Problem 48 - 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 \(x=0\). \(f(x)=\cosh x\)

Step-by-Step Solution

Answer

The linearization at \(x=0\) is \(1\); the quadratic approximation is \(1 + \frac{1}{2}x^2\).

Step 1: Understand Taylor Polynomial

The Taylor polynomial of order \( n \) for a function \( f(x) \) at \( x=a \) is given by: \[ P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \] We are finding (a) the linear approximation (order 1) and (b) the quadratic approximation (order 2) at \( x=0 \).

Step 2: Calculate Function Values

Calculate \( f(0) \), \( f'(0) \), and \( f''(0) \) for \( f(x) = \cosh x \).First, \( f(0) = \cosh(0) = 1 \).

Step 3: Differentiate for Linear Term

Find the first derivative: \[ f'(x) = \sinh x \]Therefore, \( f'(0) = \sinh(0) = 0 \).

Step 4: Differentiate for Quadratic Term

Find the second derivative: \[ f''(x) = \cosh x \]And \( f''(0) = \cosh(0) = 1 \).

Step 5: Find the Linearization (Order 1)

Substitute the calculated values into the formula for the linear approximation (Taylor polynomial of order 1):\[ P_1(x) = f(0) + f'(0)x = 1 + 0 \cdot x = 1 \]

Step 6: Find the Quadratic Approximation (Order 2)

Substitute the calculated values into the formula for the quadratic approximation (Taylor polynomial of order 2):\[ P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 1 + 0 \cdot x + \frac{1}{2}x^2 = 1 + \frac{1}{2}x^2 \]

Key Concepts

Quadratic Approximation

Quadratic approximation is a way to estimate the value of a function near a certain point using a polynomial of degree two. This method is particularly useful when you need a simple and quick estimate without requiring the entire function.
This approximation is also known as the Taylor polynomial of order two. The formula for the quadratic approximation of a function \(f(x)\) at a point \(x = a\) is:

• \( P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \)

Here, \(f(a)\), \(f'(a)\), and \(f''(a)\) represent the function's value, first derivative, and second derivative at the point \(x = a\) respectively.
For example, to find the quadratic approximation at \(x=0\) for \(f(x) = \cosh x\), we calculate the necessary derivatives and substitute them into the formula:

- \(f(0) = \cosh(0) = 1\) - \(f'(0) = \sinh(0)=0\) - \(f''(0) = \cosh(0)=1\)

Using these values, we obtain: \( P_2(x) = 1 + \frac{1}{2}x^2\).

Linearization

Linearization, or the Taylor polynomial of order one, provides a way to approximate a function using a linear equation.
This can be extremely beneficial for calculating values near a point without dealing with more complex computations.
The formula for linearization of a function \(f(x)\) at \(x = a\) is:

• \( P_1(x) = f(a) + f'(a)(x-a) \)

This equation uses the function value and its first derivative at the point \(x = a\), forming a tangent line to the function.
To find the linearization at \(x=0\) for \(f(x) = \cosh x\), calculate:

- \(f(0) = \cosh(0) = 1\) - \(f'(0) = \sinh(0) = 0\)

Substituting into the linearization formula gives \(P_1(x) = 1\). This result shows that near \(x=0\), the function is approximately constant.

Differentiation

Differentiation is the mathematical process of finding the derivative, which represents the rate of change of a function.
This is a fundamental concept used to determine how a function behaves and is essential when calculating Taylor polynomials.
The process of differentiation can involve consecutive derivatives, such as the first and second derivative, and they play critical roles in linear and quadratic approximations respectively.

• First derivative \(f'(x)\) identifies slopes at given points.
• Second derivative \(f''(x)\) gives information about the curvature or concavity of the function.

In problems involving the function \(f(x) = \cosh x\):

- The first derivative: \(f'(x) = \sinh x\)- The second derivative: \(f''(x) = \cosh x\)

Differentiation not only uncovers these properties but is also crucial in forming the terms of Taylor polynomials.

Hyperbolic Functions

Hyperbolic functions, including \(\sinh x\) and \(\cosh x\), are analogous to the trigonometric functions like sine and cosine.
They have applications in many areas, particularly where analogous behavior to circular functions is needed but within hyperbolic contexts (e.g., areas with exponential growth).
For hyperbolic functions:

• \(\cosh x = \frac{e^x + e^{-x}}{2}\)
• \(\sinh x = \frac{e^x - e^{-x}}{2}\)

In the given exercise, the function \(\cosh x\) is evaluated. At \(x=0\), it equals 1, similar to how the cosine of 0 equals 1.
Understanding these functions helps in evaluating derivatives for approximations. For instance, knowing that the derivative \(\frac{d}{dx}(\sinh x) = \cosh x\) is crucial when applying differentiation to find Taylor polynomial terms. This knowledge aids in comprehending how functions behave over intervals, especially in approximations like quadratic and linearizations.