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Chapter 10: Problem 40

If the birthrate (thousands of births per year) rises by an amount \(\frac{1-\cos t}{t}\) (above the usual level) in year \(t,\) the number of excess births during the first \(x\) years will be \(\int_{0}^{x} \frac{1-\cos t}{t} d t\). a. Find the Taylor series at 0 for \(\frac{1-\cos t}{t}\). [Hint: Modify a known series.] b. Integrate this series from 0 to \(x\), obtaining a Taylor series for the integral \(\int_{0}^{x} \frac{1-\cos t}{t} d t\) c. Estimate \(\int_{0}^{1} \frac{1-\cos t}{t} d t\) by using the first three terms of the series found in part (b) evaluated at \(x=1\)

Short Answer

Expert verified
The estimated integral is \(\frac{23}{96}\).

Step by step solution

01

Find Taylor series for Cosine Function

The Taylor series expansion for \(\cos t\) at \(t = 0\) is \(\cos t = 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \frac{t^6}{6!} + \cdots\).
02

Modify the Series for 1 - Cos t

Given \(1 - \cos t = 1 - (1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots) = \frac{t^2}{2} - \frac{t^4}{24} + \cdots\).
03

Divide by t

Divide each term in the series by \(t\): \(\frac{1 - \cos t}{t} = \frac{t}{2} - \frac{t^3}{24} + \cdots\).
04

Integrate the Series Term by Term

Integrate the series \(\int_0^x \left(\frac{t}{2} - \frac{t^3}{24} + \ldots \right) dt = \left[ \frac{t^2}{4} - \frac{t^4}{96} + \ldots \right]_0^x\).
05

Evaluate the Integral for x

Substitute \(x\) into the result: \(\frac{x^2}{4} - \frac{x^4}{96} + \cdots\).
06

Estimate the Integral for x = 1

Plug \(x = 1\) into the integrated series: \(\frac{1^2}{4} - \frac{1^4}{96} = \frac{1}{4} - \frac{1}{96}\).
07

Simplify the Result

Simplify to get \(\frac{24}{96} - \frac{1}{96} = \frac{23}{96}\).

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Key Concepts

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

Integration
Integration is the process of finding the integral of a function, which essentially means finding the area under the curve of that function.
There are different methods for integration, but in this exercise, we deal with integrating a series term by term.
  • To integrate the function \( rac{1-\cos t}{t}\), we first express it as a series by using the Taylor series expansion.
  • Once we have the series \(\frac{t}{2} - \frac{t^3}{24} + \cdots\), we integrate each term separately.
This step-by-step process allows us to handle the integration in a more manageable way, considering that directly integrating the original function could be quite complex.
By integrating each term separately and evaluating it over the desired interval, we achieve a practical solution.
Series Expansion
Series expansion refers to expressing a function as an infinite sum of terms calculated from the values of its derivatives at a certain point. This is useful in approximating functions that are otherwise difficult to handle.
  • The Taylor series is a specific type of series expansion centered around a point, often zero, which provides an approximation of a function by using its derivatives.
  • In our exercise, we use the Taylor series of the cosine function \( \cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots\) to find the series for \(1-\cos t\).
    This results in \(\frac{t^2}{2} - \frac{t^4}{24} + \cdots\).
  • Next, dividing each term by \(t\) gives us \(\frac{1 - \cos t}{t} = \frac{t}{2} - \frac{t^3}{24} + \cdots\), which can be more easily integrated term by term.
By adopting the method of series expansion, complex functions become more approachable, breaking them down into simpler polynomial forms that are easier to integrate and evaluate.
Cosine Function
The cosine function is a fundamental trigonometric function, usually written as \(\cos(t)\). It shows oscillating behavior, which is periodic with a period of \(2\pi\), and it is crucial in many areas of mathematics and physics.
  • The Taylor series expansion of the cosine function starts from \(\cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots\).
  • Such expansions are extremely helpful because they allow us to approximate cosine values using polynomials, simplifying calculations.
In the given problem, we make use of the cosine function's Taylor series to find \(1 - \cos t\) in a series form.
This enables us to take advantage of polynomial manipulation, which is easier for integration and estimation.
By understanding the cosine function and its series expansion, we can employ these approximations to make complex mathematical operations, like calculating excess births over time, more feasible.

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Most popular questions from this chapter

What happens in Newton's method if your initial estimate is actually a solution to \(f(x)=0 ?\)

a. For \(f(x)=3 x^{2}-4 x+5,\) find the second Taylor polynomial at \(x=2\) b. Multiply out the Taylor polynomial found in part (a) and show that it is equal to the original polynomial.

BUSINESS: Sinking Fund A sinking fund is an annuity designed to reach a given value at a given time in the future (often to pay off a debt or to buy new equipment). A company's \(\$ 100,000\) printing press is expected to last 8 years. If equal payments are to be made at the end of each quarter into an account paying \(8 \%\) compounded quarterly, find the size of the quarterly payments needed to yield \(\$ 100,000\) at the end of 8 years. [Hint: Solve \(\left.x+x(1.02)+x(1.02)^{2}+\cdots+x(1.02)^{31}=100,000 .\right]\)

DIVERGENCE OF GEOMETRIC SERIES Show that the geometric series $$ a+a r+a r^{2}+a r^{3}+\cdots $$ diverges if \(|r| \geq 1,\) as follows (for \(a \neq 0\) ): a. If \(r=1,\) the series becomes \(a+a+a+\cdots\). Show that this series diverges by showing that the sum of the first \(n\) terms is \(S_{n}=n \cdot a\), which does not approach a limit as \(n \rightarrow \infty(\) since \(a \neq 0)\). b. If \(r=-1,\) then the series becomes \(a-a+a-\cdots\). Show that this series diverges by showing that the partial sums \(S_{1}, S_{2}, S_{3}, S_{4}, \ldots\) take values \(a, 0, a, 0, \ldots,\) and so do not approach any limit. c. If \(|r|>1,\) use the fact that \(r^{n}\) does not approach a limit as \(n \rightarrow \infty\) to show that the partial sum formula \(a \frac{1-r^{n}}{1-r}\) does not approach a limit as \(n \rightarrow \infty\)

\(44-50 .\) Let \(a\) be a given number and suppose that the function \(f\) is \(n+1\) times differentiable between \(a\) and \(x\). Since \(a\) and \(x\) are now fixed, we will use \(t\) for a variable taking values between them. Show that the Fundamental Theorem of Integral Calculus on page 333 may be rewritten as $$ \int_{a}^{x} f^{\prime}(t) d t=f(x)-f(a) $$ and therefore $$ f(x)=f(a)+\int_{a}^{x} f^{\prime}(t) d t $$
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