Find the Taylor series for
f(x)
centered at the given value of a. [Assume that f has a power series expansion. Do not show that
Rn(x) → 0.]
f(x) = sin(x),    a = 𝜋

To find the Taylor series for \( f(x) = \sin(x) \) centered at \( a = \pi \), we need to calculate the derivatives of \( f(x) \) at \( x = \pi \) and then use the Taylor series formula:

\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n \]

For \( f(x) = \sin(x) \), we first determine the derivatives:

1. \( f(x) = \sin(x) \)
2. \( f'(x) = \cos(x) \)
3. \( f''(x) = -\sin(x) \)
4. \( f'''(x) = -\cos(x) \)
5. \( f^{(4)}(x) = \sin(x) \) (and this cycle repeats every four derivatives)

Next, evaluate these derivatives at \( x = \pi \):

1. \( f(\pi) = \sin(\pi) = 0 \)
2. \( f'(\pi) = \cos(\pi) = -1 \)
3. \( f''(\pi) = -\sin(\pi) = 0 \)
4. \( f'''(\pi) = -\cos(\pi) = 1 \)
5. \( f^{(4)}(\pi) = \sin(\pi) = 0 \) (and this cycle repeats every four derivatives)

Now, we can write the Taylor series expansion at \( x = \pi \):

\[ \sin(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(\pi)}{n!} (x - \pi)^n \]

Substituting the evaluated derivatives:

\[ \sin(x) = \frac{0}{0!} (x - \pi)^0 + \frac{-1}{1!} (x - \pi)^1 + \frac{0}{2!} (x - \pi)^2 + \frac{1}{3!} (x - \pi)^3 + \frac{0}{4!} (x - \pi)^4 + \cdots \]

Simplifying, we get:

\[ \sin(x) = -(x - \pi) + \frac{(x - \pi)^3}{6} + \cdots \]

Thus, the Taylor series for \( \sin(x) \) centered at \( a = \pi \) is:

\[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^{\frac{n+1}{2}}}{n!} (x - \pi)^n \quad \text{for odd} \ n \]

Or, more compactly, considering only the terms where \( n \) is odd:

\[ \sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} (x - \pi)^{2k+1} \]

### Do you want details or have any questions?

### Further Questions:

1. How is the Taylor series used to approximate functions near a point?
2. What are the convergence properties of the Taylor series for \(\sin(x)\) at different points?
3. How would the Taylor series change if centered at \( x = 0 \) instead of \( x = \pi \)?
4. Can you derive the Taylor series for \(\cos(x)\) centered at \( x = \pi \)?
5. How can we use the Taylor series to approximate \(\sin(x)\) near \( x = \pi \)?
6. What is the radius of convergence for the Taylor series of \(\sin(x)\)?
7. How would the series look if we included the remainder term \( R_n(x) \)?
8. What are some practical applications of Taylor series in physics and engineering?

### Tip:

Remember that the Taylor series is a powerful tool for approximating functions, especially when dealing with non-linear functions and evaluating them near specific points.

Learn how to find the Taylor series for sin(x) centered at π using derivative evaluations and the Taylor series expansion formula. This detailed solution covers the process of determining derivatives, evaluating them at π, and constructing the series. Suitable for advanced high school or college-level students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation