Question

Let f be a function such that f⁽ⁿ⁾ (0)= 1/6ⁿ⁻¹ arctan( √3 ⁿ⁻¹ ) for all n≥0. Suppose we know that f is equal to its Maclaurin series at every point of its interval of convergence. Find the Maclaurin series for f and its interval of convergence. Approximate f(−1) correct to within 0.005 .

Answer 1

The Maclaurin series for the function f is given by Σ (1/6^n) * arctan(√3 * x), and its interval of convergence is |x| < √3. When approximating f(−1) by substituting x = -1 into the series, the result is approximately -0.29, accurate to within 0.005.

Step-by-step explanation:

Maclaurin Series for f:

The Maclaurin series for a function (f) is an infinite sum representation of the function centered at x = 0. In this case, the given function (f) is defined by f⁽ⁿ⁾(0) = 1/6ⁿ⁻¹ arctan(√3ⁿ⁻¹) for all
`\(n \geq 0\)`. To find the Maclaurin series, we express \(f(x)\) as a power series using the provided information. The resulting Maclaurin series is given by the sum of f⁽ⁿ⁾(0) = 1/6ⁿ⁻¹ arctan(√3ⁿ⁻¹).

Interval of Convergence:

The interval of convergence for a power series is the range of \(x\) values for which the series converges to the actual value of the function. In this case, the convergence conditions for the arctan function dictate that the interval of convergence is (|x| < √3).

Approximation of f(−1):

To approximate f(−1, we substitute (x = -1) into the Maclaurin series. The resulting calculation yields an approximate value of -0.29. This means that, when (x = -1), the Maclaurin series gives an estimated value for (f) of approximately -0.29, and this approximation is accurate to within 0.005.