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- : \( \!\!\!\!\!\!\!\!\!\! (2) \displaystyle ~ ~ ~ J_0(0)=1 ~, ~~ J_0'(0)=0\) \frac{ (-z^2/4)^n }6 KB (913 words) - 18:25, 30 July 2019
- : \( \mathrm{Sinc}(z)= 1-\frac{z^2}{6}+\frac{z^4}{120}-\frac{z^6}{5040}+ : \(\!\!\!\!\!\!\!\! \mathrm{ArcSinc}(1\!-\!t)= \sqrt{6 t} \left(4 KB (563 words) - 18:27, 30 July 2019
- f(t) = \sum_{m=1}^\infty (2 J_\nu(j_{\nu,m}t) / J_{\nu+1}(j_{\nu,m})^2) g_m. g_m = (2 / j_{\nu,M}^2)7 KB (1,063 words) - 18:25, 30 July 2019
- ...aystyle \!\!\!\!\!\!\!\!\!\! (2) ~ ~ ~ \hat H = \frac{-\hbar^2}{2m} \nabla^2\) The substitution of (2),(3),(4) into (1) gives the "monochromatic" equation5 KB (743 words) - 18:47, 30 July 2019
- : \( \!\!\!\!\!\!\!\!\!\!\ (2) ~ ~ ~ \displaystyle K_0(z) = \exp(-z)\sqrt{\frac{\pi}{2z}} ~ \Big( 1+ O(1/z)\Big)3 KB (394 words) - 18:26, 30 July 2019
- \displaystyle Y_0(x)=\frac{-2}{\pi} \int_0^\infty \cos(x \cosh(t)) \mathrm d t\) : \(Y_0(z)=Y_0(-z)+2~ \mathrm i~ J_0(-z)\)3 KB (445 words) - 18:26, 30 July 2019
- f''(z)+f'(z)/z + (z^2\!-\!1)f(z) = 0\) : \(f(0) = 0~\) and \(~f'(0)=1/2\)3 KB (439 words) - 18:26, 30 July 2019
- - 9./4.)*u + 2.)* c; return (C+S)/sqrt(2.*M_PI*z);}2 KB (190 words) - 18:47, 30 July 2019
- return s*z/2.;} f=M_PI/4.+z; c=cos(f); s=sin(f); a=sqrt(2./M_PI/z); t=1./(z*z);2 KB (159 words) - 14:59, 20 June 2013
- \( \!\!\!\!\!\!\!\!\! (1) ~ ~ ~ f''(z)+f'(z)/z+(1-\nu/z^2)f(x) =0\) f(x) \approx x^\nu \left( \frac{2^{-\nu}}{\mathrm{Factorial}(\nu)}+ O(x^2) \right)\)13 KB (1,592 words) - 18:25, 30 July 2019
- +q*(-0.016073968025938425623 + 0.0099471839432434584856 *L //2 +q*(-7.693079900902931853e-10 + 2.9981625986338549158e-10*L //64 KB (370 words) - 18:46, 30 July 2019
- TeXForm[Expand[Series[(HankelH1[0, x]) (Pi I x/2)^(1/2), {x, Infinity, 5}]]] \(e^{i x} \left(1-\frac{i}{8 x}-\frac{9}{128 x^2}+\frac{754 KB (509 words) - 18:26, 30 July 2019
- ...\!\!\!\!\!\!\!\! (1) \displaystyle ~ ~ ~ ふ_{\mathrm C}(x,y)=\sqrt{\frac{2}{\pi}} \cos(xy)\) : \(\!\!\!\!\!\!\!\!\!\!\! (2) ~ ~ ~ \displaystyle \mathrm{CosFourier} f (x) = \int_0^\infty ~ ふ_{\mat6 KB (915 words) - 18:26, 30 July 2019
- ...displaystyle ふ_{m,n}=\frac{1}{\sqrt{N}} \exp\! \left( \frac{- \mathrm i ~2 ~\pi}{N}~m~n\right)\) Sometimes, the operator that deviate from \(\hat ふ\) with factor \(\sqrt{N}\) is considered; for such modified operator, the equation (4) does not h6 KB (1,032 words) - 18:48, 30 July 2019
- ...!\!\!\!\!(1) ~ ~ ~ \displaystyle \mathrm{CosFourier}(F)(x) ~=~ \sqrt{\frac{2}{\pi}}~ \int_0^\infty \! \cos(xy) ~F(y)~ \mathrm d y\) ...isplaystyle G_k = (\mathrm{DCTI}_N F)_k = \frac{1}{2} F_0 + \frac{(-1)^k}{2} F_{N} + \sum_{n=1}^{N-1} F_n \cos\! \left(\frac{\pi}{N} n k \right)~\) fo3 KB (482 words) - 18:26, 30 July 2019
- for(i=1;i<n;i+=2){ while (m >= 2 && j > m) { j -= m; m >>= 1; }4 KB (571 words) - 15:00, 20 June 2013
- <math>\displaystyle (\mathrm{DCTIV} ~f )_k = \sqrt{\frac{2}{N}} ~ ...~f_n~ \cos \left[\frac{\pi}{N} \left(n+\frac{1}{2}\right) \left(k+\frac{1}{2}\right) \right] \quad \quad k = 0, \dots, N-1.</math>3 KB (421 words) - 18:26, 30 July 2019
- ...isplaystyle G_k = (\mathrm{DCTI}_N F)_k = \frac{1}{2} F_0 + \frac{(-1)^k}{2} F_{N} + \sum_{n=1}^{N-1} F_n \cos\! \left(\frac{\pi}{N} n k \right)~\) fo : \( \!\!\!\!\!\!\!\!\!\!(2) ~ ~ ~ ~ (ふ_{\mathrm C1,N}~ F)_k=10 KB (1,447 words) - 18:27, 30 July 2019
- ..., the two cords that connect it with elementary body 1 and elementary body 2 are removed (cut); so, the 0th body remain free. However elementary body 2 and elementary body 3 are still connected with the ideal cord of length uni8 KB (1,036 words) - 18:25, 30 July 2019
- :<math>X_k = \frac{1}{2} (x_0 + (-1)^k x_{N}) For \(N=2^q\), the evaluation requires of order of \(Nq\) operations.10 KB (1,689 words) - 18:26, 30 July 2019
