equation solution with matrices
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I have this equation;
z=(log(x.^(-2) .* y.^3) .* 10^3 .* y.^abs(x)) ./ (factorial(4) .* 1.5 .* 2.8 ./ 3.16 ) + (2.^(1/3) .* 5 + 6 .* 2.^(-2) ./ 3.^2 .* 4 .*log(8./sqrt(x.*y)) ) ./ (exp(0.2 .* x) .* sin(y.^(-2)).^2 .* abs(cos(x-y)).^(1/3) .* 0.25)
and my x and y values are respectively like that; x=[4 2 1 3] y=[2 1 0.5 1]
how will i solve this z equation with for loop? and get z in matrix form again?
thanks guys.
回答（2 个）
John D'Errico
2016-10-22
Why do you need a loop of any kind? Did you try just evaluating that expression as is?
x = [4 2 1 3];
y = [2 1 0.5 1];
z=(log(x.^(-2) .* y.^3) .* 10^3 .* y.^abs(x)) ./ (factorial(4) .* 1.5 .* 2.8 ./ 3.16 ) + (2.^(1/3) .* 5 + 6 .* 2.^(-2) ./ 3.^2 .* 4 .*log(8./sqrt(x.*y)) ) ./ (exp(0.2 .* x) .* sin(y.^(-2)).^2 .* abs(cos(x-y)).^(1/3) .* 0.25);
So, did you try it? Why not?
The only thing that I see that might be problematic are things like:
y.^abs(x)
You need to be careful if the base can ever be negative, and the exponent a non-integer. You need to remember that a fractional power of a negative number has multiple solutions, some of which are complex.
x = [4 2 1 3];
y = [2 1 0.5 1];
z = zeros(length(x), length(y));
for ix = 1:length(x)
for iy = 1:length(y)
z = ...
end
end
Now insert "x(ix)" and "y(iy)" inside the formula.
It looks simple to create two loops, when you know already, that you want to use loops, doesn't it?
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