Problem 10 - Thomas Calculus

Question

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=\sqrt{1-x}, \quad a=0\)

Step-by-Step Solution

Answer

Taylor polynomials are 1, \(1-\frac{1}{2}x\), \(1-\frac{1}{2}x-\frac{1}{8}x^2\), \(1-\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3\).

Step 1: Understanding Taylor Polynomials

A Taylor polynomial of order n is an approximation of a function around a point a given by: \[ P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n \]The given function is \(f(x) = \sqrt{1-x}\) and needs to be expanded around \(a = 0\).

Step 2: Compute the Derivatives of the Function

The first few derivatives of \(f(x) = \sqrt{1-x}\) are required:- \( f(x) = (1-x)^{1/2} \)- First derivative: \( f'(x) = -\frac{1}{2}(1-x)^{-1/2} \)- Second derivative: \( f''(x) = -\frac{1}{4}(1-x)^{-3/2} \)- Third derivative: \( f'''(x) = \frac{3}{8}(1-x)^{-5/2} \)

Step 3: Evaluate the Derivatives at the Point a

Evaluate each derivative at \(a = 0\):- \(f(0) = \sqrt{1-0} = 1\)- \(f'(0) = -\frac{1}{2}(1-0)^{-1/2} = -\frac{1}{2}\)- \(f''(0) = -\frac{1}{4}(1-0)^{-3/2} = -\frac{1}{4}\)- \(f'''(0) = \frac{3}{8}(1-0)^{-5/2} = \frac{3}{8}\)

Step 4: Construct the Taylor Polynomials

Now use the derivatives evaluated at \(a=0\) to construct the Taylor polynomials:- Order 0: \(P_0(x) = 1\)- Order 1: \(P_1(x) = 1 - \frac{1}{2}x\)- Order 2: \(P_2(x) = 1 - \frac{1}{2}x - \frac{1}{4} \frac{x^2}{2} = 1 - \frac{1}{2}x - \frac{1}{8}x^2\)- Order 3: \(P_3(x) = 1 - \frac{1}{2}x - \frac{1}{8}x^2 + \frac{3/8}{6}x^3 = 1 - \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3\)

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that measure the rate at which a function changes at any given point. In the context of Taylor polynomials, they play a critical role in determining the coefficients of the polynomial terms. To understand this, imagine derivatives as the "slope" of a function at a particular point.- The first derivative, denoted as \( f'(x) \), gives the slope or rate of change of the original function \( f(x) \).- The second derivative, \( f''(x) \), provides information about the concavity or curvature of the function.- Higher-order derivatives, such as the third derivative \( f'''(x) \), continue to describe the function's behavior in more detail.For the function \( f(x) = \sqrt{1-x} \), evaluating these derivatives at the point \( a = 0 \) allows us to construct the Taylor polynomial. This involves calculating:\[\begin{align*} f(0) = & \ 1, \ f'(0) = & \ -\frac{1}{2}, \ f''(0) = & \ -\frac{1}{4}, \ f'''(0) = & \ \frac{3}{8}. \end{align*}\]These values help us approximate the function through polynomial expressions.

Maclaurin Series

The Maclaurin series is a special form of the Taylor series. It's an infinite sum used to approximate functions around the specific point \( a = 0 \). Think of the Maclaurin series as a powerful tool in calculus that helps break down complex functions into simpler polynomial forms.The general formula for the Maclaurin series is:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]When dealing with the Maclaurin series, understanding how functions can be expressed entirely through derivatives evaluated at 0 is crucial. For example, in calculating the Maclaurin series of \( f(x) = \sqrt{1-x} \), we use derivatives up to the desired order (for this exercise up to the third derivative) and substitute back into the series format at \( a = 0 \). This results in a polynomial approximation that is accurate near this point, allowing for easier calculations and a better understanding of the function's behavior near zero.

Polynomial Approximation

Polynomial approximation is a technique in calculus used to estimate complex functions with simpler polynomial expressions. This approach simplifies calculations and provides insight into the function's behavior close to a certain point.In the context of Taylor polynomials, the approximation is constructed using derivatives to find each term's coefficient. The nth-order Taylor polynomial is expressed as:\[ P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n \]This polynomial captures the essential features of the function \( f(x) \) over an interval near \( a \). For our specific case: - The zero-order Taylor polynomial \( P_0(x) \) is simply a constant function equal to \( f(a) \).- As we include higher order terms, like \( P_1(x), P_2(x), \) and \( P_3(x) \), the approximation becomes more accurate and starts capturing curvature.These polynomial approximations are invaluable in applied mathematics, as they allow complex functions to be broken down into potentially infinite summations, yielding usable results for analysis and computation. For example, using the third-order polynomial approximation could be helpful in many real-world applications where a function's precision near a point is essential, such as engineering or physics.