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7021 - 7030 of 13135 for Centered square numberSearch Results
Given a semicircular hump f(x) = sqrt(L^2-(x-L)^2) (1) = sqrt((2L-x)x), (2) the Fourier coefficients are a_0 = 1/2piL (3) a_n = ((-1)^nLJ_1(npi))/n (4) b_n = 0, (5) where ...
If f(x) is an odd function, then a_n=0 and the Fourier series collapses to f(x)=sum_(n=1)^inftyb_nsin(nx), (1) where b_n = 1/piint_(-pi)^pif(x)sin(nx)dx (2) = ...
The Fourier transform of the constant function f(x)=1 is given by F_x[1](k) = int_(-infty)^inftye^(-2piikx)dx (1) = delta(k), (2) according to the definition of the delta ...
F_x[cos(2pik_0x)](k) = int_(-infty)^inftye^(-2piikx)((e^(2piik_0x)+e^(-2piik_0x))/2)dx (1) = 1/2int_(-infty)^infty[e^(-2pii(k-k_0)x)+e^(-2pii(k+k_0)x)]dx (2) = ...
The Fourier transform of e^(-k_0|x|) is given by F_x[e^(-k_0|x|)](k)=int_(-infty)^inftye^(-k_0|x|)e^(-2piikx)dx = ...
The Fourier transform of the Heaviside step function H(x) is given by F_x[H(x)](k) = int_(-infty)^inftye^(-2piikx)H(x)dx (1) = 1/2[delta(k)-i/(pik)], (2) where delta(k) is ...
The Fourier transform of the generalized function 1/x is given by F_x(-PV1/(pix))(k) = -1/piPVint_(-infty)^infty(e^(-2piikx))/xdx (1) = ...
F_x[1/pi(1/2Gamma)/((x-x_0)^2+(1/2Gamma)^2)](k)=e^(-2piikx_0-Gammapi|k|). This transform arises in the computation of the characteristic function of the Cauchy distribution.
Let R(x) be the ramp function, then the Fourier transform of R(x) is given by F_x[R(x)](k) = int_(-infty)^inftye^(-2piikx)R(x)dx (1) = i/(4pi)delta^'(k)-1/(4pi^2k^2), (2) ...
Let Pi(x) be the rectangle function, then the Fourier transform is F_x[Pi(x)](k)=sinc(pik), where sinc(x) is the sinc function.
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