# Point of a function

1. Jan 13, 2010

### nhrock3

i need to prove that the in point z=0
its loren series has endless values(i dont know the proper english term)
and their presented solution
http://i48.tinypic.com/1zb7401.jpg
(click on the photo to enlarge)

i thought of just breaking the function into a series
and it will have endless members of z in negative power.

i know that they are trying to prove that we get different limit values
so there is no limit which is a sign of that type of point

but there is no limits in their solution
i cant see limits and their results
what is it?

2. Jan 13, 2010

### HallsofIvy

Staff Emeritus
You appear (your picture is difficult to read even enlarged) to be saying that
$$z_n= \frac{1}{ln(n\pi)}$$.

Now, what do you mean by its Laurent series? A Laurent series is a power series, involving negative powers, of a function around a point where the function is not analytic. Here you have "$z_n$", a sequence of points. What function are you talking about?

3. Jan 13, 2010

### nhrock3

there are 3 types of points in dont know the engish term for them
first type of point: if the limit is constant or 0
second type of point:if the limit goes to infinity
third type is: if there is no limit
my function is
f(z)=cos(exp(1/z))
i need to prove that z=0 is the third type of a point

their solution says something about limit different from different sides

but there is no limit expressions

i thought of solving it by breaking it into a series arounf point z=0
and then i will get endless members with z in negative power
which is also a sign for the third type

but i want to understand how they say limits from both sides
without presenting them?

4. Jan 14, 2010

### HallsofIvy

Staff Emeritus
Okay, a function is "analytic" at a point if it has a Taylor's series at that point and, for some neighborhood of the point, the series converges to the value of the function.

If a function is not analytic at a point, we can still write it as a "Laurent" series, with negative exponents. If the series has only a finite number of negative exponents with non-zero coefficient, the the function has a "pole" of that point (with the order of the pole given by highest negative exponent).

If the Laurent series requires an infinite number of negative exponents, then the function has an "essential singularity" at the point. You want to show that cos(e^{1/z}) has an essential singularity at z= 0.

Notice that if f(z) has a "pole of order n", then $f(z)= a_{-n}z^{-n}+ a_{-n+1}z^{-n+1}$$+ \cdot\cdot\cdot+ a_{-1}z^{-1}+ a_0+ a_1z+ a_2z^2+ \cdot\cdot\cdot$. That is, there is a "lowest power", -n. If you multiply both sides of that by $z^n$ you get $z^nf(z)= a_{-n}+ a_{-n+1}z+ \cdot\cdot\cdot+ a_{-1}z_{n-1}+ a_0z^n+$$a_1z^{n+1}+ a_2z^{n+2}+ \cdot\cdot\cdot$ which has no negative exponents. Then $\lim_{z\to 0} z^n f(z)= a_{-n}$, a non-zero constant. If there is NO n such that that is true, f has an essential singularity at z=0.

You want to show that $\lim_{z\to 0}z^n cos(e^{1/z})$ does NOT exist no matter what n is.