Problem 31 - Thomas Calculus

Question

Find the first four nonzero terms in the Maclaurin series for the functions in Exercises \(31-38 .\) $$e^{x} \sin x$$

Step-by-Step Solution

Answer

The first four nonzero terms are \(x + x^2 + \frac{x^3}{3}\).

Step 1: Identify Maclaurin Series for Individual Functions

The Maclaurin series for each function is calculated at the origin (x=0). Start by identifying the Maclaurin series for \(e^x\) and \(\sin x\). The series for \(e^x\) is given by \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\). The series for \(\sin x\) is given by \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \).

Step 2: Set Up the Product of Functions

We are asked to find the series for \(e^x \sin x\). To find this, we multiply the series of these two functions together: \((1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots)(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots)\).

Step 3: Multiply and Collect Like Terms for Nonzero Coefficients

Distribute the terms from the series of \(e^x\) with those from \(\sin x\) to find their product:- \(1 \cdot x = x\)- \(1 \cdot (-\frac{x^3}{3!}) = -\frac{x^3}{6}\)- \(x \cdot x = x^2\)- \(x \cdot (-\frac{x^3}{3!}) = -\frac{x^4}{6}\)- \(\frac{x^2}{2} \cdot x = \frac{x^3}{2}\)- \(\frac{x^3}{6} \cdot x = \frac{x^4}{6}\)Combine like terms:- For \(x\): only \(x\).- For \(x^2\): only from \(x^2\).- For \(x^3\): \(-\frac{x^3}{6} + \frac{x^3}{2} = \frac{x^3}{3}\).- For \(x^4\): \(-\frac{x^4}{6} + \frac{x^4}{6} = 0\).

Step 4: Write Down the First Four Nonzero Terms

From the collected terms: The first nonzero term is \(x\), the second is \(x^2\), the third is \(\frac{x^3}{3}\), and the fourth term is derived by continuing the pattern of the multiplication to higher degrees (such as with \(x^5\) etc).

Step 5: Verify and Conclude the Series

Upon further multiplication, you would proceed to terms contributing to higher powers such as \(x^4\) and beyond. However, since the first zero term was balance, the next nonzero term comes from continued multiplication. Therefore, the correct terms that form this Maclaurin series are \(x + x^2 + \frac{x^3}{3}\).

Key Concepts

Exponential Function

The exponential function, commonly represented as \( e^x \), is a fundamental mathematical function. It is particularly important because it describes growth in processes where the rate of change is proportional to the state of the system, such as in compound interest and population growth. In mathematics, it is essential because it forms the basis for more complex functions and operations.

The Maclaurin series expands \( e^x \) around the point 0, and the formula is:

• \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

This series allows us to approximate \( e^x \) using a polynomial form. Each term adds accuracy to the approximation. For small values of \( x \), the initial terms often provide sufficient closeness to the exponential function's value at that point.

Trigonometric Function

Trigonometric functions, like the sine function \( \sin x \), are extensively used to describe waveforms and oscillatory phenomena. In mathematics, \( \sin x \) is crucial for understanding rotational and periodic phenomena.

The Maclaurin series for \( \sin x \) is an excellent way to approximate the function. It is expressed as:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)

This series highlights how the sine function can be represented by an infinite sum of polynomial terms. Only the odd powers of \( x \) appear in the series, reflecting the function's odd nature and symmetry about the origin. Similar to the exponential function, using a Maclaurin series can simplify calculations involving trigonometric functions.

Series Expansion

Series expansion is a powerful mathematical tool that allows complex functions to be expressed as sums of polynomials. This expansion into a Maclaurin series is particularly useful because it simplifies calculations by taking only the initial terms, especially when higher precision isn't necessary.

For any function, the series expansion relies on derivatives taken at a specific point, usually zero for Maclaurin series. Each term in the series gets calculated using the function's derivatives, averaged or normalized by factorials. The principle of series expansion stands on extending basic functions, like \( e^x \) and \( \sin x \), into power series, making them easier to handle mathematically. This is incredibly helpful in solving practical problems involving calculation or approximation of values.

Product of Functions

The product of functions, such as \( e^x \) and \( \sin x \), is an operation where two functions are multiplied together to create a new function. Understanding how to multiply their series is crucial when finding the combined function's Maclaurin series.

In this problem, the product is:

• \( (1 + x + \frac{x^2}{2!} + \cdots)(x - \frac{x^3}{3!} + \cdots) \)

The multiplication involves taking each term from both series and multiplying them, and then combining like terms. This process, called distribution, can get complex quickly, so only the first few terms are typically calculated. The aim is to simplify the product into a series of its own, which can be used just like any other polynomial series. Focusing on the first nonzero terms helps in providing an accurate representation while avoiding the intricacies of process-heavy multiplication of infinite series.