Problem 50 - Applied Calculus

Question

Find the quadratic polynomial \(g(x)=a x^{2}+b x+c\) which best fits the function \(f(x)=e^{x}\) at \(x=0\), in the sense that \(g(0)=f(0)\), and \(g^{\prime}(0)=f^{\prime}(0)\), and \(g^{\prime \prime}(0)=f^{\prime \prime}(0)\) Using a computer or calculator, sketch graphs of \(f\) and \(g\) on the same axes. What do you notice?

Step-by-Step Solution

Answer

The polynomial \(g(x) = \frac{1}{2}x^2 + x + 1\) fits \(e^x\) closely around \(x = 0\).

Step 1: Set Condition for Value at x=0

Start by setting the first condition, where the value of the polynomial at \(x = 0\) must equal the value of the function \(f(x)\) at \(x = 0\). Thus, \(g(0) = c = f(0) = e^0 = 1\). This gives us \(c = 1\).

Step 2: Set Condition for First Derivative at x=0

Next, compute the first derivative of the polynomial, \(g'(x) = 2ax + b\). At \(x=0\), this should equal the first derivative of \(f(x)\), \(f'(x) = e^x\). Thus, \(g'(0) = b = f'(0) = e^0 = 1\). This gives us \(b = 1\).

Step 3: Set Condition for Second Derivative at x=0

Finally, compute the second derivative of the polynomial, \(g''(x) = 2a\). At \(x=0\), this should equal the second derivative of \(f(x)\), \(f''(x) = e^x\). Thus, \(g''(0) = 2a = f''(0) = e^0 = 1\). Solving this gives \(a = \frac{1}{2}\).

Step 4: Write the Quadratic Polynomial

Now that we have \(a = \frac{1}{2}\), \(b = 1\), and \(c = 1\), we can write the quadratic polynomial as \(g(x) = \frac{1}{2}x^2 + x + 1\).

Step 5: Graph the Functions f(x) and g(x)

Using a computer or graphing calculator, plot \(f(x) = e^x\) and \(g(x) = \frac{1}{2}x^2 + x + 1\) together. You will notice that near \(x = 0\), the graphs are very close indicating a good fit at \(x = 0\). As \(x\) moves away from 0, the graphs diverge due to the nature of exponential growth versus a quadratic curve.

Key Concepts

Understanding Best Fit for Quadratic Polynomial

When we talk about finding the 'best fit' for a quadratic polynomial, particularly in the context of this exercise, we're centered on the idea of matching the original function as closely as possible at a given point.
For instance, we need to fit a quadratic polynomial to the function \(f(x) = e^{x}\) at \(x = 0\). This is achieved by ensuring the polynomial \(g(x) = ax^2 + bx + c\) satisfies certain conditions at \(x = 0\).

• \(g(0) = f(0)\): The value of the polynomial at this point must match the value of the function.
• \(g'(0) = f'(0)\): The slope of the polynomial and the function must be the same.
• \(g''(0) = f''(0)\): The curvature of the polynomial must match that of the function.

These conditions make sure that the polynomial captures the behavior of the exponential function \(f(x)\) precisely at \(x = 0\), making it the 'best fit' in this specific framework.

The Importance of Derivative Matching

Derivative matching is crucial when approximating functions with polynomials. It involves adjusting the derivatives of the polynomial at the point of interest to match those of the original function. In our exercise, we have the following series of matches:

• First, the zero-th derivative (the value of the function) at \(x = 0\), which gave us the constant term \(c = 1\).
• Second, the first derivative at \(x = 0\). This provided the linear term coefficient \(b = 1\), aligning the slope of the polynomial with that of the function.
• Third, the second derivative at \(x = 0\), which gave us the quadratic term coefficient \(a = \frac{1}{2}\). This matched the curvature.

Matching these derivatives up to the second order ensures that the polynomial captures the initial behavior of \(f(x)\) closely around \(x = 0\). It's essential for constructing a polynomial that adequately represents the function locally.

Function Approximation with Quadratic Polynomials

Function approximation using quadratic polynomials is about using a simpler function to represent a more complex one over a certain range. By employing a quadratic polynomial, we leverage its simplicity and ease of calculation.
The polynomial is determined using derivatives and specific conditions to approximate how the given function behaves near a particular point. This lets us perform quick calculations and gets a good estimate for function values without computing the full complexity of the original function.
While the quadratic polynomial \(g(x) = \frac{1}{2}x^2 + x + 1\) can't capture all the nuances of the exponential function across its whole range, it provides a good approximation near \(x = 0\). This technique is valuable for simplifying complex models in applied mathematics, physics, and economics.

Graphing Functions for Visual Analysis

Graphing is a powerful tool for understanding function approximation. By plotting both the approximated quadratic polynomial and the original function, we gain insights visually.
In the solution, we graphed \(f(x) = e^x\) and \(g(x) = \frac{1}{2}x^2 + x + 1\). When graphed together, they nearly overlap around \(x = 0\). This shows the efficacy of our approximation at that point.
However, as \(x\) moves away from zero, the graphs start diverging because the exponential function grows much faster than the quadratic polynomial. This divergence visually demonstrates the limits of approximation; a polynomial, due to its structure, can't follow the exponential trend indefinitely.
Graphs not only validate our calculations but also provide intuition on where and why the approximation works well or fails. It's a way to check our results against real behavior, offering an interactive learning experience.