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Trigonometry and
Basic Functions

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An introduction to Trigonometry: A Math Tutorial by Johan.Claeys.
Dave's Short Trig Course by Dr. David E. Joyce (Clark University).
Planar and Spherical Trigonometry by Romuald Ireneus.
Positional Astronomy: Spherical trigonometry by Fiona Vincent.

Trigonometry, Elementary & Special Functions

(2002-01-25)   Basic Functions
What are some of the common and "special" numerical functions?

Well, people have been inventing special functions ad nauseam.  The list is quite literally endless, but we may attempt the beginning of a classification for those functions which are common enough to have a universally accepted name.  Let's start with the truly elementary functions:

  • Polynomial functions: The value y is obtained from the variable x using only a  finite  number of additions and/or multiplications involving given constants.  The simplest such functions are the  null function  (zero value, infinite degree)  and the other  constant functions  (degree 0; y = a ¹ 0).  Next are linear functions (degree 1, very rarely called monic; y = ax+b),  quadratic functions (degree 2; y = ax2+bx+c),  cubic functions, quartic (or biquadratic), quintic, sextic (rarely hexic), etc.  Specific qualifiers are virtually unused for polynomials beyond degree 6: degree 7 is heptic rather than septic [sic!]  degree 8 is octic,  9 is nonic,  10 is decic.  We're told that some people have called degree 100  hectic Just a joke!
  • Rational functions: The functions you obtain when division is allowed as well.  A rational function is the quotient of two polynomials.  The simplest of these is the reciprocal function y = 1/x.
  • Algebraic functions: The term applies to any function for which the value y and the variable x are algebraically related, which is to say that there's a two-variable polynomial P such that P(x,y) = 0.  Only a few of these functions have an explicit name and/or symbol.  The most notable is the square root function y = Öx.  Like the square root function, most algebraic functions can only be defined continuously over the complex plane as multivalued functions.  Alternately, such functions may be construed as univalued (ordinary) functions of a variable whose domain is a so-called Riemann surface for which several points may have the same projection on the complex plane.
  • Elementary transcendental functions: The simplest of these is  y = ex, the (natural) exponential function y = exp(x), a function which is equal to its own derivative (all other such functions are proportional to it).  The exponential function is defined (univalued) over the entire complex plane, and so are the other transcendental functions which may be defined directly in terms of it, including trigonometric functions (circular functions)  or  hyperbolic functions  like:
          y   =   sin(x)   =   ½ (exp(ix) - exp(-ix))
          y   =   sh(x)   =   ½ (exp(x) - exp(-x))
    Modern usage is to consider only the 3 preferred trigonometric functions (sine, cosine and tangent) whereas their 3 reciprocals (cosecant, secant and cotangent) are being deprecated.  A similar remark applies to the 3 preferred hyperbolic functions (sh, ch, th), whose reciprocals are rarely used, if ever.  Also classified as elementary transcendental functions are the inverses of the above, starting with the (natural) logarithm function, y = ln(x), which is the inverse of the exponential (x = exp(y)).  If continuity is required in the realm of complex numbers, the logarithm function may only be defined as a multivalued function.  The same thing is true of the inverse trigonometric functions (arcsin, arccos, arctg) or the inverse hyperbolic functions, which complete the modern list of elementary functions.
  • Important named combinations of elementary functions. 
    One example is the Gudermannian or hyperbolic amplitude, named after the German mathematician Christoph Gudermann (1798-1852):
    gd(x)   =   2 arctg( ex ) - p/2   =   2 arctg( th x/2 )
    The fact that  gd is  odd  is clear from the latter expression  (but obfuscated by the former one).  The derivative of gd(x) is 1/ch(x),  and the inverse of the Gudermannian is a primitive of 1/cos(x)...  In a Mercator conformal map, the distance to the equator of a point at latitude gd(u) is proportional to u. 
    sinc(x) = sin(x)/x  is the so-called  sine cardinal  or sampling function  (which arises to express  Fourier transtorms of rectangular functions).
  • Gamma function (see below).  Arguably, the most common special function, or the least "special" of them.  The other transcendental functions listed below are called "special" because you could conceivably avoid some of them by staying away from many specialized mathematical topics.  On the other hand, the Gamma function y = G(x) is most difficult to avoid.
  • Elliptic functions and elliptic integrals.
  • Exponential & logarithm integral (Ei, li), sine & cosine integral (si, ci).
  • Bessel functions.
  • Lambert's W function (multivalued); if y = W(x), then x = y exp(y).
  • Riemann's Zeta function.  A simple function with a rich structure, best known for its nontrivial zeroes  (its trivial zeroes are the even negative integers).  Infinitely many of these have been shown (by G.H. Hardy) to have a real part of ½, and billions of them have actually been found on that critical line, but it's still not known whether all of them are there, as Bernhard Riemann (1826-1866) first conjectured in 1859.  The far-reaching implications of this statement, known as the Riemann Hypothesis, make it the  most important unproved mathematical proposition  of our times.
  • ... ...

As advertised, the list is endless...

(2006-11-20)   The Trigonometric Circle Diagram
Memorize 3 basic trigonometric functions (out of 6)

Modern usage retains mostly the three trigonometric functions depicted at right:  sine (sin)  cosine (cos)  and  tangent (tan or tg).  The  cotangent function  (cot or cotg)  may be used for the reciprocal of the tangent.

The names of the reciprocals of the sine and cosine functions are deprecated.  If you must know, the secant (sec)  is the reciprocal of the cosine and the cosecant (csc or cosec) is the reciprocal of the sine function.  For the record:

tan q = sin q / cos q   sec q = 1 / cos q
cot q = cos q / sin q csc q = 1 / sin q

Each of those 6 trigonometric functions is the ratio of two sides in a right triangle where one of the acute angles is specified.  However, it's much better to consider the so-called  trigometric circle  of unit radius depicted above, which generalizes those functions  smoothly  to obtuse and/or negative values of the angle q  (e.g., the cosine of an obtuse angle is negative). 

Just memorize the above pictorial definitions of the 3 main trigonometric functions and ignore the rest  (unless you expect a quizz at school, that is).

(2002-01-29)   Trigonometry in a Nutshell
What are the basic laws used with trigonometric functions to obtain all the elements of a triangle when only some of them are known?

 Triangle A planar triangle is determined by 3 independent quantities. These could be the 3 sides (a,b,c, each of which is less than or equal to the sum of the other two), 1 angle and 2 sides, or 1 side and 2 angles. Instead of a side, you may be given some other length related to the triangle; for example, the radius r of the circumscribed circle (see figure at right). The angular data is usually (but not always) expressed directly in terms of the inside angles between sides (the angles a, b, g, which add up to p).

The Law of Sines :
2 r   =   a   =   b   =   c
Vinculum Vinculum Vinculum
sin a sin b sin g
The Law of Cosines : c2   =   a2 + b2 - 2 ab cos g
The Law of Tangents :    
 a - b   =   tg (a/2 - b/2)
Vinculum Vinculum
 a + b tg (a/2 + b/2)

The Law of Sines may well be the most useful of the three.  It delivers directly any missing quantity in all basic cases not covered by the next paragraph...

If the 3 sides are given, the Law of Cosines gives you any angle you may want (via its cosine). The Law of Cosines will also give you the missing side (c) when a, b and g are given  (the "SAS case", in high-school parlance).  Having the three sides, you could then obtain either of the missing angles by using the Law of Cosines again. However, it's more elegant and more direct to compute the missing angles with the Law of Tangents (especially, if you do not care about the value of the missing side c): Since you know a/2+b/2 (it's equal to p/2-g/2), the Law of Tangents gives you the (tangent of the) angle a/2-b/2. and the missing angles are simply the sum and the difference of a/2+b/2 and a/2-b/2.

 Law of Sines

Proofs :

The Law of Sines can be proved by remarking that, if O is the center of the circumscribed circle, one may consider an isoceles triangle like OBC which has two sides of length  r  forming an angle 2a.  The length of the base  (a)  is twice the side opposite to an angle  a  in a right triangle of hypotenuse  r.  That proof was published in 1342 by the  Provençal  rabbi  Levi ben Gerson (1288-1344)  who is variously known as "RaLBaG", Levi ben Gershon, Léon de Bagnols, Magister Leo Judaeus, or  Gersonides  [son of Gershon ben Solomon d'Arles].

The Law of Cosines is best proved (or memorized) in the modern context of vector algebra:  Consider two vectors U (from A to C) and V (from A to B), the vector  U-V goes from B to C and its square (U-V)2  is  |U|2 - 2 U.V + |V|2. The scalar product ("dot product") U.V is equal to the cosine of the angle formed by the two vectors multiplied by the product of their magnitudes.  Franciscus Vieta 

The Law of Tangents was first stated around 1580 by François Viète (Viette, or Vieta).  It has always been the least popular of the three, and it's not always found in textbooks...  It may be proved by expressing the ratio (a - b)/(a + b) in terms of sin a and sin b, using the Law of Sines. The result is then immediately obtained from the following identity for the denominator and from its counterpart for the numerator (replace b by -b).

sin a + sin b   =   2 sin(a/2 + b/2) cos(a/2 - b/2)

(2003-06-20)     Spherical Trigonometry
What are some basic relations applicable to spherical triangles?

spherical triangle  is a figure on the surface of a sphere of radius R, featuring three sides which are arcs of great circles (a "great circle" is the intersection of the sphere with a plane containing the sphere's center).  Each such figure divides the surface of the sphere into two parts, whose areas add up to 4pR.  Unless otherwise specified, the smaller part is usually considered the "inside" of the triangle, but this need not always be so...

The study of spherical triangles is often called spherical trigonometry and is about as ancient as the simpler planar trigonometry summarized above.

The internal angles of a spherical triangle always add up to more than a flat angle (of p radians).  Expressed in radians, the difference (denoted e, with 0< e <4p) is usually called the spherical excess, a term coined around 1626 by the French-born Dutch mathematician Albert Girard (1595-1632), who showed that the surface area of a spherical triangle is simply equal to:

e R 2   =   ( (a + b + g) - p ) R 2

There are some striking similarities between the two kinds of trigonometries, including the spherical Law of Sines of Abu'l-Wafa (940-998):

sin a   =   sin b   =   sin c
Vinculum Vinculum Vinculum
sin a sin b sin g

In this, a, b and c are the angular "lengths" of the sides (as seen from the sphere's center); they are the curvilinear distances along the great arcs, using R as a unit.  The spherical excess may also be expressed in terms of these quantities and the semiperimeter  s = ½ (a+b+c), using the spherical equivalent of Hero's formula:

L'Huilier's Theorem
  [tg(e / 4)]2   =   tg[s / 2]  tg[(s-a) / 2]  tg[(s-b) / 2]  tg[(s-c) / 2]  

This beautiful formula is named after the Swiss mathematician Simon L'Huilier (1750-1840) who was once a teacher of Charles-François Sturm in Geneva.  (His last name is sometimes also spelled "L'Huillier" or "Lhuilier".)

(V. R. of India. 2000-10-16)
Let   cos A + cos B = 2p   and   sin A + sin B = 2q .
Prove that     tan A/2 + tan B/2   =   2q / (p2 + q2 + p)

Let u be tan(A/2) and v be tan(B/2). We have cos(A) = (1-u2)/(1+u2), sin(A) = 2u/(1+u2), and similar relations for B and v.  Therefore:

2p   =   (1-u2 ) / (1+u2 )  +  (1-v2 ) / (1+v2 )
2q   =   2u / (1+u2 )  +  2v / (1+v2 )

Expressions such as these, which are symmetrical with respect to u and v, may be expressed in terms of the sum X=u+v and the product Y=uv. For example u2+v2 is X2-2Y and (1+u2)(1+v2) is 1+X2-2Y+Y2 or X2+(1-Y)2. The above two relations thus become:

2p (X2+(1-Y)2)   =   (1-u2)(1+v2) + (1-v2)(1+u2)
=   2-2Y2   =   2(1-Y)(1+Y)
and     2q (X2+(1-Y)2)   =   (2u)(1+v2)+(2v)(1+u2)
=   2X + 2XY   =   2X (1+Y)

Adding or subtracting these two after multiplying each by either (1-Y) or X (and removing the nonzero factor X2+(1-Y)2 which turns up) greatly simplifies this system of equations, which boils down to a linear system:

pX   =   q(1-Y)     and     p (1-Y)+ q X   =   (1+Y)

This may be rewritten

p X  +  q Y   =   q     and     q X  -  (p+1) Y   =   (1-p)

Solving for  X  gives   X [p(p+1)+q2]  =  q(p+1)+q(1-p)  =  2q,  which is the desired relation:

tan(A/2) + tan(B/2)   =   X   =   2q / (p2 + q2 + p)

Also, we may as well solve for  Y  to obtain another interesting relation:

tan(A/2) ´ tan(B/2)   =   Y   =   (p2 + q2 - p) / (p2 + q2+p)

Those two results are equivalent to the statement that the two roots of the following quadratic equation in  t  are  tan(A/2)  and  tan(B/2) :

( p 2 + q 2 + p )  t 2   -   2 q t   +   ( p 2 + q 2 - p )     =   0

brentw (Brent Watts of Hickory, NC. 2001-03-11)
Show that   | sin (x + iy) | 2   =   sin 2(x) + sinh 2(y)

The well-known relation exp(x+iy)=exp(x)(cos(y)+ i sin(y)) may be turned into a definition of the cosine and sine function (since x and y need not be real in this). In particular, exp(iz)=cos(z)+ i sin(z), so sin(z)=(exp(iz)-exp(-iz))/2i.  Therefore:

sin(x+iy)  =  [eix-y - e-ix+y ] / 2i
=  [e-y(cos(x)+ i sin(x)) - expy(cos(x)- i sin(x))]/2i
=  [cos(x)(e-y - ey ) + i sin(x)(e-y + ey )] / 2i

So far, we did not assume that x and y were real, now we do:  |z| 2  is the sum of the squares of the real and imaginary parts of z.  When z is the last of the above expressions, this translates into

| sin(x+iy) | 2  =  [cos2(x)(e-y - ey )2 + sin2(x)(e-y + ey )2 ] / 4
=  [(e-2y + e2y ) - 2 cos(2x) ] / 4

This expression can be idenfified with the given one by noticing that:

  •   sin 2 (x)   =   1/2 - cos(2x)/2
  • sinh 2 (y)   =   (e 2y + e-2y )/4 - 1/2

Therefore, the entire expression is indeed sin2(x) + sh2(y), as advertised.

(J. S. of Canada. 2000-10-15)
How do you solve these equations to exact values for x?
  1.   ln(x-2) - 3   =   ln(x+1)
  2.   sin(2x) sin(x)  +  cos(x)   =   0

1)   If   ln(x-2)-3 = ln(x+1)   then   ln((x-2)/(x+1)) = ln(e3) so we must have (x-2) = (x+1)e3   and x can only be equal to   (2+e3)/(1-e3).
    Now, however, this value of x happens to be negative (it's about -1.157) which makes it unacceptable, since both (x-2) and (x+1) should be positive (or else you can't take their logarithm).  Therefore, the original equation does not have any solutions at all!

2)   Rewrite   sin(2x)sin(x)+cos(x) = 0   as   2 cos(x)sin(x)sin(x) + cos(x) = 0, or   cos(x)[2sin2(x)+1] = 0.  As the second factor cannot be zero,  this equation boils down to   cos(x) = 0,  which has infinitely many solutions of the form   x = (k+½)p,  where k is any integer  (positive or not).

FlyingHellfish (2003-07-28)
In a broken calculator, only the 6 functions shown at right are available.  Can any positive rational number be obtained from an initial 0?

Yes.  Actually, Ö[p/q] can be obtained for any positive integers p and q, since:

If p < q, then     Ö[p/q]   =   sin arctan Ö[p/(q-p)]
If p > q, then     Ö[p/q]   =   tan arccos Ö[q/(p+q)]
=   tan arccos sin arctan Ö[q/p]
=   tan arccos sin arctan sin arctan Ö[q/(p-q)]

Using whichever relation is relevant, we may thus reduce any case to a simpler one, until we're faced with  p = q,  which we solve by pushing  cos  once.  QED

There's (almost) no need to say that the above shows that all positive rationals can be so obtained, since each of them is the square root of its square.

For example, if we wish to obtain 5/8, we observe that it's the square root of 25/64,  which is the  sin arctan  of the square root of 25/39, itself the  sin arctan  of the square root of 25/14, itself the  tan arccos sin arctan sin arctan  of the square root of 14/11, itself the  tan arccos sin arctan sin arctan  of the square root of 11/3, itself the  tan arccos sin arctan sin arctan  of the square root of 3/8, itself the  sin arctan  of the square root of 3/5, itself the  sin arctan  of the square root of 3/2, itself the  tan arccos sin arctan sin arctan  of the square root of 2, itself the  tan arccos sin arctan sin arctan  of (the square root of)  1, which is, of course, the  cos  of 0...  Only  39  keys to press on that broken calculator.

It's irrelevant whether the calculator works in degrees or in radians, since we only use trigonometric functions on angles obtained from inverse trigonometric functions, except for the initial zero angle  (either 0 degrees or 0 radians).

( John of Garland, TX. 2000-11-19)
How are the values of trigonometric functions calculated?
For example, how do we determine that sin(32°) = 0.52991?

Basically, the following relation is used:
sin(x) = x - x3/6 + x5/120 - x7/5040 + x9/362880 - ... + (-1)k x2k+1/(2k+1)! + ...

To use this for actual computations, you've got to remember that x should be expressed in radians (1° = p/180 rad). In your example, x = 32 ° = 0.558505360638... rad. The series "converges" very rapidly:

After 1 term,  S = 0.55850536063818
After 2 terms, S = 0.52946976180816
After 3 terms, S = 0.52992261296708
After 4 terms, S = 0.52991924970365
After 5 terms, S = 0.52991926427444
After 6 terms, S = 0.52991926423312
After 7 terms, S = 0.52991926423332
(no change at this precision after this)

Your computer and/or calculator uses this along with a technique called economization (the most popular of which is the Chebyshev economization) which allows a polynomial of high degree (or any reasonable function) to be very well approximated by a polynomial of lower degree.

In the case of the sine function, the convergence of the above series is so good that economization only saves you a couple of multiplications for a given precision. In some other cases (like the atan function), it is quite indispensable.

Footnote: about atan: The atan function has a nice Chebyshev expansion which allows one to bypass the intermediates step of a so-called Taylor expansion like the above. This is rather fortunate because the convergence of atan's Taylor expansion is quite lousy when x is close to 1. Modern atan routines use an economized polynomial for x between 0 and 1, and reduce the computation of atan(x) to that of atan(1/x) when x is above 1.
See the following article for more details...

(2000-11-19)   What is Chebyshev economization?

Over a finite interval, it is always possible to approximate a continuous function with arbitrary precision by a polynomial of sufficiently high degree.  In  some cases [one example is the sine function in the previous article] truncation of the function's Taylor series works well enough.  In other cases, the Taylor series may either converge too slowly or not at all  (the function may not be analytic or, if it is analytic, the radius of convergence of its Taylor series may be too small to cover comfortably the desired interval).

If a good polynomial approximation of the continuous real function  f (x)  is desired over a finite interval, the following approach may be used and is in fact the most popular one.  We may consider without loss of generality that the desired range of  x  is [-1,1] (if it's not, a linear change of variable will make it so).  Thus, a new variable q  (whose range is [0,p] )  can be introduced via the relation  cos q = x.  Either variable is a  decreasing  function of the other.

The fundamental remark is that cos(nq) is a polynomial function of cos(q). In fact, either of the following relations defines a polynomial of degree n known as the Chebyshev polynomial [of the first kind] of degree n. The symbol "T" is conventionally used for these because of alternate transliterations from Russian, like Tchebycheff or Tchebychev which are a better match for the Russian pronounciation  (the spellings "Chebychev" and "Tchebyshev" also appear).

cos(nq) = Tn(cos q)     or     ch(nq) = Tn(ch q)     [ch = hyperbolic cosine]

The trigonometric formula   cos(n+2)x = 2 cos x cos(n+1)x - cos nx   translates into a simple recurrence relation which makes Chebyshev polynomials very easy to tabulate:  Tn+2(x)   =   2x Tn+1(x) - Tn(x)

T0(x)  =   1
T1(x)  =   x
T2(x)  =   -1+2x2
T3(x)  =   -3x+4x3
T4(x)  =   1-8x2+8x4
T5(x)  =   5x-20x3+16x5
T6(x)  =   -1+18x2-48x4 +32x6
T7(x)  =   -7x+56x3-112x5 +64x7
T8(x)  =   1-32x2+160x4 -256x6+128x8

We must remark prominently that, if   y2 = x2-1   (y need not be real ),  then:

Tn(x) = [ (x+y)n + (x-y)n ] / 2

This is a consequence of  de Moivre's relation (with x = cos q  and y = i sin q ):

[ cos q + i sin q ] n   =   exp(i q) n   =   exp(i nq)   =   cos nq + i sin nq

Now,  f (cos q)  is clearly an even function of q  which is continuous when  f  is.  As such, it has a tame Fourier expansion which contains only cosines and translates into the so-called Chebyshev-Fourier expansion of  f(x):

 f (cos q) = co /2 + 
 cn cos(nq)     therefore:     f (x) = co /2 + 

The last expression is a series which is always convergent.  For "infinitely smooth" functions, it converges exponentially fast (as a function of n, the coefficient has to be smaller than the reciprocal of a polynomial of degree k+1, for any k, or else the Fourier series of the k-th derivative of  f (cos q) would not converge).   Joseph Fourier 
 (1768-1830) This is much more than what can be said about a Taylor power series...  A truncated Fourier-Chebyshev series is thus expected to give a much better approximation than a Taylor series truncated to the same order.

What is known as Chebyshev economization is often limited to the following dubious technique: Take a good polynomial approximant with many terms (possibly coming from a Taylor expansion) and express it as a linear combination of Chebyshev polynomials (whose coefficients may be obtained from the inversion formula below). This expression may be truncated at some low order to obtain a good approximation as a polynomial of lower degree.

A better approach, whenever possible, is to compute the exact Chebyshev expansion of the target function and to truncate that in order to obtain a good approximation by a polynomial of low degree...  The following inversion formula can be used for to obtain the Chebychev expansion  [watch out for the explicit halving of c]  of an  analytic  function given by its Taylor expansion:

 f (x)   =    
 an x n      =   ½ co  +  
 cn   =    2  
  a 2p+n

x    =     T1(x)
2 x2= 1+ T2(x)
4 x3  =   3 T1(x)+ T3(x)
8 x4= 3+ 4 T2(x) + T4(x)
16 x5= 10 T1(x)+ 5 T3(x) + T5(x)
32 x6= 10+ 15 T2(x) + 6 T4(x)+ T6(x)
2 n-1  x n    =    
 Tn-2k (x)     +    either... or   ½ C(n, n/2 )   if n is even.
  0   if n is odd.

The above complete inversion formula (infinite sum) is occasionally handy, but one may also always obtain the coefficients cn via the Euler formulas, which give:

cn   =     2    ó
f (x) Tn(x)   dx     =     2    ó
  f (cos q) cos(nq) dq
Vinculum Vinculum Vinculum
p Vinculum p
 Ö 1-x2

In at least one (important) case, we may even obtain the Chebyshev expansion directly by algebraic methods...  Consider the arctangent function, which gives the angle in radians between -p/2 and p/2 whose tangent equals its given [real] argument.  That function is variously abbreviated Arctg (Int'l/European), arctan (US), atg or atan (computerese). The following relation is true for small enough arguments.  [It's  true modulo p for unrestricted arguments, because of the formula giving tg(a+b) as (u+v)/(1-uv) if u and v are the respective tangents of a and b.]  This may thus be considered an algebraic relation between formal power series:

Arctg( (u+v)/(1-uv) )   =   Arctg(u) + Arctg(v)

With this in mind, we may as well use this formal identity for the complex numbers u = k [x+iÖ(1-x2 )] and v = k [x-iÖ(1-x2 )], so that  2kn Tn(x) = (u n + v n).  This turns the RHS of the above identity directly into a Chebyshev expansion where the coefficient cn is simply the coefficient of the arctangent power series multiplied by 2kn.  On the other hand, the LHS becomes Arctg(2kx/(1-k2 )).  If we let k be Ö2-1, this boils down to Arctg(x) and we have:

Arctg(x)   =     ån   [2(Ö2-1)2n+1 (-1)n / (2n+1)]   T2n+1(x)
=   2(Ö2-1)   ån   [(2Ö2-3)n / (2n+1)]   T2n+1(x)

That's [almost] all there is to it: We got the Chebyshev expansion at very little cost! How good is the convergence of this series? Well, we may first remark that it converges even if the magnitude of x exceeds unity. More precisely, when x is larger than 1, Tn(x) is asymptotically equal to half the n-th power of  x+Ö(x2-1) , a quantity which equals the reciprocal of Ö2-1 when x is Ö2. Therefore, the series converges if and only if the magnitude of x is less than (or equal to) Ö2.

More importantly, when the magnitude of x is not more than 1, a partial sum approximates the whole thing with an error smaller than the coefficient of the first discarded term. Suppose we want to use this to find a polynomial approximant of the arctangent function at a precision of about 13 significant digits (we need it only over the interval [-1,1], as we may obtain the arctangent of x for x>1 as p/2 minus the arctangent of 1/x). We find that for 2n+1=31, the relevant coefficient is about 0.88 10-13 so that the corresponding term is just about small enough to be dropped. The method will thus give the desired precision with an odd polynomial of degree 29, whose value can be computed using 16 multiplications and 14 additions. A similar accuracy would require about 10 000 000 000 000 operations with the "straight" Taylor series... Some economization, indeed!

The above "formal" computation gives the same results as the (unambiguous) relevant Euler formula for the coefficients of the Chebyshev expansion of the arctangent function. This may puzzle a critical reader, since the whole thing seems to work as long as the quantity 2k/(1-k2 ) is equal to unity, and this quadratic condition is true not only when k is Ö2-1, but also for the alternate root -(Ö2+1) as well. This latter value, however, leads to a formal Chebyshev series which diverges for any value of x...

(Mark Barnes, UK. 2000-10-24)
What can you tell me about the Gamma function? I can work out values for G(x) if x is integral or x is an integer plus one half. How can I calculate values for G(x), if x is some other value, like 2.8 or 67/9? What actually is the function?

The following intimidating definitions of the transcendental Gamma function hide its simple nature:  G(z+1) is merely the generalization of the factorial function (z!) to all real or complex values of the number z  [besides negative integers].

 Leonhard Euler 
  • Euler integral of the 2nd kind (valid only if Re(z)>0):
    G(z) = ò0¥e-t tz-1 dt
    G(z) = ò1¥e-t tz-1 dt + å (-1)n/(n!(n+z))
  • Gauss' definition: 
    G(z) = limn®¥ nzn! / (z(z+1)...(z+n))
  • Weierstrass's definition:  (g being the Euler-Mascheroni constant, namely 0.5772156649015328606065120900824024310421593359399235988... ):
    G(z) = e-g z / z Õ ez/n/(1+z/n)

G(z) has an elementary expression only when z is either a positive integer n, or a positive or negative half-integer  (½+n  or  ½-n):

(n-1)!         G(1/2 + n)
 (2n-1)!!          G(1/2 - n)
(-2)n Öp
vinculum vinculum



In this, k! ("k factorial") is the product of all positive integers less than or equal to k, whereas k!! ("k double-factorial") is the product of all such integers which have the same parity as k, namely k(k-2)(k-4)... Note that k!, is undefined (¥) when k is a negative integer (the G function is undefined at z = 0,-1,-2,-3,... as it has a simple pole at z = -n with a residue of (-1)n/n! , for any natural integer n). However, the double factorial k!! may also be defined for negative odd values of k:  The expression (-2n-1)!! = -(-1)n / (2n-1)!! ) may be obtained through the recurrence relation  (k-2)!! = k!! / k , starting with k=1.  In particular (-1)!! = 1, so that either of the above formulas does give G(1/2) = Öp , with n=0. (You may also notice that either relation holds for positive or negative values of n.)

When the real 2x is not an integer, we do not know any expression of G(x) in terms of elementary functions:

G(1/3) = 2.67893853470774763365569294097467764412868937795730...
G(1/4) = 3.62560990822190831193068515586767200299516768288006...
G(1/5) = 4.59084371199880305320475827592915200343410999829340...

The real [little known] gem which I have to offer about numerical values of the Gamma function is the so-called "Lanczos approximation formula" [pronounced "LAHN-tsosh" and named after the Hungarian mathematician Cornelius Lanczos (1893-1974), who published it in 1964]. Its form is quite specific to the Gamma function whose values it gives with superb precision, even for complex numbers. The formula is valid as long as Re(z) [the real part of z] is positive. The nominal accuracy, as I recall, is stated for  Re(z) > ½, but it's a simple application of the "reflection formula" (given below) to obtain the value for the rest of the complex plane with a similar accuracy. The Lanczos formula makes the Gamma function almost as straightforward to compute as a sine or a cosine.  Here it is:

G(z) = [1+C1/(z)+C2/(z+1)+ ... +Cn/(z+n-1) + e(z)] ´ Ö(2p) (z+p-1/2)z-1/2 / ez+p-1/2

e(z) is a small error term whose value is bounded over the half-plane described above. The values of the coefficients Ci depend on the choice of the integers p and n. For p=5 and n=6, the formula gives a relative error less than 2.2´10-10 with the following choice of coefficients: C1=76.18009173, C2= -86.50532033, C3=24.01409822, C4= -1.231739516, C5=0.00120858003, and C6= -0.00000536382.

I used this particular set of coefficients extensively for years (other sources may be used for confirmation) and stated so in my original article here.  This prompted Paul Godfrey of Intersil Corp. to share a more precise set and his own method to compute any such sets (without the fear of uncontrolled rounding errors). Paul has kindly agreed to let us post his (copyrighted) notes on the subject here.

Some of the fundamental properties of the Gamma function are:

  • Reflection formula: G(z)G(1-z) = p/sin(pz)
  • Recursion formula: G(1+z) = zG(z)
  • Exact values (when n is an integer; see above when n is negative):
    G(n) = (n-1)! and G(n+1/2) = Öp (2n)! / (n!4n)
  • Multiplication formula (also called "duplication formula" when n=2):
    G(nz) = (2p)(1/2-n/2) n(nz-1/2) [ G(z) + ... + G(z+k/n) + ... + G(z+(n-1)/n) ]

Other interesting remarks about the Gamma function include:

  • | G(ix) | 2   =   p / (x sinh px )     for x real

Louis Vlemincq  (Belgium.  2004-02-19; e-mail)
How is the equation   t + ln(t) = T ln( I / i )   solved for t and i ?

Taking the exponential of both sides makes it easy to solve for i:

t et   =   [I / i] T
i   =   I / ( t e t ) 1/T

To solve for t, you have to use Lambert's W function, one of the more common "special" functions presented above:  Apply W to both sides of the first of the above equations.  Since, by definition, W(t exp(t)) is equal to t,  we obtain:

t   =   W( [I / i] T )

This solution is valid for positive values of t  (the original equation does not make sense for negative ones).  By itself, the equation  x = t exp(t)  has 2 real solutions for t when x is between -1/e and 0 and no real solution when x is less than -1/e.

The radius of convergence of the  Taylor series  of W is 1/e (0.36787944...)


   (-n) n-1  
  z n       [ for |z| < 1/e ]


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On 2004-02-20, Louis Vlemincq wrote:
Thanks a lot for your kind, quick and learned answer.
It will be most useful to me.
 Best regards,
Louis Vlemincq,   Transmission Specialist,  BelcomLab.
BELGACOM   /   2, rue Carli, 1140 Evere   /   Belgium
visits since Dec. 6, 2000
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