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Chapter 10: Problem 9

Find the Taylor polynomials of degree \(n\) approximating the functions for \(x\) near \(0 .\) (Assume \(p\) is a constant. \()\) $$\frac{1}{\sqrt{1+x}}, \quad n=2,3,4$$

Short Answer

Expert verified
Taylor polynomials: Degree 2: \(1 - \frac{1}{2}x + \frac{3}{4}x^2\), Degree 3: \(1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3\), Degree 4: \(1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4\).

Step by step solution

01

Recall Taylor Series Formula

The Taylor series expansion of a function \(f(x)\) around \(x = a\) is given by: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots\]For this exercise, \(a=0\) and we need the expansions up to \(n=4\).
02

Derive the Function and Its Derivatives

The function given is \(f(x) = \frac{1}{\sqrt{1+x}}\). First, calculate the derivatives at \(x = 0\):- \(f(x) = (1+x)^{-1/2}\).- \(f'(x) = -\frac{1}{2}(1+x)^{-3/2}\).- \(f''(x) = \frac{3}{4}(1+x)^{-5/2}\).- \(f^{(3)}(x) = -\frac{15}{8}(1+x)^{-7/2}\).- \(f^{(4)}(x) = \frac{105}{16}(1+x)^{-9/2}\).Evaluate these derivatives at \(x=0\).
03

Simplify the Derivatives at x = 0

Evaluate the derivatives calculated in the previous step at \(x=0\):- \(f(0) = 1\).- \(f'(0) = -\frac{1}{2}\).- \(f''(0) = \frac{3}{4}\).- \(f^{(3)}(0) = -\frac{15}{8}\).- \(f^{(4)}(0) = \frac{105}{16}\).
04

Construct Taylor Polynomials

Using the derivatives at \(x=0\), construct the Taylor polynomials for different degree values:- Degree 2: \[T_2(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2\]- Degree 3: \[T_3(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3\]- Degree 4: \[T_4(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4\]
05

Present the Final Polynomials

The final Taylor polynomials approximating \(\frac{1}{\sqrt{1+x}}\) near \(x=0\) are:- Degree 2: \(T_2(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2\).- Degree 3: \(T_3(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3\).- Degree 4: \(T_4(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4\).

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Key Concepts

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

Taylor Series
The Taylor series is a powerful tool in calculus for approximating functions. It allows a function to be expressed as an infinite sum of terms. Each term is calculated from the derivatives of the function at a given point. The formula for a Taylor series expansion of a function \(f(x)\) around the value \(x = a\) is:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots\]The series provides an approximation that becomes closer to the real function as more terms are added.
  • The point \(a\) is where the series is centered.
  • Each term involves higher order derivatives of the function.
  • In practical use, only a finite number of terms are computed for a desired degree of accuracy.
For the present exercise, the series is centered at \(x = 0\) to a degree of 4. Taylor series make it easier to handle complex functions by using polynomial expressions.
Derivatives
A derivative represents the rate at which a function changes at a particular point. It's a central concept in calculus and fundamental to the formation of Taylor series. Calculating derivatives is necessary to determine each polynomial term in a Taylor series.
  • The first derivative \(f'(x)\) indicates the slope at any point on the function.
  • The second derivative \(f''(x)\) shows the rate of change of the slope.
  • Higher-order derivatives, such as third \(f^{(3)}(x)\) and fourth \(f^{(4)}(x)\), provide further information about changes in the function's behavior.
In our exercise:
  • We determined the derivatives of \(f(x) = \frac{1}{\sqrt{1+x}}\).
  • Calculated these derivatives at \(x = 0\) to construct the Taylor polynomial.
  • As each derivative is evaluated, it contributes to a specific term in the polynomial.
Understanding derivatives is crucial for constructing Taylor polynomials and interpreting how a function behaves around a certain point.
Polynomial Approximation
Polynomial approximation is an approach used to estimate more complex mathematical functions using simpler polynomials. Taylor polynomials are a type of polynomial approximation derived from the Taylor series. They offer a useful way to get close to the actual values of a function, especially near the point where the series is centered.
  • Short polynomial expressions replace a potentially infinite series.
  • Each additional term in the Taylor polynomial provides a more accurate approximation.
  • For computations or predictions, several terms usually provide enough precision.
In this exercise, the function \(\frac{1}{\sqrt{1+x}}\) is approximated using Taylor polynomials of degrees 2, 3, and 4:
  • These polynomials approximate the function near \(x = 0\).
  • As more terms are added (higher degrees), accuracy improves.
  • Taylor polynomials are especially useful when dealing with functions near their center point, \(x = a\).
By recognized practices, these polynomials help simplify complex function calculations.
Mathematical Functions
A mathematical function maps a set of inputs (domain) with corresponding outputs (range). In calculus, functions are the basis for determining behavior over intervals or specific points. A function like \(f(x) = \frac{1}{\sqrt{1+x}}\) reveals how \(f\) changes as \(x\) changes.
  • Functions can exhibit complex behavior that requires approximation for analysis.
  • Polynomial and Taylor series approximations provide simpler ways to study functions.
  • They help calculate values over an input domain and understand local behavior around specific values.
For our problem, we're focused on approximating the mathematical behavior near \(x = 0\).
  • This involves understanding how the original function behaves in this region.
  • The calculated Taylor polynomials give insight into predicted values close to the center.
  • This foundational tool in calculus supports deeper exploration of changes in rate and extent in mathematical functions.
Understanding mathematical functions is essential to leverage Taylor series and polynomial approximations effectively.

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Most popular questions from this chapter

(a) Find the first three nonzero terms of the Taylor series for \(e^{x}+e^{-x}\) (b) Explain why the graph of \(e^{x}+e^{-x}\) looks like a parabola near \(x=0 .\) What is the equation of this parabola?

using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\sqrt{1+\sin \theta}$$

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 \(C\)

Use Taylor series to explain the patterns in the digits in the following expansions: (a) \(\frac{1}{0.98}=1.02040816 \ldots\) (b) \(\left(\frac{1}{0.99}\right)^{2}=1.020304050607 \ldots\)

The pulse train of width \(c\) is the periodic function \(f\) of period \(2 \pi\) given by $$f(x)=\left\\{\begin{array}{ll}0 & -\pi \leq x<-c / 2 \\\1 & -c / 2 \leq x
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