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Sagot :
Answer:
a. [tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n + 10}}{(2n)!} + C_1[/tex]
b. [tex]\displaystyle P_5(x) = \frac{x^{10}}{10} - \frac{x^{16}}{16 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{28}}{28 \cdot 6!} + \frac{x^{34}}{34 \cdot 8!} + C_1[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Integration
- Integrals
- Indefinite Integrals
- Integration Constant C
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Sequences
Series
MacLaurin Polynomials
- MacLaurin Polynomial: [tex]\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^n(0)}{n!}x^n + \cdots[/tex]
- MacLaurin Series: [tex]\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(0)}{n!}x^n[/tex]
Power Series
- Power Series of Elementary Functions
- MacLaurin Polynomials
- Taylor Polynomials
Integration of Power Series:
- [tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} a_n(x - c)^n[/tex]
- [tex]\displaystyle \int {f(x)} \, dx = \sum^{\infty}_{n = 0} \frac{a_n(x - c)^{n + 1}}{n + 1} + C_1[/tex]
Step-by-step explanation:
*Note:
You could derive the Taylor Series for cos(x) using Taylor polynomials differentiation but usually you have to memorize it.
We are given and are trying to find the power series:
[tex]\displaystyle f(x) = \int {x^9cos(x^3)} \, dx[/tex]
We know that the power series for cos(x) is:
[tex]\displaystyle cos(x) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{2n}}{(2n)!}[/tex]
To find the power series for cos(x³), substitute in x = x³:
[tex]\displaystyle cos(x^3) = \sum^{\infty}_{n = 0} \frac{(-1)^n (x^3)^{2n}}{(2n)!}[/tex]
Simplifying it, we have:
[tex]\displaystyle cos(x^3) = \sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n}}{(2n)!}[/tex]
Rewrite the original function:
[tex]\displaystyle f(x) = \int {x^9 \sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n}}{(2n)!}} \, dx[/tex]
Rewrite the integrand by including the x⁹ in the power series:
[tex]\displaystyle f(x) = \int {\sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n + 9}}{(2n)!}} \, dx[/tex]
Integrating the power series, we have:
[tex]\displaystyle f(x) = {\sum^{\infty}_{n = 0} \frac{(-1)^n x^{6n + 10}}{(6n + 10)(2n)!}} + C_1[/tex]
To find the first 5 nonzero terms of the power series, we simply expand it as a MacLaurin polynomial:
[tex]\displaystyle P_5(x) = {\sum^4_{n = 0} \frac{(-1)^n x^{6n + 10}}{(6n + 10)(2n)!}} + C_1 = \frac{x^{10}}{10} - \frac{x^{16}}{16 \cdot 2!} + \frac{x^{22}}{22 \cdot 4!} - \frac{x^{28}}{28 \cdot 6!} + \frac{x^{34}}{34 \cdot 8!} + C_1[/tex]
Topic: AP Calculus BC (Calculus I + II)
Unit: Power Series
Book: College Calculus 10e
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