Calculators and numerical differentiation

[Albert Chan]

(11-01-2020 05:39 AM)Wes Loewer Wrote:  Instead of averaging the backwards and forwards differences, why not just check to see if f(a) exists and if it does then do the central difference method?

Try taken derivative of f(x) = |x|, at x=0.

f(0) = |0| = 0, thus exist.
Central difference slope = (|0+h| - |0-h|)/(2h) = 0

Forward difference slope = (h-0)/h = 1
Backward difference slope = (0-h)/h = -1
2 slopes does not match. f'(0) does not exist.

Oh ... Casio does not check this, d/dx(√(X²), 0) → 0

Quote:The non-CAS Nspire apparently has a bit of CAS hidden under the hood because it does not use this approximation. It appears to evaluate the derivative symbolically and then evaluates that expression numerically, keeping the CAS carefully hidden from the user.

How did you deduce there is hidden CAS under non-CAS Nspire ?
May be an example ?

Quote:Had anyone previously seen the del operator used for the backwards difference as shown in the Casio manual?

I think many text use the same symbols (or equivalent):
Ref: Fundamentals of Numerical Analysis, by Stephen Kellison

Code:

Operators:
Δf(x) = f(x+1) - f(x)       // forward difference
∇f(x) = f(x) - f(x-1)       // backward difference
Ef(x) = f(x+1)              // stepping
δf(x) = f(x+½) - f(x-½)     // central difference
µf(x) = (f(x+½)+f(x-½))/2   // average
Df(x) = f'(x)               // derivative

Or, operator notation: \(Δ=E-1,\quad ∇=1-E^{-1}\)

Combining µδ, we have the 3 points central difference slope formula

\( hD ≈ µδ = \large\left({E^{½} + E^{-½} \over 2}\right) \normalsize(E^{½}-E^{-½})
= \large{E\;-\;E^{-1} \over 2}
= \large{Δ\;+\;∇ \over 2}
\)

For more accuracy, we can add more terms: (note, there is no even powers of δ )

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

(10-30-2020 11:41 PM)Paul Dale Wrote:  The order 10 methods are a weighted summation of \( f(x \pm 1) \), \( f(x \pm 2) \), \( f(x \pm 3) \), \( f(x \pm 4) \) and \( f(x \pm 5) \).

Lets build D (order 10), without using central difference table.

XCas> c := E^(1/2) - E^(-1/2); // central difference
XCas> m := (E^(1/2) + E^(-1/2))/2; // mean
XCas> [mc, cc] := expand(simplify([m,c] .* c))       → [E/2 - 1/(E*2) , E - 2 + 1/E]

Note the symmetry of m*c, c*c. This suggested D also have similar symmetry

XCas> simplify(mc/h * horner([1/630, -1/140, 1/30, -1/6, 1], cc))

\(→ D = \large \frac {2100(E-E^{-1})\;
-\;600(E^2-E^{-2})\;
+\;150(E^3-E^{-3})\;
-\;25(E^4-E^{-4})\;
+\;2(E^5-E^{-5})}
{2520h}\)