How to do a regression which includes reciprocals?

Question

I'm trying to find an interpolating formula for a set of coefficients (I have $80$ at the moment).

I tried first to find an interpolating polynomial, but that was not useful: using the first $40$ coefficients only and extrapolate that formula to a guess for the other $40$ coefficients leads to catastrophic error.

So I tried another path: first find a linear expression which interpolates as smooth as possible (simple linear regression). Then it seems, that the reciprocals of the residuals might be best approximated by a next step of linear regression - and after some manual guesses it seems, that it is again most appropriate to use again the reciprocals of the new residuals with a linear or a quadratic curve - and possibly so on.

The idea for a regression formula were thus the following with $2$ pairs of parameters where the coefficients are indicated by $y_k$ ,$N=40$ and $k=1..N$ : $$ \sum_{k=1}^N \left( {1 \over y_k -(a_0+a_1 k) } - (b_0+b_1 k) \right)^2 \overset{!}{=} min $$

Possibly this must then be continued by a next step, where again the reciprocals of the new residuals must be approximated by a quadratic regression, so in principle I look for some algorithm which will be mechanically extensible, but that's only the next step. So just for a start:

Q: What would be the formula to determine the four parameters?

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See the requested data in a separate "answer" to keep the question better readable.

[update] P.s. : I've had similar problems casually in earlier problems (from my numbertheory fiddlings), and got in such cases stuck with some manual optimizations - so the principle of such a regression would serve me enough, please don't put too much effort in some extreme finetuning of the parameters on base of that 80 datapoints.

[update2] A bit more background to avoid misconceptions and to avoid frustration to someone willing to help. The data stem from a problem in number theory (see here in MSE) where I found a sequence of (infinitely many) points which seem to approximate to a linear or near linear decrease (linear with the index).
The problem is the following. Begin with the complex number $z_0=1$. Iterate $z_{k+1}=î ^ z_k$ (where $î$ is the imaginary unit) which process converges to a fixpoint $t$. Take the (euclidean) distance $d$ of each iterate $d_k = |z_k - t|$ . The given data $y_k$ are the $\log()$ of the $d_k$.

There is very likely no simpler function for that values available, but a functional approximation to the $y_k$ seems to be primarily linear with some unknown, diminuishing secondary distortion, again systematic.
A good approximation is enough for my purposes, but must be robust for extrapolation to infinitely many datapoints. I need not only one "practical" approximation but a method because the same problem with improving approximation to linearity occurs as well with the angular value $ \arg(z_k-t)$ for which I want then apply the same procedure.
Trying to find a working model using the ansatz of regression brought me to the above formula with four parameters (for a start) which I cannot expand/resolve for a formal solution.

Answer 1

Data which I'm working on. I use $80$ values but put it here in two portions a $40$ (vertically). I'm getting an approximation-formula from the left block of forty values and check the appropriateness by application to the full dataset. The left column of each block is the index which begins at $1$ to allow formulae with reciprocals. The data stem from a formula by which I can successively get infinitely more values (but practically limited to some $100$ or $200$ data-points)

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1. data in csv-format: file

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2. data in mathjax-format:

$$\small \begin{array} {rl|rl} 1 & -0.404219547134 & 41 & -4.96369610919 \\ 2 & -0.254635150576 & 42 & -5.08346228643 \\ 3 & -0.848826900252 & 43 & -5.19387328015 \\ 4 & -0.672497295471 & 44 & -5.30980799271 \\ 5 & -0.782648581294 & 45 & -5.42691064949 \\ 6 & -1.09661390701 & 46 & -5.53829673204 \\ 7 & -0.941305565047 & 47 & -5.65523782695 \\ 8 & -1.20932963584 & 48 & -5.77042934573 \\ 9 & -1.36432295221 & 49 & -5.88318317015 \\ 10 & -1.28358177556 & 50 & -6.00009058654 \\ 11 & -1.58603096933 & 51 & -6.11423257056 \\ 12 & -1.65776438414 & 52 & -6.22820436618 \\ 13 & -1.67153909479 & 53 & -6.34456590114 \\ 14 & -1.93014001210 & 54 & -6.45835690750 \\ 15 & -1.96844248450 & 55 & -6.57316075367 \\ 16 & -2.06003832171 & 56 & -6.68885932414 \\ 17 & -2.25865650761 & 57 & -6.80273947156 \\ 18 & -2.29610558688 & 58 & -6.91796901573 \\ 19 & -2.43438066384 & 59 & -7.03311492793 \\ 20 & -2.58368257882 & 60 & -7.14728282076 \\ 21 & -2.63902124916 & 61 & -7.26262536683 \\ 22 & -2.79434601545 & 62 & -7.37741464273 \\ 23 & -2.91134973187 & 63 & -7.49189688720 \\ 24 & -2.99089127097 & 64 & -7.60716668365 \\ 25 & -3.14368853097 & 65 & -7.72178845853 \\ 26 & -3.24410305971 & 66 & -7.83651835597 \\ 27 & -3.34517482343 & 67 & -7.95164075044 \\ 28 & -3.48674797680 & 68 & -8.06623261683 \\ 29 & -3.58231400466 & 69 & -8.18111355575 \\ 30 & -3.69791558146 & 70 & -8.29608865051 \\ 31 & -3.82709089874 & 71 & -8.41072732188 \\ 32 & -3.92498019330 & 72 & -8.52567229447 \\ 33 & -4.04770179897 & 73 & -8.64053766338 \\ 34 & -4.16709403679 & 74 & -8.75524974887 \\ 35 & -4.27042930487 & 75 & -8.87019875417 \\ 36 & -4.39467132300 & 76 & -8.98500120459 \\ 37 & -4.50802943507 & 77 & -9.09978136831 \\ 38 & -4.61704538184 & 78 & -9.21470325795 \\ 39 & -4.73962423078 & 79 & -9.32948231359 \\ 40 & -4.85032671658 & 80 & -9.44431053345 \end{array}$$

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3. data in textformat:

data

   1  -0.404219547134  41  -4.96369610919
   2  -0.254635150576  42  -5.08346228643
   3  -0.848826900252  43  -5.19387328015
   4  -0.672497295471  44  -5.30980799271
   5  -0.782648581294  45  -5.42691064949
   6   -1.09661390701  46  -5.53829673204
   7  -0.941305565047  47  -5.65523782695
   8   -1.20932963584  48  -5.77042934573
   9   -1.36432295221  49  -5.88318317015
  10   -1.28358177556  50  -6.00009058654
  11   -1.58603096933  51  -6.11423257056
  12   -1.65776438414  52  -6.22820436618
  13   -1.67153909479  53  -6.34456590114
  14   -1.93014001210  54  -6.45835690750
  15   -1.96844248450  55  -6.57316075367
  16   -2.06003832171  56  -6.68885932414
  17   -2.25865650761  57  -6.80273947156
  18   -2.29610558688  58  -6.91796901573
  19   -2.43438066384  59  -7.03311492793
  20   -2.58368257882  60  -7.14728282076
  21   -2.63902124916  61  -7.26262536683
  22   -2.79434601545  62  -7.37741464273
  23   -2.91134973187  63  -7.49189688720
  24   -2.99089127097  64  -7.60716668365
  25   -3.14368853097  65  -7.72178845853
  26   -3.24410305971  66  -7.83651835597
  27   -3.34517482343  67  -7.95164075044
  28   -3.48674797680  68  -8.06623261683
  29   -3.58231400466  69  -8.18111355575
  30   -3.69791558146  70  -8.29608865051
  31   -3.82709089874  71  -8.41072732188
  32   -3.92498019330  72  -8.52567229447
  33   -4.04770179897  73  -8.64053766338
  34   -4.16709403679  74  -8.75524974887
  35   -4.27042930487  75  -8.87019875417
  36   -4.39467132300  76  -8.98500120459
  37   -4.50802943507  77  -9.09978136831
  38   -4.61704538184  78  -9.21470325795
  39   -4.73962423078  79  -9.32948231359
  40   -4.85032671658  80  -9.44431053345

Answer 2

If I well understand the question, you have the data $(x_k\:,\:y_k)$ where $k=1$ to $k=n$ ( in your example $n=80$ and $x_k=k$ ).

As the problem is settled, this is not exactly the fitting of a function to the data but in fact the fitting of an equation to the data, which is slightly different. The equation is : $$\frac{1}{y-(a_0+a_1x)} - (b_0+b_1x)=0$$ The goal is to find approximates for $a_0\:,\:a_1\:,\:b_0\:,\:b_1$ so that the equation be at best satisfied. $$ \left(y-(a_0+a_1x)\right)(b_0+b_1x)-1=0$$ $$b_0y+b_1xy-(a_1b_0+a_0b_1)x-a_1b_1x^2=1+a_0b_0$$ Let : $\quad C_1=\frac{b_0}{1+a_0b_0}\quad;\quad C_2=\frac{b_1}{1+a_0b_0}\quad;\quad C_3=-\frac{a_1b_0+a_0b_1}{1+a_0b_0}\quad;\quad C_4=-\frac{a_1b_1}{1+a_0b_0}$

Then, compute for $k=1$ to $k=n$ : $\quad p_k=x_ky_k \quad \text{and}\quad q_k=x_k^2$

The regression with linear mean square fitting of : $$C_1y_k+C_2p_k+C_3x_k+C_4q_k\simeq1$$ will lead to $\:C_1\:,\:\:C_2\:,\:\:C_3\:,\:\:C_4\:$. Then, with the four relationships with $\:a_0\:,\:\:a_1\:,\:\:b_0\:,\:\:b_1\:$ one can compute the approximates of those four parameters.

I didn't test this method with your data because a simple linear regression seems convenient and leads to a very good fitting , or have I missed something ? :

Answer 3

Based on your initial question and using the values you give, I considered the function $$F(a_0,a_1,b_0,b_1)=\sum_{k=1}^{80} \left( {1 \over y_k -(a_0+a_1 k) } - (b_0+b_1 k) \right)^2 $$ and I just minimized it numerically.

The minimum is found for the following parameters $$a_0=36.9437\quad a_1=-0.127637$$ $$b_0=-0.0268729\quad b_1=-9.57943 \times 10^{-6}$$ for which $F_{opt}=1.06724\times 10^{-7}$