PROPERTIES OF ARCTAN - University of Florida
[Pages:7]PROPERTIES OF ARCTAN(Z)
We know from elementary calculus that the function z=tan() has an inverse =arctan(z). In differentiating z once we have-
dz = [1 + tan( )2 ] d or its equivalent
arctan(z)
=
z
0
1
dz +z
2
On setting the upper limit to 1/N with N>1. Thus-
arctan(1/ 239)
=
1 239
1
-
1 3(239)2
+
1 5(239)4
-
1 7(239)6
+ ....
However for N=1, the series just equals that of Gregory which is known to be notoriously slowly convergent-
arctan(1)
=
1
-
1 3
+
1 5
-
1 7
+
1 9
-
.....
=
4
If one takes the first hundred terms(m=100) in the Gregory series, the integral remainder will still be-
t
1
= 0
t 200 1+ t2
dt
=
0.0024999...
or some
1/ 3
percent
In general the larger N becomes the more rapidly the infinite series for arctan(z) will converge. Thus the series for (/8) =arctan{ 1/[1+sqrt(2)]} reads -
8
=
1 (1 +
2
)
1
-
(1 +
1 2)2
+
1 (1 + 2)4
- ...
which converges somewhat faster than the Gregory series.
Lets examine some of the other analytical characteristics of arctan(z). Its plot for z real looks like this-
We see that arctan(z) varies linearly with z for small z starting with value zero and becomes non-linear in its variation with increasing z, eventually approaching Pi/2 as Pi/2-1/z as z approaches infinity. The function has odd symmetry since arctan(z)=-arctan(z). Its derivative is just 1/(1+z^2) and hence represents a special case of the Witch of Agnesi ( this curve was studied by the Italian mathematician Maria Agnesi 1718-1799 and received its name due to a mistranslation of the Italian word versiero for curve by an English translator who mixed it up with the Italian word for witch). Using the multiple angle formula for tangent , one also has-
tan[arctan(
x)
+
arctan(
y)]
=
tan[arctan(x)] + tan[arctan( y)] 1 - tan[arctan(x)]tan[arctan( y)]
or the equivalent -
arctan[((1x-+xyy))] = arctan(x) + arctan( y)
On setting x=z and y= we find ?
2
=
arctan(z)
+
arctan(1z )
so that, for example, arctan(2)=/2-arctan(0.5)=/2-0.46364..= 1.1071...If x=1 and y=-1/3 one obtains the well known identity-
4
=
arctan( 12 )
+
arctan(13)
and x=1/7, y=-1/8 produces-
arctan(17) = arctan(81) + arctan(517)
Consider next the complex number z=x+iy. Writing this out in polar form yields-
x + iy =
x2
+
y2
exp[i
arctan(
y x
)]
so that-
arctan(y / x) = -i ln
x + iy
x2 + y2
This result relates the arctan to the logarithm function so that-
ln1
+ i 2
=
i
4
Looking at the near linear relation between arctan(z) and z for z ................
................
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