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Chapter 9: Problem 45

In Exercises 45鈥�48, use Taylor鈥檚 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

Expert verified
The upper bound for the error is calculated as \(\frac{(0.3)^6}{6!}\) while the actual error is given by \(\cos(0.3) - \left(1 - \frac{(0.3)^2}{2!} + \frac{(0.3)^4}{4!}\right)\)

Step by step solution

01

Apply Taylor's Theorem

Taylor's Theorem states that for a sufficiently good function, we can approximate it as a polynomial. Specifically for the cosine function, its Maclaurin series (Taylor series around 0) can be written as: \[ cos(x) = \sum_{n=0}^鈭� (-1)^n \frac{x^{2n}}{(2n)!} \] For n=0 and n=2, we obtain the approximation: \[ cos(x) \approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \] Next, we'll find the next term in the series which can be used to estimate an upper bound for the error.
02

Apply Error Bound Formula

The next term in the series would be when n=3, which is: -\ \frac{x^6}{6!}. The error bounds can be established using Taylor's theorem which states that the absolute error is less than or equal to the absolute value of the next term in the series. Therefore, an upper bound for the error can be approximated using: \[ E \leq \frac{x^6}{6!} \] Thus, plug in x = 0.3 into the above inequality to calculate the upper bound of the error.
03

Compute Actual Error

To find the actual error, subtract the approximation from the exact value: \[ E = \cos(0.3) - \left(1 - \frac{(0.3)^2}{2!} + \frac{(0.3)^4}{4!}\right) \] This gives us the actual value of the error.

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