Problem 13
Question
Arrive at the series for \(\cos x\) by differentiating the Maclaurin series for \(\sin x\).
Step-by-Step Solution
Verified Answer
The Maclaurin series for \(\cos x\) is indeed gotten by differentiating the Maclaurin series for \(\sin x\) term by term. The series we obtain from doing this is \[1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\].
1Step 1: Write the Maclaurin Series for \(\sin x\)
The Maclaurin series for \(\sin x\) is given by \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\). This is an infinite alternate series with \(x^n/n!\) as term.
2Step 2: Differentiate the Maclaurin Series for \(\sin x\)
Differentiation term by term gives the series of \(\cos x\). Following the power rule for differentiation (the derivative of \(x^n\) is \(n*x^{n-1}\)), we obtain the following series: \[1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\]
3Step 3: Identify the Derived Series with \(\cos x\)
This series obtained by term-by-term differentiation of \(\sin x\)'s Maclaurin series is the Maclaurin series of \(\cos x\). Hence we showed that differentiating the Maclaurin series for \(sin x\) gives the Maclaurin series for \(\cos x\).
Key Concepts
DifferentiationTrigonometric FunctionsPower Series
Differentiation
Differentiation is a fundamental concept in calculus, centered around determining the rate at which a function changes at any given point. To differentiate a function means to find its derivative, which provides a new function describing this rate of change. For polynomial functions, the power rule is a straightforward tool: if you have a term like \(x^n\), its derivative is \(n \cdot x^{n-1}\). This rule is exceptionally useful when working with power series.
When applying differentiation to a series, such as the Maclaurin series for \(\sin x\), each term is individually differentiated. In this context, the series is \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \]
When applying differentiation to a series, such as the Maclaurin series for \(\sin x\), each term is individually differentiated. In this context, the series is \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \]
- Apply the derivative term by term.
- Adjust the powers and coefficients according to the power rule.
Trigonometric Functions
Trigonometric functions such as \(\sin x\) and \(\cos x\) are periodic functions commonly used in various fields like physics and engineering. They relate the angles of a triangle to the lengths of its sides in a right-angled triangle. More generally, they are fundamental in describing wave patterns, oscillations, and other phenomena that repeat cyclically.
The function \(\sin x\) corresponds to the vertical coordinate of a point on the unit circle as the angle \(x\) sweeps around the circle. In contrast, \(\cos x\) gives the horizontal coordinate. They are related through their derivatives:
The function \(\sin x\) corresponds to the vertical coordinate of a point on the unit circle as the angle \(x\) sweeps around the circle. In contrast, \(\cos x\) gives the horizontal coordinate. They are related through their derivatives:
- The derivative of \(\sin x\) is \(\cos x\).
- The derivative of \(\cos x\) is \(-\sin x\).
Power Series
A power series is an infinite series of the form \(a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\), where the coefficients \(a_n\) are constants. Power series are powerful tools in mathematical analysis, allowing complex functions to be expressed as infinite sums of polynomials. They are especially useful for approximating functions and solving differential equations.
The Maclaurin series is a specific type of power series centered at zero. For trigonometric functions like \(\sin x\) and \(\cos x\), the Maclaurin series provide a polynomial approximation of the function:
The Maclaurin series is a specific type of power series centered at zero. For trigonometric functions like \(\sin x\) and \(\cos x\), the Maclaurin series provide a polynomial approximation of the function:
- For \(\sin x\): \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\)
- For \(\cos x\): \(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots\)
Other exercises in this chapter
Problem 12
The interval of convergence of a power series is \((-2,5]\). (a) What is the radius of convergence? (b) What is the center of the series?
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Determine whether the series converges or diverges. It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and Root Tests. \(\sum_{k
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(a) Find the \(n\) th degree Taylor polynomial for \(f(x)=\frac{1}{1-x}\) centered at \(x=0\). (b) How many nonzero terms of the polynomial in part (a) must be
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A power series is of the form \(\sum_{k=0}^{\infty} a_{k}(x+3)^{k} .\) Which of the intervals given below could conceivably be the interval of convergence of th
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