stuck on a simple cumsum prob

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Matlab2010
Matlab2010 2012-5-29
I have a variable x. I make a variable y. I now need to regenerate my variable xNew from y.
Its very close but not exact. Why not? what I have done wrong?
x = cumsum(randn(1000,1));
y = 0.5.*(x(3:end) - x(1:end-2));
xNew = cumsum(y);
plot(x(3:end)); hold all; plot(xNew);

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采纳的回答

Oleg Komarov
Oleg Komarov 2012-5-29
y = 0.5.*(x(3:end) - x(1:end-2));
Is linear interpolation between point 1 and 3, then between 2 and 4 and so on.
You cannot recover x from the linearly interpolated y unless you are given x(1) and x(end).
EDIT
Suppose you have 5 points only (for simplicity), then:
y2* = (x3-x1)/2;
y3* = (x4-x2)/2;
y4* = (x5-x3)/2;
I assume you know the series of y, i.e. are numbers, but x is unobserved, then you can see that the system has 5 unknowns (x1,...x5) in three equations. Thus, you can have a unique solution only if you are given x1 and x5.
Now, if you are given the x series for comparison reasons, you can guess x1 and x5 then retrieve the rest of the series and compare the distance between xNew and the original one. In an iterative fashion you can then tweak the guess and comapre again the two series, if the distance has dimished then continue tweaking the guess in the same direction until you reach distance = 0.
EDIT#2 You don't need exactly x1 and x5 but any two xi.

更多回答（1 个）

Thomas
Thomas 2012-5-29
x = cumsum(randn(1000,1));
%y = (x(2:end) - x(1:end-1));
y=[x(1); (x(2:end) - x(1:end-1))];
xNew = cumsum(y);
plot(x(2:end)); hold all; plot(xNew);
isequal(x,xNew)
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Matlab2010
Matlab2010 2012-5-29
yes. but what about "(x(3:end) - x(1:end-2));"
Thomas
Thomas 2012-5-29
Hmm. then as oleg says you need, x(1) and x(end) for each value to interpolate..

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