Colleagues, I am planning to change this thread to a category.
However, at the moment, I will post my second communication of the thread CMT.
Let us define another differential operator of infinite terms as :
e^{-D}:=∑_{j=0}^{∞}(((-1)^{j}D^{(j)})/(j!))
when j=0, we have the identity operator, and D:=(d/(dx))
Then as in my first communication post, we can question the following:
(∀ψεC^{∞}(I,ℝ))Λ(∀xεI), what will be
e^{-D}(ψ(x))=∑_{j=0}^{∞}(((-1)^{j}D^{(j)}ψ(x))/(j!))?
Consider the following example:
Example 1: Take ψ(x)=e^(x) the usual natural exponential function.
Claim: e^{-D}(ψ(x))=ψ(x-1)
Indeed,
e^{-D}(ψ(x))=∑_{j=0}^{∞}(((-1)^{j}D^{(j)}ψ(x))/(j!))
=∑_{j=0}^{∞}(((-1)^{j}D^{(j)}(e^{x}))/(j!))
=∑_{j=0}^{∞}(((-1)^{j}e^{x})/(j!))
=e^(x)∑_{j=0}^{∞}(((-1)^{j})/(j!))
=e^(x-1)=ψ(x-1)
∴ e^{-D}ψ(x)=ψ(x-1) ... which is a right translation of ψ by a unit.
One can extend this result further and write a corollary as :
Corollary: (∀kεℕ):(e^{-D})^{k}ψ(x)=ψ(x-k)-right translate of ψ by k-units.
Example 2. Let φ(x)=x³+x²+x+1.Then
e^{-D}φ(x)=∑_{j=0}^{∞}(((-1)^{j}D^{(j)}φ(x))/(j!))
=∑_{j=0}^{∞}(((-1)^{j}D^{(j)}(x³+x²+x+1))/(j!))
=x³-2x²+2x
But the expression we have at the end is precisely φ(x-1).
That is, once again we have a similar result :
e^{-D}φ(x)=φ(x-1)
Corollary: ∀ p(x) , e^{-D}p(x)=p(x-1)
Conjecture: ∀ψεC^{∞}(I,ℝ), e^{-D}ψ(x)=ψ(x-1)
Corollary to the conjecture: (∀kεℕ)(∀ψεC^{∞}(I,ℝ)),e^{-kD}ψ(x)=ψ(x-k)
Further communications will be posted on operators defined from combinations of both.
Dear Dr. Dejenie A. Lakew,
Communication in Mathematics teaching has its great roles while teaching courses where mathematics communicates about the practical engineering applications. For example: Laplace transform, State space, Graph Theory, Boolean Difference, Mod Theory, and many such which leads to realize the real concepts of the systems.
Colleagues,
The following is an interesting example on the differental operator of infinite terms that I have defined in my first communication :e^{D}:=∑_{n=0}^{∞}((D^{(n)})/(n!))
where D:=(d/(dx))
Example: e^{D}(sin x)=sin(x+1)
Proof: The function f(x)=sin x∈C^{∞}(ℝ). Therefore,
e^{D}(sin x) = ∑_{n=0}^{∞}((Dⁿ(sin x))/(n!))
= sin x∑_{n=0}^{∞}(((-1)ⁿ)/((2n)!))+cos x∑_{n=0}^{∞}(((-1)ⁿ)/((2n+1)!))
=sin xcos1+cos xsin1
But
sin xcos1+cos xsin1=sin(x+1)
and that proves the example.
In a similar way one can show that: e^{D}(cos x)=cos(x+1)
Dejenie A. Lakew
Colleagues,
This is my fourth posting in the thread : Communications in Mathematics Teaching (CMT): The posting is about a differential operator of infinite terms with odd orders--which I use superscript "odd" to denote this. I also use the superscript " $\infinity$ " to denote the number of differential terms involved. The results are interesting.
Definition: Let us define another differential operator of infinite terms as :
ℒ^{∞,odd}:=∑_{k=0}^{∞}((D^{(2k+1)})/((2k+1)!))
where D:=(d/(dx))
Then as in my first and second communications, we can question the following:
(∀ψεC^{∞}(I,ℝ))Λ(∀xεI), what will be
ℒ^{∞,odd}(ψ(x))=∑_{k=0}^{∞}((D^{(2k+1)}ψ(x))/((2k+1)!))?
Consider the following example:
Example 1: Take ψ(x)=e^{x} the usual natural exponential function.
Claim: ℒ^{∞,odd}(ψ(x))=ψ(x+1)-ψ(x-1)
Proof of claim:
ℒ^{∞,odd}(ψ(x))=∑_{k=0}^{∞}((D^{(2k+1)}ψ(x))/((2k+1)!))
=∑_{k=0}^{∞}((D^{(2k+1)}(e^{x}))/((2k+1)!))
=e^{x+1}-e^{x-1}
=ψ(x+1)-ψ(x-1)
Example 2. Let φ(x)=x³.Then
ℒ^{∞,odd}φ(x)=∑_{k=0}^{∞}((D^{(2k+1)}φ(x))/((2k+1)!))
=∑_{k=0}^{∞}((D^{(2k+1)}(x³))/((2k+1)!))
=6x²+2
But the expression we have at the end is precisely: φ(x+1)-φ(x-1)
That is :
ℒ^{∞,odd}φ(x)=φ(x+1)-φ(x-1)
In a similar argument, we can show that :
Proposition: ∀ p(x)ε℘(x) , ℒ^{∞,odd}p(x)=p(x+1)-p(x-1) , where ℘(x) is the set of all polynomial functions in x
Conjecture 1 : ∀ψεC^{∞}(I,ℝ), ℒ^{∞,odd}ψ(x)=ψ(x+1)-ψ(x-1)
Again for the differential operator:
ℒ_{k}^{∞,odd}:=∑_{j=0}^{∞}((k^{(2j+1)}D^{(2j+1)})/((2j+1)!))
where D is the differential operator above we have:
Conjecture 2: (∀kεℕ)(∀ψεC^{∞}(I,ℝ)),ℒ_{k}^{∞,odd}ψ(x)=ψ(x+k)-ψ(x-k)
N.B.:The fifth communication is about even ordered differential operator of infinite terms and will be posted shortly.
Dejenie A. Lakew
This is my fifth posting in the thread :
Communications in Mathematics Teaching ( CMT) as a series :
This is on a differential operator of even orders with infinite terms defined below :
D^{∞,even}:=∑_{k=0}^{∞}((D^{(2k)})/((2k)!))
where D:=(d/(dx))
As in my previous communications, we question the following:
(∀ψεC^{∞}(I,ℝ))Λ(∀xεI), what will be
D^{∞,even}(ψ(x))=∑_{k=0}^{∞}((D^{(2k)}ψ(x))/((2k)!))?
As my favorite example, let us consider the following:
Example 1: The natural exponential function: ψ(x)=e^(x)
Claim: D^{∞,even}(ψ(x))=ψ(x+1)+ψ(x-1)
Proof of claim: Indeed,
D^{∞,even}(ψ(x)) =∑_{k=0}^{∞}((D^{(2k)}ψ(x))/((2k)!))
=∑_{k=0}^{∞}((D^{(2k)}(e^(x)))/((2k)!))
=e^(x)(∑_{k=0}^{∞}(1/((2k)!)))
=e^(x)(∑_{k=0}^{∞}(1/((k)!))+∑_{k=0}^{∞}(((-1)^{k})/((k)!)))
=e^(x+1)+e^(x-1)
=ψ(x+1)+ψ(x-1)
Example 2. Let ξ(x)=x³.Then
D^{∞,even}ξ(x) = ∑_{k=0}^{∞}((D^{(2k)}ξ(x))/((2k)!))
=∑_{k=0}^{∞}((D^{(2k)}(x³))/((2k)!))
=3x³+6x
But the expression we have at the end is precisely: ξ(x+1)+ξ(x-1)
∴ D^{∞,even}ξ(x)=ξ(x+1)+ξ(x-1)
In a similar way:
Lemma: For the monomial : ξ(x)=xⁿ, D^{∞,even}ξ(x)=ξ(x+1)+ξ(x-1)
Proposition: ∀ p(x)ε℘(x) , D^{∞,even}p(x)=p(x+1)+p(x-1) , where ℘(x) is the set of all polynomial functions in x
Conjecture : ∀ψεC^{∞}(I,ℝ), D^{∞,even}ψ(x)=ψ(x+1)+ψ(x-1)
Also for each positive integer k, consider the differential operator of the type:
D_{k}^{∞,even}:=∑_{j=0}^{∞}((k^{(2j)}D^{(2j)})/((2j)!))
where D is the differential operator as above. Then we can write the following interesting statement:
Conjecture: (∀kεℕ)(∀ψεC^{∞}(I,ℝ)),D_{k}^{∞,even}ψ(x)=ψ(x+k)+ψ(x-k)
Note: One can show the validity of the following results:
(i) D^{∞,even}(sin(x))=2sin xcos1
(ii) D^{∞,even}(cos(x))=2cos xcos1
Remark: The differential operators I considered in five of my communications can have some sort of mapping properties. One can look at these things.
Next time I will bring some properties of the operators on trigonometric functions --who knows they might have some interesting relations.
Dejenie
In Example 1 above the claim is not displayed for some unknown reason. I therefore post it here.
Example 1: Take ψ(x)=e^(x) the usual natural exponential function.
Claim: ℒ^{∞,odd}(ψ(x))=ψ(x+1)-ψ(x-1)
Dear Dr. Dejenie A. Lakew,
Communication in Mathematics teaching has its great roles while teaching courses where mathematics communicates about the practical engineering applications. For example: Laplace transform, State space, Graph Theory, Boolean Difference, Mod Theory, and many such which leads to realize the real concepts of the systems.
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