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I am a physics guy, not a mathematician.

I am trying to solve a differential equation with variable coefficients. I am using the power series technique to solve it.

My question is: Is $\frac{\sin(x)}{x}$ an analytic function everywhere i.e. at all $x$?

ChatGPT says it is not analytic at $x=0$, while Adkins+Davidson's book says it is analytic everywhere, i.e. at all $x$.

(By the way, I will NOT believe ChatGPT.)

Various textbooks state different versions of the definition of an analytic function.

Which is the accepted definition?

Please guide me.

Gonçalo
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    In practice, removable singularities are implicitly always considered removed, so $\frac{\sin x}{x}$ is an entire function. But if one wants to be picky, the expression $\frac{\sin x}{x}$ is undefined for $x = 0$, so the expression can only give an analytic function on $\mathbb{C}\setminus {0}$. – Dermot Craddock Jun 12 '25 at 11:22
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    Under any sane definition of "analytic", for a function $f$ to be analytic on an open set $U$ it must be the case that $f$ defined on $U$. The real ambiguity is what exactly you mean by "the function $\sin x/x$". Strictly speaking, this is not a function but rather just a formula. Mathematicians are usually a bit more pedantic than physicists in this regard. To define a function, you need to specify a domain and codomain. – Joe Jun 12 '25 at 11:25
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    Just to stress the point: if we are being technical, the function $g(x)=\frac xx$ is not entire as it is undefined at $x=0$. However, if you simply define $g(0)=1$ then the extended function is constant. – lulu Jun 12 '25 at 11:28
  • clearly not about mathematics. – John B Jun 12 '25 at 11:54
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    I find the wording in some of the comments concerning. Phrases like "In practice," or "if we are being technical," might suggest that there are certain situations or scenarios in which it is acceptable to say "$f(x)=\frac{\sin(x)}{x}$ is analytic at $x=0$" or "$g(x)=\frac{x}{x}$ is an entire function." It is not. You have to deal with the removable singularity in some way, either define the function's value at that point, or mention that the singularity is removed in a canonical way, or define the function in a completely different way that does not have those problems. – Reinhard Meier Jun 12 '25 at 11:59
  • The power series $f(x)=\sum\limits_{n=0}^\infty (-1)^n x^{2n}/(2n+1)! = 1 - \frac{x^2}{6}+\frac{x^4}{120}-\cdots$ is analytic. You can also say $f(x)=\frac{\sin(x)}{x}$ when $x \not = 0$ and $f(0)=1$ and that is still analytic. – Henry Jun 12 '25 at 12:04
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    @JohnB: I am struggling to see how a question about analytic functions could not be about mathematics. – Joe Jun 12 '25 at 12:27
  • By the way, mathematicians usually distinguish between "real analytic" and "complex analytic". – Gerry Myerson Jun 12 '25 at 13:43
  • This question should be re-opened. That said, $\sin(z)/z$ has a removeable singularity at $z=0$ and once removed, renders the function entire. – Mark Viola Jun 12 '25 at 14:44
  • BTW, this function is often called sinc. – Dan Jun 12 '25 at 15:44
  • The recipe $\frac{\sin x}{x}$ defines an analytic function on $\Bbb{C} \smallsetminus {0}$. It has an analytic extension to $\Bbb{C}$ which is usually denoted "$\mathrm{sinc}$" where the removable singularity at $x = 0$ is assigned the value $1$, the continuous extension (i.e., the value of the limit of $\sin(x)/x$ as $x \rightarrow 0$ in $\Bbb{C}$) that happens to be an analytic extension. So two functions are under discussion (they differ in their domains). – Eric Towers Jun 12 '25 at 16:22
  • It is clearly a question about mathematics. – copper.hat Jun 12 '25 at 17:07
  • @Joe I don't see that a question involving chatgp is related to mathematics. – John B Jun 12 '25 at 18:41
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    @ReinhardMeier I don't find the comments concerning and I do think that there situations where it is acceptable to say that $f(x)=\frac{\sin(x)}{x}$ is analytic at $x=0$. For example, in the field of formal Laurent series (or convergent Laurent series or meromorphic functions, your preference), $\sin(x)$ and $x$ are non-zero elements and so we can divide $\sin(x)$ by $x$, this is perfectly well-defined and asking if its analytic at $x=0$ is meaningful. Similarly, in the field of rational functions $\Bbb C(x)$, $\frac{x}{x}$ unambiguosly equals $1$. – Lukas Heger Jun 23 '25 at 20:39
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    There's a difference between rigor and pedantry. And if we're being pedantic, a formula such as $\frac{\sin(x)}{x}$ is not a sufficient description for a function, because we need to specify domain and codomain. – Lukas Heger Jun 23 '25 at 20:41

2 Answers2

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Yes it is.

We should first agree that a function is analytic if and only if for every ${\displaystyle x_{0}}$ in its domain, its Taylor series about ${\displaystyle x_{0}}$ converges to the function in some neighborhood of ${\displaystyle x_{0}}$.

If we are talking about a function $f\, (x)$ such as:

$$f \, (x) : \begin{cases}\mathbb{R}_* \rightarrow C \subseteq \mathbb{R} \\x \mapsto \frac{\sin x}{x}\end{cases}$$

Then we can see very simply that it is analytical, since the only point that could cause issue (i.e. $x=0$) does not belong to the domain.

If we talk about $\mathrm{sinc} \, (x) $ such as:

$$\mathrm{sinc} \, (x) : \begin{cases}\mathbb{R} \rightarrow C \subseteq \mathbb{R} \\x \mapsto \begin{cases}\frac{\sin x}{x}, & x \neq 0 \\ 1, & x= 0\end{cases}\end{cases}$$

We only need to find its Taylor series for $x = 0$. Here it is:

$$\mathrm{sinc} \, (x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n+1)!}$$

Balfar
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Strictly speaking, $f(x) = \frac{\sin(x)}{x}$ is undefined at $x = 0$, so not analytic.

However, if you fill in the removable discontinuity, i.e.,

$$f(x) = \operatorname{sinc}(x) = \begin{cases}1 && x = 0\\\frac{\sin(x)}{x} && x \ne 0\end{cases}$$

Then $f'(x) = \frac{x \cos(x) - \sin(x)}{x^2}$ also has a removable discontinuity at $x = 0$, where $f'(x) = 0$ (which can be shown with L'Hôpital's Rule). And this process can be continued to define higher-order derivatives at zero.

Or to make things easier, just take the familiar Maclaurin series $\sin x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}$ and divide by $x$, giving:

$$f(x) = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k+1)!}$$

Special-casing the $k=0$ term to avoid having $0^0$,

$$f(x) = 1 + \sum_{k=1}^\infty \frac{(-1)^k x^{2k}}{(2k+1)!}$$

Since a well-defined Taylor series exists, $f$ is analytic.

And yes, this applies $\forall x \in \mathbb{C}$. It can be shown that

$$\operatorname{sinc}(x+iy)=\frac{ x\sin(x)\cosh(y) + y\cos(x)\sinh(y) + i(x\cos(x)\sinh(y) - y\sin(x)\cosh(y))}{x^2+y^2}$$

Which reduces to $\frac{x\sin(x)}{x^2} = \frac{\sin(x)}{x}$ when $y = 0$.

Dan
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