Is \(\frac{\sin(x)}{x}\) Analytic at \(x = 0\)?

[The Chaz]

That's basically it.
My thinking is that since it's not defined at x = 0, it is not analytic (and x = 0 is therefore a singular point).

For a further look at my work, including the original context of the question, see below:
http://www.mymathforum.com/viewtopic.php?f=22&t=14068"

Here's my first attempt at LaTex, which I have copied and pasted from some other users!

\frac{sin(x)}{x} = \sum_{n=0}^{\infty}\frac{(-1)^{n}\(x^{2n+1})}{x(2n+1)!}=... man, I give up. I can't get an "x" in the numerator! Whatever. Obviously I know how to write the power series...

I'll copy answers from this site to the other (or vice versa) as soon as they come in, as I am actively watching both sites. Thanks!

 

[Mute]

sin(x)/x has a removable singularity at x = 0, and hence is analytic.

See here: http://en.wikipedia.org/wiki/Removable_singularity

 

[The Chaz]

sin(x)/x has a removable singularity at x = 0, and hence is analytic.

See here: http://en.wikipedia.org/wiki/Removable_singularity

Thanks, Mute.
I believe you and understand the content of your post and of the wikipedia article. Interestingly (at least it's interesting to me!), my classmate and I speculated about the possibility of defining f(0) := 1.

However, given the context of the question (linear, 2nd order homog. diff eq)
xy`` + sin(x)y = 0
y`` + (sin(x)/x)y = 0
"Is the point x = 0 ordinary or singular?"

I am tempted to just say, "SINGULAR".
What do you think?

 

[jackmell]

The point 0 is an ordinary point of y''+\frac{\sin(z)}{z}y=0 since \frac{\sin(z)}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots is analytic at 0 however I'm not so sure it is for zy''+\sin(z)y=0 since division by z is valid only when z\neq 0.

 

[The Chaz]

The point 0 is an ordinary point of y''+\frac{\sin(z)}{z}y=0 since \frac{\sin(z)}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots is analytic at 0 however I'm not so sure it is for zy''+\sin(z)y=0 since division by z is valid only when z\neq 0.

When the prof explained this in class yesterday (claiming that x = 0 is ordinary and that the sinc function is analytic "at/around" x = 0), a riot almost broke out.

The reason can be identified in your post.
"sin(z)/z = (power series)". I claim that this is false and should be replaced with:
"X DIFFERENT FROM ZERO ==> sin(z)/z = (power series)"
If you can't divide by zero in the original diff eq, why would it be legal to do (an INFINITE amount of times!) in simplifying the power series?

The problem is most likely in the semantics. Finding rigorous definitions of these terms has proven to be ... rather difficult. "analytic", "at", "around", "neighborhood", etc...

Everyone agrees that for f(x) = sin(x)/x,
f(0) := 1 ==> f analytic everywhere.
(If we define f of zero to be 1, then f is analytic everywhere).
This is a true statement, but WHY?

We have a conditional statement, P ==> Q
P <=> "define f(0) := 1"
Q <=> "f analytic everywhere"

If Q (the conclusion) is true, then ( R => Q ) is true for any hypothesis 'R'.

What about the statement (~P ==> ~Q)??

 

[jostpuur]

Define two functions like this:

<br /> f_1(x) = \left\{\begin{array}{ll}<br /> \frac{\sin(x)}{x}, &amp; x\neq 0\\<br /> 1, &amp; x = 0\\<br /> \end{array}\right.<br />

<br /> f_2(x) = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots<br />

Now you want to prove that f_1(x) = f_2(x) for all x. Simply prove it for x=0 and x\neq 0 separately and it's done.

 

[jostpuur]

Strictly speaking the equation

<br /> xy&#039;&#039;(x) + \sin(x)y(x) = 0<br />

is equivalent to the equation pair

<br /> \left\{\begin{array}{l}<br /> y&#039;&#039;(x) + \frac{\sin(x)}{x} y(x) = 0, \quad x\neq 0 \\<br /> y&#039;&#039;(0),y(0)\quad \textrm{can be anything} \\<br /> \end{array}\right.<br />

If you demand that y&#039;&#039;, y must be continuous, by assumption, then the ambiguity gets dealt with. Then the original equation implies

<br /> y&#039;&#039;(0) + y(0)=0<br />

Actually I think that the continuity of y&#039;&#039; will follow anyway even if it is not assumed.