NUMERICAL EVALUATION OF A FOUR TERM ARCTAN FORMULA
ARCTAN FORMULA We have discussed earlier how one requires large values of N to obtain rapidly converging arctan (1/N) formulas for π One such formula which does this is the four term expression- π=48arctan(1/38) +80arctan(1/57) +28arctan(1/239) +96arctan(1/268) Let us demonstrate how one goes about evaluating this equality numerically
Arctangent Formulas and Pi - Grinnell College
Mathematical Assoc of America American Mathematical Monthly 121:1 August 4, 2018 2:23p m arctan˙2 tex page 2 where r is the radius of the inscribed circle Alternatively, applying Heron’s formula
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
Efficient Approximations for the Arctangent Function T
IEEE SIGNAL PROCESSING MAGAZINE [109] MAY 2006 arctan(x) ≈ π 4 x, −1 ≤ x ≤ 1 (2) This linear approximation has been used in [6] for FM demodulation due to its minimal complexity
The complex inverse trigonometric and hyperbolic functions
arccot(z) = arctan 1 z , (1) Arccot(z) = Arctan 1 z (2) Note that eqs (1) and (2) can be used as definitions of the inverse cotangent function and its principal value We now examine the principal value of the arccotangent for real-valued arguments Setting z = x, where x is real, Arccotx = 1 2 Arg x +i x − i 2
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www mathportal 2 2 2 2 2 2 2 2 2 2 2 arctan 4 0 4 4 1 2 2 4 ln 4 0 4 2 4 2 4 0 2 ax b for ac b ac b ac b ax b b ac dx for ac b ax bx c b ac ax b b ac for ac b
Derivative of arctan(x) - MIT OpenCourseWare
function of x, not of y We must now plug in the original formula for y, which was y = tan−1 x, to get y = cos2(arctan(x)) This is a correct answer but it can be simplified tremendously We’ll use some geometry to simplify it 1 x (1+x2)1/2 y Figure 3: Triangle with angles and lengths corresponding to those in the exam ple
Trigonometric Formula Sheet De nition of the Trig Functions
Trigonometric Formula Sheet De nition of the Trig Functions Right Triangle De nition Assume that: 0 <
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Mathematical Assoc. of America American Mathematical Monthly 121:1 August 4, 2018 2:23p.m. arctan2.tex page 1
Arctangent Formulas and Pi
Marc Chamberland and Eugene A. Herman
Abstract.Using both geometrical and analytical approaches, new multivariable formulas con- necting the arctangent function and the numberπare produced.1. INTRODUCTION.Since the discovery of Machin"s formula
4= 4arctan?15?
-arctan?1239? ,(1) the arctangent function has been ubiquitous in calculations ofπ. While formulas like (1) have been heavily explored [1], we seek formulas that linkπwith a linear combi- nation of arctangents of general arguments. The simplest example is the well-known equation2= arctan(x) + arctan?1x?
(2) for allx >0. Another example, a variant of an equation due to Euler, states2= arctan(x)-arctan(x-y) + arctan?x2-xy+ 1y?
for allxand wheny >0. The goal of this note is to develop arctangent formulas with several variables.2. GEOMETRY OF TRIANGLES AND TETRAHEDRA.This study started
serendipidously by considering the inscribed circle in a generaltriangle: see Figure 1. The area of the triangle can be computed in two ways. By dissecting the triangle into rr r b aa c c bFigure 1.Inscribed circle in a triangle.
three subtriangles, we find that its total areaAsatisfies A=12(a+c)r+12(a+b)r+12(b+c)r= (a+b+c)r,
January 2014]ARCTANGENT FORMULAS AND PI1
Mathematical Assoc. of America American Mathematical Monthly 121:1 August 4, 2018 2:23p.m. arctan2.tex page 2
whereris the radius of the inscribed circle. Alternatively, applying Heron"s formula to the original triangle yields A=? abc(a+b+c).Setting the two expressions equal produces
r=? abc a+b+c. Since the six angles surrounding the center of the inscribed circle sum to2π, this producesπ= arctan?
a? a+b+c abc? + arctan? b? a+b+c abc? (3) +arctan c? a+b+c abc? for alla,b,c >0. To generalize this geometric approach, one could consider an(n-1)-sphere in- scribed in a simplex inndimensions. The volume of the simplex can be calculated with the Cayley-Menger determinant. More challenging is the generalization of the angles around the sphere"s center, sometimes called solid angles"; see [3, 4]. The complexity of this approach, particularly in higher dimensions, suggests an analytic approach for finding formulas similar to equation (3).3. ARCTANGENT AND SYMMETRIC POLYNOMIALS.Some beautiful iden-
tities connect the tangent function with symmetric polynomials. Letxi= tan(θi)for i= 1,2,3,...and letek(x)denote thekth elementary symmetric polynomial in the variablesx1,x2,x3,.... The first few examples are e0(x) = 1, e1(x) =?
ix i, e2(x) =? iMathematical Assoc. of America American Mathematical Monthly 121:1 August 4, 2018 2:23p.m. arctan2.tex page 3
tan(θ1+θ2+θ3+θ4) =e1(x)-e3(x)e0(x)-e2(x) +e4(x)1-(x1x2+x1x3+x1x4+x2x3+x2x4+x3x4) +x1x2x3x4.(7)
Special choices of the angles produce formulas of the type that we seek. For ex- ample, settingθ1+θ2=π/2in equation (5) occurs exactly whenx1x2= 1. Ifx1= y1f(y1,y2)andx2=y2f(y1,y2), thenf(y1,y2) = 1/⎷
y1y2foryi>0for alli.This produces
2= arctan?y1⎷y1y2?
+ arctan?y2⎷y1y2? (8) = arctan y1 y2? + arctan? y2 y1?Of course this is equivalent to equation (2).
This approach can be generalized by lettingxi=yif(y1,y2,...,yn)fori=1,2,...,n. Setting the numerator of equation (6) equal to zero produces
π= arctan?
y 1? y1+y2+y3 y1y2y3? + arctan? y 2? y1+y2+y3 y1y2y3? +arctan y 3? y1+y2+y3 y1y2y3? ,(9) which is the same as equation (3), while setting the denominator of equation (6) equal to zero gives2= arctan?y1⎷y1y2+y2y3+y3y1?
+ arctan?y2⎷y1y2+y2y3+y3y1? +arctan ?y3 ⎷y1y2+y2y3+y3y1? .(10) Equation (7), with its numerator equal to zero, generatesπ= arctan?
y 1? y1+y2+y3+y4 y1y2y3+y1y2y4+y1y3y4+y2y3y4? +arctan y 2? y1+y2+y3+y4 y1y2y3+y1y2y4+y1y3y4+y2y3y4? +arctan y 3? y1+y2+y3+y4 y1y2y3+y1y2y4+y1y3y4+y2y3y4? +arctan y 4? y1+y2+y3+y4 y1y2y3+y1y2y4+y1y3y4+y2y3y4? .(11)January 2014]ARCTANGENT FORMULAS AND PI3
Mathematical Assoc. of America American Mathematical Monthly 121:1 August 4, 2018 2:23p.m. arctan2.tex page 4
It is natural to wonder whether there are other simple" choices offthat produce equations of the form C= arctan(y1f(y1,y2,...,yn)) + arctan(y2f(y1,y2,...,yn)) +···+ arctan(ynf(y1,y2,...,yn))(12) for some constantCand ally1,y2,...,yn>0. Theorem 1.The only equations of the form (12) satisfied by a nonzero functionf=? p q, wherepandqare polynomials, are the equations (8)-(11). Proof.Lettingxi=yif(y1,y2,...,yn), use equations (4) and (12) to write D:= tan(C) = tan(arctan(x1) +···+ arctan(xn))quotesdbs_dbs7.pdfusesText_5