TomAllen/Testing/Testing01

NB. ... script Signature1.ijs ...

gXo1=:>@(0{[)
gXo2=:>@(1{[)
gXa0=:>@(2{[)
gXb0=:>@(3{[)

gXbox0R=:>@(0{])
gXbox1R=:>@(1{])

NB. ... from 'numeric' ...

gXsteps=:{.+(1&{-{.)*(i.@>:%])@{:

NB. ... tolerant 'set zero' (see 'Essays/Tolerant Comparison') ...

gXtsz=:$@]$[0:`(I.@([>!.0|@]))`]},@]
gXts0=:(2^_44)&gXtsz

NB. ... tolerant 'equal' (see 'Essays/Tolerant Comparison') ...

gXteq=:*./@,@((gXbox0R|@:-gXbox1R)<:!.0[*gXbox0R>.&:|gXbox1R)

NB. ... verbs useful for tolerant comparison ...

gXnzmin  =:<./@:|@((0<!.0|)#])@,
gXnzmax  =:>./@:|@((0<!.0|)#])@,
gXnzcount=:+/@(0<!.0|)@,

NB. ... axes sum ...

gXsmx=:+/@(*"1)"1 _

NB. ... trig verbs ...

sin=:1&o.
cos=:2&o.
arctant1=:(1p1+_3&o.@({:%{.))`( (2p1+_3&o.@({:%{.))`(_3&o.@({:%{.))@.(>:&0@{:) )@.(>&0@{.)
arctant0=:3r2p1"_`( 1r2p1"_)@.(>&0@{:)`0:@.(=&0@{:)
arctan  =:arctant1`arctant0@.(=&0@{.)"1

\begin{gather*} \intertext {\texttt {... Bessel's equation ... } } \begin{split} x^2 \frac{d^2y}{d^2x} + x \frac{dy}{dx} + ( x^2 - p^2 ) y = 0 \end{split} \\ \intertext {\texttt {... solution of the form ... } } \begin{split} y = \sum_{n=0}^\infty a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... where ... \newline ... \newline ... $a_0 \neq 0$ ... \newline ... $a_n=0$ for all negative $n$ ... \newline ... $\rho$ is allowed to be any number ... \newline ... \newline ... so ... } } \begin{split} x^2 \frac{d^2y}{d^2x} = & \sum_{n=0}^\infty a_n (n+\rho) (n+\rho-1) x^{n+\rho} \\ x \frac{dy}{dx} = & \sum_{n=0}^\infty a_n (n+\rho) x^{n+\rho} \\ x^2 y = & \sum_{n=0}^\infty a_{n-2} x^{n+\rho} \\ - p^2 y = & \sum_{n=0}^\infty - p^2 a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} \sum_{n=0}^\infty \biggl( a_n (n+\rho) (n+\rho-1) + a_n (n+\rho) + a_{n-2} - p^2 a_n \biggr) x^{n+\rho} = 0 \end{split} \\ \begin{split} \sum_{n=0}^\infty \biggl( \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} \biggr) x^{n+\rho} = 0 \end{split} \\ \intertext {\texttt {... then ... \newline ... for all $n$ ... } } \begin{split} \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} = 0 \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} a_n = - \frac { 1 } { (n+\rho)^2 - p^2 } a_{n-2} \end{split} \\ \intertext {\texttt {... therefore ... \newline ... \newline ... $n=0$ ... } } \begin{split} \Bigl( (0+\rho)^2 - p^2 \Bigr) a_0 + 0 = 0 \end{split} \\ \begin{split} \rho = \pm p \end{split} \\ \intertext {\texttt {... and ... \newline ... (provided denominator does not vanish) ... \newline ... \newline ... $n=1$ ... } } \begin{align*} \begin{split} a_1 = & - \frac { 1 } { (1+\rho)^2 - \rho^2 } a_{-1} \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=2$ ... } } \begin{split} a_2 = & - \frac { 1 } { (2+\rho)^2 - \rho^2 } a_0 \\ = & - \frac { 1 } { 2 (2\rho+2) } a_0 \end{split} \\ \intertext {\texttt {... $n=3$ ... } } \begin{split} a_3 = & - \frac { 1 } { (3+\rho)^2 - \rho^2 } a_1 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=4$ ... } } \begin{split} a_4 = & - \frac { 1 } { (4+\rho)^2 - \rho^2 } a_2 \\ = & \frac { 1 } { (2) (4) (2\rho+2) (2\rho+4) } a_0 \end{split} \\ \intertext {\texttt {... $n=5$ ... } } \begin{split} a_5 = & - \frac { 1 } { (5+\rho)^2 - \rho^2 } a_3 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=6$ ... } } \begin{split} a_6 = & - \frac { 1 } { (6+\rho)^2 - \rho^2 } a_4 \\ = & - \frac { 1 } { (2) (4) (6) (2\rho+2) (2\rho+4) (2\rho+6) } a_0 \end{split} \\ \end{align*} \intertext {\texttt {... etc ... } } \end{gather*}



\begin{gather*} \intertext {\texttt {... Bessel's equation ... } } \begin{split} x^2 \frac{d^2y}{d^2x} + x \frac{dy}{dx} + ( x^2 - p^2 ) y = 0 \end{split} \\ \intertext {\texttt {... solution of the form ... } } \begin{split} y = \sum_{n=0}^\infty a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... where ... \newline ... \newline ... $a_0 \neq 0$ ... \newline ... $a_n=0$ for all negative $n$ ... \newline ... $\rho$ is allowed to be any number ... \newline ... \newline ... so ... } } \begin{split} x^2 \frac{d^2y}{d^2x} = & \sum_{n=0}^\infty a_n (n+\rho) (n+\rho-1) x^{n+\rho} \\ x \frac{dy}{dx} = & \sum_{n=0}^\infty a_n (n+\rho) x^{n+\rho} \\ x^2 y = & \sum_{n=0}^\infty a_{n-2} x^{n+\rho} \\ - p^2 y = & \sum_{n=0}^\infty - p^2 a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} \sum_{n=0}^\infty \biggl( a_n (n+\rho) (n+\rho-1) + a_n (n+\rho) + a_{n-2} - p^2 a_n \biggr) x^{n+\rho} = 0 \end{split} \\ \begin{split} \sum_{n=0}^\infty \biggl( \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} \biggr) x^{n+\rho} = 0 \end{split} \\ \intertext {\texttt {... then ... \newline ... for all $n$ ... } } \begin{split} \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} = 0 \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} a_n = - \frac { 1 } { (n+\rho)^2 - p^2 } a_{n-2} \end{split} \\ \intertext {\texttt {... therefore ... \newline ... \newline ... $n=0$ ... } } \begin{split} \Bigl( (0+\rho)^2 - p^2 \Bigr) a_0 + 0 = 0 \end{split} \\ \begin{split} \rho = \pm p \end{split} \\ \intertext {\texttt {... and ... \newline ... (provided denominator does not vanish) ... \newline ... \newline ... $n=1$ ... } } \begin{align*} \begin{split} a_1 = & - \frac { 1 } { (1+\rho)^2 - \rho^2 } a_{-1} \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=2$ ... } } \begin{split} a_2 = & - \frac { 1 } { (2+\rho)^2 - \rho^2 } a_0 \\ = & - \frac { 1 } { 2 (2\rho+2) } a_0 \end{split} \\ \intertext {\texttt {... $n=3$ ... } } \begin{split} a_3 = & - \frac { 1 } { (3+\rho)^2 - \rho^2 } a_1 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=4$ ... } } \begin{split} a_4 = & - \frac { 1 } { (4+\rho)^2 - \rho^2 } a_2 \\ = & \frac { 1 } { (2) (4) (2\rho+2) (2\rho+4) } a_0 \end{split} \\ \intertext {\texttt {... $n=5$ ... } } \begin{split} a_5 = & - \frac { 1 } { (5+\rho)^2 - \rho^2 } a_3 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=6$ ... } } \begin{split} a_6 = & - \frac { 1 } { (6+\rho)^2 - \rho^2 } a_4 \\ = & - \frac { 1 } { (2) (4) (6) (2\rho+2) (2\rho+4) (2\rho+6) } a_0 \end{split} \\ \end{align*} \intertext {\texttt {... etc ... } } \end{gather*}



\begin{gather*} \intertext {\texttt {... Bessel's equation ... } } \begin{split} x^2 \frac{d^2y}{d^2x} + x \frac{dy}{dx} + ( x^2 - p^2 ) y = 0 \end{split} \\ \intertext {\texttt {... solution of the form ... } } \begin{split} y = \sum_{n=0}^\infty a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... where ... \newline ... \newline ... $a_0 \neq 0$ ... \newline ... $a_n=0$ for all negative $n$ ... \newline ... $\rho$ is allowed to be any number ... \newline ... \newline ... so ... } } \begin{split} x^2 \frac{d^2y}{d^2x} = & \sum_{n=0}^\infty a_n (n+\rho) (n+\rho-1) x^{n+\rho} \\ x \frac{dy}{dx} = & \sum_{n=0}^\infty a_n (n+\rho) x^{n+\rho} \\ x^2 y = & \sum_{n=0}^\infty a_{n-2} x^{n+\rho} \\ - p^2 y = & \sum_{n=0}^\infty - p^2 a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} \sum_{n=0}^\infty \biggl( a_n (n+\rho) (n+\rho-1) + a_n (n+\rho) + a_{n-2} - p^2 a_n \biggr) x^{n+\rho} = 0 \end{split} \\ \begin{split} \sum_{n=0}^\infty \biggl( \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} \biggr) x^{n+\rho} = 0 \end{split} \\ \intertext {\texttt {... then ... \newline ... for all $n$ ... } } \begin{split} \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} = 0 \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} a_n = - \frac { 1 } { (n+\rho)^2 - p^2 } a_{n-2} \end{split} \\ \intertext {\texttt {... therefore ... \newline ... \newline ... $n=0$ ... } } \begin{split} \Bigl( (0+\rho)^2 - p^2 \Bigr) a_0 + 0 = 0 \end{split} \\ \begin{split} \rho = \pm p \end{split} \\ \intertext {\texttt {... and ... \newline ... (provided denominator does not vanish) ... \newline ... \newline ... $n=1$ ... } } \begin{align*} \begin{split} a_1 = & - \frac { 1 } { (1+\rho)^2 - \rho^2 } a_{-1} \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=2$ ... } } \begin{split} a_2 = & - \frac { 1 } { (2+\rho)^2 - \rho^2 } a_0 \\ = & - \frac { 1 } { 2 (2\rho+2) } a_0 \end{split} \\ \intertext {\texttt {... $n=3$ ... } } \begin{split} a_3 = & - \frac { 1 } { (3+\rho)^2 - \rho^2 } a_1 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=4$ ... } } \begin{split} a_4 = & - \frac { 1 } { (4+\rho)^2 - \rho^2 } a_2 \\ = & \frac { 1 } { (2) (4) (2\rho+2) (2\rho+4) } a_0 \end{split} \\ \intertext {\texttt {... $n=5$ ... } } \begin{split} a_5 = & - \frac { 1 } { (5+\rho)^2 - \rho^2 } a_3 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=6$ ... } } \begin{split} a_6 = & - \frac { 1 } { (6+\rho)^2 - \rho^2 } a_4 \\ = & - \frac { 1 } { (2) (4) (6) (2\rho+2) (2\rho+4) (2\rho+6) } a_0 \end{split} \\ \end{align*} \intertext {\texttt {... etc ... } } \end{gather*}



\begin{gather*} \intertext {\texttt {... Bessel's equation ... } } \begin{split} x^2 \frac{d^2y}{d^2x} + x \frac{dy}{dx} + ( x^2 - p^2 ) y = 0 \end{split} \\ \intertext {\texttt {... solution of the form ... } } \begin{split} y = \sum_{n=0}^\infty a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... where ... \newline ... \newline ... $a_0 \neq 0$ ... \newline ... $a_n=0$ for all negative $n$ ... \newline ... $\rho$ is allowed to be any number ... \newline ... \newline ... so ... } } \begin{split} x^2 \frac{d^2y}{d^2x} = & \sum_{n=0}^\infty a_n (n+\rho) (n+\rho-1) x^{n+\rho} \\ x \frac{dy}{dx} = & \sum_{n=0}^\infty a_n (n+\rho) x^{n+\rho} \\ x^2 y = & \sum_{n=0}^\infty a_{n-2} x^{n+\rho} \\ - p^2 y = & \sum_{n=0}^\infty - p^2 a_n x^{n+\rho} \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} \sum_{n=0}^\infty \biggl( a_n (n+\rho) (n+\rho-1) + a_n (n+\rho) + a_{n-2} - p^2 a_n \biggr) x^{n+\rho} = 0 \end{split} \\ \begin{split} \sum_{n=0}^\infty \biggl( \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} \biggr) x^{n+\rho} = 0 \end{split} \\ \intertext {\texttt {... then ... \newline ... for all $n$ ... } } \begin{split} \Bigl( (n+\rho)^2 - p^2 \Bigr) a_n + a_{n-2} = 0 \end{split} \\ \intertext {\texttt {... and ... } } \begin{split} a_n = - \frac { 1 } { (n+\rho)^2 - p^2 } a_{n-2} \end{split} \\ \intertext {\texttt {... therefore ... \newline ... \newline ... $n=0$ ... } } \begin{split} \Bigl( (0+\rho)^2 - p^2 \Bigr) a_0 + 0 = 0 \end{split} \\ \begin{split} \rho = \pm p \end{split} \\ \intertext {\texttt {... and ... \newline ... (provided denominator does not vanish) ... \newline ... \newline ... $n=1$ ... } } \begin{align*} \begin{split} a_1 = & - \frac { 1 } { (1+\rho)^2 - \rho^2 } a_{-1} \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=2$ ... } } \begin{split} a_2 = & - \frac { 1 } { (2+\rho)^2 - \rho^2 } a_0 \\ = & - \frac { 1 } { 2 (2\rho+2) } a_0 \end{split} \\ \intertext {\texttt {... $n=3$ ... } } \begin{split} a_3 = & - \frac { 1 } { (3+\rho)^2 - \rho^2 } a_1 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=4$ ... } } \begin{split} a_4 = & - \frac { 1 } { (4+\rho)^2 - \rho^2 } a_2 \\ = & \frac { 1 } { (2) (4) (2\rho+2) (2\rho+4) } a_0 \end{split} \\ \intertext {\texttt {... $n=5$ ... } } \begin{split} a_5 = & - \frac { 1 } { (5+\rho)^2 - \rho^2 } a_3 \\ = & 0 \end{split} \\ \intertext {\texttt {... $n=6$ ... } } \begin{split} a_6 = & - \frac { 1 } { (6+\rho)^2 - \rho^2 } a_4 \\ = & - \frac { 1 } { (2) (4) (6) (2\rho+2) (2\rho+4) (2\rho+6) } a_0 \end{split} \\ \end{align*} \intertext {\texttt {... etc ... } } \end{gather*}

\begin{gather*} \intertext {\texttt {... \newline ... when $\rho$ is a nonnegative integer ... \newline ... simplify by use of factorials ... \newline ... } } \begin{split} y = & a_0 x^{\rho} \biggl[ 1 - \frac { x^2 } { (2) (2\rho+2) } + \frac { x^4 } { (2) (4) (2\rho+2) (2\rho+4) } - \dotsb \biggr] \\ = & a_0 x^{\rho} \biggl[ 1 - \frac { ( \frac{x}{2} )^2 } { (1) (\rho+1) } + \frac { ( \frac{x}{2} )^4 } { (1) (2) (\rho+1) (\rho+2) } - \dotsb \biggr] \\ = & a_0 \rho! x^{\rho} \biggl[ \frac { 1 } { (0!) \rho! } - \frac { ( \frac{x}{2} )^2 } { (1!) (\rho+1)! } + \frac { ( \frac{x}{2} )^4 } { (2!) (\rho+2)! } - \dotsb \biggr] \\ = & a_0 \rho! 2^{\rho} \Bigl( \frac{x}{2} \Bigr)^\rho \biggl[ \frac { 1 } { (0!) \rho! } - \frac { ( \frac{x}{2} )^2 } { (1!) (\rho+1)! } + \frac { ( \frac{x}{2} )^4 } { (2!) (\rho+2)! } - \dotsb \biggr] \\ = & a_0 \rho! 2^{\rho} \biggl[ \frac { ( \frac{x}{2} )^\rho } { (0!) \rho! } - \frac { ( \frac{x}{2} )^{2+\rho} } { (1!) (\rho+1)! } + \frac { ( \frac{x}{2} )^{4+\rho} } { (2!) (\rho+2)! } - \dotsb \biggr] \\ = & a_0 \rho! 2^{\rho} \sum_{n=0}^\infty \frac { (-1)^n ( \frac{x}{2} )^{2n+\rho} } { n! (\rho+n)! } \\ = & a_0 \rho! 2^{\rho} J_{\rho}(x) \end{split} \\ \intertext {\texttt {... where $J_{\rho}(x)$ is called the Bessel function of order $\rho$ ... \newline ... } } \end{gather*}

NB. ... script Signature1.ijs (continued) ...

NB. ... when a0=1 ...

bSrhoG  =:>@(0{[)
bSderivG=:>@(1{[)
bSlimitG=:>@(2{[)

bSrhoR  =:>@(0{])
bSderivR=:>@(1{])
bSlimitR=:>@(2{])
bSsignR =:>@(3{])
bSxR    =:>@(4{])
bSnR    =:>@(5{])
bSmltR  =:>@(6{])
bStermR =:>@(7{])
bStotR  =:>@(8{])

bSrhoL  =:>@(0{[)
bSderivL=:>@(1{[)
bSlimitL=:>@(2{[)
bSsignL =:>@(3{[)
bSxL    =:>@(4{[)
bSnL    =:>@(5{[)
bSmltL  =:>@(6{[)
bStermL =:>@(7{[)
bStotL  =:>@(8{[)

NB. ... initial term ...

bSinitd0=:]^bSrhoG
bSinitd1=:bSrhoG*]^bSrhoG-1:
bSinitd2=:bSrhoG*(bSrhoG-1:)*]^bSrhoG-2:
bSinitd3=:bSrhoG*(bSrhoG-1:)*(bSrhoG-2:)*]^bSrhoG-3:
bSinitd4=:bSrhoG*(bSrhoG-1:)*(bSrhoG-2:)*(bSrhoG-3:)*]^bSrhoG-4:

NB. ... subsequent terms ...

bStermd0Mlt=:-@bSsignL*]
bStermd0Pwr=:(bSnL+2:)+bSrhoL
bStermd0   =:bStermd0Mlt*bSxL^bStermd0Pwr

bStermd1Mlt=:-@bSsignL*]*((bSnL+2:)+bSrhoL)
bStermd1Pwr=:(bSnL+2:)+bSrhoL-1:
bStermd1   =:bStermd1Mlt*bSxL^bStermd1Pwr

bStermd2Mlt=:-@bSsignL*]*((bSnL+2:)+bSrhoL)*((bSnL+2:)+bSrhoL-1:)
bStermd2Pwr=:(bSnL+2:)+bSrhoL-2:
bStermd2   =:bStermd2Mlt*bSxL^bStermd2Pwr

bStermd3Mlt=:-@bSsignL*]*((bSnL+2:)+bSrhoL)*((bSnL+2:)+bSrhoL-1:)*((bSnL+2:)+bSrhoL-2:)
bStermd3Pwr=:(bSnL+2:)+bSrhoL-3:
bStermd3   =:bStermd3Mlt*bSxL^bStermd3Pwr

bStermd4Mlt=:-@bSsignL*]*((bSnL+2:)+bSrhoL)*((bSnL+2:)+bSrhoL-1:)*((bSnL+2:)+bSrhoL-2:)*((bSnL+2:)+bSrhoL-3:)
bStermd4Pwr=:(bSnL+2:)+bSrhoL-4:
bStermd4   =:bStermd4Mlt*bSxL^bStermd4Pwr

NB. ... verbs ...

bSinpInit=:bSinitd0`bSinitd1`bSinitd2`bSinitd3`bSinitd4@.bSderivG
bSinp    =:(bSrhoG;bSderivG;bSlimitG;1;];0;1;(];])@bSinpInit)"1 0

bSmult=:%@((bSnR+2:)*(2*bSrhoR)+bSnR+2:)*bSmltR

bSnextTerm=:bStermd0`bStermd1`bStermd2`bStermd3`bStermd4@.bSderivL
bSnext    =:bSrhoL;bSderivL;bSlimitL;-@bSsignL;bSxL;(bSnL+2:);];bSnextTerm([;[+])bStotL

NB. ... Bessel ...

bSbssl=:bStotR"1@:(]`(] bSnext bSmult)@.((bSlimitR<|@bStermR)+.0(=!.0)bStermR)^:_"1)@:bSinp

NB. ... plot (ijs) ...

NB. ... for example ...
NB. ... using p=0.75 and p=-0.75 ...

load 'plot'
pd (];( 0.75;0;(2^_44)"_)bSbssl]) gXsteps 0.01 11.62 101
pd (];(_0.75;0;(2^_44)"_)bSbssl]) gXsteps 0.5  11.62 101
pd 'pdf'
pd 'show'

graphic01.jpg

... real solutions for \rho=0.75 and \rho=-0.75 ...

 
NB. ... execute (ijx) ...

NB. ... check derivatives ...
NB. ... for example, using p=2 ...

   (((2;1;2^_44"_)bSbssl])((2^_19)gXteq[;])((2;0;2^_44"_)bSbssl]) D.1)"0 gXsteps _11.62 11.62 21
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

   (((2;2;2^_44"_)bSbssl])((2^_14)gXteq[;])((2;1;2^_44"_)bSbssl]) D.1)"0 gXsteps _11.62 11.62 21
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

   (((2;3;2^_44"_)bSbssl])((2^_18)gXteq[;])((2;2;2^_44"_)bSbssl]) D.1)"0 gXsteps _11.62 11.62 21
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

   (((2;4;2^_44"_)bSbssl])((2^_17)gXteq[;])((2;3;2^_44"_)bSbssl]) D.1)"0 gXsteps _11.62 11.62 21
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

\begin{gather*} \intertext {\texttt {... Bessel's equation ... } } \begin{split} x^2 \frac{d^2y}{d^2x} + x \frac{dy}{dx} + ( x^2 - p^2 ) y = 0 \end{split} \\ \end{gather*}

NB. ... execute (ijx) ...

NB. ... for example ...
NB. ... using p=0.75 ...

   p1a1t1=:(]^2:)*(0.75;2;(2^_44)"_)bSbssl]
   p1a1t2=:]*(0.75;1;(2^_44)"_)bSbssl]
   p1a1t3=:((]^2:)-0.75^2:)*(0.75;0;(2^_44)"_)bSbssl]
   p1a1  =:(p1a1t1+p1a1t2+p1a1t3)"0

   (0=!.0])@((2^_33)&gXtsz@:p1a1) gXsteps _11.62 11.62 21
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

NB. ... and ...
NB. ... using p=-0.75 ...

   p1a2t1=:(]^2:)*(_0.75;2;(2^_44)"_)bSbssl]
   p1a2t2=:]*(_0.75;1;(2^_44)"_)bSbssl]
   p1a2t3=:((]^2:)-_0.75^2:)*(_0.75;0;(2^_44)"_)bSbssl]
   p1a2  =:(p1a2t1+p1a2t2+p1a2t3)"0

   (0=!.0])@((2^_31)&gXtsz@:p1a2) gXsteps _11.62 11.62 21
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

NB. ... script Signature1.ijs (continued) ...

NB. ... verbs ...
NB. ... Bessel function of order 2 and derivatives ...

bSj2D0=:((2;0;(2^_44)"_)bSbssl])`0:@.(0=!.0])"0
bSj2D1=:((2;1;(2^_44)"_)bSbssl])`0:@.(0=!.0])"0
bSj2D2=:((2;2;(2^_44)"_)bSbssl])`2:@.(0=!.0])"0
bSj2D3=:((2;3;(2^_44)"_)bSbssl])`0:@.(0=!.0])"0
bSj2D4=:((2;4;(2^_44)"_)bSbssl])`0:@.(0=!.0])"0

graphic02.jpg

... coordinates ...

\begin{gather*} \intertext {\texttt {... $x$ from $v$ ... } } \begin{align*} \begin{split} x^1 = & v^1 \cos v^2 \\ x^2 = & v^1 \sin v^2 \end{split} \\ \intertext {\texttt {... $y$ from $x$ ... } } \begin{split} y^1 = & x^1 + o1 \\ y^2 = & x^2 + o2 \end{split} \\ \intertext {\texttt {... $q$ from $y$ ... } } \begin{split} q^1 = & \sqrt { (y^1)^2 + (y^2)^2 } \\ q^2 = & \arctan { \frac{y^2}{y^1} } \end{split} \\ \end{align*} \intertext {\texttt {... } } \begin{split} t = & a_0 J2D0(b_0q^1) \cos(2q^2) \end{split} \\ \intertext {\texttt {... } } \begin{split} t = & a_0 J2D0 \biggl( b_0 q^1 \Bigl( y^i \bigl( x^i(v^1,v^2) \bigr) \Bigr) \biggr) \cos \biggl( 2 q^2 \Bigl( y^i \bigl( x^i(v^1,v^2) \bigr) \Bigr) \biggr) \end{split} \\ \end{gather*}

NB. ... script Signature1.ijs (continued) ...

NB. ... coordinates ...

cv1=:0{]
cv2=:1{]

cx1=:0{]
cx2=:1{]

cy1=:0{]
cy2=:1{]

cq1=:0{]
cq2=:1{]

NB. ... verbs ...

x1=:(cv1*cos@cv2)"1
x2=:(cv1*sin@cv2)"1
xc=:(x1,x2)"1

y1=:(cx1+gXo1)"1
y2=:(cx2+gXo2)"1
yc=:(y1,y2)"1

q1=:((*:@cy1+*:@cy2)^0.5"_)"1
q2=:arctan@(cy1,cy2)"1
qc=:(q1,q2)"1

t=:(gXa0*bSj2D0@(gXb0*cq1)*cos@(2*cq2))"1

vGen=:(0{])(,"0"0 1&gXsteps)1{]

xcC=:xc`(''"_)@.($@]-:1$0:)
ycC=:([(yc)xc)`(''"_)@.($@]-:1$0:)
qcC=:([qc[(yc)xc)`(''"_)@.($@]-:1$0:)

NB. ... execute (ijx) ...

   p1b1=:gXts0@xcC-:gXts0@([((cy1-gXo1),cy2-gXo2)"1[((cq1*cos@cq2),cq1*sin@cq2)"1 qcC)
   (0.25;0.25;2;11.62) p1b1 vGen 0 0.646447 30,:0 2p1 100
1

NB. ... plot (ijs) ...

load 'plot'
pd 'viewpoint 3 3 1'
pd 'viewsize 1 1 0.1'
pd (0.25;0.25;2;11.62) (x1;x2;[t qcC) vGen 0 0.646447 30,:0 2p1 100
pd 'pdf'
pd 'show'

graphic03.jpg

... a surface ...

last edited 2008-07-07 05:53:00 by TomAllen