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Use the Binomial Theorem to approximate the value to within \(10^{-6}\) (a) \(\frac{2}{\sqrt[3]{9}}\) (b) \(\sqrt[4]{17}\)

Short Answer

Expert verified
The approximate values using Binomial Theorem for the given expressions are:\n(a) \(\frac{2}{\sqrt[3]{9}}\) approximately equals to 2.07407\n(b) \(\sqrt[4]{17}\) approximately equals to 2.03125.

Step by step solution

01

Identify the binomial in the given expression

For (a) \(\frac{2}{\sqrt[3]{9}}\), first we need to rewrite it to match the binomial form \((1+x)^n\). So, rewrite it as \(2*(9^{-1/3}) = 2*(1 - 1/9)^{-1/3}\). Here, \(x = -1/9\) and \(n = -1/3\).\n\nFor (b) \(\sqrt[4]{17}\), rewrite it in the form \((1+x)^n\). It can be rewritten as \((16 + 1)^{1/4} = 16^{1/4}*(1 + 1/16)^{1/4}\). Here, \(x = 1/16\) and \(n = 1/4\).
02

Apply the Binomial Theorem

Use the Binomial Theorem equation, \( (1+x)^n \approx 1 + nx \), because we want the approximation to within \(10^{-6}\), only the first two terms of the Binomial expansion are needed.\n\nFor (a), after substituting, we get \(2*(1-1/3(-1/9)) = 2*(1 + 1/27) = 2*28/27\).\n\nFor (b), substitute the values of \(x\) and \(n\) in the Binomial Theorem. We get \(16^{1/4}*(1+1/4*1/16) = 2*(1+1/64) = 2*65/64\).
03

Simplify the expression

Simplify the numerical expression obtained after the application of the Binomial theorem.\n\nFor part (a), the approximation is \(2*28/27 = 56/27\). After the decimal conversion, we get approximately 2.07407.\n\nFor (b), the approximation is \(2*65/64 = 130/64\). After the decimal conversion, we get approximately 2.03125.

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Key Concepts

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

Binomial expansion
Understanding the binomial expansion is crucial for working with powers of binomials in algebra. A binomial is a two-term expression, like \(a + b\) or \(a - b\). When we raise a binomial to a power, such as \( (a + b)^n \) where \( n \) is a positive integer, we end up with a multi-term polynomial. A practical approach to expanding binomials, especially when \( n \) is large, is using the Binomial Theorem.

The Binomial Theorem tells us that any binomial raised to a power can be expanded into a series of terms that involve coefficients, which are unique to each term. These coefficients can be determined easily using Pascal's Triangle or the combinatorial formula \( \binom{n}{k} \). For example, \( (a + b)^2 \) can be expanded to \( a^2 + 2ab + b^2 \) without multiplying \(a + b\) by itself repeatedly.

In real-world applications, we might not need the whole expansion but just an approximation. For instance, if \( a \) is much larger than \( b \) and we are only interested in an estimate, the first few terms of the expansion may suffice. This approach becomes even more valuable when dealing with fractional or negative exponents, in which full expansion is not practical or possible.
Binomial Theorem
The Binomial Theorem is a powerful statement in algebra that provides a shortcut to expanding binomial expressions. It states that \( (1 + x)^n \) can be expanded into a sum of terms of the form \( \binom{n}{k} x^k \) where \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) elements out of a set of \( n \) elements without regard for order.

The theorem applies when \( n \) is any real number, not just a positive integer, which is where approximation often comes into play. For example, in the textbook problem, to find an approximation of \( \frac{2}{\sqrt[3]{9}} \) and \( \sqrt[4]{17} \) within \(10^{-6}\), the binomial theorem is used to approximate the expressions by keeping only the necessary first few terms of the expansion. This is because additional terms contribute very little to the value of the expression and can be ignored within a certain threshold of accuracy, hence, improving computational efficiency and simplicity in understanding the results.
Algebraic approximation
Algebraic approximation is a technique used to simplify complex expressions into more manageable ones by making assumptions based on the context. A common form of algebraic approximation involves recognizing when a value is very small relative to others in the expression and can be considered negligible for a rough estimate.

In the context of the Binomial Theorem, we apply algebraic approximation by retaining only the first one or two terms of the expansion when \( x \) is small and \( n \) is not a positive integer. This approximation is valid as the higher-order terms become increasingly insignificant, effectively creating a balance between accuracy and simplicity. For example, as in the given exercise, to approximate \( \frac{2}{\sqrt[3]{9}} \) or \( \sqrt[4]{17} \) we only need to calculate up to the linear term (the second term of the expansion) since this provides enough accuracy within \(10^{-6}\). Each decision to include or exclude terms is driven by the desired precision and often, in engineering and physical sciences, these approximations provide sufficiently accurate results that can be used in various calculations and designs.

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Most popular questions from this chapter

We have seen that \(\sin 1=1-\frac{1}{31}+\frac{1}{51}+\cdots .\) Determine how many terms are needed to approximate sin 1 to within \(10^{-5}\) Show that \(\sin 1=\int_{0}^{1} \cos x d x .\) Determine how many points are needed for Simpson's Rule to approximate this integral to within \(10^{-5} .\) Compare the efficiency of Maclaurin series and Simpson's Rule for this problem.

Involve the binomial expansion. Show that the Maclaurin series for \((1+x)^{y}\) is \(1+\sum_{k=1}^{\infty} \frac{r(r-1) \cdots(r-k+1)}{k !} x^{k},\) for any constant \(r\)

Determine the interval of convergence and the function to which the given power series converges. $$\sum_{k=0}^{\infty}(-1)^{k}\left(\frac{x}{2}\right)^{k}$$

Use a known Taylor series to find the Taylor series about \(c=0\) for the given function and find its radius of convergence. $$f(x)=x e^{-x^{2}}$$

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

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