# How to derive FWHM of sinc function

So this is probably a simple question, but I am unable to get my head around it. If we have $\operatorname{sinc}(2 \pi v L)$, what is the width of that $\operatorname{sinc}$ in terms of $v$ at half the maximum point.

Normally I'd just say that the maximum is $1$, thus half point is.... $1/2$

$$\operatorname{sinc}(2\pi v L) = \frac{1}{2} \\ \frac{ \sin(2 \pi vL) } { 2 \pi v L} = \frac{1}{2} \\ \sin(2\pi vL) = \pi v L$$

EDIT: How can you analytically find this? Thanks.

• Please check my edit and make sure that I did not alter the meaning of your question. – AnonSubmitter85 Jun 24 '15 at 20:16
• All ok. Thanks. – user209848 Jun 25 '15 at 9:49

I cannot resist the pleasure of providing an approximate solution for $Lv \leq 1$. This approximation was made by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ So $$\frac{\sin(x)}x\simeq \frac{16 (\pi -x) }{5 \pi ^2-4 (\pi -x) x}$$ and for a value equal to $\frac 12$, the solution is given by $$x=\frac{\pi }{2}+\sqrt{16-(\pi -4) \pi }-4\approx 1.89477$$ while the exact solution would be $\approx 1.89549426$.

Equations which mix polynomial and trigonometric functions do not show explicit solutions and numerical methods (or sophisticated approximations) should be used.

The solution being close to $\frac{3\pi}5$, we can build the simplest Pade approximant of the left hand side and using exact values for the trigonometric function of the angle get $$x=\frac{250 \left(5+\sqrt{5}\right)+3 \pi \left(100 \sqrt{10-2 \sqrt{5}}-3 \pi \left(-55+5 \sqrt{5}+3 \sqrt{2 \left(5+\sqrt{5}\right)} \pi \right)\right)}{500 \sqrt{5+2 \sqrt{5}}+75 \left(9+\sqrt{5}\right) \pi -45 \sqrt{2 \left(5+\sqrt{5}\right)} \pi ^2}$$ which is $\approx 1.89549416$

Edit

In a more general manner, if we expand $\frac{\sin(x)}x$ as its simplest approximant at $x=\theta$, the solution of $\frac{\sin(x)}x=\frac 12$ is given by $$x=\frac{-3 \theta ^2-\left(\theta ^2-2\right) \cos (2 \theta )+\theta \sin (\theta ) \left(\theta ^2+8 \cos (\theta )\right)-2}{\left(\theta ^2-2\right) \sin (\theta )-3 \theta +2 \sin (2 \theta )+2 \theta \cos (\theta )-\theta \cos (2 \theta )}$$ $\theta=\frac{29 \pi}{48}$ would have been a very good choice since the values of its trigonometric functions are known (with many radicals).

Using it, the result would have been $\approx 1.89549427$

• Just to clarify for those looking for the width of a sinc function, the value of 1.89... is the half width at half maximum of sin(x)/x. The title of the original post asked for the full width at half maximum, which would be twice this value (~3.79). In addition, the sinc function is often defined in terms of sin(pi x)/(pi x), as is the case in the original post. In this case, the FWHM is less by a factor of pi. – DanHickstein Mar 25 '16 at 15:51

I don't think there is a closed form solution, so you would be better off finding it numerically. However, if you must, you can estimate it using a Taylor series:

$$\frac{\sin(\pi L v)}{\pi L v } = a \\ \Rightarrow \pi L v = \arcsin( a \pi L v ) \\ \Rightarrow \pi L v \approx a \pi L v + \frac{1}{6}(a \pi L v)^3 + \frac{3}{40}(a \pi L v)^5 + \frac{5}{112}(a \pi L v)^7 + \cdots \\ \Rightarrow \cdots + \frac{5}{112}a^7( \pi L v)^6 + \frac{3}{40}a^5(\pi L v)^4 + \frac{1}{6}a^3( \pi L v)^2 + (a-1) = 0$$

You can find the roots of the above polynomial to solve for $v$. All but two of them will be complex and those are the values you want. However, the Taylor series for $\arcsin(x)$ quickly diverges once $\vert x \vert$ moves away from zero and this error will lead to inaccurate estimates.

• Thanks, It's pretty easy to find numerically, but I was really hoping for a closed form solution. I suppose numerical will have to suffice. – user209848 Jun 26 '15 at 20:32