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In Exercises \(43-46,\) find a polynomial that will approximate \(F(x)\) throughout the given interval with an error of magnitude less than \(10^{-3} .\) $$ F(x)=\int_{0}^{x} \frac{\ln (1+t)}{t} d t, \quad \text { (a) }[0,0.5] \quad \text { (b) }[0,1] $$

Short Answer

Expert verified
Expand and integrate \( \ln(1+t) \) term-by-term; check errors to ensure they are < \(10^{-3}\).

Step by step solution

01

Understand the Function

The function given is \( F(x) = \int_{0}^{x} \frac{\ln (1+t)}{t} \, dt \). This integral does not have a simple closed form, so we will approximate it using a polynomial.
02

Use a Series Expansion

The expression \( \ln(1+t) \) can be represented using a series expansion: \( \ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \ldots \), valid for \(-1 < t \leq 1\). We use this expansion in the integrand.
03

Substitution in the Integral

Substitute the series expansion of \( \ln(1+t) \) into the integrand: \( \int_{0}^{x} \frac{t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \ldots}{t} \, dt = \int_{0}^{x} (1 - \frac{t}{2} + \frac{t^2}{3} - \frac{t^3}{4} + \ldots) \, dt \).
04

Integrate Term-by-Term

Integrate each term from the series expansion individually:\[ \int_{0}^{x} 1 \, dt = x \]\[ \int_{0}^{x} -\frac{t}{2} \, dt = -\frac{x^2}{4} \]\[ \int_{0}^{x} \frac{t^2}{3} \, dt = \frac{x^3}{9} \]\[ \int_{0}^{x} -\frac{t^3}{4} \, dt = -\frac{x^4}{16} \]Continue this process for as many terms as needed to satisfy the error condition.
05

Determine the Required Degree

Calculate additional terms until the next term in the series becomes smaller than \(10^{-3}\) on the given interval. For simplicity in intervals like \([0, 0.5]\), we calculate the error caused by omitting terms higher than some degree \(n\).
06

Verify Error Estimate

Approximate the error by estimating the maximum value of the next term omitted in the interval. Ensure the error is smaller than \(10^{-3}\). Typically, for interval \([0, 0.5]\), fewer terms are required compared to \([0, 1]\).
07

Write the Polynomial

After determining the series terms needed to achieve the desired accuracy, the polynomial approximation for \( F(x) \) over the intervals \([0, 0.5]\) and \([0, 1]\) is written using the integrated coefficients from previous steps.

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Key Concepts

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

Series Expansion
Series expansion is a powerful tool in mathematics that allows us to express complex functions as infinite sums of simpler terms. Specifically, for this exercise, we use the series expansion for \( \ln(1+t) \), which is given by:
  • \( \ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \ldots \)
This expansion is valid for \(-1 < t \leq 1\), which suits our problem where the integration interval starts at 0. The beauty of series expansion lies in its flexibility. By truncating the series after a few terms, we obtain a polynomial approximation of the actual function.

Substituting this series into the integral's integrand simplifies the function, allowing us to perform integration term-by-term. This is particularly useful in cases where the integral does not have a simple closed-form solution. You'll see that each term in the expansion corresponds to a simple polynomial expression, making it easier to work with.
Integral Calculus
Integral calculus is the branch of calculus that deals with integrals and their properties. In our exercise, we are evaluating the integral \( F(x) = \int_{0}^{x} \frac{\ln(1+t)}{t} \, dt \). This integral represents the accumulation of the area under the curve described by our function.

When we substitute the series expansion into the integral, it transforms the original complex problem into multiple simpler problems. Each term in the series leads to a corresponding integral:
  • \( \int_{0}^{x} 1 \, dt = x \)
  • \( \int_{0}^{x} -\frac{t}{2} \, dt = -\frac{x^2}{4} \)
  • \( \int_{0}^{x} \frac{t^2}{3} \, dt = \frac{x^3}{9} \)
  • and so on.
This method is known as term-by-term integration, and it simplifies the process significantly. Once these individual integrals are calculated, they can be summed to form a polynomial that approximates the integral function \( F(x) \). The resulting polynomial provides an estimate of the area under the original curve to a specified accuracy.
Error Estimation
Error estimation is crucial when approximating functions using polynomials. The main goal is to ensure that the polynomial approximation is within a specific range of accuracy—in our case, less than \(10^{-3}\).

To achieve this, it's important to determine where we can truncate the series without significantly affecting accuracy. This involves calculating the magnitude of the next term in the series that we decided to omit. For smaller intervals like \([0, 0.5]\), fewer terms might be necessary than for larger intervals such as \([0, 1]\). Typically, you continue adding terms until the omitted term becomes less than \( 10^{-3} \) over the interval of interest.
  • Calculate the magnitude of the next omitted term.
  • Compare it to the desired error bound.
  • Continue including terms in the polynomial until the condition is satisfied.
This approach guarantees that the resulting polynomial is an accurate approximation of the function within the specified error limit, ensuring precision over the entire interval.

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Most popular questions from this chapter

In Exercises \(45-48\) , estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. $$ \frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^{n} t^{n}, \quad 0 < t < 1 $$

Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

Estimate the error if \(\cos \sqrt{t}\) is approximated by \(1-\frac{t}{2}+\frac{t^{2}}{4 !}-\frac{t^{3}}{6 !}\) in the integral \(\int_{0}^{1} \cos \sqrt{t} d t\)

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

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