# Showing the bessel function is entire

#### nacho

##### Active member
Hi,
I actually posted this problem a while back on a separate forums:
Showing the bessel function is entire

And got a response, but still cannot seem to figure out how to do this question

Given a ratio test can be used, we must first define a p(z) and q(z)

so we can see if the sum for $$\frac{p(z)}{q(z)}$$ converges,
but the denominator doesn't even have a complex variable in terms of $z$, so how is this possible?

Any help is much appreciated, this one is wrinkling my brain!

for z = 0 it converges.

for z =/ 0, we convert it to the form

$\frac{a_n+1}{a_n}$ , or do we use j?
thanks!

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#### chisigma

##### Well-known member
Hi,
I actually posted this problem a while back on a separate forums:
Showing the bessel function is entire

And got a response, but still cannot seem to figure out how to do this question

Given a ratio test can be used, we must first define a p(z) and q(z)

so we can see if the sum for $$\frac{p(z)}{q(z)}$$ converges,
but the denominator doesn't even have a complex variable in terms of $z$, so how is this possible?

Any help is much appreciated, this one is wrinkling my brain!

for z = 0 it converges.

for z =/ 0, we convert it to the form

$\frac{a_n+1}{a_n}$ , or do we use j?
thanks!
Let's start with the demonstration that $J_{n} (z)$ is an entire function, i.e. its Taylor series converges for any value of z. We have...

$\displaystyle J_{n} (z) = (\frac{z}{2})^{n}\ \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!\ (n+j)!}\ (\frac{z}{2})^{2 j}\ (1)$

... and if we compare the series in (1) with the series...

$\displaystyle e^{- (\frac{z}{2})^{2}} = \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!}\ (\frac{z}{2})^{2 j}\ (2)$

... that converges for any z we discover that also the series in (1) converges for any z, i.e. $J_{n} (z)$ is an entire function...

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
Let's start with the demonstration that $J_{n} (z)$ is an entire function, i.e. its Taylor series converges for any value of z. We have...

$\displaystyle J_{n} (z) = (\frac{z}{2})^{n}\ \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!\ (n+j)!}\ (\frac{z}{2})^{2 j}\ (1)$

... and if we compare the series in (1) with the series...

$\displaystyle e^{- (\frac{z}{2})^{2}} = \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!}\ (\frac{z}{2})^{2 j}\ (2)$

... that converges for any z we discover that also the series in (1) converges for any z, i.e. $J_{n} (z)$ is an entire function...

Kind regards

$\chi$ $\sigma$
What is the name of the second series you are comparing it to? (if it has one)
I see the main method you used here was to break a part the bessel function, so it came in the form of a known convergent series which you previously knew about.

is this a more viable approach/quicker/simpler than what the user in the previous thread opted for? which was to use a ratio test to show its convergence?