Jimmy

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7 years, 69 days

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These are replies submitted by Jimmy

Hi,@Carl Love I just want to know :Are there any other indication can get a better fitting result, such fitting well at every point and every region (or subtle deviation), but not fitting well at some region, for example fitting well at higher region (0.7~1 V), but obvious deviation at small region (0.2~0.5 V), although the residualsumofsquares is the minimum, but it can obvious see the deviation after take y-axis log scale. Maybe sacrifice some accuracy, but it can get a good fitting by eye-viewing the stack of plot, that is my opinion.

Hi,@Carl Love, thanks for your help. I trid to compare the residualmeansquare and residualsumofsquares, the result of parametervalues are the same, but the vaule of residualmeansquare is smaller than residualsumofsquares. Thanks for your support.

Hi,@Carl Love , I have a question, for curve fitting,which one is the best indication of curve fitting? Here, a residal sum of sqare was used. besides, there are standard error and standard deviation, and compare the residal sum of square, which is the best indication for curve fitting? Thank you.

HI,@Carl Love , there is a question, if I want to set one of the variables a fixed value, for example, d=0.28, not a range, how to modify to expression? Thank you.

HI,@Carl Love , there is a question, if I want to set one of the variables a fixed value, for example, d=0.28, not a range, how to modify to expression? Thank you.

Hi, thanks for your help, that is interesting. Here, why you don't have to setting the parameterranges, you don't have to specify the range of variables,but it can generate is, tau, d. My fitting result is a horizontal line (all the values are the same). This is different from previous example. Thank you so much.

Hi, thanks for your help, that is interesting. Here, why you don't have to setting the parameterranges, you don't have to specify the range of variables,but it can generate is, tau, d. My fitting result is a horizontal line (all the values are the same). This is different from previous example. Thank you so much.

Hi, thanks for your help, here I upload my dataTE.mw i.txt v.txt .

How to modify the program code that could match my desire need? Thank you.

@acer I have an idea, I want to extract the values i0,n,rs and v(v.txt) from 0.01,0.02,0.03...,2, substitute into the formula, it will return a set of i, this set of plot together with to see how the fitting result is good or not, the v is x-axis, i is y-axis, and take log(y). Then, I can see the behavior of the curve and change parameterranges to make them fitting well. How to achieve the command? Thank you.

@acer I have an idea, I want to extract the values i0,n,rs and v(v.txt) from 0.01,0.02,0.03...,2, substitute into the formula, it will return a set of i, this set of plot together with to see how the fitting result is good or not, the v is x-axis, i is y-axis, and take log(y). Then, I can see the behavior of the curve and change parameterranges to make them fitting well. How to achieve the command? Thank you.

Hi @pagan  , when I change the parameterranges, it show an warning message, 

"Warning, limiting number of major iterations has been reached"

what is the warning message mean? How to solve the warning?

Thank you.

Hi @pagan  , when I change the parameterranges, it show an warning message, 

"Warning, limiting number of major iterations has been reached"

what is the warning message mean? How to solve the warning?

Thank you.

@pagan hi, thanks for your help, parametervalues is a good tool, I try it and quickly get the result.

But the result didn't match well with my original data, how to solve the problem? Thank you very much.

@pagan hi, thanks for your help, parametervalues is a good tool, I try it and quickly get the result.

But the result didn't match well with my original data, how to solve the problem? Thank you very much.

@acer hi, thanks for your help and give me some useful concepts, this time it works, I try my previous data fitting using Mathematica (see here v.txt , i.txt ) and compare the result from Maple. Using Mathematica repeating trying and get i0=1.7*10^(-7),n=1.22,rs=117.

> with(Statistics);
> readdata("v.txt");
> V := Vector(%, datatype = float);
> readdata("i.txt");
> idata := Vector(%, datatype = float);
> dim := LinearAlgebra:-Dimension(V);
         dim := 200
>NonlinearFit(i0*(exp((1000*(v-i*rs))/(25.9*n))-1)-i, ``(V, idata), Vector(dim),
 [v, i], parameterranges = [i0 = 1.5*10^(-9) .. 2*10^(-5), rs = 1 .. 1000, n = 1 .. 5], 
output = [residualsumofsquares, leastsquaresfunction]);
           [0.1876434658e-3, 0.200000000000000e-4*exp(17.2368060494510*v-
1994.82773288428*i)-0.200000000000000e-4-i]
The picture is captured from my screen.

  i0=2*10^(-5)
from exp(17.2368060494510*v)=exp(1000*v/n*25.9) →get n=(1000/17.2368060494510*25.9)=2.24
from exp(-1994.82773288428*i)=exp(-1000*rs*i/n*25.9) →get rs=115.731
substitute the result into Mathematica and execute, compare the reult of Mathematica and
Maple, plot together,from the picture shows that a discrepancy with original data, 
although using Mathematica almost match well with original data, but it's time-consuming
 work to try and try,finally get the result, next time I will use the complete long formula 
to fitting other curve, I think I will spend a huge time to try the variables. 
I want to know how to solve the problem to make the result match well with the 
original data.

I have try to change the parameterranges, found that the residualsumofsquares will 
change, according the definition from Wikipedia, it says "In statistics, the residual
 sum of squares (RSS) is the sum of squares of residuals. 
It is also known as the sum of squared residuals (SSR) or the sum of squared errors 
of prediction (SSE). It is a measure of the discrepancy between the data and an
estimation model. A small RSS indicates a tight fit of the model to the data."
So,it is better to get a small residualsumofsquares value, how to minimize the value?
Thanks for your help, you really help me a lot.

 

 
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