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@@ -1334,6 +1334,27 @@ and Y are coerced to single-float." |
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1334
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1334
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(t
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1335
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1335
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(values rho 0)))))))
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1336
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1336
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1337
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+;; sqrt(z) = z^(1/2)
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1338
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+;; = exp(1/2*log(z))
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1339
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+;;
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1340
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+;; Some special values, with x > 0
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1341
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+;; sqrt(x + i*0) = exp(1/2*log(x+i*0))
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1342
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+;; = exp(1/2*(log(x)+i*0))
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1343
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+;; = exp(1/2*log(x)+i*0)
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1344
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+;; = exp(1/2*log(x))*exp(i*0)
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1345
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+;; = sqrt(x)*(cos 0 + i*sin(0))
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1346
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+;; = sqrt(x) + i*0
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1347
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+;; sqrt(x - i*0) = exp(1/2*log(x-i*0))
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1348
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+;; = exp(1/2*(log(x) - i*0))
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1349
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+;; = exp(1/2*log(x) - i*0)
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1350
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+;; = exp(1/2*log(x))*exp(-i*0)
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1351
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+;; = sqrt(x)*(cos(-0) + i*sin(-0))
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1352
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+;; = sqrt(x) - i*0
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1353
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+;; Kahan also gives the following for b > 0
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1354
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+;; sqrt(-b+/-i*0) = +0 +/- i*sqrt(b)
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1355
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+;; sqrt(x +/- i*inf) = +inf +/- i*inf for all finite x.
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1356
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+;; sqrt(inf +/- i*b) = +inf +/- i*0
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1357
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+;; sqrt(-inf +/- i*b) = +0 +/- i*inf
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1337
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1358
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(defun complex-sqrt (z)
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1338
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1359
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"Principle square root of Z
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1339
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1360
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@@ -1604,6 +1625,51 @@ Z may be any number, but the result is always a complex." |
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1604
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1625
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(+ z 1)
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1605
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1626
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(complex (+ (realpart z) 1) (imagpart z))))
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1606
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1627
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1628
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+;; acos(z) = 2*log(sqrt((1+z)/2) + i*sqrt((1-z)/2))/i
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1629
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+;; = pi/2 - asin(z)
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1630
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+;;
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1631
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+;; In particular for z = x > 1:
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1632
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+;; acos(x) = 2/i*log(sqrt((x+1)/2) + i*sqrt((1-x)/2))
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1633
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+;; = 2/i*log(sqrt((x+1)/2) + i*i*sqrt((x-1)/2))
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1634
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+;; = 2/i*log(sqrt((x+1)/2) - sqrt((x-1)/2))
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1635
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+;; = -2*i*log(sqrt((x+1)/2) - sqrt((x-1)/2))
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1636
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+;; = -i*log(x-sqrt(x-1)*sqrt(x+1))
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1637
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+;; Thus, acos(2) = -i*log(2-sqrt(3) = -i*1.31695...
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1638
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+;;
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1639
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+;; Similarly for z = -x, x > 1:
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1640
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+;; acos(-x) = 2/i*log(sqrt((1-x)/2) + i*sart((1+x)/2))
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1641
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+;; = 2/i*log(i*sqrt((x-1)/2) + i*sqrt((1+x)/2))
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1642
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+;; = 2/i*log(i*(sqrt((x+1)/2)+sqrt((x-1)/2)))
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1643
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+;; = 2/i*(log(sqrt((x+1)/2)+sqrt((x-1)/2)) + i*pi/2)
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1644
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+;; = -i*2*log(sqrt((x+1)/2)+sqrt((x-1)/2)) - pi
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1645
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+;; = -pi - i*log(x+sqrt(x+1)*(sqrt(x-1)))
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1646
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+;;
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1647
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+;; Thus acos(-2) = -pi - i*log(2+sqrt(3)) = -pi -i*1.31695...
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1648
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+;;
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1649
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+;; Now see what aacos(x-i0) for x > 1 is:
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1650
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+;; acos(x-i*0) = 2/i*log(sqrt((x+1)/2-i*0) + i*sqrt((1-x)/2+i*0))
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1651
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+;; = 2/i*log(sqrt((x+1)/2) -i*0+ i*(0+i*sqrt((x-1)/2)))
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1652
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+;; = 2/i*log(sqrt((x+1)/2) - sqrt((x-1)/2) - i*0)
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1653
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+;; = 2/i*[log(sqrt((x+1)/2) - sqrt((x-1)/2)) - i*0]
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1654
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+;; = -i*[2*log(sqrt((x+1)/2) - sqrt((x-1)/2)) - i*0]
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1655
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+;; = -i*log(x-sqrt(x-1)*sqrt(x+1)) + 0
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1656
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+;;
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1657
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+;; Thus acos(2 - i*0) is the same as acos(2). That is, acos is
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1658
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+;; continuous with quadrant IV for x > 1.
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1659
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+;;
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1660
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+;; For acos(-x+i0), x > 1:
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1661
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+;; acos(-x+i0) = 2/i*log(sqrt((1-x+i0)/2) + i*sqrt((1+x-i0)/2))
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1662
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+;; = 2/i*log(sqrt((1-x)/2+i0) + i*sqrt((1+x)/2-i0))
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1663
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+;; = 2/i*log((i*sqrt((x-1)/2) + 0) + i*(sqrt((1+x)/2)-i0))
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1664
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+;; = 2/i*log(i*sqrt((x-1)/2) + i*sqrt((1+x)/2) + 0)
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1665
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+;; = 2/i*log(i*(sqrt((x-1)/2 + sqrt(1+x)/2)) + 0)
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1666
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+;; = 2/i*(log(sqrt(x+sqrt(x-1)*sqrt(x+1))) + i*pi/2)
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1667
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+;; = -i*log(x+sqrt(x-1)*sqrt(x+1)) + pi
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1668
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+;; = pi - i*log(x+sqrt(x-1)*sqrt(x+1))
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1669
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+;;
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1670
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+;; Thus acos(-2+i0) = pi - i*log(sqrt(3)+2) = pi -i*1.31695, and this
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1671
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+;; equal acos(-2). This means acos is continuous with quadrant II.
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1672
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+;;
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1607
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1673
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(defun complex-acos (z)
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1608
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1674
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"Compute acos z = pi/2 - asin z
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1609
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1675
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