Equivalent form of sinc function?

Question

My trig is a bit rusty, so here goes:

I believe that the sinc function, i.e. $\frac{\sin(\theta)}{\theta}$, where $\theta = \pi x = \frac{\sin(\theta)}{N\tan(\frac{\theta}{N})}$ for sufficiently large values of $N$. I've tested this for values of $\theta$ in the range $(0, 10\pi]$ and things look promising. For the life of me I can't seem to do the trig/algebra necessary to prove that these two forms are equivalent:

$$\frac{\sin(\theta)}{\theta} \overset ?= \frac{\sin(\theta)}{N\tan(\frac{\theta}{N})}$$

I'd be very grateful if someone could shed some light upon my quandary.

Also, the secondary form (i.e. with tangent), does it go by a specific name?

TIA, Karl

Answer 1

Let us divide by $\sin(\theta)$ (assuming it's not $0$) then what I suppose you want to say is that

$$\tan(\theta/N)\sim \theta/N$$

asymptotically as $N\rightarrow\infty$.

By no means can they actually be equal, but for very large $N$, the values agree approximately.

That is indeed true since

$$\tan(x)=\frac{\sin(x)}{\cos(x)}$$

Now note that $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}\pm\cdots\sim x$$ as $x\rightarrow 0$ and

$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}\pm\cdots\sim 1$$ as $x\rightarrow 0$.

Therefore $\tan(x)\sim x$ as $x\rightarrow 0$.

Setting $x=\theta/N$ gives exactly your observation.

Answer 2

For $x \ll 1, \tan x \approx x+\frac 13x^3+\dots$, so as $N$ gets large, $N \tan(\frac \theta N) \approx \theta$ and your form is equivalent. I don't know a name for it.