Post Reply 
Accuracy of numsolve in TI, CASIO
03-26-2021, 07:35 PM
Post: #1
Accuracy of numsolve in TI, CASIO
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...
Find all posts by this user
Quote this message in a reply
03-26-2021, 09:55 PM
Post: #2
RE: Accuracy of numsolve in TI, CASIO
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
Find all posts by this user
Quote this message in a reply
03-26-2021, 10:23 PM (This post was last modified: 03-26-2021 10:34 PM by robve.)
Post: #3
RE: Accuracy of numsolve in TI, CASIO
(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

"I count on old friends" -- HP 71B,Prime|Ti VOY200,Nspire CXII CAS|Casio fx-CG50...|Sharp PC-G850,E500,2500,1500,14xx,13xx,12xx...
Visit this user's website Find all posts by this user
Quote this message in a reply
Post Reply 




User(s) browsing this thread: 1 Guest(s)