Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This evening I thought of the following question that isn't related to homework, but it's a question that seems very challenging to me, and I take some interest in it.

Let's consider the following function: $$ f(x)= \left(\frac{\sin x}{x}\right)^\frac{x}{\sin x}$$ I wonder what is the first derivative (1st, 2nd, 3rd ...) such that $\lim\limits_{x\to0} f^{(n)}(x)$ is different from $0$ or $+\infty$, $-\infty$, where $f^{(n)}(x)$ is the nth derivative of $f(x)$ (if such a case is possible). I tried to use W|A, but it simply fails to work out such limits. Maybe i need the W|A Pro version.

share|improve this question
    
Gotta bite. What is 'W|A'? –  copper.hat Jul 8 '12 at 20:51
    
@copper.hat I assume Wolfram Alpha –  Argon Jul 8 '12 at 20:51
    
@copper.hat: hi. Wolfram|Alpha –  Chris's sis Jul 8 '12 at 20:52
    
Thanks. New form of abbreviation for me. –  copper.hat Jul 8 '12 at 20:52
1  
I just want to point out that this function fails to be defined in many cases. For example, for any natural number $k\in\mathbb{N}$, let $x=\frac{(3+4k)\pi}{2}$. Then we have $$f(x)=\left(\frac{\sin(x)}{x}\right)^\frac{x}{\sin(x)}= \left(\frac{-1}{x}\right)^{\frac{x}{-1}}=\left(-\frac{1}{x}\right)^{-x}$$ which is a negative number raised to a negative power. Any positive $x$ for which $\sin(x)$ is negative, or vice versa, will have the same problem. Similarly, for any integer $k\in\mathbb{Z}$, let $x=k\pi$. Then we are raising $0$ to the $\frac{x}{0}$ power, which I would say is undefined. –  Zev Chonoles Jul 8 '12 at 21:00

2 Answers 2

up vote 3 down vote accepted

The Taylor expansion is $$f(x) = 1 - \frac{x^2}{6} + O(x^4),$$ so \begin{eqnarray*} f(0) &=& 1 \\ f'(0) &=& 0 \\ f''(0) &=& -\frac{1}{3}. \end{eqnarray*}

$\def\e{\epsilon}$

Addendum: We use big O notation. Let $$\e = \frac{x}{\sin x} - 1 = \frac{x^2}{6} + O(x^4).$$ Then \begin{eqnarray*} \frac{1}{f(x)} &=& (1+\e)^{1+\e} \\ &=& (1+\e)(1+\e)^\e \\ &=& (1+\e)(1+O(\e\log(1+\e))) \\ &=& (1+\e)(1+O(\e^2)) \\ &=& 1+\e + O(\e^2), \end{eqnarray*} so $f(x) = 1-\e + O(\e^2) = 1-\frac{x^2}{6} + O(x^4)$.

share|improve this answer
1  
Isn't that just the Taylor series for $\dfrac{\sin x}{x}$? I understand the exponent goes to 1 as $x \to 0$, but is that approach valid from the naive perspective? –  Eugene Shvarts Jul 8 '12 at 21:11
    
@EugeneShvarts: The series happen to agree to order $x^2$. There is a difference at order $x^4$. –  user26872 Jul 8 '12 at 21:13
1  
@oen Ah, okay. Did you realize this intuitively, or is there a quick technique to see this? –  Eugene Shvarts Jul 8 '12 at 21:14
1  
@oen: hold on! How did you get the Taylor expansion form?? –  Chris's sis Jul 8 '12 at 21:19
1  
@EugeneShvarts: I added something about this to the answer. –  user26872 Jul 8 '12 at 21:45

First of all, note that $$ f(x)=\left(\frac{\sin(x)}{x}\right)^{\Large\frac{x}{\sin(x)}}\tag{1} $$ is an even function. This means that all the odd terms in the power series will be zero.

Using the power series for $\log(1+x)$, we get $$ \begin{align} &\log\left(\left(1-\frac16x^2+\frac{1}{120}x^4+O\left(x^6\right)\right)^{\Large1+\frac16x^2+\frac{7}{360}x^4+O\left(x^6\right)}\right)\\ &=\left(-\frac16x^2-\frac{1}{180}x^4+O\left(x^6\right)\right)\left(1+\frac16x^2+\frac{7}{360}x^4+O\left(x^6\right)\right)\\ &=-\frac16x^2-\frac{1}{30}x^4+O\left(x^6\right)\tag{2} \end{align} $$ Then we apply the power series for $e^x$ to get $$ f(x)=1-\frac16x^2-\frac{7}{360}x^4+O\left(x^6\right)\tag{3} $$ Of course, using more terms in the power series for $\dfrac{\sin(x)}{x}$ and $\dfrac{x}{\sin(x)}$, we could get more terms for $f(x)$.

To get the derivatives at $x=0$, you can just use the fact that the Taylor series near $0$ is $$ f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\tag{4} $$ to get that $f^{(n)}(0)=0$ for all odd $n$, and $$ \begin{align} f(0)&=1\\ f''(0)&=-\frac13\\ f^{(4)}(0)&=-\frac{7}{15}\\ &\text{etc.} \end{align} $$

share|improve this answer
    
thanks for your solution! You mean Maclaurin series, right? –  Chris's sis Jul 8 '12 at 21:58
3  
A Maclaurin series is a Taylor series centered at $0$. So you can call this a Maclaurin series, but it is still a Taylor series (which is more general). –  robjohn Jul 8 '12 at 22:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.