best approximation for double numbers

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Francesco Pio
Francesco Pio 2023-7-21
评论： Walter Roberson 2023-7-21
I have the following function which has an asymptote for y = 1 :
syms f(x)
syms x
f(x) = 1 / (1 + exp(-((x + 8.4730) / 10 )));
So, so the function should have a value < 1 for all x.
Also using "format long", of course, from a certain value of x onwards, the result of the function is approximated to 1.
format long
double(f(400)) % ans = 1
Is there a way to get an approximation to the exact value for even larger x? Or should I settle for this approximation?

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采纳的回答

VBBV
VBBV 2023-7-21
Ran in:
format long
syms f(x)
syms x
f(x) = 1 / (1 + exp(-((x + 8.4730) / 10 )))
f(x) = 
vpa(f(400),100)
ans = 
0.9999999999999999981792806433707777978855866136633285567221016451791545277454316141363423975013892837
  2 个评论
VBBV
VBBV 2023-7-21
Try using vpa for large values of x
Walter Roberson
Walter Roberson 2023-7-21
Ran in:
syms f(x)
syms x
f(x) = 1 / (1 + exp(-((x + 8.4730) / 10 )))
f(x) = 
f1 = simplify(expand(1-f))
f1(x) = 
double(f1(400))
ans = 1.8207e-18
fplot(f1,[500 600])
f1n = matlabFunction(f1)
f1n = function_handle with value:
@(x)1.0./(exp(x./1.0e+1+8.473e-1)+1.0)
fplot(f1n, [500 600])
f1 gives you an idea of how quickly the value approaches 1, by showing you how quickly the difference between 1 and f falls. f1n shows that a numeric approximation (instead of a symbolic) of 1-f is still not bad at all in this kind of range.

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