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Chapter 1: Problem 55

Use a graphing calculator to evaluate each expression. $$ \left(1-\frac{1}{1000}\right)^{-1000} $$

Short Answer

Expert verified
The expression evaluates to approximately 2.71815 using a graphing calculator.

Step by step solution

01

Understand the expression

The given expression is \( \left(1 - \frac{1}{1000}\right)^{-1000} \). It involves an exponentiation of a fraction base.
02

Recognize the similarity to a known limit

Notice that the expression \( \left(1 - \frac{1}{1000}\right)^{-1000} \) is similar to the limit definition of the exponential constant \( e \), where \( \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{-n} = e \).
03

Enter the expression into the graphing calculator

On your graphing calculator, enter the expression exactly as it appears: \( \left(1 - \frac{1}{1000}\right)^{-1000} \).
04

Calculate and observe the result

After entering the expression into the calculator, evaluate it to get a result. You should find that the calculated value is approximately equal to 2.71815.

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Key Concepts

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

Exponential Constant
The exponential constant, commonly denoted as \( e \), is a fundamental mathematical constant approximately equal to 2.71828. It is crucial in fields like calculus and complex analysis.
  • The origin of \( e \) is linked to compound interest, where it resolves to continuous compounding.
  • \( e \) is transcendental, meaning it is not a solution to any polynomial equation with rational coefficients.
In mathematics, \( e \) is defined via a limit involving exponentiation, showing up in expressions like \( \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{-n} = e \).
This formula indicates how \( e \) emerges naturally when growth processes are studied.
In many applications, an accurate approximation of \( e \) is essential for calculations, as it is inherently irrational and non-repeating.
Exponentiation
Exponentiation is a mathematical operation involving two numbers: the base and the exponent. It is written as \( b^n \), where \( b \) is the base and \( n \) is the exponent.
  • When the exponent is a positive integer, it indicates how many times the base is multiplied by itself.
  • Fractional exponents correspond to roots and are essential in expressing operations like the square root or cube root.
Negative exponents represent the reciprocal of the base raised to the opposite positive exponent.
For example, \( 3^{-2} \) is equivalent to \( \frac{1}{3^2} = \frac{1}{9} \).
In our example expression \( \left(1 - \frac{1}{1000}\right)^{-1000} \), the use of a negative exponent signifies taking the reciprocal of the expression inside the parenthesis.
Limit Definition
The concept of a limit in calculus is used to describe the behavior of a function as its input approaches a particular point. The limit is essential for defining derivatives, integrals, and continuity.
  • A limit examines values that a function approaches as the input becomes arbitrarily close to a specific point.
  • Limits can be found at finite points, infinity, or nonexistent if values fluctuate too wildly.
In our example, the expression \( \left(1 - \frac{1}{1000}\right)^{-1000} \) closely resembles the limit definition of \( e \).
The limit \( \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{-n} = e \) shows how \( e \) arises as \( n \) becomes very large.
Understanding limits is foundational for tackling advanced calculus topics, providing insight into the fundamental nature of changes in a system.

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Most popular questions from this chapter

BUSINESS: Straight-Line Depreciation Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ( \({ }^{\prime \prime}\) depreciates" \()\) through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-&\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A newspaper buys a printing press for $$\$ 800,000$$ and estimates its useful life to be 20 years, after which its scrap value will be $$\$ 60,000$$. Use the formula above Exercise 63 to find a formula for the value \(V\) of the press after \(t\) years, for \(0 \leq t \leq 20\) b. Use your formula to find the value of the press after 10 years. c. Graph the function found in part (a) on a graphing calculator on the window \([0,20]\) by \([0,800,000] .\) [Hint: Use \(x\) instead of \(t\).]

How will the graph of \(y=-(x-4)^{2}+8\) differ from the graph of \(y=-x^{2} ?\) Check by graphing both functions together.

$$ \text { True or False: If } f(x)=x^{2}, \text { then } f(x+h)=x^{2}+h^{2} \text { . } $$

i. Show that the general linear equation \(a x+b y=c\) with \(b \neq 0\) can be written as \(y=-\frac{a}{b} x+\frac{c}{b}\) which is the equation of a line in slope-intercept form. ii. Show that the general linear equation \(a x+b y=c\) with \(b=0\) but \(a \neq 0\) can be written as \(x=\frac{c}{a}\), which is the equation of a vertical line. [Note: Since these steps are reversible, parts (i) and (ii) together show that the general linear equation \(a x+b y=c\) (for \(a\) and \(b\) not both zero) includes vertical and nonvertical lines.]

ATHLETICS: Juggling If you toss a ball \(h\) feet straight up, it will return to your hand after \(T(h)=0.5 \sqrt{h}\) seconds. This leads to the juggler's dilemma: Juggling more balls means tossing them higher. However, the square root in the above formula means that tossing them twice as high does not gain twice as much time, but only \(\sqrt{2} \approx 1.4\) times as much time. Because of this, there is a limit to the number of balls that a person can juggle, which seems to be about ten. Use this formula to find: a. How long will a ball spend in the air if it is tossed to a height of 4 feet? 8 feet? b. How high must it be tossed to spend 2 seconds in the air? 3 seconds in the air?
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