Chapter 30: Problem 19
Find the first three nonzero terms of the Taylor expansion for the given function and given value of \(a\) $$e^{x} \sin x \quad\left(a=\frac{\pi}{2}\right)$$
Short Answer
Expert verified
The first three nonzero terms are: \(e^{\frac{\pi}{2}}\), \(e^{\frac{\pi}{2}}(x-\frac{\pi}{2})\), and \(\frac{-2e^{\frac{\pi}{2}}}{6}(x-\frac{\pi}{2})^3\).
Step by step solution
01
Understanding Taylor Expansion
The Taylor series expansion of a function \( f(x) \) around a point \( a \) is given by \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \] To find the first three nonzero terms, we need to compute derivatives of \( e^x \sin(x) \) and evaluate them at \( a = \frac{\pi}{2} \).
02
Compute the Function Value at a
We start by evaluating the function at \( a = \frac{\pi}{2} \). \[ f\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cdot 1 = e^{\frac{\pi}{2}} \] This is our first nonzero term.
03
Compute the First Derivative
The first derivative of \( e^x \sin(x) \) is found using the product rule: \[ f'(x) = \frac{d}{dx}[e^x \sin(x)] = e^x \sin(x) + e^x \cos(x) \] Evaluate at \( a = \frac{\pi}{2} \): \[ f'\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) + e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cdot 1 + e^{\frac{\pi}{2}} \cdot 0 = e^{\frac{\pi}{2}} \] The second term is \( f'(a)(x-a) = e^{\frac{\pi}{2}} \left(x - \frac{\pi}{2}\right) \).
04
Compute the Second Derivative
The second derivative of \( e^x \sin(x) \) is:\[ f''(x) = \frac{d}{dx}[e^x \sin(x) + e^x \cos(x)] = e^x \sin(x) + e^x \cos(x) + e^x \cos(x) - e^x \sin(x) = 2e^x \cos(x) \] Evaluate at \( a = \frac{\pi}{2} \):\[ f''\left(\frac{\pi}{2}\right) = 2e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) = 2e^{\frac{\pi}{2}} \cdot 0 = 0 \] There is no contribution from the second derivative as it evaluates to zero.
05
Compute the Third Derivative
Compute the third derivative:\[ f'''(x) = \frac{d}{dx}[2e^x \cos(x)] = 2e^x \cos(x) - 2e^x \sin(x) \] Evaluate at \( a = \frac{\pi}{2} \): \[ f'''\left(\frac{\pi}{2}\right) = 2e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) - 2e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) = 0 - 2e^{\frac{\pi}{2}} = -2e^{\frac{\pi}{2}} \] Thus, the third term is \( \frac{f'''(a)}{3!}(x-a)^3 = \frac{-2e^{\frac{\pi}{2}}}{6}(x-\frac{\pi}{2})^3 \).
06
Taylor Expansion Terms
The first three nonzero terms of the Taylor expansion are:- First term: \( e^{\frac{\pi}{2}} \)- Second term: \( e^{\frac{\pi}{2}} (x - \frac{\pi}{2}) \)- Third term: \( \frac{-2e^{\frac{\pi}{2}}}{6} (x - \frac{\pi}{2})^3 \)
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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 powerful tool in calculus that allows us to approximate functions with a polynomial. This approximation is centered around a specific point, known as "a". The idea is to express a function as an infinite sum of terms calculated from the values of its derivatives at a certain point. For a function \( f(x) \), the Taylor series centered at a point \( a \) is written as follows:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]Commonly, this is used to find an approximate value of the function near \( a \). It's especially useful when dealing with functions like exponential and trigonometric functions that are otherwise difficult to evaluate precisely. In this case, we used the Taylor series to approximate \( e^{x} \sin(x) \) around \( a = \frac{\pi}{2} \).
- Start with the function value at \( a \).
- Add terms involving the function's derivatives, each multiplying a higher power of \( (x-a) \).
- The accuracy of this approximation depends on how many terms you include.
Derivatives
Derivatives are fundamental in constructing a Taylor series. They provide us with the information needed to describe how a function is changing at a certain point. The derivative of a function indicates the rate of change or the slope of the function's graph at that specific point. Calculating derivatives accurately is key:
- First Derivative: This gives the tangent line's slope at the point, capturing the function's immediate rate of change.
- Second Derivative: Provides information on the function's curvature, helping to sketch how the function is bending at the point.
- Higher Order Derivatives: These keep refining our approximation by adjusting higher power series terms.
Calculus
Calculus is an area of mathematics that deals with continuous change and is divided mainly into two branches: differential calculus and integral calculus. Here, we focus on differential calculus as it relates to finding Taylor series expansions.
Calculus allows us to compute derivatives, which are pivotal in understanding a function's behavior. We can thus break down complex functions into polynomial-like terms for easier approximation. Calculus simplifies how we handle functions involving exponentials and trigonometric expressions, as seen in \( e^{x} \sin(x) \), by:
Calculus allows us to compute derivatives, which are pivotal in understanding a function's behavior. We can thus break down complex functions into polynomial-like terms for easier approximation. Calculus simplifies how we handle functions involving exponentials and trigonometric expressions, as seen in \( e^{x} \sin(x) \), by:
- Using the product rule for differentiation, enabling us to find the derivatives required for our Taylor approximation.
- Applying calculus principles to reduce complex expressions into simpler, more manageable polynomial terms.
- Offering valuable insight into how these terms interact around particular points using Taylor expansions.
Function Evaluation
Function evaluation involves calculating the value of a function at a specific point. It's a core part of finding Taylor series terms, as each term requires evaluating derivatives at the point \( a \).
For the function \( e^{x} \sin(x) \), evaluation at \( a = \frac{\pi}{2} \) involves both direct computation and derivatives. Here's the detailed process:
For the function \( e^{x} \sin(x) \), evaluation at \( a = \frac{\pi}{2} \) involves both direct computation and derivatives. Here's the detailed process:
- Initial Function Value: We start by directly evaluating the function at \( a \). For our function, this resulted in \( e^{\frac{\pi}{2}} \).
- Subsequent Derivatives: Each derivative calculated must then be evaluated at \( a \), which feeds into the Taylor series terms. For instance, the first derivative at \( a \) contributed directly to the linear term of our expansion.
- Simplification: Often, derivative evaluations lead to simplified terms, such as zeros, which help identify non-contributing terms in the series.