Chapter 1: Problem 17

Find the Maclaurin series for the following functions. $$e^{1-\sqrt{1-x^{2}}}$$

Short Answer

Expert verified

The Maclaurin series for $e^{1-\root{1-x^2}}$ is $1 + \frac{x^2}{2} + \frac{x^4}{4} + \frac{5x^6}{48} + \cdots $.

Step by step solution

Understand the Function

The given function to find the Maclaurin series for is $e^{1-\root{1-x^{2}}}$. The Maclaurin series is a Taylor series expansion of a function about 0.

Identify Inner Function and Expand

First, focus on the inner function $1 - \root{1-x^{2}}$. We need to expand $\root{1-x^2}$ using a Taylor series about 0. The expansion is: \[\root{1-x^2} = 1 - \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^6}{16} + \text{higher order terms}\]

Substitute Inner Function Expansion

Substitute the series expansion of $\root{1-x^2}$ into the expression $1 - \root{1-x^2}$: \[1 - (1 - \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^6}{16}) = \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{16}\]

Expand the Exponential Function

Now, we need to find the Maclaurin series for $e^{u}$ where $u = \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{16}$:The exponential function $e^u$ can be expanded as \[ e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots\]

Substitute and Simplify

Substitute $u = \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{16}$ into the exponential series expansion:\[ e^{1-\root{1-x^2}} \approx 1 + \left(\frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{16}\right) + \frac{\left(\frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{16}\right)^2}{2} + \ldots\]After simplifying and collecting terms, obtain the series up to the desired order.

Collect Terms

Combine the terms to form the Maclaurin series up to the 6th power of $x$:\[ e^{1-\root{1-x^2}} \approx 1 + \frac{x^2}{2} + \left(\frac{x^4}{8} + \frac{x^2}{2} \cdot \frac{x^2}{2}\right) + \ldots \]This results in:\[ e^{1-\root{1-x^2}} \approx 1 + \frac{x^2}{2} + \frac{x^4}{4} + \frac{5x^6}{48} + \ldots\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series

The Taylor series is a way to represent a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. When expanding about zero, we call this specific case the Maclaurin series. This can be useful to approximate functions that are otherwise complicated. The basic form of a Taylor series for a function $f(x)$ about a point $a$ is:

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$

Taylor series helps in analyzing the behavior and properties of functions like solving differential equations or evaluating limits. It is instrumental in numerical mathematics, where polynomial approximations of functions are needed.

Understanding Taylor series opens the door to approximating functions with polynomials, making complex problems more manageable.

Exponential Function

The exponential function, denoted as $e^x$, is one of the most important functions in mathematics. It arises in many contexts, especially those involving growth or decay, like population growth, radioactive decay, and compounding interest. The exponential function has the unique property that its derivative is the same as the function itself:

$\frac{d}{dx} e^x = e^x$

The Maclaurin series expansion for the exponential function is given by:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

This series is invaluable because it provides a simple polynomial approximation of the exponential function for any value of $x$. For small values of $x$, the series converges quickly, and we can often use just a few terms for a good approximation.

Mastering the properties and series expansion of the exponential function is essential for any student delving into higher mathematics or physics.

Function Expansion

Function expansion refers to expressing a complex function as a series of simpler terms. The goal is to simplify complicated expressions and make them easier to work with, especially for calculations and approximations.

The Maclaurin series is a specific type of function expansion where we expand around $x = 0$. For example, consider the function $e^{1-\root{1-x^2}}$. To find its Maclaurin series, we proceed as follows:

Firstly, expand the simpler inner function $\root{1-x^2}$ using its Maclaurin series.
Substitute this expansion into our outer function to transform it into a series.
Finally, we use the Maclaurin series for the exponential function to expand $e^{1-\root{1-x^2}}$.

Combining the inner and outer expansions and simplifying the terms gives us the final series. Such expansions are vital in approximating complex functions and are widespread in many fields including physics, engineering, and economics.