Problem 32 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=2^{x}, \quad a=1\)

Step-by-Step Solution

Answer

The Taylor series is \( f(x) = 2 + 2\ln(2)(x-1) + \frac{2(\ln(2))^2}{2!}(x-1)^2 + \cdots \).

Step 1: Understand the Formula

The Taylor series for a function \( f(x) \) around the point \( x = a \) is given by \( \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \). Here, we need to find the Taylor series for \( f(x) = 2^x \) at \( a = 1 \).

Step 2: Compute Derivatives

First, we need to find the derivatives of \( f(x) = 2^x \) and evaluate them at \( x = 1 \). The function and its derivatives are: \ \ \ \( f(x) = 2^x \), \ \ \ \( f'(x) = 2^x \ln(2) \), \ \ \ \( f''(x) = 2^x (\ln(2))^2 \), and so on. Evaluating these derivatives at \( x = 1 \):\ \ \ \( f(1) = 2^1 = 2 \),\ \ \ \( f'(1) = 2 \ln(2) \),\ \ \ \( f''(1) = 2 (\ln(2))^2 \).

Step 3: Write the Series

Using the formula for the Taylor series, we substitute \( f^{(n)}(1) \) back into the series expression: \ \ \ \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^n \] Substituting the values we found:\ \ \ \( f(x) = 2 + 2\ln(2)(x-1) + \frac{2(\ln(2))^2}{2!}(x-1)^2 + \frac{2(\ln(2))^3}{3!}(x-1)^3 + \cdots \)

Key Concepts

Function derivatives

In mathematics, finding derivatives is like uncovering the rate at which a function changes as its input changes. For the function given here, \( f(x) = 2^x \), derivatives help us to understand how this exponential function behaves as \( x \) varies.

The first derivative, \( f'(x) \), of this function is determined using the rule for taking derivatives of exponential functions. The derivative of \( e^x \), a common exponential function, is \( e^x \), but since our base here is 2, we multiply the original function by \( \ln(2) \). This results in \[ f'(x) = 2^x \ln(2) \].

Continuing this pattern allows us to find higher order derivatives like \( f''(x) = 2^x (\ln(2))^2 \), which are used in the Taylor series to approximate the function around a specific point. Each derivative reflects a layer of the function's behavior, allowing the Taylor series to mimic the function's shape more precisely.

Power series

A power series is simply an infinite sum of terms in which each term is a power of the variable of interest, here denoted as \( (x - a)^n \). In the Taylor series, these power series terms help approximate the function around a point \( a \).

The Taylor series formula \( \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \) systematically builds the approximation by adding more terms from the power series.

• The term \( n! \) in the denominator accounts for the number of terms (factorial of \( n \)), which adjusts each term's contribution.
• Each power series term \( (x-a)^n \) captures the behavior of the function near \( x = a \).

This approach can be applied to many functions if derivatives at the point \( a \) exist, making it a versatile tool for function approximation.

Exponential functions

Exponential functions like \( f(x) = 2^x \) are fundamental in mathematics due to their unique rates of growth. Unlike polynomial functions that grow at power rates, exponential functions have constant ratios of growth, which means each step changes multiplicatively.

For instance, \( 2^x \) grows by doubling as \( x \) increases by 1. In math, this is defined through constant bases raised to variable powers, often transformed in derivative calculations.

• The base \( 2 \) in \( f(x) = 2^x \) dictates its rate and form.
• The natural logarithm, \( \ln(2) \), acts as a scaling factor in this context, particularly when differentiating.

Understanding these growth patterns is crucial for analyzing systems that model natural phenomena, including populations or investments compounded over time. Taylor series help represent and approximate these functions near particular points using simpler polynomial forms, effectively making complex dynamics easier to work with.