Question

In Exercises \(35-38,\) use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error. $$ \cos (0.3) \approx 1-\frac{(0.3)^{2}}{2 !}+\frac{(0.3)^{4}}{4 !} $$

Short answer

The upper bound of the error for the approximated cosine function at 0.3 using Taylor's Theorem is 0.0000605, and the exact error is approximately 0.0000531558.

Step-by-step solution

Recall Taylor's Theorem for error estimation

Taylor's Theorem states that the error, \( R_n(x) \), of an n-degree Taylor polynomial approximation of a function is given by \( R_n(x) = |f^{n+1}(c)| * |x - a|^{n+1} / (n+1)! \) for some \( c \) between \( a \) and \( x \), where \( f^{n+1}(c) \) is the \( (n+1) \)-th derivative of the function at \( c \). Here, the function is \( \cos(x) \) and \( a \) is the point at which the function is approximated, 0 in this case. The value of \( x \) is 0.3.

Calculate higher order derivative and estimate error bounds

For the cosine function, the higher order derivatives have a feature that alternate between \( \sin(x) \) and \( \-\cos(x) \), while changing their sign. As the series expansion used in the approximation goes up to the fourth degree, errors begin from the fifth derivative. Because \( \cos(0.3) \) is positive, the worst-case error (upper bound) \( |R_4(0.3)| \) occurs when \( \sin(c) \) takes strictly its maximum value, 1. So, \( |R_4(0.3)| = |1|*|0.3 - 0|^5 / 5! = 0.3^5 / 120 = 0.0000605 \).

Find the exact value of the error

The exact error can be calculated using the exact value of \( \cos(0.3) \) and comparing it to the approximated value given by our series. The precise value of \( cos(0.3) \) can be computed via a calculator as \( \cos(0.3) = 0.955336489125...\) The approximate value from series is \( 1 - 0.3^2/2 + 0.3^4/24 = 0.9552833333... \). Finally, finding the difference between these two values, \( 0.955336489125 - 0.9552833333 = 0.0000531558...\), gives the exact error.