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Approximating sin x Using maclaurin polynomials

From [link] b., the Maclaurin polynomials for sin x are given by

p 2 m + 1 ( x ) = p 2 m + 2 ( x ) = x x 3 3 ! + x 5 5 ! x 7 7 ! + + ( −1 ) m x 2 m + 1 ( 2 m + 1 ) !

for m = 0 , 1 , 2 , .

  1. Use the fifth Maclaurin polynomial for sin x to approximate sin ( π 18 ) and bound the error.
  2. For what values of x does the fifth Maclaurin polynomial approximate sin x to within 0.0001?
  1. The fifth Maclaurin polynomial is
    p 5 ( x ) = x x 3 3 ! + x 5 5 ! .

    Using this polynomial, we can estimate as follows:
    sin ( π 18 ) p 5 ( π 18 ) = π 18 1 3 ! ( π 18 ) 3 + 1 5 ! ( π 18 ) 5 0.173648.

    To estimate the error, use the fact that the sixth Maclaurin polynomial is p 6 ( x ) = p 5 ( x ) and calculate a bound on R 6 ( π 18 ) . By [link] , the remainder is
    R 6 ( π 18 ) = f ( 7 ) ( c ) 7 ! ( π 18 ) 7

    for some c between 0 and π 18 . Using the fact that | f ( 7 ) ( x ) | 1 for all x , we find that the magnitude of the error is at most
    1 7 ! · ( π 18 ) 7 9.8 × 10 −10 .
  2. We need to find the values of x such that
    1 7 ! | x | 7 0.0001 .

    Solving this inequality for x , we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as | x | < 0.907 .
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Use the fourth Maclaurin polynomial for cos x to approximate cos ( π 12 ) .

0.96593

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Now that we are able to bound the remainder R n ( x ) , we can use this bound to prove that a Taylor series for f at a converges to f .

Representing functions with taylor and maclaurin series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Finding a taylor series

Find the Taylor series for f ( x ) = 1 x at x = 1 . Determine the interval of convergence.

For f ( x ) = 1 x , the values of the function and its first four derivatives at x = 1 are

f ( x ) = 1 x f ( 1 ) = 1 f ( x ) = 1 x 2 f ( 1 ) = −1 f ( x ) = 2 x 3 f ( 1 ) = 2 ! f ( x ) = 3 · 2 x 4 f ( 1 ) = −3 ! f ( 4 ) ( x ) = 4 · 3 · 2 x 5 f ( 4 ) ( 1 ) = 4 !.

That is, we have f ( n ) ( 1 ) = ( −1 ) n n ! for all n 0 . Therefore, the Taylor series for f at x = 1 is given by

n = 0 f ( n ) ( 1 ) n ! ( x 1 ) n = n = 0 ( −1 ) n ( x 1 ) n .

To find the interval of convergence, we use the ratio test. We find that

| a n + 1 | | a n | = | ( −1 ) n + 1 ( x 1 ) n + 1 | | ( −1 ) n ( x 1 ) n | = | x 1 | .

Thus, the series converges if | x 1 | < 1 . That is, the series converges for 0 < x < 2 . Next, we need to check the endpoints. At x = 2 , we see that

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

diverges by the divergence test. Similarly, at x = 0 ,

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

diverges. Therefore, the interval of convergence is ( 0 , 2 ) .

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Find the Taylor series for f ( x ) = 1 2 at x = 2 and determine its interval of convergence.

1 2 n = 0 ( 2 x 2 ) n . The interval of convergence is ( 0 , 4 ) .

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We know that the Taylor series found in this example converges on the interval ( 0 , 2 ) , but how do we know it actually converges to f ? We consider this question in more generality in a moment, but for this example, we can answer this question by writing

f ( x ) = 1 x = 1 1 ( 1 x ) .

That is, f can be represented by the geometric series n = 0 ( 1 x ) n . Since this is a geometric series, it converges to 1 x as long as | 1 x | < 1 . Therefore, the Taylor series found in [link] does converge to f ( x ) = 1 x on ( 0 , 2 ) .

We now consider the more general question: if a Taylor series for a function f converges on some interval, how can we determine if it actually converges to f ? To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for f at a , the n th partial sum is given by the n th Taylor polynomial p n . Therefore, to determine if the Taylor series converges to f , we need to determine whether

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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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