Iterated Binomial Sums and their Associated Iterated Integrals - - PowerPoint PPT Presentation

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Iterated Binomial Sums and their Associated Iterated Integrals

Jakob Ablinger

joint work with J. Bl¨ umlein, C. Raab and C. Schneider SFB F050 Algorithmic and Enumerative Combinatorics Research Institute for Symbolic Computation Johannes Kepler University Linz

Outline 2

◮ Iterated Binomial Sums ◮ Iterated Integrals over Square-Root Valued Alphabets ◮ Mellin Transformation of D-finite Functions ◮ Inverse Mellin Transfromation ◮ Asymptotic Expansions of Nested Sums ◮ Generating functions for Iterated Integrals

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Nested Sums

Nested Sums 4

n

• i1=1

s1(i1)

i1

• i2=1

s2(i2)

i2

• i3=1

s3(i3) · · ·

ik−1

• ik=1

sk(ik)

Nested Sums 4

n

• i1=1

s1(i1)

i1

• i2=1

s2(i2)

i2

• i3=1

s3(i3) · · ·

ik−1

• ik=1

sk(ik)

◮ si(j) = (±1)j jci , ci ∈ N

harmonic sums

Nested Sums 4

n

• i1=1

s1(i1)

i1

• i2=1

s2(i2)

i2

• i3=1

s3(i3) · · ·

ik−1

• ik=1

sk(ik)

◮ si(j) = (±1)j jci , ci ∈ N

harmonic sums

◮ si(j) = xij jci , xi ∈ R∗; ci ∈ N

S-sums

Nested Sums 4

n

• i1=1

s1(i1)

i1

• i2=1

s2(i2)

i2

• i3=1

s3(i3) · · ·

ik−1

• ik=1

sk(ik)

◮ si(j) = (±1)j jci , ci ∈ N

harmonic sums

◮ si(j) = xij jci , xi ∈ R∗; ci ∈ N

S-sums

◮ si(j) = (±1)j (aij+bi)ci , ai, ci ∈ N; bi ∈ N0

cyclotomic sums

Nested Sums 4

n

• i1=1

s1(i1)

i1

• i2=1

s2(i2)

i2

• i3=1

s3(i3) · · ·

ik−1

• ik=1

sk(ik)

◮ si(j) = (±1)j jci , ci ∈ N

harmonic sums

◮ si(j) = xij jci , xi ∈ R∗; ci ∈ N

S-sums

◮ si(j) = (±1)j (aij+bi)ci , ai, ci ∈ N; bi ∈ N0

cyclotomic sums

◮ si(j) = xij jci

2i

i

di, xi ∈ R∗; ci ∈ N; di ∈ Z binomial sums

Nested Sums 5

Definition (Iterated Binomial Sums (B-Sums))

For ai, ci ∈ N, bi ∈ N0, xi ∈ R∗, di ∈ Z and n ∈ N we define S(a1,b1,c1,d1,x1),...,(ak,bk,ck,dk,xk) (n) =

• n≥i1≥···≥ik≥1

x1i1 (a1i1 + b1)c1 2i1 i1 d1 · · · xkik (akik + bk)ck 2ik ik dk k is called the depth and w = k

i=1 ci is called the weight of the

iterated binomial sum S(a1,b1,c1,d1,x1),...,(ak,bk,ck,dk,xk) (n).

Nested Sums 5

Definition (Iterated Binomial Sums (B-Sums))

For ai, ci ∈ N, bi ∈ N0, xi ∈ R∗, di ∈ Z and n ∈ N we define S(a1,b1,c1,d1,x1),...,(ak,bk,ck,dk,xk) (n) =

• n≥i1≥···≥ik≥1

x1i1 (a1i1 + b1)c1 2i1 i1 d1 · · · xkik (akik + bk)ck 2ik ik dk k is called the depth and w = k

i=1 ci is called the weight of the

iterated binomial sum S(a1,b1,c1,d1,x1),...,(ak,bk,ck,dk,xk) (n). S(2,1,3,−1, 1

4 ),(1,0,3,1,1) (n)

=

n

• i=1

( 1

4)i i j=1

(2j

j )

j3

(2i + 1)32i

i

Nested Sums 5

Definition (Iterated Binomial Sums (B-Sums))

For ai, ci ∈ N, bi ∈ N0, xi ∈ R∗, di ∈ Z and n ∈ N we define S(a1,b1,c1,d1,x1),...,(ak,bk,ck,dk,xk) (n) =

• n≥i1≥···≥ik≥1

x1i1 (a1i1 + b1)c1 2i1 i1 d1 · · · xkik (akik + bk)ck 2ik ik dk k is called the depth and w = k

i=1 ci is called the weight of the

iterated binomial sum S(a1,b1,c1,d1,x1),...,(ak,bk,ck,dk,xk) (n). S(2,1,3,−1, 1

4 ),(1,0,3,1,1) (n)

=

n

• i=1

( 1

4)i i j=1

(2j

j )

j3

(2i + 1)32i

i

• S(1,0,2,0,1),(1,0,3,0,1),(1,0,1,0,−1) (n)

=

n

• i=1

i

j=1 j

k=1 (−1)k k

j3

i2

Quasi Shuffle Algebra 6

Sp (n) Sq (n) =

• r=p

∃

q

Sr (n) + sums of lower depth here p ∃ q represent all merges of p and q in which the relative

• rders of the elements of p and q are preserved.

Quasi Shuffle Algebra 6

Sp (n) Sq (n) =

• r=p

∃

q

Sr (n) + sums of lower depth here p ∃ q represent all merges of p and q in which the relative

• rders of the elements of p and q are preserved.

Sa1,a2 (n) Sb1,b2 (n) = Sa1,a2,b1,b2 (n) + Sa1,b1,a2,b2 (n) +Sa1,b1,b2,a2 (n) + Sb1,b2,a1,a2 (n) +Sb1,a1,b2,a2 (n) + Sb1,a1,a2,b2 (n) +sums of lower depth

Quasi Shuffle Algebra 7

n

• i=1

2i

i

• i
• n
• i=1

2i

i

i

j=1 (−1)j

(2j

j )j2

• 2i + 1
• =

Quasi Shuffle Algebra 7

n

• i=1

2i

i

• i
• n
• i=1

2i

i

i

j=1 (−1)j

(2j

j )j2

• 2i + 1
• =

n

• i=1

2i

i

i

j=1 (−1)j j

k=1 (2k k ) k

(2j

j )j2

2i + 1 +

n

• i=1

2i

i

i

j=1

(2j

j )

j

k=1 (−1)k

(2k

k )k2

j

2i + 1 +

n

• i=1

2i

i

i

j=1

(2j

j )

j

k=1 (−1)k

(2k

k )k2

2j+1

i −

n

• i=1

2i

i

2 i

j=1 (−1)j

(2j

j )j2

i +2

n

• i=1

2i

i

2 i

j=1 (−1)j

(2j

j )j2

2i + 1 −

n

• i=1

2i

i

i

j=1 (−1)j j3

2i + 1

Quasi Shuffle Algebra 8

S(1,0,1,1,1) (n) S(2,1,1,1,1),(1,0,2,−1,0) (n) =

Quasi Shuffle Algebra 8

S(1,0,1,1,1) (n) S(2,1,1,1,1),(1,0,2,−1,0) (n) = S(2,1,1,1,1),(1,0,2,−1,0),(1,0,1,1,1) (n) +S(2,1,1,1,1),(1,0,1,1,1),(1,0,2,−1,0) (n) +S(1,0,1,1,1),(2,1,1,1,1),(1,0,2,−1,0) (n) −S(1,0,1,1,2),(1,0,2,−1,−1) (n) +2 S(2,1,1,1,2),(1,0,2,−1,−1) (n) −S(2,1,1,1,2),(1,0,3,−1,0) (n)

9

Iterated Integrals

Iterated Integrals 10

x f1(y1) y1 f2(y2) y2 f3(y3) · · · yk−1 fk(yk)dyk · · · dy3dy2dy1

Iterated Integrals 10

x f1(y1) y1 f2(y2) y2 f3(y3) · · · yk−1 fk(yk)dyk · · · dy3dy2dy1

◮ fi(y) = 1 y−ai , ai ∈ {−1, 0, 1}

harmonic polylogarithms

Iterated Integrals 10

x f1(y1) y1 f2(y2) y2 f3(y3) · · · yk−1 fk(yk)dyk · · · dy3dy2dy1

◮ fi(y) = 1 y−ai , ai ∈ {−1, 0, 1}

harmonic polylogarithms

◮ fi(y) = 1 y−ai , ai ∈ R

multiple polylogarithms

Iterated Integrals 10

x f1(y1) y1 f2(y2) y2 f3(y3) · · · yk−1 fk(yk)dyk · · · dy3dy2dy1

◮ fi(y) = 1 y−ai , ai ∈ {−1, 0, 1}

harmonic polylogarithms

◮ fi(y) = 1 y−ai , ai ∈ R

multiple polylogarithms

◮ fi(y) = yji Φai(y), ji ∈ N

cyclotomic polylogarithms

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

integral representation (inv. Mellin transform)

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

integral representation (inv. Mellin transform) Mellin transform

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

integral representation (inv. Mellin transform) Mellin transform

S−1,2 (∞) S1,2

• 1

2 , 1; ∞

• S(2,1,−1) (∞)

n → ∞

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

integral representation (inv. Mellin transform) Mellin transform

S−1,2 (∞) S1,2

• 1

2 , 1; ∞

• S(2,1,−1) (∞)

n → ∞

H−1,1(1) H(4,1),(0,0)(1) H2,3(c)

x → 1 x → 1 x → c ∈ R

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

integral representation (inv. Mellin transform) Mellin transform

S−1,2 (∞) S1,2

• 1

2 , 1; ∞

• S(2,1,−1) (∞)

n → ∞

H−1,1(1) H(4,1),(0,0)(1) H2,3(c)

x → 1 x → 1 x → c ∈ R power series expansion

Connection between these structures 11

H-Sums

S−1,2 (n)

S-Sums

S1,2

• 1

2 , 1; n

• C-Sums

S(2,1,−1) (n)

H-Logs

H−1,1(x)

C-Logs

H(4,1),(0,0)(x)

M-Logs

H2,3(x)

integral representation (inv. Mellin transform) Mellin transform

S−1,2 (∞) S1,2

• 1

2 , 1; ∞

• S(2,1,−1) (∞)

n → ∞

H−1,1(1) H(4,1),(0,0)(1) H2,3(c)

x → 1 x → 1 x → c ∈ R power series expansion

Iterated Integrals 12

x f1(y1) y1 f2(y2) y2 f3(y3) · · · yk−1 fk(yk)dyk · · · dy3dy2dy1

◮ fi(y) = 1 y−ai , ai ∈ {−1, 0, 1}

harmonic polylogarithms

◮ fi(y) = 1 y−ai , ai ∈ R

multiple polylogarithms

◮ fi(y) = yji Φai(y), ji ∈ N

cyclotomic polylogarithms

Iterated Integrals 13

fa(x) := sign(1 − a − 0) x − a , f{a1,...,ak}(x) := fa1(x)1/2 . . . fak(x)1/2 k ≥ 2, f(a0,{a1,...,ak})(x) := fa0(x)fa1(x)1/2 . . . fak(x)1/2 k ≥ 1, f({a1,...,ak},j)(x) := xjf(a1,...,ak)(x) j ∈ {1, . . . , k − 2}. Restricting to at most two root-singularities we are left with the following cases: fa(x) := sign(1 − a − 0) x − a , f(a,{b})(x) := fa(x)

• fb(x),

f{a,b}(x) :=

• fa(x)
• fb(x),

f(a,{b,c})(x) := fa(x)

• fb(x)
• fc(x).

Iterated Integrals 14

Examples

H(2)(x) = x 1 2 − xdx

Iterated Integrals 14

Examples

H(2)(x) = x 1 2 − xdx H(2,{−4})(x) = x 1 (2 − x)√x + 4dx

Iterated Integrals 14

Examples

H(2)(x) = x 1 2 − xdx H(2,{−4})(x) = x 1 (2 − x)√x + 4dx H(2,{−4,4})(x) = x 1 (2 − x)√x + 4√4 − xdx

Iterated Integrals 14

Examples

H(2)(x) = x 1 2 − xdx H(2,{−4})(x) = x 1 (2 − x)√x + 4dx H(2,{−4,4})(x) = x 1 (2 − x)√x + 4√4 − xdx H(2,{−4}),(4,{−1})(x) = x 1 (2 − x)√x + 4 y 1 (4 − y)√1 + ydydx

Iterated Integrals 14

Examples

H(2)(x) = x 1 2 − xdx H(2,{−4})(x) = x 1 (2 − x)√x + 4dx H(2,{−4,4})(x) = x 1 (2 − x)√x + 4√4 − xdx H(2,{−4}),(4,{−1})(x) = x 1 (2 − x)√x + 4 y 1 (4 − y)√1 + ydydx H(2),({−1}),({−1})(x) = x 1 (2 − x) y 1 √1 + y y 1 √1 + z dzdydx

Shuffle Algebra 15

Hp(x)Hq(x) =

• r=p

∃

q

Hr(x) here p ∃ q represent all merges of p and q in which the relative

• rders of the elements of p and q are preserved.

Shuffle Algebra 15

Hp(x)Hq(x) =

• r=p

∃

q

Hr(x) here p ∃ q represent all merges of p and q in which the relative

• rders of the elements of p and q are preserved.

Ha1,a2(x)Hb1,b2(x) = Ha1,a2,b1,b2(x) + Ha1,b1,a2,b2(x) +Ha1,b1,b2,a2(x) + Hb1,b2,a1,a2(x) +Hb1,a1,b2,a2(x) + Hb1,a1,a2,b2(x) There are no additional algebraic relations among the iterated integrals if the alphabet is chosen carefully.

16

Mellin transform

Mellin transform of D-finite functions 17

◮ Let K be a field of characteristic 0.

Mellin transform of D-finite functions 17

◮ Let K be a field of characteristic 0. ◮ A function f = f(x) is called D-finite if there exist

pd(x), pd−1(x), . . . , p0(x) ∈ K[x] (not all pi = 0) such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0.

Mellin transform of D-finite functions 17

◮ Let K be a field of characteristic 0. ◮ A function f = f(x) is called D-finite if there exist

pd(x), pd−1(x), . . . , p0(x) ∈ K[x] (not all pi = 0) such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0.

◮ A sequence (fn)n≥0 ∈ KN is called P-finite if there exist

pd(n), pd−1(n), . . . , p0(n) ∈ K[n] (not all pi = 0) such that pd(n)fn+d + · · · + p1(n)fn+1 + p0(n)fn = 0.

Mellin transform of D-finite functions 17

◮ Let K be a field of characteristic 0. ◮ A function f = f(x) is called D-finite if there exist

pd(x), pd−1(x), . . . , p0(x) ∈ K[x] (not all pi = 0) such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0.

◮ A sequence (fn)n≥0 ∈ KN is called P-finite if there exist

pd(n), pd−1(n), . . . , p0(n) ∈ K[n] (not all pi = 0) such that pd(n)fn+d + · · · + p1(n)fn+1 + p0(n)fn = 0.

◮ If f(x) is D-finite, then the coefficients fn of the formal

power series expansion f(x) =

∞

• n=0

fnxn form a P-finite sequence.

Mellin transform of D-finite functions 17

◮ Let K be a field of characteristic 0. ◮ A function f = f(x) is called D-finite if there exist

pd(x), pd−1(x), . . . , p0(x) ∈ K[x] (not all pi = 0) such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0.

◮ A sequence (fn)n≥0 ∈ KN is called P-finite if there exist

pd(n), pd−1(n), . . . , p0(n) ∈ K[n] (not all pi = 0) such that pd(n)fn+d + · · · + p1(n)fn+1 + p0(n)fn = 0.

◮ If f(x) is D-finite, then the coefficients fn of the formal

power series expansion f(x) =

∞

• n=0

fnxn form a P-finite sequence.

◮ The generating function of a P-finite sequence (fn)n≥0 is

D-finite.

D-finite to P-finite 18

◮ Assume that f(x) =

• n≥0

fnxn is D-finite such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. (1)

D-finite to P-finite 18

◮ Assume that f(x) =

• n≥0

fnxn is D-finite such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. (1)

◮ It is easy to check that

xkf(j)(x) =

• n≥0

j

• i=1

(n + i − k)fn+j−kxn (2)

D-finite to P-finite 18

◮ Assume that f(x) =

• n≥0

fnxn is D-finite such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. (1)

◮ It is easy to check that

xkf(j)(x) =

• n≥0

j

• i=1

(n + i − k)fn+j−kxn (2)

◮ Transform (1) according to this relation.

D-finite to P-finite 18

◮ Assume that f(x) =

• n≥0

fnxn is D-finite such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. (1)

◮ It is easy to check that

xkf(j)(x) =

• n≥0

j

• i=1

(n + i − k)fn+j−kxn (2)

◮ Transform (1) according to this relation. ◮ Equate coefficients of same powers of x on both sides.

D-finite to P-finite 18

◮ Assume that f(x) =

• n≥0

fnxn is D-finite such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. (1)

◮ It is easy to check that

xkf(j)(x) =

• n≥0

j

• i=1

(n + i − k)fn+j−kxn (2)

◮ Transform (1) according to this relation. ◮ Equate coefficients of same powers of x on both sides. ◮ We get a linear recurrence equation with polynomial

coefficients, satisfied by (fn)n≥0.

D-finite to P-finite 19 Consider f(x) = x τ1 1 (1 + τ1) (1 − τ2)dτ2dτ1 =

• n≥0

fnxn.

D-finite to P-finite 19 Consider f(x) = x τ1 1 (1 + τ1) (1 − τ2)dτ2dτ1 =

• n≥0

fnxn. We can derive the differential equation: (x + 1)(x − 1)f ′′′(x) + (3x − 1)f ′′(x) + f ′(x) = 0.

D-finite to P-finite 19 Consider f(x) = x τ1 1 (1 + τ1) (1 − τ2)dτ2dτ1 =

• n≥0

fnxn. We can derive the differential equation: (x + 1)(x − 1)f ′′′(x) + (3x − 1)f ′′(x) + f ′(x) = 0. Expanding leads to x2f ′′′(x) − f ′′′(x) + 3xf ′′(x) − f ′′(x) + f ′(x) = 0.

D-finite to P-finite 19 Consider f(x) = x τ1 1 (1 + τ1) (1 − τ2)dτ2dτ1 =

• n≥0

fnxn. We can derive the differential equation: (x + 1)(x − 1)f ′′′(x) + (3x − 1)f ′′(x) + f ′(x) = 0. Expanding leads to x2f ′′′(x) − f ′′′(x) + 3xf ′′(x) − f ′′(x) + f ′(x) = 0. Using (2) results in:

∞

• n=0

(n + 1)fn+1xn + 3

∞

• n=0

n(n + 1)fn+1xn +

∞

• n=0

(n − 1)n(n + 1)fn+1xn −

∞

• n=0

(n + 1)(n + 2)fn+2xn −

∞

• n=0

(n + 1)(n + 2)(n + 3)fn+3xn = 0

D-finite to P-finite 19 Consider f(x) = x τ1 1 (1 + τ1) (1 − τ2)dτ2dτ1 =

• n≥0

fnxn. We can derive the differential equation: (x + 1)(x − 1)f ′′′(x) + (3x − 1)f ′′(x) + f ′(x) = 0. Expanding leads to x2f ′′′(x) − f ′′′(x) + 3xf ′′(x) − f ′′(x) + f ′(x) = 0. Using (2) results in:

∞

• n=0

(n + 1)fn+1xn + 3

∞

• n=0

n(n + 1)fn+1xn +

∞

• n=0

(n − 1)n(n + 1)fn+1xn −

∞

• n=0

(n + 1)(n + 2)fn+2xn −

∞

• n=0

(n + 1)(n + 2)(n + 3)fn+3xn = 0 Hence (n + 1)3fn+1 − (n + 2)(n + 1)fn+2 − (n + 2)(n + 3)(n + 1)fn+3 = 0 holds for (fn)n≥0.

Mellin transform of D-finite functions 20

Let f(x) be a D-finite function such that M[f(x)](n) := 1 xnf(x)dx exists and let pi(x) ∈ K[x] such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0.

Mellin transform of D-finite functions 20

Let f(x) be a D-finite function such that M[f(x)](n) := 1 xnf(x)dx exists and let pi(x) ∈ K[x] such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. Since we have M[xmf(p)(x)](n) = (−1)p(n + m)! (n + m − p)! M[f(x)](n + m − p) +

p−1

• i=0

(−1)i(n + m)! (n + m − i)! f(p−1−i)(1).

Mellin transform of D-finite functions 20

Let f(x) be a D-finite function such that M[f(x)](n) := 1 xnf(x)dx exists and let pi(x) ∈ K[x] such that pd(x)f(d)(x) + · · · + p1(x)f′(x) + p0(x)f(x) = 0. Since we have M[xmf(p)(x)](n) = (−1)p(n + m)! (n + m − p)! M[f(x)](n + m − p) +

p−1

• i=0

(−1)i(n + m)! (n + m − i)! f(p−1−i)(1). We can conclude:

Proposition

If the Mellin transform of a D-finite function is defined i.e., the integral 1

0 xnf(x)dx exist, then it is P-finite.

Mellin transform of D-finite functions 21

Given a D-finite function f(x). Find an expression F(n) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have M[f(x)](n) := 1 xnf(x)dx = F(n).

Mellin transform of D-finite functions 21

Given a D-finite function f(x). Find an expression F(n) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have M[f(x)](n) := 1 xnf(x)dx = F(n). Method:

• 1. Compute a D-finite differential equation for f(x).

Mellin transform of D-finite functions 21

Given a D-finite function f(x). Find an expression F(n) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have M[f(x)](n) := 1 xnf(x)dx = F(n). Method:

• 1. Compute a D-finite differential equation for f(x).
• 2. Use the proposition above to compute a P-finite recurrence

for M[f(x)](n).

Mellin transform of D-finite functions 21

Given a D-finite function f(x). Find an expression F(n) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have M[f(x)](n) := 1 xnf(x)dx = F(n). Method:

• 1. Compute a D-finite differential equation for f(x).
• 2. Use the proposition above to compute a P-finite recurrence

for M[f(x)](n).

• 3. Compute initial values for the recurrence.

Mellin transform of D-finite functions 21

Given a D-finite function f(x). Find an expression F(n) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have M[f(x)](n) := 1 xnf(x)dx = F(n). Method:

• 1. Compute a D-finite differential equation for f(x).
• 2. Use the proposition above to compute a P-finite recurrence

for M[f(x)](n).

• 3. Compute initial values for the recurrence.
• 4. Solve the recurrence to get a closed form representation for

M[f(x)](n).

Mellin transform of D-finite functions 22

We want to compute the Mellin transform of f(x) := x √1 − τ 1 + τ dτ.

Mellin transform of D-finite functions 22

We want to compute the Mellin transform of f(x) := x √1 − τ 1 + τ dτ. We find that (−3 + x)f(x)′ + 2(−1 + x)(1 + x)f(x)′′ = 0

Mellin transform of D-finite functions 22

We want to compute the Mellin transform of f(x) := x √1 − τ 1 + τ dτ. We find that (−3 + x)f(x)′ + 2(−1 + x)(1 + x)f(x)′′ = 0 which leads to the recurrence 6 1 √1 − τ 1 + τ dτ = −2(n − 1)n M[f(x)](n − 2) + 3n M[f(x)](n − 1) +(n + 1)(2n + 3) M[f(x)](n).

Mellin transform of D-finite functions 22

We want to compute the Mellin transform of f(x) := x √1 − τ 1 + τ dτ. We find that (−3 + x)f(x)′ + 2(−1 + x)(1 + x)f(x)′′ = 0 which leads to the recurrence 6 1 √1 − τ 1 + τ dτ = −2(n − 1)n M[f(x)](n − 2) + 3n M[f(x)](n − 1) +(n + 1)(2n + 3) M[f(x)](n). Initial values can be computed easily

Mellin transform of D-finite functions 22

We want to compute the Mellin transform of f(x) := x √1 − τ 1 + τ dτ. We find that (−3 + x)f(x)′ + 2(−1 + x)(1 + x)f(x)′′ = 0 which leads to the recurrence 6 1 √1 − τ 1 + τ dτ = −2(n − 1)n M[f(x)](n − 2) + 3n M[f(x)](n − 1) +(n + 1)(2n + 3) M[f(x)](n). Initial values can be computed easily and solving the recurrence leads to M[f(x)](n) = (−1)n

• 4n+1

(2n + 1)(2n + 3) 2n

n

+ 1

√1−τ 1+τ dτ − 2

n + 1

• −

4(−1)n n

i=1 4i (2i+1)(2i

i )

n + 1 + 1

√1−τ 1+τ dτ

n + 1 .

Mellin transform of D-finite functions 23

◮ we have a general method to express the Mellin transform of

nested integrals which are D-finite to indefinite nested sums and products.

Mellin transform of D-finite functions 23

◮ we have a general method to express the Mellin transform of

nested integrals which are D-finite to indefinite nested sums and products.

◮ we can exploit the structure of our nested integrals instead.

Mellin transform of D-finite functions 23

◮ we have a general method to express the Mellin transform of

nested integrals which are D-finite to indefinite nested sums and products.

◮ we can exploit the structure of our nested integrals instead. ◮ we get direct rewrite rules to compute the Mellin transform.

Mellin transform of D-finite functions 23

◮ we have a general method to express the Mellin transform of

nested integrals which are D-finite to indefinite nested sums and products.

◮ we can exploit the structure of our nested integrals instead. ◮ we get direct rewrite rules to compute the Mellin transform.

1−ε xNf(x)dx = 1 N + 1

• (1 − ε)N+1f(1 − ε) −

1−ε dxxN+1f ′(x)

Mellin transform of D-finite functions 23

◮ we have a general method to express the Mellin transform of

nested integrals which are D-finite to indefinite nested sums and products.

◮ we can exploit the structure of our nested integrals instead. ◮ we get direct rewrite rules to compute the Mellin transform.

1−ε xNf(x)dx = 1 N + 1

• (1 − ε)N+1f(1 − ε) −

1−ε dxxN+1f ′(x)

• 1−ε

xNf(x) √x − a dx = (4a)N (2N + 1) 2N

N

• 1−ε

dx f(x) √x − a +2

N

• i=1

2i

i

• (4a)i

√ 1 − a − ε(1 − ε)if(1 − ε) − 1−ε dxxi√ x − af ′(x)

• .

Mellin transform of D-finite functions 24

M

• H∗

h1,...,hk(x)

• (N)

= 1 N + 1 M

• xh1(x)H∗

h2,...,hk(x)

• (N)

M

• H∗

h1,...,hk(x)

√x

• (N)

= 1 N + 1/2 M √xh1(x)H∗

h2,...,hk(x)

• (N)

M

• H∗

h1,...,hk(x)

√x − a

• (N)

= (4a)N (2N + 1) 2N

N

• 1

dx H∗

h1,...,hk(x)

√x − a + +2

N

• i=1

2i

i

• (4a)i M

√ x − ah1(x)H∗

h2,...,hk(x)

• (i)
• M
• H∗

h1,...,hk(x)

• x(x − a)
• (N)

= a 4 N 2N N 1 dx H∗

h1,...,hk(x)

• x(x − a)

+ +

N

• i=1

(4/a)i i 2i

i

M

• x − a

x h1(x)H∗

h2,...,hk(x)

• (i)

Some Properties of the Mellin transform 25

◮ It inherits the linearity from the integral.

Some Properties of the Mellin transform 25

◮ It inherits the linearity from the integral. ◮ Shifts in N correspond to multiplication by powers of x, i.e.,

M[f(x)](N + k) = M[xkf(x)](N) .

Some Properties of the Mellin transform 25

◮ It inherits the linearity from the integral. ◮ Shifts in N correspond to multiplication by powers of x, i.e.,

M[f(x)](N + k) = M[xkf(x)](N) .

◮ As a consequence we have the following summation formula N

• i=1

ci M[f(x)](i) = cN M

• x

x − 1

c

f(x)

• (N)− M
• x

x − 1

c

f(x)

• (0) .

Some Properties of the Mellin transform 25

◮ It inherits the linearity from the integral. ◮ Shifts in N correspond to multiplication by powers of x, i.e.,

M[f(x)](N + k) = M[xkf(x)](N) .

◮ As a consequence we have the following summation formula N

• i=1

ci M[f(x)](i) = cN M

• x

x − 1

c

f(x)

• (N)− M
• x

x − 1

c

f(x)

• (0) .

◮ Furthermore, the following properties are immediate, where

a > 0: M [ln(x)mf(x)] (N) = dm dNm M[f(x)](N), M[f(ax)](N) = 1 aN+1 M[f(x)θ(a − x)](N), a ≤ 1, M[f(xa)](N) = 1 a M[f(x)] N + 1 − a a

• .

Some Properties of the Mellin transform 26

◮ The Mellin-convolution of two real functions is defined by

f(x) ∗ g(x) = 1 dx1 1 dx2δ(x − x1x2)f(x1)g(x2) . The Mellin transform obeys the relation M[f(x) ∗ g(x)](N) = M[f(x)](N) · M[g(x)](N) .

Some Properties of the Mellin transform 26

◮ The Mellin-convolution of two real functions is defined by

f(x) ∗ g(x) = 1 dx1 1 dx2δ(x − x1x2)f(x1)g(x2) . The Mellin transform obeys the relation M[f(x) ∗ g(x)](N) = M[f(x)](N) · M[g(x)](N) .

◮ The Mellin transformation for functions with +-prescription

M[[f(x)]+](N) = 1 dx(xN − 1)f(x) . has similar properties.

27

Inverse Mellin transform

Inverse Mellin transform 28

Aim: represent our nested sums in terms of Mellin transforms in the form c0 +

k

• j=1

cN

j M[fj(x)](N),

(3) where the constants cj and functions fj(x) do not depend on N.

Inverse Mellin transform 28

Aim: represent our nested sums in terms of Mellin transforms in the form c0 +

k

• j=1

cN

j M[fj(x)](N),

(3) where the constants cj and functions fj(x) do not depend on N.

◮ achieved by virtue of the properties of the Mellin transform.

Inverse Mellin transform 28

Aim: represent our nested sums in terms of Mellin transforms in the form c0 +

k

• j=1

cN

j M[fj(x)](N),

(3) where the constants cj and functions fj(x) do not depend on N.

◮ achieved by virtue of the properties of the Mellin transform. ◮ as starting point we only need the following basic integral

representations: 1 N = M 1 x

• (N)

2N N

• =

4N π M

• 1
• x(1 − x)
• (N)

1 N 2N N

• =

1 4N M

• 1

x√1 − x

• (N).

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

◮ Starting from the innermost sum we move outwards.

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum N

• ij=1

aj(ij)

ij

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (4) we first set up an integral rep. for aj(N) of the form (3).

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum N

• ij=1

aj(ij)

ij

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (4) we first set up an integral rep. for aj(N) of the form (3).

◮ This may require the computation of Mellin convolutions.

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum N

• ij=1

aj(ij)

ij

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (4) we first set up an integral rep. for aj(N) of the form (3).

◮ This may require the computation of Mellin convolutions. ◮ We obtain an integral representation of the same form of

aj(N)

N

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (5) by Mellin convolution with the result for the inner sums.

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum N

• ij=1

aj(ij)

ij

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (4) we first set up an integral rep. for aj(N) of the form (3).

◮ This may require the computation of Mellin convolutions. ◮ We obtain an integral representation of the same form of

aj(N)

N

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (5) by Mellin convolution with the result for the inner sums.

◮ By the summation property we obtain an integral

representation for (4).

Inverse Mellin transform 29

We obtain integral rep. for nested sums as follows:

◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum N

• ij=1

aj(ij)

ij

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (4) we first set up an integral rep. for aj(N) of the form (3).

◮ This may require the computation of Mellin convolutions. ◮ We obtain an integral representation of the same form of

aj(N)

N

• ij+1=1

aj+1(ij+1) · · ·

ik−1

• ik=1

ak(ik) (5) by Mellin convolution with the result for the inner sums.

◮ By the summation property we obtain an integral

representation for (4).

◮ Repeat until the outermost sum has been processed.

Inverse Mellin transform 30

N

• i=1

1 i

• 2i

i

• =

1 dx( x

4 )N − 1

x − 4 1 √1 − x

N

• i=1

1 i

• 2i

i

• (−1)i

= 1 π 1 dx(−4x)N − 1 x + 1

4

H∗

w1(x) N

• i=1

1 i2

• 2i

i

• i
• j=1
• 2j

j

• (−2)j

= − 1 dx(−2x)N − 1 x + 1

2

• ln(x) + H∗

w28(x)

6 √ 2

• −2

3 1 dx( x

4 )N − 1

x − 4 H∗

w3(x) .

31

Asymptotic Expansion of Nested Sums

Asymptotic Expansion of Nested Sums 32

We say that the function f : R → R is expanded in an asymptotic series f(x) ∼

∞

• k=0

ak xk , x → ∞, where ak are constants, if for all K ≥ 0 RK(x) = f(x) −

K

• k=0

ak xk = o 1 xK

• , x → ∞.

Asymptotic Expansion of Nested Sums 33

Why do we need these expansions of nested sums? E.g.,

◮ for limits of the form

lim

n→∞ n

• S2(n) − ζ2 − S2,2(n) + 7 ζ2

2

10

• ◮ for the approximation of the values of analytic continued

nested sums at the complex plane S2,−3(−2.5 + 2i)

Asymptotic Expansion of Nested Sums 34

Basic Idea

S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

Asymptotic Expansion of Nested Sums 34

Basic Idea

ϕ(x) S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

Asymptotic Expansion of Nested Sums 34

Basic Idea

ϕ(x) − → ϕ(1 − x) =

∞

• k=0

akxk S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

Asymptotic Expansion of Nested Sums 34

Basic Idea

ϕ(x) − → ϕ(1 − x) =

∞

• k=0

akxk S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

∞

• k=0

ak+1k! n(n + 1) . . . (n + k)

Asymptotic Expansion of Nested Sums 34

Basic Idea

ϕ(x) − → ϕ(1 − x) =

∞

• k=0

akxk S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

∞

• k=0

ak+1k! n(n + 1) . . . (n + k) =

∞

• k=1

bk nk

Asymptotic Expansion of Nested Sums 34

Basic Idea

ϕ(x) − → ϕ(1 − x) =

∞

• k=0

akxk S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

∞

• k=0

ak+1k! n(n + 1) . . . (n + k) =

∞

• k=1

bk nk b1 = a1 bk =

k−1

• l=0

(−1)lSl+1

k−l+1ak−l(k − l)!

Asymptotic Expansion of Nested Sums 34

Basic Idea

ϕ(x) − → ϕ(1 − x) =

∞

• k=0

akxk S−1,3(n) = (−1)n 1 xn

• H1,0,0(x)

1 + x dx

• + const

∞

• k=0

ak+1k! n(n + 1) . . . (n + k) =

∞

• k=1

bk nk b1 = a1 bk =

k−1

• l=0

(−1)lSl+1

k−l+1ak−l(k − l)!

S−1,3(n) ∼ (−1)n

• − 1

4n3 + 5 8n4 − 5 8n5 − 5 16n6

• + 3 log(2) ζ3

4 +(−1)n 1 2n − 1 4n2 + 1 8n4 − 1 4n6

• ζ3 − 19 ζ2

2

40

Asymptotic Expansion of Nested Sums 35

◮ this basic idea can be turned into an algorithm

S−1,3(n) ∼ (−1)n

• − 1

4n3 + 5 8n4 − 5 8n5 − 5 16n6

• + 3 log(2) ζ3

4 + (−1)n 1 2n − 1 4n2 + 1 8n4 − 1 4n6

• ζ3 − 19 ζ22

40

Asymptotic Expansion of Nested Sums 35

◮ this basic idea can be turned into an algorithm

S−1,3(n) ∼ (−1)n

• − 1

4n3 + 5 8n4 − 5 8n5 − 5 16n6

• + 3 log(2) ζ3

4 + (−1)n 1 2n − 1 4n2 + 1 8n4 − 1 4n6

• ζ3 − 19 ζ22

40

◮ the algorithm can be extended to cyclotomic harmonic sums

S(2,1,2),(1,0,1)(n) ∼ −4 log(2)S(2,1,−1)(∞)2 − 8 log(2)S(2,1,−1)(∞) − 4 log(2) + 1 9n3 − 1 4n + 7ζ3 4 +

• − 11

48n3 + 1 4n2 − 1 4n

• (log(n) + γ)

Asymptotic Expansion of Nested Sums 35

◮ this basic idea can be turned into an algorithm

S−1,3(n) ∼ (−1)n

• − 1

4n3 + 5 8n4 − 5 8n5 − 5 16n6

• + 3 log(2) ζ3

4 + (−1)n 1 2n − 1 4n2 + 1 8n4 − 1 4n6

• ζ3 − 19 ζ22

40

◮ the algorithm can be extended to cyclotomic harmonic sums

S(2,1,2),(1,0,1)(n) ∼ −4 log(2)S(2,1,−1)(∞)2 − 8 log(2)S(2,1,−1)(∞) − 4 log(2) + 1 9n3 − 1 4n + 7ζ3 4 +

• − 11

48n3 + 1 4n2 − 1 4n

• (log(n) + γ)

◮ extension to a subset of the S-Sums:

S2,1(1, 1 3)(n) ∼ −S1,2 1 3, 1; ∞

• + 3−n

3 2n3 − 3 4n2 + 1 2n

• H0,3(1) + 3−n
• − 3

2n3 + 3 4n2 − 1 2n

• S2

1 3; ∞

• +
• 3−n
• − 3

2n3 + 1 2n2 + 3n

• − 1

6n3 + 1 2n2 − 1 n

• + ζ2
• S1

1 3; ∞

• +S3

1 3; ∞

• + H3(1)3−n

3 2n3 − 1 2n2

• + 3−n

4n3

Asymptotic Expansion of Nested Sums 35

◮ this basic idea can be turned into an algorithm

S−1,3(n) ∼ (−1)n

• − 1

4n3 + 5 8n4 − 5 8n5 − 5 16n6

• + 3 log(2) ζ3

4 + (−1)n 1 2n − 1 4n2 + 1 8n4 − 1 4n6

• ζ3 − 19 ζ22

40

◮ the algorithm can be extended to cyclotomic harmonic sums

S(2,1,2),(1,0,1)(n) ∼ −4 log(2)S(2,1,−1)(∞)2 − 8 log(2)S(2,1,−1)(∞) − 4 log(2) + 1 9n3 − 1 4n + 7ζ3 4 +

• − 11

48n3 + 1 4n2 − 1 4n

• (log(n) + γ)

◮ extension to a subset of the S-Sums:

S2,1(1, 1 3)(n) ∼ −S1,2 1 3, 1; ∞

• + 3−n

3 2n3 − 3 4n2 + 1 2n

• H0,3(1) + 3−n
• − 3

2n3 + 3 4n2 − 1 2n

• S2

1 3; ∞

• +
• 3−n
• − 3

2n3 + 1 2n2 + 3n

• − 1

6n3 + 1 2n2 − 1 n

• + ζ2
• S1

1 3; ∞

• +S3

1 3; ∞

• + H3(1)3−n

3 2n3 − 1 2n2

• + 3−n

4n3

◮ extension to binomial sums: n

• i1=1

i1

i2=1 (−1)i2(2i2

i2 )

i3

2

2i1

i1

• 1 + 2i1
• ∼

4−nc1 √n√π

• −1

3 + 25 72n − 683 1152n2 + 35425 27648n3 − 9101113 2654208n4 + 79879765 7077888n5 − 15245392063 339738624n6

• (−1)n

1 5n4 − 16 25n5 + 67 125n6

• + c2

Asymptotic Expansion of Nested Sums 35

◮ this basic idea can be turned into an algorithm

S−1,3(n) ∼ (−1)n

• − 1

4n3 + 5 8n4 − 5 8n5 − 5 16n6

• + 3 log(2) ζ3

4 + (−1)n 1 2n − 1 4n2 + 1 8n4 − 1 4n6

• ζ3 − 19 ζ22

40

◮ the algorithm can be extended to cyclotomic harmonic sums

S(2,1,2),(1,0,1)(n) ∼ −4 log(2)S(2,1,−1)(∞)2 − 8 log(2)S(2,1,−1)(∞) − 4 log(2) + 1 9n3 − 1 4n + 7ζ3 4 +

• − 11

48n3 + 1 4n2 − 1 4n

• (log(n) + γ)

◮ extension to a subset of the S-Sums:

S2,1(1, 1 3)(n) ∼ −S1,2 1 3, 1; ∞

• + 3−n

3 2n3 − 3 4n2 + 1 2n

• H0,3(1) + 3−n
• − 3

2n3 + 3 4n2 − 1 2n

• S2

1 3; ∞

• +
• 3−n
• − 3

2n3 + 1 2n2 + 3n

• − 1

6n3 + 1 2n2 − 1 n

• + ζ2
• S1

1 3; ∞

• +S3

1 3; ∞

• + H3(1)3−n

3 2n3 − 1 2n2

• + 3−n

4n3

◮ extension to binomial sums: n

• i1=1

i1

i2=1 (−1)i2(2i2

i2 )

i3

2

2i1

i1

• 1 + 2i1
• ∼

4−nc1 √n√π

• −1

3 + 25 72n − 683 1152n2 + 35425 27648n3 − 9101113 2654208n4 + 79879765 7077888n5 − 15245392063 339738624n6

• (−1)n

1 5n4 − 16 25n5 + 67 125n6

• + c2

Asymptotic Expansion of Nested Sums 36

Why do we need these expansions of nested sums? E.g.,

◮ for limits of the form

lim

n→∞ n

• S2(n) − ζ2 − S2,2(n) + 7 ζ2

2

10

• = ζ2 − 1

◮ for the approximation of the values of analytic continued

nested sums at the complex plane S2,−3(−2.5 + 2i) = −0.795096 − 0.105476i

37

Generating Functions for Iterated Integrals

Generating Functions for Iterated Integrals 38

Given an iterated integral f(x). Find a generating series i.e., find (fn)n≥0 such that f(x) =

∞

• n=0

fnxn

Generating Functions for Iterated Integrals 38

Given an iterated integral f(x). Find a generating series i.e., find (fn)n≥0 such that f(x) =

∞

• n=0

fnxn

◮ Make use of the D-finitness of our integrals.

Generating Functions for Iterated Integrals 38

Given an iterated integral f(x). Find a generating series i.e., find (fn)n≥0 such that f(x) =

∞

• n=0

fnxn

◮ Make use of the D-finitness of our integrals. ◮ Compute a D-finite differential equation for f(x).

Generating Functions for Iterated Integrals 38

Given an iterated integral f(x). Find a generating series i.e., find (fn)n≥0 such that f(x) =

∞

• n=0

fnxn

◮ Make use of the D-finitness of our integrals. ◮ Compute a D-finite differential equation for f(x). ◮ Compute a P-finite recurrence for fn.

Generating Functions for Iterated Integrals 38

Given an iterated integral f(x). Find a generating series i.e., find (fn)n≥0 such that f(x) =

∞

• n=0

fnxn

◮ Make use of the D-finitness of our integrals. ◮ Compute a D-finite differential equation for f(x). ◮ Compute a P-finite recurrence for fn. ◮ Compute initial values for the recurrence.

Generating Functions for Iterated Integrals 38

Given an iterated integral f(x). Find a generating series i.e., find (fn)n≥0 such that f(x) =

∞

• n=0

fnxn

◮ Make use of the D-finitness of our integrals. ◮ Compute a D-finite differential equation for f(x). ◮ Compute a P-finite recurrence for fn. ◮ Compute initial values for the recurrence. ◮ Solve the recurrence to get a closed form for fn

Generating Functions for Iterated Integrals 39

Example

We want to compute the power series expansion of f(x) := x √1 − y 1 + y dy.

Generating Functions for Iterated Integrals 39

Example

We want to compute the power series expansion of f(x) := x √1 − y 1 + y dy. We find that (x − 3)f(x)′ + 2(x − 1)(x + 1)f(x)′′ = 0

Generating Functions for Iterated Integrals 39

Example

We want to compute the power series expansion of f(x) := x √1 − y 1 + y dy. We find that (x − 3)f(x)′ + 2(x − 1)(x + 1)f(x)′′ = 0 which leads to the recurrence (i(2i − 1))fi − 3(i + 1)fi+1 − 2(i + 1)(2 + i)fi+2 = 0.

Generating Functions for Iterated Integrals 39

Example

We want to compute the power series expansion of f(x) := x √1 − y 1 + y dy. We find that (x − 3)f(x)′ + 2(x − 1)(x + 1)f(x)′′ = 0 which leads to the recurrence (i(2i − 1))fi − 3(i + 1)fi+1 − 2(i + 1)(2 + i)fi+2 = 0. Computing initial values and solving the recurrence leads to f(x) =

∞

• i=1

(−x)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• .

40

Nested Sums at Infinity Iterated Integrals at One

41

◮ Using the generating function we can convert the generalized

harmonic polylogarithms at one to binomial sums at infinity.

x √1 − y 1 + y dy =

∞

• i=1

(−x)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

41

◮ Using the generating function we can convert the generalized

harmonic polylogarithms at one to binomial sums at infinity.

x √1 − y 1 + y dy =

∞

• i=1

(−x)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• 1

√1 − y 1 + y dy =

∞

• i=1

(−1)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

41

◮ Using the generating function we can convert the generalized

harmonic polylogarithms at one to binomial sums at infinity.

x √1 − y 1 + y dy =

∞

• i=1

(−x)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• 1

√1 − y 1 + y dy =

∞

• i=1

(−1)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• =

∞

• i=1

(−1)i i

j=1 (−4)−j(2j

j )

2j−1

i +

∞

• i=1

4−i2i

i

• i

−2

∞

• i=1

4−i2i

i

• 2i − 1 −

∞

• i=1

(−1)i i

41

◮ Using the generating function we can convert the generalized

harmonic polylogarithms at one to binomial sums at infinity.

x √1 − y 1 + y dy =

∞

• i=1

(−x)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• 1

√1 − y 1 + y dy =

∞

• i=1

(−1)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• =

∞

• i=1

(−1)i i

j=1 (−4)−j(2j

j )

2j−1

i +

∞

• i=1

4−i2i

i

• i

−2

∞

• i=1

4−i2i

i

• 2i − 1 −

∞

• i=1

(−1)i i

◮ Relations between the constants due to the shuffle algebra of

the iterated integrals.

41

◮ Using the generating function we can convert the generalized

harmonic polylogarithms at one to binomial sums at infinity.

x √1 − y 1 + y dy =

∞

• i=1

(−x)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• 1

√1 − y 1 + y dy =

∞

• i=1

(−1)i i

j=1

• − 1

4

j(2j

j )

2j−1

i −

• − 1

4

i2i

i

• i(2i − 1)

− 1 i

• =

∞

• i=1

(−1)i i

j=1 (−4)−j(2j

j )

2j−1

i +

∞

• i=1

4−i2i

i

• i

−2

∞

• i=1

4−i2i

i

• 2i − 1 −

∞

• i=1

(−1)i i

◮ Relations between the constants due to the shuffle algebra of

the iterated integrals.

◮ Relations between the constants due to the quasi shuffle

algebra of the nested sums.

Summary 42

Nested Binomial Sums

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets integral representation

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets integral representation Mellin transform

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets integral representation Mellin transform Nested Binomial Sums at ∞ n → ∞

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets integral representation Mellin transform Nested Binomial Sums at ∞ n → ∞ Iterated Integrals at 1 x → 1

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets integral representation Mellin transform Nested Binomial Sums at ∞ n → ∞ Iterated Integrals at 1 x → 1 generating function

Summary 42

Nested Binomial Sums Iterated Integrals

• ver

Root-valued Alphabets integral representation Mellin transform Nested Binomial Sums at ∞ n → ∞ Iterated Integrals at 1 x → 1 generating function

Summary 43

All the algorithms mentioned in this talk and many more are implemented in the package HarmonicSums which is available at http://www.risc.jku.at/research/combinat/software/