! 2 = ( ) k 2 k 1 1 x 1 k! e 2 Hence, the the Taylor Series of e 2x is ( 1) k 2x 1 e k! ( ) k. ( ) k

TAYLOR SERIES Problem 13 f ( x) = sin( x) at a = 4 Solution Taylor Series of f x form ( f k ) ( ) = sin( x) at a = 4 is of the ( 4) k k! x 4 general expression of the k th. We need to find the derivative of sin( x).

TAYLOR SERIES Solution(cont d) We derive sin x f ( x) = sin( x), f 1 ( In general, f k ) ( ) until a pattern is found. ( )( x) = cos ( ( x), f 2 ) sin( x) if k = 4n cos ( x) if k = 4n + 1 ( x) = sin( x) if k = 4n + 2 cos ( x) if k = 4n + 3 ( x) = sin( x)

TAYLOR SERIES Solution(cont d) ( ) ( ) and odd order derivatives are either cos( x) In other words, even order derivatives are either sin x or -sin x or -cos ( x). So the Taylor Series at a = 4 can be written as ( ) ( 1) k sin 4 x ( 2k)! 4 2k ( ) k cos ( 4) + 1 x ( 2k + 1)! 4 2k+1

TAYLOR SERIES Solution(cont d) Since, at a = 4, sin( 4) = cos( 4) = 1 2, the Taylor Series can be simplified to ( 1) k x 2 ( 2k)! 4 2k + ( 1) k x 2 ( 2k + 1)! 4 2k+1.

TAYLOR SERIES Problem 14 f ( x) = 10 x at a = 1 Solution Taylor Series of f x ( f k ) ( 1) ( ) k ( ) = 10 x at a = 1 is of the form x 1. We need to find the general k! expression of the k th derivative of 10 x.

TAYLOR SERIES Solution(cont d) We derive 10 x f x ( ) ( ) = 10 x, f 1 ( ) = ln 10 ( )( x) = ln k 10 ( 1) = ln k 10 In general, f k ( Therefore f k ) until a pattern is found. x ( ( ) 10 x, f 2 ) x ( )10 x. ( )10. ( ) = ln 2 10 ( )10 x

TAYLOR SERIES Solution(cont d) After inserting the general expression of the k th derivative evaluated at 1 we obtain, ( f k )( 1) ( x 1) k ln k ( 10)10 = x 1 k! k! ( ) k

TAYLOR SERIES Problem 15 f ( x) = ln( 1 + x) at a = 2 Solution Taylor Series of f x ( f k ) ( ) = ln( 1 + x) at a = 2 is of ( 2) ( x + 2) k the form. We need to find the k! general expression of the k th derivative of ln( 1 + x).

TAYLOR SERIES Solution(cont d) We derive ln( x + 1) until a pattern is found. f ( x) = ln( ( x + 1), f 1 ) ( x) = 1 x + 1, f 2 ( ) ( x) = 1 ( x + 1) 2 ( In general, f k ) ( x) = ( 1) k ( x + 1). Therefore f k ( k) ( 2) = 1.

TAYLOR SERIES Solution(cont d) After inserting the general expression of the k th derivative evaluated at -2 we obtain ( f k )( 2) ( x + 2) k 1 = x + 2 k! k! ( ) k.