Problem 13 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(11-28 .\) $$\frac{x^{2}}{2}-1+\cos x$$

Step-by-Step Solution

Answer

The Taylor series at \(x=0\) for the function is \(\frac{x^4}{24} - \frac{x^6}{720} + \cdots\)."

Step 1: Identify the Known Taylor Series

We know the Taylor series for the cosine function centered at \(x = 0\) is given by \(\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\). This series will be used directly as part of the solution.

Step 2: Separate the Function Components

Our function is \(\frac{x^2}{2} - 1 + \cos x\). We will deal with each part of this function separately: \(\frac{x^2}{2}\), \(-1\), and \(\cos x\).

Step 3: Write Each Component as a Series

The expression \(\frac{x^2}{2}\) can be written as a power series \(\frac{x^2}{2} = \frac{1}{2}x^2\cdot1\). The constant \(-1\) remains as \(-1\). The Taylor series for \(\cos x\) is \(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots\).

Step 4: Combine the Series

Start with combining the series for each part: \(\frac{x^2}{2} = 0 + \frac{1}{2}x^2 + 0x^4 + 0x^6 + \cdots\), \(-1 = -1\), and the known series for \(\cos x\). Add them term-by-term: \[1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\] + \[-1\] + \[0 + \frac{1}{2}x^2 + 0x^4 + \cdots\].

Step 5: Simplify the Combined Series

Cancel or combine coefficients for the powers of \(x\). We have: constant term: \(1 - 1 = 0\), \(x^2\) term: \(-\frac{1}{2}x^2 + \frac{1}{2}x^2 = 0\), the \(x^4\) term remains \(\frac{x^4}{24}\), and higher-order terms follow similarly without contributions from \(\frac{x^2}{2} or -1\).

Step 6: Write Final Taylor Series

The Taylor series at \(x = 0\) for the function \(\frac{x^2}{2} - 1 + \cos x\) is simply \(\frac{x^4}{24} - \frac{x^6}{720} + \cdots\).

Key Concepts

Power Series

A power series is a way to represent a function as an infinite sum of terms by using powers of a variable, say, \(x\). Think of it as a long polynomial that can go on indefinitely. Power series are incredibly useful because they allow us to express complex functions in simpler terms.
They are written in the form:\[\sum_{n=0}^{finity} a_n (x-c)^n\]Here, \(a_n\) represents the coefficients of the series, and \(c\) is the center of the series. When \(c=0\), we refer to this as a Maclaurin series, a specific kind of Taylor series.

• Power series can approximate functions to a very high degree of accuracy.
• They make it easier to perform calculus operations like integration and differentiation.
• Understanding how to manipulate power series terms is crucial for solving problems like finding Taylor or Maclaurin series.

By applying the concept of power series, we can manipulate the given function \(\frac{x^2}{2} - 1 + \cos x\) into a Taylor series form at \(x=0\).

Cosine Function

The cosine function, known from trigonometry, is more than just a way to determine the adjacent component of an angle in a right triangle. It is also an essential function in calculus and series representation.
In the context of power series, specifically Taylor and Maclaurin series, the cosine function has a well-defined expression:\[\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\]Essentially, the cosine function is represented by an alternating series of even powers of \(x\).

• The cosine power series converges for all \(x\), making it reliable for many applications.
• It starts with \(1\) and alternates between subtracting and adding terms based on the power of \(x\).
• Understanding the cosine series is crucial when you need to integrate it into more complex functions, as demonstrated in the exercise.

In the given solution, the cosine series part makes it possible to combine and simplify the function into a Taylor series centered at \(x=0\).

Maclaurin Series

The Maclaurin series is a special case of the Taylor series where the expansion point is at \(x=0\). This makes calculations somewhat simpler, especially when dealing with symmetric functions like \(\cos x\).
Maclaurin series have the general form:\[\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]For functions like cosine, this series is quite efficient since many higher-order derivatives are zero or alternate in a predictable pattern.

• Maclaurin series help simplify and approximate functions around \(x=0\).
• This form is particularly useful when direct function evaluation is complex or impossible.
• Used widely in mathematics, engineering, and physics, Maclaurin series simplify real-world problem solving.

The solution to the exercise takes advantage of the Maclaurin series for the cosine function and combines it with simpler polynomials to derive the final series representation of the entire function at \(x=0\).