bisection iteration wither different values than 0
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How to solv the eq f(y)=0.2? instead of f(y)=0.
d=0.250; % I SUPPOSED d=1.3
f=@(y) 0.46-0.5*cos(3.14*(y/d))+0.004*cos(2*3.14*(y/d))
format long
eps_abs = 1e-5;
eps_step = 1e-5;
a = 0; % Initial Guess to your function such that f(a)>0.
b = 1; % Initial Guess to your function such that f(b)<0.
while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) ) >= eps_abs ) )
c = (a + b)/2;
if ( f(c) == 0 )
break;
elseif ( f(a)*f(c) < 0 )
b = c;
else
a = c;
end
end
采纳的回答
Walter Roberson
2013-3-10
0 个投票
To solve f(y) = 0.2, define f1(y) = f(y) - 0.2, and then solve for f1(y) = 0.
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2013-3-10
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