Question

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

Short answer

The Taylor polynomial is \( P_4(x) = \ln(2) + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{24}(x-1)^3 - \frac{1}{64}(x-1)^4 \).

Step-by-step solution

Understand the Function and Requirements

We have the function \(f(x) = \ln(x+1)\), and we need to find the Taylor polynomial of degree 4 centered at \(x = 1\). To do this, we must calculate the first four derivatives of the function.

Evaluate the Function and Derivatives at c=1

Calculate \(f(1) = \ln(2)\). Now find the derivatives: 1. \(f'(x) = \frac{1}{x+1}\), so \(f'(1) = \frac{1}{2}\).2. \(f''(x) = -\frac{1}{(x+1)^2}\), so \(f''(1) = -\frac{1}{4}\).3. \(f'''(x) = \frac{2}{(x+1)^3}\), so \(f'''(1) = \frac{2}{8} = \frac{1}{4}\).4. \(f^{(4)}(x) = -\frac{6}{(x+1)^4}\), so \(f^{(4)}(1) = -\frac{6}{16} = -\frac{3}{8}\).

Construct the Taylor Polynomial of Degree 4

The general formula for the Taylor polynomial is: \[ P_n(x) = f(c) + f'(c)\frac{(x-c)^1}{1!} + f''(c)\frac{(x-c)^2}{2!} + f'''(c)\frac{(x-c)^3}{3!} + f^{(4)}(c)\frac{(x-c)^4}{4!} \]Substitute the calculated values:\[P_4(x) = \ln(2) + \frac{1}{2}(x-1) - \frac{1}{4}\frac{(x-1)^2}{2} + \frac{1}{4}\frac{(x-1)^3}{6} - \frac{3}{8}\frac{(x-1)^4}{24}\]

Simplify the Polynomial Expression

Simplify each term:- \(P_4(x) = \ln(2) + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{24}(x-1)^3 - \frac{1}{64}(x-1)^4\)The Taylor polynomial of degree 4 for \(f(x) = \ln(x+1)\) at \(x = 1\) is:\[ P_4(x) = \ln(2) + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{24}(x-1)^3 - \frac{1}{64}(x-1)^4 \]

Key concepts

Derivative Calculations

Derivatives are essential in creating Taylor polynomials because they provide information on how the function behaves around a point. When calculating derivatives, you're essentially finding the rate at which a function changes. This is crucial for determining the coefficients in the polynomial.
For the function \( f(x) = \ln(x+1) \), we first calculate its derivatives:

• The first derivative is \( f'(x) = \frac{1}{x+1} \). It represents the slope or the rate of change of the function. For \( x = 1 \), this translates to \( f'(1) = \frac{1}{2} \), suggesting that at this point, the function rises half a unit per one-unit change in \( x \).
• The second derivative, \( f''(x) = -\frac{1}{(x+1)^2} \), gives us information about the concavity of the curve. A negative value at \( x = 1 \) means the curve is concave down there. The calculation for this is \( f''(1) = -\frac{1}{4} \).
• For higher derivatives, like the third (\( f'''(x) = \frac{2}{(x+1)^3} \)) and the fourth (\( f^{(4)}(x) = -\frac{6}{(x+1)^4} \)), they provide information on how the curvature and shape of the curve changes beyond immediate vicinity of the point \( x = 1 \).

These derivatives are used to build up the polynomial step by step, each one adding more detail based on its behavior.

Polynomial Approximation

Polynomial approximation involves using polynomials to approximate more complex functions. The idea is to get a polynomial that acts like the function near a point.
With Taylor polynomials, this involves taking the function value and its derivatives, and forming a polynomial that matches these values at a particular point \( c \). This makes the polynomial a good approximation of the function nearby.

• The Taylor polynomial we develop is based at \( x = c \), which in this case is \( x = 1 \). This is why it's called a 'centered' approximation.
• The degree of the polynomial \( n \) is how many terms of the polynomial we include. The higher the degree, the more accurately it could represent the function.

The specific formula looks like: \[ P_n(x) = f(c) + f'(c)\frac{(x-c)^1}{1!} + f''(c)\frac{(x-c)^2}{2!} + \ldots + f^{(n)}(c)\frac{(x-c)^n}{n!} \] The Taylor polynomial becomes an approximately equivalent representation of the function near \( x = c \), where each derivative adjusts for accuracy.

Natural Logarithm

The natural logarithm, represented as \( \ln(x) \), is considered special because it uses the base \( e \), Euler's number. It's ubiquitous in science and engineering because it simplifies differentiation and integration.
When you consider \( f(x) = \ln(x+1) \), every term of the Taylor polynomial provides an insight or summary of the natural logarithm's behavior near any given point.

• The natural logarithm is continuous and differentiable everywhere in its domain \( x > -1 \). For \( \ln(x+1) \), \( x = 0 \) becomes \( \ln(1)\) which equals zero, setting a convenient starting point for approximating.
• This logarithm function increases but at a decreasing rate, pointing towards its concavity, which is reflected by the second derivative \( f''(x) \).

Leveraging this specific function in our Taylor polynomial showcases the power of these techniques to dissect and understand subtle variations of more complex mathematical behaviors.