PROPERTIES OF ARCTAN - University of Florida [PDF] production électricité particulier
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1 0 12
51
In general the larger N becomes the more rapidly the infinite series for arctan(z) will converge. Thus the series for (ʌ/8) =arctan{ 1/[1+sqrt(2)]} reads - ...)21(1 )21(11)21(1 8 42
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- or the equivalent - )arctan()arctan(])1()(arctan[yxxyyx
xyiyxiyx so that- 22
ln)/arctan( yxiyxixy This result relates the arctan to the logarithm function so that- 421ln
ii Looking at the near linear relation between arctan(z) and z for z<<1 suggests that arctan(1/N)=m*arctan(1/(m*N) +small correction of the order 1/N^3 for large N. This is indeed the case. By looking at the imaginary part of- )ln()ln()()(ln
iNpiNpiNiN pp one finds- ))34(1arctan()21arctan(2)1arctan( 2 NNNN )118278arctan()31arctan(3)1arctan( 24
NNN NN and- 242
NNNN NN We next solve an integral in terms of arctan to get-
1 0 2 t ttdt It is also possible to manipulate the original integral form for artctan(z) into a variety of different range integrals. Consider the substitutions t=u/N and
22
22
42
arctan 42
0 22/1
0 2 )]11()[cosh(111)1arctan( vuN t N vdv NN uNduNdttN Expanding the term in the denominator of the last integral leads to an alternate series for arctan(1/N). In compact form, it reads- 022
2 )1()!12(!4
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PROPERTIES OF ARCTAN(Z)
We know from elementary calculus that the function z=tan(ș) has an inverse ș=arctan(z). In differentiating z once we have- z zdzzequivalentitsorddz 0221)arctan(])tan(1[
On setting the upper limit to 1/N with N<1 we find the infinite series expansion for arctan given by- 01212/1
0 01 )12(1)1()1()/1arctan( nnnnN z nnNndzzN or the equivalent- dttt NnN N tm m n nn /1 0 221 0 12
1)12()1()/1arctan(
This series will converge quite rapidly when N>>1. Thus- ....)239(71 )239(51 )239(3112391)239/1arctan(642 However for N=1, the series just equals that of Gregory which is known to be notoriously slowly convergent-4.....91
7151
311)1arctan(
If one takes the first hundred terms(m=100) in the Gregory series, the integral remainder will still be- percentsomeordttt t3/1...0024999.01
1 0 2200In general the larger N becomes the more rapidly the infinite series for arctan(z) will converge. Thus the series for (ʌ/8) =arctan{ 1/[1+sqrt(2)]} reads - ...)21(1 )21(11)21(1 8 42
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- or the equivalent - )arctan()arctan(])1()(arctan[yxxyyx
On setting x=z and y= we find -
)1arctan()arctan(2zz S so that, for example, arctan(2)=ʌ/2-arctan(0.5)=ʌ/2-0.46364..= 1.1071...If x=1 and y=-1/3 one obtains the well known identity- )31arctan()21arctan(4 S and x=1/7, y=-1/8 produces- )571arctan()81arctan()71arctan( Consider next the complex number z=x+iy. Writing this out in polar form yields- )]arctan(exp[ 22xyiyxiyx so that- 22
ln)/arctan( yxiyxixy This result relates the arctan to the logarithm function so that- 421ln
ii Looking at the near linear relation between arctan(z) and z for z<<1 suggests that arctan(1/N)=m*arctan(1/(m*N) +small correction of the order 1/N^3 for large N. This is indeed the case. By looking at the imaginary part of- )ln()ln()()(ln
221121
21iNpiNpiNiN pp one finds- ))34(1arctan()21arctan(2)1arctan( 2 NNNN )118278arctan()31arctan(3)1arctan( 24
NNN NN and- 242
NNNN NN We next solve an integral in terms of arctan to get-
Therefore one finds-
)117arctan(72)]73arctan()75[arctan(72 431 0 2 t ttdt It is also possible to manipulate the original integral form for artctan(z) into a variety of different range integrals. Consider the substitutions t=u/N and
Nt=tanh(v). These produce the integrals-
2222
22
42
arctan 42
44()2(1
bacbat bac abac abtdt a cbtatdt f 022210 22/1
0 2 )]11()[cosh(111)1arctan( vuN t N vdv NN uNduNdttN Expanding the term in the denominator of the last integral leads to an alternate series for arctan(1/N). In compact form, it reads- 022
2 )1()!12(!4