Problem 3 - Calculus

Question

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

Step-by-Step Solution

Answer

The Taylor polynomials are: \(P_2(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2\), \(P_3(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3\), \(P_4(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4\).

Step 1: Understand the Function

Identify the main function we are working with, which is \(f(x) = \sqrt{1+x}\). We need to find its Taylor polynomials about the point \(x = 0\).

Step 2: Determine the Derivatives

Find the first few derivatives of \(f(x) = \sqrt{1+x}\) up to the fourth derivative:1. \(f(x) = \sqrt{1+x}\)2. \(f'(x) = \frac{1}{2(1+x)^{1/2}}\)3. \(f''(x) = -\frac{1}{4(1+x)^{3/2}}\)4. \(f'''(x) = \frac{3}{8(1+x)^{5/2}}\)5. \(f^{(4)}(x) = -\frac{15}{16(1+x)^{7/2}}\)

Step 3: Evaluate Derivatives at \(x = 0\)

Substitute \(x = 0\) into each derivative to find their values:1. \(f(0) = 1\)2. \(f'(0) = \frac{1}{2}\)3. \(f''(0) = -\frac{1}{4}\)4. \(f'''(0) = \frac{3}{8}\)5. \(f^{(4)}(0) = -\frac{15}{16}\)

Step 4: Formulate the Taylor Polynomial\(n=2\)

For the Taylor polynomial up to degree 2, use the formula:\[ P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 \]Thus, \[ P_2(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 \]

Step 5: Formulate the Taylor Polynomial\(n=3\)

For the Taylor polynomial up to degree 3, include the terms up to the third derivative:\[ P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 \]Thus, \[ P_3(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 \]

Step 6: Formulate the Taylor Polynomial\(n=4\)

For the Taylor polynomial up to degree 4, use the fourth derivative:\[ P_4(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \frac{f^{(4)}(0)}{24}x^4 \]Thus, \[ P_4(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 \]

Key Concepts

Understanding Derivatives

The concept of derivatives is fundamental in calculus and helps us find the rate at which a function is changing at any given point. In the context of Taylor polynomials, derivatives play a crucial role in determining the coefficients of each term in the polynomial.

When we work with the function \(f(x) = \sqrt{1+x}\), we calculate its derivatives to approximate the behavior of the function around a specific point, here being \(x=0\).

• The first derivative, \(f'(x) = \frac{1}{2(1+x)^{1/2}}\), gives us the slope of the tangent to the curve at any point \(x\).
• The second derivative, \(f''(x) = -\frac{1}{4(1+x)^{3/2}}\), tells us about the curvature or concavity of the function. If it's positive, the function is concave up, if negative, it's concave down.
• Further derivatives provide us an even deeper understanding of how the function behaves locally.

Function Approximation through Taylor Polynomials

Function approximation is a powerful technique used to estimate complex functions with simpler polynomials. Taylor polynomials, in particular, allow us to approximate functions near a given point using information obtained from the function's derivatives.

The Taylor polynomial of degree \(n\) for a function \(f(x)\) about \(x = 0\) can be constructed by using:

• The value of the function at the point, i.e., \(f(0)\).
• The derivatives \(f'(0), f''(0), ..., f^{(n)}(0)\) evaluated at that point.

This polynomial provides a local approximation of the function, which often simplifies calculations and visualizations. The higher the degree \(n\), the more accurate the approximation tends to be.

Exploring the Maclaurin Series

A Maclaurin series is a special case of the Taylor series, where the approximation is centered around \(x = 0\). This simplifies the mathematical process and computations. For \(f(x) = \sqrt{1+x}\), the Maclaurin series allows us to write the function as an infinite sum of derivatives evaluated at zero, multiplied by powers of \(x\) divided by factorials:

• For degree 2: \(P_2(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2\).
• For degree 3: \(P_3(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3\).
• For degree 4: \(P_4(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4\).

These polynomials represent the truncated form of the Maclaurin series for the function, providing a balance between complexity and accuracy in calculations.