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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 $$k=\frac{\sqrt{2-\sqrt{3}}}{2}?$$

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What's $cn(x,k)$? Also, you seem to have unbalanced parentheses in the denominator of $f(x)$. –  Ilmari Karonen Jul 24 '11 at 22:52
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What's $\mathrm{cn}(x,k)$? The elliptic function usuallty denoted $\mathrm{cn}(x,k)$. –  GEdgar Jul 24 '11 at 23:07
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@Ilmari: Jacobi elliptic functions. –  t.b. Jul 24 '11 at 23:11
    
Thanks, I guess the elliptic-functions tag should've clued me in. –  Ilmari Karonen Jul 24 '11 at 23:15
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2 Answers 2

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 $k^2 < 1$).

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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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Or 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... –  anon Jul 25 '11 at 2:53
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