Showing that the limit, as x approaches infinity, of arctan(x
Showing that the limit, as x approaches infinity, of arctan(x) is Pi/2 Powers of 2 x arctan(x)-----1 0 7853982 0 78539816 = Pi / 4 2 1 1071487 4 1 3258177 8 1 4464413 16 1 5083775 32 1 5395565 64 1 5551726 3 14159265 = Pi 128 1 5629840 256 1 5668901 1 57079633 = Pi / 2 512 1 5688432
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PROPERTIES OF ARCTAN - University of Florida
This result relates the arctan to the logarithm function so that- 2 4 1 ln π i i = + Looking at the near linear relation between arctan(z) and z for z
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Inverse arcsin arccos arctan arcsec arccsc arccot Pythagorean identity The basic relationship between the sine and the cosine is the Pythagorean trigonometric identity: where cos2 θ means (cos(θ))2 and sin2 θ means (sin(θ))2 This can be viewed as a version of the Pythagorean theorem, and follows from the equation x2 + y2 = 1 for the unit
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limite finie en a Démonstration Si f admet une limite finie en a notée L alors, en notant ε la fonction x f x L֏ ()− On a f x L x L o()= + = +ε() x a→ (1) et () 0 x a lim xε → =: f admet un développement limité à l’ordre 0 en a Réciproquement si f admet un développement limité à l’ordre 0 en a alors () 0 0x a (1) x a f x
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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
1)1arctan(nnnNnn
NN N and produces the identity- ...!92!4 !72!3 !52!2 !32!11242322212
Also using the variable substitution u=w/sqrt(w^2+1) yields the symmetric form- f f w w N wwdw NNN ]11)1[()1()1(2)/1arctan( 2222so that- 0022
21)(cosh()21)(cosh(2
1)5.0(
vv vdv vdv wwdw It seems that this last integral in w can form the starting point for an AGM approach for finding precise values of ʌ. It can also be expanded as the series- ...97531!47531!3
531!231!1
1!0 2 S which shows an interesting pattern but is unfortunately only slowly convergent. A much more rapidly convergent series is found for larger N. Indeed, we have in general that- 0
12002/32122202/322
)]12...(531[)1(!2 )1()1()1)(1(111 )1(1)/1arctan(nnn n wnnnNNn wdwNNwNwdw
NNNThis yields at N=239 the result-
01212nnnnnn which is accurate to 43 places when adding up just the first nine terms in the infinite series. By telescoping the arctan(1/N) series terms by two, one finds the even faster convergent form- )31616()58()34(
212()1()
21(!4)/1arctan( 22
0 222
nnnNn nNnnN Nnnn Also using our earlier discussed four term arctan formula for ʌ ( see-NUMERICAL EVALUATION OF PI BY A FOUR TERM ARCTAN FORMULA) we have that- )57122(13384 )3250(9120 )1445(3648 )!2()!1(!4 41
1 nnnnnn nnn with each additional term taken in this series improving the accuracy of ʌ by about
3 places. Note the summation procedure requires no taking of roots and simply
involves summation, multiplication, and division of integers. Arctan(z) also relates to the hypergeometric series. Matching term by term of the infinite series for F(a,b,c,x) with the the first infinite series expansion for arctan given earlier, one has- 1 022)1)(1(2);2/3;1,2/1()arctan(ttztdtzzzFz
Also it follows that the second order differential equation-0)53()1(2
22wdzdwzdzwdzz has- )arctan(1zzw as a solution. Finally, one integration of arctan yields- )1ln(21)arctan()arctan( 2 zzzz which is easily verified by differentiating both sides.