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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