Problem 15 - Thomas Calculus

Question

Find the Maclaurin series for the functions. \(\sin 3 x\)

Step-by-Step Solution

Answer

The Maclaurin series for \( \sin 3x \) is \( 3x - \frac{9}{2}x^3 + \frac{81}{40}x^5 - \cdots \).

Step 1: Recall the General Maclaurin Series Formula

The Maclaurin series of a function \( f(x) \) is given by \( f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \cdots \). We need to find the derivatives of \( \sin 3x \) and evaluate them at \( x = 0 \).

Step 2: Find the Derivatives of \( \sin 3x \)

Calculate the derivatives of \( \sin 3x \):- \( f(x) = \sin 3x \)- \( f'(x) = 3\cos 3x \)- \( f''(x) = -9\sin 3x \)- \( f'''(x) = -27\cos 3x \)- Continue this pattern for more derivatives as needed.

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

Now evaluate each derivative at \( x = 0 \).- \( f(0) = \sin 0 = 0 \)- \( f'(0) = 3\cos 0 = 3 \)- \( f''(0) = -9\sin 0 = 0 \)- \( f'''(0) = -27\cos 0 = -27 \)

Step 4: Write the Maclaurin Series Using Derivatives

Substitute the evaluated derivatives into the Maclaurin series formula:\[ \sin 3x = 0 + 3x + \frac{0}{2!}x^2 - \frac{27}{3!}x^3 + \cdots \]This simplifies to:\[ \sin 3x = 3x - \frac{27}{6}x^3 + \cdots \]Reducing further gives:\[ \sin 3x = 3x - \frac{27}{6}x^3 + \cdots = 3x - \frac{9}{2}x^3 + \cdots \]

Step 5: Construct the General Term and Final Series

Observe the pattern of the series to represent it generally. The series can be written as:\[ \sin 3x = \sum_{n=0}^{\infty} (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!} = 3x - \frac{9}{2}x^3 + \frac{81}{40}x^5 - \cdots \]

Key Concepts

Taylor series

The Taylor series is an essential mathematical concept that helps express a complicated function as an infinite sum of terms. Each of these terms is derived from the function's derivatives evaluated at a specific point. When the point of expansion is zero, the Taylor series becomes a Maclaurin series. This makes it particularly useful for approximating functions like sine, cosine, and exponential functions around zero. Maclaurin series is especially convenient for trigonometric functions due to their periodic nature, allowing us to simplify their behavior in a understandable form.

derivative

The derivative of a function is a fundamental concept in calculus, used to measure how a function changes as its input changes. It's like the 'rate of change,' similar to speed in motion. Calculating higher-order derivatives gives us more information about the function's behavior. For the function \(3x\sin 3x\):

• The first derivative with respect to \(x\) is \(3\cos 3x\), indicating the rate of change in the original function's slope.
• The second derivative, \(-9\sin 3x\), helps to understand the curvature, or how the slope is changing.
• Higher derivatives continue this way, alternating between sine and cosine with increasing coefficients and signs.

By evaluating these derivatives at \(x = 0\), we gather the necessary information for constructing the series terms.

trigonometric functions

Trigonometric functions such as sine and cosine are pivotal in both pure and applied mathematics. They describe periodic phenomena, such as sound waves, light waves, and other sinusoids. In the context of the Maclaurin series, \(\sin 3x\) can be broken down into a series of simpler polynomial terms derived from its derivatives. The sine function inherently repeats at regular intervals, which is effectively captured by its series representation.

infinite series

An infinite series is a sum of an infinite number of terms. Each term in the series is derived from a structured pattern, usually based on a function's derivatives in the context of Taylor or Maclaurin series. The beauty of an infinite series is that it allows approximation of functions that might be difficult to express in a simple closed form. For \(\sin 3x\), the Maclaurin series creates an infinite series:

• Starting with \(3x\) as its initial term.
• Followed by higher powers like \(-(9/2)x^3\), showing alternating signs and increasing powers of \(x\).

The summation notation, as \(\sum_{n=0}^{\infty} (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!}\), captures this pattern, enabling us to visualize \(\sin 3x\) as a series extending infinitely.