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Find the binomial series for the functions in Exercises \(11-14\) . $$ (1+x)^{4} $$

Short Answer

Expert verified
The binomial series for \((1+x)^4\) is \(1 + 4x + 6x^2 + 4x^3 + x^4\).

Step by step solution

01

Understand the Binomial Theorem

The Binomial Theorem states that for any positive integer \(n\), \((1+x)^n\) can be expanded as \(\sum_{k=0}^{n} \binom{n}{k} x^k\). We will use this to expand \((1+x)^4\).
02

Apply the Binomial Theorem

Using the Binomial Theorem for \((1+x)^4\), we know \(n=4\). The expansion becomes \(\sum_{k=0}^{4} \binom{4}{k} x^k\).
03

Calculate the Binomial Coefficients

Calculate each binomial coefficient using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) for \(k=0, 1, 2, 3, 4\). These are \(\binom{4}{0} = 1\), \(\binom{4}{1} = 4\), \(\binom{4}{2} = 6\), \(\binom{4}{3} = 4\), and \(\binom{4}{4} = 1\).
04

Write the Binomial Series

Plug the coefficients into the expanded series: \((1+x)^4 = 1 \cdot x^0 + 4 \cdot x^1 + 6 \cdot x^2 + 4 \cdot x^3 + 1 \cdot x^4\).
05

Simplify the Binomial Series

Simplify the series to get \((1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4\). This is the binomial series for the given function.

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Key Concepts

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

Binomial Expansion
The binomial expansion is a way to express a binomial expression that is raised to a power in the form of a series. The binomial expression \((1+x)^n\) can be expanded into a sum of terms using the Binomial Theorem. This theorem helps us break down the expression into a series that involves powers of \(x\) and specific coefficients.

The process involves determining the terms in the series using the formula: \[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \] This formula showcases that each term in the expansion is derived by multiplying a binomial coefficient, \(\binom{n}{k}\), with a corresponding power of \(x\).
  • The process becomes especially useful in simplifying large powers of binomial expressions into more manageable terms.
  • It also aids in solving combinatorial problems and simplifying polynomial expressions in algebraic exercises.
Understanding binomial expansion can give you a foundational toolset in algebra, calculus, and various applications in mathematics.
Binomial Coefficients
Binomial coefficients are numerical factors in the terms of a binomial expansion. These coefficients are represented by \(\binom{n}{k}\), which is read as "n choose k." These coefficients tell us how many different ways we can choose \(k\) elements from \(n\) elements without regard to order.

You can calculate them using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(!\) is the factorial function, meaning the product of all positive integers up to that number.
  • For example, \(\binom{4}{2} = 6\) indicates that there are 6 ways to choose 2 items from a group of 4.
  • These coefficients not only play a critical role in the expansion of binomials but also appear in various areas of mathematics, including permutations and combinations.
In our specific binomial expansion problem with \((1+x)^4\), the coefficients were 1, 4, 6, 4, and 1, systematically determining the multiplier of each corresponding term in the expansion.
Polynomial Expansion
Polynomial expansion involves expressing a power of a binomial expression as a sum of terms in the form of a polynomial. Each term consists of a binomial coefficient and a power of a variable, typically denoted as \(x\).

In the case of the binomial \((1+x)^4\), the polynomial expansion results in the expression: \[ (1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \] This expanded form is a polynomial where each term is constructed from the binomial coefficients and respective powers of \(x\).

  • Polynomial expansions like this simplify handling and manipulating expressions raised to large powers.
  • This concept is a vital part of algebra and calculus because it bridges the formulation of polynomial expressions and the computation of their behaviors.
Through polynomial expansion, complex expressions become easier to work with, solving or approximating problems more conveniently.

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Most popular questions from this chapter

Euler's constant Graphs like those in Figure 11.8 suggest that as \(n\) increases there is little change in the difference between the sum $$1+\frac{1}{2}+\cdots+\frac{1}{n}$$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem 9 , show that $$\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n$$ or $$0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1$$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1\()\) , the numbers \(a_{n}\) defined in part (a) converge: $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma$$ The number \(\gamma,\) whose value is \(0.5772 \ldots,\) is called Euler's constant. In contrast to other special numbers like \(\pi\) and \(e,\) no other expression with a simple law of formulation has ever been found for \(\gamma .\)

Is it true that if \(\sum_{n=1}^{\infty} a_{n}\) is a divergent series of positive numbers then there is also a divergent series \(\sum_{n=1}^{\infty} b_{n}\) of positive numbers with \(b_{n}< a_{n}\) for every \(n ?\) Is there a "smallest" divergent series of positive numbers? Give reasons for your answers.

Outline of the proof of the Rearrangement Theorem (Theo- rem 17\()\) a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n},\) and let \(s_{k}=\sum_{n=1}^{k} a_{n}\) . Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\) $$\sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \quad\( and \)\quad\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2}$$ since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\},\) there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3},\) then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1} .\) Therefore, if \(n \geq N_{3}\) $$\begin{aligned}\left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned}$$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\)

Which of the series 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.) $$ \sum_{n=1}^{\infty} \frac{8 \tan ^{-1} n}{1+n^{2}} $$

Series for tan \(^{-1} x\) for \(|x|>1\) Derive the series $$ \begin{array}{l}{\tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1} \\ {\tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1}\end{array} $$ by integrating the series $$ \frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots $$ in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x .\)

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