Taylor series


a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = 2ˣ, a = 1

Verified step by step guidance

1. Recall the definition of the Taylor series of a function \(f(x)\) centered at \(a\): \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
2. Identify the function and the center: here, \(f(x) = 2^x\) and \(a = 1\). We will need to find the derivatives of \(f(x)\) and evaluate them at \(x=1\).
3. Find the first four derivatives of \(f(x)\): - \(f(x) = 2^x\) - \(f'(x) = 2^x \ln(2)\) - \(f''(x) = 2^x (\ln(2))^2\) - \(f^{(3)}(x) = 2^x (\ln(2))^3\) - \(f^{(4)}(x) = 2^x (\ln(2))^4\)
4. Evaluate each derivative at \(x = 1\): - \(f(1) = 2^1\) - \(f'(1) = 2^1 \ln(2)\) - \(f''(1) = 2^1 (\ln(2))^2\) - \(f^{(3)}(1) = 2^1 (\ln(2))^3\) - \(f^{(4)}(1) = 2^1 (\ln(2))^4\)
5. Write the first four nonzero terms of the Taylor series using the formula: \[f(x) \approx \sum_{n=0}^3 \frac{f^{(n)}(1)}{n!} (x - 1)^n = \sum_{n=0}^3 \frac{2 \cdot (\ln(2))^n}{n!} (x - 1)^n,\] which explicitly is: \[2 + 2 \ln(2)(x-1) + \frac{2 (\ln(2))^2}{2!} (x-1)^2 + \frac{2 (\ln(2))^3}{3!} (x-1)^3.\]

Key Concepts

Taylor Series Definition

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point. Each term involves the nth derivative evaluated at the center point, multiplied by (x - a)^n and divided by n!. This series approximates the function near the center a.

Derivatives of Exponential Functions

For the function f(x) = 2^x, derivatives involve the natural logarithm of the base. Specifically, the nth derivative of 2^x is (ln 2)^n times 2^x. Understanding this pattern is essential to compute the terms of the Taylor series accurately.

Evaluating the Series at the Center Point

To find the Taylor series terms centered at a = 1, each derivative must be evaluated at x = 1. These values are then used in the formula for each term, ensuring the series accurately approximates the function near x = 1.