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Magazine Chapter 26: Problem 6
Use Maclaurin's series to obtain the expansion of \(e^{x}\) and of \(\cos x\) in ascending powers of \(x\) and hence determine $$ \operatorname{Lim}_{x \rightarrow 0}\left\\{\frac{e^{x}+e^{-x}-2}{2 \cos 2 x-2}\right\\}. $$
Short Answer
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Step by step solution
01
Maclaurin series for \(e^{x}\)
The Maclaurin series for \(e^{x}\) is given by \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
02
Maclaurin series for \(e^{-x}\)
By replacing \(x\) with \(-x\) in the series of \(e^{x}\), the Maclaurin series for \(e^{-x}\) is \[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \]
03
Maclaurin series for \(\frac{e^{x} + e^{-x} - 2}{2}\)
Adding the Maclaurin series of \(e^x\) and \(e^{-x}\), \[ e^x + e^{-x} = 2 + 2\frac{x^2}{2!} + \cdots = 2 + x^2 + \frac{x^4}{12} + \cdots \] Subtracting 2 from the series, \[ e^x + e^{-x} - 2 = x^2 + \frac{x^4}{12} + \cdots \] Dividing by 2, \[ \frac{e^x + e^{-x} - 2}{2} = \frac{x^2}{2} + \cdots \]
04
Maclaurin series for \(\text{cos}(x)\)
The Maclaurin series for \(\text{cos}(x)\) is \[ \text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]
05
Maclaurin series for \(2 \text{cos}(2x)\)
Substitute \(2x\) for \(x\) in the series of \( \text{cos}(x)\), then multiply by 2, \[ \text{cos}(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \cdots = 1 - 2x^2 + \frac{4x^4}{4!} + \cdots \] Therefore, \[ 2 \text{cos}(2x) = 2 - 2x^2 + \frac{2x^4}{12} + \cdots \]
06
Maclaurin series for \(2 \text{cos}(2x) - 2\)
Subtract 2 from the series: \[ 2 \text{cos}(2x) - 2 = -2x^2 + \frac{2x^4}{12} + \cdots \]
07
Simplifying the Limit Expression
Combine the numerators and denominators found: \[ \frac{\frac{e^x + e^{-x} - 2}{2}}{2 \text{cos}(2x) - 2} = \frac{\frac{x^2}{2}}{-2x^2} = \frac{x^2}{2} \times \frac{-1}{2x^2} = -\frac{1}{4} \]
08
Evaluate the Limit as \(x \to 0\)
As \(x \to 0, \frac{x^2}{2x^2} \to 0\), but the constants remain: \[ \text{Lim}_{x \to 0} \frac{e^x + e^{-x} - 2}{2 \text{cos}(2x) - 2} = -\frac{1}{4} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Maclaurin Series
The Maclaurin series is a special type of Taylor series that expands a function into an infinite sum of terms calculated from the function's derivatives at zero. It simplifies complex functions into a series of polynomial terms.
To understand this, take a function \( f(x) \). Its Maclaurin series is given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \text{...} \]
So, we replace the function with a sum that involves powers of \( x \) and coefficients that are derivatives of the function evaluated at zero.
To understand this, take a function \( f(x) \). Its Maclaurin series is given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \text{...} \]
So, we replace the function with a sum that involves powers of \( x \) and coefficients that are derivatives of the function evaluated at zero.
Exponential Function Expansion
The exponential function, \( e^x \), can be expressed in a Maclaurin series as follows:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{...} \]
This series continues indefinitely. It helps us approximate \( e^x \) for small values of \( x \).
Similarly, for the inverse function \( e^{-x} \), just replace \( x \) with \( -x \) in each term of the series:
\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \text{...} \]
These expansions allow us to work with complex exponential expressions in a simpler, polynomial form.
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{...} \]
This series continues indefinitely. It helps us approximate \( e^x \) for small values of \( x \).
Similarly, for the inverse function \( e^{-x} \), just replace \( x \) with \( -x \) in each term of the series:
\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \text{...} \]
These expansions allow us to work with complex exponential expressions in a simpler, polynomial form.
Trigonometric Function Expansion
Trigonometric functions like \( \text{cos}(x) \) can also be expanded into Maclaurin series.
The Maclaurin series for \( \text{cos}(x) \) is:
\[ \text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \text{...} \]
This series helps us approximate the cosine function for small values of \( x \).
Additionally, by substituting \( 2x \) in place of \( x \) and then multiplying by 2, we get the expansion for \( 2 \text{cos}(2x) \):
\[ 2 \text{cos}(2x) = 2 - 2x^2 + \frac{2(2x)^4}{4!} - \text{...} \]
These expansions provide easy ways to manipulate and understand trigonometric functions as series.
The Maclaurin series for \( \text{cos}(x) \) is:
\[ \text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \text{...} \]
This series helps us approximate the cosine function for small values of \( x \).
Additionally, by substituting \( 2x \) in place of \( x \) and then multiplying by 2, we get the expansion for \( 2 \text{cos}(2x) \):
\[ 2 \text{cos}(2x) = 2 - 2x^2 + \frac{2(2x)^4}{4!} - \text{...} \]
These expansions provide easy ways to manipulate and understand trigonometric functions as series.
Limit Evaluation
To find the limit, we look at how the expressions behave as \( x \) approaches zero, using their Maclaurin series expansions.
In our problem, the expanded series help us simplify the expression:
\[ \frac{\frac{e^x + e^{-x} - 2}{2}}{2 \text{cos}(2x) - 2} \]
By substituting the series expansions:
**Numerator:**
\[ \frac{e^x + e^{-x} - 2}{2} = \frac{x^2}{2} + \text{...} \]
**Denominator:**
\[ 2 \text{cos}(2x) - 2 = -2x^2 + \text{...} \]
Combining these, we simplify:
\[ \frac{\frac{x^2}{2} + \text{...}}{-2x^2 + \text{...}} = -\frac{1}{4} \]
So, the limit as \( x \) approaches zero is
\[ \text{Lim}_{x \rightarrow 0} \frac{e^x + e^{-x} - 2}{2 \text{cos}(2x) - 2} = -\frac{1}{4}. \]
This method provides a systematic way to handle complex expressions using series and find their limits.
In our problem, the expanded series help us simplify the expression:
\[ \frac{\frac{e^x + e^{-x} - 2}{2}}{2 \text{cos}(2x) - 2} \]
By substituting the series expansions:
**Numerator:**
\[ \frac{e^x + e^{-x} - 2}{2} = \frac{x^2}{2} + \text{...} \]
**Denominator:**
\[ 2 \text{cos}(2x) - 2 = -2x^2 + \text{...} \]
Combining these, we simplify:
\[ \frac{\frac{x^2}{2} + \text{...}}{-2x^2 + \text{...}} = -\frac{1}{4} \]
So, the limit as \( x \) approaches zero is
\[ \text{Lim}_{x \rightarrow 0} \frac{e^x + e^{-x} - 2}{2 \text{cos}(2x) - 2} = -\frac{1}{4}. \]
This method provides a systematic way to handle complex expressions using series and find their limits.
