/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Thinking of exercises \(9 a-9 c\... [FREE SOLUTION] | 91Ó°ÊÓ

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Thinking of exercises \(9 a-9 c\) as numbers of the form \(\sqrt[4]{16+\Delta x},\) denote the errors as \(e(\Delta x)\) (where \(\Delta x=0.04\) \(\Delta x=0.08\) and \(\Delta x=0.16\) ). Based on these three computations, determine a constant \(c\) such that \(e(\Delta x) \approx c(\Delta x)^{2}\)

Short Answer

Expert verified
The constant \( c \) can be determined by calculating errors for given \( \Delta x \), formulating a formula for error, solving system of equations and then verifying the derived constant.

Step by step solution

01

Calculate Errors

First, compute \( e(\Delta x) \) for each given \( \Delta x \). It is calculated as \( e(\Delta x) = |\sqrt[4]{16+\Delta x} - 2| \) for \( \Delta x = 0.04, 0.08, 0.16 \).
02

Formulate Error Equation

Create a temporary formula \( e(\Delta x) = c(\Delta x)^{2} \).
03

Determine Constant \( c \)

Now with the help of calculated errors from step 1 and the error equation from step 2, a system of equations can be created. Solve the equations to find value of \( c \).
04

Verify Constant \( c \)

Check if the determined \( c \) satisfies the conditions \( e(\Delta x) \approx c(\Delta x)^{2} \) for all values of \( \Delta x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Approximation
Numerical approximation is a technique used to estimate the value of a function when it cannot be computed exactly. This is especially important in calculus where certain values might be difficult or time-consuming to determine through direct calculation. For the exercise, we're approximating a fourth root, which could otherwise be an intensive calculation.

When dealing with small changes in a variable, denoted as \( \Delta x \), numerical approximations allow us to predict changes in a function's value. This is useful in everyday applications like physics, engineering, and finance, where exact values aren't necessary but estimates need to be fast and fairly accurate.

To perform a numerical approximation, you'll often use methods like linearization or polynomial approximations. In our exercise, we consider deviations from the exact fourth root of 16, using small values of \( \Delta x \). This exercise is a great opportunity to see how approximate methods can closely predict function values, opening the door to solving more complex real-world problems.
Taylor Series
A Taylor series is a mathematical tool used to approximate functions with an infinite sum of terms calculated from the function's derivatives at a single point. This series is remarkably powerful because it provides a way to approximate previously complex functions using polynomials, which are easier to work with both analytically and numerically.

In the context of the exercise, even though we are not explicitly mentioned, Taylor series could be leveraged for a more detailed error approximation. For example, if you were to expand \( \sqrt[4]{16+\Delta x} \) using Taylor series centered around \( \Delta x = 0 \), it would provide an elaborate polynomial approximation. The leading terms of this expansion could help better understand the error characteristics for small \( \Delta x \).
  • The Taylor series expansion allows one to include higher order terms for better accuracy.
  • The degree of the Taylor polynomial decides the closeness of approximation to the actual function. More terms, better accuracy.
When calculating errors, as we see in our exercise with \( e(\Delta x) \), understanding how the approximation behaves through Taylor series concepts can guide us in formulating effective error predictions.
Calculus Error Estimation
In calculus, error estimation refers to the process of determining the potential error magnitude introduced when using an approximation method. It's an essential part of numerical calculus, as it helps provide bounds on how far off an approximation might be from the true value.

Error estimation can be done in various ways, but for our type of problem, it often involves comparing the exact value with the approximated value and determining the difference. In the exercise provided, errors were calculated as \( e(\Delta x) = |\sqrt[4]{16+\Delta x} - 2| \) for different \( \Delta x \) values, showing how close the approximation gets as \( \Delta x \) changes.
  • Recognizing errors is crucial for improving approximation methods or switching to more accurate ones for desired preciseness.
  • Often, results are expressed relative to \( \Delta x^n \), highlighting how error changes with different step sizes.
Understanding these errors helps analysts and engineers determine if an approximation is sufficient for their needs or if adjustments in methodology are required. This adds strategic value to the calculations allowing better resource management in situations where precision is paramount.

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Most popular questions from this chapter

A water wave of length \(L\) meters in water of depth \(d\) meters has velocity \(v\) satisfying the equation $$v^{2}=\frac{4.9 L}{\pi} \frac{e^{2 \pi d / L}-e^{-2 \pi d / L}}{e^{2 \pi d / L}+e^{-2 \pi d / L}}$$ Treating \(L\) as a constant and thinking of \(v^{2}\) as a function \(f(d)\) use a linear approximation to show that \(f(d) \approx 9.8 d\) for small values of \(d .\) That is, for small depths, the velocity of the wave is approximately \(\sqrt{9.8 d}\) and is independent of the wavelength \(L\)

Prove that any fourth-order polynomial must have at least one local extremum and can have a maximum of three local extrema. Based on this information, sketch several possible graphs of fourth-order polynomials.

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Boyle's law for a gas at constant temperature is \(P V=c\) where \(P\) is pressure, \(V\) is volume and \(c\) is a constant. Assume that both \(P\) and \(V\) are functions of time. Show that \(P^{\prime}(t) / V^{\prime}(t)=-c / V^{2}\)

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