Problem 33 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=\cos (2 x+(\pi / 2)), \quad a=\pi / 4\)

Step-by-Step Solution

Answer

The Taylor series is \( 2(x-\pi/4) - \frac{4}{3}(x-\pi/4)^3 + \dots \).

Step 1: Find the Derivative of the Function

First, we need to find the first few derivatives of \( f(x) = \cos(2x + \frac{\pi}{2}) \). The first derivative \( f'(x) \) is found using the chain rule:\[ f'(x) = -\sin(2x + \frac{\pi}{2}) \cdot 2 = -2\sin(2x + \frac{\pi}{2}). \]

Step 2: Evaluate Derivatives at \(x = a\)

Calculate the value of \( f(x) \) and its derivatives at \( x = \pi/4 \):- \( f(a) = \cos(2(\pi/4) + \frac{\pi}{2}) = \cos(\pi + \frac{\pi}{2}) = \cos(\frac{3\pi}{2}) = 0. \)- \( f'(a) = -2\sin(2(\pi/4) + \frac{\pi}{2}) = -2\sin(\pi + \frac{\pi}{2}) = -2(-1) = 2. \)

Step 3: Find the Next Derivatives at \(x = a\)

Compute higher-order derivatives:- The second derivative is \( f''(x) = -2\cdot2\cos(2x + \frac{\pi}{2}) = -4\cos(2x + \frac{\pi}{2}) \).- \( f''(a) = -4\cos(2(\pi/4) + \frac{\pi}{2}) = -4\cos(\frac{3\pi}{2}) = 0. \)- The third derivative is \( f'''(x) = 8\sin(2x + \frac{\pi}{2}) \).- \( f'''(a) = 8\sin(\frac{3\pi}{2}) = 8(-1) = -8. \)

Step 4: Use the Taylor Series Formula

The Taylor series for a function \( f \) centered at \( a \) is:\[ T(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots \]Substituting the calculated derivatives:\[ T(x) = 0 + \frac{2}{1!}(x-\pi/4) + \frac{0}{2!}(x-\pi/4)^2 + \frac{-8}{3!}(x-\pi/4)^3 + \dots \]Thus, \[ T(x) = 2(x-\pi/4) - \frac{8}{6}(x-\pi/4)^3 + \dots \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used to find the derivative of composite functions. In essence, when you have a function inside another function, the chain rule helps you differentiate it.
Here's how it works:

• Identify the outer function and the inner function within the composite.
• Differentiate the outer function while leaving the inner function unchanged.
• Multiply the result by the derivative of the inner function.

Applying this to the original function, \( f(x) = \cos(2x + \frac{\pi}{2}) \), we identify the outer function as \( \cos(u) \) and the inner function as \( u = 2x + \frac{\pi}{2} \).
To differentiate, we found the derivative of \( \cos(u) \), which is \( -\sin(u) \), and then multiplied by the derivative of \( u \), which is 2. This gives \( f'(x) = -2\sin(2x + \frac{\pi}{2}) \). Remember, the chain rule is crucial when dealing with function compositions.

Cosine Function

The cosine function, \( \cos(x) \), is one of the basic trigonometric functions. It models the cosine wave found in periodic phenomena like light and sound waves.
In mathematics, the cosine function is continuous and differentiable, making it suitable for calculus operations.
Characteristics of the cosine function include:

• Periodicity: The cosine function has a period of \( 2\pi \). This means \( \cos(x + 2\pi) = \cos(x) \).
• Amplitude: The function oscillates between -1 and 1.
• Symmetry: The cosine function is even, which means \( \cos(-x) = \cos(x) \).

Cosine's derivatives alternate between \( -\sin(x) \) and \( -\cos(x) \).
In our problem, the \( \cos(2x + \frac{\pi}{2}) \) function was differentiated by considering these properties. Here, the shifted and compressed cosine graph changes how its trigonometric properties are applied in calculus.

Derivative Calculation

Derivative Calculation involves finding how a function behaves at an instantaneous point using calculus. In other words, it's about how the function changes as its input changes.
To compute the first derivative for \( f(x) = \cos(2x + \frac{\pi}{2}) \), derivative rules and trigonometric identities are used. The chain rule comes into play here to manage the composite aspect of the function.
The steps for derivative calculation include:

• Identify the function in layers, such as inner and outer functions.
• Apply the necessary rules: basic derivative rules, product/quotient rules, and trigonometric identities.
• Simplify the result where possible.

Higher derivatives follow the same principles but require repeated applications.
In the exercise, we calculated derivatives up to the third order. Each step calculated becomes a part of the Taylor series expansion, encoding the function's behavior at and near the center point.

Higher-order Derivatives

Higher-order derivatives are successive derivatives of a function. They reveal deeper insights into the function’s behavior, like its curvature.
For the given function, we calculated the second and third derivatives:

• Second Derivative: Gives insights about the concavity of the function (whether it curves up or down). For \( f(x) = \cos(2x + \frac{\pi}{2}) \), it becomes \( f''(x) = -4\cos(2x + \frac{\pi}{2}) \). When evaluated at \( x = \pi/4 \), it equals 0, indicating no curvature at this point.
• Third Derivative: Relates to the change of the curvature. Calculated as \( f'''(x) = 8\sin(2x + \frac{\pi}{2}) \), its value \( f'''(a) = -8 \) indicates the rate of concavity change.

These derivatives are crucial in constructing Taylor series expansions as each derivative contributes to the approximation.
In higher-order approximations, these derivatives introduce terms that better capture complex behaviors of the function around the expansion point.