Math Problem Statement

sin x centered at x= π/4

Solution

To find the Taylor series expansion of the function sin(x)\sin(x) centered at x=π4x = \frac{\pi}{4}, we will express sin(x)\sin(x) as a Taylor series around x=π4x = \frac{\pi}{4}.

Taylor Series Formula

The Taylor series of a function f(x)f(x) centered at x=ax = a is given by:

f(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \cdots

In this case, f(x)=sin(x)f(x) = \sin(x) and a=π4a = \frac{\pi}{4}.

Step-by-Step Calculation

  1. Compute the derivatives of sin(x)\sin(x):
f(x) &= \sin(x), \\ f'(x) &= \cos(x), \\ f''(x) &= -\sin(x), \\ f'''(x) &= -\cos(x), \\ f^{(4)}(x) &= \sin(x), \quad \text{(and repeats in cycles)} \end{align*}$$ 2. **Evaluate the derivatives at $$x = \frac{\pi}{4}$$:** $$\begin{align*} f\left(\frac{\pi}{4}\right) &= \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \\ f'\left(\frac{\pi}{4}\right) &= \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \\ f''\left(\frac{\pi}{4}\right) &= -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \\ f'''\left(\frac{\pi}{4}\right) &= -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \\ f^{(4)}\left(\frac{\pi}{4}\right) &= \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}. \end{align*}$$ 3. **Construct the Taylor series:** Substitute these values into the Taylor series formula: $$\sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} \cdot (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12} \cdot (x - \frac{\pi}{4})^3 + \cdots$$ ### Simplified Taylor Series for $$\sin(x)$$ around $$x = \frac{\pi}{4}$$: $$\sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12} (x - \frac{\pi}{4})^3 + \cdots$$ This gives the Taylor expansion up to the third degree. You can continue the series by calculating higher-order derivatives and including more terms as needed. Would you like to see more details or have any questions about this process? ### Additional Related Questions: 1. How do you derive the Taylor series for other trigonometric functions like $$\cos(x)$$ around different points? 2. What is the radius of convergence for the Taylor series of $$\sin(x)$$? 3. How can you use Taylor series to approximate $$\sin(x)$$ for small values of $$x$$? 4. What is the significance of choosing a different center point for a Taylor series expansion? 5. How do you determine the error or remainder of a Taylor series approximation? #### Tip: When approximating a function using a Taylor series, the more terms you include, the closer the approximation will be to the original function, especially near the center point $$x = a$$.

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Math Problem Analysis

Mathematical Concepts

Calculus
Taylor Series
Trigonometry

Formulas

Taylor series expansion: f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3 + ...
Derivatives of sin(x): f'(x) = cos(x), f''(x) = -sin(x), f'''(x) = -cos(x), ...

Theorems

Taylor's Theorem

Suitable Grade Level

Undergraduate Calculus