Telephone Numbers and the Umbral Calculus

April Chow1 Iris Gray1 Lawrence Mouillé2 James Pickett3

1Louisiana State University Baton Rouge, LA

2University of Louisiana at Lafayette Lafayette, LA

3University of South Alabama Mobile, AL

SMILE @ LSU 2013, July 12, 2013 t4 Possible Conversations t4 Possible Conversations There are 76 ways to have these conversations.

t6 Possible Conversations t6 Possible Conversations

There are 76 ways to have these conversations. # of users Ways of Calling 0 1 1 1 2 2 3 4 4 10 5 26 6 76 7 232 8 764 9 2620 10 9496

Calculation of t0 to t10 Calculation of t0 to t10

# of users Ways of Calling 0 1 1 1 2 2 3 4 4 10 5 26 6 76 7 232 8 764 9 2620 10 9496 Proof. Either user 1 is making a call or user 1 is not making a call.

• If user 1 is not making a call, there are tn−1 ways that the network of n − 1 remaining users can be conﬁgured. Hence, the tn−1 term. • If user 1 is making a call, we may choose who he is calling n − 1 different ways. In each n − 1 case, there are n − 2 users remaining. Hence, the (n − 1)tn−2 term.

Recursion Relation of tn

The telephone numbers satisfy the following recursion:

tn = tn−1 + (n − 1)tn−2, where t0 and t1 are both equal to 1. • If user 1 is not making a call, there are tn−1 ways that the network of n − 1 remaining users can be conﬁgured. Hence, the tn−1 term. • If user 1 is making a call, we may choose who he is calling n − 1 different ways. In each n − 1 case, there are n − 2 users remaining. Hence, the (n − 1)tn−2 term.

Recursion Relation of tn

The telephone numbers satisfy the following recursion:

tn = tn−1 + (n − 1)tn−2, where t0 and t1 are both equal to 1.

Proof. Either user 1 is making a call or user 1 is not making a call. • If user 1 is making a call, we may choose who he is calling n − 1 different ways. In each n − 1 case, there are n − 2 users remaining. Hence, the (n − 1)tn−2 term.

Recursion Relation of tn

The telephone numbers satisfy the following recursion:

tn = tn−1 + (n − 1)tn−2, where t0 and t1 are both equal to 1.

Proof. Either user 1 is making a call or user 1 is not making a call.

• If user 1 is not making a call, there are tn−1 ways that the network of n − 1 remaining users can be conﬁgured. Hence, the tn−1 term. Recursion Relation of tn

The telephone numbers satisfy the following recursion:

tn = tn−1 + (n − 1)tn−2, where t0 and t1 are both equal to 1.

Proof. Either user 1 is making a call or user 1 is not making a call.

Theorem

b n c X2 n! t = n 2k k!(n − 2k)! k=0 Theorem The EGF of tn is

∞ 2 X tn n x+ x T (x) = x = e 2 . n! k=0

Exponential Generating Functions

An exponential generating function, or EGF of an, a sequence, can be deﬁned as the following power series:

∞ X xn a n! n k=0 Exponential Generating Functions

An exponential generating function, or EGF of an, a sequence, can be deﬁned as the following power series:

∞ X xn a n! n k=0

Theorem The EGF of tn is

∞ 2 X tn n x+ x T (x) = x = e 2 . n! k=0 ∞ X t + • n 1 xn = T 0(x) − 1. n! n=1 ∞ X tn • xn = T (x) − 1. n! n=1 ∞ X n t − • n 1 xn = xT 0(x). n! n=1

The EGF of tn

Proof. Now, recall that tn+1 = tn + ntn−1. In terms of the summation:

∞ ∞ ∞ X t + X tn X n t − n 1 xn = xn + n 1 xn n! n! n! n=1 n=1 n=1 ∞ X t + • n 1 xn = T 0(x) − 1. n! n=1 ∞ X tn • xn = T (x) − 1. n! n=1 ∞ X n t − • n 1 xn = xT 0(x). n! n=1

The EGF of tn

Proof. Now, recall that tn+1 = tn + ntn−1. In terms of the summation:

∞ ∞ ∞ X t + X tn X n t − n 1 xn = xn + n 1 xn n! n! n! n=1 n=1 n=1 ∞ X tn • xn = T (x) − 1. n! n=1 ∞ X n t − • n 1 xn = xT 0(x). n! n=1

The EGF of tn

Proof. Now, recall that tn+1 = tn + ntn−1. In terms of the summation:

∞ ∞ ∞ X t + X tn X n t − n 1 xn = xn + n 1 xn n! n! n! n=1 n=1 n=1

∞ X t + • n 1 xn = T 0(x) − 1. n! n=1 ∞ X n t − • n 1 xn = xT 0(x). n! n=1

The EGF of tn

Proof. Now, recall that tn+1 = tn + ntn−1. In terms of the summation:

∞ ∞ ∞ X t + X tn X n t − n 1 xn = xn + n 1 xn n! n! n! n=1 n=1 n=1

∞ X t + • n 1 xn = T 0(x) − 1. n! n=1 ∞ X tn • xn = T (x) − 1. n! n=1 The EGF of tn

Proof. Now, recall that tn+1 = tn + ntn−1. In terms of the summation:

∞ ∞ ∞ X t + X tn X n t − n 1 xn = xn + n 1 xn n! n! n! n=1 n=1 n=1

∞ X t + • n 1 xn = T 0(x) − 1. n! n=1 ∞ X tn • xn = T (x) − 1. n! n=1 ∞ X n t − • n 1 xn = xT 0(x). n! n=1 Now, T 0(x) = (1 + x)T (x), where T (0) = 1 2 x+ x =⇒ T (x) = e 2

The EGF of tn (cont.)

Proof (Cont.) Solving for T (x) :

T 0(x) − 1 = T (x) − 1 + xT 0(x). The EGF of tn (cont.)

Proof (Cont.) Solving for T (x) :

T 0(x) − 1 = T (x) − 1 + xT 0(x).

Now, T 0(x) = (1 + x)T (x), where T (0) = 1 2 x+ x =⇒ T (x) = e 2 Applications of Hermite Polynomials: • Probability; • Numerical Analysis; • Combinatorics; • Physics.

Hermite Polynomials

The set of Hermite polynomials is a polynomial sequence. • Numerical Analysis; • Combinatorics; • Physics.

The set of Hermite polynomials is a polynomial sequence. Applications of Hermite Polynomials: • Probability; • Combinatorics; • Physics.

Hermite Polynomials

The set of Hermite polynomials is a polynomial sequence. Applications of Hermite Polynomials: • Probability; • Numerical Analysis; • Physics.

Hermite Polynomials

The set of Hermite polynomials is a polynomial sequence. Applications of Hermite Polynomials: • Probability; • Numerical Analysis; • Combinatorics; Hermite Polynomials

The set of Hermite polynomials is a polynomial sequence. Applications of Hermite Polynomials: • Probability; • Numerical Analysis; • Combinatorics; • Physics. Hermite Polynomials

Deﬁnition

bn/2c X (−1)k H (u) = n! (2u)n−2k n k!(n − 2k)! k=0 H0(u) to H10(u)

n Hn(u) 0 1 1 2u 2 2(−1 + 2u2) 4u3 3 6(−2u + 3 ) 1 2 2u4 4 24( 2 − 2u + 3 ) 4u3 4u5 5 120(u − 3 + 15 ) 1 2 2u4 4u6 6 720(− 6 + u − 3 + 45 ) u 2u3 4u5 8u7 7 5040(− 3 + 3 − 15 + 315 ) 1 u2 u4 4u6 2u8 8 40320( 24 − 3 + 3 − 45 + 315 ) u 2u3 2u5 8u7 4u9 9 362880( 12 − 9 + 15 − 315 + 2835 ) 1 u2 u4 2u6 2u8 4u10 10 3628800(− 120 + 12 − 9 + 45 − 315 + 14175 ) i 1 −i √ H1 √ = 1 = t1 2 2

i 2 −i √ H2 √ = 2 = t2 2 2

i 3 −i √ H3 √ = 4 = t3 2 2

Relating Hn and tn

Observe i 0 −i √ H0 √ = 1 = t0 2 2 i 2 −i √ H2 √ = 2 = t2 2 2

i 3 −i √ H3 √ = 4 = t3 2 2

Relating Hn and tn

Observe i 0 −i √ H0 √ = 1 = t0 2 2

i 1 −i √ H1 √ = 1 = t1 2 2 i 3 −i √ H3 √ = 4 = t3 2 2

Relating Hn and tn

Observe i 0 −i √ H0 √ = 1 = t0 2 2

i 1 −i √ H1 √ = 1 = t1 2 2

i 2 −i √ H2 √ = 2 = t2 2 2 Relating Hn and tn

Observe i 0 −i √ H0 √ = 1 = t0 2 2

i 1 −i √ H1 √ = 1 = t1 2 2

i 2 −i √ H2 √ = 2 = t2 2 2

i 3 −i √ H3 √ = 4 = t3 2 2 Relating Hn and tn

Theorem

i n −i √ Hn √ = tn 2 2 bn/2c X n! = 2k k!(n − 2k)! k=0

= tn

Relating Hn and tn

Proof.

n bn/2c n−2k i −i in X (−1)k −i √ Hn √ = √ n! 2 √ ( )n k!(n − 2k)! 2 2 2 k=0 2 = tn

Relating Hn and tn

Proof.

n bn/2c n−2k i −i in X (−1)k −i √ Hn √ = √ n! 2 √ ( )n k!(n − 2k)! 2 2 2 k=0 2

bn/2c X n! = 2k k!(n − 2k)! k=0 Relating Hn and tn

Proof.

n bn/2c n−2k i −i in X (−1)k −i √ Hn √ = √ n! 2 √ ( )n k!(n − 2k)! 2 2 2 k=0 2

bn/2c X n! = 2k k!(n − 2k)! k=0

= tn We wish to solve for an in terms of bk , and will use the umbrae.

n X n Bn = Ak k k=0

The Umbral Method

Suppose an is given and we deﬁne another sequence

n X n b = a n k k k=0 The Umbral Method

Suppose an is given and we deﬁne another sequence

n X n b = a n k k k=0

We wish to solve for an in terms of bk , and will use the umbrae.

n X n Bn = Ak k k=0 =⇒ Bn = (1 + A)n

=⇒ B = 1 + A

=⇒ A = B − 1

=⇒ An = (B − 1)n

n X n =⇒ An = Bk (−1)n−k k k=0 n X n =⇒ a = b (−1)n−k n k k k=0

n X n Bn = Ak k k=0 =⇒ B = 1 + A

=⇒ A = B − 1

=⇒ An = (B − 1)n

n X n =⇒ An = Bk (−1)n−k k k=0 n X n =⇒ a = b (−1)n−k n k k k=0

n X n Bn = Ak k k=0

=⇒ Bn = (1 + A)n =⇒ A = B − 1

=⇒ An = (B − 1)n

n X n =⇒ An = Bk (−1)n−k k k=0 n X n =⇒ a = b (−1)n−k n k k k=0

n X n Bn = Ak k k=0

=⇒ Bn = (1 + A)n

=⇒ B = 1 + A =⇒ An = (B − 1)n

n X n =⇒ An = Bk (−1)n−k k k=0 n X n =⇒ a = b (−1)n−k n k k k=0

n X n Bn = Ak k k=0

=⇒ Bn = (1 + A)n

=⇒ B = 1 + A

=⇒ A = B − 1 n X n =⇒ An = Bk (−1)n−k k k=0 n X n =⇒ a = b (−1)n−k n k k k=0

n X n Bn = Ak k k=0

=⇒ Bn = (1 + A)n

=⇒ B = 1 + A

=⇒ A = B − 1

=⇒ An = (B − 1)n n X n =⇒ a = b (−1)n−k n k k k=0

n X n Bn = Ak k k=0

=⇒ Bn = (1 + A)n

=⇒ B = 1 + A

=⇒ A = B − 1

=⇒ An = (B − 1)n

n X n =⇒ An = Bk (−1)n−k k k=0 n X n Bn = Ak k k=0

=⇒ Bn = (1 + A)n

=⇒ B = 1 + A

=⇒ A = B − 1

=⇒ An = (B − 1)n

n X n =⇒ An = Bk (−1)n−k k k=0 n X n =⇒ a = b (−1)n−k n k k k=0 Vectors in V are polynomials of the form

d X k p(A) = ck A k=0

n Deﬁne the linear functional L : V → R such that L(A ) = an.

Then, d d X k X k L(p(A)) = L( ck A ) = ck L(A ) k=0 k=0

• A is the umbra of an. n • Many formulas are easier to derive if we change an to A for some variable A. After some algebraic manipulation , n we change A back to an.

Consider the vector space V with the basis {1, A, A2,... }. n Deﬁne the linear functional L : V → R such that L(A ) = an.

Then, d d X k X k L(p(A)) = L( ck A ) = ck L(A ) k=0 k=0

• A is the umbra of an. n • Many formulas are easier to derive if we change an to A for some variable A. After some algebraic manipulation , n we change A back to an.

Consider the vector space V with the basis {1, A, A2,... }.

Vectors in V are polynomials of the form

d X k p(A) = ck A k=0 Then, d d X k X k L(p(A)) = L( ck A ) = ck L(A ) k=0 k=0

• A is the umbra of an. n • Many formulas are easier to derive if we change an to A for some variable A. After some algebraic manipulation , n we change A back to an.

Consider the vector space V with the basis {1, A, A2,... }.

Vectors in V are polynomials of the form

d X k p(A) = ck A k=0

n Deﬁne the linear functional L : V → R such that L(A ) = an. • A is the umbra of an. n • Many formulas are easier to derive if we change an to A for some variable A. After some algebraic manipulation , n we change A back to an.

Consider the vector space V with the basis {1, A, A2,... }.

Vectors in V are polynomials of the form

d X k p(A) = ck A k=0

n Deﬁne the linear functional L : V → R such that L(A ) = an.

Then, d d X k X k L(p(A)) = L( ck A ) = ck L(A ) k=0 k=0 n • Many formulas are easier to derive if we change an to A for some variable A. After some algebraic manipulation , n we change A back to an.

Consider the vector space V with the basis {1, A, A2,... }.

Vectors in V are polynomials of the form

d X k p(A) = ck A k=0

n Deﬁne the linear functional L : V → R such that L(A ) = an.

Then, d d X k X k L(p(A)) = L( ck A ) = ck L(A ) k=0 k=0

• A is the umbra of an. Consider the vector space V with the basis {1, A, A2,... }.

Vectors in V are polynomials of the form

d X k p(A) = ck A k=0

n Deﬁne the linear functional L : V → R such that L(A ) = an.

Then, d d X k X k L(p(A)) = L( ck A ) = ck L(A ) k=0 k=0

• A is the umbra of an. n • Many formulas are easier to derive if we change an to A for some variable A. After some algebraic manipulation , n we change A back to an. n X n Applying these two we get, L(Bn) = L(Ak ) k k=0 n X n = L Ak = L (1 + A)n k k=0

The Umbral Method

n X n 1 We have b = a , and want to solve for a in n k k n k=0 terms of bk . n n 2 Recall: L(A ) = an and L(B ) = bn n X n = L Ak = L (1 + A)n k k=0

The Umbral Method

n X n 1 We have b = a , and want to solve for a in n k k n k=0 terms of bk . n n 2 Recall: L(A ) = an and L(B ) = bn n X n Applying these two we get, L(Bn) = L(Ak ) k k=0 = L (1 + A)n

The Umbral Method

n X n 1 We have b = a , and want to solve for a in n k k n k=0 terms of bk . n n 2 Recall: L(A ) = an and L(B ) = bn n X n Applying these two we get, L(Bn) = L(Ak ) k k=0 n X n = L Ak k k=0 The Umbral Method

n X n 1 We have b = a , and want to solve for a in n k k n k=0 terms of bk . n n 2 Recall: L(A ) = an and L(B ) = bn n X n Applying these two we get, L(Bn) = L(Ak ) k k=0 n X n = L Ak = L (1 + A)n k k=0 d d d X k X k X k L(p(B)) = L ck B = ck L(B ) = ck L[(1 + A) ] k=0 k=0 k=0 = L(p(1 + A))

The Umbral Method

1 We determined L(Bn) = L[(1 + A)n] 2 Let p(B) be a polynomial in B. d d X k X k = ck L(B ) = ck L[(1 + A) ] k=0 k=0 = L(p(1 + A))

The Umbral Method

1 We determined L(Bn) = L[(1 + A)n] 2 Let p(B) be a polynomial in B. d X k L(p(B)) = L ck B k=0 d X k = ck L[(1 + A) ] k=0 = L(p(1 + A))

The Umbral Method

1 We determined L(Bn) = L[(1 + A)n] 2 Let p(B) be a polynomial in B. d d X k X k L(p(B)) = L ck B = ck L(B ) k=0 k=0 = L(p(1 + A))

The Umbral Method

1 We determined L(Bn) = L[(1 + A)n] 2 Let p(B) be a polynomial in B. d d d X k X k X k L(p(B)) = L ck B = ck L(B ) = ck L[(1 + A) ] k=0 k=0 k=0 The Umbral Method

1 We determined L(Bn) = L[(1 + A)n] 2 Let p(B) be a polynomial in B. d d d X k X k X k L(p(B)) = L ck B = ck L(B ) = ck L[(1 + A) ] k=0 k=0 k=0 = L(p(1 + A)) p(B) = (B − 1)n

L(B − 1)n = L((1 + A − 1)n) = L(An)

n X n L Bk (−1)n−k = L(An) k k=0 n X n b (−1)n−k = a k k n k=0

The Umbral Method

L(p(B)) = L(p(1 + A)) for any polynomial p. L(B − 1)n = L((1 + A − 1)n) = L(An)

n X n L Bk (−1)n−k = L(An) k k=0 n X n b (−1)n−k = a k k n k=0

The Umbral Method

L(p(B)) = L(p(1 + A)) for any polynomial p. p(B) = (B − 1)n n X n L Bk (−1)n−k = L(An) k k=0 n X n b (−1)n−k = a k k n k=0

The Umbral Method

L(p(B)) = L(p(1 + A)) for any polynomial p. p(B) = (B − 1)n

L(B − 1)n = L((1 + A − 1)n) = L(An) n X n b (−1)n−k = a k k n k=0

The Umbral Method

L(p(B)) = L(p(1 + A)) for any polynomial p. p(B) = (B − 1)n

L(B − 1)n = L((1 + A − 1)n) = L(An)

n X n L Bk (−1)n−k = L(An) k k=0 The Umbral Method

L(p(B)) = L(p(1 + A)) for any polynomial p. p(B) = (B − 1)n

L(B − 1)n = L((1 + A − 1)n) = L(An)

n X n L Bk (−1)n−k = L(An) k k=0 n X n b (−1)n−k = a k k n k=0 • Now we derive the EGF of the Hermite polynomials using umbral methods • If we deﬁne the umbra M such that L(M2n+1) = 0 and 2n (−1)n(2n)! n L(M ) = n! then Hn(u) = L[(M + 2u) ] • Then we have,

∞ X Hn(u) 2 2 xn = L(e(M+2u)x ) = e2ux L(eMx ) = e2ux e−x = e2ux−x n! n=0

EGF of Hn(u))

Theorem

∞ X Hn(u) 2 F(u, x) = xn = e2ux−x n! n=0 • If we deﬁne the umbra M such that L(M2n+1) = 0 and 2n (−1)n(2n)! n L(M ) = n! then Hn(u) = L[(M + 2u) ] • Then we have,

∞ X Hn(u) 2 2 xn = L(e(M+2u)x ) = e2ux L(eMx ) = e2ux e−x = e2ux−x n! n=0

EGF of Hn(u))

Theorem

∞ X Hn(u) 2 F(u, x) = xn = e2ux−x n! n=0

• Now we derive the EGF of the Hermite polynomials using umbral methods • Then we have,

∞ X Hn(u) 2 2 xn = L(e(M+2u)x ) = e2ux L(eMx ) = e2ux e−x = e2ux−x n! n=0

EGF of Hn(u))

Theorem

∞ X Hn(u) 2 F(u, x) = xn = e2ux−x n! n=0

• Now we derive the EGF of the Hermite polynomials using umbral methods • If we deﬁne the umbra M such that L(M2n+1) = 0 and 2n (−1)n(2n)! n L(M ) = n! then Hn(u) = L[(M + 2u) ] EGF of Hn(u))

Theorem

∞ X Hn(u) 2 F(u, x) = xn = e2ux−x n! n=0

• Now we derive the EGF of the Hermite polynomials using umbral methods • If we deﬁne the umbra M such that L(M2n+1) = 0 and 2n (−1)n(2n)! n L(M ) = n! then Hn(u) = L[(M + 2u) ] • Then we have,

∞ X Hn(u) 2 2 xn = L(e(M+2u)x ) = e2ux L(eMx ) = e2ux e−x = e2ux−x n! n=0 • Notice any similarities? Choosing u and x0 for F(u, x0) appropriately we can achieve equality. Those choices would be, u = √−i and x0 = √ix 2 2 • Proof.

2 −i ix ix −i ix 2 √ √ − √ x+ 1 x2 F √ , √ = e 2 2 2 = e 2 = T (x) 2 2

• Let’s take a look at the EGFs tn and Hn(u), they are

∞ X Hn(u) 0 0 2 F(u, x0) = (x0)n = e2ux −(x ) n! n=0 and, ∞ 2 X tn n x+ x T (x) = x = e 2 n! n=0 • Proof.

2 −i ix ix −i ix 2 √ √ − √ x+ 1 x2 F √ , √ = e 2 2 2 = e 2 = T (x) 2 2

• Let’s take a look at the EGFs tn and Hn(u), they are

∞ X Hn(u) 0 0 2 F(u, x0) = (x0)n = e2ux −(x ) n! n=0 and, ∞ 2 X tn n x+ x T (x) = x = e 2 n! n=0

• Notice any similarities? Choosing u and x0 for F(u, x0) appropriately we can achieve equality. Those choices would be, u = √−i and x0 = √ix 2 2 • Let’s take a look at the EGFs tn and Hn(u), they are

∞ X Hn(u) 0 0 2 F(u, x0) = (x0)n = e2ux −(x ) n! n=0 and, ∞ 2 X tn n x+ x T (x) = x = e 2 n! n=0

• Notice any similarities? Choosing u and x0 for F(u, x0) appropriately we can achieve equality. Those choices would be, u = √−i and x0 = √ix 2 2 • Proof.

2 −i ix ix −i ix 2 √ √ − √ x+ 1 x2 F √ , √ = e 2 2 2 = e 2 = T (x) 2 2 • ∞ H √−i ∞ X n ix n X tn 2 √ = xn. n! n! n=0 2 n=0 Hence, for all n

−i Hn √ ix n t 2 √ = n xn. n! 2 n! Therefore, i n −i √ Hn √ = tn. 2 2

Relating Hn and tn

• Writing the EGFs in their power series representations with F √−i , √ix = T (x) we have 2 2 Relating Hn and tn

• Writing the EGFs in their power series representations with F √−i , √ix = T (x) we have 2 2 • ∞ H √−i ∞ X n ix n X tn 2 √ = xn. n! n! n=0 2 n=0 Hence, for all n

−i Hn √ ix n t 2 √ = n xn. n! 2 n! Therefore, i n −i √ Hn √ = tn. 2 2 Further Applications

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Further Applications

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Further Applications

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Further Applications

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0

2 where α = x , β = √12xu , z is a solution of (1+24xu)3/2 1+24xu 12αz2 + (24αβ − 1)z + 12αβ2 = 0, and φ = 2βz − 8αβ3 − 24αβ2z + +2z2 − 24αβz2 − 8αz3.

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0

2 where α = x , β = √12xu , z is a solution of (1+24xu)3/2 1+24xu 12αz2 + (24αβ − 1)z + 12αβ2 = 0, and φ = 2βz − 8αβ3 − 24αβ2z + +2z2 − 24αβz2 − 8αz3. So,

∞ n X x 3 H (u) = Le(M+2u) x 3n n! n=0 3 2 3 2 = e8u x L(e6uxM exM +12u xM ).

Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof. For any positive integer ω,

∞ n X x ω H (u) = Le(M+2u) x . ωn n! n=0 Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof. For any positive integer ω,

∞ n X x ω H (u) = Le(M+2u) x . ωn n! n=0

So,

∞ n X x 3 H (u) = Le(M+2u) x 3n n! n=0 3 2 3 2 = e8u x L(e6uxM exM +12u xM ). Thus,

∞ n X x 3 2 3 2 H (u) = e8u x L(e6uxM exM +12xu M ) 3n n! n=0 3 8u x x 12u2x e M3+ √ M = √ L e (1+24ux)3/2 1+24ux . 1 + 24ux

Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.) If f (M) is a power series and γ is a function not dependent on M, then 2 1 M LeγM f (M) = √ L f √ . 1 + 4γ 1 + 4γ Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.) If f (M) is a power series and γ is a function not dependent on M, then 2 1 M LeγM f (M) = √ L f √ . 1 + 4γ 1 + 4γ

Thus,

∞ n X x 3 2 3 2 H (u) = e8u x L(e6uxM exM +12xu M ) 3n n! n=0 3 8u x x 12u2x e M3+ √ M = √ L e (1+24ux)3/2 1+24ux . 1 + 24ux Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.) For α and β functions not dependent on M,

2 −(β+z) +φ M3α 3 e L(eαM +βM ) = Le (1−24α(β+z))3/2 , p1 − 24α(β + z) where φ = 2βz − 8αβ3 − 24αβ2z + 2z2 − 24αβz2 − 8αz3, and z is a solution of 12αz2 + (24αβ − 1)z + 12αβ2 = 0. Hence,

3 ∞ n 8u x x 12u2x x e M3+ √ M X (1+24ux)3/2 1+24ux H3n(u) = √ L e n! 1 + 24ux n=0 3 2 e8u x−(β+z) +φ M3α = √ L e (1−24α(β+z))3/2 . 1 + 24uxp1 − 24α(β + z)

Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.)

2 −(β+z) +φ M3α 3 e L(eαM +βM ) = Le (1−24α(β+z))3/2 . p1 − 24α(β + z) Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.)

2 −(β+z) +φ M3α 3 e L(eαM +βM ) = Le (1−24α(β+z))3/2 . p1 − 24α(β + z) Hence,

3 ∞ n 8u x x 12u2x x e M3+ √ M X (1+24ux)3/2 1+24ux H3n(u) = √ L e n! 1 + 24ux n=0 3 2 e8u x−(β+z) +φ M3α = √ L e (1−24α(β+z))3/2 . 1 + 24uxp1 − 24α(β + z) we have

∞ 3 2 3 xn e8u x−(β+z) +φ M α X (1−24α(β+z))3/2 H3n(u) = √ L e n! + p − α(β + ) n=0 1 24ux 1 24 z 8u3x−(β+z)2+φ e 3 3/4 = L eM α/(1−48αβ) . (1 + 48ux)1/4

Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.) 2 Substituting α = x and β = √12xu , (1+24xu)3/2 1+24xu Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) . 3n n! (1 + 48ux)1/4 n=0

Proof (Cont.) 2 Substituting α = x and β = √12xu , we have (1+24xu)3/2 1+24xu

∞ 3 2 3 xn e8u x−(β+z) +φ M α X (1−24α(β+z))3/2 H3n(u) = √ L e n! + p − α(β + ) n=0 1 24ux 1 24 z 8u3x−(β+z)2+φ e 3 3/4 = L eM α/(1−48αβ) . (1 + 48ux)1/4 Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0 Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0 we have

∞ 3 3 3/4 X 1 M α n L e(M α)/(1−48αβ) = L · n! (1 − 48αβ)3/4 n=0

∞ X αn = L M3n . n!(1 − 48αβ)3n/4 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof. Using the series expansion of ex , Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof. Using the series expansion of ex , we have

∞ 3 3 3/4 X 1 M α n L e(M α)/(1−48αβ) = L · n! (1 − 48αβ)3/4 n=0

∞ X αn = L M3n . n!(1 − 48αβ)3n/4 n=0 ∞ n 3 3/4 X α L e(M α)/(1−48αβ) = L M3n n!(1 − 48αβ)3n/4 n=0 ∞ X αn = LM3n . n!(1 − 48αβ)3n/4 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof (Cont.) Since L is linear over umbrae, Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof (Cont.) Since L is linear over umbrae,

∞ n 3 3/4 X α L e(M α)/(1−48αβ) = L M3n n!(1 − 48αβ)3n/4 n=0 ∞ X αn = LM3n . n!(1 − 48αβ)3n/4 n=0 ∞ n 3 3/4 X α L e(M α)/(1−48αβ) = LM3n n!(1 − 48αβ)3n/4 n=0 ∞ X α2n = LM6n (2n)! (1 − 48αβ)3n/2 n=0 ∞ X (−1)3n(6n)! α2n = · . (3n)! (2n)!(1 − 48αβ)3n/2 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof (Cont.) 2n+1 2n (−1)n(2n)! Since L M = 0 and L M = n! , Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof (Cont.) 2n+1 2n (−1)n(2n)! Since L M = 0 and L M = n! ,

∞ n 3 3/4 X α L e(M α)/(1−48αβ) = LM3n n!(1 − 48αβ)3n/4 n=0 ∞ X α2n = LM6n (2n)! (1 − 48αβ)3n/2 n=0 ∞ X (−1)3n(6n)! α2n = · . (3n)! (2n)!(1 − 48αβ)3n/2 n=0 we have

∞ 3n 2n 3 3/4 X (−1) (6n)! α L e(M α)/(1−48αβ) = · (3n)! (2n)!(1 − 48αβ)3n/2 n=0 ∞ X (−1)n (6n)! x2n = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof (Cont.) Substituting for α and β, Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Proof (Cont.) Substituting for α and β, we have

∞ 3n 2n 3 3/4 X (−1) (6n)! α L e(M α)/(1−48αβ) = · (3n)! (2n)!(1 − 48αβ)3n/2 n=0 ∞ X (−1)n (6n)! x2n = . (3n)! (1 + 48ux)3n/2 (2n)! n=0 Lemma (1 & 2)

∞ 3 2 ∞ X xn e8u x−(β+z) +φ X (−1)n (6n)! x2n H (u) = . 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0

Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x).

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x).

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (1 & 2)

∞ 3 2 ∞ X xn e8u x−(β+z) +φ X (−1)n (6n)! x2n H (u) = . 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Lemma (1 & 2)

∞ 3 2 ∞ X xn e8u x−(β+z) +φ X (−1)n (6n)! x2n H (u) = . 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0

Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x). we have

2 2 3 1 + 72xu + 864x u − (1 + 48xu) 2 8xu3 − (β + z)2 + φ = 864x2 = 8v 3x + 144v 4x2 √ where v = ( 1 + 48ux − 1)/(24x).

Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x). Proof. Substituting for β, z, and φ using Mathematica, Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x). Proof. Substituting for β, z, and φ using Mathematica, we have

2 2 3 1 + 72xu + 864x u − (1 + 48xu) 2 8xu3 − (β + z)2 + φ = 864x2 = 8v 3x + 144v 4x2 √ where v = ( 1 + 48ux − 1)/(24x). 3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x

Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x). Proof (Cont.) Since 8xu3 − (β + z)2 + φ = 8v 3x + 144v 4x2, Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x). Proof (Cont.) Since 8xu3 − (β + z)2 + φ = 8v 3x + 144v 4x2,

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x).

Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0 Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x).

Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0 Lemma (1)

∞ n 8u3x−(β+z)2+φ X x e 3 3/4 H (u) = L eM α/(1−48αβ) , 3n n! (1 + 48ux)1/4 n=0

Lemma (2)

∞ n 2n 3 3/4 X (−1) (6n)! x L e(M α)/(1−48αβ) = . (3n)! (1 + 48ux)3n/2 (2n)! n=0

Lemma (3)

3 2 3 4 2 e8u x−(β+z) +φ = e8v x+144v x , √ where v = ( 1 + 48ux − 1)/(24x). Further Applications

Theorem

∞ 3 4 2 ∞ X xn e8v x+144v x X (−1)n (6n)! x2n H (u) = , 3n n! (1 + 48ux)1/4 (3n)! (1 + 48ux)3n/2 (2n)! n=0 n=0 √ where v = ( 1 + 48ux − 1)/(24x). Corollary

√ ∞ 3 4 2 ∞ X xn e−4w x 2−72w x X (6n)! x2n t = , 3n n! (1 + 24x)1/4 (3n)! (1 + 24x)3n/2 (2n)! 2n n=0 n=0

where √ √ 2( 1 + 24x − 1) w = . 24x

Further Applications

Recall that i n −i √ Hn √ = tn 2 2 Further Applications

Recall that i n −i √ Hn √ = tn 2 2

Corollary

√ ∞ 3 4 2 ∞ X xn e−4w x 2−72w x X (6n)! x2n t = , 3n n! (1 + 24x)1/4 (3n)! (1 + 24x)3n/2 (2n)! 2n n=0 n=0 where √ √ 2( 1 + 24x − 1) w = . 24x References

I.M. Gessel. Applications of the Classical Umbral Calculus. Algebra univers., 49 (2003), 397–434. V. De Angelis. Class Lecture, The Umbral Calculus. Louisiana State University, Baton Rouge, LA. Summer 2013. Acknowledgements

Ben Warren and Dr. Valerio De Angelis