A. [5 purtos] Determine los cuatro primeros términos (no nulos), de la serie de Taylor de la función $$ f(x)=\sin (2 x) $$ alrededor del punto $$ a=\pi $$. B. [3 purtos] Escriba la serie de Taytor usando la notación de Sumatoria.

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Alright, I need to determine the first four non-zero terms of the Taylor series for the function $$ f(x) = \sin(2x) $$ centered at $$ a = \pi $$. Let's break this down step by step.

First, recall that the Taylor series expansion of a function $$ f(x) $$ around a point $$ a $$ is given by:
$$
f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots
$$
So, I need to compute the derivatives of $$ f(x) = \sin(2x) $$ at $$ x = \pi $$ and then plug them into the series.

Let's start by finding the derivatives:
1. $$ f(x) = \sin(2x) $$
2. $$ f'(x) = 2\cos(2x) $$
3. $$ f''(x) = -4\sin(2x) $$
4. $$ f'''(x) = -8\cos(2x) $$
5. $$ f^{(4)}(x) = 16\sin(2x) $$
6. And so on, following the pattern.

Now, evaluate these derivatives at $$ x = \pi $$:
1. $$ f(\pi) = \sin(2\pi) = 0 $$
2. $$ f'(\pi) = 2\cos(2\pi) = 2 \times 1 = 2 $$
3. $$ f''(\pi) = -4\sin(2\pi) = -4 \times 0 = 0 $$
4. $$ f'''(\pi) = -8\cos(2\pi) = -8 \times 1 = -8 $$
5. $$ f^{(4)}(\pi) = 16\sin(2\pi) = 16 \times 0 = 0 $$
6. $$ f^{(5)}(\pi) = 32\cos(2\pi) = 32 \times 1 = 32 $$
7. $$ f^{(6)}(\pi) = -64\sin(2\pi) = -64 \times 0 = 0 $$
8. $$ f^{(7)}(\pi) = -128\cos(2\pi) = -128 \times 1 = -128 $$
9. $$ f^{(8)}(\pi) = 256\sin(2\pi) = 256 \times 0 = 0 $$
10. $$ f^{(9)}(\pi) = 512\cos(2\pi) = 512 \times 1 = 512 $$

Now, plug these into the Taylor series formula:
$$
f(x) = 0 + 2(x - \pi) + \frac{0}{2!}(x - \pi)^2 + \frac{-8}{3!}(x - \pi)^3 + \frac{0}{4!}(x - \pi)^4 + \frac{32}{5!}(x - \pi)^5 + \cdots
$$
Simplifying, the non-zero terms are:
$$
f(x) = 2(x - \pi) - \frac{8}{6}(x - \pi)^3 + \frac{32}{120}(x - \pi)^5 - \cdots
$$
Which simplifies further to:
$$
f(x) = 2(x - \pi) - \frac{4}{3}(x - \pi)^3 + \frac{4}{15}(x - \pi)^5 - \cdots
$$
So, the first four non-zero terms are:
1. $$ 2(x - \pi) $$
2. $$ -\frac{4}{3}(x - \pi)^3 $$
3. $$ \frac{4}{15}(x - \pi)^5 $$
4. $$ -\frac{16}{105}(x - \pi)^7 $$

Now, for part B, I need to write the Taylor series using summation notation. The general form of the Taylor series is:
$$
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n
$$
Given that the non-zero terms occur at odd powers of $$ (x - \pi) $$, the series can be expressed as:
$$
f(x) = \sum_{n=0}^{\infty} \frac{f^{(2n+1)}(\pi)}{(2n+1)!}(x - \pi)^{2n+1}
$$
Substituting the derivatives:
$$
f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1}}{(2n+1)!}(x - \pi)^{2n+1}
$$
This represents the Taylor series for $$ \sin(2x) $$ centered at $$ \pi $$.

A. Los cuatro primeros términos no nulos de la serie de Taylor de $$ f(x) = \sin(2x) $$ alrededor de $$ a = \pi $$ son: 1. $$ 2(x - \pi) $$ 2. $$ -\frac{4}{3}(x - \pi)^3 $$ 3. $$ \frac{4}{15}(x - \pi)^5 $$ 4. $$ -\frac{16}{105}(x - \pi)^7 $$ B. La serie de Taylor utilizando la notación de Sumatoria es: $$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1}}{(2n+1)!}(x - \pi)^{2n+1} $$

A. [5 purtos] Determine los cuatro primeros términos (no nulos), de la serie de Taylor de la función \( f(x)=\sin (2 x) \) alrededor del punto \( a=\pi \). B. [3 purtos] Escriba la serie de Taytor usando la notación de Sumatoria.

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