How can we find the Maclaurin series for the function f(x) = cos(x^2) and use it to approximate f(80)?

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Finding the Maclaurin Series for f(x) = cos(x2)

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(x2), we first recall the Taylor series expansion for cosine:

cos(x) = ∑n=0∞ 
rac{(-1)nx2n}{(2n)!}

Now, by substituting x2 for x, we get:

cos(x2) = ∑n=0∞ 
rac{(-1)n(x2)2n}{(2n)!} = ∑n=0∞ 
rac{(-1)nx4n}{(2n)!}

Explicit Form of the Maclaurin Series

The explicit form of the Maclaurin series for f(x) = cos(x2) is thus:

f(x) = 1 - rac{x4}{2!} + rac{x8}{4!} - rac{x12}{6!} + rac{x16}{8!} - rac{x20}{10!} + …

Using the Series to Determine f(80)

Now that we have the Maclaurin series, we can use it to calculate f(80) = cos(802) = cos(6400).

To use the Maclaurin series, we will evaluate a few terms to see how accurately we can approximate this value.

  • First term: 1
  • Second term: - rac{804}{2!} = - rac{804}{2} = - rac{40960000}{2} = -20480000
  • Third term: + rac{808}{4!} = + rac{808}{24} = + rac{167772160000000}{24} ext{ (Large number, but for approximation, we keep it simple)}

Given that these series converge for small values of "x", the values become impractically large as x = 80. Instead, we can analyze the behavior of cos(x) as an oscillatory function ranging from -1 to 1. Therefore:

The value of cos(6400) will oscillate between -1 and 1, and calculating the Maclaurin series provides diminishing returns as x increases.

Conclusion

In summary, while the Maclaurin series offers a mathematical method to approximate f(80) or cos(6400), it highlights an important aspect of series expansions: with large inputs, particularly in trigonometric functions, simpler evaluations can often provide the needed insight without extensive computation.

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