(1 point) match each of the maclaurin series with the function it represents.

1. $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$

2. $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$

3. $\sum_{n=0}^{\infty} \frac{x^n}{n!}$

4. $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$

a. $\arctan(x)$
b. $\sin(x)$
c. $e^x$
d. $\cos(x)$

Question

(1 point) match each of the maclaurin series with the function it represents.

1. $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$

2. $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$

3. $\sum_{n=0}^{\infty} \frac{x^n}{n!}$

4. $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$

a. $\arctan(x)$
b. $\sin(x)$
c. $e^x$
d. $\cos(x)$

Accepted Answer

{ "quality": "clear", "question_count": 4, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Maclaurin Series", "Power Series Representation", "Taylor Series" ], "current_concepts": [ "Maclaurin Series", "Power Series Representation", "Taylor Series" ] } ## Identify standard Maclaurin series expansions $$ \begin{aligned} \arctan(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \ \sin(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \ e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \ \cos(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \end{aligned} $$ ## Match given series with functions $$ \begin{aligned} 1. &\quad \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \implies \text{A. } \arctan(x) \ 2. &\quad \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \implies \text{B. } \sin(x) \ 3. &\quad \sum_{n=0}^{\infty} \frac{x^n}{n!} \implies \text{C. } e^x \ 4. &\quad \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \implies \text{D. } \cos(x) \end{aligned} $$ | No. | Answer | |---|---| | 1 | A | | 2 | B | | 3 | C | | 4 | D | { "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Calculus", "Maclaurin Series" ] }

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Question

Explanation:

Identify standard Maclaurin series expansions

Match given series with functions

Answer: