Question

Find the Taylor polynomial of degree \(n\), at \(x=c\), for the given function. $$f(x)=\sqrt{x}, \quad n=4, \quad c=1$$

Short answer

The Taylor polynomial is \(P_4(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 - \frac{5}{128}(x-1)^4\).

Step-by-step solution

Determine the Derivatives

To find the Taylor polynomial, we need the derivatives of the function up to the 4th order evaluated at the given center, \(c=1\). If \(f(x) = \sqrt{x}\), then its derivatives are calculated as follows:- \(f(x) = x^{1/2}\), so \(f'(x) = \frac{1}{2}x^{-1/2}\)- \(f''(x) = -\frac{1}{4}x^{-3/2}\)- \(f'''(x) = \frac{3}{8}x^{-5/2}\)- \(f''''(x) = -\frac{15}{16}x^{-7/2}\)

Evaluate Derivatives at x = c

Now evaluate these derivatives at \(x = 1\):- \(f(1) = \sqrt{1} = 1\)- \(f'(1) = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}\)- \(f''(1) = -\frac{1}{4}(1)^{-3/2} = -\frac{1}{4}\)- \(f'''(1) = \frac{3}{8}(1)^{-5/2} = \frac{3}{8}\)- \(f''''(1) = -\frac{15}{16}(1)^{-7/2} = -\frac{15}{16}\)

Construct the Taylor Polynomial

Using the formula for the Taylor polynomial, \[ P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \frac{f''''(c)}{4!}(x-c)^4 \]Substitute \(c = 1\) and our evaluated derivatives to find:- When \(n = 4\),\[ P_4(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{4}\cdot\frac{1}{2}(x-1)^2 + \frac{3}{8}\cdot\frac{1}{6}(x-1)^3 - \frac{15}{16}\cdot\frac{1}{24}(x-1)^4 \]

Simplify the Expression

Simplify each term of the polynomial:\[ P_4(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 - \frac{5}{128}(x-1)^4 \]This is the Taylor polynomial of degree 4 for \(f(x) = \sqrt{x}\) centered at \(c = 1\).

Key concepts

Taylor series

A Taylor series is a powerful tool in mathematics that allows you to approximate complex functions using polynomials. This series is expressed as an infinite sum of terms calculated from the derivatives of a function at a single point. The formula for a Taylor series centered at a point \(c\) is:

• \( f(c) \)
• \( f'(c)(x-c) \)
• \( \frac{f''(c)}{2!}(x-c)^2 \)
• \( \frac{f'''(c)}{3!}(x-c)^3 \)
• \( \cdots \)

The Taylor series breaks down a function into these smaller pieces, which are more manageable and easier to analyze. By summing these up to a finite number of terms, we obtain a Taylor polynomial, which can be a convenient approximation of the original function. This is especially useful when calculating values of functions that are difficult to compute otherwise.
The main aim of using a Taylor series is to approximate functions locally around a point \(c\), leading to great versatility in analysis and engineering applications.

derivatives

Derivatives are the backbone of the Taylor series, as they describe how a function changes at any given point. In the context of our problem, where we have the function \( f(x) = \sqrt{x} \), finding successive derivatives helps us build each term of the Taylor polynomial.

• The first derivative, \( f'(x) \), represents the slope of the tangent to the curve at point \( x \).
• The second derivative, \( f''(x) \), is related to the curvature or concavity of the function.
• Higher-order derivatives, like \( f'''(x) \) and \( f''''(x) \), provide further granularity into the function's behavior.

When calculating these derivatives, each successive derivative is typically more complex. For the function \( f(x) = x^{1/2} \), the derivatives become:

• \( f'(x) = \frac{1}{2}x^{-1/2} \)
• \( f''(x) = -\frac{1}{4}x^{-3/2} \)
• \( f'''(x) = \frac{3}{8}x^{-5/2} \)
• \( f''''(x) = -\frac{15}{16}x^{-7/2} \)

By evaluating these derivatives at the specified center of expansion, we can construct the Taylor polynomial needed for approximation.

degree of polynomial

The degree of a Taylor polynomial is crucial as it indicates how many terms of the Taylor series are included in the polynomial approximation. In our problem, we aim to find a Taylor polynomial of degree \( n = 4 \) for the function \( f(x) = \sqrt{x} \).
The degree represents:

• The number of derivatives used in constructing the polynomial.
• The complexity or accuracy of the approximation—the higher the degree, the more accurate the polynomial is around the center \( c \).

For this specific example, using a polynomial of degree 4 means we incorporate up to the 4th derivative of the function. This ensures the polynomial captures more information about the function's behavior near the center, \( c = 1 \).
Mathematically, the polynomial matches the function's values and derivatives up to that degree, making higher-degree polynomials better suited for functions with rapid changes.

center of expansion

The center of expansion in Taylor series, also known as the point of expansion, is the specific point \( c \) around which the function is approximated. This point significantly influences the accuracy of the Taylor polynomial.

• Choosing a point close to where you need the approximation ensures better accuracy.
• In the given problem, \( c = 1 \) was chosen as the center for \( f(x) = \sqrt{x} \).

Evaluating the function and its derivatives at this point allows us to build a polynomial that mirrors the function's behavior closely around \( x = 1 \). The closer the values of \( x \) are to \( c \), the more the polynomial approximation reflects the actual function values.
By using \( c = 1 \) for the Taylor series, we capitalize on the function's simplicity at this point, making calculations more straightforward and intuitive. This is particularly helpful when dealing with functions that change rapidly, as the polynomial will perform better near this center.