Question

Use the alternating series estimation theorem to estimate the range of values of x for which the given approximation is accurate to within the stated error. check your answer graphically. (round your answers to three decimal places.) sin(x) ≈ x − x3 6 |error| < 0.000001

Answer 1

The expansion series of sin(x) about 0 is
$sin(x)=x/1!-x^3/3!+x^5/5!-x^7/7!+x^9/9!-...$  ...............(1)

Using the given approximation S(x)=x-x^3/3!           ................(2)
Therefore the error term for the approximation is 
error=(2)-(1)
$=-(+x^5/5!-x^7/7!+x^9/9!-..)$
$=-x^5/5!+x^7/7!-x^9/9!-..$

The alternative series theorem says we need only to ensure that the absolute value of the first term dropped (x^5/5!) is less than the error limit, in order to ensure that the sum will be below the error limit
This means that
|x^5/5!| <  0.000001

Solving for x:
x^5=5!*0.000001=0.000120
x=(0.000120)^(1/5)=0.164375

Thus the approximation of sin(x) ≈ x-x^3/3! has an absolute error below 0.000001 for the interval [-0.164375,+0.164375].


Check:
sin(0.164375)-(0.164375)^3/3!=9.993513*10^(-7) < 0.000001

Answer 2

The approximation sin (x) = x - x^3/6 has an absolute error below 0.000001 for the interval [-0.164375,+0.164375].

What is alternating series?

In mathematics, the alternating series is an infinite series of a term in the form of

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}}{\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}} \\{\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}a_{n}}{\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}a_{n}}$

The expansion series of sin(x)

$\rm sin (x) = \frac{x}{1!} -\frac{x^{3} }{3!} +\frac{x^{5} }{5!} -\frac{x^{7} }{7!} +..$  (a)

the given approximation $\rm sin (x) = \frac{x}{1!} -\frac{x^{3} }{3!}$     (b)

So, the approximation error term =(b)-(a)

       $\rm = -[ +\frac{x^{5} }{5!} -\frac{x^{7} }{7!} +\frac{x^{9} }{9!}-....]\\\rm = -\frac{x^{5} }{5!} +\frac{x^{7} }{7!} -\frac{x^{9} }{9!}+....$

The alternative series theorem states that we need to ensure that the absolute value of the first term $\rm \frac{x^{5} }{5!}$less than the error limit, For that the sum will be below the error limit, which means

$\rm \frac{x^{5} }{5!}$ <  0.000001

Solving for x, we get

x $\rm x^{5} = 5! \times 0.000001\\ = 0.000120\\\rm = 0.000120^{1/5} \\ = 0.164375$

Therefore, the approximation   $\rm sin (x) = \frac{x}{1!} -\frac{x^{3} }{3!}$ has an absolute error below 0.000001 for the interval [-0.164375,+0.164375].

by checking

$\rm sin(0.164375)-(0.164375)^{3/3!} \\= 9.993513\times10^{-7} < 0.000001$

Learn more about alternating series;

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