Answer:
[tex]\displaystyle \sin^3(x) = \sum^{\infty}_{n = 0} \bigg( \frac{(-1)^nx^{2n + 1}}{(2n + 1)!} \bigg)^3[/tex]
General Formulas and Concepts:
Calculus
Sequences
Series
Taylor Polynomials
- General Form: [tex]\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \cdots + \frac{f^n(c)}{n!}(x - c)^n + \cdots[/tex]
Power Series
- Power Series of Elementary Functions
Step-by-step explanation:
*Note:
You could derive the Taylor Series for sin(x) using Taylor polynomials using differentiation but usually you have to memorize it.
We are given and are trying to find the nth derivative of:
[tex]\displaystyle \sin^3(x)[/tex]
We know that a Taylor Polynomial infinitely differentiates a function in the form of a polynomial.
We know that the power series for sin(x) is:
[tex]\displaystyle \sin(x) = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}[/tex]
To find the power series for sin³x, we simply raise the entire series by 3:
[tex]\displaystyle \sin^3(x) = \sum^{\infty}_{n = 0} \bigg( \frac{(-1)^nx^{2n + 1}}{(2n + 1)!} \bigg)^3[/tex]
And we have our final answer.
Topic: AP Calculus BC (Calculus I + II)
Unit: Power Series
