How to obtain the original matrix of a cumsum()?

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if we have a matrix B = cumsum(A);
B=[0.102493854430154 0.107645445153016 0.109982543018989 0.111250846129182 0.112023941002529
0.252983718693220 0.267537140617764 0.274291923474296 0.277936513591507 0.280166731981901
0.446994905411664 0.475357155580891 0.488884534539884 0.496152338879191 0.500676903345327
0.681174961701521 0.727902668424241 0.750721436460152 0.763062453240230 0.770697210445687
0.954490719298870 1.02481028893085 1.05975275214650 1.07879511204532 1.09055541700809];
how to obtain the "cumdiff" of the matrix B and obtain A.
Answer is:
A=[0.102493854430154,0.107645445153016,0.109982543018989,0.111250846129182,0.112023941002529;0.150489864263066,0.159891695464748,0.164309380455307,0.166685667462325,0.168142790979372;0.194011186718444,0.207820014963127,0.214592611065588,0.218215825287684,0.220510171363426;0.234180056289857,0.252545512843350,0.261836901920268,0.266910114361039,0.270020307100360;0.273315757597349,0.296907620506609,0.309031315686344,0.315732658805087,0.319858206562400];

采纳的回答

John D'Errico
John D'Errico 2018-3-21
Just use diff! Then append the first element.
A = rand(1,10);
B = cumsum(A);
C = [B(1), diff(B)];
  3 个评论
John D'Errico
John D'Errico 2020-5-28
Then it is time for you to learn about floating point numbers, about how computations are done using computers in floating point arithmetic. And... why you should NEVER trust the least significant bits of a number done in such a computation, UNLESS you know enough about what you are doing that you fully understand why not to trust those least significant bits. And even then, don't trust those least significant bits.
dW(1,:)
ans =
-0.11486 0.0044322 0.027686 -0.069706 -0.17065 -0.23663 0.21574 -0.041219 0.093753
>> diff(W(1,:))
ans =
-0.11486 0.0044322 0.027686 -0.069706 -0.17065 -0.23663 0.21574 -0.041219 0.093753
Are they identically the same? No.
abs(dW(1, :) - diff(W(1, :), 1, 2))
ans =
0 4.3368e-18 6.9389e-18 1.3878e-17 0 0 0 0 0
Are they significantly different? No. NEVER test for exact equality in something like that.
Giacomo Tabarelli
Giacomo Tabarelli 2020-5-29
Ok I know about floating point numbers. My question is why there is such a difference? Is there a way to avoid this errors? I need this to compute SDE convergence errors. I found problems in the order of convergence and investigating I found this. Any suggestion?

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