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I have 30 data points that I have digitised from the red dashed line in the graph below. My goal is to find an approximate equation to represent the line.

I have tried to get a 6 degree polynomial trendline through Excel for these points, but if I then plot the trendline equation in Excel or Wolfram, the numbers are very clearly incorrect. I have tried several other websites to get a more accurate polynomial expression but so far I have not been able to get an equation that looks anything like the original graph and I am wondering what I could be doing wrong.

enter image description here

The below points are what I have been pasting into various regression analysis calculator websites (for example Desmos). I have then retyped the 6 degree polynomial from these websites into Excel and fed a couple of x coordinates into them. I have tried 8 different sites now and I can't replicate the line in the graph. Can anyone recommend a different approach or provide some advice?

14510.28    500.01
23609.49    443.79
36769.24    426.06
63088.75    407.6
85648.33    393.39
130767.49   377.42
198446.22   361.99
294324.44   347.57
432245.5    333.84
647381.85   320.45
891025.3    310.05
1113844.52  303.03
1359458.84  296.97
1614059.8   291.76
1891078.16  287.06
2166769.49  283.11
2454404.12  279.59
2730758.96  276.68
3024738.48  273.68
3301051.85  271.28
3583046.59  269.03
3875657.6   266.85
4158315.85  264.97
4442467.02  263.13
4733585.11  261.35
5015579.85  259.88
5314535.74  258.56
5590849.11  257.29
5889805     255.94
6036275.7   255.91
david_10001
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  • why degree 6? have you tried higher degrees? – Albert Jun 18 '18 at 09:56
  • The function doesn't look like a polynomial would be the first choice to approximate it. Have you tried fitting a sum of inverse powers? Or perhaps just subtract out the singularity at the origin and approximate the rest with a polynomial. Do you have any information (other than the data in the graph) about the type of the singularity at the origin? – joriki Jun 18 '18 at 11:05
  • Unfortunately the graph is all I have. But I did end up taking out a couple of the plot values from the start and end of the line and have gotten the equation 1289.3859x^(-0.1039) which fits it pretty well. It seems some of the initial values were throwing it of. – david_10001 Jun 18 '18 at 13:35

2 Answers2

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Looking at the data, I thought that something as simple as $$y=a+\frac b {x^c}$$ could be reasonable (this is very similar to what you wrote in a comment).

A nonlinear regression leads to $R^2=0.999822$ which is quite good and the parameters are $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 101.314 & 26.3362 & \{47.1788,155.449\} \\ b & 1605.95 & 157.906 & \{1281.37,1930.53\} \\ c & 0.14960 & 0.01667 & \{0.11535,0.18386\} \\ \end{array}$$

If, as you did, we remove the $a$ term, we get $R^2=0.999775$ (almost the same) and the parameters are $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ b & 1306.83 & 24.8952 & \{1255.65,1358.00\} \\ c & 0.10489 & 0.00145 & \{0.10192,0.10786\} \\ \end{array}$$ very close to what you wrote in your comment.

In order to better represent the left part of the plot, we could consider $$y=\frac b {(x-d)^c}$$ which would lead to $R^2=0.999914$ and $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ b & 1166.87 & 21.3244 & \{1122.96,1210.79\} \\ c & 0.09707 & 0.00132 & \{0.09435,0.09978\} \\ d & 7658.20 & 743.875 & \{6126.16,9190.24\} \\ \end{array}$$

Claude Leibovici
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  1. Polynomial regression is based on Taylor's expansion at the origin, hence as further your data goes from the $0$, the less accurate the model.

  2. Try to fit an exponential model instead of the polynomial, i.e., $$ \mathbb{E}[y_i|x]=\beta_0 \exp\{-(\mathbf{x}^T \beta)\}, $$
    then $$ \ln\mathbb{E}[y_i|x] = \ln \beta_0-\mathbf{x}^T\beta, $$ now estimate the parameters and then exponentiation the predicted values. These values will still be biased, and if the bias severe you can use the OLS' values as initial values to the non-linear regression.

V. Vancak
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  • It looks more like an inverse power than an exponential to me. – joriki Jun 18 '18 at 12:34
  • Maybe you are right. He models time to failure, hence the first intuitive go-to model is commonly an exponential. I haven't bothered to fit the model to his data, so I cannot say anything concrete – V. Vancak Jun 18 '18 at 12:44
  • But not time to failure as a random variable (which might be expected to be exponentially distributed), but the relationship between stress amplitude and (presumably average) time to failure (where I wouldn't know what to expect). – joriki Jun 18 '18 at 13:13