Texas Instruments TI-92+, Tip List

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( √ (abs(expr(ffun)/(d3/(h 1 ^2)))))/ 1 0 → hh
if hh=0: .0 1→ hh
endif
©Initialize for solution loop
(expr(fphh)-expr(fmhh))/(2*hh) → amat[ 1 , 1 ]
big → err
©Loop to estimate derivative
for i,2,ntab
hh/con → hh
(expr(fphh)-expr(fmhh))/(2*hh) → amat[ 1 ,i]
con2 → fac
for j,2,i
(amat[j- 1 ,i]*fac-amat[j- 1 ,i- 1 ])/(fac- 1 ) → amat[j,i]
con2*fac → fac
max(abs(amat[j,i]-amat[j- 1 ,i]),abs(amat[j,i]-amat[j- 1 ,i- 1 ])) → errt
if errt ≤ err then
errt → err
amat[j,i] → dest
endif
endfor
if abs(amat[i,i]-amat[i- 1 ,i- 1 ]) ≥ safe*err:exit
endfor
return {dest,err}
endfunc
nder1() is called with the function name as a string, the evaluation point, and the initial step size. If the
step size is "auto", then nder1() tries to find a good step size based on the function and evaluation
point. nder1() returns a list with two elements. The first element is the derivative estimate, and the
second element is the error estimate.
nder1() is called as
nder 1 (fname,x,h)
where fname is the name of the function as a string in double quotes. x is the point at which to find the
derivative. h is a parameter which specifies a starting interval. If h is "auto" instead of a number, then
nder1() will try to automatically select a good value for h.
For example, to find the derivative of tan(x) at x = 1.0, with an automatic step size, use
nder 1 ("tan", 1 ,"auto")
which returns {3.42551882077, 3.7E-11}. The derivative estimate is 3.4255..., and the error estimate is
3.7E-11. To find the same derivative with a manual step size of 0.1, use
nder 1 ("tan", 1 ,. 1 )
which returns {3.42551882081,1.4E-12}.
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