Problem 1 - Active Calculus

Question

Find the Taylor polynomials of degree \(n\) approximating \(\cos (3 x)\) for \(x\) near 0 : For \(n=2, P_{2}(x)=\) For \(n=4, P_{4}(x)=\) For \(n=6, P_{6}(x)=\)

Step-by-Step Solution

Answer

For \(n=2\): \(P_2(x) = 1 - \frac{9}{2}x^2\). For \(n=4\): \(P_4(x) = 1 - \frac{9}{2} x^2 + \frac{81}{24} x^4\). For \(n=6\): \(P_6(x) = 1 - \frac{9}{2} x^2 + \frac{81}{24} x^4 - \frac{729}{720} x^6.\)

Step 1: Identify the function and its derivatives

The function is \(\text{cos}(3x)\). To find the Taylor polynomials, compute the derivatives at \(x = 0\).

Step 2: Compute the first few derivatives

Calculate the derivatives of \(\text{cos}(3x)\): \(\text{First derivative: } f'(x) = -3\text{sin}(3x)\), \(\text{Second derivative: } f''(x) = -9\text{cos}(3x)\). Continue this process up to the 6th derivative.

Step 3: Evaluate derivatives at \(x = 0\)

Evaluate each derivative at \(x = 0\). For example, \(f(0) = \text{cos}(0) = 1\), \(f'(0) = 0\), \(f''(0) = -9\). Continue for higher-order derivatives as needed.

Step 4: Formulate the Taylor polynomial for \(n = 2\)

Using the formula \(P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 \), the Taylor polynomial of degree 2 is \(P_2(x) = 1 - \frac{9}{2}x^2\).

Step 5: Formulate the Taylor polynomial for \(n = 4\)

The Taylor polynomial of degree 4 includes up to the fourth derivative: \(P_4(x) = 1 - \frac{9}{2} x^2 + \frac{81}{24} x^4 \).

Step 6: Formulate the Taylor polynomial for \(n = 6\)

The Taylor polynomial of degree 6 involves up to the sixth derivative: \(P_6(x) = 1 - \frac{9}{2} x^2 + \frac{81}{24} x^4 - \frac{729}{720} x^6. \)

Key Concepts

Derivatives

Derivatives measure how a function changes as its input changes. For a function like \(\text{cos}(3x)\), we need to find several derivatives to build the Taylor polynomial. The first derivative is the rate of change of the original function. Here, \(f'(x) = -3\text{sin}(3x)\). For the second derivative, we take the derivative of the first derivative. Thus, \(f''(x) = -9\text{cos}(3x)\). This continues for higher derivatives.
Remember, evaluating these derivatives at specific points like \(x = 0\) helps in forming the Taylor series.

Taylor Series

A Taylor series approximates a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For a function \(f(x)\) centered at \(x=0\), the Taylor series is:
\[ P(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... \]
In this exercise with \(\text{cos}(3x)\), computing terms up to the 6th derivative helps build the Taylor polynomials of degrees 2, 4, and 6, which are easier to calculate.

Function Approximation

Function approximation involves estimating a function using simpler functions. Taylor polynomials are one way to approximate functions, especially for values near a designated point. Here, \(x=0\). For example, the 2nd-degree polynomial for \( \text{cos}(3x) \):
\[ P_2(x) = 1 - \frac{9}{2} x^2 \]
This approximation gets better as more terms are included, like in the 4th and 6th-degree polynomials:

• \( P_4(x) = 1 - \frac{9}{2}x^2 + \frac{81}{24}x^4 \)
• \( P_6(x) = 1 - \frac{9}{2}x^2 + \frac{81}{24}x^4 - \frac{729}{720}x^6 \)

Trigonometric Functions

Trigonometric functions like \(\text{cos}(x)\) and \( \text{sin}(x) \) are fundamental in mathematics. They appear in various fields, including physics and engineering. Here, the function \( \text{cos}(3x) \) is used. When we compute its Taylor polynomial, we use its derivatives, all of which involve \( \text{cos}(3x) \) and \( \text{sin}(3x) \).
The cosine function is especially useful because it is periodic and symmetric, simplifying calculations.