Inverse Tangent

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The inverse tangent is the multivalued function tan^(-1)z (Zwillinger 1995, p. 465), also denoted arctanz (Abramowitz and Stegun 1972, p. 79; Harris and Stocker 1998, p. 311; Jeffrey 2000, p. 124) or arctgz (Spanier and Oldham 1987, p. 333; Gradshteyn and Ryzhik 2000, p. 208; Jeffrey 2000, p. 127), that is the inverse function of the tangent. The variants Arctanz (e.g., Bronshtein and Semendyayev, 1997, p. 70) and Tan^(-1)z are sometimes used to refer to explicit principal values of the inverse cotangent, although this distinction is not always made (e.g,. Zwillinger 1995, p. 466).

ArcTan

The inverse tangent function tan^(-1)x is plotted above along the real axis.

Worse yet, the notation arctanz is sometimes used for the principal value, with Arctanz being used for the multivalued function (Abramowitz and Stegun 1972, p. 80). Note that in the notation tan^(-1)z (commonly used in North America and in pocket calculators worldwide), tanz denotes the tangent and -1 the inverse function, not the multiplicative inverse.

The principal value of the inverse tangent is implemented as ArcTan[z] in the Wolfram Language. In the GNU C library, it is implemented as atan(double x).

InverseTangentBranchCut

The inverse tangent is a multivalued function and hence requires a branch cut in the complex plane, which the Wolfram Language's convention places at (-iinfty,-i] and [i,iinfty). This follows from the definition of tan^(-1)z as

 tan^(-1)z=1/2i[ln(1-iz)-ln(1+iz)].
(1)

In the Wolfram Language (and in this work), this branch cut definition determines the range of tan^(-1)x for real x as (-pi/2,pi/2). Care must be taken, however, as other branch cut definitions can give different ranges (most commonly, (0,pi)).

ArcTanReImAbs
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The inverse tangent function tan^(-1)z is plotted above in the complex plane.

tan^(-1)z has the special values

tan^(-1)(-infty)=-1/2pi
(2)
tan^(-1)(-i)=-iinfty
(3)
tan^(-1)0=0
(4)
tan^(-1)i=iinfty
(5)
tan^(-1)infty=1/2pi.
(6)

The derivative of tan^(-1)z is

 d/(dz)tan^(-1)z=1/(1+z^2)
(7)

and the indefinite integral is

 inttan^(-1)zdz=ztan^(-1)z-1/2ln(1+z^2)+C.
(8)
InverseTangentYXReImInverseTangentYXContours

The complex argument of a complex number z=x+iy is often written as

 theta=tan^(-1)(y/x),
(9)

where theta, sometimes also denoted phi, corresponds to the counterclockwise angle from the positive real axis, i.e., the value of theta such that x=costheta and y=sintheta. Plots of tan^(-1)(y/x) are illustrated above for real values of x and y.

InverseTangentReImAbs
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A special kind of inverse tangent that takes into account the quadrant in which z lies and is returned by the FORTRAN command ATAN2(y, x), the GNU C library command atan2(double y, double x), and the Wolfram Language command ArcTan[x, y], and is often restricted to the range -pi<theta<=pi. In the degenerate case when x=0,

 phi={-1/2pi   if y<0; undefined   if y=0; 1/2pi   if y>0.
(10)

The usual tan^(-1)z has the Maclaurin series of

tan^(-1)z=sum_(n=0)^(infty)((-1)^nz^(2n+1))/(2n+1)
(11)
=z-1/3z^3+1/5z^5-1/7z^7+...
(12)

(OEIS A033999 and A005408). A more rapidly converging form due to Euler is given by

 tan^(-1)x=sum_(n=0)^infty(2^(2n)(n!)^2)/((2n+1)!)(x^(2n+1))/((1+x^2)^(n+1))
(13)

for real x (Castellanos 1988). This is related to the formula of Euler given by

 tan^(-1)x=y/x(1+2/3y+(2·4)/(3·5)y^2+(2·4·6)/(3·5·7)y^3+...),
(14)

where

 y=(x^2)/(1+x^2).
(15)

The inverse tangent formulas are connected with many interesting approximations to pi

tan^(-1)(1+x)=pi/4+isum_(n=1)^(infty)((-1-i)^n-(i-1)^n)/(2^(n+1)n)x^n
(16)
=1/4pi+1/2x-1/4x^2+1/(12)x^3-1/(40)x^5+1/(48)x^6-1/(112)x^7+...
(17)

(OEIS A075553 and A075554).

The inverse tangent satisfies

 tan^(-1)z=cot^(-1)(1/z)
(18)

for z!=0,

 tan^(-1)z=-tan^(-1)(-z)
(19)

for all complex z,

tan^(-1)x=1/2pi-cos^(-1)(x/(sqrt(x^2+1)))
(20)
=sin^(-1)(x/(sqrt(x^2+1)))
(21)
=csc^(-1)((sqrt(x^2+1))/x)
(22)

for all real x, where equality for the last equation is understood to be in the limit as x->0, and

tan^(-1)x={-1/2pi-tan^(-1)(1/x) for x<0; 1/2pi-tan^(-1)(1/x) for x>0
(23)
={-1/2pi+cot^(-1)(-x) for x<0; 1/2pi+cot^(-1)(-x) for x>0
(24)
={-1/2pi-cot^(-1)x for x<0; 1/2pi-cot^(-1)x for x>0
(25)
={-cos^(-1)(1/(sqrt(x^2+1))) for x<0; cos^(-1)(1/(sqrt(x^2+1))) for x>0
(26)
={-sec^(-1)(sqrt(x^2+1)) for x<0; sec^(-1)(sqrt(x^2+1)) for x>0.
(27)

In terms of the hypergeometric function,

 tan^(-1)z=z_2F_1(1,1/2;3/2;-z^2)
(28)

for complex z, and

 tan^(-1)x=x/(1+x^2)_2F_1(1,1;3/2;(x^2)/(1+x^2))
(29)

for real x (Castellanos 1988).

Castellanos (1986, 1988) also gives some curious formulas in terms of the Fibonacci numbers,

tan^(-1)x=sum_(n=0)^(infty)((-1)^nF_(2n+1)t^(2n+1))/(5^n(2n+1))
(30)
=5sum_(n=0)^(infty)((-1)^nF_(2n+1)^2)/((2n+1)(u+sqrt(u^2+1))^(2n+1))
(31)
=sum_(n=0)^(infty)((-1)^n5^(n+2)F_(2n+1)^3)/((2n+1)(v+sqrt(v^2+5))^(2n+1)),
(32)

where

t=(2x)/(1+sqrt(1+(4x^2)/5))
(33)
u=5/(4x)(1+sqrt(1+(24)/(25)x^2)),
(34)

and v is the largest positive root of

 8xv^4-100v^3-450xv^2+875v+625x=0.
(35)

The inverse tangent satisfies the addition formula

 tan^(-1)x+tan^(-1)y=tan^(-1)((x+y)/(1-xy))
(36)

for -1<x,y<1, as well as the more complicated formula

 tan^(-1)(1/a)=2tan^(-1)(1/(2a))-tan^(-1)(1/(4a^3+3a))
(37)

valid for all complex a. An additional identity known to Euler is given by

 tan^(-1)(1/(a-b))=tan^(-1)(1/a)+tan^(-1)(b/(a^2-ab+1))
(38)

for (a>b ^ a>0) or (a<b ^ a<0). Another interesting inverse tangent identity attributed to Charles Dodgson (Lewis Carroll) by Lehmer (1938b; Bromwich 1991, Castellanos 1988) is

 tan^(-1)(p+r)+tan^(-1)(p+q)-tan^(-1)p=1/2pi,
(39)

where

 1+p^2=qr
(40)

and p,q,r>0.

The inverse tangent has continued fraction representations

 tan^(-1)x=x/(1+(x^2)/(3+(4x^2)/(5+(9x^2)/(7+(16x^2)/(9+...)))))
(41)

(Lambert 1770; Lagrange 1776; Wall 1948, p. 343; Olds 1963, p. 138) and

 tan^(-1)x=x/(1+(x^2)/(3-x^2+(9x^2)/(5-3x^2+(25x^2)/(7-5x^2+...))))
(42)

due to Euler and sometimes known as Euler's continued fraction (Borwein et al. 2004, p. 30).

To find tan^(-1)x numerically, the following arithmetic-geometric mean-like algorithm can be used. Let

a_0=(1+x^2)^(-1/2)
(43)
b_0=1.
(44)

Then compute

a_(i+1)=1/2(a_i+b_i)
(45)
b_(i+1)=sqrt(a_(i+1)b_i),
(46)

and the inverse tangent is given by

 tan^(-1)x=lim_(n->infty)x/(sqrt(1+x^2)a_n)
(47)

(Acton 1990).

An inverse tangent tan^(-1)n with integral n is called reducible if it is expressible as a finite sum of the form

 tan^(-1)n=sum_(k=1)f_ktan^(-1)n_k,
(48)

where f_k are positive or negative integers and n_i are integers <n. tan^(-1)m is reducible iff all the prime factors of 1+m^2 occur among the prime factors of 1+n^2 for n=1, ..., m-1. A second necessary and sufficient condition is that the largest prime factor of 1+m^2 is less than 2m. Equivalent to the second condition is the statement that every Gregory number t_x=cot^(-1)x can be uniquely expressed as a sum in terms of t_ms for which m is a Størmer number (Conway and Guy 1996). To find this decomposition, write

 arg(1+in)=argproduct_(k=1)(1+n_ki)^(f_k),
(49)

so the ratio

 r=(product_(k=1)(1+n_ki)^(f_k))/(1+in)
(50)

is a rational number. Equation (50) can also be written

 r^2(1+n^2)=product_(k=1)(1+n_k^2)^(f_k).
(51)

Writing (◇) in the form

 tan^(-1)n=sum_(k=1)f_ktan^(-1)n_k+ftan^(-1)1
(52)

allows a direct conversion to a corresponding inverse cotangent formula

 cot^(-1)n=sum_(k=1)f_kcot^(-1)n_k+ccot^(-1)1,
(53)

where

 c=2-f-2sum_(k=1)f_r.
(54)

Todd (1949) gives a table of decompositions of tan^(-1)n for n<=342. Conway and Guy (1996) give a similar table in terms of Størmer numbers.

Arndt and Gosper give the remarkable inverse tangent identity

 sin(sum_(k=1)^(2n+1)tan^(-1)a_k)=((-1)^n)/(2n+1)(sum_(k=1)^(2n+1)product_(j=1)^(2n+1)[a_j-tan((pi(j-k))/(2n+1))])/(sqrt(product_(j=1)^(2n+1)(a_j^2+1))).
(55)

There is an amazing set of BBP-type formulas for tan^(-1)(4/5):

tan^(-1)(4/5)=1/(131072)sum_(k=0)^(infty)1/(1048576^k)[(262144)/(40k+2)-(163840)/(40k+5)-(65536)/(40k+6)+(16384)/(40k+10)-(4096)/(40k+14)-(5120)/(40k+15)+(1024)/(40k+18)-(256)/(40k+22)+(160)/(40k+25)+(64)/(40k+26)-(16)/(40k+30)+4/(40k+34)+5/(40k+35)-1/(40k+38)]
(56)
=1/(131072)sum_(k=0)^(infty)1/(1048576^k)[(393216)/(40k+4)+(163840)/(40k+5)-(131072)/(40k+6)-(163840)/(40k+8)+(24576)/(40k+12)-(8192)/(40k+14)-(15360)/(40k+15)-(10240)/(40k+16)-(1024)/(40k+20)-(512)/(40k+22)-(640)/(40k+24)-(160)/(40k+25)+(96)/(40k+28)-(32)/(40k+30)-(40)/(40k+32)+(15)/(40k+35)+6/(40k+36)-2/(40k+38)]
(57)
=1/(131072)sum_(k=0)^(infty)1/(1048576^k)[(262144)/(40k+1)-(262144)/(40k+3)-(65536)/(40k+5)-(327680)/(40k+6)+(65536)/(40k+7)-(163840)/(40k+8)+(16384)/(40k+9)-(40960)/(40k+10)-(16384)/(40k+11)-(4096)/(40k+13)-(20480)/(40k+14)-(16384)/(40k+15)-(10240)/(40k+16)+(1024)/(40k+17)-(1024)/(40k+19)-(2560)/(40k+20)-(256)/(40k+21)-(1280)/(40k+22)+(256)/(40k+23)-(640)/(40k+24)+(64)/(40k+25)-(64)/(40k+27)-(16)/(40k+29)-(40)/(40k+30)+(16)/(40k+31)-(40)/(40k+32)+4/(40k+33)+(16)/(40k+35)-1/(40k+37)-5/(40k+38)+1/(40k+39)]
(58)
=1/(262144)sum_(k=0)^(infty)1/(1048576^k)[(262144)/(40k+3)+(262144)/(40k+4)+(131072)/(40k+6)-(65536)/(40k+7)+(81920)/(40k+10)+(16384)/(40k+11)+(16384)/(40k+12)+(8192)/(40k+14)-(4096)/(40k+15)+(1024)/(40k+19)+(1024)/(40k+20)+(512)/(40k+22)-(256)/(40k+23)+(64)/(40k+27)+(64)/(40k+28)-(48)/(40k+30)-(16)/(40k+31)+4/(40k+35)+4/(40k+36)+2/(40k+38)-1/(40k+39)],
(59)

the finding one of which is a given as a problem by Bailey et al. (2006, p. 225).

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