cosx2nEn=o(-1)";00s x =(2n)! Find the smallest n in order to use this series to approximate cos 5° correct to 0.000 001,then find the approximation. (hint: use radians, not degrees)

Answered: Find the smallest n in order to use…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Find the smallest n in order to use this series to approximate cos 5° correct to 0.000 001,
then find the approximation. (hint: use radians, not degrees)

Expert Solution

Step 1

Let s be the sum of the infinite series, s_n be a partial sum of n terms, and R_n be the remainder terms. Then,

$s = s_{n} + R_{n} = \sum_{i = 0}^{n} f (i) + \sum_{j = n + 1}^{\infty} f (j)$

Step 2

If the error must be less than 0.000 001, then the remainder sum must be less than 0.000 001. We can estimate the remainder using integral.

$\int \frac{x^{2 n}}{(2 n)!} d x = \frac{x^{2 n + 1}}{(2 n + 1) \times (2 n)!} = \frac{x^{2 n + 1}}{(2 n + 1)!} \leq 0.000 001$

The angle is given in degrees. Convert it to radians.

$5 ° = 5 \times \frac{π}{180} = \frac{π}{36}$

Substitute in the inequality above.

$\frac{{(\frac{π}{36})}^{2 n + 1}}{(2 n + 1)!} \leq 0.000 001$

We must find an n for which the above inequality is true.

Use trial-and-error method:

$\begin{array}{l} n = 0 \\ \Rightarrow \frac{{(\frac{π}{36})}^{1}}{1!} \approx 0.0872 > 0.000 001 \\ n = 1 \\ \Rightarrow \frac{{(\frac{π}{36})}^{3}}{3!} \approx 0.0011 > 0.000 001 \\ n = 2 \\ \Rightarrow \frac{{(\frac{π}{36})}^{5}}{5!} \approx 0.000 000 04 \leq 0.000 001 \end{array}$

Therefore, n = 2 is the smallest value of n for which the approximation is correct to 0.000 001.

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