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Chapter 2: Problem 17

PROPERTY 4. \(E(-x)=\frac{1}{E(x)}=(E(x))^{-1}\)

Short Answer

Expert verified
The property \(E(-x) = \frac{1}{E(x)}\) holds true based on the definitions of the exponential function and negative exponents. Taking the exponent to be the negative of a number is the same as taking the reciprocal of the exponential function with the exponent being the positive of that number.

Step by step solution

01

Explain the Exponential Function

The exponential function, denoted as \(E(x)\), is defined as a mathematical function that has the form \(a^x\), where \(a\) is a constant and \(x\) is variable. More commonly, this function uses the number \(e\) (approximately equal to 2.71828) as the base. Thus, the exponential function is often written as \(e^x\).
02

Define the Properties

Properties of exponential functions are used to simplify and solve equations. The property discussed in this exercise states that \(E(-x) = \frac{1}{E(x)} = (E(x))^{-1}\). This indicates that taking the exponent to be the negative of a number \(x\) is the same as taking the reciprocal of the exponential function with the exponent being the positive of that number.
03

Justify the Property

The property \(E(-x) = \frac{1}{E(x)}\) can be justified using the definition of the exponential function. Since \(E(x)\) is defined as \(e^x\), then \(E(-x)\) is equal to \(e^{-x}\). By the definition of negative exponents, \(e^{-x}\) is the same as \(\frac{1}{e^x}\), which can be written as \(\frac{1}{E(x)}\). Therefore, \(E(-x) = \frac{1}{E(x)}\).

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Key Concepts

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

Understanding the Exponential Function
An exponential function is a mathematical expression written in the form of a^x, where a represents a non-zero constant known as the base, and x is the exponent or power. Exponential functions exhibit a rapid increase or decrease in value as x changes, which is a unique characteristic distinguishing them from other types of functions.

For example, if we consider a to be the special number e (approximately 2.71828), known as Euler's number, we commonly write the exponential function as e^x. This particular function has widespread applications across different domains such as finance, computer science, and natural sciences due to its unique property where the rate of change is proportional to the value of the function itself.

Exponential functions also follow certain properties that make calculations easier. For instance, multiplying two exponential expressions with the same base can be simplified by adding the exponents. Understanding these properties is crucial for working with exponential growth or decay problems.
Demystifying Negative Exponents
Negative exponents can often be a source of confusion, but their concept is quite straightforward. The negative sign in an exponent, such as in a^{-x}, indicates the reciprocal of the base raised to the corresponding positive exponent. In other words, a^{-x} is the same as 1/a^x.

Knowing this property is incredibly useful when dealing with exponential functions as it simplifies expressions and equations that might otherwise seem complex. It allows us to transform division problems into multiplication, streamlining computations and making it easier to follow algebraic rules. For example, e^{-x}, using Euler's number as the base, is equivalent to 1/e^x, thus simplifying the expression to a form that is more intuitive to work with.
The Reciprocal of an Exponential Function
The reciprocal of an exponential function, such as (E(x))^{-1}, essentially inverts the function's value. If we have an exponential function E(x) = e^x, taking the reciprocal means we're looking for 1/e^x or e^{-x}. This reciprocal relationship is fundamental when dealing with exponential decay processes or merely trying to find the inverse value of an exponential growth.

Understanding that E(-x) is the same as (E(x))^{-1} is a powerful tool, for it allows us to effortlessly transition between an exponential function and its inverse without cumbersome computations. This simplification relies on our grasp of negative exponents, reinforcing the interconnectedness of these concepts within the realm of algebra.

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Most popular questions from this chapter

Evaluate the following integrals as series. (a) \(\int_{x=0}^{1} e^{x^{2}} \mathrm{~d} x\) (b) \(\int_{x=0}^{1} \frac{1}{1+x^{4}} \mathrm{~d} x\) (c) \(\int_{x=0}^{1} \sqrt[3]{1-x^{3}} \mathrm{~d} x\)

(a) Show that $$ \int_{x=0}^{1 / 2} \sqrt{x-x^{2}} \mathrm{~d} x=\sum_{n=0}^{\infty} \frac{(-1)^{n} \prod_{j=0}^{n-1}\left(\frac{1}{2}-j\right)}{\sqrt{2} n !(2 n+3) 2^{n}} $$ and use this to show that $$ \pi=16\left(\sum_{n=0}^{\infty} \frac{(-1)^{n} \prod_{j=0}^{n-1}\left(\frac{1}{2}-j\right)}{\sqrt{2} n !(2 n+3) 2^{n}}\right) $$ (b) We now have two series for calculating \(\pi:\) the one from part (a) and the one derived earlier, namely $$ \pi=4\left(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\right) $$

(a) Show that if \(y=\sum_{n=0}^{\infty} a_{n} x^{n}\) satisfies the differential equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=-y\), then $$ a_{n+2}=\frac{-1}{(n+2)(n+1)} a_{n} $$ and conclude that $$ y=a_{0}+a_{1} x-\frac{1}{2 !} a_{0} x^{2}-\frac{1}{3 !} a_{1} x^{3}+\frac{1}{4 !} a_{0} x^{4}+\frac{1}{5 !} a_{1} x^{5}-\frac{1}{6 !} a_{0} x^{6}-\frac{1}{7 !} a_{1} x^{7}+\cdots . $$ (b) Since \(y=\sin x\) satisfies \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=-y,\) we see that $$ \sin x=a_{0}+a_{1} x-\frac{1}{2 !} a_{0} x^{2}-\frac{1}{3 !} a_{1} x^{3}+\frac{1}{4 !} a_{0} x^{4}+\frac{1}{5 !} a_{1} x^{5}-\frac{1}{6 !} a_{0} x^{6}-\frac{1}{7 !} a_{1} x^{7}+\cdots $$ for some constants \(a_{0}\) and \(a_{1}\). Show that in this case \(a_{0}=0\) and \(a_{1}=1\) and obtain $$ \sin x=x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}-\frac{1}{7 !} x^{7}+\cdots=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1} $$

(a) Show that if \(y=\sum_{n=0}^{\infty} a_{n} x^{n}\) satisfies the differential equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=-y,\) then $$ a_{n+2}=\frac{-1}{(n+2)(n+1)} a_{n} $$ and conclude that $$ y=a_{0}+a_{1} x-\frac{1}{2 !} a_{0} x^{2}-\frac{1}{3 !} a_{1} x^{3}+\frac{1}{4 !} a_{0} x^{4}+\frac{1}{5 !} a_{1} x^{5}-\frac{1}{6 !} a_{0} x^{6}-\frac{1}{7 !} a_{1} x^{7}+\cdots $$ (b) Since \(y=\sin x\) satisfies \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=-y,\) we see that $$ \sin x=a_{0}+a_{1} x-\frac{1}{2 !} a_{0} x^{2}-\frac{1}{3 !} a_{1} x^{3}+\frac{1}{4 !} a_{0} x^{4}+\frac{1}{5 !} a_{1} x^{5}-\frac{1}{6 !} a_{0} x^{6}-\frac{1}{7 !} a_{1} x^{7}+\cdots $$ for some constants \(a_{0}\) and \(a_{1}\). Show that in this case \(a_{0}=0\) and \(a_{1}=1\) and obtain $$ \sin x=x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}-\frac{1}{7 !} x^{7}+\cdots=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1} $$ (b) Let \(s(x, N)=\sum_{n=0}^{N} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1}\) and \(c(x, N)=\sum_{n=0}^{N} \frac{(-1)^{n}}{(2 n) !} x^{2 n}\) and use a computer algebra system to plot these for \(-4 \pi \leq x \leq 4 \pi, N=\) \(1,2,5,10,15 .\) Describe what is happening to the series as \(N\) becomes larger.

Without using Taylor's Theorem, represent the following functions as power series expanded about 0 (i.e., in the form \(\left.\sum_{n=0}^{\infty} a_{n} x^{n}\right)\). (a) \(\ln \left(1-x^{2}\right)\) (b) \(\frac{x}{1+x^{2}}\) (c) \(\arctan \left(x^{3}\right)\) (d) \(\ln (2+x)\) [Hint: \(\left.2+x=2\left(1+\frac{x}{2}\right)\right]\)
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