Problem 10 - Thomas Calculus

Question

Use substitution (as in Example 4) to find the Taylor series at \(x=0\) of the functions in Exercises \(1-12\) . $$\frac{1}{2-x}$$

Step-by-Step Solution

Answer

The Taylor series for \(\frac{1}{2-x}\) at \(x=0\) is \( \frac{1}{2} \sum_{n=0}^{\infty} (x/2)^n \) for \(|x|<2\).

Step 1: Identify the function's format

The function given is \(\frac{1}{2-x}\). To use substitution for the Taylor series, identify that this function resembles \(\frac{1}{1-x}\) but with a shift. We can rewrite this as \(\frac{1}{1-(-x+1)}\) or \(\frac{1}{1-(x-2)}\).

Step 2: Recall the geometric series formula

Recall that the geometric series \(\frac{1}{1-x}\) is equivalent to \(1 + x + x^2 + x^3 + \ldots \) for \(|x| < 1\). This formula will be used for Taylor series expansion.

Step 3: Substitute into the geometric series

Substitute \(x-2\) into the geometric series formula. Replacing \(x\) with \(x-2\) in the expansion \(1 + x + x^2 + x^3 + \ldots\) results in \(1 + (x-2) + (x-2)^2 + (x-2)^3 + \ldots\).

Step 4: Simplify the series

Simplify each term in the series:- \(1\) remains \(1\).- \((x-2)\) is simply \(x-2\).- \((x-2)^2\) expands to \(x^2 - 4x + 4\).- \((x-2)^3\) expands to \(x^3 - 6x^2 + 12x - 8\), and so on.The first few terms are: \[1 - (x-1) + (x-1)^2 + (x-1)^3 + \ldots\].

Step 5: Write the Taylor series around x=0

Thus, the Taylor series at \(x=0\) for \(\frac{1}{2-x}\) is:\[1 + (x-1) + (x-1)^2 + (x-1)^3 + \ldots\]but as this approaches 1, multiply throughout by -1 and simplify to obtain:\[ \frac{1}{2-x} = \frac{1}{2} \sum_{n=0}^{\infty} (x/2)^n \text{ for } |x|<2\]

Key Concepts

Geometric Series

A geometric series is a sequence of numbers in which each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. It appears in many areas of mathematics because it describes situations where something is increasing or decreasing by a constant factor.
The formula for the sum of an infinite geometric series is given by \[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \]This is valid for \(|x| < 1\), meaning that the absolute value of the common ratio must be less than one for the series to converge.

• The series starts with the first term as 1.
• Each subsequent term is a power of x, the common ratio.

Understanding geometric series is crucial when working on problems involving series expansion, such as in Taylor series.

Substitution Method

The substitution method in mathematical series is a powerful technique for simplifying complex expressions or series. In the given problem, we substituted the variable in the geometric series formula, \( \frac{1}{1-x} \), with \( x-2 \).
This helps us transform our function \( \frac{1}{2-x} \) into a format where it is easier to apply the series expansion. The key idea here is:

• Identify a familiar series pattern, like the geometric series.
• Replace the variable with the expression that matches your function.

By substituting \( x-2 \) for x, we're effectively shifting the function to align with a known series format, making further manipulation and simplification easier.

Series Expansion

Series expansion, especially Taylor series, allows functions to be expressed as an infinite sum of terms calculated from the values of a function's derivatives at a single point.
This concept is used to approximate more complex functions using polynomials, which are easier to analyze and compute.
In this exercise, we expanded the function \( \frac{1}{2-x} \) using a geometric series format. The series begins with terms:

• 1 (the initial term of the series)
• (x-2) (the substitution)
• (x-2)^2, which expands more into a polynomial form as \( x^2 - 4x + 4 \)

This series continues indefinitely, providing a polynomial approximation for the function near the chosen point.

Mathematical Series

Mathematical series involve the sum of sequences, where each term is derived from applying a definite rule or formula. There are various types of series based on their characteristics and applications.
A series can be finite or infinite, and understanding the type is crucial for manipulating them correctly.

• Geometric series, as we've seen, is when each term is a constant multiple of the previous term.
• Arithmetic series, on the other hand, have a constant difference between terms.

In calculus, series are used to represent functions that might otherwise be too complex to handle directly, enabling powerful calculations, such as differentiation and integration.

Functions

Functions are fundamental in mathematics, representing relationships between sets of data. In calculus, functions describe how one quantity changes with respect to another.
For instance, the given function \( \frac{1}{2-x} \) can be analyzed and simplified using Taylor series or other series expansions to provide insights into its behavior for small changes in x. Functions are often expressed as:

• Polynomials, like \( x^2 - 4x + 4 \)
• Rational functions, such as \( \frac{1}{2-x} \)

Understanding how to manipulate and interpret these functions through series and substitutions is key to solving many mathematical problems.

Calculus

Calculus is a branch of mathematics that studies continuous change. Its two main branches—differential and integral calculus—deal with the calculation of derivatives and integrals, respectively.
In the context of Taylor series expansion, calculus helps in approximating functions using derivatives.

• Derivatives measure how a function changes as its input changes, which is pivotal in forming a Taylor series.
• Integrals provide cumulative measures, which can also be useful in series development.

By combining the concepts of derivatives and integrals, calculus offers tools such as Taylor series to approximate functions that might otherwise be difficult to handle analytically.