Problem 52 - Calculus

Question

Decide if the statements are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. The Taylor series for \(x^{3} \cos x\) about \(x=0\) has only odd powers.

Step-by-Step Solution

Answer

True, the Taylor series of \(x^3 \cos x\) contains only odd powers.

Step 1: Understand the Function

The function provided is \(x^3 \cos x\). We are asked to evaluate the Taylor series expansion of this function around \(x = 0\) and check if it consists only of odd powers of \(x\).

Step 2: Recall the Maclaurin Series for \(\cos x\)

The Maclaurin (Taylor series at \(x=0\)) for \(\cos x\) is given by:\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \]Notice that all terms in this series are even powers of \(x\).

Step 3: Multiply \(x^3\) by the Series for \(\cos x\)

To find the Taylor series of \(x^3 \cos x\), multiply each term of the \(\cos x\) series by \(x^3\):\[ x^3 \cdot (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots) \]This results in: \[ x^3 - \frac{x^5}{2!} + \frac{x^7}{4!} - \frac{x^9}{6!} + \cdots \]

Step 4: Check for Powers of \(x\)

Note that after multiplying, the resulting series consists of terms:\ }\(x^3, -\frac{x^5}{2!}, \frac{x^7}{4!}, -\frac{x^9}{6!}, \ldots\) which are all odd powers of \(x\).

Step 5: Conclude

Since we only get terms with odd powers (\(x^3, x^5, x^7, \ldots\)), the Taylor series for \(x^3 \cos x\) about \(x=0\) indeed has only odd powers.

Key Concepts

Maclaurin series

When we talk about Taylor series expansions at the point where \(x = 0\), we specifically refer to what is known as the Maclaurin series. This is a special case of the more general Taylor series, centered around a point, typically zero.
In a Maclaurin series, a function is expressed as an infinite sum of terms calculated from the values of its derivatives at zero. If we have a function \(f(x)\), its Maclaurin series is given by:

• \( f(0) + f'(0) \cdot x + \frac{f''(0)}{2!} \cdot x^2 + \cdots \)
• In general form: \( \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \)

It's important to understand that the terms in a Maclaurin series expansion are dependent on the derivatives of the function at zero, which means the shape and behavior of the series' expansion are dictated by these values.
For example, the Maclaurin series for \(\cos x\) contains even power terms only because the odd derivatives of \(\cos x\) evaluated at zero are zero.

power series

A power series is an infinite series of the form:

• \( \sum_{n=0}^{\infty} a_n (x-c)^n \)

In a power series, \(c\) is the center of the series, and \(a_n\) are the coefficients. This type of series provides a powerful way to approximate functions.
Power series can represent various functions, sometimes even when no simple closed-form expression for the function exists. Looking at a specific example, the Maclaurin series—when you set \(c = 0\)—is a form of power series.
Power series are incredibly useful because they allow us to work with functions using algebraic manipulation, making it easier to perform calculus operations on them like differentiation and integration.
For the function \(x^3 \cos x\), its Taylor series about zero (Maclaurin series) uses the properties of \(\cos x\) and the power of \(x^3\) to create a sequence of terms that only contains odd powers of \(x\), highlighting how different mathematical functions can be represented and manipulated.

convergence of series

Convergence of series is a critical concept when dealing with infinite series like a Taylor or Maclaurin series. For a series to be useful in representing a function, it needs to converge, meaning the terms of the series approach a finite value as you add more and more terms.
In this context, each series has its radius of convergence, which tells us the interval in which the series will converge to the true value of the function.
What's fascinating is that not all series converge everywhere. The interval of convergence must be determined carefully, often using tests like the ratio test or root test to ascertain how far you can go with \(x\) values while ensuring convergence.
For the series representing \(x^3 \cos x\), it's essential to note that it converges within the same radius as the \(\cos x\) series, which is because its behavior is dependent on the function \(\cos x\) itself. Understanding convergence ensures the series correctly approximates the function across an intended domain.