Chapter 9: Problem 4
Find the first four nonzero terms of the Taylor series for the functions. $$(1-2 x)^{1 / 2}$$
Short Answer
Expert verified
The Taylor series is \(1 - x + \frac{x^2}{2} - \frac{x^3}{4}\).
Step by step solution
01
Identify the Function Format
The given function is \((1-2x)^{1/2}\), which resembles the format \((1 + u)^n\). In this case, \(u = -2x\) and \(n = \frac{1}{2}\).
02
Use Binomial Series Expansion
The Taylor series expansion for \((1 + u)^n\) centered at 0 is given by:\[(1 + u)^n = 1 + \frac{n}{1!}u + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\]
03
Substitute and Calculate Each Term
Substitute \(n = \frac{1}{2}\) and \(u = -2x\) into the binomial series formula:1. The first term is simply 1.2. The second term is \(\frac{1}{2}(-2x) \Rightarrow -x\).3. The third term is \(\frac{1}{2}\cdot \frac{-1}{2} \cdot \frac{(-2x)^2}{2!} = \frac{1}{2}\cdot -\frac{1}{2} \cdot\frac{4x^2}{2} = \frac{x^2}{2}\).4. The fourth term is \(\frac{1}{2}\cdot -\frac{1}{2}\cdot -\frac{3}{2} \cdot \frac{(-2x)^3}{3!} = \frac{1}{2} \cdot \frac{1}{4} \cdot (-8x^3) = -\frac{x^3}{4}\).
04
Verify and Write Out the Taylor Series
Combine all the terms to write out the Taylor series:\[(1-2x)^{1/2} \approx 1 - x + \frac{x^2}{2} - \frac{x^3}{4}\]These are the first four non-zero terms of the Taylor series expansion.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Series Expansion
When we look at a function such as \((1-2x)^{1/2}\), one way to find its Taylor series is through the concept of the binomial series expansion. This expansion helps us express functions of the form \((1 + u)^n\) as an infinite series. For the binomial series, we use the formula:
For the given function \((1-2x)^{1/2}\), we set \(u = -2x\) and \(n = \frac{1}{2}\). By applying these to the binomial series formula, we can determine and evaluate each term separately, simplifying calculations by handling one part at a time. This method is particularly useful in calculus for approximating functions and predicting their behavior locally around a point, typically 0 in basic applications.
- \((1 + u)^n = 1 + \frac{n}{1!}u + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)
For the given function \((1-2x)^{1/2}\), we set \(u = -2x\) and \(n = \frac{1}{2}\). By applying these to the binomial series formula, we can determine and evaluate each term separately, simplifying calculations by handling one part at a time. This method is particularly useful in calculus for approximating functions and predicting their behavior locally around a point, typically 0 in basic applications.
Calculus
Calculus is a branch of mathematics focused on change and motion; it centers on concepts such as derivatives, integrals, limits, and series. The Taylor series is a quintessential application of calculus principles, as it provides detailed approximations of complex functions.
The Taylor series, named after mathematician Brook Taylor, uses derivatives to represent a function as an infinite sum of terms. The terms are calculated using the initial function and its derivatives at a single point. This representation is vital: it allows difficult functions to be expressed as an easily computable sum of simpler powers.
The Taylor series, named after mathematician Brook Taylor, uses derivatives to represent a function as an infinite sum of terms. The terms are calculated using the initial function and its derivatives at a single point. This representation is vital: it allows difficult functions to be expressed as an easily computable sum of simpler powers.
- The Taylor series at a point \(a\) for a function \(f(x)\) is \(f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots\)
Power Series
A power series is an infinite series of the form \(\sum_{n=0}^{\infty} a_n (x - c)^n\), where each term is a power of \((x - c)\). This concept is widely important in calculus and mathematical analysis. It allows us to express many mathematical functions that aren't polynomials as sums of infinitely many polynomial terms.
Using a power series can simplify complex calculations, letting us inspect and manipulate functions in a "tame" form of polynomial terms. In practice, to create a power series for a function, we find the coefficients \(a_n\) through derivatives.
Using a power series can simplify complex calculations, letting us inspect and manipulate functions in a "tame" form of polynomial terms. In practice, to create a power series for a function, we find the coefficients \(a_n\) through derivatives.
- Each term in the series corresponds to a derivative of the function at the point \(c\), divided by its factorial, times \((x-c)^n\).