Chapter 1: Problem 2

Use power series to evaluate the function at the given point. Observe in each case that, even with a hand calculator available, it is easier or necessary to use series. \(e^{-x^{2} / 2}-1-\ln \cos x\) at \(x=0.0024\)

Short Answer

Expert verified

At x=0.0024, the function evaluates to approximately 0.00000288.

Step by step solution

Understand the problem

The goal is to evaluate the function \(e^{-x^{2} / 2}-1-\ln \cos x\) at a given point, specifically at \(x=0.0024\).

Recall the power series expansions

Recall the power series expansions for three functions:1. The exponential function: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\)2. The logarithm function: \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\)3. The cosine function: \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\)

Expand the functions using power series

Expand each function at \(x=0\):For \(e^{-x^2/2}\): \[ e^{-\frac{x^2}{2}} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\]At \(x=0.0024\), using only the first two non-zero terms for simplicity: \[ e^{-\frac{(0.0024)^2}{2}} \approx 1 - \frac{(0.0024)^2}{2!} = 1 - \frac{0.00000576}{2} = 1 - 0.00000288 \]So, \( e^{-\frac{x^2}{2}} \approx 1 - 0.00000288\).

Expand the logarithm and cosine functions

For \(\ln \cos x\), use the series for the cosine function first:\[ \cos x \approx 1 - \frac{x^2}{2!}\]Then, apply the logarithm series:\[ \ln (1-x^2/2) \approx -\frac{x^2}{2}\] So, for \(x = 0.0024\), \[ \ln \cos (0.0024) \approx -\frac{(0.0024)^2}{2} = -0.00000288\]

Combine the terms to get the final result

Now, combine all approximations:\[ e^{-\frac{(0.0024)^2}{2}} - 1 - \ln \cos(0.0024) \]Substitute the approximations:\[ (1 - 0.00000288) - 1 - (-0.00000288) = 0.00000288\]

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Key Concepts

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

Exponential Function

The exponential function, denoted as \(e^x\), is fundamental in mathematics and various applications, such as modeling growth processes. It is defined as the sum of an infinite series:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
In the context of this problem, we need the power series expansion for \(e^{-x^2 / 2}\). Substituting \(x^2 / 2\) for \(x\) in the series expansion, you get:
\[ e^{-\frac{x^2}{2}} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \]
For small values of \(x\), like \(x = 0.0024\), higher-order terms become negligible. Evaluating the first two non-zero terms for simplicity, we find:
\[ e^{-\frac{(0.0024)^2}{2}} \approx 1 - \frac{(0.0024)^2}{2} \approx 1 - 0.00000288 \]
This simplification shows how power series can make complicated calculus problems manageable by focusing on dominant terms.

Logarithm Function

The logarithm function, specifically the natural logarithm \(\ln(x)\), is vital for solving problems involving multiplication processes and growth scales. Its power series expansion around 0 is:
\[ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \]
In this problem, \(\ln \cos x\) is used. Given the series expansion for \(\cos x\):
\[ \cos x \approx 1 - \frac{x^2}{2!} \]
We substitute and approximate \(\cos x\) for small \(x\), here \(x = 0.0024\), into \(\ln(1 - x^2/2)\):
\[ \ln(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2} \]
At \(x = 0.0024\), this becomes:
\[ \ln(\cos(0.0024)) \approx -0.00000288 \]
This simplification, using the power series, shows the power of series expansions in approximating complex logarithmic expressions.

Cosine Function

The cosine function, \(\cos x\), frequently appears in trigonometry and wave-related applications. The function can be represented by a power series:
\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]
This series representation helps simplify and compute the function's value without direct evaluation for small \(x\). For \(x = 0.0024\), consider the first two terms:
\[ \cos(0.0024) \approx 1 - \frac{(0.0024)^2}{2} \approx 1 - 0.00000288 \]
By approximating \(\cos x\) for small angles, the power series transformation of \(\cos x\) makes it easier to evaluate logarithmic and complex trigonometric functions. Combining series simplifications allows for efficient and accurate results without a calculator.