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Use the identity \(\frac{1}{1-y}=\sum_{n=0}^{\infty} y^{n}\) to express the function as a geometric series in the indicated term. $$ \frac{x}{1+x} \text { in } x $$

Short Answer

Expert verified
The function \( \frac{x}{1+x} \) is expressed as \( \sum_{n=0}^{\infty} (-1)^n x^{n+1} \).

Step by step solution

01

Identify the Form of the Function

The function given is \( \frac{x}{1+x} \). We aim to express this function as a geometric series using the identity \( \frac{1}{1-y}=\sum_{n=0}^{\infty} y^{n} \). First, notice that the denominator has the form \(1+y\). To use the identity, we need to rewrite the expression in terms of \(1-y\).
02

Rewrite the Function in a Usable Form

Rewrite \( \frac{x}{1+x} \) as \( x \cdot \frac{1}{1+(-x)} \). Notice that now the expression has the structure \( 1-y \), where \( y = -x \).
03

Apply the Geometric Series Identity

Using the identity \( \frac{1}{1-y} = \sum_{n=0}^{\infty} y^n \), substitute \( y = -x \). The expression \( \frac{1}{1+(-x)} \) becomes \( \sum_{n=0}^{\infty} (-x)^n \).
04

Multiply by the Factor Outside the Series

The function \( \frac{x}{1+x} = x \cdot \frac{1}{1+(-x)} \). Using the result of Step 3, we have \( x \cdot \sum_{n=0}^{\infty} (-x)^n \). Simplify this to get \( \sum_{n=0}^{\infty} x(-x)^n = \sum_{n=0}^{\infty} (-1)^n x^{n+1} \).
05

Simplify the Expression

The expression \( \sum_{n=0}^{\infty} (-1)^n x^{n+1} \) is the geometric series representation of \( \frac{x}{1+x} \) in \(x\). For each term in the series, the exponent of \(x\) is \(n+1\), and the sign alternates based on \((-1)^n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Geometric Series Identity
The geometric series identity is a powerful tool in mathematics, simplifying complex expressions into a series. The identity states: \( \frac{1}{1-y} = \sum_{n=0}^{\infty} y^{n} \). This means that any function resembling the form \( \frac{1}{1-y} \) can be expressed as an infinite series expansion of powers of \( y \).

To use this identity, it's important to recognize expressions that can be manipulated to fit into this form. Inverting the denominator and identifying \( y \) are key steps. Once in the right form, you can effortlessly expand the function into an infinite series. This method is not only useful for simplification but also crucial in mathematical analysis and calculus.
Power Series Expansion and Its Applications
A power series is an infinite sum of terms in the form of \( a_n(y)^n \). It builds expressions as sums of powers of a variable, typically around zero. They are incredibly handy in approximating functions and solving differential equations.

Using the geometric series identity, any function that fits the \( \frac{1}{1-y} \) form can be expanded into a power series. This transformation provides a way to approximate complex functions with polynomially computable components, which is very useful when exact calculation is too difficult or impossible.

Power series expansion gives engineers and scientists a valuable method to model natural phenomena mathematically. Additionally, power series expansions are foundational in Taylor and Maclaurin series, which approximate functions at specific points.
Mathematical Expressions and Their Rearrangement
Mathematical expressions can often be rewritten to reveal underlying structures or relationships. In the case of \( \frac{x}{1+x} \), rearranging the expression involves recognizing that the denominator can be restructured as \( 1+(-x) \), enabling us to utilize the geometric series identity.

This rearrangement highlights the importance of flexibility when dealing with mathematical expressions. It may appear complex initially, but adjusting terms and factors can simplify and illuminate the solution.

Student Tip: When facing challenging expressions, always look for opportunities to refactor or rearrange terms. Understanding these manipulations is key to successfully applying identities and solving problems.
Series Convergence and Its Implications
Series convergence is an essential concept in the realm of infinite series. It addresses whether the series sum approaches a finite value as the number of terms goes to infinity. For a series \( \sum_{n=0}^{\infty} y^n \) to converge, the variable \( y \) must satisfy \(|y| < 1 \).

In the exercise's solution, convergence ensures that our expanded series accurately represents \( \frac{x}{1+x} \) within the radius of convergence. This concept is not only theoretical but practical, ensuring the validity and applicability of our series expansions.

Understanding convergence is fundamental when applying infinite series in real-world scenarios. It guarantees that mathematical models remain robust and accurate.
  • Practical Tip: Always check convergence conditions before employing an infinite series expansion in applications.
  • Recognize the impact of convergence on the functionality and reliability of mathematical models.

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