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Chapter 10: Problem 7

Find the first four terms of the binomial series for the functions. \begin{equation}\left(1+x^{3}\right)^{-1 / 2}\end{equation}

Short Answer

Expert verified
1 \(- \frac{1}{2}x^3 + \frac{3}{8}x^6 - \frac{5}{16}x^9\).

Step by step solution

01

Identify the Binomial Series Formula

The binomial series expansion formula is given by \[ (1 + x)^n = \sum_{k=0}^{fty} \binom{n}{k} x^k \] where \( \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \) is the generalized binomial coefficient. In this problem, our expression is \((1+x^3)^{-1/2}\), so \( n = -1/2 \) and \( x \) is replaced by \( x^3 \).
02

Apply the Formula for Each Term

We need to find the first four terms of the series. These terms correspond to \( k = 0, 1, 2, 3 \). Using the binomial coefficient formula, compute each term.1. First Term \( (k=0) \): \[ \binom{-1/2}{0} (x^3)^0 = 1 \]2. Second Term \( (k=1) \): \[ \binom{-1/2}{1} (x^3) = \left( -\frac{1}{2} \right) x^3 = -\frac{1}{2}x^3 \]3. Third Term \( (k=2) \): \[ \binom{-1/2}{2} (x^3)^2 = \frac{-1/2(-3/2)}{2!}(x^6) = \frac{3}{8}x^6 \]4. Fourth Term \( (k=3) \): \[ \binom{-1/2}{3} (x^3)^3 = \frac{-1/2(-3/2)(-5/2)}{3!}(x^9) = -\frac{5}{16}x^9 \]
03

Write Down the First Four Terms

Compile the results from Step 2 to write down the first four terms of the series:\[ 1 - \frac{1}{2}x^3 + \frac{3}{8}x^6 - \frac{5}{16}x^9 \]
04

Verification

Double-check each step to ensure the binomial coefficients and power of each term are calculated correctly. Ensure that calculations follow the pattern \( \binom{n}{k} x^{3k} \), confirming the logic and arithmetic are correctly applied.

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Key Concepts

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

Binomial Series
The Binomial Series is a representation of an expression raised to a power as a sum of terms. Specifically, it expands the expression \((1 + x)^n\) into a series:
  • This series helps in approximating expressions that are generally difficult to compute.
  • It is particularly useful when dealing with fractional or negative powers.
For the expression \((1+x^3)^{-1/2}\), we use the binomial series to find the first few terms. Here, the expansion simplifies dealing with the negative fractional exponent by expressing it as an infinite series, allowing us to calculate each term individually using known formulas.
Generalized Binomial Coefficient
The Generalized Binomial Coefficient extends the idea of coefficients beyond non-negative integers. Normally, binomial coefficients \(\binom{n}{k}\) involve integers, but for the binomial series with noninteger powers:
  • The coefficient is calculated using the expression: \(\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}\).
  • This formula works regardless of the value of \(n\), allowing us to expand series with any real number exponent.
In our specific case, with \(n = -1/2\), these coefficients allow us to generate each term of the binomial series, contributing to expanding the expression \((1+x^3)^{-1/2}\).
Series Expansion
Series Expansion breaks an expression into a sum of terms with increasing powers of a variable. For binomial series, the series does not need to be finite and instead can be expressed to any desired number of terms following a pattern:
  • The expression \((1 + x^3)^{-1/2}\) is expanded into terms \(k = 0, 1, 2, 3\), providing terms like \(1, -\frac{1}{2}x^3, \frac{3}{8}x^6, -\frac{5}{16}x^9\).
  • This process illustrates how a complex expression can be approximated by simpler terms.
The series expansion is vital for understanding functions in calculus, as it paves the way for calculating values at specific points or under limits.
Binomial Coefficient Formula
The Binomial Coefficient Formula is central to calculating each term in a binomial series. It defines how we derive coefficients for expanded series terms:
  • This formula is: \(\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}\).
  • While traditional binomial coefficients consider natural numbers, this formula's beauty is its application to fractional or negative exponents.
In deriving terms for \((1+x^3)^{-1/2}\), we calculate each \(\binom{-1/2}{k}\) for \(k = 0, 1, 2,\) and \(3\). This method ensures that each term reflects the ongoing power changes in the series.

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Most popular questions from this chapter

a. Use the binomial series and the fact that \begin{equation} \frac{d}{d x} \sin ^{-1} x=\left(1-x^{2}\right)^{-1 / 2} \end{equation} \begin{equation} \begin{array}{l}{\text { to generate the first four nonzero terms of the Taylor series }} \\ {\text { for sin }^{-1} x . \text { What is the radius of convergence? }}\end{array} \end{equation} b. Series for \(\cos ^{-1} x\) Use your result in part (a) to find the first five nonzero terms of the Taylor series for \(\cos ^{-1} x .\)

Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for sin \(^{2} x .\) Then differentiate this series to obtain the Maclaurin series for 2 \(\sin x \cos x\) . Check that this is the series for \(\sin 2 x\) .

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \operatorname{sech} n $$

Prove that $$ \lim _{n \rightarrow \infty} x^{1 / n}=1,(x>0) $$

For a sequence \(\left\\{a_{n}\right\\}\) the terms of even index are denoted by \(a_{2 k}\) and the terms of odd index by \(a_{2 k+1} .\) Prove that if \(a_{2 k} \rightarrow L\) and \(a_{2 k+1} \rightarrow L,\) then \(a_{n} \rightarrow L\)
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