The zero crossings of the unnormalized sinc are at non-zero integer multiples of π, while zero crossings of the normalized sinc occur at non-zero integers.
The local maxima and minima of the unnormalized sinc correspond to its intersections with the cosine function. That is, sin(ξ)/ξ = cos(ξ) for all points ξ where the derivative of sin(x)/x is zero and thus a local extremum is reached. This follows from the derivative of the sinc function: d d x sinc ( x ) = { cos ( x ) − sinc ( x ) x , x ≠ 0 0 , x = 0 . {\displaystyle {\frac {d}{dx}}\operatorname {sinc} (x)={\begin{cases}{\dfrac {\cos(x)-\operatorname {sinc} (x)}{x}},&x\neq 0\\0,&x=0\end{cases}}.}
The first few terms of the infinite series for the x coordinate of the n-th extremum with positive x coordinate are x n = q − q − 1 − 2 3 q − 3 − 13 15 q − 5 − 146 105 q − 7 − ⋯ , {\displaystyle x_{n}=q-q^{-1}-{\frac {2}{3}}q^{-3}-{\frac {13}{15}}q^{-5}-{\frac {146}{105}}q^{-7}-\cdots ,} where q = ( n + 1 2 ) π , {\displaystyle q=\left(n+{\frac {1}{2}}\right)\pi ,} and where odd n lead to a local minimum, and even n to a local maximum. Because of symmetry around the y axis, there exist extrema with x coordinates −xn. In addition, there is an absolute maximum at ξ0 = (0, 1).
The normalized sinc function has a simple representation as the infinite product: sin ( π x ) π x = ∏ n = 1 ∞ ( 1 − x 2 n 2 ) {\displaystyle {\frac {\sin(\pi x)}{\pi x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}}}\right)}
and is related to the gamma function Γ(x) through Euler's reflection formula: sin ( π x ) π x = 1 Γ ( 1 + x ) Γ ( 1 − x ) . {\displaystyle {\frac {\sin(\pi x)}{\pi x}}={\frac {1}{\Gamma (1+x)\Gamma (1-x)}}.}
Euler discovered9 that sin ( x ) x = ∏ n = 1 ∞ cos ( x 2 n ) , {\displaystyle {\frac {\sin(x)}{x}}=\prod _{n=1}^{\infty }\cos \left({\frac {x}{2^{n}}}\right),} and because of the product-to-sum identity10
∏ n = 1 k cos ( x 2 n ) = 1 2 k − 1 ∑ n = 1 2 k − 1 cos ( n − 1 / 2 2 k − 1 x ) , ∀ k ≥ 1 , {\displaystyle \prod _{n=1}^{k}\cos \left({\frac {x}{2^{n}}}\right)={\frac {1}{2^{k-1}}}\sum _{n=1}^{2^{k-1}}\cos \left({\frac {n-1/2}{2^{k-1}}}x\right),\quad \forall k\geq 1,} Euler's product can be recast as a sum sin ( x ) x = lim N → ∞ 1 N ∑ n = 1 N cos ( n − 1 / 2 N x ) . {\displaystyle {\frac {\sin(x)}{x}}=\lim _{N\to \infty }{\frac {1}{N}}\sum _{n=1}^{N}\cos \left({\frac {n-1/2}{N}}x\right).}
The continuous Fourier transform of the normalized sinc (to ordinary frequency) is rect(f): ∫ − ∞ ∞ sinc ( t ) e − i 2 π f t d t = rect ( f ) , {\displaystyle \int _{-\infty }^{\infty }\operatorname {sinc} (t)\,e^{-i2\pi ft}\,dt=\operatorname {rect} (f),} where the rectangular function is 1 for argument between −1/2 and 1/2, and zero otherwise. This corresponds to the fact that the sinc filter is the ideal (brick-wall, meaning rectangular frequency response) low-pass filter.
This Fourier integral, including the special case ∫ − ∞ ∞ sin ( π x ) π x d x = rect ( 0 ) = 1 {\displaystyle \int _{-\infty }^{\infty }{\frac {\sin(\pi x)}{\pi x}}\,dx=\operatorname {rect} (0)=1} is an improper integral (see Dirichlet integral) and not a convergent Lebesgue integral, as ∫ − ∞ ∞ | sin ( π x ) π x | d x = + ∞ . {\displaystyle \int _{-\infty }^{\infty }\left|{\frac {\sin(\pi x)}{\pi x}}\right|\,dx=+\infty .}
The normalized sinc function has properties that make it ideal in relationship to interpolation of sampled bandlimited functions:
Other properties of the two sinc functions include:
The normalized sinc function can be used as a nascent delta function, meaning that the following weak limit holds:
lim a → 0 sin ( π x a ) π x = lim a → 0 1 a sinc ( x a ) = δ ( x ) . {\displaystyle \lim _{a\to 0}{\frac {\sin \left({\frac {\pi x}{a}}\right)}{\pi x}}=\lim _{a\to 0}{\frac {1}{a}}\operatorname {sinc} \left({\frac {x}{a}}\right)=\delta (x).}
This is not an ordinary limit, since the left side does not converge. Rather, it means that
lim a → 0 ∫ − ∞ ∞ 1 a sinc ( x a ) φ ( x ) d x = φ ( 0 ) {\displaystyle \lim _{a\to 0}\int _{-\infty }^{\infty }{\frac {1}{a}}\operatorname {sinc} \left({\frac {x}{a}}\right)\varphi (x)\,dx=\varphi (0)}
for every Schwartz function, as can be seen from the Fourier inversion theorem. In the above expression, as a → 0, the number of oscillations per unit length of the sinc function approaches infinity. Nevertheless, the expression always oscillates inside an envelope of ±1/πx, regardless of the value of a.
This complicates the informal picture of δ(x) as being zero for all x except at the point x = 0, and illustrates the problem of thinking of the delta function as a function rather than as a distribution. A similar situation is found in the Gibbs phenomenon.
We can also make an immediate connection with the standard Dirac representation of δ ( x ) {\displaystyle \delta (x)} by writing b = 1 / a {\displaystyle b=1/a} and
lim b → ∞ sin ( b π x ) π x = lim b → ∞ 1 2 π ∫ − b π b π e i k x d k = 1 2 π ∫ − ∞ ∞ e i k x d k = δ ( x ) , {\displaystyle \lim _{b\to \infty }{\frac {\sin \left(b\pi x\right)}{\pi x}}=\lim _{b\to \infty }{\frac {1}{2\pi }}\int _{-b\pi }^{b\pi }e^{ikx}dk={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{ikx}dk=\delta (x),}
which makes clear the recovery of the delta as an infinite bandwidth limit of the integral.
All sums in this section refer to the unnormalized sinc function.
The sum of sinc(n) over integer n from 1 to ∞ equals π − 1/2:
∑ n = 1 ∞ sinc ( n ) = sinc ( 1 ) + sinc ( 2 ) + sinc ( 3 ) + sinc ( 4 ) + ⋯ = π − 1 2 . {\displaystyle \sum _{n=1}^{\infty }\operatorname {sinc} (n)=\operatorname {sinc} (1)+\operatorname {sinc} (2)+\operatorname {sinc} (3)+\operatorname {sinc} (4)+\cdots ={\frac {\pi -1}{2}}.}
The sum of the squares also equals π − 1/2:1112
∑ n = 1 ∞ sinc 2 ( n ) = sinc 2 ( 1 ) + sinc 2 ( 2 ) + sinc 2 ( 3 ) + sinc 2 ( 4 ) + ⋯ = π − 1 2 . {\displaystyle \sum _{n=1}^{\infty }\operatorname {sinc} ^{2}(n)=\operatorname {sinc} ^{2}(1)+\operatorname {sinc} ^{2}(2)+\operatorname {sinc} ^{2}(3)+\operatorname {sinc} ^{2}(4)+\cdots ={\frac {\pi -1}{2}}.}
When the signs of the addends alternate and begin with +, the sum equals 1/2: ∑ n = 1 ∞ ( − 1 ) n + 1 sinc ( n ) = sinc ( 1 ) − sinc ( 2 ) + sinc ( 3 ) − sinc ( 4 ) + ⋯ = 1 2 . {\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}\,\operatorname {sinc} (n)=\operatorname {sinc} (1)-\operatorname {sinc} (2)+\operatorname {sinc} (3)-\operatorname {sinc} (4)+\cdots ={\frac {1}{2}}.}
The alternating sums of the squares and cubes also equal 1/2:13 ∑ n = 1 ∞ ( − 1 ) n + 1 sinc 2 ( n ) = sinc 2 ( 1 ) − sinc 2 ( 2 ) + sinc 2 ( 3 ) − sinc 2 ( 4 ) + ⋯ = 1 2 , {\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}\,\operatorname {sinc} ^{2}(n)=\operatorname {sinc} ^{2}(1)-\operatorname {sinc} ^{2}(2)+\operatorname {sinc} ^{2}(3)-\operatorname {sinc} ^{2}(4)+\cdots ={\frac {1}{2}},}
∑ n = 1 ∞ ( − 1 ) n + 1 sinc 3 ( n ) = sinc 3 ( 1 ) − sinc 3 ( 2 ) + sinc 3 ( 3 ) − sinc 3 ( 4 ) + ⋯ = 1 2 . {\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}\,\operatorname {sinc} ^{3}(n)=\operatorname {sinc} ^{3}(1)-\operatorname {sinc} ^{3}(2)+\operatorname {sinc} ^{3}(3)-\operatorname {sinc} ^{3}(4)+\cdots ={\frac {1}{2}}.}
The Taylor series of the unnormalized sinc function can be obtained from that of the sine (which also yields its value of 1 at x = 0): sin x x = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n + 1 ) ! = 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + ⋯ {\displaystyle {\frac {\sin x}{x}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n+1)!}}=1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+\cdots }
The series converges for all x. The normalized version follows easily: sin π x π x = 1 − π 2 x 2 3 ! + π 4 x 4 5 ! − π 6 x 6 7 ! + ⋯ {\displaystyle {\frac {\sin \pi x}{\pi x}}=1-{\frac {\pi ^{2}x^{2}}{3!}}+{\frac {\pi ^{4}x^{4}}{5!}}-{\frac {\pi ^{6}x^{6}}{7!}}+\cdots }
Euler famously compared this series to the expansion of the infinite product form to solve the Basel problem.
The product of 1-D sinc functions readily provides a multivariate sinc function for the square Cartesian grid (lattice): sincC(x, y) = sinc(x) sinc(y), whose Fourier transform is the indicator function of a square in the frequency space (i.e., the brick wall defined in 2-D space). The sinc function for a non-Cartesian lattice (e.g., hexagonal lattice) is a function whose Fourier transform is the indicator function of the Brillouin zone of that lattice. For example, the sinc function for the hexagonal lattice is a function whose Fourier transform is the indicator function of the unit hexagon in the frequency space. For a non-Cartesian lattice this function can not be obtained by a simple tensor product. However, the explicit formula for the sinc function for the hexagonal, body-centered cubic, face-centered cubic and other higher-dimensional lattices can be explicitly derived14 using the geometric properties of Brillouin zones and their connection to zonotopes.
For example, a hexagonal lattice can be generated by the (integer) linear span of the vectors u 1 = [ 1 2 3 2 ] and u 2 = [ 1 2 − 3 2 ] . {\displaystyle \mathbf {u} _{1}={\begin{bmatrix}{\frac {1}{2}}\\{\frac {\sqrt {3}}{2}}\end{bmatrix}}\quad {\text{and}}\quad \mathbf {u} _{2}={\begin{bmatrix}{\frac {1}{2}}\\-{\frac {\sqrt {3}}{2}}\end{bmatrix}}.}
Denoting ξ 1 = 2 3 u 1 , ξ 2 = 2 3 u 2 , ξ 3 = − 2 3 ( u 1 + u 2 ) , x = [ x y ] , {\displaystyle {\boldsymbol {\xi }}_{1}={\tfrac {2}{3}}\mathbf {u} _{1},\quad {\boldsymbol {\xi }}_{2}={\tfrac {2}{3}}\mathbf {u} _{2},\quad {\boldsymbol {\xi }}_{3}=-{\tfrac {2}{3}}(\mathbf {u} _{1}+\mathbf {u} _{2}),\quad \mathbf {x} ={\begin{bmatrix}x\\y\end{bmatrix}},} one can derive15 the sinc function for this hexagonal lattice as sinc H ( x ) = 1 3 ( cos ( π ξ 1 ⋅ x ) sinc ( ξ 2 ⋅ x ) sinc ( ξ 3 ⋅ x ) + cos ( π ξ 2 ⋅ x ) sinc ( ξ 3 ⋅ x ) sinc ( ξ 1 ⋅ x ) + cos ( π ξ 3 ⋅ x ) sinc ( ξ 1 ⋅ x ) sinc ( ξ 2 ⋅ x ) ) . {\displaystyle {\begin{aligned}\operatorname {sinc} _{\text{H}}(\mathbf {x} )={\tfrac {1}{3}}{\big (}&\cos \left(\pi {\boldsymbol {\xi }}_{1}\cdot \mathbf {x} \right)\operatorname {sinc} \left({\boldsymbol {\xi }}_{2}\cdot \mathbf {x} \right)\operatorname {sinc} \left({\boldsymbol {\xi }}_{3}\cdot \mathbf {x} \right)\\&{}+\cos \left(\pi {\boldsymbol {\xi }}_{2}\cdot \mathbf {x} \right)\operatorname {sinc} \left({\boldsymbol {\xi }}_{3}\cdot \mathbf {x} \right)\operatorname {sinc} \left({\boldsymbol {\xi }}_{1}\cdot \mathbf {x} \right)\\&{}+\cos \left(\pi {\boldsymbol {\xi }}_{3}\cdot \mathbf {x} \right)\operatorname {sinc} \left({\boldsymbol {\xi }}_{1}\cdot \mathbf {x} \right)\operatorname {sinc} \left({\boldsymbol {\xi }}_{2}\cdot \mathbf {x} \right){\big )}.\end{aligned}}}
This construction can be used to design Lanczos window for general multidimensional lattices.16
Some authors, by analogy, define the hyperbolic sine cardinal function.171819
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