Problem 45 - 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 \(x=0\). \(f(x)=\ln (\cos x)\)

Step-by-Step Solution

Answer

Linearization at x=0 is 0; Quadratic approximation at x=0 is \(-\frac{1}{2}x^2\).

Step 1: Understanding Linearization

The linearization of a function is given by the Taylor polynomial of order 1, which approximates the function using its first derivative. The formula for linearization at point \(x = a\) is: \[ L(x) = f(a) + f'(a)(x-a) \]In this problem, we're finding the linearization at \(x=0\) for \(f(x) = \ln (\cos x)\).

Step 2: Calculate Function Value at x=0

To find the linearization, first calculate \(f(0)\):\[ f(x) = \ln(\cos x) \]\[ f(0) = \ln(\cos 0) = \ln 1 = 0 \]

Step 3: Find the First Derivative

Calculate the first derivative \(f'(x)\):\[ f'(x) = \frac{d}{dx} \ln(\cos x) = \frac{-\sin x}{\cos x} = -\tan x \]Evaluate this derivative at \(x=0\):\[ f'(0) = -\tan(0) = 0 \]

Step 4: Formulate Linearization at x=0

Using the results from the previous steps, the linearization of \(f(x)\) at \(x=0\) is:\[ L(x) = f(0) + f'(0)(x-0) = 0 + 0 \times x = 0 \]

Step 5: Understanding Quadratic Approximation

The quadratic approximation is the Taylor polynomial of order 2, which includes both the linear and quadratic terms. The formula at \(x=a\) is:\[ Q(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \]

Step 6: Find the Second Derivative

Calculate the second derivative \(f''(x)\):\[ f''(x) = \frac{d}{dx}(-\tan x) = -\sec^2 x \]Evaluate this at \(x=0\):\[ f''(0) = -\sec^2 0 = -1 \]

Step 7: Formulate Quadratic Approximation at x=0

Using the values calculated in the previous steps, the quadratic approximation of \(f(x)\) at \(x=0\) is:\[ Q(x) = f(0) + f'(0)(x-0) + \frac{f''(0)}{2}(x-0)^2 = 0 + 0 \times x + \frac{-1}{2}x^2 \]\[ Q(x) = -\frac{1}{2}x^2 \]

Key Concepts

Quadratic Approximation

Let's talk about quadratic approximation. This is a technique in calculus where we use a polynomial (Taylor polynomial of order 2) to approximate the behavior of a function near a specific point, say \(x = a\). It involves using a quadratic polynomial, which means it includes a term with \(x^2\).For a function \(f(x)\), the quadratic approximation at \(x = a\) is given by:\[Q(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2\]- **\(f(a)\):** The function value at the point- **\(f'(a)\times(x-a)\):** The linear part, using the first derivative- **\(\frac{f''(a)}{2}(x-a)^2\):** The quadratic part, using the second derivative divided by 2The power of quadratic approximation lies in its ability to capture more complexity of the function's curvature compared to a linear approximation. In our example with \(f(x) = \ln(\cos x)\) at \(x = 0\), the quadratic approximation was calculated to be \(-\frac{1}{2}x^2\). This approach helps provide a better fit to the function around \(x = 0\), especially when the function has significant curvature changes.

Linearization

Linearization provides a simplified way to approximate a function. It uses a linear (straight line) function to estimate the value of a more complex function near a certain point. It is essentially a Taylor polynomial of order 1.For any differentiable function \(f(x)\), the linear approximation or linearization near the point \(x = a\) looks like this:\[L(x) = f(a) + f'(a)(x-a)\]- **\(f(a)\):** This value gives the function's starting point at \(x=a\)- **\(f'(a)\times(x-a)\):** This is the slope or first derivative, showing how the function behaves near this pointIn our problem, we found the linearization of \(f(x) = \ln(\cos x)\) at \(x=0\). Both \(f(0)\) and \(f'(0)\) were zeros, leading to the linearization formula \(L(x) = 0\). When the slope at the point is zero, the linear approximation results in a flat line at that point, indicating little to no change in function value right there.

Derivatives

Derivatives are fundamental in calculus, representing how a function changes its value as input changes—a concept known as the rate of change. They play a crucial role in both linear and quadratic approximations.- **First Derivative \(f'(x)\):** It tells us the slope of the tangent to the function at any point \(x\). The first derivative indicates how steeply or gently the function is rising or falling. - Example: For \(f(x) = \ln(\cos x)\), the first derivative \(f'(x) = -\tan x\) was used to find both linearization and quadratic approximations. Evaluating at \(x=0\), we got \(f'(0) = 0\).- **Second Derivative \(f''(x)\):** This captures the curvature, indicating how the slope itself is changing. - Example: We found \(f''(x) = -\sec^2 x\) for the same function, used in the quadratic approximation. It was evaluated at \(x=0\) to give \(f''(0) = -1\), showing the concavity of the function at that point.Understanding derivatives helps in analyzing the behavior and the shape of a function graph, which is why they are so valuable for approximations and optimization problems.