/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 The coefficient of \(x^{4}\) in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 67

The coefficient of \(x^{4}\) in the expansion of \(\left.\mid \sqrt{1+x^{2}}-x\right\\}^{-1}\) in ascending powers of \(x\), when \(|x|<1\), is a. 0 b. \(\frac{1}{2}\) c. \(-\frac{1}{2}\) d. \(-\frac{1}{8}\)

Short Answer

Expert verified
The coefficient of \(x^4\) is \(-\frac{1}{2}\).

Step by step solution

01

Simplify Given Expression

The expression given is \( \left( \sqrt{1+x^2} - x \right)^{-1} \). We need to express it in a form that allows easy expansion. Use the identity \( \sqrt{1+x^2} = \left( 1 + x^2 \right)^{1/2} \). Thus the expression becomes \( \left( (1 + x^2)^{1/2} - x \right)^{-1} \).
02

Use Binomial Expansion for \( (1+x^2)^{1/2} \)

Expand \( (1+x^2)^{1/2} \) using the binomial series: \
\( (1 + x^2)^{1/2} = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots \). \
This expansion is valid as long as \(|x| < 1\).
03

Find Reciprocal Series Expansion

Now we need to find the series expansion of the reciprocal. Let \( f(x) = \left( (1+x^2)^{1/2} - x \right) \). Then \( f(x) = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 - x + \cdots \), which simplifies to \(1 - x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots \).\
We are looking for \((1 - f(x))^{-1}\).
04

Use Geometric Series for Reciprocal

Using the formula for the expansion of a geometric series \( (1-y)^{-1} = 1 + y + y^2 + \cdots \), replace \(y\) with \(x - \frac{1}{2}x^2 + \frac{1}{8}x^4 - \cdots \). Start expanding \
1. \(1 + (x - \frac{1}{2}x^2 + \frac{1}{8}x^4)\).\
2. Focus on terms resulting in \(x^4\): \
- From \(y^4\), (only \((x)^4)\), which is \(x^4\), leading to coefficient of 1.\
- From other combinations \((y^2 \, \text{and} \, y^3)\), check powers resulting in \(x^4\).\
Add contributions: \(-\frac{1}{2} \cdot x^2 \cdot x^2 = -\frac{1}{2}x^4\), impacting total contribution as \(-\frac{1}{2}\).
05

Collect Terms for Coefficient of \(x^4\)

Sum all \(x^4\) contributions computed from earlier steps. Factor all \(1 + (x-\frac{1}{2}x^2 + \cdots) + (-\frac{1}{2}x^2)^2 \cdots\). The expression \(-\frac{1}{2}x^4\) stands alone, hence coefficient is \(-\frac{1}{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Binomial Series
The binomial series is a powerful mathematical tool used to expand expressions of the form \( (1 + x)^n \), where \( n \) can be any real number. It is particularly useful for approximating expressions when \( |x| < 1 \).
  • The basic idea is to expand the expression into an infinite series.
  • Each term in the series involves coefficients called binomial coefficients, commonly represented using combinations or the notation \( \binom{n}{k} \).
  • For instance, the expansion of \( (1 + x)^n \) yields \( 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots \).
In the original problem, the expression \( (1 + x^2)^{1/2} \) is expanded using this series. The binomial series simplifies complex expressions into a more manageable form, which is crucial for subsequent steps like applying geometric or other series expansions.
Geometric Series
A geometric series is a series of terms that have a constant ratio between successive terms. Its standard form is \( 1 + r + r^2 + r^3 + \cdots \), valid when \( |r| < 1 \).
  • It's commonly used to quickly sum infinite sequences that would otherwise be complex to compute one by one.
  • The sum can be found using the formula \( \frac{1}{1-r} \) as it provides a concise way to find the total for the entire infinite sequence.
In the exercise, to address the problem of finding coefficients in expanded terms, the geometric series comes into play when we need to deal with expressions like \( (1 - f(x))^{-1} \). By treating \( f(x) \) as \( y \), the geometric series expansion helps in opening up the reciprocal expression into manageable parts, perfect for extracting specific term coefficients such as that of \( x^4 \).
Reciprocal Function Expansion
Reciprocal function expansion involves expanding a given function where one term is expressed as the inverse or reciprocal of another. It's commonly used in contexts where you have an expression like \( (a - b)^{-1} \).
  • This expansion forms the basis for deciphering how different terms in a function contribute to a power series.
  • When combined with the binomial and geometric series, it allows for complex expressions to be rewritten as infinite series, unveiling insights such as the specific coefficients of terms like \( x^4 \).
The key in this technique is to correctly identify the central term and manipulate it consistently through expansions. Here, by realizing the form of \( (1 - f(x))^{-1} \), it guides us in discerning the contributions of various power terms efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The value of \({ }^{15} C_{0}^{2}-{ }^{15} C_{1}^{2}+{ }^{15} C_{2}^{2}-\cdots-{ }^{15} C_{13}^{2}\) is a. 15 b. \(-15\) c. 0 d. 51

In the expansion of \(\left(1+3 x+2 x^{2}\right)^{6}\), the coefficient of \(x^{11}\) is a. 144 b. 288 c. 216 d. 576

If \(n\) be a positive integer, prove that $$ \begin{aligned} &\begin{array}{l} 1-2 n+\frac{2 n(2 n-1)}{2 !}-\frac{2 n(2 n-1)(2 n-2)}{3 !}+\cdots \\ +(-1)^{n-1} \frac{2 n(2 n-1) \cdots(n+2)}{(n-1) !} \\ =(-1)^{n+1}(2 n) ! / 2(n !)^{2} \end{array} \end{aligned} $$

If \(p=(8+3 \sqrt{7})^{n}\) and \(f=p-[p]\), where \([\cdot]\) denotes the greatest integer function, then the value of \(p(1-f)\) is equal to a. 1 b. 2 c. \(2^{n}\) d. \(2^{2 n}\)

If \((1-x)^{-11}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a x^{r}+\cdots\), then \(a_{0}+a_{1}+a_{2}+\cdots+a_{r}\) is equal to a. \(\frac{n(n+1)(n+2) \cdots(n+r)}{r !}\) b. \(\frac{(n+1)(n+2) \cdots(n+r)}{r !}\) c. \(\frac{n(n+1)(n+2) \cdots(n+r-1)}{r !}\) d. none of these
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.