0
$\begingroup$

I was playing with series with Wolfram Alpha online calculator when I've considered the plot of the function $$f(x)=\frac{(\sin x)^{\cos x}}{(\cos x)^{\sin x}},\tag{1}$$ over real numbers in $[0,\frac{\pi}{4}]$. And after I wondered next question.

Question. What about the asymptotic behaviour of $$\sum_{1\leq n\leq x}\frac{(\sin \left(\frac{\pi}{4n}\right))^{\cos \left(\frac{\pi}{4n}\right)}}{(\cos \left(\frac{\pi}{4n}\right))^{\sin \left(\frac{\pi}{4n}\right)}}\tag{2}$$ as $x\to\infty$? I am looking (as curiosity) a big/little oh statement, or well an asymptotic equivalence $\sim$ as $x\to\infty$. Since my series seems tedious and artificious only is required, if you want, that you explain a good strategy to get such asymptotic behaviour. Many thanks.

I know the Taylor series for the sine and cosine functions, and the Euler–Maclaurin formula, see this Wikipedia, but even this seems complicated to get. Finally one can write our sequence as

$$\sum_{1\leq n\leq x}\frac{\operatorname{exp}\left(\cos \left(\frac{\pi}{4n}\right)\log(\sin \left(\frac{\pi}{4n}\right))\right)}{\operatorname{exp}\left(\sin \left(\frac{\pi}{4n}\right)\log (\cos \left(\frac{\pi}{4n}\right))\right)}.\tag{3}$$

$\endgroup$
1
  • $\begingroup$ Everyone in the spirit or my question the feedback of previous answers is enough to me. Many thanks to those users that provided an answer. $\endgroup$ – user243301 Feb 5 '18 at 20:12
0
$\begingroup$

For small $t>0,$

$$\cos t < (\cos t)^{\sin t} < 1.$$

Thus as $t\to 0^+,$ $(\cos t)^{\sin t}\to 1.$ This takes care of the denominators of you terms.

As for the numerators, it seems likely that $(\sin t)^{\cos t}\sim t$ as $t\to 0^+.$ The intuition here being that $\sin t \sim t$ and $\cos t \sim 1.$ That's not a proof of course. However, $(\sin t)^{\cos t}\sim t$ does hold; see if you can prove it.

It follows that the terms of your series are asymptotic to $\dfrac{\pi}{4 n}.$ From this it follows that the sum of your terms from $1$ to $N$ is asymptotic to

$$\sum_{n=1}^{N} \frac{\pi}{4 n} = \frac{\pi}{4}\sum_{n=1}^{N} \frac{1}{n}\sim \frac{\pi}{4}\,\ln N.$$

$\endgroup$
1
  • $\begingroup$ Many thansk to you and Robert for these details. I know that using L'Hôpital as $t\to0^{+}$ our limit is $\frac{0}{0}$ thus we compute the limit $\frac{\frac{d}{dt} (\sin t)^{\cos t}}{1}$ and from here working (I think that is required one more time the same trick combined to logarithmic differentiation) we prove that our limit is $1$, and thus the asymptotic equivalence that you've proposed to me as exercise. $\endgroup$ – user243301 Feb 5 '18 at 19:55
0
$\begingroup$

According to Maple,

$$ f(x) = \frac{\sin(x)^{\cos(x)}}{\cos(x)^{\sin(x)}} = x + \left(-\frac{1}{6} - \frac{\ln(x)}{2}\right) x^3 + O(x^4) $$

so $$\sum_{n=1}^N f\left(\frac{\pi}{4n}\right) = \sum_{n=1}^N \frac{\pi}{4n} + O(1) = \frac{\pi}{4} \log(N) + O(1)$$

$\endgroup$
1
  • $\begingroup$ Many thanks, now I am going to read your answer. I am going to wait some days if there are more feedback/answers, but yours seems very clear. $\endgroup$ – user243301 Feb 5 '18 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy