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[T] Suppose that a set of standardized test scores is normally distributed with mean μ = 100 and standard deviation σ = 10 . Set up an integral that represents the probability that a test score will be between 90 and 110 and use the integral of the degree 10 Maclaurin polynomial of 1 2 π e x 2 / 2 to estimate this probability.

The probability is p = 1 2 π ( a μ ) / σ ( b μ ) / σ e x 2 / 2 d x where a = 90 and b = 100 , that is, p = 1 2 π −1 1 e x 2 / 2 d x = 1 2 π −1 1 n = 0 5 ( −1 ) n x 2 n 2 n n ! d x = 2 2 π n = 0 5 ( −1 ) n 1 ( 2 n + 1 ) 2 n n ! 0.6827 .

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[T] Suppose that a set of standardized test scores is normally distributed with mean μ = 100 and standard deviation σ = 10 . Set up an integral that represents the probability that a test score will be between 70 and 130 and use the integral of the degree 50 Maclaurin polynomial of 1 2 π e x 2 / 2 to estimate this probability.

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[T] Suppose that n = 0 a n x n converges to a function f ( x ) such that f ( 0 ) = 1 , f ( 0 ) = 0 , and f ( x ) = f ( x ) . Find a formula for a n and plot the partial sum S N for N = 20 on [ −5 , 5 ] .


This graph is a wave curve symmetrical about the origin. It has a peak at y = 1 above the origin. It has lowest points at -3 and 3.
As in the previous problem one obtains a n = 0 if n is odd and a n = ( n + 2 ) ( n + 1 ) a n + 2 if n is even, so a 0 = 1 leads to a 2 n = ( −1 ) n ( 2 n ) ! .

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[T] Suppose that n = 0 a n x n converges to a function f ( x ) such that f ( 0 ) = 0 , f ( 0 ) = 1 , and f ( x ) = f ( x ) . Find a formula for a n and plot the partial sum S N for N = 10 on [ −5 , 5 ] .

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Suppose that n = 0 a n x n converges to a function y such that y y + y = 0 where y ( 0 ) = 1 and y ( 0 ) = 0 . Find a formula that relates a n + 2 , a n + 1 , and a n and compute a 0 , ... , a 5 .

y = n = 0 ( n + 2 ) ( n + 1 ) a n + 2 x n and y = n = 0 ( n + 1 ) a n + 1 x n so y y + y = 0 implies that ( n + 2 ) ( n + 1 ) a n + 2 ( n + 1 ) a n + 1 + a n = 0 or a n = a n 1 n a n 2 n ( n 1 ) for all n · y ( 0 ) = a 0 = 1 and y ( 0 ) = a 1 = 0 , so a 2 = 1 2 , a 3 = 1 6 , a 4 = 0 , and a 5 = 1 120 .

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Suppose that n = 0 a n x n converges to a function y such that y y + y = 0 where y ( 0 ) = 0 and y ( 0 ) = 1 . Find a formula that relates a n + 2 , a n + 1 , and a n and compute a 1 , ... , a 5 .

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The error in approximating the integral a b f ( t ) d t by that of a Taylor approximation a b P n ( t ) d t is at most a b R n ( t ) d t . In the following exercises, the Taylor remainder estimate R n M ( n + 1 ) ! | x a | n + 1 guarantees that the integral of the Taylor polynomial of the given order approximates the integral of f with an error less than 1 10 .

  1. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than 1 100 .
  2. Compare the accuracy of the polynomial integral estimate with the remainder estimate.

[T] 0 π sin t t d t ; P s = 1 x 2 3 ! + x 4 5 ! x 6 7 ! + x 8 9 ! (You may assume that the absolute value of the ninth derivative of sin t t is bounded by 0.1 . )

a. (Proof) b. We have R s 0.1 ( 9 ) ! π 9 0.0082 < 0.01 . We have 0 π ( 1 x 2 3 ! + x 4 5 ! x 6 7 ! + x 8 9 ! ) d x = π π 3 3 · 3 ! + π 5 5 · 5 ! π 7 7 · 7 ! + π 9 9 · 9 ! = 1.852... , whereas 0 π sin t t d t = 1.85194... , so the actual error is approximately 0.00006 .

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[T] 0 2 e x 2 d x ; p 11 = 1 x 2 + x 4 2 x 6 3 ! + x 22 11 ! (You may assume that the absolute value of the 23 rd derivative of e x 2 is less than 2 × 10 14 . )

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The following exercises deal with Fresnel integrals .

The Fresnel integrals are defined by C ( x ) = 0 x cos ( t 2 ) d t and S ( x ) = 0 x sin ( t 2 ) d t . Compute the power series of C ( x ) and S ( x ) and plot the sums C N ( x ) and S N ( x ) of the first N = 50 nonzero terms on [ 0 , 2 π ] .


This graph has two curves. The first one is a solid curve labeled Csub50(x). It begins at the origin and is a wave that gradually decreases in amplitude. The highest it reaches is y = 1. The second curve is labeled Ssub50(x). It is a wave that gradually decreases in amplitude. The highest it reaches is 0.9. It is very close to the pattern of the first curve with a slight shift to the right.
Since cos ( t 2 ) = n = 0 ( −1 ) n t 4 n ( 2 n ) ! and sin ( t 2 ) = n = 0 ( −1 ) n t 4 n + 2 ( 2 n + 1 ) ! , one has S ( x ) = n = 0 ( −1 ) n x 4 n + 3 ( 4 n + 3 ) ( 2 n + 1 ) ! and C ( x ) = n = 0 ( −1 ) n x 4 n + 1 ( 4 n + 1 ) ( 2 n ) ! . The sums of the first 50 nonzero terms are plotted below with C 50 ( x ) the solid curve and S 50 ( x ) the dashed curve.

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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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