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Is (sinx)/x analytic at x = 0?

  1. May 18, 2010 #1
    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:

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

    [tex]\frac{sin(x)}{x} = \sum_{n=0}^{\infty}\frac{(-1)^{n}\(x^{2n+1})}{x(2n+1)!}=[/tex].... 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!
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 18, 2010 #2


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  4. May 18, 2010 #3
    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???
  5. May 19, 2010 #4
    The point [itex]0[/itex] is an ordinary point of [itex]y''+\frac{\sin(z)}{z}y=0[/itex] since [itex]\frac{\sin(z)}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots[/itex] is analytic at [itex]0[/itex] however I'm not so sure it is for [itex]zy''+\sin(z)y=0[/itex] since division by z is valid only when [itex]z\neq 0[/itex].
    Last edited: May 19, 2010
  6. May 20, 2010 #5
    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)??
  7. May 20, 2010 #6
    Define two functions like this:

    f_1(x) = \left\{\begin{array}{ll}
    \frac{\sin(x)}{x}, & x\neq 0\\
    1, & x = 0\\

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

    Now you want to prove that [itex]f_1(x) = f_2(x)[/itex] for all [itex]x[/itex]. Simply prove it for [itex]x=0[/itex] and [itex]x\neq 0[/itex] separately and it's done.
  8. May 20, 2010 #7
    Strictly speaking the equation

    xy''(x) + \sin(x)y(x) = 0

    is equivalent to the equation pair

    y''(x) + \frac{\sin(x)}{x} y(x) = 0, \quad x\neq 0 \\
    y''(0),y(0)\quad \textrm{can be anything} \\

    If you demand that [itex]y'', y[/itex] must be continuous, by assumption, then the ambiguity gets dealt with. Then the original equation implies

    y''(0) + y(0)=0

    Actually I think that the continuity of [itex]y''[/itex] will follow anyway even if it is not assumed.
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