Question

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. $$ \frac{\sqrt{x}}{1-x^{3 / 2}} \text { in } \sqrt{x} $$

Short answer

Expressed as \( \sum_{n=0}^{\infty} x^{(3n+1)/2} \).

Step-by-step solution

Identify the Function

We are given the function \( \frac{\sqrt{x}}{1-x^{3/2}} \) and we need to express this function as a geometric series in the variable \( \sqrt{x} \).

Recognize the Geometric Series Identity

The geometric series identity is \( \frac{1}{1-y} = \sum_{n=0}^{\infty} y^n \), where \( |y| < 1 \). We need to express the denominator \( 1-x^{3/2} \) in the form \( 1-y \) to use this identity.

Replace with Given Terms

In the denominator \( 1-x^{3/2} \), identify \( y = x^{3/2} \), so \( \frac{1}{1-x^{3/2}} \) is in the form of the geometric series.

Form Geometric Series

Using \( y = x^{3/2} \), express the geometric series: \( \sum_{n=0}^{\infty} (x^{3/2})^n = \sum_{n=0}^{\infty} x^{3n/2} \).

Combine with Numerator

The given function has a numerator \( \sqrt{x} \). Multiply the series by this numerator: \( \sqrt{x} \sum_{n=0}^{\infty} x^{3n/2} = \sum_{n=0}^{\infty} x^{(3n/2) + 1/2} \).

Simplify the Series Expression

The expression becomes \( \sum_{n=0}^{\infty} x^{(3n+1)/2} \), which is the desired geometric series in terms of \( \sqrt{x} \).

Key concepts

Geometric Series Identity

In mathematics, a geometric series is a series with a constant ratio between successive terms. The geometric series identity is a powerful tool that helps us express functions in series form. This identity is given by:

• \(\frac{1}{1-y} = \sum_{n=0}^{\infty} y^n\)

This formula works when \(|y| < 1\). With this identity, functions can be written as an infinite sum, allowing for easier manipulation and integration in more complex problems.
A geometric series is particularly useful in calculus for expressing functions in a form that can be differentiated or integrated term by term. This is a cornerstone of calculus problem solving, offering insights into function behavior and solutions to differential equations.

Function Transformation

Function transformation involves altering a function to change its appearance or form. When tackling calculus problems, transforming functions to a particular form can simplify the problem.
The key transformation in this exercise is recognizing that the function \(\frac{\sqrt{x}}{1-x^{3/2}}\) can be rearranged using the geometric series identity. By identifying \(y = x^{3/2}\), we transform the denominator \(1-x^{3/2}\) into the form required by the geometric series identity. This transformation is vital because it allows us to express more complex functions as a series readily.
Function transformations are not limited to geometric series; they can include scaling, shifting, rotating, and reflecting. Such changes make complex functions more manageable and applicable to various calculus techniques.

Series Representation

Series representation is a method of expressing functions as the sum of an infinite series. In this exercise, we sought to express the given function in terms of \(\sqrt{x}\) using a geometric series.
Series representation provides a way to break down functions into their constituent parts, which can be analyzed separately. Here, the function \(\frac{1}{1-x^{3/2}}\) was represented as an infinite series \(\sum_{n=0}^{\infty} x^{3n/2}\). When multiplied by the numerator \(\sqrt{x}\), each term in the series takes the form \(x^{(3n+1)/2}\).
Capturing a function as a series is crucial for performing operations like differentiation or integration, as it allows treating each term independently. This approach simplifies many problems in calculus, both theoretically and practically.

Calculus Problem Solving

Calculus is about analyzing and solving problems concerning change and motion. Problem solving in calculus often requires breaking down complex functions into simpler parts. Using series representation and function transformation makes this feasible.
For the function \(\frac{\sqrt{x}}{1-x^{3/2}}\), understanding the calculus techniques of series expansion enables us to represent the given function as an infinite series. This step-wise approach turns a challenging problem into manageable steps.
These simplified series are then easier to work with using calculus operations like differentiation and integration, leading to simplified solutions or easier interpretations of the behaviour of functions.

• Expressing functions as series.
• Transforming functions for simplification.
• Using series representations for calculus operations.

By doing this, students learn to use calculus not just for solving equations but for understanding more about the function's inherently dynamic nature.