Let us give the power series
f(x) = c0 + c1 x + c2 xx + c3 xxx +....
From this define a modified series with capital F
F(x) = c0 + c1 x +c2 x(x+h) + c3 x(x+h)(x+2h) +...
This is really a function of two variables, but concéntrate on the x variable.
The derivative of the original series is Df
Df = c1 + 2 c2 x +...
We call now the transformation of the derivative series TDF
The point of interest is now
F(x+h) - F(x) = h TDF(x+h)
We get the right hand side by first calculating the transformation of the derivative function Df, to get TDF and then this series
is evaluated not at x but at x+h.
These operations are not commutative.
This shows the connection between the continium and the discrete in
concrete form. Hopefully the result is correct.
If h tends to zero one just gets Df=Df
@ Juan
Can you please present your questions more precisely?
I see one of your questions: What is the name of the operator which assignes function F to f. Answer: I don't know.
My questions: What is the definition of TDF? Is the equation "F(x+h) - F(x) = h TDF(x+h)" a fact or a conjecture?
Regards, Joachim
What is the definition of TDF?
First take the derivative of f(x), get Df(x)
Then apply the same transformation to the derivative function, as I Illustrated for the originl function f
Finally evaluate the result at (x+h)
This is what I call TDF. Maybe there is a better notation . It could have been TDf
instead. Transformed derivative evaluated at x+h
This is an identity
If you start from f(x) = xxx
you will get 3h(x+h)(x+2h) on both sides of the identity.
Im looking for references where this is applied or used
in some maner, ie orbit calculation, numerical analysis etc.
OK, Thanks. Let me suggest the following notation of the involved two operations on formal series, which may make the formulas for iterations more readable:
D(f)(x) := a1 + 2 a2 x + 3 a3 xx + ... if f=a0 + a1 x + a2 xx + ...
Th(g)(x) := b0 + b1 x + b2 x(x+h) +... if g=b0 + b1 x + b2 xx + ...
Then your propositions read that Th tends to the identity and
(Th(f)(x+h) - Th(f)(x)) / h= Th(D(f))(x+h), which tends to D(f)(x) as h\to 0 .
To the best of my knowledge this formula is a new one with the use of divided differences (here, of order 1). Therefore, the orbits of the iterations [Th \circ D]n should be somehow related to divided differences of higher orders. Nothing more comes now to my memory:(
Best, Joachim
Has this issue something to do with the shift operators (Hildebrand)?
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