Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

The author of a biology text claimed that the smallest positive solution to \(x=1-e^{-(1+k) x}\) is approximately \(x=2 k\), provided \(k\) is very small. Show how she reached this conclusion and check it for \(k=0.01\).

Step-by-Step Solution

Answer

The approximation \(x = 2k\) is reasonable for small \(k\), as shown by substituting \(k=0.01\).

Step 1: Understanding the equation structure

We start with the equation \(x=1-e^{-(1+k) x}\). This is a transcendental equation that describes a relationship between \(x\) and \(k\). The task is to understand how \(x\) is related to \(k\) when \(k\) is very small.

Step 2: Calculate approximate solution assumption

The claim is that the smallest positive solution is approximately \(x=2k\). Let's assume \(x = 2k\) and substitute it into the original equation to verify it is a reasonable approximation.

Step 3: Substitute x in the equation

Substitute \(x = 2k\) into the equation: \[ x = 1 - e^{-(1+k)x} \Longrightarrow 2k = 1 - e^{-(1+k)2k}. \] We need to verify whether \(2k\) satisfies the equation for small \(k\).

Step 4: Simplify using series expansion

For small \(k\), the exponential function can be expanded using the Taylor series: \[ e^{-(1+k)2k} \approx 1 - (1+k)2k + \frac{((1+k)2k)^2}{2} + \cdots \]Thus, \(1 - e^{-(1+k)2k} \approx (1+k)2k\), and we check if this is close to \(2k\).

Step 5: Check the approximation for k=0.01

Let's substitute \(k = 0.01\) into the approximation: \[ 2k = 2 \times 0.01 = 0.02. \]Substitute into the equation:\[ 0.02 \approx 1 - e^{-0.0402}. \] Calculate \(e^{-0.0402} \approx 0.9604\), thus \(1 - 0.9604 = 0.0396\), close to \(0.02\).

Step 6: Conclude the validation

The small value of \(0.0396\) doesn't match \(0.02\) exactly but demonstrates that \(x = 2k\) is a valid approximation for very small \(k\). The deviation is due to the nonlinearity present in the full exact solution.

Key Concepts

Transcendental Equation

In calculus, a transcendental equation is one that includes a variable positioned inside a transcendental function such as an exponential, logarithmic, sine, or cosine. These functions transcend algebraic equations because they cannot be simplified to a finite series of algebraic operations. The equation \( x = 1 - e^{-(1+k)x} \) in the original exercise is a perfect example of a transcendental equation.

• It involves the exponential function, which introduces complexity in finding analytical solutions.
• Solving them typically requires approximate methods or numerical solutions.

For small values of parameters, like \( k \), approximations become especially useful to simplify the analyses, as they allow us to bypass solving the equation directly by using clever guesswork and checks.

Series Expansion

Series expansion, particularly Taylor series, is a fundamental method in solving transcendental equations approximately. When \( k \) is small, complex functions can be simplified using their series expansion.
The Taylor series expands a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For the exponential function \( e^x \), the expansion can be useful:

• For \( e^{-(1+k)2k} \), approximate with the first few terms: \( 1 - (1+k)2k + \frac{((1+k)2k)^2}{2} + \ldots \)
• This approximation makes it easier to solve complex transcendental equations by reducing them to simpler polynomial equations.

The series helps us understand how the function behaves around certain points, aiding in checks or proving that an assumption is solid, as seen in our exercise.

Approximation Techniques

Approximation techniques are invaluable in calculus, especially when dealing with complicated functions involved in transcendental equations. The exercise demonstrates how assuming \( x = 2k \) simplifies our calculations when \( k \) is small.

• An initial approximation is just an educated guess based on intuition or prior experience of the behavior of the function.
• In this exercise, substituting \( x = 2k \) into the expanded series gives us a close predictive value.

Checking such an approximation against known values, like with \( k = 0.01 \), reveals how closely it mirrors the actual solution. Importantly, understanding where discrepancies arise further informs when an approximation is fine to use and when consideration of higher accuracy is necessary.