Problem 16 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(13-30\) . $$\sin x-x+\frac{x^{3}}{3 !}$$

Step-by-Step Solution

Answer

The Taylor series at \( x=0 \) is \( \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \).

Step 1: Identify Terms to Expand

The function to find the Taylor series for is \( f(x) = \sin x - x + \frac{x^3}{3!} \). We need the Taylor series for each part of this expression centered around \( x = 0 \).

Step 2: Expand \( \sin x \) Using Taylor Series

The Taylor series for \( \sin x \) centered at \( x = 0 \) is: \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \)

Step 3: Combine Terms with Specific Function

Since the function is \( \sin x - x + \frac{x^3}{3!} \), substitute the Taylor series expansion of \( \sin x \) into the expression: \[ \sin x - x + \frac{x^3}{3!} = \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right) - x + \frac{x^3}{3!} \].

Step 4: Simplify the Expression

This equals \[ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots - x + \frac{x^3}{3!} \]. Notice the \( x \) and \( -x \) terms cancel out, as do the \( -\frac{x^3}{3!} \) and \( \frac{x^3}{3!} \).

Step 5: Write the Final Simplified Series

The terms that remain in the series are the higher-order terms from the \( \sin x \) expansion: \( \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \). Thus, the Taylor series centered at \( x = 0 \) is \( \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \).

Key Concepts

Power Series

A power series is an infinite sum of terms expressed as \[ \sum_{n=0}^{\infty} a_n (x-c)^n \] where:

• \( a_n \) are coefficients based on the function you are expanding.
• \( c \) is the center of the series, or the point around which you are expanding the function.
• \( x \) is the variable in the function.

For Taylor series, these coefficients \( a_n \) come from the derivatives of the function evaluated at \( x=c \). Power series allow us to represent complicated functions in a simpler polynomial form.
They are incredibly useful for approximating functions, particularly around the point \( x=0 \) (Maclaurin series). This method helps solve complex problems in calculus by converting them into manageable algebraic expressions.
Understanding power series is crucial for grasping more advanced calculus concepts, such as series expansion and analytic functions.

Sine Function

The sine function \( \sin x \) is one of the fundamental trigonometric functions. It describes periodic phenomena like waves or circular motion. When dealing with the sine function analytically, Taylor series come in handy.
For \( \sin x \), the Taylor series at \( x=0 \) (also known as the Maclaurin series) is:\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]This pattern continues indefinitely using odd-powered terms of \( x \) with alternating signs.
Each term of this series contributes to increasing the accuracy of the approximation, particularly close to the expansion point \( x=0 \).

• The more terms you include, the closer the series resembles \( \sin x \).
• A simple series like this can also be helpful in solving differential equations or simplifying complex integrals.

Series Expansion

Series expansion refers to the process of expressing a function as an infinite sum of terms. Taylor and Maclaurin series are specific types of series expansions.
They convert a function into a polynomial form by creating an infinite series that matches the function's values or derivatives at certain points. This makes it possible to approximate and analyze functions that might otherwise be unsolvable.
For example, the function \( \sin x - x + \frac{x^3}{3!} \) is broken down using the series expansion of \( \sin x \). By individually expanding each component, we combine terms and simplify to form a new series:

• Start with expressing individual functions as a power series.
• Combine these series by carrying out polynomial addition or subtraction.
• Simplify using algebraic operations to cancel terms where possible.

This methodology not only simplifies calculations but also provides insights into the function's behavior near specific points.

Function Simplification

Function simplification involves reducing complex expressions into simpler, more digestible forms. By applying processes like term cancellation and re-arranging terms, you can significantly simplify equations.
In our context, starting with the function \( \sin x - x + \frac{x^3}{3!} \), according to its series expansion:

• The terms \( x \) and \( -x \) cancel each other out.
• Similarly, \( -\frac{x^3}{3!} \) and \( \frac{x^3}{3!} \) cancel each other too.

This leaves higher-order terms from the \( \sin x \) series expansion:\[ \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]Simplifying functions in this manner not only makes calculations easier but also offers a clearer understanding of the core function behavior.Such simplifications are particularly helpful in physics and engineering, where approximations may yield practical solutions without extensive calculations.