Problem 22 - Calculus

Question

using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\sqrt{1+\sin \theta}$$

Step-by-Step Solution

Answer

The series is \( 1 + \frac{\theta}{2} - \frac{\theta^2}{8} - \frac{\theta^3}{48} \).

Step 1: Identify the Known Taylor Series

First, recall the Taylor series expansion for the function \( \sin \theta \) about \( 0 \), which is given by:\[\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\]Use this series to express \( 1 + \sin \theta \).

Step 2: Substitute into the Taylor Series

Substitute \( \sin \theta \) into \( \sqrt{1 + x} \)'s Taylor series: \[1 + \sin \theta = 1 + \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} + \cdots\]This is our new expression for \( 1 + \sin \theta \).

Step 3: Recall the Taylor Series for \( \sqrt{1 + x} \)

The Taylor series for \( \sqrt{1 + x} \) about 0 is:\[1 + \frac{x}{2} - \frac{x^2}{8} + \cdots\]Now, substituting \( x = \theta - \frac{\theta^3}{6} + \cdots \) into this series.

Step 4: Substitute and Simplify the Series

Substitute \( x = \theta - \frac{\theta^3}{6} \) into the series:\[\sqrt{1 + x} \approx 1 + \frac{\theta - \frac{\theta^3}{6}}{2} - \frac{\left(\theta - \frac{\theta^3}{6}\right)^2}{8} + \cdots\]Simplify these terms to find the first four nonzero terms.

Step 5: Calculate and Summarize

After simplification, calculate the Taylor series terms:1. The first term is \( 1 \).2. The second term is \( \frac{\theta}{2} \).3. The third term is \( -\frac{\theta^2}{8} \).4. The fourth term is \( -\frac{\theta^3}{48} \).Thus, the first four nonzero terms are:\[1 + \frac{\theta}{2} - \frac{\theta^2}{8} - \frac{\theta^3}{48}\]

Step 6: Conclusion: Present the Taylor Series

The first four nonzero terms of the Taylor series for \( \sqrt{1 + \sin \theta} \) about 0 are:\[1 + \frac{\theta}{2} - \frac{\theta^2}{8} - \frac{\theta^3}{48}\]

Key Concepts

Sin Function

The sine function, represented as \( \sin \theta \), is a fundamental trigonometric function used to describe wave patterns like sound waves and light. In the realm of calculus, we often work with its Taylor series representation.
The Taylor series expansion is a powerful tool to express complex functions as infinite sums of polynomial terms. For \( \sin \theta \), the Taylor series about 0 is:

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

This expansion lets us approximate the sine function using a few terms when \( \theta \) is small. It captures the wave's cyclical nature by including both linear and higher-degree terms.
In the exercise, we use this series as part of the process to approximate \( \sqrt{1 + \sin \theta} \) by first expressing \( \sin \theta \) and then substituting it into another series. This demonstrates the versatility and connectivity of mathematical concepts through series expansions.

Square Root Function

The square root function, especially when combined with algebraic or trigonometric functions, often leads to more complicated expressions. Finding the Taylor series for such functions requires us to carefully substitute and simplify.
For a simple square root function like \( \sqrt{1 + x} \), we can approximate it using a Taylor series. The Taylor series for \( \sqrt{1 + x} \) around 0 is:

• \( \sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots \)

This approximation captures how the function behaves near \( x = 0 \) and is useful for simplifying calculations. By substituting \( x \) with more complex expressions, such as what we derived for \( 1 + \sin \theta \), we can further expand the utility of this series.
In our exercise, substituting \( x = \theta - \frac{\theta^3}{6} \) (from the \( 1 + \sin \theta \) expression) into the square root function shows the interplay between algebra and calculus, making complex calculations manageable.

Series Expansion

Series expansion is the process of expressing a function as an infinite sum of terms, often polynomials. This helps simplify problems involving complex functions by breaking them down into more manageable parts.
The beauty of series expansion lies in its ability to approximate functions that are otherwise difficult to work with. For instance, in the Taylor series, each term becomes a polynomial of increasing power, making it easier to compute function values, derive properties, or approximate solutions:

• Series expansion provides an approximate form of functions using known, simpler functions.


• The focus is typically around a point, such as 0, which makes calculations straightforward near that point.
• This method can be extended to many functions beyond simple polynomials, like trigonometric or exponential functions.

In this specific exercise, expanding \( \sqrt{1 + \sin \theta} \) required utilizing Taylor series for both the sine and square root functions. By doing so, we found manageable polynomial terms that approximate the behavior of our complex function. Mastery of series expansions is crucial for tackling higher-level calculus problems dealing with limits, derivatives, and integrals of intricate functions.