MATH SOLVE

9 months ago

Q:
# 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

Accepted Solution

A:

The expansion series of sin(x) about 0 is

[tex]sin(x)=x/1!-x^3/3!+x^5/5!-x^7/7!+x^9/9!-...[/tex] ...............(1)

Using the given approximation S(x)=x-x^3/3! ................(2)

Therefore the error term for the approximation is

error=(2)-(1)

[tex]=-(+x^5/5!-x^7/7!+x^9/9!-..)[/tex]

[tex]=-x^5/5!+x^7/7!-x^9/9!-..[/tex]

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

[tex]sin(x)=x/1!-x^3/3!+x^5/5!-x^7/7!+x^9/9!-...[/tex] ...............(1)

Using the given approximation S(x)=x-x^3/3! ................(2)

Therefore the error term for the approximation is

error=(2)-(1)

[tex]=-(+x^5/5!-x^7/7!+x^9/9!-..)[/tex]

[tex]=-x^5/5!+x^7/7!-x^9/9!-..[/tex]

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