Problem 46 - Thomas Calculus

Question

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a .\) Find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=e^{\sin x}\)

Step-by-Step Solution

Answer

(a) Linearization: \(1 + x\); (b) Quadratic approximation: \(1 + x + \frac{x^2}{2}\).

Step 1: Find Function Value at x=0

To begin, determine the value of the function \( f(x) = e^{\sin x} \) at \( x = 0 \). Since \( \sin(0) = 0 \), it follows that \( f(0) = e^0 = 1 \).

Step 2: Calculate the First Derivative

Next, we calculate the first derivative of \( f(x) \). Using the chain rule, \( f'(x) = \frac{d}{dx} e^{\sin x} = e^{\sin x} \cdot \cos x \). Evaluating this at \( x = 0 \), we get \( f'(0) = e^0 \cdot \cos(0) = 1 \cdot 1 = 1 \).

Step 3: Linearization at x=0

The linear approximation (or linearization) of \( f \) at \( x = 0 \) is given by the Taylor polynomial of order 1: \( L(x) = f(0) + f'(0) \cdot (x - 0) = 1 + 1 \cdot x = 1 + x \).

Step 4: Calculate the Second Derivative

Calculate the second derivative of \( f(x) \): \( f''(x) = \frac{d}{dx} (e^{\sin x} \cdot \cos x) \). Applying the product rule and chain rule, it is \( f''(x) = e^{\sin x} \cdot (\cos^2 x - \sin x) \). Evaluating at \( x = 0 \), \( f''(0) = e^0 \cdot (\cos^2(0) - \sin(0)) = 1 \cdot (1 - 0) = 1 \).

Step 5: Quadratic Approximation at x=0

The quadratic approximation of \( f \) at \( x = 0 \) uses the Taylor polynomial of order 2: \( Q(x) = f(0) + f'(0) \cdot x + \frac{1}{2}f''(0) \cdot x^2 = 1 + x + \frac{1}{2} \cdot 1 \cdot x^2 = 1 + x + \frac{x^2}{2} \).

Key Concepts

Taylor polynomial

The Taylor polynomial is a powerful mathematical tool for approximating functions. When you can't solve a complex function directly, Taylor polynomials come to the rescue by helping you estimate the value of the function at a specific point with great simplicity. The most basic version of a Taylor polynomial is the order 1 polynomial, which is also called linearization.
Taylor polynomials of higher order can give even better approximations. For example, a quadratic approximation utilizes a Taylor polynomial of order 2 to give a closer estimate. Quadratic approximation adds one more term to the linear approximation, bringing in a second derivative term that can more closely mirror the curve of the function over a small region near the point of approximation.
In the exercise above, the aim is to approximate the function \(f(x) = e^{\sin x}\) at \(x = 0\) using both linear and quadratic approximations. Knowing this helps us understand that Taylor polynomials give a series of increasingly precise approximations depending on the number of derivatives taken into account.

derivatives

Derivatives are a fundamental concept in calculus used to observe how a function changes as its input changes. The derivative tells us the slope of the tangent line to the function at any given point, essentially providing the rate of change at that point.
In the process of quadratic approximation, derivatives play a crucial role. To find a Taylor polynomial, we need the function value, its first derivative, and sometimes higher-order derivatives like the second derivative, all evaluated at the point of interest.
For the function \(f(x) = e^{\sin x}\), identified in the exercise, the first derivative \(f'(x)\) is computed using the chain rule to be \(e^{\sin x} \cdot \cos x\), and at \(x = 0\), the value is 1. The second derivative \(f''(x)\) incorporates the product rule and chain rule, resulting in \(e^{\sin x} \cdot (\cos^2 x - \sin x)\), which evaluates to 1 at \(x = 0\). By using these derivatives, we can form both the linear and quadratic approximations of the function.

linearization

Linearization is the simplest form of approximation using a Taylor polynomial. It only involves the value of the function and its first derivative at a specific point. The goal is to create a line, hence 'linear,' that matches the slope of the function at that point. This line can then be used to estimate the function's value near that point without actually computing more complex evaluations.
For instance, in the given exercise, linearization of \(f(x) = e^{\sin x}\) at \(x = 0\) uses the first-order Taylor polynomial: \( L(x) = f(0) + f'(0) \cdot x\). Given \(f(0) = 1\) and \(f'(0) = 1\), the linear approximation is simply \( L(x) = 1 + x\).
This linear polynomial is an excellent first step towards approximating \(f(x)\) near \(x = 0\). While it might not fully capture all the intricacies of the function over a wider interval, it gives a very good point-wise estimate when dealing with very small changes around the point of interest.