Problem 32 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=\sqrt{x+1}, \quad a=0\)

Step-by-Step Solution

Answer

The Taylor series is \( 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots \).

Step 1: Find the derivatives of f

First, identify the function provided: \( f(x) = \sqrt{x+1} = (x+1)^{1/2} \). Compute the first few derivatives at \( x = 0 \). 1. First derivative: \( f'(x) = \frac{1}{2}(x+1)^{-1/2} \)2. Second derivative: \( f''(x) = -\frac{1}{4}(x+1)^{-3/2} \)3. Third derivative: \( f'''(x) = \frac{3}{8}(x+1)^{-5/2} \)Continue this process for more derivatives as needed.

Step 2: Evaluate derivatives at x = a = 0

Now, substitute \( x = 0 \) into each derivative to evaluate them at this point.1. \( f(0) = \sqrt{1} = 1 \)2. \( f'(0) = \frac{1}{2} \cdot 1^{-1/2} = \frac{1}{2} \)3. \( f''(0) = -\frac{1}{4} \cdot 1^{-3/2} = -\frac{1}{4} \)4. \( f'''(0) = \frac{3}{8} \cdot 1^{-5/2} = \frac{3}{8} \)These values will be used in forming the Taylor series.

Step 3: Write the Taylor series formula

The Taylor series for a function \( f(x) \) about \( a = 0 \) is given by:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] For this exercise, \( a = 0 \), so it simplifies to:\[\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]

Step 4: Substitute the evaluated derivatives into the series

Substitute the derivative values from Step 2 back into the Taylor series formula:- The constant term: \( f(0) = 1 \)- The \( x \) term: \( \frac{f'(0)}{1!} x = \frac{1}{2}x \)- The \( x^2 \) term: \( \frac{f''(0)}{2!} x^2 = -\frac{1}{8}x^2 \)- The \( x^3 \) term: \( \frac{f'''(0)}{3!} x^3 = \frac{1}{16}x^3 \)So the series begins: \[ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots \]

Step 5: Write the general term and conclude the Taylor series

Recognize the pattern from the evaluated terms: The nth term of the series can be written as:\[ \sum_{n=0}^{\infty} \frac{(-1)^{n} \cdot (2n)!}{(1-2n)(2^n)(n!)^2} x^n \]Thus, the Taylor series generated by \(f(x) = \sqrt{x+1}\) at \(x = 0\) is\[ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots \]

Key Concepts

Derivatives

Derivatives are fundamental in calculus, serving as tools to understand how functions change. In essence, the derivative of a function at a specific point indicates the rate at which the function's value changes as the input changes. For instance, when we take the function \( f(x) = \sqrt{x+1} = (x+1)^{1/2} \), we compute its derivatives to help construct the Taylor series.

To start, the first derivative \( f'(x) = \frac{1}{2}(x+1)^{-1/2} \) calculates how \( f(x) \) changes initially. Following this, higher-order derivatives, such as the second \( f''(x) = -\frac{1}{4}(x+1)^{-3/2} \) and third \( f'''(x) = \frac{3}{8}(x+1)^{-5/2} \) derivatives, provide nuanced insights into the function's behavior. Evaluating these derivatives at \( x = 0 \) delivers specific values that are crucial for constructing the Taylor series. This evaluation includes:

• \( f(0) = 1 \)
• \( f'(0) = \frac{1}{2} \)
• \( f''(0) = -\frac{1}{4} \)
• \( f'''(0) = \frac{3}{8} \)

These calculated derivatives at specific points encapsulate the function's behavior around \( x = 0 \). Knowing how to derive and use these derivatives enables strong foundational understanding in constructing series like the Taylor series.

Maclaurin Series

The Maclaurin series is a specific case of the Taylor series, centered at \( a = 0 \). It's an excellent tool for expressing a function as an infinite sum of terms calculated from the derivatives at zero. When the function \( f(x) \) is smooth enough over its domain, we can use its Maclaurin series to approximate values around \( x = 0 \).

The Maclaurin series formula is given by:\[\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]
This series expansion allows us to express the function \( \sqrt{x+1} \) using derivatives evaluated at \( x = 0 \). We observed the derivatives calculated previously, leading to terms in our series like:

• Constant term: \( f(0) = 1 \)
• Linear: \( \frac{f'(0)}{1!} x = \frac{1}{2}x \)
• Quadratic: \( \frac{f''(0)}{2!} x^2 = -\frac{1}{8}x^2 \)
• Cubic: \( \frac{f'''(0)}{3!} x^3 = \frac{1}{16}x^3 \)

These allow for a straightforward approximation of the function near zero, providing a powerful way to approximate complex functions without drastic computational methods.

Function Expansion

Function expansion is a method in calculus that enables us to express a function as a series; in this case, a Taylor or Maclaurin series. This expansion can be crucial for approximating functions that are otherwise difficult to compute directly. The series representation gives us a way to understand the function's behavior by examining the coefficients in front of each term.

The process involves several steps. First, we determine how many derivatives we need, which inform the terms of our series. Each derivative contributes a term based on its specific value and factorial division. By expanding the function \( f(x) = \sqrt{x+1} \) at \( a = 0 \), we achieved the representation:
\[ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots \]
This series reflects an approximate version of our original function, suitable for values of \( x \) near zero. The expansion reveals important aspects of \( f(x) \), such as how rapidly the function increases or decreases around \( x = 0 \). Understanding function expansion thus enables us to analyze and solve problems involving complex functions with relative ease.