-2
$\begingroup$

$$2\Delta\frac{\sin(2\pi\tilde{\nu}\Delta)}{2\pi\tilde{\nu}\Delta},$$ $\tilde{\nu}$ is wave number (1/cm), $\Delta$ some constant (no units).

Is this just a sinc function scaled by $2\Delta$, or does it have a specific name?

$\endgroup$
2
  • 1
    $\begingroup$ I mean, it's also $\sin (2 \pi \tilde{\nu} \Delta)$ scaled by $1/\pi \tilde{\nu}$. It kind of depends on whether you're viewing it as a function of $\tilde{\nu}$ or of $\Delta$. $\endgroup$ Commented Feb 6, 2023 at 21:42
  • $\begingroup$ ”no units” $\Delta$ needs to have units. How are you going to take the sine of some number of inverse centimeters? $\endgroup$
    – Ghoster
    Commented Feb 6, 2023 at 22:49

1 Answer 1

3
$\begingroup$

His name is George, please take good care of him.

If you are doing numerical computations and want a name for your computer library, sin(x)/x may have precision issues for $x≈0$ which a well-coded sinc(x) can avoid.

Naming this function "George" is a little silly, as at least one downvoter seems to have pointed out. However, if a function is included in a software library under its own name, that's a hint that there may be some computational issue which you would prefer not to discover by rolling your own version. The classic example is the hypotenuse function hypot, which has an overflow issue. Suppose you want hypot(3e200,4e200) == 5e200. The naïve computation $\sqrt{a^2+b^2}$ overflows, because 9e400 doesn't fit in a double-precision number. But the equivalent $b\sqrt{1+(a/b)^2}$ does not overflow, and also has better accuracy when $a\ll b$.

The question of whether your function has a name is a reasonable question.

$\endgroup$
4
  • 1
    $\begingroup$ You know what, I'll take it $\endgroup$
    – Lyu
    Commented Feb 7, 2023 at 0:50
  • $\begingroup$ It might help if you clarify the relevance of the answer part of your answer. Right how, it is hard to identify the signal after the noise. $\endgroup$ Commented Feb 7, 2023 at 16:57
  • 1
    $\begingroup$ @NorbertSchuch Good point. $\endgroup$
    – rob
    Commented Feb 8, 2023 at 1:25
  • $\begingroup$ Much better :) Only the thing with the downvoter is no longer up to date. $\endgroup$ Commented Feb 8, 2023 at 17:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.