Finding the Maclaurin Series for f(x) = cos(x^3)
The Maclaurin series is a special case of the Taylor series centered at 0. To find the Maclaurin series for the function f(x) = cos(x^3), we start by recalling the Taylor series expansion for the cosine function:
cos(x) = ∑ n=0 (−1)n / (2n)! * x2n
We can substitute x^3 for x in this expansion:
cos(x^3) = ∑ n=0 (−1)n / (2n)! * (x^3)2n
This simplifies to:
cos(x^3) = ∑ n=0 (−1)n / (2n)! * x6n
This indicates that the Maclaurin series for cos(x^3) is:
f(x) = 1 – rac{x^6}{2!} + rac{x^{12}}{4!} – rac{x^{18}}{6!} + rac{x^{24}}{8!} – …
Calculating f(6) using the Maclaurin Series
To estimate f(6) using the series, we will substitute x = 6 into the Maclaurin series we derived:
f(6) = 1 – rac{6^6}{2!} + rac{6^{12}}{4!} – rac{6^{18}}{6!} + rac{6^{24}}{8!} – …
Now we calculate the first few terms:
- First term: 1
- Second term: - rac{6^6}{2!} = - rac{46656}{2} = -23328
- Third term: + rac{6^{12}}{4!} = + rac{2176782336}{24} = 90532597.33
- Fourth term: - rac{6^{18}}{6!} = - rac{101559956668416}{720} = -140504025928.56
Adding these values together will give us:
f(6) ≈ 1 – 23328 + 90532597.33 – 140504025928.56
This calculation yields an estimated value of f(6), though to achieve a more accurate tail, we might need to compute more terms of the series.
It’s important to note that the Maclaurin series approximates the function well for values of x close to 0. For larger values, such as 6, this may yield less accurate results, and thus further terms may need to be calculated for better precision.