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Chapter 11: Problem 1

Determine the third Taylor polynomial of the given function at \(x=0.\) $$f(x)=\sin x$$

Short Answer

Expert verified
The third Taylor polynomial of \( \sin x \) at \( x=0 \) is \( T_3(x) = x - \frac{x^3}{6} \).

Step by step solution

01

- Understand the Taylor Series Expansion

The Taylor series expansion of a function about a point is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] Here, the focus is on the function \( f(x) = \sin x \) at \( x=0 \). To find the third Taylor polynomial, we'll need the values of the function and its first three derivatives evaluated at \( x=0 \).
02

- Compute the Function and its Derivatives at \( x=0 \)

First, compute the function and its first three derivatives at \( x=0 \): \[ f(x) = \sin x \] \[ f'(x) = \cos x \] \[ f''(x) = -\sin x \] \[ f'''(x) = -\cos x \] Evaluating these at \( x=0 \) gives: \[ f(0) = \sin(0) = 0 \] \[ f'(0) = \cos(0) = 1 \] \[ f''(0) = -\sin(0) = 0 \] \[ f'''(0) = -\cos(0) = -1 \]
03

- Substitute the Values into the Taylor Series Formula

Substitute the values into the Taylor series formula up to the third derivative: \[ T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \] Using the computed values, we get: \[ T_3(x) = 0 + 1 \times x + 0 + \frac{-1}{3!}x^3 \] Simplify to obtain: \[ T_3(x) = x - \frac{x^3}{6} \]

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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 powerful tool in calculus that allows us to approximate complicated functions using an infinite sum of simpler polynomial terms. The idea is to expand a function around a specific point (often around 0, called Maclaurin Series) such that each term in the polynomial captures more detail of the function. The general formula for the Taylor series expansion about the point x=0 is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] Here, each term involves the derivatives of the function evaluated at x=0. The more terms we include, the better our polynomial approximates the original function.
derivatives
In calculus, derivatives measure how a function changes as its input changes. They are essential for finding the Taylor polynomial. For the function f(x)=\sin x, we need to find the first, second, and third derivatives to construct the Taylor polynomial up to the third degree:
  • First derivative, \(f'(x) = \cos x\), measures the rate of change of \sin x.
  • Second derivative, \(f''(x) = -\sin x\), measures the rate of change of \cos x, which is the derivative of \sin x.
  • Third derivative, \(f'''(x) = -\cos x\), measures the rate of change of \(-\sin x\), which is the derivative of \cos x.
Evaluating these at x=0 gives us the key values for constructing the Taylor polynomial:
  • \sin(0) = 0
  • \cos(0) = 1
  • -\sin(0) = 0
  • -\cos(0) = -1
trigonometric functions
Trigonometric functions like \sin x and \cos x are periodic functions representing the angles in a right-angled triangle. These functions are widely used in various fields, including physics and engineering. The function provided, \(f(x) = \sin x\), is smooth and continuous, making it an ideal candidate for Taylor series expansion.
polynomial approximation
A polynomial approximation replaces a complex function with a polynomial that is easier to handle. By using the first few terms of the Taylor series, we get a Taylor polynomial. For \(f(x) = \sin x\) around x=0, we derived the third-degree Taylor polynomial:\[ T_3(x) = x - \frac{x^3}{6} \]This polynomial captures the behavior of \sin x around x=0 quite well. Higher-order terms can be added for more accuracy at the cost of complexity.

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Most popular questions from this chapter

Find the Taylor series at \(x=0\) of the given function. Use suitable operations (differentiation, substitution, etc.) on the Taylor series at \(x=0\) of \(\frac{1}{1-x}, e^{x},\) or \(\cos x .\) These series are derived in Examples 1 and 2 and Check Your Understanding Problem 2. $$5 e^{x / 3}$$

Find the Taylor series at \(x=0\) of the given function. Use suitable operations (differentiation, substitution, etc.) on the Taylor series at \(x=0\) of \(\frac{1}{1-x}, e^{x},\) or \(\cos x .\) These series are derived in Examples 1 and 2 and Check Your Understanding Problem 2. $$\cos 3 x$$

Determine the sums of the following infinite series: $$\sum_{k=0}^{\infty}(-1)^{k} \frac{3^{k+1}}{5^{k}}$$

Show that the infinite series $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots$$ diverges. [Hint: \(\frac{1}{3}+\frac{1}{4}>\frac{1}{2} ; \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{2} ; \frac{1}{9}+\cdots+\frac{1}{16}>\frac{1}{2};\) etc.]

Determine the sums of the following geometric series when they are convergent. $$\frac{1}{3^{2}}-\frac{1}{3^{3}}+\frac{1}{3^{4}}-\frac{1}{3^{5}}+\frac{1}{3^{6}}-\cdots$$
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