Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Compute the third degree Taylor polynomial generated by \(\sin x\) at \(x=\frac{\pi}{4}\).

Step-by-Step Solution

Answer

The third degree Taylor polynomial generated by \(\sin x\) at \(x=\frac{\pi}{4}\) is \(P_3(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} \cdot (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{48} \cdot (x - \frac{\pi}{4})^3\).

Step 1: Compute the derivatives of the function

Calculate the first three derivatives of the function \(\sin x\). The first derivative \(f'(x)\) is \(\cos x\). The second derivative \(f''(x)\) is \(-\sin x\). And the third derivative \(f'''(x)\) is \(-\cos x\).

Step 2: Evaluate the derivatives at \(x=\frac{\pi}{4}\)

Replace \(x\) in the derivatives computed in the previous step with \(\frac{\pi}{4}\).You get:\(f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\)\(f''(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\)\(f'''(\frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\)

Step 3: Substituting the values into the Taylor series formula

Put the values obtained in the previous step into the Taylor series formulaYou get:\(P_3(x) = f(\frac{\pi}{4}) + f'(\frac{\pi}{4}) \cdot (x - \frac{\pi}{4}) + \frac{f''(\frac{\pi}{4}) \cdot (x - \frac{\pi}{4})^2}{2!} + \frac{f'''(\frac{\pi}{4}) \cdot (x - \frac{\pi}{4})^3}{3!}\)Calculating these:\(P_3(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot (x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4} \cdot (x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{48} \cdot (x - \frac{\pi}{4})^3\)

Key Concepts

sin x Series Expansion

Understanding how to expand the sine function, or any trigonometric function for that matter, into a power series is crucial for calculative applications in mathematics and physics. The series expansion of \textbf{sin x} is given by the infinite sum of powers of x, combined with factorial denominators and alternating signs. This expansion is a part of the Maclaurin series, which is a special case of the Taylor series around zero.

In general, the sine function can be expanded as: \[\begin{equation}\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\end{equation}\]Each term in this series involves a higher odd power of x and is derived from the function's derivatives. Why odd powers? Because the sine function is an odd function, exhibiting symmetry about the origin. This characteristic is reflected in its series expansion. When computing a Taylor polynomial for sine, such as the third-degree polynomial in our example, we're essentially truncating this infinite series to a finite number of terms—giving us an approximation of the sine function around a specific point.

Taylor Series Formula

The Taylor series formula is a powerful tool that approximates complex functions using polynomials. The approximate polynomial is constructed so that its value and the first few derivatives at a specific point match those of the original function. The general formula for a Taylor series around a point 'a' is given by:\[\begin{equation}P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x - a)^k\end{equation}\]Where:

• \( P_n(x) \) is the nth degree Taylor polynomial.
• \( f^{(k)}(a) \) represents the kth derivative of the function evaluated at point 'a'.
• \( k! \) is the factorial of k, representing the product of all positive integers up to k.
• \( x \) is the variable in which the function is approximated.
• \( a \) is the point around which the approximation is centered.

The 'nth degree' refers to the highest power of (x - a) that appears in the polynomial. For instance, in our exercise, we're looking for the third degree Taylor polynomial for \(\sin x\) centered at \(x = \frac{\pi}{4}\), which means up to the third derivative of the sine function is used, and the highest power will be \((x - \frac{\pi}{4})^3\).

Derivatives of Trigonometric Functions

Working with the derivatives of trigonometric functions is fundamental when applying the Taylor series formula. Because trigonometric functions are periodic and exhibit specific symmetries, their derivatives follow a cyclical pattern. The sine function (\textbf{sin x}), specifically, has a derivative sequence that repeats every four steps:\[\begin{equation} f(x) = \sin x \Rightarrow f'(x) = \cos x \Rightarrow f''(x) = -\sin x \Rightarrow f'''(x) = -\cos x \Rightarrow f^{(4)}(x) = \sin x \end{equation}\]Knowing this repeating pattern allows us to calculate the derivatives needed for the Taylor series quickly. Considering the derivative involves taking the rate of change at an instant, for trigonometric functions, these derivatives translate to slopes and curvatures at specific points along their curves. In our example, we computed the first three derivatives of \(\sin x\) at \(\frac{\pi}{4}\) to construct the third degree Taylor polynomial. These derivatives provided the coefficients for each term of the polynomial, each coefficient capturing a different aspect of the sine curve's behavior around the point \(\frac{\pi}{4}\).