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© 2000-2017   Gérard P. Michon, Ph.D.

Power Series


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Complex Variables, Complex Analysis   by  John H. Mathews  (2000).
Complex Variables, Contour Integration  by  Joceline Lega  (1998).

Wikipedia :   Power Series   |   Formal Power Series   |   Taylor Series   |   Analytic Continuation
The Bieberbach conjecture (1916) was proven by Louis de Branges in 1985.


Power Series and Analytic Continuations

 Brook Taylor 
 Brook Taylor 
 (1685-1731) (2009-01-07)   Taylor's Expansion   (1712, 1715)
Smooth functions as sums of power series.

Brook Taylor (1685-1731) invented the  calculus of finite differences  and came up with the fundamental technique of  integration by parts.

Nowadays, we call  Taylor's Theorem  several variants of the following expansion of a smooth function  f  about a regular point  a, in terms of a polynomial whose coefficients are determined by the successive derivatives of the function at that point:

f (a + x)   =   f (a)  +  f ' (a) x  +  f '' (a) x/2  +  ...  +  f (n) (a) xn / n!  +  ...

The variants of Taylor's theorem differ by the distinct explicit expressions that can be substituted for the trailing ellipsis, which stands for the difference between the value of the function and that of Taylor's polynomial.

Taylor published two versions of his theorem in 1715.  In a letter to his friend John Machin (1680-1751)  dated July 26, 1712, Taylor gave Machin credit for the idea.  Several variants or precursors of the theorem had also been discovered independently by  James Gregory (1638-1675), Isaac Newton (1643-1727), Gottfried Leibniz (1646-1716), Abraham de Moivre (1667-1754) and Johann Bernoulli (1667-1748).  A Taylor expansion about the origin  (a = 0)  is often called a  Taylor-Maclaurin expansion,  in honor of Colin Maclaurin (1698-1746) who focused on that special case in 1742.

 Lagrange (1736-1813) (2015-04-19)   Basing  calculus  on Taylor's expansions  (1772)
Lagrange's strict  algebraic  interpretation of differential calculus.

Taylor's theorem  was brought to great prominence in 1772 by Joseph-Louis Lagrange (1736-1813) who declared it the basis for differential calculus  (he made this part of his own lectures at  Polytechnique  in 1797).

Arguably, this was a rebuttal to religious concerns which had been raised in 1734  (The Analyst)  by George Berkeley (1685-1753)  Bishop of Cloyne (1734-1753)  about the  infinitesimal  foundations of Calculus.

The mathematical concepts behind differentiation and/or integration are so pervasive that they can be introduced or discussed outside of the historical context which originally gave birth to them, one century before Lagrange.

The starting point of Lagrange is an exact expression,  valid for any  polynomial  f  of degree  n  or less,  in any commutative  ring :

f (a + x)   =   f (a)  +  D1 f (a) x  +  D2 f (a) x2  +  ...  +  Dn f (a) xn

In the ordinary interpretation of Calculus  [over any field of characteristic zero]  the following relation holds, for any polynomial  f :

D0 f (a)   =   f (a)         Dk f (a)   =   f (k) (a) / k!

However, the above expansion remains true in very general circumstances when neither the reciprocal of  k!  nor higher-order derivatives need be defined.  Lagrange's general definition of   Dk f (a)   is strictly based on the binomial theorem.  Dk f  is always a polynomial of degree  n-k.

 Come back later, we're
 still working on this one...

"Théorie des fonctions analytiques contenant les principes du calcul différentiel, dégagés de toute considération d'infiniment petits ou d'évanouissants, de limites ou de fluxions et réduits à l'analyse algébrique des quantités finies"     by  Joseph-Louis Lagrange  (1797)  Journal de l'École polytechnique, 9, III, 52, p. 49

(2008-12-23)   Radius of Convergence of a Complex Power Series
A complex power series converges inside a disk and diverges outside of it  (the situation at different points of the boundary circle may vary).

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 still working on this one...

Brian Keiffer (Yahoo! 2011-08-07)   Formal properties of  exp  series.
Defining   exp (x) = Sn xn/n!   and  e = exp (1)  prove that  exp (x) = e x

In their  open  disk of convergence  (i.e., circular boundary excluded)  power series are  absolutely convergent  series.  So, in that domain, the sum of the series is unchanged by modifying the order of the terms  (commutativity)  and/or  grouping them together  (associativity).  This allows us to establish directly the following fundamental property  (using the binomial theorem):

exp (x)  exp (y)   =   exp (x+y)

Such manipulations are disallowed for convergent series that are not  absolutely convergent  (which is to say that the series consisting of the absolute values of the terms diverges).  Rearranging the terms of any such real series can make it converge to any arbitrary limit !

exp (x)  exp (y)    =    (
n = 0
xn/n! )(
n = 0
yn/n! )    =   
(xn/n!) (ym/m!)

xk yn-k
k! (n-k)!
   =    exp (x+y)

This lemma shows immediately that  exp (-x) = (exp x)-1.  Then, by induction on the absolute value of the integer  n, we can establish that:

exp (n x)   =   (exp x) n

With  m = n  and  y = n x,  this gives   exp (y)   =   (exp y/m)m .   So :

exp (y / m)   =   (exp y) 1/m

Chaining those two results, we obtain, for any rational  q = n/m

exp (q y)   =   (exp y) q

By continuity, the result holds for any real  q = x.  In particular, with y = 1:

exp (x)   =   (exp 1) x   =   e x   QED

(2008-12-23)   Analytic Continuation   (Weierstrass, 1842)
Power series that coincide wherever their disks of convergence overlap.

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 still working on this one...

Loosely speaking,  analytic continuations  can make sense of  divergent series  in a consistent way.  Consider, for example, the classic summation formula for the  geometric series,  which converges when  |z| < 1 :

1  +  z  +  z +  z +  z +  ...  +  z +  ...   =   1 / (1-z)

The right-hand-side always makes sense, unless  z = 1.  It's thus tempting to equate it  formally  to the left-hand-side, even when the latter diverges!  This viewpoint has been shown to be consistent.  It makes perfect sense of the following "sums" of  divergent series  which may otherwise look like monstrosities  (respectively obtained for  z = -1, 2, 3) :

-  1  +  1  -  1  +  ...  +  (-1) +  ...   =   ½
1  +  2  +  4  +  8  +  16  +  ...  +  2 +  ...   =   -1
1  +  3  +  9  +  27  +  81  +  ...  +  3 +  ...   =   - ½

Divergent geometric series   |   1 + 2 + 4 + 8 +...   |   1 - 2 + 4 - 8 +...   |   Borel summation (1899)
Adding Past Infinity  &  Taming Infinity  by  Henry Reich  (Minute Physics)

Dimitrina Stavrova  (2008-12-22; e-mail)   Decimated Power Series
What is the sum of  8n / (3n)!  over all natural integers  n ?

Answer :   1/3 ( e2  +  2 cos (Ö3) / e )   =   2.423641733185364535425...

That's a special case  (for  z = 2,  k = 3,  an = 1/n! )  of the following problem:

For an integer  k  and a known series   f (z)  =  ån az n ,  find the value of:

    fk (z)  =  å n  a kn z kn    

The key is to introduce a primitive  kth  root of unity, like  w = exp (2pi / k).

1  +  w  +  w2  +  ...  +  wk-1   =   0             wk   =   1

This lets the quantity   1  +  wj  +  ...  +  w(k-1) j   be  k  when  j  is a multiple of  k  and  vanish  otherwise.  Equating corresponding coefficients of  az j , we obtain:

    f (z)  +  f (w z)  +  f (w2 z)  +  ...  +  f (wk-1 z)     =     k  fk (z)    

For  f (z)  =  e z   this gives the advertised result as  f3 (2)  in the form:

1/3 [ exp (2)  +  exp (2w)  +  exp (2w2 ) ]     where   w  =  ½ (-1 + i Ö3 )

On 2008-12-26,  Dimitrina Stavrova  wrote:   [edited summary]
I am greatly impressed by the quick and accurate generalization of my question, which gave me a deeper understanding of the related material.  Thank you for creating such a great site!
Dimitrina Stavrova, Ph.D.
Sofia, Bulgaria

Thanks for the kind words, Dimitrina.

å  xn / 3n!   in closed form

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