When I try to calculate the function $f(x)=1-\operatorname{sinc}x$ for small values of $x$ I get large relative errors due to catastrophic cancellation. I want an accurate way to calculate $f(x)$ without using a series expansion or an iterative method. (For my purposes, accurate means always within about 5 ULPs).

I can easily calculate $\operatorname{sinc}x$. The problem is calculating $1-\operatorname{sinc}x$. I am looking for a way that does not simply compute a truncated series like this:

$$ \begin{aligned} f(x) &= 1-\operatorname{sinc}x \\ &= 1-\frac{\sin x}{x} \\ &= 1-\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-...\right) \\ &= \frac{x^2}{3!}-\frac{x^4}{5!}+... \\ \end{aligned} $$

Notice the problematic $1-1$ in the second-last line above.

I tried to reformulate $f(x)$ in terms of standard functions by applying trigonometric identities. After finding the common denominator $x$, there are no identities that bridge the $x$ and $\sin x$ terms in the numerator.

I also tried an approach similar to Kahan's $\operatorname{log1p}()$ and $\operatorname{expm1}()$ which rely on matching the loss rates between numerator and denominator, but I still get large errors. Maybe that only works for functions involving $\log$ and $\exp$.

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    $\begingroup$ What if you define $ f(x) = \frac{{x^2 }}{{3!}} - \frac{{x^4 }}{{5!}} + \cdots $ and use it for small $x$ without any preceeding steps involving $1-1$? You can also try a Padé approximation of this convergent series. $\endgroup$
    – Gary
    Apr 22, 2021 at 13:49
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    $\begingroup$ Presumably, the reason you want to avoid series is that you think it takes too long. Even if your language supports $\sin(x)$ it can take a long time compared to a multiply. I would just set a threshold and compute $\sinc(x)$ if $x$ greater than the threshold, then use the series when it is less. If the threshold is small, it won't take many terms of the series to get the accuracy you want. $\endgroup$ Apr 22, 2021 at 13:52
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    $\begingroup$ Instead of the direct series, you can use a Padé approximation, which often converges faster. It is the ratio of polynomials. The fact that sinc doesn't have poles makes me suspect that it won't help much. $\endgroup$ Apr 22, 2021 at 14:09
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    $\begingroup$ @Gary Good points also, but I was hoping for a clever identity or perhaps an error "balancing trick" like in Kahan. $\endgroup$ Apr 22, 2021 at 14:10
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    $\begingroup$ Disappointing. It looks just like 1-cos(x) = x^2/2! - x^4/4! + x^6/6! - ... which does have a nice formula 2*(sin(x/2)^2). Is 1-cos(x) just lucky, and 1-sinc(x) is unlucky? $\endgroup$
    – Don Hatch
    Jun 8, 2022 at 7:17

1 Answer 1


As other contributors, $[2n+2,2n]$ Padé approximants are probably the best approximation.

For example, the simplest $$f(x)= 1-\frac{\sin x}{x} \sim\frac{x^2(420 -11 x^2)}{60 \left(x^2+42\right)}$$ gives an error of $1.10\times 10^{-8}$ for $x=\frac \pi 6$ and $1.70\times 10^{-12}$ for $x=\frac \pi{24} $.

  • $\begingroup$ Claude Leibovici That works very well for my purposes. Thanks all! RossMillikan, Gary $\endgroup$ Apr 24, 2021 at 11:49

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