Recursive Method returning answers backwards
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Consider my code for Euler's Method below. If you run the code, you will see the correct values be stored in the matrix "vals", but after the return statement, the "vals" that is returned is from the first iteration. Why does MATLAB do this?
f = @(x,y)(x^2*(2+y));
ans = euler(1, 0.1, f, 0.0, 1.0, [])
function vals = euler(y,h,f,start,finish,vals)
if (finish - h < start)
y;
return
else
y = y + h*f(start,y)
start = start + h
vals = [vals,y]
euler(y,h,f,start,finish, vals)
end
end
1 个评论
John D'Errico
2023-5-2
编辑：John D'Errico
2023-5-2
MATLAB does what you tell it to do. It is not DOING anything to you. Just your code, code written in error. Why you are usign recursion, I don't know, since Euler is simple to write with a loop. Read the answer from James.
采纳的回答
James Tursa
2023-5-2
编辑：James Tursa
2023-5-2
This is a very convoluted way to program the Euler method. There is no need for recursion here, just use simple loops instead.
Regarding why you aren't getting the results you expected, it has to do with how you update vals. Your very first call to Euler is this:
ans = euler(1, 0.1, f, 0.0, 1.0, [])
which returns the vals from the very first call via this line:
vals = [vals,y]
but then in all your subsequent calls to euler( ) you throw away the returned vals:
% you don't save the returned vals, so this line essentially accomplishes nothing
euler(y,h,f,start,finish, vals)
Maybe changing this euler( ) call to this might get you what you expected:
vals = euler(y,h,f,start,finish, vals)
But, seriously, this code is convoluted and hard to follow because of the recursion. It would be best to rewrite it as a simple loop.
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