Taylor Monomials - in Time Scales - Kansas State University · Taylor Monomials in Time Scales...

Taylor Monomialsin Time Scales

Pauline Ballesteros, Juan Batista, Samantha Bell

Department of MathematicsKansas State UniversityManhattan, KS 66506

Mathematics REUJuly 24, 2012

P. Ballesteros, J. Batista, S. Bell Taylor Monomials

Time Scales

Definition

A Time Scale T is an arbitrary, non-empty closed subset of R.

Examples: R, N, [0, 1], and [−5, 0] ∪ [3, 4] ∪ [8]

Q, (0, 1), and (3, 8] are not time scales.


Time Scales

Definition


Examples: R, N, [0, 1], and [−5, 0] ∪ [3, 4] ∪ [8]



Time Scales

Definition


Examples: R, N, [0, 1], and [−5, 0] ∪ [3, 4] ∪ [8]



Operators

Definition

The forward jump operatorσ : T→ T is defined by

σ(t) := inf {s ∈ T : s > t}.

Definition

The backward jump operatorρ : T→ T is defined by

ρ(t) := sup{s ∈ T : s < t}.

left scattered ρ(t) < t

right scattered σ(t) > t

left dense ρ(t) = t

right dense σ(t) = t

isolated ρ(t) < t < σ(t)

ρ2(t) ρ(t) t σ(t) σ2(t)


Operators

Definition


σ(t) := inf {s ∈ T : s > t}.

Definition


ρ(t) := sup{s ∈ T : s < t}.






ρ2(t) ρ(t) t σ(t) σ2(t)


Operators

Definition


σ(t) := inf {s ∈ T : s > t}.

Definition


ρ(t) := sup{s ∈ T : s < t}.






ρ2(t) ρ(t) t σ(t) σ2(t)


Delta Derivatives

Definition

Let f : T→ R and t ∈ T. Then f ∆(t) the real number, if it exists,such that the following holds: ∀ε > 0 ∃δε > 0 such that if|s − t| < δε and s ∈ T then

|f (σ(t))− f (s)− f ∆(t)(σ(t)− s)| ≤ ε|σ(t)− s|.

Theorem

Assume f : T→ R is a continuous function and fix t ∈ Tκ.If t is right dense, then

f ∆(t) = lims→t

f (t)− f (s)

t − s= f ′(t).

If t is right scattered, then

f ∆(t) =f (σ(t))− f (t)

µ(t).


Delta Derivatives

Definition

Let f : T→ R and t ∈ T. Then f ∆(t) the real number, if it exists,such that the following holds: ∀ε > 0 ∃δε > 0 such that if|s − t| < δε and s ∈ T then

|f (σ(t))− f (s)− f ∆(t)(σ(t)− s)| ≤ ε|σ(t)− s|.

Theorem

Assume f : T→ R is a continuous function and fix t ∈ Tκ.If t is right dense, then

f ∆(t) = lims→t

f (t)− f (s)

t − s= f ′(t).

If t is right scattered, then

f ∆(t) =f (σ(t))− f (t)

µ(t).


Nabla Derivatives

Definition

Let f : T→ R and t ∈ T. Then f ∇(t) is the real number, if itexists, such that the following holds:∀ε > 0 ∃δε > 0 such that if|s − t| < δε and s ∈ T then

|f (ρ(t))− f (s)− f ∇(t)(ρ(t)− s)| ≤ ε|ρ(t)− s|.

Theorem

Assume f : T→ R is a continuous function and fix t ∈ Tκ.If t is left dense, then

f ∇(t) = lims→t

f (t)− f (s)

t − s= f ′(t).

If t is left scattered, then

f ∇(t) =f (t)− f (ρ(t))

ν(t).


Nabla Derivatives

Definition

Let f : T→ R and t ∈ T. Then f ∇(t) is the real number, if itexists, such that the following holds:∀ε > 0 ∃δε > 0 such that if|s − t| < δε and s ∈ T then

|f (ρ(t))− f (s)− f ∇(t)(ρ(t)− s)| ≤ ε|ρ(t)− s|.

Theorem

Assume f : T→ R is a continuous function and fix t ∈ Tκ.If t is left dense, then

f ∇(t) = lims→t

f (t)− f (s)

t − s= f ′(t).

If t is left scattered, then

f ∇(t) =f (t)− f (ρ(t))

ν(t).


Existance of Pre-Antiderivatives

A function f : T→ R is:

regulated if right hand limits are finite at right dense pointsand left hand limits are finite at left dense points.

right dense continuous if it is regulated and continuous atright dense points (denoted as Crd(T)).

pre-differentiable on the region D if Tκ\D is countable and Dcontains no right scattered elements.

Theorem

Let f be regulated. Then there exists a pre-differentiable F withregion of differentiation D such that

F∆(t) = f (t) ∀t ∈ D.







Theorem


F∆(t) = f (t) ∀t ∈ D.







Theorem


F∆(t) = f (t) ∀t ∈ D.







Theorem


F∆(t) = f (t) ∀t ∈ D.







Theorem


F∆(t) = f (t) ∀t ∈ D.


Existance of Antiderivatives

Definition

Let f : T→ R be a regulated function and F be apre-differentiable function of f . Define the indefinite integral of fby ∫

f (t)∆t = F (t) + C .

Define the Cauchy integral by∫ s

rf (t)∆t = F (s)− F (r). ∀r , s ∈ T

Theorem

Every rd-continuous function has an antiderivative. If t0 ∈ T, then

F (t) :=∫ ttof (τ)∆τ for t ∈ T is an antiderivative of f .


Existance of Antiderivatives

Definition

Let f : T→ R be a regulated function and F be apre-differentiable function of f . Define the indefinite integral of fby ∫

f (t)∆t = F (t) + C .

Define the Cauchy integral by∫ s

rf (t)∆t = F (s)− F (r). ∀r , s ∈ T

Theorem

Every rd-continuous function has an antiderivative. If t0 ∈ T, then

F (t) :=∫ ttof (τ)∆τ for t ∈ T is an antiderivative of f .


Delta Integration

Theorem

If f ∈ Crd and t ∈ Tκ, then∫ σ(t)

tf (τ)∆τ = µ(t)f (t).

Corollary

If [a, b] consists of only isolated points, then

∫ b

af (t)∆t =

ρ(b)∑t=a

µ(t)f (t) : a < b

0 : a = b

−ρ(a)∑t=b

µ(t)f (t) : a > b.


Delta Integration

Theorem

If f ∈ Crd and t ∈ Tκ, then∫ σ(t)

tf (τ)∆τ = µ(t)f (t).

Corollary


∫ b

af (t)∆t =

ρ(b)∑t=a

µ(t)f (t) : a < b

0 : a = b

−ρ(a)∑t=b

µ(t)f (t) : a > b.


Nabla Integration

Theorem

If f ∈ Cld and t ∈ Tκ, then∫ t

ρ(t)f (τ)∇τ = ν(t)f (t).

Corollary


∫ b

af (t)∇t =

b∑t=σ(a)

ν(t)f (t) : a < b

0 : a = b

−a∑

t=σ(b)

ν(t)f (t) : a > b.


Nabla Integration

Theorem

If f ∈ Cld and t ∈ Tκ, then∫ t

ρ(t)f (τ)∇τ = ν(t)f (t).

Corollary


∫ b

af (t)∇t =

b∑t=σ(a)

ν(t)f (t) : a < b

0 : a = b

−a∑

t=σ(b)

ν(t)f (t) : a > b.


Integrals

Proof.

We have ∫ tρ(t) f (τ)∇τ = F (t)− F (ρ(t))

= ν(t)F∇(t)= ν(t)f (t).

For a < b, ∫ ba f (τ)∇τ =

ρ(b)∑t=a

∫ σ(t)

tf (τ)∇τ

=

ρ(b)∑t=a

ν(σ(t))f (σ(t))

=b∑

t=σ(a)

ν(t)f (t).


Integrals

Proof.

We have ∫ tρ(t) f (τ)∇τ = F (t)− F (ρ(t))

= ν(t)F∇(t)= ν(t)f (t).

For a < b, ∫ ba f (τ)∇τ =

ρ(b)∑t=a

∫ σ(t)

tf (τ)∇τ

=

ρ(b)∑t=a

ν(σ(t))f (σ(t))

=b∑

t=σ(a)

ν(t)f (t).


Taylor Formula

Taylor’s Formula on R:

Theorem

Assume f ∈ Cn+1(R). Fix s. Then

f (t) =n∑

k=0

f (k)′(s)hk(t, s) +

∫ t

shn+1(t, τ)f (n+1)′(τ)dτ

where

h0(t, s) = 1,

hk+1(t, s) =∫ ts hk(τ, s)dτ, k ≥ 0.

When T = Rhk(t, s) = (t−s)k

k!


Taylor Formula


Theorem


f (t) =n∑

k=0

f (k)′(s)hk(t, s) +

∫ t


where

h0(t, s) = 1,

hk+1(t, s) =∫ ts hk(τ, s)dτ, k ≥ 0.


k!


Taylor Formula


Theorem


f (t) =n∑

k=0

f (k)′(s)hk(t, s) +

∫ t


where

h0(t, s) = 1,

hk+1(t, s) =∫ ts hk(τ, s)dτ, k ≥ 0.


k!


Taylor Formula


Theorem


f (t) =n∑

k=0

f (k)′(s)hk(t, s) +

∫ t


where

h0(t, s) = 1,

hk+1(t, s) =∫ ts hk(τ, s)dτ, k ≥ 0.


k!


Taylor Formula cont.

Taylor’s Formula on T for ∆ derivatives:

Theorem

Assume f ∈ Cn+1rd (T). Fix s ∈ T. Then

f (t) =n∑

k=0

f ∆k(s)hk∆(t, s) +

∫ t

sh(n+1)∆(t, σ(τ))f ∆n+1

(τ)∆τ

where

h0∆(t, s) = 1,

h(k+1)∆(t, s) =∫ ts hk∆(τ, s)∆τ, k ≥ 0.

hk∇(t, s) defined similarly.




Theorem


f (t) =n∑

k=0

f ∆k(s)hk∆(t, s) +

∫ t

sh(n+1)∆(t, σ(τ))f ∆n+1

(τ)∆τ

where

h0∆(t, s) = 1,

h(k+1)∆(t, s) =∫ ts hk∆(τ, s)∆τ, k ≥ 0.





Theorem


f (t) =n∑

k=0

f ∆k(s)hk∆(t, s) +

∫ t

sh(n+1)∆(t, σ(τ))f ∆n+1

(τ)∆τ

where

h0∆(t, s) = 1,

h(k+1)∆(t, s) =∫ ts hk∆(τ, s)∆τ, k ≥ 0.





Theorem


f (t) =n∑

k=0

f ∆k(s)hk∆(t, s) +

∫ t

sh(n+1)∆(t, σ(τ))f ∆n+1

(τ)∆τ

where

h0∆(t, s) = 1,

h(k+1)∆(t, s) =∫ ts hk∆(τ, s)∆τ, k ≥ 0.



T = hZ = {0,±h,±2h,±3h, ...}

Example ([1])

For T = hZ and t > s,

hk∆(t, s) =

∏(k−1)i=0 (t − ih − s)

k!

Compute hk∆ for small k .

Identify a pattern.

Verify by Taylor’s Formula.


T = hZ = {0,±h,±2h,±3h, ...}

Example ([1])


hk∆(t, s) =

∏(k−1)i=0 (t − ih − s)

k!

Compute hk∆ for small k .

Identify a pattern.

Verify by Taylor’s Formula.


How to Find h2∇(t, s)

h1∇(t, s) = t − s

h2∇(t, s) =

∫ t

sh1∇(τ, s)∇τ

=t∑

σ(s)

h1∇(τ, s)ν(τ)

=t∑

σ(s)

(τ − s) · 1

=k∑

τ=1

τ

=k(k + 1)

2

=(t − s)(t − s + 1)

2


hk∇(t, s)

Example (B3)


hk∇(t, s) =

∏k−1i=0 (t + ih − s)

k!

Recall:

hk∆(t, s) =

∏(k−1)i=0 (t − ih − s)

k!

Differ by sign on ih

hk∆(t, s) is 0 for large k

hk∇(t, s) 6= 0 for large k


hk∇(t, s)

Example (B3)


hk∇(t, s) =

∏k−1i=0 (t + ih − s)

k!

Recall:

hk∆(t, s) =

∏(k−1)i=0 (t − ih − s)

k!





hk∇(t, s)

Example (B3)


hk∇(t, s) =

∏k−1i=0 (t + ih − s)

k!

Recall:

hk∆(t, s) =

∏(k−1)i=0 (t − ih − s)

k!





T = qZ

Example ([1])

For T = qZ and t > s,

hk∆(t, s) =k−1∏m=0

t − qmsm∑j=0

qj

Example (B3)

hk∇(t, s) =k−1∏m=0

tqm − sm∑j=0

qj

Similar to hZ.


When t < s...

Define gk∆(t, s), gk∇(t, s) to be Taylor monomials when t < s.

Example (B3)

In hZ,gk∆(t, s) =

gk∇(t, s) =(−1)k

∏ki=0(s − t − ih)

(k − 1)!


Comparisons

T = hZ

hk∆(t, s) =∏(k−1)

i=0 (t−ih−s)k! hk∇(t, s) =

∏k−1i=0 (t+ih−s)

k!

gk∆(t, s) =(−1)k

∏k−1i=0 (s+ih−t)k! gk∇(t, s) =

(−1)k∏k

i=0(s−t−ih)(k−1)!

T = qZ

hk∆(t, s) =∏k−1

m=0t−qmsm∑j=0

qjhk∇(t, s) =

∏k−1m=0

tqm−sm∑j=0

qj

gk∆(t, s) =∏k−1

m=0 s−qmtm∑j=0

qjgk∇(t, s) =

∏k−1m=0 sq

m−km∑j=0

qj


Main Theorem

Theorem

Let T be an isolated or dense time scale, s ∈ Tκ, t ∈ Tκ, witht > s. Then

hk∆(t, s) = (−1)kgk∇(s, t)

if

(hk∆(t, s))(µ(t)− ν(s)) = µ(t)hk∆(t, ρ(s))− ν(s)hk∆(σ(t), s)


Proof of Main Theorem

Proof

We prove this by induction

Fix s, t ∈ T, with s < t.Base Case: h0∆(t, s) ≡ 1 ≡ g0∇(s, t) ∀t, s ∈ TAssume k th step: hk∆(t, s) = (−1)kgk∇(s, t)Need: hk+1∆(t, s) = (−1)k+1gk+1∇(s, t)

Show this by existence/uniqueness of solutions to differentialequations:

Note that hk+1∆(s, s) = (−1)k+1gk+1∇(s, s) = 0Consider the following differential equation:

(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof





(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof


Fix s, t ∈ T, with s < t.

Base Case: h0∆(t, s) ≡ 1 ≡ g0∇(s, t) ∀t, s ∈ TAssume k th step: hk∆(t, s) = (−1)kgk∇(s, t)Need: hk+1∆(t, s) = (−1)k+1gk+1∇(s, t)



(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof


Fix s, t ∈ T, with s < t.Base Case: h0∆(t, s) ≡ 1 ≡ g0∇(s, t) ∀t, s ∈ T

Assume k th step: hk∆(t, s) = (−1)kgk∇(s, t)Need: hk+1∆(t, s) = (−1)k+1gk+1∇(s, t)



(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof


Fix s, t ∈ T, with s < t.Base Case: h0∆(t, s) ≡ 1 ≡ g0∇(s, t) ∀t, s ∈ TAssume k th step: hk∆(t, s) = (−1)kgk∇(s, t)

Need: hk+1∆(t, s) = (−1)k+1gk+1∇(s, t)



(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof





(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof





(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof




Note that hk+1∆(s, s) = (−1)k+1gk+1∇(s, s) = 0

Consider the following differential equation:

(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s



Proof





(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s


Proof Continued

Proof.

Computing the derivative gives

(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s

⇔ hk+1∆(t, s)− hk+1∆(t, ρ(s))

ν(s)= (−1)k+1gk∇(s, t)

Inserting the inductive hypothesis and inserting the definitionof hk∆(t, s):

hk+1δ(t, s)− hk+1δ(t, ρ(s))

ν(s)= −(hk∆(σ(t), s)− hk∆(t, s))

µ(t)

Algebraic rearrangement gives



Proof Continued

Proof.


(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s

⇔ hk+1∆(t, s)− hk+1∆(t, ρ(s))

ν(s)= (−1)k+1gk∇(s, t)



ν(s)= −(hk∆(σ(t), s)− hk∆(t, s))

µ(t)




Proof Continued

Proof.


(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s

⇔ hk+1∆(t, s)− hk+1∆(t, ρ(s))

ν(s)= (−1)k+1gk∇(s, t)



ν(s)= −(hk∆(σ(t), s)− hk∆(t, s))

µ(t)




Proof Continued

Proof.


(hk+1∆(t, s))∇s = ((−1)k+1gk+1∇(s, t))∇s

⇔ hk+1∆(t, s)− hk+1∆(t, ρ(s))

ν(s)= (−1)k+1gk∇(s, t)



ν(s)= −(hk∆(σ(t), s)− hk∆(t, s))

µ(t)




Future Research

Taylor Monomials on non-isolated Time Scales.

Fourier and Laplacian Analysis on Time Scales.

Multivariate Taylor Expansions.


Future Research





Future Research





Future Research





References

Reagan J. Higgins and Allan Peterson. Cauchy Functions andTaylor’s Formula for Time Scales. Department ofMathematics, University of Nebraska-Lincoln.

M. Bohner and A. Peterson, Dynamic Equations On TimeScales: An Introduction With Applications, Birkhauser,Boston, 2001.


Special Thanks To...

National Science Foundation

Dr. Nathan Pennington


????

Thank you!

Any questions?


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