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Chapter 30: Problem 17

Calculate the value of each of the given functions. Use the indicated number of terms of the appropriate series. Compare with the value found directly on a calculator. $$1.032^{6}\quad(3)$$

Short Answer

Expert verified
The approximate value using 3 terms is 1.20736. Direct calculator value is 1.21868.

Step by step solution

01

Understand the Function and Series

We need to compute \(1.032^6\). The problem indicates to use 3 terms of the series expansion. A common way to approximate powers is using the binomial expansion. For small values of \(x\), the binomial series is \( (1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots \). We will apply this for \(x = 0.032\) and \(n = 6\).
02

Apply the Binomial Series

Apply the formula: \[ (1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 \] For \(x = 0.032\) and \(n = 6\), plug in these values: \[ 1.032^6 \approx 1 + 6 \times 0.032 + \frac{6 \times 5}{2} \times (0.032)^2 \] This simplifies to \[ 1 + 0.192 + \frac{30}{2} \times 0.001024 \].
03

Perform the Calculation

Calculate each part:- \(6 \times 0.032 = 0.192\)- \( (0.032)^2 = 0.001024\)- \( \frac{6 \times 5}{2} \times 0.001024 = 0.01536 \)Now add these results:- \( 1 + 0.192 + 0.01536 = 1.20736\)
04

Compare with Calculator

Using a calculator, compute \(1.032^6\) directly. You should get approximately \(1.21868\). Compare this with the approximation \(1.20736\). The binomial expansion gives a moderately close approximation with just 3 terms.

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Key Concepts

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

Understanding Binomial Expansion
The binomial expansion is a powerful tool in calculus. It allows us to approximate expressions to a simpler form. Especially when dealing with powers of small numbers. The formula \((1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots\) gives us an approximation by expanding the function into a series. In this scenario, we are interested in raising \(1.032\) to the power of 6. Here, \(x = 0.032\) and \(n = 6\). By substituting these values into the series expansion, we can approximate the result without intensive computation.
  • The term \(n\) signifies the power we are expanding to, here \(6\).
  • \(x\) represents the small increment from 1, which is \(0.032\).
Using only three terms of the series, the approximation is fairly accurate, showcasing the usefulness of the binomial series in quick calculations.
Series Approximation Simplified
Series approximation is a method of estimating the value of a function by breaking it into a series. Each term in that series brings us closer to the actual value. For small numbers like \(0.032\), a series gives a reasonable approximation without needing full precision.The process involves
  • identifying the function and its expansion form,
  • using only a few terms for simplicity, and
  • by systematically adding each term for a cumulative result.
In our example, three terms provide an approximation close to the exact calculation. As a result, it's a computationally light method for calculating powers in everyday problems, especially useful in fields like physics and engineering.
Calculator Comparison Insight
Comparing the results of the series approximation with those of a calculator highlights the balance between precision and simplicity. Calculators give us high precision instantly, for example, \(1.032^6 \approx 1.21868\).Here's how to see differences:
  • The binomial approximation yields \(1.20736\) using just three terms.
  • The calculator delivers \(1.21868\), a more precise figure with many more decimals.
While calculators provide immediate accuracy, the series method is excellent for quick estimates and enhances conceptual understanding of processes, making it invaluable in learning and practical applications.
Essential Power Calculation
Power calculations, or raising a number to a power, form the backbone of many mathematical operations. Using series approximations to tackle power problems simplifies complex calculations into manageable steps, beneficial for quick answers. In power calculation:
  • Each term in the series represents a sequential operation building towards the power.
  • Fewer terms lead to a faster but less precise estimate.
Such approximations offer valuable insight, especially for manual computations or educational purposes, where understanding the process is as valuable as the result itself. This blend of accuracy and simplicity aids students and professionals alike in honing their mathematical skills effectively.

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Most popular questions from this chapter

Find the indicated series by the given operation. Show that by integrating term by term the expansion for \(\cos x,\) the result is the expansion for \(\sin x\)

Solve the given problems. In the analysis of the optical paths of light from a narrow slit \(S\) to a point \(P\) as shown in Fig. \(30.7,\) the law of cosines is used to obtain the equation \(c^{2}=a^{2}+(a+b)^{2}-2 a(a+b) \cos \frac{s}{a}\) where \(s\) is part of the circular arc \(\overrightarrow{A B}\). By using two nonzero terms of the Maclaurin expansion of \(\cos \frac{s}{a}, \operatorname{sim}\)plify the right side of the equation. (In finding the expansion, let \(x=\frac{s}{a}\) and then substitute back into the expansion.)

Solve the given problems. The charge \(q\) on a capacitor in a certain electric circuit is given by \(q=c e^{-a t} \sin 6 a t,\) where \(t\) is the time. By multiplication of series, find the first four nonzero terms of the expansion for \(q\)

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Solve the given problems as indicated. Referring to Chapter \(19,\) we see that the sum of the first \(n\) terms of a geometric sequence is \(S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \quad(r \neq 1)\) $$(19.6)$$ where \(a_{1}\) is the first term and \(r\) is the common ratio. We can visualize the corresponding infinite series by graphing the function \(f(x)=a_{1}\left(1-r^{x}\right) /(1-r)(r \neq 1)\) (or using a calculator that can graph a sequence). The graph represents the sequence of partial sums for values where \(x=n,\) since \(f(n)=S_{n}\) Use a calculator to visualize the first five partial sums of the series \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\) What value does the infinite series approach? (Remember: Only points for which \(x\) is an integer have real meaning.)
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