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Accuracy of numsolve in TI, CASIO - lrdheat - 03-26-2021 07:35 PM
I notice on my TI-30X Pro MathPrint...if I solve d/dx to =0 for (x^2 +7x +3)/ ^2, the solution found is most accurate when choosing an epsilon of 1E-05. Why is this the optimum choice? On my CASIO fx-991EX, the solution is more accurate, last digit shown is correctly rounded...wonder what epsilon is used, and what is different in it’s solve methodology...
RE: Accuracy of numsolve in TI, CASIO - Albert Chan - 03-26-2021 09:55 PM
Assuming calculator use central-difference derivative formula, this might answer your question.
Is there a general formula for estimating the step size h in numerical differentiation formulas ?
Example, estimate (ln(x))' at x=2:
lua> function D(x,h) return (log(x+h)-log(x-h))/(2*h) end
lua> for i=4,8 do print(i, D(2, 10^-i)) end
4 0.5000000004168335
5 0.5000000000088269
6 0.5000000000143777
7 0.49999999973682185
8 0.49999999696126407
Interestingly, optimal h for this example is also about 1e-5:
At the cost of more computation, we can use bigger h, and extrapolate for slope.
(similar to Romberg's integration, extrapolate from raw trapezoids, or rectangles)
lua> h = 1e-3
lua> d1 = D(2,h)
lua> d2 = D(2,h/2)
lua> d1, d2, d2+(d2-d1)/3
0.500000041666615 0.5000000104167235 0.5000000000000929
RE: Accuracy of numsolve in TI, CASIO - robve - 03-26-2021 10:23 PM
(03-26-2021 09:55 PM)Albert Chan Wrote: Assuming calculator use central-difference derivative formula, this might answer your question.
Is there a general formula for estimating the step size h in numerical differentiation formulas ?
Just my 2c.
This is a very good question. With numerical differentiation and finite difference stencils I've always used the cube-root of machine epsilon (MachEps). On 12 digit machines, this is 1E-4 and for 15 digit machines (no surprise) this is 1E-5.
As Albert says, 1E-5 should be about optimal.
See also step size that explains the difficulty of choosing a step size. Let me add that an approximate to the optimal step can be empirically established.
- Rob
PS. (edit) you may also want to scale h with the magnitude of the point(s) you're differentiating, otherwise you will end up with a slope that is closer to zero. For a 10 digit machine, let's take 1E-3, then what you want to do when differentiating at point A is something like this:
h=1E-3
IF abs(A)>1 THEN h=h*abs(A)
The points X are at \( A\pm h \), but due to rounding you may want to do the following to get to X-A and then adjust H so it is exact:
X=A-h, h=A-X