Problem 57 - Calculus Single Variable

Question

A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x)=\ln (x), c=e, \Delta x=3-e, N=2 $$

Step-by-Step Solution

Answer

N-step approx. 1.1011, one-step approx. 1.0291, direct value 1.0986. N-step is more accurate.

Step 1: Define the Increment Method

The method of increments uses the formula \( f(c + \Delta x) \approx f(c) + f'(c) \cdot \Delta x \). First, we need to identify \( f(c), \Delta x \), and find \( f'(x) \).

Step 2: Identify \( f(c) \) and \( f'(x) \)

Given \( f(x) = \ln(x) \), the derivative is \( f'(x) = \frac{1}{x} \). We have \( c = e \), so \( f(c) = f(e) = \ln(e) = 1 \). Using \( f'(e) = \frac{1}{e} \).

Step 3: Apply Increment Method for One-Step Approximation

Calculate \( f(e + 3-e) \approx f(e) + f'(e) \cdot (3-e) = 1 + \frac{1}{e} \cdot (3-e) \). Substitute approximate values: \( e \approx 2.718 \). \( f(e + 3-e) \approx 1 + \left(\frac{1}{2.718}\right) \cdot (3-2.718) \approx 1 + 0.1033 \cdot 0.282 \approx 1.0291 \).

Step 4: Calculate Step Size \( h \)

Set \( h = \frac{\Delta x}{N} = \frac{3-e}{2} \approx \frac{0.282}{2} = 0.141 \).

Step 5: Estimate \( y_1 \) using Increments

Use \(c = e\), \( f(c) = 1 \), and \( h = 0.141 \): \( y_1 = f(e + h) \approx 1 + \frac{1}{e} \cdot 0.141 \approx 1 + 0.0518 \approx 1.0518 \).

Step 6: Estimate \( y_2 \) using \( y_1 \) as the base

Use \( c + h = e + 0.141\), \( y_1 = 1.0518 \), and \( h = 0.141 \): \( y_2 = y_1 + \frac{1}{c+h} \cdot 0.141 \approx 1.0518 + \frac{1}{2.859} \cdot 0.141 \approx 1.0518 + 0.0493 \approx 1.1011 \).

Step 7: Direct Calculation of \( f(c+\Delta x) \)

Calculate \( f(3) = \ln(3) \approx 1.0986 \) using a calculator or computer. This value is the exact answer.

Step 8: Compare One-step and N-step Approximations

The one-step approximation of \( f(c+\Delta x) \) is \( 1.0291 \) and the N-step approximation using \( N=2 \) steps is \( 1.1011 \). The direct calculation gives \( 1.0986 \). The N-step approximation is more accurate than the one-step approximation.

Key Concepts

Increment Approximation

Increment approximation is a mathematical concept where we estimate the value of a function at a certain point by using its value and the slope of the function at a nearby point. This is often done using the function's derivative, much as we use a tangent line to approximate the curve of a graph. For the function \( f(x) \), we express this as:

• \( f(c + \Delta x) \approx f(c) + f'(c) \cdot \Delta x \)

This formula means we are estimating the function's value at \( c + \Delta x \) by taking its known value at \( c \) and adding the product of the derivative \( f'(c) \) and the increment \( \Delta x \).
This approach relies on the assumption that over a small interval, the function behaves linearly, making the derivative a good measure of change.

Step Size

The step size \( h \) is a crucial component in increment approximations and other numerical methods. It is essentially the size of the increment we take to reach the next point at which the function is to be evaluated. It's calculated as:

• \( h = \frac{\Delta x}{N} \)

In our example, \( \Delta x = 3 - e \) and \( N = 2 \), so the step size \( h \) becomes \( \frac{3 - e}{2} \).
Choosing an appropriate step size is key for balancing accuracy and computational effort. A small \( h \) can give a more accurate result but requires more computational steps, whereas a larger \( h \) reduces steps but may decrease accuracy.

One-step vs N-step Approximation

Using the method of increments can be tailored in terms of the number of steps, each affecting the accuracy of our approximation of a function.

• **One-step Approximation**: You use a single application of the increment approximation, which involves larger steps and, as seen in our scenario, can lead to less accurate results.
• **N-step Approximation**: Involves multiple small steps and as in the example, was more accurate than the one-step method.
  This method sequentially builds finer increments (\( h \)) to approximate the function, enhancing precision by minimizing linear approximation errors with each smaller step.

The smaller increments allow the method to accommodate and correct approximation errors effectively, thus offering better alignment with the real function's output.

Natural Logarithm Function

The natural logarithm function, represented by \( \ln(x) \), is a fundamental logarithmic function, where the base is the number \( e \) (approximately 2.718). In calculus, it's frequently used due to its unique properties:

• **It is the inverse of the exponential function**: This means \( \ln(e^x) = x \).
• **Its derivative is straightforward**: The derivative of \( \ln(x) \) is \( \frac{1}{x} \), simplifying calculations in many contexts.

Given this function, we can effectively harness its derivative in the method of increments, like in our calculations where we used \( f'(x) = \frac{1}{x} \) to estimate values of \( f(x) = \ln(x) \) at points slightly shifted from \( e \).
This explains the robust performance and accuracy in mathematical approximations and simulations when it comes to exponential growth and decay, hence its prevalence in analytical approaches.