Problem 1 - Thomas Calculus

Question

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=e^{2 x}, \quad a=0\)

Step-by-Step Solution

Answer

The Taylor polynomials are: \(P_0(x) = 1\), \(P_1(x) = 1 + 2x\), \(P_2(x) = 1 + 2x + 2x^2\), \(P_3(x) = 1 + 2x + 2x^2 + \frac{4}{3}x^3\).

Step 1: Identify the Function and Point

We are given the function \(f(x) = e^{2x}\) and the point \(a = 0\). We aim to find the Taylor polynomials of different orders centered around this point.

Step 2: Calculate Derivatives

To find Taylor polynomials, we first need to calculate the derivatives of \(f(x)\) at \(x = 0\). - The first derivative is \(f'(x) = 2e^{2x}\).- The second derivative is \(f''(x) = 4e^{2x}\).- The third derivative is \(f'''(x) = 8e^{2x}\).- The function itself is \(f(x) = e^{2x}\).

Step 3: Evaluate Derivatives at a

Evaluate each derived function at \(x = 0\).- \(f(0) = e^0 = 1\).- \(f'(0) = 2e^0 = 2\).- \(f''(0) = 4e^0 = 4\).- \(f'''(0) = 8e^0 = 8\).

Step 4: Formulate the Taylor Polynomials

Use the general form of the Taylor polynomial to create polynomials of different orders.- Order 0 polynomial: \(P_0(x) = 1\).- Order 1 polynomial: \(P_1(x) = 1 + 2x\).- Order 2 polynomial: \(P_2(x) = 1 + 2x + \frac{4}{2!}x^2 = 1 + 2x + 2x^2\).- Order 3 polynomial: \(P_3(x) = 1 + 2x + 2x^2 + \frac{8}{3!}x^3 = 1 + 2x + 2x^2 + \frac{4}{3}x^3\).

Step 5: Final Results

We have the following Taylor polynomials:- \(P_0(x) = 1\)- \(P_1(x) = 1 + 2x\)- \(P_2(x) = 1 + 2x + 2x^2\)- \(P_3(x) = 1 + 2x + 2x^2 + \frac{4}{3}x^3\)

Key Concepts

Derivatives

Derivatives are the building blocks when working with Taylor polynomials. They help us understand the rate of change of a function at a specific point. In the context of Taylor polynomials, derivatives allow us to approximate a function using just a few terms. The Taylor polynomial is constructed using derivatives evaluated at a particular point, which in this exercise is given as \( a = 0 \). Derivatives are calculated successively, each taking the rate of change of the previous derivative.
To construct a Taylor series, you start with the function and compute successive derivatives:

• The first derivative \( f'(x) = 2e^{2x} \) shows how rapidly the original function is changing.


• The second derivative \( f''(x) = 4e^{2x} \) provides the acceleration or the second order rate of change.


• The third derivative \( f'''(x) = 8e^{2x} \) further refines the rate at which changes accumulate.

Once these derivatives are calculated, they are evaluated at \( x = 0 \), which simplifies to constants that form the coefficients of the Taylor polynomial.

Exponential Functions

An exponential function is a mathematical function of the form \( f(x) = e^{kx} \), where \( e \) is Euler's number, a constant approximately equal to 2.71828. In this exercise, the function \( f(x) = e^{2x} \) is an example of an exponential function. Exponential functions are ubiquitous in mathematics and the sciences due to their unique property of self-derivation: the derivative of an exponential function is proportional to itself.
This unique property of exponential functions simplifies the process of finding Taylor polynomials because its derivatives follow a predictable pattern. For \( f(x) = e^{2x} \):

• Every derivative is a multiple of \( e^{2x} \).


• The sequence of derivatives increase their intensity in multiples of 2, reflecting the power of \( x \) in the original function.

Evaluating these derivatives at \( a = 0 \) simplifies each term significantly, resulting in a pattern that can systematically expand into a Taylor series.

Evaluation of Polynomials

The evaluation of polynomials is a key process involved in generating Taylor polynomials. It involves computing the value of polynomials formed by derivatives at a given point, in this case, \( x = 0 \). The result of these evaluations forms the coefficients of the Taylor polynomial.
For example, in this exercise:

• Evaluating \( f(0) = e^{0} = 1 \) gives the constant term.


• Evaluating \( f'(0) = 2 \times e^{0} = 2 \) provides the linear term coefficient.


• Evaluating \( f''(0) = 4 \times e^{0} = 4 \) results in the quadratic term coefficient, but divided by factorial 2, simplifies to 2.
• Similarly, \( f'''(0) = 8 \times e^{0} = 8 \) divided by factorial 3, converts to \( \frac{8}{6} \) or \( \frac{4}{3} \).

These coefficients, derived from evaluating derivatives at the specified point, are then used to construct the Taylor polynomials, which are approximations of the function \( e^{2x} \) around \( a = 0 \). The polynomial increases in accuracy as we add more terms, effectively capturing the behavior of the function.