Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find a cubic polynomial, \(p(x)=p_{0}+p_{1} x+p_{2} x^{2}+p_{3} x^{3}\) that matches \(f(x)=e^{-x}\) at \(a=0\)

Step-by-Step Solution

Answer

The polynomial is \( p(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \).

Step 1: Understand the Problem

We need to find a cubic polynomial \( p(x) = p_0 + p_1 x + p_2 x^2 + p_3 x^3 \) that approximates \( f(x) = e^{-x} \) near \( x = 0 \). The polynomial should match the function and its first three derivatives at \( x = 0 \).

Step 2: Calculate Function Value at a Point

Calculate \( f(0) \) for the function \( f(x) = e^{-x} \). We have:\[ f(0) = e^{0} = 1 \]

Step 3: Derive Function

Compute the first three derivatives of \( f(x) = e^{-x} \):1. First derivative: \( f'(x) = -e^{-x} \)2. Second derivative: \( f''(x) = e^{-x} \)3. Third derivative: \( f'''(x) = -e^{-x} \)

Step 4: Evaluate Derivatives at a Point

Evaluate each derivative at \( x = 0 \):1. \( f'(0) = -1 \)2. \( f''(0) = 1 \)3. \( f'''(0) = -1 \)

Step 5: Use Taylor Series Expansion

The Taylor series expansion of \( f(x) \) around \( x = 0 \) is given by:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 \]Substitute the values from previous steps:\[ f(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \]

Step 6: Identify Polynomial Coefficients

The polynomial \( p(x) = p_0 + p_1 x + p_2 x^2 + p_3 x^3 \) now takes the same form as the Taylor series:- \( p_0 = 1 \)- \( p_1 = -1 \)- \( p_2 = \frac{1}{2} \)- \( p_3 = -\frac{1}{6} \)

Step 7: Write the Final Polynomial

Thus, the polynomial \( p(x) \) that matches \( f(x) = e^{-x} \) at \( x = 0 \) is:\[ p(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \]

Key Concepts

Cubic Polynomial

A cubic polynomial is a type of mathematical expression involving a variable raised to the third power. It takes the general form \[ p(x) = p_0 + p_1 x + p_2 x^2 + p_3 x^3 \]where:

• \( p_0 \) is the constant term,
• \( p_1 x \) is the linear term,
• \( p_2 x^2 \) is the quadratic term, and
• \( p_3 x^3 \) is the cubic term.

This expression describes the behavior of a curve in a plane. By varying these coefficients \( p_0, p_1, p_2, \) and \( p_3 \), the shape and position of the curve can be modified.
Cubic polynomials are the simplest type of polynomials that can describe curves with inflection points—places where the curve changes direction in its curvature. These polynomials are widely used for interpolation and approximation because of their simplicity and flexibility.
Let's dive deeper into how derivatives interact with these polynomials, as it is through derivatives that we gain further insights into the polynomial's properties.

Derivatives

Derivatives measure how a function changes as its input changes. When applied to our cubic polynomial, derivatives reveal the rate at which values of the polynomial output change for small changes in the input variable, \(x\).
For any polynomial \( p(x) = p_0 + p_1 x + p_2 x^2 + p_3 x^3 \), the derivatives are as follows:

• The first derivative, \( p'(x) \), is obtained by applying the power rule \((nx^{n-1})\) on each term, leading to

\[ p'(x) = p_1 + 2p_2 x + 3p_3 x^2 \]This expression gives us the slope of the tangent to the curve at any point \(x\). It tells us how steep the curve is at that point.

• The second derivative, \( p''(x) \), measures the curvature or concavity of the polynomial and is given by:

\[ p''(x) = 2p_2 + 6p_3 x \]This derivative helps us understand how the rate of change of the slope is itself changing.

• The third derivative, \( p'''(x) \), provides information about the variation in curvature.

\[ p'''(x) = 6p_3 \]In the context of Taylor series expansion, calculating derivatives at a point gives us coefficients for each term in the series. For example, the function \( f(x) = e^{-x} \) at \( x = 0 \) involves these derivatives directly in forming its Taylor polynomial.

Polynomial Approximation

Polynomial approximation involves using a polynomial to estimate a function’s behavior over a specified range. The Taylor series is one powerful tool for achieving such approximations, particularly when functions are too complex to work with directly.
In our exercise, we aim to approximate the exponential function \( f(x) = e^{-x} \) using a cubic polynomial around \( x = 0 \). By determining the derivatives at this point, we establish the coefficients of the polynomial that mimic the original function’s value and behavior locally.
Taylor polynomial is a finite sum used to approximate a function near a point \(a\). For practical purposes:

• We focus only on a few terms since subsequent terms become increasingly less significant.
• The approximation is given by matching function values and their derivatives at the point of interest.

The Taylor polynomial of \( f(x) = e^{-x} \) at \( x = 0 \) is obtained by plugging in the value and derivatives computed previously:\[ f(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \]This polynomial captures not just the function value at \( x = 0 \) but also reflects how it changes, capturing the immediate behavior with great accuracy. Polynomial approximations are invaluable for simplifying calculations and gaining insights into functions that are challenging to analyze directly.