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Chapter 10: Problem 25

Using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$e^{t} \cos t$$

Short Answer

Expert verified
The first four non-zero terms are \(1 + t + \frac{t^2}{2} - \frac{t^3}{2}\).

Step by step solution

01

Recall Taylor Series for Individual Functions

The Taylor series for the exponential function is given by: \( e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \frac{t^4}{4!} + \dots \). The Taylor series for \( \cos t \) is: \( \cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \).
02

Apply Product of Taylor Series

To find the Taylor series for \( e^t \cos t \), multiply the series for \( e^t \) and \( \cos t \). Focus on finding terms up to \( t^4 \), as those will help in identifying the first four non-zero terms after multiplication.
03

Multiply Terms for e^t and cos t

Multiply the first few terms of each series:- \(1 \times 1 = 1\)- \(1 \times -\frac{t^2}{2} + t \times 1 = t - \frac{t^2}{2}\)- \(1 \times \frac{t^4}{24} + t \times -\frac{t^2}{2} + \frac{t^2}{2} \times 1 = \frac{t^2}{2} - \frac{t^3}{2} + \frac{t^4}{24}\) Continue multiplying terms and grouping like powers of \(t\).
04

Combine Like Terms and Identify First Four Non-Zero Terms

By combining like terms, after sorting based on powers of \(t\), the expansion becomes:- Constant term: \(1\)- Linear term: \(t\)- Quadratic term: \(\frac{t^2}{2}\)- Quartic term: \(-\frac{t^3}{2}\) Hence, the first four non-zero terms are \(1 + t + \frac{t^2}{2} - \frac{t^3}{2}\).
05

Verify Terms in Series Expansion

Check each term by expanding manually to ensure no mistakes: \(1 + t - \frac{t^2}{2} - \frac{t^3}{2} + \frac{t^4}{24}\) for accuracy, ensuring alignment with desired non-zero terms.

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Key Concepts

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

Taylor expansion
Taylor expansion is a powerful mathematical tool. It allows us to express functions as infinite sums of terms calculated from the values of their derivatives at a single point. For many functions, this method makes complex calculations simpler.
In general, the Taylor series of a function \( f(t) \) around the point \( t = 0 \) is expressed as:
  • \( f(t) = f(0) + f'(0)t + \frac{f''(0)}{2!}t^2 + \frac{f'''(0)}{3!}t^3 + \cdots \)
For the purpose of the exercise, the Taylor series is used to approximate both the exponential and trigonometric functions independently. This strategy allows us to later predict the behavior of their product. Taylor series provide clear insights into how these functions behave near the origin.
Product of series
When dealing with the product of two functions, like \( e^t \) and \( \cos t \), we need to find the product of their respective Taylor series. This gives us an approximation for complex products.
Each term in the series of \( e^t \) is multiplied by each term in the series of \( \cos t \). It's essential to consider terms up to the desired power. In our example, this meant terms up to \( t^4 \).
  • Multiply the first terms to get the leading constants.
  • Combine terms with the same power of \( t \).
  • Only focus on the first few terms for practical simplification.
By doing so, we identify non-zero terms up to the fourth degree, simplifying the process with the Taylor expansion technique.
Exponential function
The exponential function \( e^t \) is one of the most significant functions in mathematics. Its Taylor series expansion is beautifully simple and provides valuable insights.
The series for \( e^t \) around 0 is:
  • \( e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots \)
Each term in this series is derived from the derivative of \( e^t \) at \( t=0 \). Since all derivatives of \( e^t \) are \( e^t \) itself, its Taylor series expansion is straightforward. This function's behavior is essential for modeling growth and decay processes in physics and finance. Understanding its expansion sets the stage for studying more complex interactions, like its product with trigonometric functions.
Trigonometric functions
Trigonometric functions like \( \cos t \), have expansive roles in modeling oscillations, waves, and rotations. Their Taylor series provide an essential tool for approximation.
For \( \cos t \), the Taylor series is:
  • \( \cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots \)
This alternating series reveals even power terms only. The series captures the periodic nature of cosine. By understanding this series, we can accurately model periodic behavior around zero. Simplifying trigonometric functions using Taylor series helps solve differential equations and other complex problems in engineering and physics.

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Most popular questions from this chapter

Find the Taylor polynomial of degree \(n\) for \(x\) near the given point \(a\) $$\frac{1}{1+x}, \quad a=2, \quad n=4$$

Decide if the statements are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(L_{1}(x)\) is the linear approximation to \(f_{1}(x)\) near \(x=0\) and \(L_{2}(x)\) is the linear approximation to \(f_{2}(x)\) near \(x=0,\) then \(L_{1}(x)+L_{2}(x)\) is the linear approximation to \(f_{1}(x)+f_{2}(x)\) near \(x=0.\)

Find \(f(42)\) for \(f(x)=\int_{1}^{g(x)}(1 / t) d t\) where $$ g(x)=\lim _{b \rightarrow-\infty} \int_{b}^{x} h(t) d t, \quad \text { and } \quad h(x)=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} $$.

Decide if the statements are true or false. Give an explanation for your answer. The linear approximation to \(f(x)\) near \(x=-1\) shows that if \(f(-1)=g(-1)\) and \(f^{\prime}(-1)

A pendulum consists of a mass, \(m,\) swinging on the end of a string of length \(l .\) With the angle between the string and the vertical represented by \(\theta,\) the motion satisfies the differential equation $$\theta^{\prime \prime}+\frac{g}{l} g \sin \theta=0$$. (a) For small swings, we can replace sin \(\theta\) by its lowest nonzero Taylor approximation. What does the differential equation become? (b) If the amplitude of the oscillation is \(\theta_{0},\) the solutions to the original differential equation are oscillations with \(^{7}\) $$\text { Period }=2 \pi \sqrt{\frac{l}{g}}\left(1+\frac{1}{16} \theta_{0}^{2}+\cdots\right)$$. The solutions to the approximate differential equation are oscillations with $$\text { Period }=2 \pi \sqrt{\frac{l}{g}}$$. If \(\theta_{0}=20^{\circ},\) by what percentage is the more accurate estimate of the period obtained using the solution to the original equation up to the \(\theta_{0}^{2}\) -term different to the approximate estimate using the solution of the approximate equation?
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