Post Reply 
Calculators and numerical differentiation
11-01-2020, 11:43 PM
Post: #6
RE: Calculators and numerical differentiation
(11-01-2020 05:39 PM)Albert Chan Wrote:  For more accuracy, we can add more terms: (note, there is no even powers of δ Smile)

\( hD ≡ µδ \large\left(1 - {δ^2\over 6} + {δ^4\over 30} - {δ^6\over 140} + {δ^8\over 630}
- {δ^{10}\over 2772} + {δ^{12}\over 12012} - {δ^{14}\over 51480} \;+\; ... \right) \)

We can show Df(0) = f'(0) does not require calculating f(0).
In other words, operator form will not have a constant term, the "1" operator.

From previous post, we have µδ = (E-1/E)/2, δδ = (E+1/E) - 2
Doing "operator" mathematics, with x = log(E), we have:

µδ = sinh(x)
δδ = 2*cosh(x) - 2

Hyperbolics identities:
(1): cosh(z1)*cosh(z2) = (cosh(z1 - z2) + cosh(z1 + z2)) / 2
(2): sinh(z1)*cosh(z2) = (sinh(z1 - z2) + sinh(z1 + z2)) / 2

hD = sinh(x) * (k1 + k2*cosh(x) + k3*cosh(x)^2 + k4*cosh(x)^3 + ...)       // apply (1)
     = sinh(x) * (k1' + k2'*cosh(x) + k3'*cosh(2x) + k4'*cosh(3x) + ... )       // apply (2)
     = k1''*sinh(x) + k2''*sinh(2x) + k3''*sinh(3x) + k4''*sinh(4x) + ...

sinh(nx) = (En - E-n)/2

→ this explained why En coefs = negative of E-n coefs.
→ RHS terms will not generate constant term (i.e., no "1" operator)

→ D does not require calculating f(0)
→ same for D^odd_powers, since RHS is still linear combinations of sinh's.
Find all posts by this user
Quote this message in a reply
Post Reply 


Messages In This Thread
RE: Calculators and numerical differentiation - Albert Chan - 11-01-2020 11:43 PM



User(s) browsing this thread: