What is the maximum possible error generated in using the 5th degree Maclaurin polynomial for sin(x) for x in the interval [−0.3, 0.3]?
a). The absolute value of 0.3 raised to the 7th power over 6 factorial
b). The absolute value of 0.3 raised to the 6th power over 6 factorial
c). The value cannot be determined.
d). There is no error. The value will be exact.
take a look at Lagrange's formula, and it should be clear.
Can you please take me through it? I'm having trouble plugging the values into the formula.
|En(x)| ≤ M/(n+1)! * |x - a|^(n+1)
Please! I don't know what x, a, or M are.
To find the maximum possible error generated when using the 5th degree Maclaurin polynomial for sin(x), we can use Taylor's theorem with Lagrange remainder. The Lagrange remainder formula tells us that the error term for a given polynomial is equal to the next term in the series divided by the factorial of the degree of the polynomial multiplied by a value of x raised to the power of the degree plus one.
In this case, since we are using the 5th degree Maclaurin polynomial for sin(x), the error term would be equal to the 6th degree term in the series divided by 6 factorial, multiplied by x raised to the power of 6. The 6th degree term can be found by differentiating the sin(x) function 6 times.
Now let's calculate the error term for x = 0.3:
The 6th degree term in the Maclaurin series for sin(x) is (-1)^6 * (0.3)^6 / 6!, which simplifies to (0.3)^6 / 720.
So, the maximum possible error generated when using the 5th degree Maclaurin polynomial for sin(x) in the interval [-0.3, 0.3] is given by the absolute value of (0.3)^6 / 720.
Therefore, the correct answer is option b). The absolute value of 0.3 raised to the 6th power over 6 factorial.