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 discovered8 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 identity9
∏ 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.
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 − 1/2:1011
∑ 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 1/2:12 ∑ 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 derived13 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 derive14 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.15
Some authors, by analogy, define the hyperbolic sine cardinal function.161718
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