Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The expression you say the original one reduces to is incorrect. – DonAntonio Jul 29 '12 at 3:27

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 $0$.

share|cite|improve this answer
I think $$g(z)=\frac{4\sin^4z/2}{z^4}=\frac{1}{4}\left(\frac{\sin z/2}{z/2}\right)^4$$ – DonAntonio Jul 29 '12 at 2:42
@DonAntonio: Yes, at first I assumed the OP's formula was correct, but noticed that there there was a typo in the formula given/ – André Nicolas Jul 29 '12 at 2:46

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

share|cite|improve this answer

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...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.