Texas Instruments TI-92+, Tip List

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nder2("tan", /2.01)
✜
which returns 16374.242. The absolute error is about 1.01E-4, and the relative error is 6.19E-9.
The table below demonstrates the improvement in the derivative estimate as h decreases, for tan(x)
evaluated at x = 1.
3E-7 3.4255 185 1E-7
2E-8 3.4255 188 1E-6
9.5E-8 3.4255 1882 1E-5
9.356E-6 3.4255 1891 5 1E-4
9.359E-4 3.4255 2827 135 1E-3
(none) 3.4264 6416 009 1E-2
difference in f'(x) f'(x) h
nder2() starts with h = 0.01, and divides h by 10 for each new estimate, so that the steps for h are
1E-2, 1E-3, ... 1E-14. Since the difference in f'(x) starts increasing at h = 1E-7, nder2() returns the
value for h = 1E-6.
More accurate results with Ridders' method: nder1()
Ridders' method (Advances in Engineering Software, vol. 4, no. 2, 1982) is based on extrapolating the
central difference formula to h=0. This method also has the important advantage that it returns an
estimate of the error, as well. This program, nder2(), implements the algorithm:
nder 1 (ff,xx,hh)
func
©("f",x,"auto" or h), return {f'(x),err}
©Based on Ridders' algorithm
©6jun00/dburkett@infinet.com
local con,con2,big,ntab,safe,i,j,err,errt,fac,amat,dest,fphh,fmhh,ffun,h 1 ,d3
© Initialize constants
1 .4 → con
con*con → con2
1 900 → big
1 0 → ntab
2 → safe
newmat(ntab,ntab) → amat
© Build function strings
ff&"(xx+hh)" → fphh
ff&"(xx-hh)" → fmhh
ff&"(xx)" → ffun
© Find starting hh if hh="auto"
if hh="auto" then
if xx=0 then: .0 1→ h 1
else: xx/ 1 000 → h 1
endif
expr(ff&"(xx+h 1 )")-2*expr(ffun)+expr(ff&"(xx-h 1 )") → d3
if d3=0: .0 1→ d3
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