0

I have a question for you. I was asked to find the Maclaurin series of $\ln(\sin x/x)$ and to evaluate its convergence. After finding the power series, I've applied the ratio test and I've found that the series converges for $|-x^2/6+x^4/120|<1$. When I solve the system of inequalities, I find that it is actually impossible, because the first inequality is verified for every real value of x, while the second one has no solution. How can it be? Where do I go wrong?

enter image description here

Carmen Medugno
  • 1
  • 1
    Welcome to Maths SX! Which system of inequalities? – Bernard Jul 20 '21 at 14:08
  • The inequality |-x^2/6 + x^4/120| < 1 generates two inequalities which form a system
    1. -x^2/6 + x^4/120 > -1
    2. -x^2/6 + x^4/120 < 1
     – Carmen Medugno Jul 20 '21 at 14:11
  • Welcome to MSE. Please, learn to use MathJax to type the math in your questions. Also, share your work so we can help you. For example, how did you arrive to $|-x^2/6+x^4/120|<1$? – jjagmath Jul 20 '21 at 14:12
  • Here's how to ask a good question – jjagmath Jul 20 '21 at 14:13
  • I'm sorry, I'm new here and I'm still learning how to use MathJax. Anyway, I got this result by finding the power series of sin(x) and putting it into ln(sinx/x), and then finding the power series of ln (1+y), where y=-x^2/6 + x^4/120 (I was asked to stop at the fourth order). – Carmen Medugno Jul 20 '21 at 14:17
  • Hello :) I'm not sure, if your series is right. Checking by wolfram alpha yields $-\frac{1}{6} x^2-\frac{1}{180}x^4$. This is not an answer. For all $x\in \mathbb R$ the series $-1+\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k}$ converges to $\frac{\sin(x)}{x}-1$. For $|y|<1$ the series $\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{k}y^{k}$ converges to $\ln(1+y)$. So, i guess we need to find the $x\in \mathbb R$ with $|\frac{\sin(x)}{x}-1|<1$. – Jochen Jul 20 '21 at 14:40
  • Thank you so much! :) – Carmen Medugno Jul 20 '21 at 14:50
  • Given that if $f(x)$ has a power series convergent in $|x|<R$ then so does it's derivative, and vice-versa, why not look at the derivative first: it is $\cot x-\frac{1}{x}$. It has power series expansion, see here https://math.stackexchange.com/questions/1952632/laurent-series-for-cot-z . I think (but may be quite wrong) that this converges for $|x|<\frac{\pi}{2}$. – ancient mathematician Jul 20 '21 at 15:59
  • Perhaps @Claude Leibovici might be able to give a complete answer. – ancient mathematician Jul 20 '21 at 16:05

2 Answers2

2

The firsts terms of the function $\log(\sin x / x)$ are $-\frac{1}{6}x^2 -\frac{1}{180} x^4 + \cdots$, so you may want to check your calculations.

That mistake apart, knowing only a finite numbers of terms of the series you can't determine the radius of convergence. To apply the ratio test, for example, you need to calculate $\lim\limits_{n\to \infty} \frac{a_{n+1}}{a_n}$, so it's of no use to know that $a_2 = -\frac{1}{6}$ and $a_4 = -\frac{1}{180}$.

You'll need some other way to determine the convergence radius.

jjagmath
  • 18,214
  • What if I get the general term of the series by putting $-x^2/6 - x^4/4 $ into the power series of $ln(1+y)$, where $y=-x^2/6 - x^4/4$? If I proceed in this way, I get an expression for the general term of the series, which is $[((-1)^n(-1)) (-x^2/6-x^4/4)^n]/n$ – Carmen Medugno Jul 20 '21 at 14:46
0

First of all, $f(x) = \ln \left( \frac{\sin x}{x} \right)$ technically is undefined at $x = 0$ because directly plugging in $x = 0$ gives us $\ln \left( \frac{0}{0} \right)$, so we can't give a Maclaurin series without specifying that we're talking about an extension $\bar{f}$ which satisfies $\bar{f}(x) = f(x)$ for $x \neq 0$ and $\bar{f}(0) := 0$. By abuse of notation, we're going to write $f(x)$ when we really mean $\bar{f}$ throughout the rest of this discussion.

Second of all, the radius of convergence of any power series on $\Bbb{C}$ is the distance to the nearest singularity in the complex plane. Since the function $\operatorname{sinc(x)}$ defined by $\operatorname{sinc}(x) := \frac{\sin x}{x}$ for $x \neq 0$ and $\operatorname{sinc}(0) := 1$ is entire, this means the only singularities of our $f(x)$ must happen at the zeroes of $\operatorname{sinc}(x)$, which are at nonzero integer multiples of $\pi$. Our Maclaurin series, whatever its coefficients, therefore has a radius of convergence of $\pi$, the distance from the origin to the nearest singularity of $\operatorname{sinc}(x)$.

Rivers McForge
  • 5,787