3
$\begingroup$

I am trying to find the type of singularity the function $f(z) = \exp\left(\frac{(\cos z-1)^2}{z^4}\right).$ has at $z = 0$.

The expression reduces to $\exp\left(\frac{\sin^4(z/2)}{z^4/2}\right).$ The function has removable singularity at $z=0$ as $\lim_{z\rightarrow 0}zf(z)=0.$

Am I right?

  • 0
    The expression you say the original one reduces to is incorrect.2012-07-29

3 Answers 3

3

Let our function be $\exp(g(z))$, where $g(z)=\frac{4\sin^4(z/2)}{z^4}$. Note that $g(z)=\frac{1}{4}\left(\frac{\sin(z/2)}{z/2}\right)^4.$ But $\lim_{z\to 0}\frac{\sin(z/2)}{z/2}=1$ It follows that $\lim_{z\to 0} \,\exp(g(z))=\exp(1/4).$ So although there is a singularity at $0$, it is indeed removable by defining a new function which agrees with our function everywhere (except $0$ of course) and is $\exp(1/4)$ at $0$.

Remark: The limit of $zf(z)$ is indeed $0$, but that has no real connection with the fact that $f(z)$ has a removable singularity at $0$.

  • 0
    @DonAntonio: Yes, at first I assumed the OP's formula was correct, but noticed that there there was a typo in the formula given/2012-07-29
1

I don't understand:

$\cos z=\cos\left(\frac{z}{2}+\frac{z}{2}\right)=\cos^2\frac{z}{2}-\sin^2\frac{z}{2}=1-2\sin^2\frac{z}{2}\Longrightarrow (\cos z-1)^2=4\sin^4\frac{z}{2}\Longrightarrow$ $\Longrightarrow e^\frac{(\cos z-1)^2}{z^4}=e^{\frac{\sin^4z/2}{\left(z^2/2\right)^2}}=e^{g(z)}\,,\,\,\text{with}$

$g(z)=\frac{4\sin^4z/2}{z^4}=\frac{1}{4}\left(\frac{\sin z/2}{z/2}\right)^4$

Thus, $e^{g(z)}\xrightarrow [z\to 0]{} e^{1/4}\Longrightarrow \lim_{z\to 0}\,\,z\,e^{g(z)}=0$

1

Using the power series representation: $\cos(z)-1 = -z^2/2 +z^4/4! - \cdots$ $(\cos(z)-1)^2 = z^4/4 -z^6/4! +z^8/320 - \cdots$ $(\cos(z)-1)^2/z^4 = 1/4 -z^2/4! +z^4/320 - \cdots$ $\lim_{z \to 0} ((\cos(z)-1)^2/z^4) = 1/4 $ $\lim_{z \to 0} f(z) = \exp(1/4) $ Perhaps it is also needed to do statements on the convergence of the occuring series, but I think it is sufficient to say one time, that $\cos(x)$ is entire...