Problem 5 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The Taylor polynomials are: \(T_0(x) = \frac{1}{2}\), \(T_1(x) = \frac{1}{2} - \frac{1}{4}(x-2)\), \(T_2(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2\), and \(T_3(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3\).

Step 1: Understanding the Taylor Polynomial Formula

The Taylor polynomial of degree \(n\) for a function \(f(x)\) centered at \(a\) is given by:\[T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]

Step 2: Calculate the Function and Its Derivatives at \(a = 2\)

Firstly, compute the value of the function and its derivatives at \(x = 2\):1. \(f(x) = \frac{1}{x} \rightarrow f(2) = \frac{1}{2} \)2. First derivative \(f'(x) = -\frac{1}{x^2} \rightarrow f'(2) = -\frac{1}{4}\)3. Second derivative \(f''(x) = \frac{2}{x^3} \rightarrow f''(2) = \frac{1}{4}\)4. Third derivative \(f'''(x) = -\frac{6}{x^4} \rightarrow f'''(2) = -\frac{3}{8} \)

Step 3: Formulate the Taylor Polynomial of Order 0

The Taylor polynomial of order 0 is simply the value of the function at \(a\):\[ T_0(x) = f(2) = \frac{1}{2} \]

Step 4: Formulate the Taylor Polynomial of Order 1

Using the Taylor polynomial formula until the first derivative term:\[ T_1(x) = \frac{1}{2} - \frac{1}{4}(x-2) \]

Step 5: Formulate the Taylor Polynomial of Order 2

Include up to the second derivative:\[ T_2(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{4} \cdot \frac{1}{2} (x-2)^2 = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 \]

Step 6: Formulate the Taylor Polynomial of Order 3

Include up to the third derivative:\[ T_3(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{3}{8} \cdot \frac{1}{6} (x-2)^3 = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 \]

Key Concepts

Understanding Derivatives

When you're first learning calculus, derivatives are one of the most fundamental concepts you'll encounter. They represent the rate at which a function is changing at any given point. If you imagine a curve as a path, then the slope of the tangent to that curve is its derivative. This is crucial because it helps us understand the behavior of functions and how they respond to changes in input.

For example, in our exercise, we found several derivatives of the function \(f(x) = \frac{1}{x}\). Here’s the breakdown:

• The first derivative \(f'(x) = -\frac{1}{x^2}\) shows the rate of change of \(f\). A negative sign indicates that the function is decreasing as \(x\) increases.
• The second derivative \(f''(x) = \frac{2}{x^3}\) provides insight into the concavity of the graph. A positive second derivative indicates the graph is convex, or "smiling" upwards.
• The third derivative \(f'''(x) = -\frac{6}{x^4}\) continues this analysis. Each derivative level gives us additional layers of understanding about how the function behaves.

Through derivatives, we not only measure immediate rates of change but also accumulate information for higher-level understanding in polynomial approximations like Taylor series.

Polynomial Approximation Through Taylor Series

Taylor polynomials act as approximations of more complex functions. They enable us to express functions that might be otherwise difficult to work with in simpler polynomial forms.

The idea of polynomial approximation is about taking a function at a specific point \(a\) and building up a polynomial that approximates this function using derivatives.

In our exercise, Taylor polynomials were developed step by step:

• Zero Order (\(T_0(x)\)): It's the most basic approximation - just a constant, identical to the function's value at point \(a\), which was \(\frac{1}{2}\) in our case.
• First Order (\(T_1(x)\)): This adds the first derivative, providing a linear approximation that considers the function's immediate slope.
• Second Order (\(T_2(x)\)): With the inclusion of the second derivative, it delivers a quadratic approximation, capturing both slope and curvature.
• Third Order (\(T_3(x)\)): By adding the third derivative, this polynomial becomes more accurate, incorporating changes in concavity as well.

Learn how using Taylor polynomials can simplify complex calculations. By approximating functions, you can apply these simpler forms for calculations, integrations, and solving equations with less computational effort.

Calculus and Its Power

Calculus might seem challenging at first, but it offers powerful tools to explain phenomena in the scientific world. It's not just about solving equations – it's a way to model and predict real-world behaviors, from physics to economics.

The process of taking derivatives and forming Taylor polynomials is a cornerstone of understanding how calculus is applied. Each derivative used in the Taylor polynomial provides more insight and precision into the function’s behavior.

Applications in Real Life

• Predicting movement and changes, such as how a car accelerates or decelerates.
• Optimizing areas like maximizing profits in economics or minimizing costs in engineering.
• Modeling growth rates in biology or spreading diseases in epidemiology.

Calculus links simple algebraic operations to complex real-world phenomena seamlessly. By understanding how we derive and use functions, like we did with the Taylor polynomials in our exercise, we gain new perspectives on problem-solving with mathematics.