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

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(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.

Taylor's theorem  was brought to great prominence in 1772 by Joseph-Louis Lagrange (1736-1813) who declared it the basis for differential calculus.

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(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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Brian Keiffer (Yahoo! 2011-08-07)   Formal properties of  exp  series.
Defining   exp (x) = S_n 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).  In fact, rearranging the terms of any such real series can make it converge to any arbitrary limit.

=

¥ å n=0

n å k=0

xk yn-k k! (n-k)!

=

¥ å n=0

(x+y)n n!

=    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  

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

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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...

This is a special case  (for  z = 2,  k = 3,  a_n = ¹/_n! )  of the following problem:

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

fk (z)  =  å n  a kn z kn

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

1  +  w  +  w²  +  ...  +  w^k-1   =   0             w^k   =   1

This lets the quantity   1  +  w^j  +  ...  +  w^{(k-1) j}   be  k  when  j  is a multiple of  k  and  vanish  otherwise.  Equating corresponding coefficients of  a_j z ^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