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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.

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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 – user 1618033 Jul 8 '12 at 20:52
Thanks. New form of abbreviation for me. – copper.hat Jul 8 '12 at 20:52
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
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*}


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 $f(x) = 1-\e + O(\e^2) = 1-\frac{x^2}{6} + O(x^4)$.

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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
@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
@oen: hold on! How did you get the Taylor expansion form?? – user 1618033 Jul 8 '12 at 21:19
@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 $$ \begin{align} f(0)&=1\\ f''(0)&=-\frac13\\ f^{(4)}(0)&=-\frac{7}{15}\\ &\text{etc.} \end{align} $$

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thanks for your solution! You mean Maclaurin series, right? – user 1618033 Jul 8 '12 at 21:58
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

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