Problem 8 - Calculus

Question

using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\frac{z}{e^{z^{2}}}$$

Step-by-Step Solution

Answer

The first four nonzero terms are \(z, -z^3, \frac{z^5}{2}, -\frac{z^7}{6}\).

Step 1: Recall the Taylor Series for \(e^x\)

The Taylor series for \(e^x\) centered at 0 is given by \[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\] We will use this series to find the expansion for \(e^{-z^2}\).

Step 2: Find the Taylor Series for \(e^{-z^2}\)

Substitute \(-z^2\) into the Taylor series of \(e^x\):\[e^{-z^2} = 1 - \frac{z^2}{1!} + \frac{z^4}{2!} - \frac{z^6}{3!} + \cdots\]This provides the exponential series for \(e^{-z^2}\) centered at 0.

Step 3: Multiply \(z\) by the Series for \(e^{-z^2}\)

Multiply \(z\) by each term in the series for \(e^{-z^2}\):\[z \cdot e^{-z^2} = z \cdot \left(1 - \frac{z^2}{1!} + \frac{z^4}{2!} - \frac{z^6}{3!} + \cdots\right)\]This results in:\[z - z^3 + \frac{z^5}{2} - \frac{z^7}{6} + \cdots\]

Step 4: Extract the First Four Nonzero Terms

The first four nonzero terms of the function \(\frac{z}{e^{z^2}}\) are the first four terms from our multiplication series:\[z, \quad -z^3, \quad \frac{z^5}{2}, \quad -\frac{z^7}{6}\] These are the terms required in the solution.

Key Concepts

Exponential Function

The exponential function, denoted as \( e^x \), is one of the most important functions in mathematics. It describes exponential growth and has unique properties that make it vital in calculus, complex analysis, and differential equations. The base of the natural exponential function is Euler's number \( e \), approximately equal to 2.71828. This function can be defined as its own derivative, meaning:

• The rate of change of \( e^x \) with respect to \( x \) is \( e^x \) itself.

It can be expanded into a series called the Taylor series. For functions like \( e^x \), this series takes the form:

• \( e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

This series expansion is extremely useful for approximating the function in cases where it's difficult to compute directly.

Polynomial Approximation

Polynomial approximation is a technique used to simplify complex functions by representing them as a polynomial. In calculus, one of the most powerful tools for polynomial approximation is the Taylor series. This allows functions to be represented as an infinite sum of terms calculated from the derivatives of the function at a single point. For example, in the context of the exponential function used here, \( e^{-z^2} \) can be approximated by replacing \( x \) with \( -z^2 \) in the Taylor expansion of \( e^x \).
This gives the series:

• \( e^{-z^2} = 1 - \frac{z^2}{1!} + \frac{z^4}{2!} - \frac{z^6}{3!} + \cdots \)

Using polynomial approximation, functions that are otherwise complex can become more manageable in analysis, especially when only a few terms are required for a close approximation.

Series Expansion

Series expansion is a powerful concept in mathematical analysis and calculus. It involves expressing functions as a sum of terms in a series, which can often give an approximate solution. Through series expansion, complex functions like \( \frac{z}{e^{z^2}} \) can be approached by examining only the early terms of their Taylor series.

• In our example, once \( e^{-z^2} \) is expanded into a series, multiplying its terms by \( z \) allows us to derive the series for \( \frac{z}{e^{z^2}} \).

To focus on the first few non-zero terms:

• \( z \cdot (1 - \frac{z^2}{1!} + \frac{z^4}{2!} - \frac{z^6}{3!} + ...) = z - z^3 + \frac{z^5}{2} - \frac{z^7}{6} + ... \)

The beauty of series expansion is its flexibility, allowing us to extract valuable insights about a function's behavior without needing its full definition.