Back to Taylor series

Exercise 9

Avg difficulty 2.5

As we will see in more detail in the next section, we can use known Taylor series to obtain other Taylor series, and we preview that idea in this exercise.

  1. Calculate the first four derivatives of sin⁡(x2)\sin(x^2) and hence find the fourth order Taylor polynomial for sin⁡(x2)\sin(x^2) centered at a=0a=0.

  2. Part (a) demonstrates the direct approach to finding Taylor polynomials and series. Next we utilize a known Taylor series to make the process simpler. Recall that the Taylor series centered at 0 for f(x)=sin⁡(x)f(x) = \sin(x) is T(x)=∑k=0∞(−1)kx2k+1(2k+1)!\begin{gathered} T(x) = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} \end{gathered}.

    1. Substitute x2x^2 for xx in the Taylor series T(x)T(x) in Equation. Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should result in the Taylor series for sin⁡(x2)\sin(x^2) centered at 0.

    2. For what interval of xx-values should we expect the Taylor series for sin⁡(x2)\sin(x^2) to converge?

Boelkins, Active Calculus, §8.4 Taylor series, Exercise 9

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.