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Magazine Chapter 3: Problem 49
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 $$g(0)=f(0), \text { and } g^{\prime}(0)=f^{\prime}(0), \text { and } g^{\prime \prime}(0)=f^{\prime \prime}(0)$$. Using a computer or calculator, sketch graphs of \(f\) and \(g\) on the same axes. What do you notice?
Short Answer
Expert verified
The polynomial is \( g(x) = \frac{1}{2}x^2 + x + 1 \). It closely approximates \( e^x \) near \( x=0 \).
Step by step solution
01
Evaluate the function and its derivatives at x=0
Given the function \( f(x) = e^x \), we need to evaluate \( f(0) \), \( f'(0) \), and \( f''(0) \). \Calculate these values: - \( f(0) = e^0 = 1 \)- \( f'(x) = e^x \) so \( f'(0) = e^0 = 1 \)- \( f''(x) = e^x \) so \( f''(0) = e^0 = 1 \).
02
Define the polynomial and its derivatives
The quadratic polynomial is \( g(x) = ax^2 + bx + c \). Calculate its derivatives:- \( g'(x) = 2ax + b \)- \( g''(x) = 2a \).
03
Set conditions for matching the function and derivatives at x=0
We need to set up the equations based on the condition that the polynomial and its derivatives match those of the function \( f(x) = e^x \) at \( x=0 \):- \( g(0) = c = f(0) = 1 \)- \( g'(0) = b = f'(0) = 1 \)- \( g''(0) = 2a = f''(0) = 1 \).
04
Solve for coefficients a, b, and c
From the equations derived:- From \( c = 1 \), we have \( c = 1 \).- From \( b = 1 \), we find \( b = 1 \).- From \( 2a = 1 \), we solve for \( a \) to get \( a = \frac{1}{2} \).
05
Write the quadratic polynomial
Substitute \( a = \frac{1}{2} \), \( b = 1 \), and \( c = 1 \) back into the polynomial:\[ g(x) = \frac{1}{2}x^2 + x + 1 \].
06
Analyze the graphs of f(x) and g(x)
Using a calculator or computer software, plot both \( f(x) = e^x \) and \( g(x) = \frac{1}{2}x^2 + x + 1 \) on the same axes. Notice that the quadratic polynomial approximates \( e^x \) near \( x=0 \), capturing the same initial value, slope, and curvature at that point.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
A Taylor Series is a way to represent a function using an infinite sum of terms calculated from its derivatives at a single point. It's like building a bridge from polynomials to more complex functions. Here鈥檚 a simplified breakdown:
- A Taylor series approximates a function around a particular point, often referred to as 'a'.
- The series uses derivatives at 'a' to create terms of a polynomial.
- In simpler terms, you expand a function into a sum of terms of increasing powers of (x - a).
Derivative Match
The concept of a Derivative Match involves making sure the polynomial used for approximation not only passes through a point on the original function but also has the same slope and curvature at that point. Here鈥檚 how it works:
- The function value: both the original function and the quadratic polynomial must be the same at the point of approximation.
- The first derivative: the rate of change (slope) must be the same for both the polynomial and the original function at that point.
- The second derivative: the measure of curvature should match, ensuring the polynomial curvatures similarly to the original function.
Polynomial Fit
Polynomial Fit refers to using a polynomial to represent a section of a curve from a more complex function, ideally providing a good approximation over that section. In our case, we use a quadratic polynomial for a perfect fit around x=0.Here's a bit of a breakdown:
- "Fit" means the polynomial aligns closely with the actual data of the function in the area of interest. Here, the exponential function around x=0.
- Quadratic Polynomial: Our target polynomial g(x) = \( \frac{1}{2}x^2 + x + 1 \) fits because it closely follows the exponential function鈥檚 behavior near the origin.
- Through fitting, parameters a, b, and c in the polynomial are adjusted to minimize the difference between the function and polynomial values (often using sequential derivative matches).
Exponential Function
An Exponential Function is expressed in the form \( f(x) = e^x \), where e is a mathematical constant approximately equal to 2.71828. Its characteristics include:
- The rate of change grows exponentially, not linearly.
- It tends to increase rapidly, particularly after x is a positive value.
- Its graph has a characteristic J-curve, starting slowly for small x and increasingly rising steeply as x increases.
- In our quadratic approximation, using the derivatives of \( f(x) = e^x \) helps establish the fit through initial values, slopes, and curvatures.
Graphical Analysis
Graphical Analysis allows us to visually compare the relationship between the function we are approximating and the polynomial fit. This visual inspection is crucial for understanding how well our polynomial fits the function.For instance:
- When plotted, the quadratic polynomial \( g(x) = \frac{1}{2}x^2 + x + 1 \) will closely follow \( f(x) = e^x \) around x=0, showing a good fit at that point.
- This match is evident where both graphs have the same initial value and rate of change.
- As x moves away from 0, the polynomial will diverge from the exponential. This is why the approximation is best near x=0.
- Graphical tools make it easier to compare and note where approximations succeed or fail visually.
