[HELP] A Classical Numerical Computing Question

Question

Xiao Tang 2012-9-8

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采纳的回答： Matt Fig

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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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Answer 1

Matt Fig 2012-9-8

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编辑：Matt Fig 2012-9-8

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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 2012-9-8

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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.