Generalized Factorials and Taylor Expansionscourse1.winona.edu/eerrthum/UndergradResearch/... ·...

Factorials Examples Taylor Series Expansions Extensions

Generalized Factorials and Taylor Expansions

Michael R. PillaWinona State University

April 16, 2010

Michael R. Pilla Generalizations of the Factorial Function

Factorials

Recall the definition of a factorial

n! = n·(n−1)···2·1 =

(n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Define n!S = (an − a0)(an − a1) · · · (an − an−1).

We have a nice formula.

It works on subsets of N.

It is not theoreticallyuseful.

What about the order ofthe subset?

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Factorials

n! = n·(n−1)···2·1 = (n−0)(n−1)(n−2)···(n−(n−2))(n−(n−1)).

For S = {a0, a1, a2, a3, ...} ⊆ N,

Generalized Factorials

Bhargava: Let’s look at prime factorizations and play a gamecalled p-ordering for each prime p.

If the ordering works for all primes simultaneosly, then we canachieve nice formulas.

Theorem

If S = {ai} is p-ordered for all primes simultaneously then

n!S = |(an − a0)(an − a1) · · · (an − an−1)|.

If a subset cannot be simultaneously ordered, the formulas areugly (if and when they exist) and the factorials difficult tocalculate.

Goal: Extend factorials to “nice”, “natural” subsets of N thathave closed formulas.

Theorem

n!S = |(an − a0)(an − a1) · · · (an − an−1)|.

Theorem

n!S = |(an − a0)(an − a1) · · · (an − an−1)|.

Theorem

n!S = |(an − a0)(an − a1) · · · (an − an−1)|.

Theorem

n!S = |(an − a0)(an − a1) · · · (an − an−1)|.

Arithmetic Sets

Example

Consider the arithmetic set A = {2, 9, 16, 23, ...} (i.e. allintegers 2 mod 7). This is p-ordered for all primessimultaneously. Thus

1!A = (9− 2) = 7

= 711!

2!A = (16− 2)(16− 9) = 98

= 722!

3!A = (23− 2)(23− 9)(23− 16) = 2058

= 733!

n!A = 7nn!

Example

Let S = aN + b of all integers b mod a. The natural ordering isp-ordered for all primes simultaneously. Thus

n!aN+b = ann!

Arithmetic Sets

Example

1!A = (9− 2) = 7

= 711!

2!A = (16− 2)(16− 9) = 98

= 722!

3!A = (23− 2)(23− 9)(23− 16) = 2058

= 733!

n!A = 7nn!

Example

n!aN+b = ann!

Arithmetic Sets

Example

1!A = (9− 2) = 7

= 711!

2!A = (16− 2)(16− 9) = 98

= 722!

3!A = (23− 2)(23− 9)(23− 16) = 2058

= 733!

n!A = 7nn!

Example

n!aN+b = ann!

Arithmetic Sets

Example

1!A = (9− 2) = 7

= 711!

2!A = (16− 2)(16− 9) = 98

= 722!

3!A = (23− 2)(23− 9)(23− 16) = 2058

= 733!

n!A = 7nn!

Example

n!aN+b = ann!

Arithmetic Sets

Example

1!A = (9− 2) = 7 = 711!

2!A = (16− 2)(16− 9) = 98 = 722!

3!A = (23− 2)(23− 9)(23− 16) = 2058 = 733!

n!A = 7nn!

Example

n!aN+b = ann!

Arithmetic Sets

Example

1!A = (9− 2) = 7 = 711!

2!A = (16− 2)(16− 9) = 98 = 722!

3!A = (23− 2)(23− 9)(23− 16) = 2058 = 733!

n!A = 7nn!

Example

n!aN+b = ann!

Set of Squares

Example

The set Z2 = {0, 1, 4, 9, 16, ...} of square numbers admits asimultaneous p-ordering. Thus

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Set of Squares

Example

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Set of Squares

Example

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Set of Squares

Example

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Set of Squares

Example

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Set of Squares

Example

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Set of Squares

Example

1!Z2 = (1− 0) = 1

2!Z2 = (4− 0)(4− 1) = 12

3!Z2 = (9− 0)(9− 1)(9− 4) = 360

4!Z2 = (16− 0)(16− 1)(16− 4)(16− 9) = 20160

n!Z2 = (n2 − 0)(n2 − 1)(n2 − 4) · · · (n2 − (n − 1)2)

= (n − 0)(n + 0)(n − 1)(n + 1)...(n − (n − 1))(n + (n − 1))

2(2n − 1)(2n − 2) · · · (n)(n − 1) · · · (1)

=(2n)!

Twice Triangulars (Squares Modified)

Example

Likewise, one can show the set2T = {n2 + n | n ∈ N} = {0, 2, 6, 12, 20, ...} admits asimultaneous p-ordering. Thus

1!2T = (2− 0) = 2

2!2T = (6− 0)(6− 2) = 24

3!2T = (12− 0)(12− 2)(12− 6) = 720

4!2T = (20− 0)(20− 2)(20− 6)(20− 12) = 40320

n!2T = (2n)!

Example

1!2T = (2− 0) = 2

2!2T = (6− 0)(6− 2) = 24

3!2T = (12− 0)(12− 2)(12− 6) = 720

4!2T = (20− 0)(20− 2)(20− 6)(20− 12) = 40320

n!2T = (2n)!

Example

1!2T = (2− 0) = 2

2!2T = (6− 0)(6− 2) = 24

3!2T = (12− 0)(12− 2)(12− 6) = 720

4!2T = (20− 0)(20− 2)(20− 6)(20− 12) = 40320

n!2T = (2n)!

Example

1!2T = (2− 0) = 2

2!2T = (6− 0)(6− 2) = 24

3!2T = (12− 0)(12− 2)(12− 6) = 720

4!2T = (20− 0)(20− 2)(20− 6)(20− 12) = 40320

n!2T = (2n)!

Example

1!2T = (2− 0) = 2

2!2T = (6− 0)(6− 2) = 24

3!2T = (12− 0)(12− 2)(12− 6) = 720

4!2T = (20− 0)(20− 2)(20− 6)(20− 12) = 40320

n!2T = (2n)!

Example

1!2T = (2− 0) = 2

2!2T = (6− 0)(6− 2) = 24

3!2T = (12− 0)(12− 2)(12− 6) = 720

4!2T = (20− 0)(20− 2)(20− 6)(20− 12) = 40320

n!2T = (2n)!

A Geometric Progression

Example

Consider the geometric progressionG = {5 · 3n} = {5, 15, 45, 135, 405, ...}. This set admits asimultaneous p-ordering and thus

1!G = (15− 5) = 10

2!G = (45− 5)(45− 15) = 1200

3!G = (135− 5)(135− 15)(135− 45) = 1404000

4!G = (405−5)(405−15)(405−45)(405−135) = 15163200000

Example

1!G = (15− 5) = 10

2!G = (45− 5)(45− 15) = 1200

3!G = (135− 5)(135− 15)(135− 45) = 1404000

4!G = (405−5)(405−15)(405−45)(405−135) = 15163200000

Example

1!G = (15− 5) = 10

2!G = (45− 5)(45− 15) = 1200

3!G = (135− 5)(135− 15)(135− 45) = 1404000

4!G = (405−5)(405−15)(405−45)(405−135) = 15163200000

Example

1!G = (15− 5) = 10

2!G = (45− 5)(45− 15) = 1200

3!G = (135− 5)(135− 15)(135− 45) = 1404000

4!G = (405−5)(405−15)(405−45)(405−135) = 15163200000

Example

1!G = (15− 5) = 10

2!G = (45− 5)(45− 15) = 1200

3!G = (135− 5)(135− 15)(135− 45) = 1404000

4!G = (405−5)(405−15)(405−45)(405−135) = 15163200000

A General Geometric Progression

Example

The geometric progression aqN gives a simultaneous p-ordering.Thus

n!aqN = (aqn − a)(aqn − aq) · · · (aqn − aqn−1)

= anqn(1− q−n)(1− q1−n) · · · (1− q−2)(1− q−1))

= (−aq)nq−n(n+1)

2 (1− q)(1− q2) · · · (1− qn)

= (−aq)nq−n(n+1)

2 (q : q)n

where (q : q)n is the q-Pochhammer symbol.

Example

= anqn(1− q−n)(1− q1−n) · · · (1− q−2)(1− q−1))

= (−aq)nq−n(n+1)

2 (1− q)(1− q2) · · · (1− qn)

= (−aq)nq−n(n+1)

2 (q : q)n

Example

= anqn(1− q−n)(1− q1−n) · · · (1− q−2)(1− q−1))

= (−aq)nq−n(n+1)

2 (1− q)(1− q2) · · · (1− qn)

= (−aq)nq−n(n+1)

2 (q : q)n

Example

= anqn(1− q−n)(1− q1−n) · · · (1− q−2)(1− q−1))

= (−aq)nq−n(n+1)

2 (1− q)(1− q2) · · · (1− qn)

= (−aq)nq−n(n+1)

2 (q : q)n

Example

= anqn(1− q−n)(1− q1−n) · · · (1− q−2)(1− q−1))

= (−aq)nq−n(n+1)

2 (1− q)(1− q2) · · · (1− qn)

= (−aq)nq−n(n+1)

2 (q : q)n

Non-Simultaneous Sets

Example

While it can be shown that the primes do NOT admit asimultaneous p-ordering, Bhargava gives an intricate formulato calculate them.

n!P =∏

pb n−1

p−1c+b n−1

p(p−1)c+b n−1

p2(p−1)c+···

{n!P} = {1, 2, 24, 48, 5670, ...}

Example

The set of cubes N3 does not admit a simultaneous p-orderingand has no nice formula.

{n!N3} = {1, 2, 504, 504, 35280, ...}

Example

n!P =∏

pb n−1

p−1c+b n−1

p(p−1)c+b n−1

p2(p−1)c+···

{n!P} = {1, 2, 24, 48, 5670, ...}

Example

{n!N3} = {1, 2, 504, 504, 35280, ...}

Example

n!P =∏

pb n−1

p−1c+b n−1

p(p−1)c+b n−1

p2(p−1)c+···

{n!P} = {1, 2, 24, 48, 5670, ...}

Example

{n!N3} = {1, 2, 504, 504, 35280, ...}

Example

n!P =∏

pb n−1

p−1c+b n−1

p(p−1)c+b n−1

p2(p−1)c+···

{n!P} = {1, 2, 24, 48, 5670, ...}

Example

{n!N3} = {1, 2, 504, 504, 35280, ...}

Example

n!P =∏

pb n−1

p−1c+b n−1

p(p−1)c+b n−1

p2(p−1)c+···

{n!P} = {1, 2, 24, 48, 5670, ...}

Example

{n!N3} = {1, 2, 504, 504, 35280, ...}

Summary of Examples

N ! n!N = n!

aN + b ! n!aN+b = ann!

2T ! n!2T = (2n)!

Z2 ! n!Z2 = (2n)!2

aqN ! n!aqN = (−aq)nq−n(n+1)

2 (q : q)n

P ! n!P =∏

p p(stuff )

Summary of Examples

N ! n!N = n!

2T ! n!2T = (2n)!

Z2 ! n!Z2 = (2n)!2

2 (q : q)n

P ! n!P =∏

p p(stuff )

Summary of Examples

N ! n!N = n!

2T ! n!2T = (2n)!

Z2 ! n!Z2 = (2n)!2

2 (q : q)n

P ! n!P =∏

p p(stuff )

Summary of Examples

N ! n!N = n!

2T ! n!2T = (2n)!

Z2 ! n!Z2 = (2n)!2

2 (q : q)n

P ! n!P =∏

p p(stuff )

Summary of Examples

N ! n!N = n!

2T ! n!2T = (2n)!

Z2 ! n!Z2 = (2n)!2

2 (q : q)n

P ! n!P =∏

p p(stuff )

Summary of Examples

N ! n!N = n!

2T ! n!2T = (2n)!

Z2 ! n!Z2 = (2n)!2

2 (q : q)n

P ! n!P =∏

p p(stuff )

Question

Recall that one can write ex as a Taylor series. That is

ex =∞∑

Question: What is the !S -analogue to this equation?

As it turns out, this is the wrong question to ask.

Question

Recall that one can write ex as a Taylor series.

That is

ex =∞∑

Question

ex =∞∑

Question

ex =∞∑

Question

ex =∞∑

Right Question

Raising to the power of m, we can write (ex )m as follows:

(ex )m = emx =∞∑

n!xn = 1 +

2!x2 +

3!x3 + · · ·

Right Question: What is the !S -analogue of this equation?

1) The numerator of each “coefficient” is a polynomial in m.

2) The denominator of each “coefficient” is a factorial.

Right Question

(ex )m = emx =∞∑

n!xn = 1 +

2!x2 +

3!x3 + · · ·

Right Question

(ex )m = emx =∞∑

n!xn = 1 +

2!x2 +

3!x3 + · · ·

Right Question

(ex )m = emx =∞∑

n!xn = 1 +

2!x2 +

3!x3 + · · ·

(aN+ b) -analogue

Example

( aa−x )m = (1− x

a )−m

= 1 + ma x + m(m−1)

2a2x2 + m(m−1)(m−2)

+ m(m−1)(m−2)(m−3)24a4

x4 + m(m−1)(m−2)(m−3)(m−4)120a5

x5 + · · ·

=∑∞

n=0PaN+b,n(m)

n!aN+bxn

Notice our numerators are polynomials in m and our denominatorsare ann! = n!aN+b.

(aN+ b) -analogue

Example

( aa−x )m = (1− x

a )−m

= 1 + ma x + m(m−1)

2a2x2 + m(m−1)(m−2)

+ m(m−1)(m−2)(m−3)24a4

x4 + m(m−1)(m−2)(m−3)(m−4)120a5

x5 + · · ·

=∑∞

n=0PaN+b,n(m)

n!aN+bxn

(aN+ b) -analogue

Example

( aa−x )m = (1− x

a )−m

= 1 + ma x + m(m−1)

2a2x2 + m(m−1)(m−2)

+ m(m−1)(m−2)(m−3)24a4

x4 + m(m−1)(m−2)(m−3)(m−4)120a5

x5 + · · ·

=∑∞

n=0PaN+b,n(m)

n!aN+bxn

(aN+ b) -analogue

Example

( aa−x )m = (1− x

a )−m

= 1 + ma x + m(m−1)

2a2x2 + m(m−1)(m−2)

+ m(m−1)(m−2)(m−3)24a4

x4 + m(m−1)(m−2)(m−3)(m−4)120a5

x5 + · · ·

=∑∞

n=0PaN+b,n(m)

n!aN+bxn

2T -analogue

Example

cosm(√x) =

1− m2 x + m+3m(m−1)

−15m(m−1)+m+15m(m−1)(m−2)720 x3 + · · ·

=∑∞

n=0P2T,n(m)

n!2Txn

Note that by allowing multiplication by scalars, the Z2-analogue is

2cosm(√x) =

∞∑n=0

PZ2,n(m)

2T -analogue

Example

cosm(√x) = 1− m

2 x + m+3m(m−1)24 x2

−15m(m−1)+m+15m(m−1)(m−2)720 x3 + · · ·

=∑∞

n=0P2T,n(m)

n!2Txn

2cosm(√x) =

∞∑n=0

PZ2,n(m)

2T -analogue

Example

cosm(√x) = 1− m

2 x + m+3m(m−1)24 x2

−15m(m−1)+m+15m(m−1)(m−2)720 x3 + · · ·

=∑∞

n=0P2T,n(m)

n!2Txn

2cosm(√x) =

∞∑n=0

PZ2,n(m)

2T -analogue

Example

cosm(√x) = 1− m

2 x + m+3m(m−1)24 x2

−15m(m−1)+m+15m(m−1)(m−2)720 x3 + · · ·

=∑∞

n=0P2T,n(m)

n!2Txn

2cosm(√x) =

∞∑n=0

PZ2,n(m)

P -analogue

Example

(− ln(1− x)

= 1 + m2 x + m(3m+5)

24 x2 + m(m2+5m+6)48 x3 + · · ·

=∞∑

PP,n(m)

Notice that the coefficients in the denominator seem to be thesame as the factorials for the set of primes.

It turns out that this is so (Chabert, 2005).

P -analogue

Example

(− ln(1− x)

= 1 + m2 x + m(3m+5)

24 x2 + m(m2+5m+6)48 x3 + · · ·

=∞∑

PP,n(m)

P -analogue

Example

(− ln(1− x)

= 1 + m2 x + m(3m+5)

24 x2 + m(m2+5m+6)48 x3 + · · ·

=∞∑

PP,n(m)

Summary of !S-analogues

N ! n!N = n! !

(ex )m

aN + b ! n!aN+b = ann! !

( aa−x )m

2T ! n!2T = (2n)! !

cosm(√x)

Z2 ! n!Z2 = (2n)!2 !

2cosm(√x)

2 (q : q)n !

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

aN + b ! n!aN+b = ann! !

( aa−x )m

2T ! n!2T = (2n)! !

cosm(√x)

Z2 ! n!Z2 = (2n)!2 !

2cosm(√x)

2 (q : q)n !

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

aN + b ! n!aN+b = ann! ! ( aa−x )m

2T ! n!2T = (2n)! !

cosm(√x)

Z2 ! n!Z2 = (2n)!2 !

2cosm(√x)

2 (q : q)n !

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

2T ! n!2T = (2n)! ! cosm(√x)

Z2 ! n!Z2 = (2n)!2 !

2cosm(√x)

2 (q : q)n !

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

2T ! n!2T = (2n)! ! cosm(√x)

Z2 ! n!Z2 = (2n)!2 ! 2cosm(

2 (q : q)n !

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

2T ! n!2T = (2n)! ! cosm(√x)

Z2 ! n!Z2 = (2n)!2 ! 2cosm(

2 (q : q)n !

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

2T ! n!2T = (2n)! ! cosm(√x)

Z2 ! n!Z2 = (2n)!2 ! 2cosm(

2 (q : q)n ! ?

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

N ! n!N = n! ! (ex )m

2T ! n!2T = (2n)! ! cosm(√x)

Z2 ! n!Z2 = (2n)!2 ! 2cosm(

2 (q : q)n ! ?

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? !

(tan−1(x))m

N ! n!N = n! ! (ex )m

2T ! n!2T = (2n)! ! cosm(√x)

Z2 ! n!Z2 = (2n)!2 ! 2cosm(

2 (q : q)n ! ?

P ! n!P =∏

p p(stuff ) !

(− ln(1−x)

? ! n!? ! (tan−1(x))m

Future Work

Conjecture A

Every subset of N corresponds to a function.Issues:

They probably won’t be classical functions.

Difficulty: Finding the correct polynomials in the numerator.

Is it theoretically useful to allow for constants such as for Z2?

Conjecture B

Every analytic function

(with conditions?)

corresponds to a subsetof N.Issues:

What are the conditions?

Future Work

Conjecture A

Every subset of N corresponds to a function.

Issues:

Conjecture B

(with conditions?)

corresponds to a subsetof N.

Issues:

Future Work

Conjecture A

Conjecture B

(with conditions?)

Issues:

Future Work

Conjecture A

Conjecture B

(with conditions?)

Issues:

Future Work

Conjecture A

Conjecture B

(with conditions?)

Issues:

Future Work

Conjecture A

Conjecture B

(with conditions?)

Issues:

Future Work

Conjecture A

Conjecture B

(with conditions?)

corresponds to a subsetof N.Issues:

Future Work

Conjecture A

Conjecture B

Every analytic function (with conditions?) corresponds to a subsetof N.Issues:

Future Work

Conjecture A

Conjecture B

Every analytic function (with conditions?) corresponds to a subsetof N.Issues:

If The Conjectures Are True...

−a ln(a− x) // ?

a− x

��

aN + boo

(a− x)2// ?

Thanks

References

Bhargava, M. (2000). The factorial function andgeneralizations. The American Mathematical Monthly,107(9), 783-799.

Chabert, J.L. (2007). Integer-valued polynomials on primenumbers and logarithm power expansion. EuropeanJournal of Combinatorics, 28(3), 754-761.

Thanks Professor Eric Errthum!

Questions

Thanks

References

Questions

Thanks

References

Questions

Generalized Factorials and Taylor Expansionscourse1.winona.edu/eerrthum/UndergradResearch/... ·...

Documents

VARIOUS PROBLEMS CONCERNING FACTORIALS, BINOMIAL

Factorials to N digits

Products of Factorials in Smarandache Type Expressions

4-2 Factorials and Permutations

Factorials in APSIM Why use Factorials? Basics of Factorial Configuration Using Met components Using Soil Components Using Manager Components Advanced

Blocking and Confounding Fractional Factorials The … fileLecture 7: Fractional Factorials EE290H F05 Spanos 1 Blocking and Confounding Fractional Factorials The concept of design

AN EFFICIENT APPROACH TO CALCULATE FACTORIALS IN …

Part 1 Module 5 Factorials, Permutations, and Combinations

Factorials and Stirling numbers in the algebra of formal Laurent … · 2017-02-26 · DISCRETE MATHEMATICS ELSEVIER Discrete Mathematics 132 (1994) 197-213 Factorials and Stirling

FACTORIAL DESIGNS F Terms for Factorials F Types of Factorial Designs F Notation for Factorials F Types of Effects F Looking at Tables of Means F Looking

Generalized Taylor formulae, computations in real closed valued

10.1 - The Fundamental Counting Principal - Permutations - Factorials

Factorials permutations and_combinations_using_excel

Introductions to Factorial - Wolfram ResearchIntroductions to Factorial Introduction to the factorials and binomials General The factorials and binomials have a very long history connected

More factorials (pdf)

Optimal Designs of Two-Level Factorials when N 1 and · APPROVAL Name: Kunasekaran Nirmalkanna Degree: Master of Science Title of Project: Optimal Designs of Two-Level Factorials

Factorials, Blocking & Repeated Measuresusers.ece.utexas.edu/~perry/education/382c/L10.pdf · Factorials, Blocking & Repeated Measures Dewayne E Perry ENS 623. Perry@ece.utexas.edu

2-3 Factorials and the Binomial Theorem (Presentation)

Blocking and Confounding Fractional Factorials The concept of design …ee290h/fa05/Lectures/PDF... · · 2005-10-02Blocking and Confounding Fractional Factorials The concept of

Unbalanced Designs in Factorials Type I, II, III SShomepage.stat.uiowa.edu/...unbalanced_factorials.pdf · Unbalanced factorials: Type I, II, III SS But if we have an unbalanced factorial