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taylor series

pantboio

Member
Nov 20, 2012
45
Consider the function $$f(z)=e^{\frac{1}{1-z}}$$ It has an essential singularity at $z_0=1$ and hence it can be expanded in a Laurent series at $z_0$. But i'm interested in Taylor expansion. The function is analytic in the unit open disc at the origin, so i'm looking for $a_n$ where $f(z)=\displaystyle\sum_{n=0}^\infty a_nz^n$ for $|z|<1$. How can i find $a_n$'s?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Consider the function $$f(z)=e^{\frac{1}{1-z}}$$ It has an essential singularity at $z_0=1$ and hence it can be expanded in a Laurent series at $z_0$. But i'm interested in Taylor expansion. The function is analytic in the unit open disc at the origin, so i'm looking for $a_n$ where $f(z)=\displaystyle\sum_{n=0}^\infty a_nz^n$ for $|z|<1$. How can i find $a_n$'s?
They will be the coefficients from the Maclaurin expansion (Taylor series about \(z_0=0\) )

CB
 

chisigma

Well-known member
Feb 13, 2012
1,704
Consider the function $$f(z)=e^{\frac{1}{1-z}}$$ It has an essential singularity at $z_0=1$ and hence it can be expanded in a Laurent series at $z_0$. But i'm interested in Taylor expansion. The function is analytic in the unit open disc at the origin, so i'm looking for $a_n$ where $f(z)=\displaystyle\sum_{n=0}^\infty a_nz^n$ for $|z|<1$. How can i find $a_n$'s?
The coefficient of order n of the Taylor Series of $f(z)$ is given by...

$\displaystyle a_{n} = \frac{1}{n!}\ \frac{d^{n}}{d z^{n}} e^ {\frac{1}{1-z}}\ \text{in}\ z=0$ (1)

... so that we have ...


$\displaystyle f(z)= e^ {\frac{1}{1-z}}\ \implies a_{0}=e$

$\displaystyle f^{(1)}(z)= e^ {\frac{1}{1-z}}\ \frac{1}{(1-z)^{2}} \implies a_{1}=e$

$\displaystyle f^{(2)}(z)= e^ {\frac{1}{1-z}}\ \{\frac{1}{(1-z)^{4}} + \frac{2}{(1-z)^{3}}\ \} \implies a_{2}=e\ \frac{3}{2}$

$\displaystyle f^{(3)}(z)= e^ {\frac{1}{1-z}}\ \{\frac{6}{(1-z)^{4}} + \frac{6}{(1-z)^{5}} + \frac{1}{(1-z)^{6}}\ \} \implies a_{2}=e\ \frac{13}{6}$

Of course it is possible to proceed... may be that with little more efforts a general formula for the $a_{n}$ can be found...

Kind regards

$\chi$ $\sigma$
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
The coefficient of order n of the Taylor Series of $f(z)$ is given by...

$\displaystyle a_{n} = \frac{1}{n!}\ \frac{d^{n}}{d z^{n}} e^ {\frac{1}{1-z}}\ \text{in}\ z=0$ (1)

... so that we have ...


$\displaystyle f(x)= e^ {\frac{1}{1-z}}\ \implies a_{0}=e$

$\displaystyle f^{(1)}(x)= e^ {\frac{1}{1-z}}\ \frac{1}{(1-z)^{2}} \implies a_{1}=e$

$\displaystyle f^{(2)}(x)= e^ {\frac{1}{1-z}}\ \{\frac{1}{(1-z)^{4}} + \frac{2}{(1-z)^{3}}\ \} \implies a_{2}=e\ \frac{3}{2}$

$\displaystyle f^{(3)}(x)= e^ {\frac{1}{1-z}}\ \{\frac{1}{(1-z)^{4}} + \frac{2}{(1-z)^{3}} + \frac{4}{(1-z)^{5}} + \frac{6}{(1-z)^{3}}\ \} \implies a_{2}=e\ \frac{13}{6}$

Of course it is possible to proceed... may be that with little more efforts a general formula for the $a_{n}$ can be found...

Kind regards

$\chi$ $\sigma$
(you have a typo, you have a power of 3 where there should be a power of 6, also you are confusing the use of x and z)

That was the easy part which I am sure the OP could do for themselves. In fact this is a purely mechanisable process; Maxima gives the first nine terms:

\[f(z)= e+e\,z+\frac{3\,e\,{z}^{2}}{2}+\frac{13\,e\,{z}^{3}}{6}+\frac{73\,e\,{z}^{4}}{24}+\frac{167\,e\,{z}^{5}}{40}+\frac{4051\,e\,{z}^{6}}{720}+\frac{37633 \, e \, {z}^{7}}{5040}+\frac{43817\,e\,{z}^{8}}{4480}+...\]

Finding a general formula for \(a_n\) is the tricky part, though induction looks promising in this case.

CB
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Maxima gives the first nine terms:

\[f(z)= e+e\,z+\frac{3\,e\,{z}^{2}}{2}+\frac{13\,e\,{z}^{3}}{6}+\frac{73\,e\,{z}^{4}}{24}+\frac{167\,e\,{z}^{5}}{40}+\frac{4051\,e\,{z}^{6}}{720}+\frac{37633 \, e \, {z}^{7}}{5040}+\frac{43817\,e\,{z}^{8}}{4480}+...\]
The coefficient of $z^n$ is $\dfrac{eA_n}{n!}$, where $A_n$ is the $n$th term in the Sloane sequence A000262. That link gives a number of other contexts in which the same sequence occurs. It also gives a formula for $A_n$ in terms of a certain confluent hypergeometric function.
 

pantboio

Member
Nov 20, 2012
45
(you have a typo, you have a power of 3 where there should be a power of 6, also you are confusing the use of x and z)

That was the easy part which I am sure the OP could do for themselves. In fact this is a purely mechanisable process; Maxima gives the first nine terms:

\[f(z)= e+e\,z+\frac{3\,e\,{z}^{2}}{2}+\frac{13\,e\,{z}^{3}}{6}+\frac{73\,e\,{z}^{4}}{24}+\frac{167\,e\,{z}^{5}}{40}+\frac{4051\,e\,{z}^{6}}{720}+\frac{37633 \, e \, {z}^{7}}{5040}+\frac{43817\,e\,{z}^{8}}{4480}+...\]

Finding a general formula for \(a_n\) is the tricky part, though induction looks promising in this case.

CB
i was looking for a a general formula for $a_n$, in order to compute the limit
$$\lim_{n\rightarrow\infty}\frac{a_n}{a_{n+1}}$$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
i was looking for a a general formula for $a_n$, in order to compute the limit
$$\lim_{n\rightarrow\infty}\frac{a_n}{a_{n+1}}$$
The Sloane link gives the recurrence relation $A_n = (2n-1)A_{n-1} - (n-1)(n-2)A_{n-2}$, from which $\dfrac{A_n}{A_{n-1}} \approx n+1$. If you then put $a_n= \dfrac{eA_n}{n!}$, you get $\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}} = 1.$
 

pantboio

Member
Nov 20, 2012
45
The Sloane link gives the recurrence relation $A_n = (2n-1)A_{n-1} - (n-1)(n-2)A_{n-2}$, from which $\dfrac{A_n}{A_{n-1}} \approx n+1$. If you then put $a_n= \dfrac{eA_n}{n!}$, you get $\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}} = 1.$
unfortunately, this is the opposite of what i was hoping, since i'm looking for a function $f$, holomorphic in an open set containing the closed unit disc, with an essential singularity at $z_0$ with $|z_0|=1$, with taylor expansion $\sum a_nz^n$ in the unit open disk and such that $\lim_{n\rightarrow\infty}\frac{a_n}{a_{n+1}}\neq z_0$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
i'm looking for a function $f$, holomorphic in an open set containing the closed unit disc, with an essential singularity at $z_0$ with $|z_0|=1$, with taylor expansion $\sum a_nz^n$ in the unit open disk and such that $\lim_{n\rightarrow\infty}\frac{a_n}{a_{n+1}}\neq z_0$
Interesting! You can get an example of this in a rather trivial way with the function $\exp\Bigl(\dfrac1{1-z^2}\Bigr)$. This, being an analytic function of $z^2$, will have $a_n=0$ whenever $n$ is odd. So $\frac{a_n}{a_{n+1}}$ will be alternately undefined and 0, and $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ will not exist.

But I'm guessing that what you want is for the limit $z_0 = \displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ to exist, but for $z_0$ not to be an essential singularity. My first attempt was the function $(1-z)\exp\Bigl(\dfrac1{1-z^2}\Bigr)$. For that function, $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = -1$, but unfortunately $-1$ is an essential singularity. Similarly, the function $(1+iz-z^2-iz^3)\exp\Bigl(\dfrac1{1-z^4}\Bigr)$ has $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = i$, but $i$ is an essential singularity. In the same way, for any root of unity $z_0$ you can construct a function analytic in the unit disc, with $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = z_0$, but with an essential singularity at $z_0$.

Those examples somehow give the impression of something that is meant to happen. It makes me wonder whether instead of looking for a counterexample you should be looking for a theorem. Conjecture: If $\displaystyle f(z) = \sum_{n=0}^\infty a_nz^n$ is analytic in the open unit disc, with an essential singularity somewhere on the unit circle, and the limit $z_0 = \displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then $f(z)$ has an essential singularity at $z_0.$
 
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pantboio

Member
Nov 20, 2012
45
Interesting! You can get an example of this in a rather trivial way with the function $\exp\Bigl(\dfrac1{1-z^2}\Bigr)$. This, being an analytic function of $z^2$, will have $a_n=0$ whenever $n$ is odd. So $\frac{a_n}{a_{n+1}}$ will be alternately undefined and 0, and $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ will not exist.

But I'm guessing that what you want is for the limit $z_0 = \displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ to exist, but for $z_0$ not to be an essential singularity. My first attempt was the function $(1-z)\exp\Bigl(\dfrac1{1-z^2}\Bigr)$. For that function, $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = -1$, but unfortunately $-1$ is an essential singularity. Similarly, the function $(1+iz-z^2-iz^3)\exp\Bigl(\dfrac1{1-z^4}\Bigr)$ has $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = i$, but $i$ is an essential singularity. In the same way, for any root of unity $z_0$ you can construct a function analytic in the unit disc, with $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = z_0$, but with an essential singularity at $z_0$.

Those examples somehow give the impression of something that is meant to happen. It makes me wonder whether instead of looking for a counterexample you should be looking for a theorem. Conjecture: If $\displaystyle f(z) = \sum_{n=0}^\infty a_nz^n$ is analytic in the open unit disc, with an essential singularity somewhere on the unit circle, and the limit $z_0 = \displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then $f(z)$ has an essential singularity at $z_0.$
Thank you for all explanations and examples. What i was originally requested to prove was the following: let $f$ be analytic in the closed unit disc, except for an isolated singularity (of not specified type) on the unit circle. Then prove that $\lim\frac{a_n}{a_{n+1}}=z_0$. I started from the case $z_0$= simple pole, and i was able to prove that, and adapting the same argument i could prove the same for a pole of arbitrary order. Then i disprove the fact for $z_0$ removable. Finally i tried to prove/disprove for essential. But you gave me that counterexample, hence the fact is definitely false if $z_0$ is not a pole. Maybe my professor simply forgot to specify that the singularity should be a pole.
 

chisigma

Well-known member
Feb 13, 2012
1,704
If f(*) is analytic in $z=0$ and has a singularity in $z_{0}$ on the unit circle and its Taylor series $\displaystyle \sum_{n=0}^{\infty} a_{n}\ z^{n}$ converges for any $|z|<1$ and diverges in $z_{0}$, then for the ratio test it must be $\displaystyle \ |\frac{a_{n+1}}{a_{n}}\ z|<1$ for all $|z|<1$. Combining all these results we obtain...

$\displaystyle \lim_{n \rightarrow \infty} \lim_{z \rightarrow z_{0}\ , |z|<1} \frac{a_{n+1}}{a_{n}}\ z = 1 \implies \lim_{n \rightarrow \infty} \frac{a_{n}}{a_{n+1}} = z_{0}$ (1)

If the Taylor series converges in $z=z_{0}$ it has to be examined if the (1) is valid or not...

Kind regards

$\chi$ $\sigma$
 
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pantboio

Member
Nov 20, 2012
45
Interesting! You can get an example of this in a rather trivial way with the function $\exp\Bigl(\dfrac1{1-z^2}\Bigr)$. This, being an analytic function of $z^2$, will have $a_n=0$ whenever $n$ is odd. So $\frac{a_n}{a_{n+1}}$ will be alternately undefined and 0, and $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ will not exist.

But I'm guessing that what you want is for the limit $z_0 = \displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ to exist, but for $z_0$ not to be an essential singularity. My first attempt was the function $(1-z)\exp\Bigl(\dfrac1{1-z^2}\Bigr)$. For that function, $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = -1$, but unfortunately $-1$ is an essential singularity. Similarly, the function $(1+iz-z^2-iz^3)\exp\Bigl(\dfrac1{1-z^4}\Bigr)$ has $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = i$, but $i$ is an essential singularity. In the same way, for any root of unity $z_0$ you can construct a function analytic in the unit disc, with $\displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}} = z_0$, but with an essential singularity at $z_0$.

Those examples somehow give the impression of something that is meant to happen. It makes me wonder whether instead of looking for a counterexample you should be looking for a theorem. Conjecture: If $\displaystyle f(z) = \sum_{n=0}^\infty a_nz^n$ is analytic in the open unit disc, with an essential singularity somewhere on the unit circle, and the limit $z_0 = \displaystyle\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then $f(z)$ has an essential singularity at $z_0.$
I realized just now that $\exp\Bigl(\dfrac1{1-z^2}\Bigr)$ has two essential singularities on the unit circle, while i'm searching for a counterexample with only one essential singularity