Math Problem Statement

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

Solution

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

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

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

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

Next, evaluate these derivatives at x=πx = \pi:

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

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

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

Substituting the evaluated derivatives:

sin(x)=00!(xπ)0+11!(xπ)1+02!(xπ)2+13!(xπ)3+04!(xπ)4+\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)=(xπ)+(xπ)36+\sin(x) = -(x - \pi) + \frac{(x - \pi)^3}{6} + \cdots

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

sin(x)=n=0(1)n+12n!(xπ)nfor odd n\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 nn is odd:

sin(x)=k=0(1)k(2k+1)!(xπ)2k+1\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)\sin(x) at different points?
  3. How would the Taylor series change if centered at x=0x = 0 instead of x=πx = \pi?
  4. Can you derive the Taylor series for cos(x)\cos(x) centered at x=πx = \pi?
  5. How can we use the Taylor series to approximate sin(x)\sin(x) near x=πx = \pi?
  6. What is the radius of convergence for the Taylor series of sin(x)\sin(x)?
  7. How would the series look if we included the remainder term Rn(x)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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Taylor Series
Trigonometric Functions

Formulas

Taylor series expansion formula
Sine function derivative formulas

Theorems

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Suitable Grade Level

Advanced High School or College