Problem 4 - Active Calculus

Question

Find the first four terms of the Taylor series for the function \(\cos (x)\) about the point \(a=\) \(-\pi / 4\). (Your answers should include the variable \(\mathrm{x}\) when appropriate.) \(\cos (x)=\) \(+\ldots\)

Step-by-Step Solution

Answer

The first four terms of the Taylor series for \(\cos(x)\) about \(a = -\pi/4\) are \[T(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x + \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x + \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x + \frac{\pi}{4})^3\]

Step 1: Identify the function and the point of expansion

The given function is \(\cos(x)\) and the point of expansion is \(a = -\pi / 4\). So \(f(x) = \cos(x)\) and \(a = -\pi / 4\).

Step 2: Find the function's derivatives

The first few derivatives of \(\cos(x)\) are:\[f'(x) = -\sin(x)\]\[f''(x) = -\cos(x)\]\[f'''(x) = \sin(x)\]\[f''''(x) = \cos(x)\]

Step 3: Evaluate the derivatives at the point \(a = -\pi / 4\)

Next, evaluate each derivative at \(x = -\pi / 4\):\[f(-\pi / 4) = \cos(-\pi / 4) = \frac{\sqrt{2}}{2}\]\[f'(-\pi / 4) = -\sin(-\pi / 4) = \frac{\sqrt{2}}{2}\]\[f''(-\pi / 4) = -\cos(-\pi / 4) = -\frac{\sqrt{2}}{2}\]\[f'''(-\pi / 4) = \sin(-\pi / 4) = -\frac{\sqrt{2}}{2}\]

Step 4: Construct the Taylor series

Using the Taylor series formula \[T(x) = f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \ldots\]substitute \(-\pi / 4\) for \(a\) and the evaluated derivatives to get:\[T(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x + \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x + \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x + \frac{\pi}{4})^3\]

Key Concepts

Trigonometric Functions

Trigonometric functions are a cornerstone in mathematics, especially in the study of periodic phenomena such as waves and circular motion. Functions like sine, cosine, and tangent relate the angles of a triangle to the lengths of its sides. The cosine function, denoted as \(\text{cos}(x)\), is particularly vital. It tells us the horizontal position of a point on the unit circle as it moves around. For a given angle \(x\), \(\text{cos}(x)\) provides the x-coordinate of the corresponding point on the circle. Beyond geometry, trigonometric functions have applications in physics, engineering, and computer science.
Understanding the behavior of these functions is crucial for solving many real-world problems.

Derivatives

Derivatives measure how a function changes as its input changes. In simpler terms, they tell us the slope of the function at any given point. For the cosine function, its derivatives reveal a cyclic pattern:

• The first derivative of \(\text{cos}(x)\) is \(-\text{sin}(x)\).
• The second derivative brings us back to \(-\text{cos}(x)\).
• As we calculate higher-order derivatives, we alternate between sine and cosine functions. This recursive nature allows us to predict subsequent derivatives easily.
  
• The third derivative is \( \text{sin}(x) \).
• The fourth derivative circles back to \( \text{cos}(x) \).
  Using derivatives is essential in creating Taylor series and understanding how functions behave near a point.

Series Expansion

Series expansion lets us express complex functions as infinite sums of simpler terms. With Taylor series, we can represent functions as power series around a particular point, called the expansion point. For the cosine function, using Taylor series helps us approximate its values near any point we choose. The Taylor series for \(\text{cos}(x)\) around a point \(a\) is given by:
\[\text{T}(x) = f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \text{...} \].

This formula uses the function's derivatives at the expansion point. Calculating the first few terms of this series gives a good approximation of the function near that point. The more terms we include, the more accurate our approximation becomes. This is incredibly useful in physics and engineering, where exact solutions can be complex or impossible to find.

Cosine Function

The cosine function, denoted as \(\text{cos}(x)\), is a fundamental trigonometric function. It is periodic with a period of \ (2\text{π}) \, meaning \(\text{cos}(x + 2\text{π}) = \text{cos}(x)\).
The Taylor series for \(\text{cos}(x)\) at a point \(a\) uses its derivatives evaluated at that point. In our example, we expanded around \(a = -\text{π}/4\).
Calculating the derivatives of \(\text{cos}(x)\) at \(a = -\text{π}/4\):

• \(f(-\text{π}/4) = \text{cos}(-\text{π}/4) = \frac{\text{√}2}{2}\)
• \(f'(-\text{π}/4) = -\text{sin}(-\text{π}/4) = \frac{\text{√}2}{2}\)
• \(f''(-\text{π}/4) = -\text{cos}(-\text{π}/4) = -\frac{\text{√}2}{2}\)
• \(f'''(-\text{π}/4) = \text{sin}(-\text{π}/4) = -\frac{\text{√}2}{2}\).

Substituting these values into the Taylor series formula gives us an excellent approximation of the cosine function near \(a = -\text{π}/4\).