Problem 10 - Calculus

Question

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

Step-by-Step Solution

Answer

The Taylor polynomials are: \( P_2(x) = 1 + px + \frac{p(p-1)}{2}x^2 \), \( P_3(x) = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 \), \( P_4(x) = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 + \frac{p(p-1)(p-2)(p-3)}{24}x^4 \).

Step 1: Understanding the Taylor Series

The Taylor series of a function \( f(x) \) expanded around \( x = a \) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). Here, we are expanding around \( x = 0 \), so \( a = 0 \).

Step 2: Find Derivatives of \((1+x)^p\)

Start by finding the derivatives necessary for the Taylor polynomials:- \( f(x) = (1+x)^{p} \)- First derivative: \( f'(x) = p(1+x)^{p-1} \)- Second derivative: \( f''(x) = p(p-1)(1+x)^{p-2} \)- Third derivative: \( f'''(x) = p(p-1)(p-2)(1+x)^{p-3} \)- Fourth derivative: \( f''''(x) = p(p-1)(p-2)(p-3)(1+x)^{p-4} \)

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

Evaluate the derivatives obtained in Step 2 at \( x = 0 \):- \( f(0) = (1+0)^p = 1 \)- \( f'(0) = p(1+0)^{p-1} = p \)- \( f''(0) = p(p-1)(1+0)^{p-2} = p(p-1) \)- \( f'''(0) = p(p-1)(p-2)(1+0)^{p-3} = p(p-1)(p-2) \)- \( f''''(0) = p(p-1)(p-2)(p-3)(1+0)^{p-4} = p(p-1)(p-2)(p-3) \)

Step 4: Formulate Taylor Polynomials for \( n=2\), \(n=3\), and \(n=4\)

Now, using the formula for the Taylor series, write the polynomials:- **For \( n=2 \):** \[ P_2(x) = 1 + px + \frac{p(p-1)}{2}x^2 \]- **For \( n=3 \):** \[ P_3(x) = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 \]- **For \( n=4 \):** \[ P_4(x) = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 + \frac{p(p-1)(p-2)(p-3)}{24}x^4 \]

Key Concepts

Taylor Series

The Taylor Series is a tool used to express functions as infinite sums of derivatives at a single point, focused around that point. For any function \( f(x) \) expanded around \( x = a \), its Taylor series starts as follows: \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). In simpler terms, this series provides a way to approximate complex functions with polynomials, which are easier to analyze and compute.
In this particular exercise, we are interested in expanding around \( x = 0 \), making the calculations straightforward, as the series becomes \( f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \). This expression helps in finding a polynomial that closely represents the behavior of the function around a specific point, which is very useful in mathematics and various applications where exact functions might be difficult to work with.

Derivative Evaluation

Derivative evaluation is the process of finding the rate at which a function changes at any given point. In our task, we need derivatives of the function \((1+x)^{p}\) to include them in the Taylor series. Derivatives here help in understanding how the function behaves and curves around \( x = 0 \).
We compute the derivatives as follows:

• First derivative: \( f'(x) = p(1+x)^{p-1} \)
• Second derivative: \( f''(x) = p(p-1)(1+x)^{p-2} \)
• Third derivative: \( f'''(x) = p(p-1)(p-2)(1+x)^{p-3} \)
• Fourth derivative: \( f''''(x) = p(p-1)(p-2)(p-3)(1+x)^{p-4} \)

Evaluating these derivatives at \( x = 0 \) gives the coefficients for the Taylor polynomial. This step is crucial as each derivative changes the polynomial's degree, contributing to more precise approximations.

Polynomial Approximation

In mathematics, polynomial approximation involves representing a function using a polynomial. This is incredibly valuable because polynomials are simpler to handle in many contexts, such as calculations or graphing. In the given exercise, we derive polynomial approximations of \( n = 2, 3, \) and \( 4 \) degrees.
The polynomial approximation gives us these equations:

• For \( n=2 \): \( P_2(x) = 1 + px + \frac{p(p-1)}{2}x^2 \)
• For \( n=3 \): \( P_3(x) = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 \)
• For \( n=4 \): \( P_4(x) = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 + \frac{p(p-1)(p-2)(p-3)}{24}x^4 \)

These approximations allow us to capture the essence of the function using a finite series. As \( n \) increases, the polynomial becomes a better approximation of the actual function.

Higher Order Derivatives

Higher-order derivatives are derivatives beyond the first order and are crucial for understanding the finer behavior of functions. These derivatives

• provide insight into the curvature and rate of change of the rate of change,
• and how these properties evolve as more derivatives are considered.

In Taylor polynomials, each additional derivative provides coefficients for higher powers of \( x \), allowing a closer approximation.
By computing higher-order derivatives at \( x = 0 \), we can note how:

• \( f''(0) = p(p-1) \)
• \( f'''(0) = p(p-1)(p-2) \)
• \( f''''(0) = p(p-1)(p-2)(p-3) \)

Adding more terms and utilizing these higher-order derivatives, we refine the polynomial to increasingly mimic the function's behavior around the point of expansion.