Finding Maclaurin expansion and interval of convergence

In summary, the Maclaurin series for ##f(x) = \log (\cos x)## is ##-\frac{x^2}{2} + \frac{x^4}{12} - \frac{x^6}{720} + O (x^8)## and the interval of convergence can be found by considering the interval of convergence for ##\log (1+y)##. This is a valid expansion as verified by Mathematica. Alternatively, one can find the series for ##f'(x) = -\tan x## and integrate term by term to find the series for ##f(x)##.
  • #1
Mr Davis 97
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Homework Statement


Find the Maclaurin series and inverval of convergence for ##f(x) = \log (\cos x)##

Homework Equations

The Attempt at a Solution


I used the fact that ##\log (\cos x) = \log (1+ (\cos x - 1))##, and the standard expansions for ##\cos x## and ##\log (x+1)## to get that ##\displaystyle \log (\cos x) = -\frac{x^2}{2} + \frac{x^4}{12} - \frac{x^6}{720} + O (x^8)##. How do I find the interval of convergence for this? Also, how do I know that this is the valid expansion?
 
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  • #2
Mr Davis 97 said:
How do I find the interval of convergence for this
Try working from the outside in: what is the interval for log( 1+y)?
 
  • #3
Mr Davis 97 said:

Homework Statement


Find the Maclaurin series and inverval of convergence for ##f(x) = \log (\cos x)##

Homework Equations

The Attempt at a Solution


I used the fact that ##\log (\cos x) = \log (1+ (\cos x - 1))##, and the standard expansions for ##\cos x## and ##\log (x+1)## to get that ##\displaystyle \log (\cos x) = -\frac{x^2}{2} + \frac{x^4}{12} - \frac{x^6}{720} + O (x^8)##. How do I find the interval of convergence for this? Also, how do I know that this is the valid expansion?
According to Mathematica, the series is
$$-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17 x^8}{2520}-\frac{31 x^{10}}{14175}+O\left(x^{11}\right)$$
It might be easier to find the series for ##f'(x) = -\tan x## and then integrate term by term.
 

1. What is a Maclaurin expansion?

A Maclaurin expansion is a mathematical series representation of a function around the point x = 0. It is similar to a Taylor series expansion, but is specifically centered at x = 0. It is named after the Scottish mathematician Colin Maclaurin.

2. How do you find the Maclaurin expansion of a function?

To find the Maclaurin expansion of a function, you can use the formula:
f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...
where f'(0), f''(0), etc. represent the derivatives of the function evaluated at x = 0. Alternatively, you can use known Maclaurin expansions for common functions, such as sine, cosine, and exponential functions.

3. What is the interval of convergence for a Maclaurin series?

The interval of convergence for a Maclaurin series is the range of values for which the series converges. It can be determined by using the ratio test, which compares the absolute value of the (n+1)th term to the nth term of the series. The series will converge within a radius of convergence, which may include the endpoints or may be infinite.

4. How do you determine the radius of convergence for a Maclaurin series?

The radius of convergence for a Maclaurin series can be determined by using the ratio test, which compares the absolute value of the (n+1)th term to the nth term of the series. The radius of convergence is the distance from the center point (x = 0) to the nearest point where the series diverges. It can be represented by the inequality |x| < R, where R is the radius of convergence.

5. Why is it important to find the Maclaurin expansion and interval of convergence for a function?

Knowing the Maclaurin expansion and interval of convergence for a function allows us to approximate the function with polynomials and determine the range of values for which the approximation is valid. This can be useful in many areas of science, including physics, engineering, and economics, as well as in solving differential equations and finding solutions to complex problems. Additionally, finding the Maclaurin expansion and interval of convergence can help us understand the behavior of a function and make predictions about its values at different points.

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