/*! 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 24 Show that $$ x^{4} \approx a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that $$ x^{4} \approx a^{4}+4 a^{3}(x-a) $$ for \(x\) close to \(a\).

Short Answer

Expert verified
The approximation is derived using the first two terms in the Taylor expansion for \(x^4\) around \(a\).

Step by step solution

01

Use Taylor Expansion

Start by using the Taylor series expansion for a function around a point. For the function \( f(x) = x^4 \) around \( a \), the Taylor series is: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f''''(a)}{4!}(x-a)^4 + \cdots \]
02

Calculate Derivatives

Calculate the derivatives needed for the Taylor series. The first few derivatives are: - \( f(x) = x^4 \), so \( f(a) = a^4 \).- \( f'(x) = 4x^3 \), so \( f'(a) = 4a^3 \).- \( f''(x) = 12x^2 \), so \( f''(a) = 12a^2 \).- \( f'''(x) = 24x \), so \( f'''(a) = 24a \).- \( f''''(x) = 24 \), so \( f''''(a) = 24 \).
03

Write First-Two Terms of Expansion

Substitute the derivatives into the Taylor series formula, focusing on the first two terms: \[ x^4 = a^4 + 4a^3(x-a) + \frac{12a^2}{2}(x-a)^2 + \frac{24a}{6}(x-a)^3 + \frac{24}{24}(x-a)^4 + \cdots \]
04

Simplify Higher Order Terms

Since we are showing the approximation and terms are higher powers of \(x-a\), we can neglect these higher order terms for \(x\) close to \(a\):\[ x^4 \approx a^4 + 4a^3(x-a) \]
05

Summary of Approximation

Finally, the approximation for \(x^4\) when \(x\) is close to \(a\) is: \[ x^4 \approx a^4 + 4a^3(x-a) \] This shows the local linear approximation of \(x^4\) around \(a\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Local Linear Approximation
Local linear approximation essentially simplifies a function near a certain point by using only the initial terms in its Taylor series expansion. In a sense, it "linearizes" the function around the point of interest, which is very useful when dealing with small changes.

Let's consider you have a smooth function like a curve. When you zoom in around a specific point on this curve, it starts to look more and more like a straight line. This is the essence of local linear approximation.
  • The point we're focusing on is usually called the "base point," and in this example, it's the point \( a \).
  • We make use of the function's value and its derivative at this point to create a line that closely follows the curve at \( a \).
The approximation becomes: \( x^4 \approx a^4 + 4a^3(x-a) \), where the first term \( a^4 \) is the function value at \( a \), and the second term uses the derivative to show how the function changes as \( x \) moves slightly away from \( a \). This linear approach minimizes complexity, which is especially beneficial in practical calculations.
Derivatives Calculation
Calculating derivatives is a key aspect of building Taylor series and, subsequently, local linear approximations. Derivatives describe how a function changes as its input changes.

For the function \( f(x) = x^4 \), we evaluate it at \( x = a \):
  • The 0th derivative (the function itself) gives \( f(a) = a^4 \).
  • The 1st derivative \( f'(x) = 4x^3 \) gives us \( f'(a) = 4a^3 \), showing how sharply the function tilts at point \( a \).
  • The 2nd derivative, \( f''(x) = 12x^2 \), tells us about the curvature, resulting in \( f''(a) = 12a^2 \).
  • We go further to find \( f'''(x) = 24x \) and so on, computing up to the 4th derivative \( f''''(x) = 24 \).
Once derivatives are evaluated at \( a \), they enable us to form the terms in the Taylor series. Each order of the derivative contributes a higher-degree term based on powers of \( (x-a) \). This fits perfectly for creating an approximation that closely mirrors the true function around \( a \).
Higher Order Terms Neglect
When performing Taylor series expansions, especially for approximations like in this exercise, higher order terms are often neglected. This means we only focus on a few of the first terms.

Higher order terms are the elements in the Taylor series that involve powers of \( (x - a) \) beyond what's needed for the approximation.
  • The precisions of these terms diminish when \( x \) is very close to \( a \).
  • Since these powers grow higher as we expand beyond quadratic terms, they contribute less to the value of the function, particularly for small disruptions around \( a \).
By truncating the series after the linear term, \( x^4 \approx a^4 + 4a^3(x-a) \), the rest of the terms (\( (x-a)^2 \) and above) are considered negligible, allowing us a simpler, more practical representation of \( x^4 \) around \( a \). This balance of accuracy and simplicity is the power of neglecting higher order terms.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.