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Find the first four terms of the binomial series for the functions. $$\left(1-\frac{x}{3}\right)^{4}$$

Short Answer

Expert verified
The first four terms are: \(1 - \frac{4x}{3} + \frac{2x^2}{3} - \frac{4x^3}{27}\).

Step by step solution

01

Recall the Binomial Theorem

The Binomial Theorem states that for any positive integer \( n \) and any \( x \), \( (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \). In this case, the function is \( (1 - \frac{x}{3})^4 \). We can treat \( (-\frac{x}{3}) \) as \( x \) in the theorem.
02

Apply the Binomial Theorem

Use the theorem to expand \( \left(1 - \frac{x}{3}\right)^{4} \) into a series. First, identify \( n = 4 \) and \( x = -\frac{x}{3} \). Therefore, the expansion is: \[(1 - \frac{x}{3})^4 = \sum_{k=0}^{4} \binom{4}{k} (-\frac{x}{3})^k.\]
03

Calculate the Binomial Coefficients

Compute the binomial coefficients \( \binom{4}{k} \) for \( k = 0, 1, 2, 3, 4 \). These are:- \( \binom{4}{0} = 1 \)- \( \binom{4}{1} = 4 \)- \( \binom{4}{2} = 6 \)- \( \binom{4}{3} = 4 \)- \( \binom{4}{4} = 1 \)
04

Substitute and Simplify Each Term

For each \( k \), substitute \( \binom{4}{k} \) and \((-\frac{x}{3})^k \) into the series expansion:1. For \( k=0 \): \( \binom{4}{0} (-\frac{x}{3})^0 = 1 \)2. For \( k=1 \): \( \binom{4}{1} (-\frac{x}{3})^1 = -\frac{4x}{3} \)3. For \( k=2 \): \( \binom{4}{2} (-\frac{x}{3})^2 = \frac{6x^2}{9} = \frac{2x^2}{3} \)4. For \( k=3 \): \( \binom{4}{3} (-\frac{x}{3})^3 = -\frac{4x^3}{27} \)
05

Write the Final Expression

Combine the terms to get the first four terms of the expansion:\[(1 - \frac{x}{3})^4 = 1 - \frac{4x}{3} + \frac{2x^2}{3} - \frac{4x^3}{27}.\]

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Key Concepts

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

Binomial Coefficients
Binomial coefficients are special numbers that appear in the expansion of any power of a binomial expression. These coefficients are represented as \( \binom{n}{k} \), read as "n choose k". They tell us how many ways we can choose \( k \) items from a total of \( n \) items, without considering the order of selection.

Understanding binomial coefficients requires knowing about factorials. A factorial, noted by \( n! \), is the product of all positive integers up to \( n \). For instance, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Binomial coefficients are calculated as follows:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
In our example, calculating the coefficients \( \binom{4}{k} \) for different values of \( k \) gives us: 1, 4, 6, 4, and 1. These coefficients determine the weights of each term in the binomial expansion.
Series Expansion
Series expansion is a way to represent complicated expressions in simpler, more manageable terms. By expressing functions as series, we break them down into a sum of terms that follow a specific pattern.

For the function \( (1 - \frac{x}{3})^4 \), using series expansion helps us write it in a sequence of polynomials. Each term in this kind of series is determined using binomial coefficients and corresponding powers of \( x \). This method is particularly useful in calculus and numerical analysis to approximate functions.

In this problem, the series expansion of \( (1 - \frac{x}{3})^4 \) gives us the first four terms:
  • 1st term: 1 (for \( k=0 \))
  • 2nd term: \( -\frac{4x}{3} \) (for \( k=1 \))
  • 3rd term: \( \frac{2x^2}{3} \) (for \( k=2 \))
  • 4th term: \( -\frac{4x^3}{27} \) (for \( k=3 \))
These terms capture the behavior of the original function up to the third power of \( x \).
Polynomial Expansion
Polynomial expansion involves expressing a power of a binomial expression as a sum of terms, each containing a power of \( x \). This is directly related to the binomial theorem and series expansion concepts.

Using polynomial expansion, we can express \( (1 - \frac{x}{3})^4 \) as a polynomial with variable powers of \( x \). This involves applying the binomial theorem which provides a clear formula for expansion. Expanding simplifies the problem by converting a potentially complex expression into a more straightforward polynomial form.

The final result of polynomial expansion for this expression is:
  • \( 1 - \frac{4x}{3} + \frac{2x^2}{3} - \frac{4x^3}{27} \).
Each term is a polynomial of decreasing significance but collectively, they represent the entire function accurately within the scope of the expansion. Polynomial expansions are crucial for simplifying equations in algebra and higher-level mathematics.

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Most popular questions from this chapter

a. Show that if two power series \(\Sigma_{n=0}^{\infty} a_{n} x^{n}\) and \(\Sigma_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !)\) b. Show that if \(\Sigma_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n\)

A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient's system decays exponentially at a constant hourly rate of \(k=0.12\) a. How many milligrams of the drug are in the patient's system just before the second tablet is taken? Just before the third tablet is taken? b. In the long run, after taking the medication for at least six months, what quantity of drug is in the patient's body just before taking the next regularly scheduled morning tablet?

Assume that each sequence converges and find its limit. \(2,2+\frac{1}{2}, 2+\frac{1}{2+\frac{1}{2}}, 2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}, \dots\)

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) \(a_{n}=\frac{n^{41}}{19^{n}}\)

Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\).

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