The 4th degree Maclaurin polynomial for f(x)=cosx isT4(x)=1−12x2+124x4. Determine the maximum error resulting from using this polynomial to estimate cos(0.1) . Select the correct answer below:|R4(0.1)|≤0.8333 |R4(0.1)|≤0.00008333 |R4(0.1)|≤0.00000008333 |R4(0.1)|≤0.008333 FEEDBACKMORE INSTRUCTION
Question
The 4th degree Maclaurin polynomial for f(x)=cosx isT4(x)=1−12x2+124x4. Determine the maximum error resulting from using this polynomial to estimate cos(0.1) . Select the correct answer below:|R4(0.1)|≤0.8333 |R4(0.1)|≤0.00008333 |R4(0.1)|≤0.00000008333 |R4(0.1)|≤0.008333 FEEDBACKMORE INSTRUCTION
Solution
The error of a Taylor (or Maclaurin) polynomial is given by the remainder term, which can be calculated using the formula:
|Rn(x)| ≤ (M / (n+1)!) * |x - a|^(n+1)
where M is the maximum value of the (n+1)th derivative of the function on the interval from a to x.
In this case, we are looking for the error of the 4th degree Maclaurin polynomial for f(x) = cos(x) at x = 0.1. The Maclaurin series is a Taylor series expansion of a function about 0, so a = 0.
The (n+1)th derivative of cos(x) is sin(x) or cos(x) or -sin(x) or -cos(x), depending on the value of n. The maximum value of these functions on the interval from 0 to 0.1 is 1.
So, M = 1, n = 4, a = 0, and x = 0.1.
|R4(0.1)| ≤ (1 / (4+1)!) * |0.1 - 0|^(4+1) |R4(0.1)| ≤ (1 / 120) * (0.1)^5 |R4(0.1)| ≤ 0.00000008333
So, the maximum error resulting from using this polynomial to estimate cos(0.1) is |R4(0.1)| ≤ 0.00000008333.
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