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Using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\sqrt{1-2 x}$$

Short Answer

Expert verified
\( \sqrt{1-2x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{2} + \cdots \)

Step by step solution

01

Recognize and Simplify the Function

We are given the function \( f(x) = \sqrt{1 - 2x} \). Recognize that this can be expressed in terms of binomial series expansion, similar to \((1-x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\). Here, take \( n = \frac{1}{2} \) and substitute \( x \) with \( -2x \).
02

Compute the Binomial Series Expansion

Using the binomial series formula with \( n = \frac{1}{2} \), compute the terms:\[ (1 - 2x)^{1/2} = 1 + \frac{1}{2}(-2x) + \frac{1}{2}\left(\frac{1}{2} - 1\right)\frac{(-2x)^2}{2} + \frac{1}{2}\left(\frac{1}{2} - 1\right)\left(\frac{1}{2} - 2\right)\frac{(-2x)^3}{6} + \cdots \]
03

Simplify and Calculate Each Term

Let's calculate each term:- First term: \( 1 \)- Second term: \( \frac{1}{2}(-2x) = -x \)- Third term: \( \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \frac{(-2x)^2}{2} = \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \frac{4x^2}{2} = \frac{x^2}{2} \)- Fourth term: \( \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right) \cdot \frac{(-2x)^3}{6} = \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right) \cdot \left(-8x^3\right)/6 = -\frac{x^3}{2} \).
04

Assemble the First Four Nonzero Terms

Combine the terms calculated: \[ \sqrt{1-2x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{2} + \cdots \].Thus, the first four nonzero terms of the Taylor series are: \( 1, -x, \frac{x^2}{2}, -\frac{x^3}{2} \).

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Key Concepts

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

Binomial Series Expansion
Binomial series expansion is a powerful tool used in calculus to break down expressions with exponents into an infinite series. This method is particularly useful for functions that are difficult to handle in their original form, such as radicals or roots.

The binomial series expansion for a general expression \( (1-x)^n \) is given by:
  • \( 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots \)
Here, \( n \) can be any real number, and each term's coefficient is derived from the previous terms by factoring in \( n \) and its preceding integers, then dividing by the factorial of the term's degree. For fractional exponents, this helps simplify expressions like \( \sqrt{1-2x} \) as seen in the exercise.

By comprehending the framework of binomial expansion, we can compute terms individually, which helps in realizing the full Taylor series for complex functions.
Function Simplification
Simplifying functions is a significant step in calculus problem-solving, especially when working with Taylor or binomial series expansions. The main goal is to transform a given function into a more manageable form that closely resembles standard series templates.

For the function \( f(x) = \sqrt{1 - 2x} \), the simplification involves expressing it in the form of \( (1-x)^n \) with \( n = \frac{1}{2} \). Here, recognize that root expressions can resemble binomial expressions by substituting variables appropriately, thus fitting the structure needed for expansion.

This step enables us to clearly identify terms in the series and understand each part's contribution, making it easier to compute subsequent terms. By focusing on simplifying functions, students can build strong analytical skills and demystify complicated problems.
Nonzero Terms
In any series expansion, especially related to Taylor or binomial series, nonzero terms are critical because they represent the meaningful components that actually contribute to the function's behavior.

While expanding \( \sqrt{1-2x} \) through the binomial series, identifying the first few nonzero terms helps approximate the function. In this case, the first four nonzero terms are derived directly from the expansion methodology: \( 1, -x, \frac{x^2}{2}, -\frac{x^3}{2} \).

Each nonzero term reflects a different degree of the polynomial, and early terms usually provide a more precise approximation when close to the center of expansion. Understanding nonzero terms allows us to gauge how each piece affects the function, thus illustrating the Taylor series' role in function approximation.
Calculus Problem Solving
Solving calculus problems often requires a multi-step approach, where understanding and applying fundamental techniques are crucial. Taylor series representation and binomial expansion are key methods used in tackling such problems.

The process begins with recognizing potential expansions for simplification purposes, as demonstrated with \( \sqrt{1-2x} \). This is followed by simplifying the function to match mathematical series formats like \( (1-x)^n \).
  • Computations involve expansion and simplification of terms.
  • Later, these terms are evaluated to yield nonzero contributions to the sequence.
  • Finally, students compile these terms to form a coherent solution.
This problem-solving strategy in calculus nurtures critical thinking and enhances the ability to break down complex problems systematically. Mastery of these techniques is a stepping stone towards tackling higher-level mathematics with confidence.

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Most popular questions from this chapter

By recognizing each series in Problems \(43-51\) as a Taylor series evaluated at a particular value of \(x,\) find the sum of each of the following convergent series. $$1-\frac{100}{2 !}+\frac{10000}{4 !}+\dots+\frac{(-1)^{n} \cdot 10^{2 n}}{(2 n) !}+\cdots$$

Resonance in electric circuits leads to the expression $$\left(\omega L-\frac{1}{\omega C}\right)^{2}$$ where \(\omega\) is the variable and \(L\) and \(C\) are constants. (a) Find \(\omega_{0},\) the value of \(\omega\) making the expression zero. (b) In practice, \(\omega\) fluctuates about \(\omega_{0},\) so we are interested in the behavior of this expression for values of \(\omega\) near \(\omega_{0} .\) Let \(\omega=\omega_{0}+\Delta \omega\) and expand the expression in terms of \(\Delta \omega\) up to the first nonzero term. Give your answer in terms of \(\Delta \omega\) and \(L\) but not \(C\)

(a) From a graph of \(f(x)=\ln (2 x+4)\) decide if the tangent line approximation to \(f\) at \(x=0\) has a positive or negative slope. (b) Find the tangent line approximation. (c) Find the second and third Taylor approximations to \(f\) about \(x=0\)

Decide if the statements in Problems \(65-71\) are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(f\) has the following Taylor series about \(x=0,\) then \(f^{(7)}(0)=-8\) $$f(x)=1-2 x+\frac{3}{2 !} x^{2}-\frac{4}{3 !} x^{3}+\cdots$$ (Assume the pattern of the coefficients continues.)

Decide if the statements in Problems \(65-71\) are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(f\) is an even function, then the Taylor series for \(f\) near \(x=0\) has only terms with even exponents.

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