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Chapter 8: Problem 46

In each of Exercises 43-48, write Newton's binomial series for the given expression up to and including the \(x^{3}\) term. $$ 1 /(1+x) $$

Short Answer

Expert verified
The series is \(1 - x + x^2 - \frac{1}{2}x^3\).

Step by step solution

01

Recall the Binomial Series Formula

The binomial series expansion for \((1 + x)^{-n}\) is given by \(1 - nx + \frac{n(n+1)}{2!}x^2 - \frac{n(n+1)(n+2)}{3!}x^3 + \ldots\). For our expression, we want the series for \(1/(1+x)\), which can be rewritten as \((1+x)^{-1}\).
02

Substitute into the Binomial Formula

Substitute \(n = 1\) (since we have \((1+x)^{-1}\)) into the binomial series formula: \[1 - 1x + \frac{1(1+1)}{2!}x^2 - \frac{1(1+1)(1+2)}{3!}x^3 + \ldots \]
03

Evaluate the Terms Up to \(x^3\)

Evaluate each term of the series up to \(x^3\):- Zero order term: \(1\)- First order term: \(-x\)- Second order term: \(\frac{1 \times 2}{2}x^2 = x^2\)- Third order term: \(-\frac{1 \times 2 \times 3}{6}x^3 = -\frac{1}{2}x^3\)
04

Write the Final Series Expression

Combine all the evaluated terms to write the series up to the \(x^3\) term:\[1 - x + x^2 - \frac{1}{2}x^3\]

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Key Concepts

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

Newton's Binomial Theorem
Newton's Binomial Theorem is a generalization of the Binomial Theorem. It's especially useful for expanding expressions raised to any power, including negative and fractional powers. Unlike the basic Binomial Theorem, which handles only non-negative integers, Newton's version allows a broader range of applications. This is useful for dealing with series expansions.
  • The binomial expression typically takes the form \((a + b)^n\).
  • When \(n\) is a real number (not just whole numbers), we use Newton's Binomial Theorem.
  • The series expansion becomes: \[(1 + x)^{-n} = 1 - nx + \frac{n(n+1)}{2!}x^2 - \frac{n(n+1)(n+2)}{3!}x^3 + \ldots\]
In the current exercise, we're exploring the series for \((1+x)^{-1}\). Here, \(n = -1\), which means we're using it for a negative power.
This allows us to express functions like \(1/(1+x)\) in an expanded form that's easier to handle analytically, similar to a polynomial.
Power Series Expansion
A power series is an infinite sum of terms in the form of \(a_nx^n\). This expression allows us to represent functions as infinite polynomials, enabling analysis and computation in calculus and potentially solving differential equations.
  • A power series centered at zero is written as \(f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots\).
  • The coefficients \(a_n\) are typically determined by the function that the series represents.
  • Power series are closely related to Taylor and Maclaurin series, which are specialized forms of power series.
In our example, expanding \(1/(1+x)\) using Newton's Binomial Theorem results in an approximation: \[1 - x + x^2 - \frac{1}{2}x^3\] up to the \(x^3\) term. This helps visually approximate and analyze the behavior of complex rational functions like \(1/(1+x)\).
Taylor Series Expansion
The Taylor Series is a specific type of power series, which represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point. Named after the mathematician Brook Taylor, it allows us to approximate functions around a given point.
  • The general formula for a Taylor Series of a function \(f\) centered at \(a\) is: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots\]
  • When the series is centered at zero, it becomes a Maclaurin series: \[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots\]
  • For functions to be effectively approximated by Taylor series, they must be infinitely differentiable at the point of expansion.
In the exercise, the expansion of \(1/(1+x)\) using the binomial form aligns with its Maclaurin series. We approximated it up to the \(x^3\) term: \[1 - x + x^2 - \frac{1}{2}x^3\]. This expansion provides insight into the function's local behavior at \(x=0\), which is highly useful in practical applications of calculus.

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Most popular questions from this chapter

In each of Exercises 49-54, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} $$

Fix a nonzero real number \(\alpha\). Define \(f(x)=(1+x)^{\alpha}\) and $$ g(x)=\sum_{n=0}^{\infty}\left(\begin{array}{l} \alpha \\ n \end{array}\right) x^{n} $$ Our goal is to show that \(f(x)=g(x)\) for \(x \in(-1,1)\). We know from the text that the series certainly converges on that interval. Now complete the following steps: a. Prove that $$ (n+1) \cdot\left(\begin{array}{c} \alpha \\ n+1 \end{array}\right)+n \cdot\left(\begin{array}{l} \alpha \\ n \end{array}\right)=\alpha \cdot\left(\begin{array}{l} \alpha \\ n \end{array}\right) $$ b. Prove that $$ (1+x) \cdot g^{\prime}(x)=\alpha \cdot g(x), \quad x \in(-1,1) $$ c. Prove that $$ (1+x) \cdot f^{\prime}(x)=\alpha \cdot f(x), \quad x \in(-1,1) $$ d. Notice that \(f(0)=g(0)=1\) e. Solve the equation \((1+x) y^{\prime}=\alpha \cdot y\) by rewriting it as $$ \frac{y^{\prime}}{y}=\frac{\alpha}{1+x} $$ f. Find the unique solution to the equation \((1+x) y^{\prime}=\) \(\alpha \cdot y\) satisfying \(y(0)=1\). Conclude that \(f=g\).

Use the Uniqueness Theorem to determine the coefficients \(\left\\{a_{n}\right\\}\) of the solution \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the given initial value problem. \(d y / d x=1+x+y \quad y(0)=0\)

In each of Exercises 55-60, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-3 x}-6 x}{x^{3}} $$

It is known that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{1}{6} \pi^{2}\) and \(\sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{1}{90} \pi^{4}\). Let \(f(x, y)=\sum_{n=1}^{\infty} \frac{x+y \cdot(\pi n)^{2}}{n^{4}} .\) Describe the solution set \(\left\\{(x, y) \in \mathbb{R}^{2}: f(x, y)=0\right\\}\).
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