Having problem with empty matrix...Can anyone help?

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Prasenjit Dewanjee
Prasenjit Dewanjee 2016-11-20
回答： Star Strider 2016-11-20
here is the code..
disp('dy/dx=1+xy');
h=0.2;
xfinal=0.6;
y(1)=2;
x(1)=0;
f=@(x,y) 1+x*y;
for i=1:ceil(xfinal/h)
x(i+1)=x(i)+h;
k1=f((x(i)),y(i));
k2=f((x(i)+.5*h),(y(i)+.5*h*k1));
k3=f((x(i)+.5*h),(y(i)+.5*h*k2));
k4=f((x(i)+h),(y(i)+k3*h));
y(i+1)=y(i)+((1/6)*(k1+(2*k2)+(2*k3)+k4)*h);
end
plot(x,y);
xlabel('x');
ylabel('y');
grid on;
idx1=find(x==0.2);
y1=y(idx1)
idx2=find(x==0.4);
y2=y(idx2)
idx3=find((x==0.6))
y3=y(idx3)disp('dy/dx=1+xy');
it returns an empty matrix error for y3...what should i do?

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采纳的回答

Star Strider
Star Strider 2016-11-20
The problem is that you are encountering floating point approximation error.
If you do the comparison:
x_error = x(4) - 0.6
x_error =
111.0223e-018
The solution is to allow for the tolerance. The easiest way is to change your ‘idx3’ assignment to:
idx3=find((x<=0.6),1,'last')
which works.
For a thorough discussion, see Why is 0.3 - 0.2 - 0.1 (or similar) not equal to zero?

更多回答（1 个）

Walter Roberson
Walter Roberson 2016-11-20
Do not compare floating point numbers for exact equality. Compare them with a tolerance instead. Two different ways of calculating the same result can give different results with floating point. For example 1 + 1E200 - 1E200 will not give 1

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