Problem 37 - Thomas Calculus

Question

Use the Taylor series generated by \(e^{x}\) at \(x=a\) to show that $$e^{x}=e^{a}\left[1+(x-a)+\frac{(x-a)^{2}}{2 !}+\cdots\right].$$

Step-by-Step Solution

Answer

Using the Taylor series of \( e^{x} \) at \( x = a \), we confirm: \( e^{x} = e^{a}[1 + (x-a) + \frac{(x-a)^2}{2!} + \cdots] \).

Step 1: Recall the Taylor Series Formula

The Taylor series for a function \( f(x) \) around a point \( x = a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] where \( f'(a), f''(a), \) and so on represent the derivatives of \( f \) evaluated at \( x = a \).

Step 2: Apply the Formula for \( e^{x} \)

For the exponential function \( e^{x} \), all derivatives are \( e^{x} \). Therefore, \( f(a) = f'(a) = f''(a) = \cdots = e^{a} \). Substituting these into the Taylor series formula gives: \[ e^{x} = e^{a} + e^{a}(x-a) + \frac{e^{a}}{2!}(x-a)^2 + \frac{e^{a}}{3!}(x-a)^3 + \cdots \]

Step 3: Factor Out \( e^{a} \)

Notice that there is a common factor of \( e^{a} \) in all terms on the right side of the equation. Factoring \( e^{a} \) out gives: \[ e^{x} = e^{a} \left[ 1 + (x-a) + \frac{(x-a)^2}{2!} + \frac{(x-a)^3}{3!} + \cdots \right] \] This shows that the expansion is as required.

Key Concepts

Understanding the Exponential Function

The exponential function, denoted as \( e^{x} \), is a fundamental mathematical concept that appears frequently in calculus, complex analysis, and various applied fields. At its core, the exponential function is a mathematical function that describes a constant rate of growth or decay.

• One of the key properties of \( e^{x} \) is that its derivative is equal to itself. This makes it unique and simplifies many calculus operations for functions involving \( e^{x} \).
• Another important property is that \( e^{x} \) is always positive, meaning it never crosses the x-axis when graphed.

These properties make the exponential function crucial for modeling real-world phenomena like population growth or radioactive decay.Interestingly, while \( e \) is just a number approximately equal to 2.71828, \( e^{x} \) defines a curve with its special properties of growth and self-derivative, setting a base for understanding more complex functions.

Explaining Derivatives in Context

Derivatives provide a way to understand how a function changes as its input changes. In a simple sense, a derivative tells you the slope of a function at any point. It's the rate at which the function's output is changing relative to changes in input.

• For any function \( f(x) \), the first derivative, \( f'(x) \), gives the slope of the tangent line to the curve at any point \( x \).
• Higher-order derivatives, like the second \( f''(x) \) or third \( f'''(x) \), provide information about the curvature and concavity of the function.

In our context, knowing the derivatives of \( e^{x} \)—which are remarkably equal to \( e^{x} \) itself—helps when applying the Taylor series to expand into more complex forms.
Differentiation is a vital tool in calculus, allowing us to dissect and analyze functions in a precise manner.

Role of Series Expansion

Series expansion is a method to represent a function as an infinite sum of terms. This technique allows us to approximate functions in a form that is often simpler to analyze or compute with.

• The most common type of series expansion is the Taylor series, which uses a function's derivatives at a specific point to build the expansion.
• In general, the Taylor series of a function \( f(x) \) at point \( x = a \) is \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]

For the exponential function \( e^{x} \), using its derivatives (which are all \( e^{x} \)) in the Taylor series, we express \( e^{x} \) as: \[ e^{x} = e^{a} + e^{a}(x-a) + \frac{e^{a}}{2!}(x-a)^2 + \frac{e^{a}}{3!}(x-a)^3 + \cdots \]This simplification, and subsequent factoring of \( e^{a} \), shows its utility in mathematics by providing a manageable representation of complex functions.