Question

Show that the approximate relative error \((d f) / f\) of a product \(f=g h\) is the sum of the approximate relative errors of the factors.

Short answer

The approximate relative error of the product function is the sum of the approximate relative errors of its factors: \( \frac{df}{f} = \frac{dg}{g} + \frac{dh}{h} \).

Step-by-step solution

Understand the Problem

The goal is to show that for a product function defined as \( f = gh \), the approximate relative error \( \frac{df}{f} \) is the sum of the approximate relative errors of \( g \) and \( h \).

Define the Product Function

Let \( f = gh \), where \( g \) and \( h \) are functions of the same variable.

Find the Differential of f

Using the product rule for differentiation, find the differential of \( f \). \[ df = g \frac{dh}{dh} + h \frac{dg}{dg} \rightarrow df = g \frac{dh}{dg} + h \frac{dg}{dh}\]

Express the Relative Error

Rearrange the differential \( df \) in terms of the relative errors of \( g \) and \( h \): \[ \frac{df}{f} = \frac{g \frac{dh}{dg} + h \frac{dg}{dh}}{gh} \]

Simplify the Expression

Separate the terms in the numerator to simplify the expression: \[ \frac{df}{f} = \frac{g \frac{dh}{dg}}{gh} + \frac{h \frac{dg}{dh}}{gh} \rightarrow \frac{df}{f} = \frac{\frac{dh}{dg}}{h} + \frac{\frac{dg}{dh}}{g} \]

Combine Terms

Combine the terms to show the sum of the relative errors: \[ \frac{df}{f} = \frac{dh}{h} + \frac{dg}{g} \]

Key concepts

product differentiation

Product differentiation in calculus involves finding the rate at which a product of two or more functions changes. When you have two functions, say \( g \) and \( h \), and you multiply them to get a new function \( f \), written as \( f = gh \), we use the product rule to differentiate \( f \).

Here’s what the product rule says:

• \( df = g \frac{dh}{dh} + h \frac{dg}{dg} \)

This rule helps us find the differential of \( f \), which is how much \( f \) changes when both \( g \) and \( h \) change slightly.

Remember, the differential \( df \) can be thought of as an 'instantaneous' change in \( f \), given tiny changes in \( g \) and \( h \).

Now, why is this important? Because we often use differentials to approximate changes in complex expressions!

relative error

Relative error measures how much the value of a quantity deviates from its expected value, relative to the value itself. It's a handy measure to understand the accuracy of a measurement.

Specifically, the relative error of a product \( f = gh \), can be shown as the sum of the relative errors of \( g \) and \( h \). Here is how you derive it:

Take the differential \( df \) we found using the product rule. Divide it by \( f \), which is \( gh \), to express relative error:

• \(\frac{df}{f} = \frac{g \frac{dh}{dg} + h \frac{dg}{dh}}{gh} \)

Then, simplify the fractions inside the expression:

• \(\frac{df}{f} = \frac{\frac{dh}{dg}}{h} + \frac{\frac{dg}{dh}}{g} \)

This step-by-step breakdown shows that the relative error \( \frac{df}{f} \) is equal to the sum of the relative errors of \( g \) and \( h \):

• \(\frac{df}{f} = \frac{dg}{g} + \frac{dh}{h} \)

This means if you know the relative errors of the individual factors, you can easily find the relative error of their product.

calculus

Calculus is a branch of mathematics focused on studying rates of change (derivatives) and accumulation of quantities (integrals). It is a powerful tool in mathematics with wide applications in science and engineering.

In our context, we're using calculus to handle products of functions and to study their behavior as they change. The differential approach we took comes from a fundamental concept in calculus.

There are two main types of calculus:

• Differential Calculus: It deals with the concept of a derivative, which tells us how a quantity changes as its input changes.
• Integral Calculus: It focuses on the concept of an integral, which aggregates quantities across continuous domains (like finding areas under curves).

When dealing with errors and approximations, differential calculus is particularly useful. The derivative or differential gives an instantaneous rate of change, and by understanding differentials, we can get insightful approximations.

This is why understanding calculus, particularly differential calculus, is essential in fields that require precise measurement and error analysis, such as physics, engineering, and economics.