Método recorrente de Newton–Raphson

# usado no artigo do Quora https://n9.cl/8d2x4

newton-raphson()
{

  local f d p x args
  . <(printf '%q\n' $@)

  if [[ ! ( $f && $d && $p && $x ) ]]
  then
    args='f=<f(x)> d=<df(x)/dx> p=<#díg> x=<chute inicial>'
    echo Uso: $FUNCNAME $args >&2
    return 1
  fi

  BC_LINE_LENGTH=0 bc -l << EOT

    pr    = $p
    x     = $x
    scale = pr

    phi = (sqrt(5)+1)/2
    pi  = 4*a(1)

    define p(x,y){ return e(l(x)*y) }
    define r(x  ){ return x*180/pi }
    define f(x  ){ return $f }
    define d(x  ){ return $d }
    define a(x  ){ if (x>0) return x else return -x }

    print "\nx0 = ", x, "\n"

    n    = 0
    diff = 1
    while( diff > 10^-pr )
    {
      scale = pr+2
      x0    = x
      x     = x0-f(x0)/d(x0)
      diff  = a(x0-x)
      n    += 1
      scale = pr
      print "x", n, " = ", x/1, \
            "  |x", n, "-x", n-1, "| = ", diff/1, "\n"
    }

    print "\n  x = ", x/1, "\n pi = ", pi, \
          "\npi6 = ", f(x)+pi, "\ndif = ", f(x), "\n\n"

EOT

}

newton-raphson  f='p(phi,x)+p(2,s(r(2)))+p(2,c(r(2)))-7/460-pi'  d='p(phi,x)*l(phi)'  p=20  x=-4
# x0 = -4
# x1 = -4.56240513234095781898  |x1-x0| = .56240513234095781898
# x2 = -4.65373961837263842635  |x2-x1| = .09133448603168060736
# x3 = -4.65580654146557042174  |x3-x2| = .00206692309293199538
# x4 = -4.65580757005703567658  |x4-x3| = .00000102859146525484
# x5 = -4.65580757005729023781  |x5-x4| = .00000000000025456123
# x6 = -4.65580757005729023781  |x6-x5| = 0
#
#   x = -4.65580757005729023781
#  pi = 3.14159265358979323844
# pi6 = 3.14159265358979323839
# dif = -.00000000000000000005


f='p(phi,x)+p(2,s(r(2)))+p(2,c(r(2)))-7/460'
x=-4.65580757005729023781

BC_LINE_LENGTH=0 bc -l <<< "
  define p(x,y) { return e(l(x)*y) }
  define r(x)   { return x*180/pi  }
  scale=50
  x=$x
  phi=(sqrt(5)+1)/2
  pi=4*a(1)
  $f
"
# 3.14159265358979323895386758911237016045350271747023