computing error on least square fitting

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Sumera Yamin
Sumera Yamin 2020 年 8 月 13 日
編集済み: Sumera Yamin 2020 年 8 月 16 日
Hi, i slove a system of equations (Ax-b) using least square method. i get an output with x like [2.5; -11.1; 0.8; 0.5]. the status flag is zero with system converging at iteration 2 and relative residual of 0.019. I want to calculate the error on my fit i--e with which certainity my solution is accurate. Can i claim that the residual which is norm of (Ax-b)/b means that my fit has an error of 1.9%?if not how can i calculate error on my fit?
  2 件のコメント
Alan Stevens
Alan Stevens 2020 年 8 月 15 日
0.019 is 1.9% not 19%.
Sumera Yamin
Sumera Yamin 2020 年 8 月 15 日
oh yes obviously, thx for correction

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回答 (1 件)

Bruno Luong
Bruno Luong 2020 年 8 月 15 日
Can i claim that the residual which is norm of (Ax-b)/b
No make it
norm(A*x-b) / norm(b)
  3 件のコメント
Bruno Luong
Bruno Luong 2020 年 8 月 15 日
If you want an unnambiguous mathematical statement, just state exactly what mean:
norm(A*x-b) / norm(b) is approximatively 0.019
At your place I would say in the speaking language
The fit has a relative l2-norm residual of 1.9%.
The fit error usually designates the difference between the true and the estimated fit (parameters). So to me you shouldn't use the word "error."
Sumera Yamin
Sumera Yamin 2020 年 8 月 15 日
Sorry i am not getting hang of it. " The fit has a relative l2-norm residual of 1.9%." how would i interpret this statement in terms of accuracy of my solution.
"The fit error usually designates the difference between the true and the estimated fit (parameters)" how can i calculate fit error?

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