Problem 56 - 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. $$ f(x)=\sqrt{x}, c=4, \Delta x=0.5, N=5 $$

Step-by-Step Solution

Answer

The N-step estimate \( y_5 = 2.12 \) is closer to the actual value \( \sqrt{4.5} \approx 2.121 \) than the one-step estimate \( y_1 = 2.025 \).

Step 1: Understand Given Parameters

We are given a function \( f(x) = \sqrt{x} \), a point \( c = 4 \), an increment \( \Delta x = 0.5 \), and \( N = 5 \). The task is to estimate \( f(c+\Delta x) = f(4.5) \) using the method of increments.

Step 2: Calculate Step Size

We need to calculate the step size \( h \) using the formula \( h = \frac{\Delta x}{N} = \frac{0.5}{5} = 0.1 \). This is the increment for each step.

Step 3: Initial Estimate Using Increments

Start by estimating \( f(c+h) = f(4.1) \) using the increment formula: \( f(c+h) \approx f(c) + f'(c) \cdot h \). First compute \( f'(x) = \frac{1}{2 \sqrt{x}} \) and \( f'(4) = \frac{1}{2 \cdot 2} = \frac{1}{4} \). So, \( f(4.1) \approx \sqrt{4} + \frac{1}{4} \cdot 0.1 = 2 + 0.025 = 2.025 \). Let this be \( y_1 = 2.025 \).

Step 4: Calculate Next Estimate Using y_1

Using \( y_1 = 2.025 \) as \( f(c+h) \), estimate \( f(c+2h) = f(4.2) \). We have \( f(4.2) \approx y_1 + f'(4.1) \cdot h = 2.025 + \frac{1}{2\sqrt{4.1}} \cdot 0.1 \). Compute \( \frac{1}{2\sqrt{4.1}} \approx 0.247 \), so \( f(4.2) \approx 2.025 + 0.0247 = 2.0497 \). Let this be \( y_2 = 2.0497 \).

Step 5: Proceed with Further Steps

Repeat the procedure for \( f(c+3h) = f(4.3) \), \( f(c+4h) = f(4.4) \), and \( f(c+5h) = f(4.5) \) using \( y_{n-1} \) as the base value. For \( y_3 \): \( f(4.3) \approx 2.0497 + \frac{1}{2\sqrt{4.2}} \cdot 0.1 \approx 2.0739 \). For \( y_4 \): \( f(4.4) \approx 2.0739 + \frac{1}{2\sqrt{4.3}} \cdot 0.1 \approx 2.097 \). For \( y_5 \), compute \( f(4.5) \approx 2.097 + \frac{1}{2\sqrt{4.4}} \cdot 0.1 \approx \) 2.120.

Step 6: Direct Calculation of f(4.5)

Using a calculator, compute \( f(4.5) = \sqrt{4.5} \approx 2.121 \).

Step 7: Compare Estimates

The one-step estimate for \( f(4.5) \) was \( y_1 = 2.025 \), while the multi-step estimate \( y_5 \approx 2.12 \), which compares to the direct calculation \( \sqrt{4.5} \approx 2.121 \). The multi-step estimate is closer.

Key Concepts

Step Size in Function Estimation

The concept of "step size" is crucial when using the method of increments to estimate a function value. Step size, denoted as \( h \), is defined by the formula \( h = \frac{\Delta x}{N} \), where \( \Delta x \) is the total increment and \( N \) is the number of steps. In our example, \( \Delta x = 0.5 \) and \( N = 5 \), thereby calculating the step size as \( h = 0.1 \).
This means with each step, we will increment our base point by 0.1. Using smaller step sizes usually results in more accurate approximations as each step is closer to real value changes in the function.

• Step Size (\( h \)) = Total Increment (\( \Delta x \)) / Number of Steps (\( N \))
• Smaller step sizes tend to increase accuracy.
• For our problem, \( h = 0.1 \).

Understanding this ensures precision in incremental function estimation.

Function Estimation through Method of Increments

Estimating function values at particular points can often be done using methods like the method of increments. This is particularly useful for functions that are difficult to solve directly. This technique estimates the next function value, \( f(c+h) \), by considering the current function value \( f(c) \) and adjusting it based on an increment.
To do this, you add the product of the derivative of the function at the current point and the step size to the current function value. For example, to estimate \( f(4.1) \) given that \( f(x)=\sqrt{x} \) at \( x=4 \), you compute:

• Find \( f(c) = \sqrt{4} = 2 \).
• Calculate the derivative \( f'(x) = \frac{1}{2 \sqrt{x}} \).
• Estimate \( f(4.1) \approx f(4) + f'(4) \times 0.1 = 2 + 0.025 = 2.025 \).

This approach makes it manageable to estimate values near known values without advanced computation.

Derivative of Square Root Function

To carry out incremental approximation accurately, understanding how to derive the square root function is pivotal. The derivative of \( f(x) = \sqrt{x} \) is \( f'(x) = \frac{1}{2\sqrt{x}} \). This derivative tells us how the function changes as \( x \) changes. It provides the rate of change of the function and is used in each step of the incremental approximation process.
For example, at \( x = 4 \), the derivative is:

• \( f'(4) = \frac{1}{2 \cdot 2} = \frac{1}{4} \).

This helps us to know exactly how much the function value will change when \( x \) increases by the step size, aiding our estimation of the next point.

Incremental Approximation Process

Using the method of increments involves a step-by-step process where each step builds on the previous estimate. This process systematically refines the approximation.
Here's how it works:

• Start with an initial approximation \( f(c+h) \), using \( f(c) \) and its derivative \( f'(c) \).
• Continue with subsequent steps, using the most recent estimate as your base point.
• At each step, calculate \( f(c+nh) \) using the previous estimate and the derivative at the current point.
• Example: \( y_1 = 2.025 \), then \( y_2 \approx 2.0497 \), and so forth until \( y_5 \).

In conclusion, by taking multiple incremental steps, you collect a series of increasingly accurate estimations, demonstrating the power of the incremental approach over a single, large step.