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BUSINESS: Advertising Five days after the end of an advertising campaign, a company's daily sales (as a proportion of its peak sales) will be \(e^{-0.25}\). Approximate \(e^{-0.25}\) by using the second Taylor polynomial at \(x=0\) for \(e^{x}\) (rounding to two decimal places). Check your answer by using the key on your calculator.

Short Answer

Expert verified
The approximation using the second Taylor polynomial is 0.78.

Step by step solution

01

Understanding the Taylor Polynomial

The second Taylor polynomial of a function provides an approximation by using the function value and derivatives up to the second order. For the function \(e^x\), the Taylor series expansion at \(x=0\) is \(e^x \approx 1 + x + \frac{x^2}{2}\).
02

Substituting into the Polynomial

To approximate \(e^{-0.25}\), substitute \(x = -0.25\) into the second Taylor polynomial: \(e^{-0.25} \approx 1 + (-0.25) + \frac{(-0.25)^2}{2}\).
03

Calculating Each Term

First, calculate the linear term: \(-0.25\). Next, find the square: \((-0.25)^2 = 0.0625\). Then, divide by 2: \(\frac{0.0625}{2} = 0.03125\).
04

Adding the Terms Together

Now, add together all components: \(1 - 0.25 + 0.03125 = 0.78125\).
05

Rounding the Result

Finally, round the result of the sum to two decimal places: \(0.78\).
06

Verifying with a Calculator

Use a calculator to find \(e^{-0.25}\) directly. This should be approximately equal to the rounded polynomial value calculated, which is \(0.78\).

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Key Concepts

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

Taylor Series
The Taylor Series is a way to approximate complex functions using polynomials. It's like breaking down a difficult puzzle into simpler pieces. For any function, the Taylor series represents it as an infinite sum of terms calculated from its derivatives at a specific point. In simple terms, you're using basic arithmetic to estimate something much bigger and more complex. This process can provide increasingly accurate approximations as you include more terms.

For example, the Taylor series expansion for the exponential function at a point zero is given as:
  • \(f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!}\) and so on.
What's fascinating is that each derivative of \(e^x\) is \(e^x\), which makes its Taylor series very approachable.

Using just the first few terms is often enough to approximate well, especially near zero.
Exponential Function
When you think about exponential functions, think about rapid growth or decay. The function \(e^x\) is a fundamental form in mathematics, often used in growth models like populations or investments. In this context, it is also vital for modeling decays such as radioactive decay or cooling.

The exponential function has a neat property: its rate of change (derivative) is proportional to its current value. This property makes \(e^x\) very predictable and useful in calculus and throughout various fields like physics and finance.

Specifically, when substituting \(x\) values into this function, it changes the outcome, which can model various real-world phenomena effectively. These traits make it one of the most important functions in mathematics.
Approximation Methods
Approximation Methods can be essential tools when dealing with complex functions that don't have simple solutions. These methods help us calculate values that we otherwise can't find exactly.

One commonly used method is employing a Taylor polynomial, which gives a practical way to estimate values of functions by adding a few lower-order terms. As demonstrated in the original exercise, using the second Taylor polynomial provides a quick and uncomplicated estimate for \(e^{-0.25}\).

The full expansion was:
  • Start with \(1\)
  • Add \(x\), where in this exercise \(x = -0.25\)
  • Include the quadratic term \(\frac{x^2}{2}\)
The combined value gives an approximation, easy to carry out even without technology.

This method showcases how powerful simple mathematical ideas can be in solving complex real-world problems. It's important to remember that while an approximation might not be exact, it could be more than sufficient depending on the context.

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Most popular questions from this chapter

GENERAL: Chessboards It is said that the game of chess was invented by the Grand Vizier Sissa Ben Dahir for the Indian King Shirham. When the king offered him any reward, Sissa asked for one grain of wheat to be placed on the first square of the chessboard, two on the second, four on the third, and so on, doubling each time for each of the 64 squares. Estimate the total number of grains of wheat required. (It is also said that the king, upon learning that the total was many times the world production of wheat, executed Sissa for making a fool out of him.)

Find the Taylor series at \(x=0\) for each function by modifying one of the Taylor series from this section. [Hint: For Exercise \(26,\) use two series from this section, writing the terms in order of increasing powers of \(x\).] \(\frac{x-\sin x}{x^{3}}\)

GENERAL: Million Dollar Lottery Most state lottery jackpots are paid out over time (often 20 years), so the "real" cost to the state is the sum of the present values of the payments. Find the cost of a "million dollar" lottery by summing the present values of 240 monthly payments of \(\$ 4167,\) beginning now, if the interest rate is \(6 \%\) compounded monthly. [Hint: The present value of \(\$ 4167\) in \(k\) months is \(4167 /(1+0.005)^{k} .\)

For a convergent infinite series \(S=c_{1}+c_{2}+c_{3}+\cdots,\) let \(S_{n}\) be the sum of the first \(n\) terms: \(S_{n}=c_{1}+c_{2}+c_{3}+\cdots+c_{n}\). a. Show that \(S_{n}-S_{n-1}=c_{n}\). b. Take the limit of the equation in part (a) as \(n \rightarrow \infty\) to show that \(\lim _{n \rightarrow \infty} c_{n}=0 .\) This proves the \(n\) th term test for infinite series: In a convergent infinite series, the \(n\) th term must approach zero as \(n \rightarrow \infty\). Or equivalently: If the \(n\) th term of a series does not approach zero as \(n \rightarrow \infty,\) then the series diverges.

As the pace of change in modern society quickens, popular fashions may fluctuate increasingly rapidly. Suppose that sales (above a certain minimum level) for a fashion item are \(\cos t^{2}\) in year \(t,\) so that extra sales during the first \(x\) years are \(\int_{0}^{x} \cos t^{2} d t \quad\) (in thousands). a. Find the Taylor series at 0 for \(\cos t^{2}\). [Hint: Modify a known series.] b. Integrate this series from 0 to \(x\), obtaining a Taylor series for the integral \(\int_{0}^{x} \cos t^{2} d t\) c. Estimate \(\int_{0}^{1} \cos t^{2} d t\) by using the first three terms of the series found in part (b) evaluated at \(x=1\).

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