Given $f(x) = \frac{1 - \mathrm{cn}(x,k)}{{\sqrt3}(1+\mathrm{cn}(x,k)) - 1 + \mathrm{cn}(x,k)}$ what would be $\lim_{x\to 0} f(x)$ and $\lim_{x\to\infty} f(x)$ when $k=\frac{\sqrt{2-\sqrt{3}}}{2}?$
Limits of a function involving $\mathrm{cn}(x,k)$
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0Thanks, I guess the `elliptic-functions` tag should've clued me in. – 2011-07-24
2 Answers
For $x=0$, use the Taylor series: $ \mathrm{cn}(x,k) = 1 - \frac{1}{2} x^{2} + \Biggl(\frac{1}{24} + \frac{k^{2}}{6}\Biggr) x^{4} - \Biggl(\frac{1}{720} + \frac{11 k^{2}}{180} + \frac{k^{4}}{45}\Biggr) x^{6} + \operatorname{O} \bigl(x^{8}\bigr) $ to conclude that $ \lim_{x\to 0}\frac{1 - \mathrm{cn} (x,k)}{\sqrt{3}(1 + \mathrm{cn} (x,k)) - 1 + \mathrm{cn} (x,k)} = \lim_{x\to0}\left[\frac{\sqrt{3}}{12} x^{2} + \Biggl(\frac{\sqrt{3}}{72} + \frac{1}{48} - \frac{\sqrt{3} k^{2}}{36}\Biggr) x^{4} + \operatorname{O} \bigl(x^{6}\bigr)\right] = 0 $
But for $x \to \infty$ ... it looks like your expression has an essential singularity at $x=\infty$ since your denominator has a sequence of zeros that go to infinity.
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0Or you could just plug $\mathrm{cn}(0,k)\equiv 1$ straight into the expression for $x=0$ and use $f(x)$'s periodicity to prove $\lim_{x\to\infty}$ doesn't exist... – 2011-07-25
Here's a L'Hôpital-free way of doing things:
We can rearrange the expression a bit:
$f(x)=\left({\sqrt3}\frac{1+\mathrm{cn}(x,k)}{1 - \mathrm{cn}(x,k)} - 1\right)^{-1}$
use this,
$f(x)=\left({\sqrt3}\mathrm{cd}^2\left(\frac{x}{2},k\right)\mathrm{ns}^2\left(\frac{x}{2},k\right) - 1\right)^{-1}$
and transform back
$f(x)=\frac{\mathrm{sn}^2\left(\frac{x}{2},k\right)}{\sqrt3 \mathrm{cd}^2\left(\frac{x}{2},k\right)- \mathrm{sn}^2\left(\frac{x}{2},k\right)}$
and since $\mathrm{sn}(0,k)$ is $0$ while $\mathrm{cd}(0,k)$ is $1$, $f(0)=0$.
Here's another way to do
$\lim_{x\to 0} \left({\sqrt3}\frac{1+\mathrm{cn}(x,k)}{1 - \mathrm{cn}(x,k)} - 1\right)^{-1}$
Remember that $\mathrm{cn}(x,k)=\cos(\mathrm{am}(x,k))$, where $\mathrm{am}(x,k)$ (the Jacobian amplitude) is the inverse of the incomplete elliptic integral of the first kind; that is,
$\phi=\mathrm{am}(x,k) \quad\leftrightarrow\quad x=F(\phi,k)$
where
$F(\phi,k)=\int_0^\phi\frac{\mathrm dt}{\sqrt{1-k^2\sin^2t}}$
We can then determine that $\mathrm{am}(0,k)=0$, so we can perform an appropriate substitution and consider the limit
$\lim_{t\to 0} \left({\sqrt3}\frac{1+\cos\,t}{1-\cos\,t}-1\right)^{-1}$
thus,
$\lim_{t\to 0} \left(\sqrt3\,\cot^2\frac{t}{2}-1\right)^{-1}=\lim_{t\to 0} \frac{\sin^2\frac{t}{2}}{\sqrt3\,\cos^2\frac{t}{2}-\sin^2\frac{t}{2}}=0$
The procedure for the limit to infinity is similar (as $\phi\to\infty$, $F(\phi,k)\to\infty$ for $k^2 < 1$).