Problem 15 - Calculus: One and Several Variables

Question

Determine the \(n\) th Taylor polynomial \(P_{n}\) for the function \(f\). $$f(x)=e^{r x}, \quad r \text { a real number. }$$

Step-by-Step Solution

Answer

The nth Taylor polynomial of the function \(f(x) = e^{rx}\), where r is a real number, is given by: $$P_{n}(x) = \sum_{k=0}^{n} \frac{r^k}{k!} x^{k}$$

Step 1: First, we need to find the first n derivatives of the given function \(f(x) = e^{rx}\). We'll denote \(f^{(k)}(x)\) as the k-th derivative of the function with respect to x. #Step 2: Evaluate the derivatives at x = 0 #

Next, we need to evaluate these first n derivatives at x = 0, as we are expanding around 0. This will give us \(f^{(k)}(0)\) for k = 0, 1, ..., n. #Step 3: Form the nth Taylor polynomial#

Step 2: Finally, we can use the evaluated derivatives and the general formula for Taylor polynomials to form the nth Taylor polynomial for the given function. This polynomial will be given by: $$P_{n}(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^{k}$$ Now, let's find the nth Taylor polynomial for the given function \(f(x) = e^{rx}\), with r being a real number: #step 4: Calculate the derivatives of f(x)#

The first few derivatives of \(f(x) = e^{rx}\) are: $$f'(x) = re^{rx}$$ $$f''(x) = r^2e^{rx}$$ $$f'''(x) = r^3e^{rx}$$ We can generalize that the kth derivative is: $$f^{(k)}(x) = r^{k} e^{rx}$$ #Step 5: Evaluate the derivatives at x = 0 #

Step 3: Now, we will evaluate these derivatives at x = 0: $$f^{(k)}(0) = r^{k} e^{r(0)} = r^{k}$$ #Step 6: Form the nth Taylor polynomial#

Finally, using the general formula for Taylor polynomial, the nth Taylor polynomial of the given function is: $$P_{n}(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^{k} = \sum_{k=0}^{n} \frac{r^k}{k!} x^{k}$$ Thus, the nth Taylor polynomial of the function \(f(x) = e^{rx}\) is given by: $$P_{n}(x) = \sum_{k=0}^{n} \frac{r^k}{k!} x^{k}$$

Key Concepts

Derivative

The concept of the derivative is central to calculus and involves understanding how a function changes as its input changes. The derivative of a function gives the rate at which one quantity changes with respect to another. For the function given in the exercise, which is an exponential function of the form \(f(x) = e^{rx}\), the derivative involves some specific steps: - The first derivative, \(f'(x)\), tells us the slope of the tangent line at any point \(x\).- This derivative is calculated as \(re^{rx}\). Here, \(r\) is a constant and represents the rate of growth or decay depending on its value.Higher-order derivatives can be found by repeatedly taking the derivative of the derivative:

• The second derivative, \(f''(x)\), is \(r^2e^{rx}\).
• The third derivative, \(f'''(x)\), is \(r^3e^{rx}\).

In general, the \(k\)-th derivative of an exponential function \(f(x) = e^{rx}\) is \(f^{(k)}(x) = r^{k} e^{rx}\). This pattern shows the power of \(r\) increases with each differentiation. Understanding this pattern is crucial when working with polynomial approximations.

Exponential Function

Exponential functions are mathematical functions of the form \(f(x) = e^{rx}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828, and \(r\) is a real number.Exponential functions are unique because: - They increase (or decrease) at rates proportional to their current value, leading to rapid growth (or decay).- Their derivatives are proportional to the original function, making them easy to differentiate and integrate.In the context of Taylor polynomials, understanding the behavior of exponential functions is crucial since they appear frequently in scientific and financial models due to their properties of continuous growth/decay.In our exercise, the function \(f(x) = e^{rx}\) under consideration is an exponential function where:

• The choice of \(r\) determines the scaling factor of the growth/decay rate.
• When \(r > 0\), the function represents exponential growth.
• When \(r < 0\), it represents exponential decay.

Remembering these exponential characteristics helps when applying derivatives and constructing polynomials.

Series Expansion

The series expansion in calculus is a way of expressing functions as sums of simpler terms. Taylor series allow functions to be written as infinite sums of terms calculated from the values of the function's derivatives at a certain point, usually around 0.For example, if you have a function \(f(x)\), its Taylor series expansion around \(x = 0\) is given as:\[P_{n}(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^{k}\]This expansion uses:

• The derivatives \(f^{(k)}(x)\) to determine how the function behaves near the chosen point.
• The factorial in the denominator \(k!\) to manage the accumulation of larger terms typical of factorial growth.

In our exercise, the Taylor polynomial for the exponential function \(f(x) = e^{rx}\) can be simplified because its derivatives are particularly straightforward, involving powers of \(r\):- When evaluated at \(x = 0\), these derivatives simplify the expansion, leveraging the exponential function's self-similarity.This result highlights how the series expansion, in the form of a Taylor polynomial, can approximate the function using easily calculated terms, a powerful tool for analyzing functions.

Calculus

Calculus is the mathematical study of continuous change, and it is foundational to many areas of mathematics and science due to its broad applications to modeling and solving real-world problems. Key to calculus are two main concepts:

• **Differentiation**, which investigates how a function changes and is the process of finding a derivative.
• **Integration**, which is the reverse of differentiation and calculates the total accumulation of quantities.

In examining our exercise, calculus comes into play by taking derivatives to construct a Taylor polynomial. This process involves understanding not just *how* values change, but *how they accumulate* to form the entirety of the function—critical in approximations and modeling. Taylor polynomials, as a product of calculus, help us approximate more complex functions using a set of simpler polynomial terms, bridging the gap between simple algebraic calculations and complex functional behavior—ultimately underscoring calculus's role in practical problem-solving.