Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ \tan x, a=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

Taylor series: \( 1 + 2(x-\frac{\pi}{4}) + 2(x-\frac{\pi}{4})^2 + \frac{8}{3}(x-\frac{\pi}{4})^3 \).

Step 1: Understand Taylor Series Expansion

The Taylor series expansion of a function \( f(x) \) about a point \( a \) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). We need to find this expansion for \( \tan x \) at \( a = \frac{\pi}{4} \) up to the term \( (x-a)^3 \).

Step 2: Calculate Function and Derivatives at \( a \)

Start by calculating \( \tan(\frac{\pi}{4}) = 1 \). For derivatives: \( f'(x) = \sec^2 x \). So, \( f'(\frac{\pi}{4}) = 2 \). Next, \( f''(x) = 2 \sec^2 x \tan x \), giving \( f''(\frac{\pi}{4}) = 4 \). Lastly, \( f'''(x) = 4 \sec^2x + 4\sec^2 x \tan^2 x \), leading to \( f'''(\frac{\pi}{4}) = 16 \).

Step 3: Plug into Taylor Series Formula

Using these values, the Taylor series expansion can be constructed: \[ \tan x = 1 + 2(x-\frac{\pi}{4}) + \frac{4}{2!}(x-\frac{\pi}{4})^2 + \frac{16}{3!}(x-\frac{\pi}{4})^3 \].

Step 4: Simplify Taylor Series

Simplify each term: The \((x-\frac{\pi}{4})^1\) term is \(2(x-\frac{\pi}{4})\), the \((x-\frac{\pi}{4})^2\) term simplifies to \(2(x-\frac{\pi}{4})^2\), and the \((x-\frac{\pi}{4})^3\) term becomes \(\frac{8}{3}(x-\frac{\pi}{4})^3\). Thus, the series is \[ 1 + 2(x-\frac{\pi}{4}) + 2(x-\frac{\pi}{4})^2 + \frac{8}{3}(x-\frac{\pi}{4})^3 \].

Key Concepts

Derivatives

In calculus, a derivative represents how a function changes as its input changes. It's like finding the rate at which something is happening. For the Taylor series expansion of a function, derivatives play a crucial role. Each term in a Taylor series involves a derivative of the function at a specific point. In this exercise, we find the derivatives of the tangent function, specifically calculated at the point \(a = \frac{\pi}{4}\). Here's a simple breakdown of the derivatives involved:

• The first derivative of \(\tan(x)\) is \(\sec^2(x)\), which gives the rate of change of the tangent function. For \(x = \frac{\pi}{4}\), \(\sec^2(\frac{\pi}{4}) = 2\).
• The second derivative is \(2\sec^2(x)\tan(x)\), indicating how the rate itself is changing. Calculating this at \(x = \frac{\pi}{4}\) yields \(4\).
• The third derivative, which is more complex, is \(4\sec^2(x) + 4\sec^2(x) \tan^2(x)\), and at \(x = \frac{\pi}{4}\), the value is \(16\).

Derivatives thus form the foundation for constructing the series, each one allowing us to incrementally build the function's local behavior around \(x = \frac{\pi}{4}\).

Trigonometric Functions

Trigonometric functions describe the relationship between angles and lengths in right triangles. Functions like tangent (\(\tan\)), sine (\(\sin\)), and cosine (\(\cos\)) are essential in both geometry and calculus. In this exercise, we focus on the tangent function, which expresses the ratio of the opposite side to the adjacent side in a right triangle.The point \(a = \frac{\pi}{4}\) has special significance because it equates to 45 degrees, where the values of many trigonometric functions are well-known. Here, \(\tan(\frac{\pi}{4}) = 1\). Knowing this helps simplify calculations in the Taylor series expansion. Furthermore, derivative calculations require the use of other trigonometric functions like secant (\(\sec\)), which is the reciprocal of cosine (\(\cos\)). This illustrates how understanding trigonometric identities is vital for calculus, particularly when working with derivative evaluations of trigonometric functions.

Series Expansion

Series expansion is a powerful mathematical tool for expressing functions as the sum of an infinite series. The Taylor series is a type of series expansion that gives an approximation of a function around a specific point \(a\). For the function \(\tan(x)\), the series around \(x = \frac{\pi}{4}\) uses derivatives calculated at that point, expanding the function as a polynomial. Each term in the polynomial represents how closely the series approximates the function near that point. Unlike a simple polynomial, a Taylor series can be infinitely long, but we often truncate it to make it usable for practical purposes.In this case, the Taylor series expands up to the term \((x-a)^3\), creating a polynomial:

• Constant term: \(1\)
• \((x-a)^1\) term: \(2(x-\frac{\pi}{4})\)
• \((x-a)^2\) term: \(2(x-\frac{\pi}{4})^2\)
• \((x-a)^3\) term: \(\frac{8}{3}(x-\frac{\pi}{4})^3\)

This results in the simple expression \(1 + 2(x-\frac{\pi}{4}) + 2(x-\frac{\pi}{4})^2 + \frac{8}{3}(x-\frac{\pi}{4})^3\), making complex functions easier to handle and understand within a local context.