Problem 39 - Calculus

Question

Padé approximants are rational functions used to approximate more complicated functions. In this problem, you will derive the Padé approximant to the exponential function. (a) Let \(f(x)=(1+a x) /(1+b x),\) where \(a\) and \(b\) are constants. Write down the first three terms of the Taylor series for \(f(x)\) about \(x=0\) (b) By equating the first three terms of the Taylor series about \(x=0\) for \(f(x)\) and for \(e^{x},\) find \(a\) and \(b\) so that \(f(x)\) approximates \(e^{x}\) as closely as possible near \(x=0\)

Step-by-Step Solution

Answer

\( a = 1 + \frac{1}{\sqrt{2}}, b = \frac{1}{\sqrt{2}} \)

Step 1: Find the Taylor series of \( f(x) \)

The given function is \( f(x) = \frac{1 + ax}{1 + bx} \). To find its Taylor series around \( x = 0 \), perform a binomial expansion on the denominator. Using the identity \( \frac{1}{1+u} = 1 - u + u^2 - u^3 + \cdots \) for \( |u| < 1 \), we have \( \frac{1}{1 + bx} \approx 1 - bx + b^2x^2 \). Thus, \( f(x) = (1 + ax)(1 - bx + b^2x^2) = 1 + ax - bx + (a - ab)x^2 + \text{higher order terms} \). Therefore, the first three terms are \( 1 + (a-b)x + (a - ab)x^2 \).

Step 2: Find the Taylor series of \( e^x \)

The Taylor series for \( e^x \) around \( x = 0 \) is \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \). We take the first three terms for our comparison which gives us \( 1 + x + \frac{x^2}{2} \).

Step 3: Match coefficients to find \( a \) and \( b \)

To approximate \( e^x \) with \( f(x) \) as closely as possible near \( x = 0 \), equate the first three terms of the Taylor series for \( f(x) \) and \( e^x \). 1. From the linear term comparison: \( a - b = 1 \). 2. From the quadratic term comparison: \( a - ab = \frac{1}{2} \). Solving these equations will give values for \( a \) and \( b \).

Step 4: Solve the system of equations

We have the system:1. \( a - b = 1 \) 2. \( a(1 - b) = \frac{1}{2} \) Substitute \( a = b + 1 \) from the first equation into the second:\((b+1)(1-b) = \frac{1}{2}\)\(b(1-b) + 1-b = \frac{1}{2}\) \( -b^2 + b + 1 - b = \frac{1}{2}\) \( -b^2 + 1 = \frac{1}{2} \) \( -b^2 = -\frac{1}{2}\) \( b^2 = \frac{1}{2} \) \( b = \pm \frac{1}{\sqrt{2}} \) For practical, real, and straightforward positive solution use \( b = \frac{1}{\sqrt{2}} \). Substituting back into \( a = b + 1 \) gives \( a = 1 + \frac{1}{\sqrt{2}} \).

Key Concepts

Taylor Series

The Taylor series is a way to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. In simpler terms, it helps break down complex, non-linear functions into a series of polynomials centered around a specific value, usually zero. This method is particularly useful because it allows for the approximation of a function with a polynomial, making it easier to analyze and calculate.

• A Taylor series around the point 0, also known as a Maclaurin series, takes the form: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \,\ldots \)
• Each term in the series involves higher derivatives of the function, divided by the factorial of the degree of that term, scaled by powers of \(x\).
• The more terms we include, the more accurate our approximation becomes.

In the context of this exercise, finding the Taylor series of both \(f(x) = \frac{1 + ax}{1 + bx}\) and the exponential function \(e^x\) around \(x = 0\) allows us to set up a comparison. By matching the coefficients of these expansions, we find the best possible approximation for the exponential function using a rational function.

Exponential Function

The exponential function, denoted as \(e^x\), is fundamental in mathematics due to its unique properties. It describes growth processes, compound interest, and appears before us in many natural phenomena. One of its distinctive features is that it is its own derivative, meaning that both \(e^x\) and its derivative, \(\frac{d}{dx}e^x\), are the same.

• The Taylor series for \(e^x\) around \(x = 0\) is given by: \(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\)
• This series is an infinite sum, yet in practice, we often stop after a few terms due to diminishing returns in accuracy versus computational complexity.
• The exponential function plays a crucial role in various mathematical areas such as calculus, complex analysis, and in describing natural processes like population growth or radioactive decay.

In this exercise, comparing the series expansion of \(e^x\) to a simpler function \(f(x)\) allows us to derive approximations that retain critical characteristics of the exponential function, using a simpler model.

Binomial Expansion

Binomial expansion is a method used in algebra to expand expressions that are raised to any power. For a positive integer \(n\), the binomial theorem gives us a way to expand powers of the form \((a + b)^n\). However, it is also applicable to expressions like \(\frac{1}{1+bx}\) using the series expansion formula.

• For \(|u| < 1\), one can expand \(\frac{1}{1+u}\) into a series: \(1 - u + u^2 - \ldots\)
• This technique simplifies the function, making it easier to combine with other expressions in series expansions or approximations.
• In the context of our original function \(f(x) = \frac{1 + ax}{1 + bx}\), applying a binomial expansion to the denominator allows us to express the function as a polynomial, facilitating the comparison with the Taylor series of \(e^x\).

Using the binomial expansion allows us to deal with the non-polynomial nature of \(\frac{1}{1+bx}\) by converting it into a series, which can then be multiplied with the numerator to yield a cleaner series form for approximation purposes.