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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. 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 |