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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.

(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!  +  R_n (x)

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.

Other variants of Taylor's theorem differ by the distinct explicit expressions which can be given for the so-called  remainder  R_n .

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). 

The term  Taylor series  was apparently coined by  Simon Lhuilier (1785).

Taylor's theorem
 
Taylor's theorem  by  Robin Whitty  (Theorem of the Day #217).

(2019-11-25)   Several expressions for the Taylor remainder  R_n (x)
Due to Lagrange, Cauchy (1821), Young (1910) and Schlömilch (1923).

The  aforementioned  difference  R_n (x)  between the value of a function  f  and that of its Taylor polynomial  (at order n)  has a nice  exact  expression:

Taylor Remainder   (Brook Taylor, 1715)

Rn (x)   =    ò  x      0 (x-t)n n!   f  (n+1) (a+t)   dt

Proof :   If  n > 0 ,  we use  Taylor's own  integration by parts  to obtain:

R_n (x)   =   R_n-1 (x)  -  f  ⁽ⁿ⁾ (a)  xⁿ / n!

By  induction  on  n,  the general formula then follows from the trivial case  (n = 0)  which is just the  fundamental theorem of calculus:

R₀ (x)   =   f (a+x)  -  f (a)    

Taylor-Lagrange Formula :

Lagrange Remainder   (Joseph-Louis Lagrange)

$ q,   | q | ≤ | x | ,   Rn (x)   =    (x-q)n n!   f  (n+1) (a+q)   dt

Proof :   That's the result of applying the  mean-value theorem  to Taylor's original expression  (1715)  for  R_n(x)  as given in the previous section.  

Taylor-Cauchy Formula

Taylor-Young formula (1910) :   R_n (x)  is  negligible  compared to  xⁿ  as  x  tends to  0.  A formulation due to  William Henry Young (1863-1942).

R_n (x)   <<   x ⁿ

Mathworld :   Lagrange remainder   |   Cauchy remainder   |   Schlömilch remainder
 
Interpolation and Taylor's Theorem  (Mathematics StackExchange, 2014-04-14).

(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  +  ...  +  D_n 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)         D_k f (a)   =   f ^(k) (a) / k!

However, the expressions below are true even when neither the reciprocal of  k!  nor higher-order derivatives are defined,  over any commutative ring.

Lagrange's definitions of   D_k f (a)   are just based on the binomial theorem.  D_k f  is simply a polynomial of degree  k.  No divisions are needed.

The following manipulations are limited to the case when  f  is a polynomial whose order is at most  n.  So only finitely many terms are involved in the data and in the results.  With infinitely many terms, the convergence of neither would be guaranteed.

"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
 
Lagrange's algebraic basis for differential calculus (48:26)  by  N.J. Wildberger  (2013-08-14).

(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).

That disk is called the  disk of convergence.  Its radius is the  radius of convergence  and its boundary is the  circle of convergence.

The result advertised above is often called  Abel's power series theorem.  Although it was known well before him,  Abel is credited for making this part of a general discussion which includes the status of points  on  the circumference of the circle of convergence.  The main tool for that is  another  theorem due to Abel,  discussed in the  next section.

Radius of convergence

(2018-06-01)   Stolz Sector
Slice of the disk of convergence with its apex on the boundary.

Stolz angle  by  Andrzej Kozlowski   (Wolfram Demonstrations Project).
Stolz region.  Question of  Daniel  answered by  Robert Israel   (StackExchange, 2012-09-02).
 
Otto Stolz (1842-1905, PhD 1864)

(2021-08-08)   Composition of two  (formal)  power series.

Lagrange inversion formula   |   Composition of two power series  (Math Stack Exchange, 2017-05-27).   |   Composition of Power Series  by  John Armstrong  (The Unapologetic Mathematician).

(2019-12-06)   Lagrange Inversion Formula
Lagrange-Bürmann Inversion Formula.

Lagrange inversion formula   |   Joseph-Louis Lagrange (1736-1813)
Lagrange-Bürmann formula   |   Hans Heinrich Bürmann (c.1770-1817)

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, unless it's at infinity)  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 !

=

¥ 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  

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

In the realm of real or  complex  numbers,  two polynomials which coincide at infinitely many distinct points are necessarily equal  (HINT:  as a polynomial with infinitely many roots, their difference must be zero).

This result on polynomials doesn't have an immediate generalization to  analytic functions  for the simple reason that there are  analytic functions  with infinitely many zeroes.  The  sine  function is one example of an analytic function with infinitely  discrete  seroes. 

However,  an analytic function defined on a nonempty open domain can be extended in only one way to a larger open domain of definition which doesn't encircle any point outside the previous one.  Such an extension of an analytic function is called an  analytic continuation  thereof.

Divergent Series :

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  +  z²  +  z³  +  z⁴  +  ...  +  zⁿ  +  ...   =   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  +  ...  +  (-1)ⁿ  +  ...   =   ½
1  +  2  +  4  +  8  +  16  +  ...  +  2ⁿ  +  ...   =   -1
1  +  3  +  9  +  27  +  81  +  ...  +  3ⁿ  +  ...   =   - ½

Analytic continuation   |   Proof of uniqueness   |   Identity theorem
 
Several complex variables   |   Sheaf cohomology
 
Divergent geometric series   |   1 + 2 + 4 + 8 +...   |   1 - 2 + 4 - 8 +...   |   Borel summation (1899)
Adding Past Infinity  &  Taming Infinity  by  Henry Reich  (Minute Physics)
Visualizing analytic continuation (20:27)  by  Grant Sanderson (2016-12-09).
Analytic Continuation and the Zeta Function (49:33)  by  Zetamath (2021-12-16).

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,  a_n = ¹/_n! )  of this 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 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

The quantity   1  +  w^j  +  ...  +  w^{(k-1) j}   is  k  when  j  is a multiple of  k  and  vanishes  otherwise.  Matching 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

(2021-07-22)   Periodic Decomposition of a Power Series
A slight generalization of the technique introduced  above.

Instead of retaining only the terms of a power series whose indices are multiples of a given modulus  k,  we may wish to keep only indices whose  residues modulo  k  are a prescribed remainder  r.  Thus,  we're now after:

f k,r (z)  =  å n  a kn+r z kn+r

That can be worked out with our  previous result  (the special case  r = 0)  by applying it to the function  z^(k-r) f (z).  Using  w = exp(2pi/k),  we have:

z^(k-r) f (z)  +  (w z)^(k-r) f (w z)  +...+  (w^k-1 z)^(k-r) f (w^k-1 z)   =   k  z^(k-r) f _k,r (z)

Dividing both sides by  k z^(k-r),  using  w^k = 1,  we obtain the desired result:

In the example  k = 4  for  f = exp,  we have  w = i  and,  therefore:

f _4,r (z)   =   ¼  [ e^z  +  (-i)^r e^iz  +  (-1)^r e^-z  +  i^r e^-iz ]

That translates into four equations,  for  r = 0, 1, 2 or 3:

f _4,0 (z)   =   ¼  [ e^z  +  e^iz  +  e^-z  +  e^-iz ]   =   ½  [ ch z  +  cos z ]
f _4,1 (z)   =   ¼ [ e^z  -  i e^iz  -  e^-z  +  i e^-iz ]   =   ½  [ sh z  +  sin z ]
f _4,2 (z)   =   ¼  [ e^z  -  e^iz  +  e^-z  -  e^-iz ]   =   ½  [ ch z  -  cos z ]
f _4,3 (z)   =   ¼  [ e^z  +  i e^iz  -  e^-z  -  i e^-iz ]   =   ½  [ sh z  -  sin z ]

The whole machinery may be an overkill in this case,  where the above four relations are fairly easy to obtain directly from the expansions of cos, ch, sin and sh.  However,  that's a good opportunity to introduce the methodology which is needed in less trivial cases with other roots of unity...

Let's use this to compute the  deformed exponential  exp_q (z)  when  q = -1.

expq (z)

=

¥ å n = 0

zn n!

q n(n-1)/2

When  q = -1,  the value  of  q ^(n-1)n/2  depends only on what  n  is modulo 4:

n mod 4	0	1	2	3
(-1) (n-1) n/2	+1	+1	-1	-1

Therefore,  exp_-1 (z)   =   f_4,0 (z)  +  f_4,1 (z)  -  f_4,2 (z)  -  f_4,3 (z).  So:

exp-1 (z)   =   cos z  +  sin z   =   Ö2  sin (z+p/4)

The technique applies when  q  is a root of unity.  For example,  with  q = i  the series splits according to the residue of the index  n  modulo 8:

n mod 8	0	1	2	3	4	5	6	7
i (n-1) n/2	+1	+1	+i	-i	-1	-1	-i	+i

exp_i (z)    =     f_8,0  +  f_8,1  +  i f_8,2  -  i f_8,3  -  f_8,4  -  f_8,5  -  i f_8,6  +  i f_8,7

where   f 8,r  =  1/8

7 å m = 0

e-mr ip/4 exp ( z em ip/4 )

After a fairly tedious computation,  this boils down to:

expi (z)    =    ½ [ e (7z+1) i p/4  +  e 5z i p/4  -  e (3z+1) i p/4  +  e z i p/4 ]

The case  k = 3  (w = e^2ip/3 )  is much simpler,  entailing only a 3-way split:

n mod 3	0	1	2
w (n-1) n/2	+1	+1	w

exp_w (z)    =     f_3,0  +  f_3,1  +  w f_3,2

f _3,0 (z)   =   ¹/₃  [ e^z  +  e^{z w}  +  e^{z w*} ]
f _3,1 (z)   =   ¹/₃  [ e^z  +  w* e^zw  +  w  e^{z w*} ]
f _3,2 (z)   =   ¹/₃  [ e^z  +  w e^zw  +  w*  e^{z w*} ]

exp_w (z)    =   ¹/₃  [ (2+w) e^z  +  (1+2w*) e^zw  +  (2+w)  e^{z w*} ]

Period  m  (along  n)  of  exp ( n(n-1)pi/k )     (A022998)

k	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
m	1	2	3	8	5	12	7	16	9	20	11	24	13	28	15	32

Numericana :   Deformed exponential

(2021-10-05)   Finite-Difference Calculus   (FDC)
Applying the methods of calculus to discrete sequences.

Difference Operator  D   (discrete derivative) :

D f (n)   =   f (n+1) - f (n)

Like the usual  differential operator  (d)  this is a  linear  operator,  as are all  iterated  difference operators  D^k  recursively defined,  for  k ≥ 0 :

D⁰ f   =   f
D^k+1 f   =   D^k (D f )   =   D (D^k f )

Unlike the differential operator  d  of infinitesimal calculus,  the above difference operator  D  yields ordinary finite quantities whose products can't be neglected;  there's a third term in the corresponding  product rule :

D (uv)   =   (D u) v   +   u (D v)   +   (D u) (D v)

Falling Powers   (falling factorials) :

The number of ways to pick a sequence of  m  objects out of  n  possible choices  (allowing repetitions)  is  n^m,  pronounced  n  to the power of  m.

When objects already picked are disallowed,  the result is denoted  n^m  and called  n to the falling power of m.  It's the product of  m  decreasing factors:

n^m   =   (n)_m   =   n (n-1) (n-2) ... (n+1-m)

As usual,  that's  1  when  m = 0,  because it's the product of zero factors.  Falling powers are closely related to  choice numbers:

C(n,m)   =   nCm   =   C	n m	=	æ è	n m	ö ø	=	n!	=	nm	=	(n)m
							(n-m)! m!		m!		m!

Falling powers are to FDC what powers are to infinitesimal calculus since:

D n^m   =   m  n^m-1

Iterating this relation yields the pretty formula:

D^k n^m   =   m^k  n^m-k

James Gregory (1638-1675)

Gregory-Newton forward-difference formula :

f (n)   =    ¥ å k = 0   æ è n k ö ø   Dk f (0)

When  n  is a natural integer,  the right-hand side is a  finite  sum,  as all  binomial coefficients  with  k > n  vanish.

Proof :   By  induction  on  n  (the case  n = 0  being trivial):

Assuming the formula holds for a given  n,  we apply it to  D f  and obtain:

Note the zero leading term (k = 0) in the re-indexed rightmost sum.  We may add this finite sum termwise to the previous expansion of  f (n)  to obtain:

f (n+1)   =

¥ å k = 0

[

æ è

n k-1

ö ø

+

æ è

n k

ö ø

]

Dk f (0)

=

¥ å k = 0

æ è

n+1 k

ö ø

Dk f (0)

This says that the formula holds for  n+1    

Falling powers,  make the above look like  a  Taylor-MacLaurin  expansion:

f (n)   =   f (0)  +  Df (0) n  +  D² f (0) n²/2  +  ...  +  D^k f (0) n^k/k!  +  ...

When one  D^k  is zero,  so are all the subsequent ones and the above right-hand-side gives directly  f  as a polynomial function of  n.

Encyclopedia of Mathematics   |   Wikipedia   |   MathWorld
 
Finite Differences?  by  John D. Cook  (2009-02-01).
 
Calculus of Finite Differences  by  Andreas Klappenecker  (2009-02-01).
 
Newton Gregory backward Interpolation Formula (13:42)  by  Tessy Cyriac  (2020-04-18).
 
What comes next? (47:10)  by  Burkard Polster  (Mathologer, 2021-10-02).