[HELP] A Classical Numerical Computing Question
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f(x) = (exp(x)-1)/x; g(x) = (exp(x)-1)/log(exp(x))
Analytically, f(x) = g(x).
When x is approaching to 0, both f(x) and g(x) are approaching to 1. However, g(x) works better than f(x).
% Compute y against x
for k = 1:15
x(k) = 10^(-k);
f(k) =(exp(x(k))-1)/x(k);
De(k) = log(exp(x(k)));
g(k)= (exp(x(k))-1)/De(k);
end
% Plot y
plot(1:15,f,'r',1:15,g,'b');
f(x) actually diverges when x approaches to 0.But shouldn't them be the same??
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Matt Fig
2012년 9월 8일
편집: Matt Fig
2012년 9월 8일
No, they shouldn't be the same at the fringes. This is an example of why we often have to look for more stable ways of doing in floating point arithmetic what is analytically simple. Look how f oscillates:
f = @(x) (exp(x)-1)./x;
g = @(x) (exp(x)-1)./log(exp(x));
x = 0:1e-13:1e-7; % Try with x = 0:2e-14:1e-7;
ax(1) = subplot(1,2,1);
plot(x,f(x),'r')
title('(exp(x)-1)./x')
ax(2) = subplot(1,2,2);
plot(x,g(x),'b');
title('(exp(x)-1)./log(exp(x))')
L = get(gca,{'xlim','ylim'});
axis(ax(1),[L{:}])
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Matt Fig
2012년 9월 8일
I think what we have is a case of near perfect cancellation of errors. Take a look:
x = 1e-12:1e-12:1e-9; % Double values
X = vpa(1/10^12:1/10^12:1/10^9,80); % Symbolic values
syms Z
D = abs(X - log(exp(x)));
E = abs(X-x);
plot(D)
figure
plot(E) % Very little difference
F = abs((exp(x)-1) - subs(exp(Z)-1,X));
figure
plot(F) % Notice similarity to D!
G = F-D;
max(double(G)) % Cancellation of errors.
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