Problem 31 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=e^{x}, \quad a=2\)

Step-by-Step Solution

Answer

The Taylor series for \( e^x \) at \( x=2 \) is \( \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n. \)

Step 1: Understand the Taylor Series Formula

The Taylor series of a function \( f \) at a point \( a \) is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] where \( f^{(n)}(a) \) is the \( n \)-th derivative of \( f \) evaluated at \( a. \)

Step 2: Evaluate Derivatives of \( f(x) = e^x \)

The function \( f(x) = e^x \) has derivatives that repeat as \( f^{(n)}(x) = e^x \) for all \( n. \) Thus, \( f^{(n)}(a) = e^a \) for all \( n. \) Since we're evaluating at \( a = 2, \) then \( f^{(n)}(2) = e^2 \) for all \( n. \)

Step 3: Substitute into the Taylor Series Formula

Substitute \( f^{(n)}(2) = e^2 \) into the Taylor series formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n \] This represents the Taylor series for \( e^x \) centered at \( x = 2. \)

Key Concepts

Derivatives of Functions

In calculus, understanding the derivative of a function is crucial. The derivative measures how a function changes as its input changes. For the exponential function \( f(x) = e^x \), the derivative is unique in the sense that it remains the same after differentiation. That is, each derivative of \( f(x) = e^x \) is also \( e^x \).
This property makes exponential functions quite straightforward to work with when finding derivatives multiple times.

• The first derivative \( f'(x) \) is \( e^x \).
• The second derivative \( f''(x) \) is \( e^x \).
• In general, all the higher order derivatives \( f^{(n)}(x) \) will be \( e^x \).

This principle simplifies the process of constructing a Taylor series, as the derivatives at the point of expansion \( a \) are simply \( e^a \). This repetition of the exponential function is an important property in calculus when dealing with series, especially Taylor series.

Exponential Function

The exponential function \( f(x) = e^x \) is one of the most important and fascinating functions in mathematics. Its base, \( e \), is an irrational constant approximately equal to 2.71828. The exponential function has unique properties that make it very useful: it grows rapidly and has a distinctive self-similarity in its derivatives.
The function \( e^x \) can be applied in various contexts such as compound interest, population growth, and natural processes that follow an exponential trend.

• It is defined in the real domain and continues to grow exponentially as \( x \) increases.
• As mentioned, its derivative is itself, which makes it very useful in differential equations and series expansions.

The exponential function's 'self-derivative' nature particularly impacts its Taylor series expansion. At any point \( a \), the function and each of its derivatives will evaluate to \( e^a \), significantly influencing the terms of the series.

Series Expansion

A series expansion is a way of writing a function as an infinite sum of terms. The Taylor series is a specific type of series expansion used to approximate functions locally around a point \( a \). The general formula for the Taylor series of \( f(x) \) at \( a \) is:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \]This means each term in the series is composed of derivatives computed at \( a \), multiplied by powers of \( x-a \), divided by factorials. When dealing with the exponential function \( e^x \), the simplicity of its derivatives allows us to write:\[ f(x) = \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n \]

• This expansion centers the series at \( x=2 \).
• Each term \( \frac{e^2}{n!} (x-2)^n \) involves calculating powers of \( (x-2) \) and dividing by \( n! \), making it straightforward computationally once \( e^2 \) is known.

This representation is not only exact but captures the essence of how Taylor series can approximate functions smoothly across an interval. It's a powerful tool for understanding local behavior of functions.