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 is approximately \(1 - \frac{e}{2}x^2 \).
Step by step solution
01
Write down the function and recall the Maclaurin series
Consider the function: \[ f(x) = e^{1 - \sqrt{1-x^{2}}} \]Recall the Maclaurin series expansion for a function \( f(x) \) is given by \[ f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + ... \].
02
Calculate \(f(0)\)
Substitute \( x = 0 \) into the function:\[ f(0) = e^{1 - \sqrt{1-0^2}} = e^{1 - 1} = e^{0} = 1 \] Thus, \( f(0) = 1 \).
03
Find the first derivative \(f'(x)\)
Apply the chain rule to differentiate: \[ f'(x) = e^{1 - \sqrt{1-x^{2}}} \cdot \left( -\frac{x}{\sqrt{1 - x^2}} \right) \] Evaluate at \( x = 0 \): \[ f'(0) = e^{1 - \sqrt{1-0^2}} \cdot \left( -\frac{0}{\sqrt{1 - 0^2}} \right) = 0 \] Thus, \( f'(0) = 0 \).
04
Calculate the second derivative \(f''(x)\)
Differentiate \( f'(x) \) again using the product rule and substitution where necessary: \[ f''(x) = \frac{d}{dx} \left( e^{1 - \sqrt{1-x^{2}}} \cdot \left( -\frac{x}{\sqrt{1 - x^2}} \right) \right) \] This requires careful application of the product and chain rules, but for simplicity, find \( f''(x) \) and then evaluate at \( x = 0 \): \[ f''(0) = -e^{1} = -e \]
05
Assemble the Maclaurin series
Collect the terms calculated for the Maclaurin series: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + ... \] Substitute the known values: \[ f(x) = 1 + 0 \cdot x + \frac{-e}{2}x^2 \] Thus, the Maclaurin series up to the second term is: \[ f(x) \approx 1 - \frac{e}{2}x^2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
A Taylor series is a way to represent a function as an infinite sum of terms based on the function's derivatives evaluated at a single point, typically around zero (Maclaurin series) or any other point. This concept is instrumental in approximating complex functions using polynomials, making calculations easier.
For a function \( f(x) \), the Taylor series expansion around a point \( a \) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]
You can remember the steps involved:
For a function \( f(x) \), the Taylor series expansion around a point \( a \) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]
You can remember the steps involved:
- First, determine the point \( a \).
- Then, compute the function's value at \( a \), noted as \( f(a) \).
- Next, find the derivatives of the function (first, second, third, etc.).
- Evaluate each derivative at the point \( a \).
- Finally, use the evaluated derivatives to create the series.
Derivative Calculus
Derivative calculus is a branch of mathematics that deals with the rate of change of functions. It helps to find the slopes of curves, optimize functions, and solve real-life problems such as finding acceleration from velocity.
When calculating a derivative, you're essentially finding how much a function changes when you change its input slightly. Mathematically, the first derivative of a function \( f(x) \) is given by:
\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
The second derivative, noted as \( f''(x) \), and further higher-order derivatives, can be found by differentiating the function repeatedly.
When calculating a derivative, you're essentially finding how much a function changes when you change its input slightly. Mathematically, the first derivative of a function \( f(x) \) is given by:
\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
The second derivative, noted as \( f''(x) \), and further higher-order derivatives, can be found by differentiating the function repeatedly.
- For a given function \( f(x) \), the first derivative \( f'(x) \) describes its slope.
- The second derivative \( f''(x) \) provides information about the concavity or curvature of the function.
- Higher-order derivatives, such as \( f'''(x) \), can indicate inflection points and more complex behavior.
Exponential Function
The exponential function, denoted as \( e^x \), is one of the most important functions in mathematics. It is unique because its derivative is the same function. In other words, the rate of change of \( e^x \) is \( e^x \) itself, making it invaluable in calculus.
Mathematically, the exponential function is defined as:
\[ e^x = \sum_{{n=0}}^{\infty} \frac{x^n}{n!} \]
This shows that \( e^x \) can be expressed as an infinite series, much like a Taylor series. The exponential function has several key properties:
In the context of the Maclaurin series, understanding the exponential function helps simplify and approximate more complex functions through expansion.
Mathematically, the exponential function is defined as:
\[ e^x = \sum_{{n=0}}^{\infty} \frac{x^n}{n!} \]
This shows that \( e^x \) can be expressed as an infinite series, much like a Taylor series. The exponential function has several key properties:
- It grows rapidly as \( x \) increases.
- The function is always positive, regardless of the value of \( x \).
- Multiplying two exponential functions together follows the rule \( e^a \cdot e^b = e^{a+b} \).
In the context of the Maclaurin series, understanding the exponential function helps simplify and approximate more complex functions through expansion.