Problem 56 - Thomas Calculus

Question

The series $$ e^{\mathrm{x}}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\frac{x^{5}}{5 !}+\cdots $$ converges to \(e^{x}\) for all \(x\) $$ \begin{array}{l}{\text { a. Find a series for }(d / d x) e^{x} \text { . Do you get the series for } e^{x} ?} \\ {\text { Explain your answer. }}\\\\{\text { b. Find a series for } \int e^{x} d x . \text { Do you get the series for } e^{x} ? \text { Ex- }} \\ {\text { plain your answer. }} \\\ {\text { c. Replace } x \text { by }-x \text { in the series for } e^{x} \text { to find a series that con- }} \\ {\text { verges to } e^{-x} \text { for all } x \text { . Then multiply the series for } e^{x} \text { and } e^{-x}} \\\ {\text { to find the first six terms of a series for } e^{-x} \cdot e^{x}}\end{array} $$

Step-by-Step Solution

Answer

Yes, both differentiation and integration return the series for \(e^x\), and multiplying the series verifies \(e^{x} \cdot e^{-x} = 1\).

Step 1: Differentiate the Series Term-by-Term

To find the derivative of the series representing \(e^x\), we differentiate term by term. The series is:\[ f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \]Differentiating each term, we get:\[ f'(x) = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \frac{4x^3}{4!} + \frac{5x^4}{5!} + \cdots \]Simplifying, this becomes:\[ f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \]Notice, this is the same as the original series for \(e^x\). Thus, \((d/dx)e^x = e^x\).

Step 2: Integrate the Series Term-by-Term

To find the integral of the series representing \(e^x\), we integrate term by term:\[ \int (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots) \, dx \]Integrating each term, we have:\[ x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots \]This simplifies to:\[ x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots + C \]This series, when compared term-wise, is the same as the original series except for the constant \(C\). Thus, \(\int e^x \, dx = e^x + C\).

Step 3: Replace \(x\) with \(-x\) to find series for \( e^{-x} \)

Substitute \(-x\) for \(x\) in the exponential series:\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots \]This new series represents \(e^{-x}\) for all \(x\).

Step 4: Multiply Series for \(e^{x}\) and \(e^{-x}\)

Multiply the series for \(e^x\) and \(e^{-x}\):\[ (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots) \times (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots) \]Compute the first few terms:- Constant term: \(1 \times 1 = 1\)- Coefficient of \(x\): \(1 \times (-x) + x \times 1 = 0\)- Coefficient of \(x^2\): \(1 \times \frac{x^2}{2} - x \times x + \frac{x^2}{2} \times 1 = 0\)- Coefficient of \(x^3\): Similar pattern leads to zero.In general, all coefficients for powers of \(x\) other than the constant term cancel out, confirming \(e^{-x} \cdot e^{x} = 1\).

Step 5: Conclusion and Verification

From these steps, it follows that:- The series for \((d/dx)e^x\) is \(e^x\), confirming \(d/dx(e^x) = e^x\).- The integral of \(e^x\) returns to \(e^x\) with an added constant \(C\).- Replacing \(x\) with \(-x\) in the series for \(e^x\) gives \(e^{-x}\).- Multiplying \(e^x\) and \(e^{-x}\) yields 1, consistent with their expected product.

Key Concepts

Exponential Function

The exponential function, denoted as \(e^x\), is a fundamental mathematical function that plays a key role in many areas of mathematics. The series expansion of the exponential function, given by

• \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\),

is an infinite series. It resembles a polynomial with infinitely many terms, each term being a fraction where the coefficients involve factorials.
This series is based on expanding the function around zero and is known as a Taylor series expansion.
The remarkable property of this series is that it converges to the actual function for all real numbers \(x\).
The exponential function is central not only in pure mathematics but also in applied fields such as physics, engineering, and finance. It is particularly useful because of its unique properties related to differentiation and integration.

Differentiation

Differentiation is the process of finding the derivative of a function, and it plays a vital role in calculus.
In the context of the exponential function \(e^x\), differentiating the Taylor series term by term, we observe:

• Each term follows the pattern: \(\frac{n \cdot x^{n-1}}{n!}\).
• As a result, differentiating \(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\) leads to the same series

This unique characteristic, where the derivative of \(e^x\) is \(e^x\) itself, is what makes the exponential function particularly elegant.
The simple rule that \(\frac{d}{dx} e^x = e^x\) simplifies many differential calculus problems and is used extensively in real-world applications, where growth or decay processes follow exponential laws.

Integration

Integration is the reverse process of differentiation and involves finding the original function from its derivative.
For the exponential function \(e^x\), integrating the series term by term, we find:

• \(\int (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots) \, dx = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + C\),

where \(C\) is an arbitrary constant of integration.
Just like its differentiation, the integration of \(e^x\) delivers a function closely related to \(e^x\) itself, differing only by a constant. This makes \(e^x\) highly adaptable in solving various integral calculus problems.
It's especially valuable in areas such as solving differential equations that model real-world phenomena.

Convergence of Series

When we talk about the convergence of a series, we refer to the series' behavior as it sums up infinitely many terms.
The exponential Taylor series for \(e^x\)

• \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\)

converges for all real numbers \(x\).
This means that as you include more terms into the sum, the value gets closer to the actual exponential function value.
A powerful result is that the product \(e^x \cdot e^{-x} = 1\), confirmed by multiplying their series expansions. The convergence of these series not only offers insights into the behavior of exponential functions but also ensures that series operations like differentiation, integration, and more complex substitutions lead to valid and valuable outcomes.
Understanding convergence is critical, especially when approximating functions in practical applications, where exact calculations may not be possible.