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The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.

Key concepts

  • The binomial series is the Maclaurin series for f ( x ) = ( 1 + x ) r . It converges for | x | < 1 .
  • Taylor series for functions can often be derived by algebraic operations with a known Taylor series or by differentiating or integrating a known Taylor series.
  • Power series can be used to solve differential equations.
  • Taylor series can be used to help approximate integrals that cannot be evaluated by other means.

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

( 1 + x 2 ) −1 / 3

( 1 + x 2 ) −1 / 3 = n = 0 ( 1 3 n ) x 2 n

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( 1 2 x ) 2 / 3

( 1 2 x ) 2 / 3 = n = 0 ( −1 ) n 2 n ( 2 3 n ) x n

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In the following exercises, use the substitution ( b + x ) r = ( b + a ) r ( 1 + x a b + a ) r in the binomial expansion to find the Taylor series of each function with the given center.

x 2 + 2 at a = 0

2 + x 2 = n = 0 2 ( 1 / 2 ) n ( 1 2 n ) x 2 n ; ( | x 2 | < 2 )

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2 x x 2 at a = 1 ( Hint: 2 x x 2 = 1 ( x 1 ) 2 )

2 x x 2 = 1 ( x 1 ) 2 so 2 x x 2 = n = 0 ( −1 ) n ( 1 2 n ) ( x 1 ) 2 n

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( x 8 ) 1 / 3 at a = 9

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x at a = 4

x = 2 1 + x 4 4 so x = n = 0 2 1 2 n ( 1 2 n ) ( x 4 ) n

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x at x = 9

x = n = 0 3 1 3 n ( 1 2 n ) ( x 9 ) n

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In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most 1 / 1000 .

[T] ( 15 ) 1 / 4 using ( 16 x ) 1 / 4

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[T] ( 1001 ) 1 / 3 using ( 1000 + x ) 1 / 3

10 ( 1 + x 1000 ) 1 / 3 = n = 0 10 1 3 n ( 1 3 n ) x n . Using, for example, a fourth-degree estimate at x = 1 gives ( 1001 ) 1 / 3 10 ( 1 + ( 1 3 1 ) 10 −3 + ( 1 3 2 ) 10 −6 + ( 1 3 3 ) 10 −9 + ( 1 3 4 ) 10 −12 ) = 10 ( 1 + 1 3.10 3 1 9.10 6 + 5 81.10 9 10 243.10 12 ) = 10.00333222... whereas ( 1001 ) 1 / 3 = 10.00332222839093 .... Two terms would suffice for three-digit accuracy.

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In the following exercises, use the binomial approximation 1 x 1 x 2 x 2 8 x 3 16 5 x 4 128 7 x 5 256 for | x | < 1 to approximate each number. Compare this value to the value given by a scientific calculator.

[T] 1 2 using x = 1 2 in ( 1 x ) 1 / 2

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[T] 5 = 5 × 1 5 using x = 4 5 in ( 1 x ) 1 / 2

The approximation is 2.3152 ; the CAS value is 2.23 .

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[T] 3 = 3 3 using x = 2 3 in ( 1 x ) 1 / 2

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[T] 6 using x = 5 6 in ( 1 x ) 1 / 2

The approximation is 2.583 ; the CAS value is 2.449 .

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Integrate the binomial approximation of 1 x to find an approximation of 0 x 1 t d t .

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[T] Recall that the graph of 1 x 2 is an upper semicircle of radius 1 . Integrate the binomial approximation of 1 x 2 up to order 8 from x = −1 to x = 1 to estimate π 2 .


1 x 2 = 1 x 2 2 x 4 8 x 6 16 5 x 8 128 + . Thus

−1 1 1 x 2 d x = x x 3 6 x 5 40 x 7 7 · 16 5 x 9 9 · 128 + | −1 1 2 1 3 1 20 1 56 10 9 · 128 + error = 1.590 ... whereas π 2 = 1.570 ...

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In the following exercises, use the expansion ( 1 + x ) 1 / 3 = 1 + 1 3 x 1 9 x 2 + 5 81 x 3 10 243 x 4 + to write the first five terms (not necessarily a quartic polynomial) of each expression.

( 1 + 4 x ) 1 / 3 ; a = 0

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( 1 + 4 x ) 4 / 3 ; a = 0

( 1 + x ) 4 / 3 = ( 1 + x ) ( 1 + 1 3 x 1 9 x 2 + 5 81 x 3 10 243 x 4 + ) = 1 + 4 x 3 + 2 x 2 9 4 x 3 81 + 5 x 4 243 +

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( 3 + 2 x ) 1 / 3 ; a = −1

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( x 2 + 6 x + 10 ) 1 / 3 ; a = −3

( 1 + ( x + 3 ) 2 ) 1 / 3 = 1 + 1 3 ( x + 3 ) 2 1 9 ( x + 3 ) 4 + 5 81 ( x + 3 ) 6 10 243 ( x + 3 ) 8 +

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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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